A Problem- Solving Approach to Aquatic Chemistry [2 ed.] 9781119884347, 9781119884354, 9781119884361

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ISTUDY

A Problem-­Solving Approach to Aquatic Chemistry

A Problem-­Solving Approach to Aquatic Chemistry

Second Edition

James N. Jensen

University at Buffalo, NY, US

Copyright © 2023 John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/ permission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Further, readers should be aware that websites listed in this work may have changed or disappeared between when this work was written and when it is read. Neither the publisher nor authors shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services or for technical support, please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317) 572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic formats. For more information about Wiley products, visit our web site at www.wiley.com. Library of Congress Cataloging-­in-­Publication Data Names: Jensen, James N., author. Title: A problem-solving approach to aquatic chemistry / James N. Jensen, SUNY—Buffalo, NY, US. Description: Second edition. | Hoboken : Wiley, [2023] | Includes bibliographical references and index. Identifiers: LCCN 2022022564 (print) | LCCN 2022022565 (ebook) | ISBN 9781119884347 (cloth) | ISBN 9781119884354 (adobe pdf) | ISBN 9781119884361 (epub) Subjects: LCSH: Water chemistry. | Environmental chemistry. Classification: LCC GB855 .J46 2023 (print) | LCC GB855 (ebook) | DDC 551.4801/54—dc23/eng20221004 LC record available at https://lccn.loc.gov/2022022564 LC ebook record available at https://lccn.loc.gov/2022022565 Cover Design: Wiley Cover Image: © Nick Brundle Photography/Getty Images

B R I E F CO N T E N T S

Preface

Part I

1

Fundamental Concepts

Getting Started with the Fundamental Concepts

xix

3

2

Concentration Units

13

3

Thermodynamic Basis of Equilibrium

35

4

Manipulating Equilibrium Expressions

69

Part II Solving Chemical Equilibrium Problems 5

Getting Started with ­Solving Equilibrium Problems

6

Setting Up Chemical Equilibrium Calculations

105

7

Algebraic Solutions to Chemical Equilibrium ­Problems

131

8

Graphical Solutions to Chemical Equilibrium ­Problems

155

9

Computer Solutions to Chemical Equilibrium ­Problems

183

95

Part III Acid–Base Equilibria in Homogenous Aqueous Systems 10

Getting Started with Acid–Base Equilibrium in ­ Homogenous Aqueous ­Systems

209

11

Acids and Bases

219

12

Acid–Base Titrations

249

13

Alkalinity and Acidity

287

Part IV Other Equilibria in Homogenous ­ Aqueous ­Systems 14

 etting Started with Other Equilibria in Homogeneous G Aqueous Systems

319

15

Complexation

327

16

Oxidation and Reduction

371 v

vi   Brief contents

Part V Heterogeneous Systems

17

Getting Started with ­Heterogeneous Systems

425



18

Gas–Liquid Equilibria

431



19

Solid–Liquid Equilibria

463

Part VI Beyond Dilute Solutions at Equilibrium

20

 etting Started with Beyond Dilute Solutions at G ­Equilibrium

501



21

T hermodynamics Revisited: The Effects of Ionic Strength, Temperature, and Pressure

509



22

Aquatic Chemistry of ­Surfaces

535



23

Chemical Kinetics of Aquatic Systems

559



24

 utting It All Together: Integrated Case Studies in P Aquatic Chemistry

593

Appendices Appendix A: Background Information

617

Appendix B: Equilibrium Revisited

625

Appendix C: Summary of Procedures

633

Appendix D: Selected Equilibrium Constants

641

Appendix E: Animations and Example Spreadsheet Files

653

Appendix F: Nanoql SE

661

Index Biographical Index

669 677

CO N T E N T S

Preface

xix

Part I Fundamental Concepts 1 Getting Started with the Fundamental Concepts  3 1.1  Introduction  3 1.2  Why Calculate Chemical Species Concentrations at Equilibrium?  3 1.3  Primary Variables: Importance of pH and pe  6 1.4  Properties of Water  7 1.5  Part I Roadmap  9 1.6  Chapter Summary  9 1.7 Part I Case Study: Can Methylmercury be Formed Chemically in Water?  10 Chapter Key Ideas  11 Chapter Glossary  11 Historical Note: S.P.L. Sørensen and the p in pH  11 Chapter References  12

2 Concentration Units  13 2.1   Introduction  13 2.2   Units Analysis  13 2.3   Molar Concentration Units  14 2.4   Mass Concentration Units  19 2.5   Dimensionless Concentration Units  24 2.6   Equivalents  25 2.7   Review of Units Interconversion  26 2.8   Common Concentration Units in the Gas Phase  27 2.9   Common Concentration Units in the Solid Phase  28 2.10  Activity  28 2.11  Chapter Summary  30 2.12 Part I Case Study: Can Methylmercury Be Formed Chemically in Water?  30

vii

viii   Contents Chapter Key Ideas  31 Chapter Glossary  31 Historical Note: Amadea Avogadro and Avogadro’s Number  32 Problems  33 Chapter References  34

3 Thermodynamic Basis of Equilibrium  35 3.1   Introduction  35 3.2   Thermodynamic Properties  36 3.3  Why Do We Need Thermodynamics to Calculate Species Concentrations?  39 3.4   Thermodynamic Laws  42 3.5   Gibbs Free Energy  45 3.6  Properties of Thermodynamic Functions  48 3.7  Changes in Thermodynamic Properties During Chemical Reactions  50 3.8   Relating Gibbs Free Energy to Species Concentrations  55 3.9   Chemical Equilibrium and the Equilibrium Constant  60 3.10  Chapter Summary  62 3.11 Part I Case Study: Can Methylmercury Be Formed Chemically in Water?  63 Chapter Key Ideas  63 Chapter Glossary  64 Historical Note: Josiah Willard Gibbs  66 Problems  67 Chapter References  68

4 Manipulating Equilibrium Expressions  69 4.1  Introduction  69 4.2  Chemical and Mathematical Forms of Equilibria  69 4.3  Units of Equilibrium Constants  73 4.4  Reversing Equilibria  75 4.5  Effects of Stoichiometry  76

4.6  Adding Equilibria  78 4.7  Creating Equilibria  81 4.8  Chapter Summary  87 4.9 Part I Case Study: Can Methylmercury Be Formed Chemically in Water?  87 Chapter Key Ideas  88 Chapter Glossary  89

Contents   ix

Historical Note: Henri-­Louis Le Châtelier and Le Châtelier’s Principle  89 Problems  90 Chapter References  91

Part II Solving Chemical Equilibrium Problems 5 Getting Started with ­Solving Equilibrium Problems  95 5.1  Introduction  95 5.2  A Framework for Solving Chemical Equilibrium Problems  95 5.3  Introduction to Defining the Chemical System  97 5.4  Introduction to Enumerating Chemical Species  98 5.5 Introduction to Defining the Constraints on Species ­Concentrations  98 5.6  Part II Roadmap  100 5.7  Chapter Summary  100 5.8  Part II Case Study: Have You Had Your Zinc Today?  101 Chapter Key Ideas  101 Chapter Glossary  101 Historical Note: “Active Mass” and Familial Relations  102 Chapter References  103

6 Setting Up Chemical Equilibrium Calculations  105 6.1  Introduction  105 6.2  Defining the Chemical System  105 6.3  Enumerating Chemical Species  106 6.4  Defining the Constraints on Species Concentrations  112 6.5  Review of Procedures for Setting up Equilibrium Systems  120 6.6  Concise Mathematical Form for Equilibrium Systems  121 6.7  Chapter Summary  122 6.8  Part II Case Study: Have You Had Your Zinc Today?  123 Chapter Key Ideas  126 Chapter Glossary  126 Historical Note: Salts of the Ocean  127 Problems  129 Chapter References  130

7 Algebraic Solutions to Chemical Equilibrium ­Problems  131 7.1  Introduction  131 7.2  Background on Algebraic Solutions  131

x   Contents 7.3  Method of Substitution  133 7.4  Method of Approximation  139 7.5  Chapter Summary  148 7.6  Part II Case Study: Have You Had Your Zinc Today?  148 Chapter Key Ideas  152 Historical Note: What’s in a Name?  152 Problems  153

8 Graphical Solutions to Chemical Equilibrium ­Problems  155 8.1  Introduction  155 8.2  Log Concentration and pC-­pH Diagrams  156 8.3  Using pC-­pH Diagrams with More Complex Systems  162 8.4  Special Shortcuts for Monoprotic Acids  167 8.5  When Graphical Methods Fail: The Proton Condition  171 8.6  Chapter Summary  177 8.7  Part II Case Study: Have You Had Your Zinc Today?  178 Chapter Key Ideas  179 Chapter Glossary  180 Historical Note: Who Was First?  180 Problems  181 Chapter Reference  182

9 Computer Solutions to Chemical Equilibrium ­Problems  183 9.1  Introduction  183 9.2  Chapter Problem  183 9.3  Spreadsheet Solutions  184 9.4  Equilibrium Calculation Software  188 9.5  Nanoql SE  190 9.6  The Tableau Method and Other Equilibrium Calculation Apps  192 9.7  Visual MINTEQ  201 9.8  Chapter Summary  202 9.9  Part II Case Study: Have You Had Your Zinc Today?  202 Chapter Key Ideas  203 Chapter Glossary  203 Historical Note: ALGOL to VBA  203 Problems  204 Chapter References  205

Contents   xi

Part III Acid–Base Equilibria in Homogenous Aqueous Systems 10 Getting Started with Acid–Base Equilibrium in ­Homogenous Aqueous ­Systems  209 10.1  Introduction  209 10.2  Homogeneous Systems  209 10.3  Types of Reactions in Homogeneous Systems  211 10.4  The Wonderful World of Acids and Bases  212 10.5  Part III Roadmap  215 10.6  Chapter Summary  215 10.7  Part III Case Study: Acid Rain  215 Chapter Key Ideas  216 Chapter Glossary  216 Historical Note: “An Evil of the Highest Magnitude”  217 Chapter References  218

11 Acids and Bases  219 11.1  Introduction  219 11.2  Definitions of Acids and Bases   219 11.3  Acid and Base Strength   223 11.4  Polyprotic Acids  228 11.5  Alpha Values (Distribution Functions)  236 11.6  Chapter Summary  239 11.7  Part II Case Study: Acid Rain  239 Chapter Key Ideas  241 Chapter Glossary  242 Historical Note: Why Is a Base a Base?  242 Problems  243 Addendum: A Surprising Exact Solution  245 Chapter References  248

12 Acid–Base Titrations  249 12.1  Introduction  249 12.2  Principles of Acid–Base Titrations  250 12.3  Equivalence Points  255 12.4  Titration of Polyprotic Acids  265 12.5  Buffers  269 12.6 Interpretation of Acid–Base Titration Curves with Complex Mixtures  277

xii   Contents 12.7  Chapter Summary  279 12.8  Part III Case Study: Acid Rain  280 Chapter Key Ideas  282 Chapter Glossary  283 Historical Note: Mohr about Titrations  284 Problems  285 Chapter References  286

13 Alkalinity and Acidity  287 13.1  Introduction  287 13.2  Alkalinity and the Acid Neutralizing Capacity  287 13.3  Alkalinity and the Charge Balance   290 13.4  Characteristics of Alkalinity and Acidity   292 13.5  Using the Definitions of Alkalinity to Solve Problems  302 13.6  Effects of Other Weak Acids and Bases on Alkalinity  308 13.7  Chapter Summary  310 13.8  Part III Case Study: Acid Rain  310 Chapter Key Ideas  311 Chapter Glossary  312 Historical Note: Can You Pass the Litmus Test?  313 Problems  314 Chapter References  316

Part IV Other Equilibria in Homogenous ­Aqueous ­Systems 14 Getting Started with Other Equilibria in Homogeneous Aqueous Systems  319 14.1  Introduction  319 14.2  Electron-­Sharing Reactions  319 14.3  Electron Transfer  321 14.4  Part IV Roadmap  323 14.5  Chapter Summary  323 14.6 Part IV Case Study: Which Form of Copper Plating Should You Use?  323 Chapter Key Ideas  324 Historical Note: Hauptvalenz and Nebenvalenz  324 Chapter References  325

Contents   xiii

15 Complexation  327 15.1  Introduction  327 15.2  Metals  327 15.3  Ligands  330 15.4  Equilibrium Calculations with Complexes  335 15.5  Systems with Several Metals and Ligands  345 15.6  Applications of Complexation Chemistry  357 15.7  Chapter Summary  361 15.8 Part IV Case Study: Which Form of Copper Plating Should You Use?  362 Chapter Key Ideas  364 Chapter Glossary  365 Historical Note: British Anti-­Lewisite – A WMD-­Inspired Ligand  366 Problems  368 Chapter References  369

16 Oxidation and Reduction  371 16.1   Introduction  371 16.2   A Few Definitions  371 16.3   Balancing Redox Reactions  374 16.4   Which Redox Reactions Occur?  383 16.5  Redox Thermodynamics and Oxidant and Reductant Strength  386 16.6   Manipulating Half Reactions  393 16.7  Algebraic Equilibrium Calculations in Systems Undergoing Electron Transfer  396 16.8  Graphical Representations of Systems Undergoing Electron Transfer  399 16.9   Applying Redox Equilibrium Calculations to the Real World  413 16.10  Chapter Summary  414 16.11 Part IV Case Study: Which Form of Copper Plating Should You Use?  415 Chapter Key Ideas  417 Chapter Glossary  418 Historical Note: Walther Hermann Nernst  419 Problems  420 Chapter References  422

xiv   Contents

Part V  Heterogeneous Systems 17 Getting Started with ­Heterogeneous Systems  425 17.1  Introduction  425 17.2  Equilibrium Exchange Between Gas and Aqueous Phases  426 17.3  Equilibrium Exchange Between Solid and Aqueous Phases  427 17.4  Part V Roadmap  428 17.5  Chapter Summary  428 17.6  Part V Case Study: The Killer Lakes  428 Chapter Key Ideas  429 Historical Note: “A Spirit Case and a Gasogene”  429 Chapter References  430

18 Gas–Liquid Equilibria  431 18.1  Introduction  431 18.2  Raoult’s Law and Henry’s Law  431 18.3  Equilibrium Calculations Involving Gas–Liquid Equilibria  438 18.4  Dissolved Carbon Dioxide  449 18.5  Chapter Summary  456 18.6  Part V Case Study: The Killer Lakes  456 Chapter Key Ideas  457 Chapter Glossary  458 Historical Note: A Brief History of Carbon Dioxide  459 Problems  460 Chapter References  462

19 Solid–Liquid Equilibria  463 19.1  Introduction  463 19.2  Saturation and the Activity of Pure Solids  463 19.3  Equilibrium Calculations with Solid–Liquid Equilibria  466 19.4  Factors Affecting Metal Solubility  474 19.5  Solubility of Calcium Carbonate  480 19.6  Models for the Acid–Base Chemistry of Natural Waters  484 19.7  Chapter Summary  491 19.8  Part V Case Study: The Killer Lakes  491 Chapter Key Ideas  492 Chapter Glossary  493 Historical Note: Black Smokers and White Smokers  493 Problems  494

Contents   xv

Addendum: Information Requirements  497 Chapter References  498

Part VI Beyond Dilute Solutions at Equilibrium 20 Getting Started with Beyond Dilute Solutions at ­Equilibrium  501 20.1  Introduction  501 20.2  Extensions to Nonideal and Nonstandard Conditions  502 20.3  The Strange World of Surfaces  503 20.4  Nonequilibrium Conditions  504 20.5  Integrated Case Studies  504 20.6  Part VI Roadmap  505 20.7  Chapter Summary  505 Chapter Key Ideas  506 Chapter Glossary  506 Historical Note: “Harcourt, Come to Me!”  506 Chapter References  507

21 Thermodynamics Revisited: The Effects of Ionic Strength, Temperature, and Pressure  509 21.1  Introduction  509 21.2  Effects of Ionic Strength  510 21.3  Effects of Temperature on Equilibrium Constants  522 21.4  Effects of Pressure on Equilibrium Constants  528 21.5  Chapter Summary  529 Chapter Key Ideas  530 Chapter Glossary  531 Historical Note: Jacobus Henricus van’t Hoff  531 Problems  532 Chapter References  534

22 Aquatic Chemistry of ­Surfaces  535 22.1  Introduction  535 22.2  Nomenclature  535 22.3  Isotherms and Ion Exchange  538 22.4  Introduction to Surface Complexation Modeling  543 22.5  Surface Complexation Modeling  546 22.6  Chapter Summary  552

xvi   Contents Chapter Key Ideas  553 Chapter Glossary  553 Historical Note: From “Cat’s Cradle” to the “Swiss Model” to Surface Complexation Modeling  554 Problems  555 Addendum: The Freundlich Isotherm and Adsorption Equilibria  556 Chapter References  557

23 Chemical Kinetics of Aquatic Systems  559 23.1  Introduction  559 23.2  The Need for Chemical Kinetics  560 23.3  Reaction Rates  561 23.4  Common Rate Expressions  569 23.5  More Complex Kinetic Forms  577 23.6  Effects of Temperature and Ionic Strength on Kinetics  582 23.7  Chapter Summary  587 Chapter Key Ideas  587 Chapter Glossary  588 Historical Note: Arrhenius, Chick, and Foote  589 Problems  590 Chapter References  592

24 Putting It All Together: Integrated Case Studies in Aquatic Chemistry  593 24.1  Introduction  593 24.2  Integrated Case Study 1: Metal Finishing  594 24.3  Integrated Case Study 2: Oxidation of Fe(+II) by Oxygen  598 24.4 Integrated Case Study 3: Inorganic Mercury Chemistry in Natural Waters  603 24.5  Integrated Case Study 4: Phosphate Buffers  607 24.6  Integrated Case Study 5: Global Climate Change  610 24.7  Chapter Summary  613 Historical Note: Stumm and Morgan  614 Chapter References  614

APPENDIX A: Background Information  617 A.1  Introduction  617 A.2  Chemical Principles  617 A.3  Mathematical Principles  619 A.4  Spreadsheet Skills  620

Contents   xvii

Chapter Key Ideas  623 Chapter Glossary  623 Useful Physical Constants and Conversions  623

APPENDIX B: Equilibrium Revisited  625 B.1  Introduction  625 B.2  Equilibrium and Steady State  625 B.3  Energy Minimization and Algebraic Solutions  628 Chapter Key Ideas  631 Chapter Glossary  631

APPENDIX C: Summary of Procedures  633 C.1   Oxidation States and Balancing Reactions  633 C.2   Setting Up Chemical Equilibrium Systems (Section 6.5)  634 C.3   Algebraic Solution Techniques  635 C.4   Graphical Solutions  635 C.5   Computer Solutions: Tableau Method (Section 9.6.6)  637 C.6   Acid–Base Titrations  638 C.7   Complexation (Section 15.4.4)  638 C.8   Ionic Strength Effects (Section 21.2.7)  639 C.9   Surface Complexation Modeling Method (Section 22.5.4)  639 C.10  Chemical Kinetics (Section 23.3.4)  639

APPENDIX D: Selected Equilibrium Constants  641 Chapter References  651

APPENDIX E: Animations and Example Spreadsheet Files  653 E.1  Introduction to Animations  653 E.2 Variation of the Equilibrium pH of a Monoprotic Acid Solution with the Total Acid Concentration and Ka  653 E.3  How to Draw pC-­pH Diagrams for Monoprotic Acids  654 E.4 Equilibrium pH During the Titration of a Monoprotic Acid with a Strong Base  656 E.5  Spreadsheet Examples  657

xviii   Contents

APPENDIX F: Nanoql SE  661 F.1  Introduction  661 F.2  Entering Your System  661 F.3  How to Solve Systems and Vary System Parameters  663 F.4  Nanoql SE Examples  666 Chapter Reference  668

Index  669 Biographical Index  677

P R E FAC E

FOR THE STUDENT It is my honor and privilege to help you on your journey through aquatic chemistry with the second edition of this book. I hope that this text will deepen your appreciation of water. As with the first edition, my goal is that you come to look at rain, a stream, or tap water and admire the wealth of chemistry in every drop. Each chapter of the text is built to help you learn. Overall, the text is designed to mimic the Socratic method of teaching. You will encounter Thoughtful Pause1 boxes throughout the text. Please stop and try to answer the questions in the Thoughtful Pauses before continuing to read. You will find over 215 important definitions and over 300 Key Ideas highlighted in the margins and summarized at the end of each chapter. Do you learn better through examples? The text contains over 180 worked examples. To help you integrate the material, each part of the text contains a case study that is developed through several chapters. The last chapter of the text contains five integrative case studies. These are large, practical problems that combine many of the tools you will learn to use throughout the text. The first edition of this text was accompanied by a user-­friendly equilibrium calculation app called Nanoql. In this edition, the app has been ported to Excel as Nanoql SE (spreadsheet edition). The text provides instructions on how to use Nanoql SE, along with a number of examples. To give you an appreciation for the fascinating history of the ideas presented in the text, the number of Historical Notes in the second edition has been greatly expanded. You will travel from the battlefields of World War II to deep-­sea trenches. You will meet famous long-­ bearded chemists and an underappreciated female scientist who provided the first experimental evidence for global climate change. Along the way, you will explore the connections between aquatic chemistry and Sherlock Holmes, Lewis Carroll, and Kurt Vonnegut. And, of course, what textbook on water chemistry would be complete without haikus? Each part of the text is summarized in seventeen syllables to aid in reinforcing the main ideas.

FOR THE INSTRUCTOR The text continues to be appropriate for a one-­semester course in water chemistry (or in an environmental chemistry course emphasizing aquatic systems) for advanced undergraduates, first-­year graduate students, or self-­studying professionals. This edition features extended coverage of the effects of global climate change on the aquatic environment and surface complexation modeling. The number of end-­of-­chapter problems has been nearly doubled to over 340 problems in this edition. In addition, the Nanoql equilibrium 1  Professor James J. Morgan (who you will meet throughout the text, but especially in Chapter  24), referred to them as “Thoughtful Paws” boxes.

xix

xx   Preface calculation app has been updated to a more accessible Excel-­based version. Numerous other spreadsheet solutions are developed in the text.

ACKNOWLEDGMENTS Once again, I wish to acknowledge my great debt to the aquatic chemistry teachers I have had. Since the first edition, we lost Jim Morgan and Phil Singer. Their legacy lives on in countless labs, treatment plants, and classrooms through the world, and also in the pages of this book. I also am indebted to my colleagues at the University at Buffalo for their steadfast support. They have made me a better teacher. Numerous comments from my students and other readers of the first edition have influenced and improved the text. I am deeply appreciative to the staff at Wiley for their guidance and support. In particular, I am thankful for the efforts of Paul Sayer, Kimberly Monroe-­Hill, Cheryl Ferguson, Michael New, and Judit Anbu Hena Daniel in bringing this book to fruition. James N. Jensen Buffalo, NY October 2022

PA R T I

Fundamental Concepts Chapter 1: Getting Started with the Fundamental Concepts Chapter 2: Concentration Units Chapter 3: Thermodynamic Basis of Equilibrium Chapter 4: Manipulating Equilibrium Expressions

The fundamental things apply As time goes by. –Herman Hupfeld (1931) Part I Haiku Use molar units: Equilibrium constants are dimensionless. ...the beginning is the most important part of any work... –Plato (c. 375 bce)

CHAPTER 1

Getting Started with the Fundamental Concepts 1.1  INTRODUCTION The first part of this text reviews fundamental concepts that must be mastered prior to learning how to calculate and interpret species concentrations in aquatic systems. In this chapter, the motivation for studying chemical species and a few general principles concerning aquatic systems are presented. In Section 1.2, the motivation for why engineers and scientists are interested in individual chemical species concentrations at equilibrium will be discussed. Important water quality parameters, called primary variables, are introduced in Section  1.3. It is impossible to study water chemistry without a little knowledge of the structure of water. A few of the unique properties of water will be explored in Section 1.4, especially as they relate to the chemical reactions that occur in water. In Section 1.5, a road map for Part I of the text is presented and discussed. Finally, the Part I case study is presented at the conclusion of this chapter. Before beginning Part I of the text, you are urged to review the chemistry background material in Appendix A (Section A.2).

1.2  WHY CALCULATE CHEMICAL SPECIES

CONCENTRATIONS AT EQUILIBRIUM?

1.2.1  Overview The bulk of this book is dedicated to the calculation of species concentrations at equilibrium. The focus here is on chemical species that undergo chemical reactions; in other words, reactive species. More specifically, the emphasis here is on chemical species which react with water. Reactions with water are called hydrolysis reactions (from the Greek hydōr water + lyein to loosen). When substances react with water, numerous other compounds can be formed. Indeed, the richness of aquatic chemistry stems from the large number of substances that react not only with water but also with the products of myriad other hydrolysis reactions. This richness is illustrated in Figure 1.1. Inputs of chemical species (from aqueous discharge, runoff, atmospheric deposition, and dissolution from sediments) react with water to form hydrolysis products. The hydrolysis products and input chemicals react further to increase the complexity of aquatic systems.

hydrolysis reactions: reactions with water Key idea: Hydrolysis reactions produce a wealth of dissolved chemical species

3

4  

Chapter 1  Getting Started with the Fundamental Concepts atmospheric species

inputs + H2O

hydrolysis products other products

FIGURE 1.1  Complexity of Aquatic Systems (rain cloud image: OpenClipart-­Vectors/Pixabay)

So why so much interest in calculating the equilibrium concentrations of chemical species? This question is really two questions. First, why calculate the concentrations of individual chemical species? Second, why calculate species concentrations at equilibrium?

1.2.2  Importance of Individual Chemical Species

Key idea: The ability to calculate the concentrations of individual chemical species is critically important in analyzing many environmental problems

Throughout this text, you will see that knowing the concentrations of individual chemical species is critically important in analyzing many environmental problems. At first glance, this statement may not make sense. After all, many environmental regulations are based on total concentrations of classes of compounds rather than on the concentrations of individual species. Should you be more concerned about the total amount of mercury or phenol or ammonia than about individual species stemming from the hydrolysis of mercury, phenol, or ammonia? In fact, you will find that individual species frequently are more important. Three general examples will illustrate this point. First, adverse impacts on human health and ecosystem viability may be due to only one or several of a large number of related hydrolysis products. A prime example is the transition metals (such as mercury, copper, zinc, cadmium, iron, and lead), in which toxicity varies dramatically among the hydrolysis products. Another example is cyanide. Hydrogen cyanide (HCN) is much more toxic to humans than cyanide ion (CN–). Second, the success of engineered treatment systems may depend on knowledge of the concentrations of key individual species. Since hydrolysis products vary in their physical, chemical, and biochemical properties, the design and operation of treatment processes depend on quantitative models for the concentrations of individual chemical species. For example, the addition of gaseous chlorine to wastewater for disinfection results in the formation of many chemical species (including HOCl, OCl–, NH2Cl, and NHCl2), each of which differs in its ability to inactivate (i.e., kill) microorganisms. Third, individual species vary greatly in how readily they cross cell membranes or cell walls and are assimilated by aquatic biota. Thus, understanding the cycling of trace nutrients in the aquatic environment (and humankind’s impact on nutrient cycling) requires knowledge of the concentrations of individual chemical species. As an example of the importance of the concentrations of individual chemical species, consider the soup created when copper sulfate crystals, CuSO4(s),1 are added 1  In this example, (s) represents a solid species and (aq) represents a dissolved species in water. See Appendix A (Section A.2.2) for a review of this nomenclature.

1.2  |  Why Calculate Chemical Species Concentrations at Equilibrium?    5

to a reservoir for algae control. The CuSO4(s) dissolves in water to form a copper-­ containing ion (called the aquo cupric ion) and sulfate. The structure of the aquo cupric ion is usually abbreviated as Cu2+. The Cu2+ ions thus formed react very quickly with water to form a number of hydrolysis products, including CuOH+, Cu(OH)2(aq), Cu(OH)3–, Cu(OH)42–, and Cu2(OH)22+. Under certain chemical conditions, copper may precipitate as CuO(s). As you spread the copper sulfate from the back of a boat, carbon dioxide in the atmosphere is equilibrating with the reservoir water to form its own hydrolysis products. The hydrolysis products of carbon dioxide are H2CO3, HCO3–, and CO32–. The aquo cupric ion will react to some extent with the hydrolysis products of carbon dioxide to form CuCO3(aq), Cu(CO3)22–, and perhaps even solids containing copper and carbonate (CO32–). By adding one copper compound to a natural water body, you may be faced with accounting for as many as 10 copper-­containing species even in a relatively simple chemical model. Of course, the real world is even more complex. The reservoir water contains many more species that can react with copper than just hydroxide (OH–) and carbonate. A realistic model for copper in the reservoir would have to include the reactions of Cu2+ with (among other chemical species) chloride, amino acids, ammonia, particulates, and microorganisms. In reality, the act of throwing copper sulfate crystals into Key idea: the reservoir will produce dozens of chemical species containing copper. Doses depend Why should you care that copper sulfate forms many copper-­containing species on both the in a lake? Remember that copper is added to kill algae. It is well-­established that required copper toxicity to algae is due almost entirely to one chemical species: Cu2+ (Jackson concentration of the target and Morgan 1978). Thus, to determine the copper sulfate dose, you must be able to individual species and the calculate the concentration of Cu2+ after a certain amount of copper sulfate is added chemistry of the water to the reservoir. Since the Cu2+ concentration usually is exceedingly small, this is akin to counting needles of Cu2+ in this haystack of copper-­containing species. In practice, you would back-­calculate the copper sulfate dose required to achieve a required level of Cu2+. Even if two reservoirs had the same amounts of algae, different water chemistries in the reservoirs may lead to very different copper sulfate doses to achieve the same Cu2+ concentration. The chemistry of the water determines how the required concentration of one species (Cu2+) is translated back into a copper sulfate dose. The process of relating a dose to a required concentration of an individual chemical species is illustrated in Figure 1.2. The arrows in Figure 1.2 indicate chemical reactions that must be included in a mathematical model to allow for the determination of the copper sulfate dose. In this text, you will learn the tools to make quantitative decisions to solve similar problems in the aquatic environment.

You must add this much copper sulfate...

x kg CuSO4(s) y µg/L Cu2+ Cu2+ + SO42– hydrolysis products

…to get this much aquo cupric ion

FIGURE 1.2  Qualitative Relationship Between the Dose Required and End Species Concentrations Desired

6  

Chapter 1  Getting Started with the Fundamental Concepts

1.2.3  Importance of Equilibrium

chemical kinetics: the study of chemical reaction rates Key idea: The equilibrium state is a useful approximation of many aquatic chemical systems

An entire chapter of this book (Chapter  3) is devoted to developing the thermodynamic basis of equilibrium. For the present, you can think of the equilibrium state as the condition in which the concentrations of all chemical species do not change with time. To impose equilibrium on a chemical system, the interesting and important time-­ dependent nature of chemical concentrations are excluded. The study of the rates of chemical reaction is called chemical kinetics and is covered in ­Chapter 23. Why constrain the discussion mainly to the equilibrium state here, with shorter ­coverage of chemical kinetics? There are two reasons for focusing on equilibrium. First, many of the chemical reactions you will examine in this text are fast. For example, the reaction of H+ and OH– to form water occurs on the time scale of 10–5 s at natural water conditions (Morgan and Stone 1985). Second, equilibrium models give insight into chemical systems, even when kinetics are known to be important. As an example, equilibrium chemistry provides a good framework to understand coagulation chemistry in drinking water treatment. Chemical kinetics also play a role in coagulation, but equilibrium models remain useful. Thus, the focus on equilibrium is somewhat constraining but not overly restrictive.

1.3  PRIMARY VARIABLES: IMPORTANCE

OF pH AND pe

primary variable: a species concentration (or function of a species concentration) that controls the chemistry of a system pH: a primary variable controlling H+ (proton) transfer and defined by –log(H+ activity) pe: a primary variable controlling electron transfer and defined by –log(e– activity)

As discussed in Section 1.2, it is important to know the concentrations of individual chemical species in aquatic systems. Some species may be of greater or lesser importance, but a small number of chemical species often control the chemistry of an aquatic system. “Controlling the chemistry” means that certain chemical species play a dominant role in determining the concentrations of other chemical species. The concentrations of the controlling species are expressed in parameters called primary variables.2 The two most important primary variables in water are pH and pe. The primary variable pH is defined as the negative of the log of the activity of the ion H+.3 Similarly, pe is defined as the negative of the log of the activity of the electron, e–. (See the Historical Note at the end of this chapter for the story of what the p in pH and pe means.) Since the hydrogen atom contains one proton and one electron in its most abundant state, H+ is a proton. Thus, the transfer of H+ is sometimes called proton transfer or acid–base chemistry. Electron transfer is also called oxidation-­reduction chemistry or redox chemistry. Acid–base chemistry is discussed in more detail in Chapter 11. Redox chemistry is discussed in more detail in Chapter 16. As noted in Appendix A, a low pH corresponds to high activity of H+. Conditions of high H+ activity are called acidic conditions. Thus, low pH (high H+ activity) means acidic conditions. High pH (low H+ activity) corresponds to basic conditions. Low pe (high e– activity) corresponds to reducing conditions, and high pe (low e– activity) corresponds to oxidizing conditions.

 More formally, a primary variable is the single variable that determines the ratios of species concentrations (see Sillén 1959). Primary variables also are called master variables. 3  Activity is an idealized concentration. It is discussed further in Section 2.10 and Chapter 3. The characteristics of the p function are explored in Appendix A (Section A.3). 2

1.4  |  Properties of Water    7

1.4  PROPERTIES OF WATER 1.4.1  Introduction Water is a unique substance. In freshman chemistry courses, you probably learned of the periodic nature of the properties of the elements. Elements in the same column of the periodic table share similar chemical properties. Yet water, H2O, has very different physical and chemical properties than other dihydrogen complexes of elements in oxygen’s column of the periodic table. For example, water is a liquid under standard temperature and pressure conditions, but H2S is a gas. The unique properties of water stem from the large differences between the affinity of oxygen and hydrogen for electrons. Oxygen has a much higher affinity for electrons than hydrogen, resulting in a highly polarized bond between O and H. In fact, the oxygen atoms in water are partially negatively charged and the hydrogen atoms are partially positively charged, as illustrated in Figure  1.3. As a result of this polar bond and the difference in partial charge, oxygen atoms in water have the ability to form weak, but important, chemical bonds with more than two hydrogen atoms. These weak bonds are called hydrogen bonds. Without exaggeration, hydrogen bonds influence every molecular interaction in the aquatic environment. Consider, for example, the influence of hydrogen bonds on ice.4 The structure of ice is different than the structure of water. In ice, all the water molecules participate in four hydrogen bonds arranged in a tetrahedral shape. The shape results in an open structure for ice and a corresponding density of 0.92 g/cm3. The density of ice is less than the density of liquid water at 0°C, so ice floats. Hydrogen bonds reduce the O-­H bond length by only about 4% in ice compared to water. This small change is responsible for the open structure and low density of ice. If not for this impact of hydrogen bonds, ice would sink and water bodies would have frozen solid during the ice ages. If hydrogen bonds were slightly different, life on Earth might not have survived long enough to allow you to read this book. Hydrogen bonds affect the three-­dimensional shape of water. The bond distance between oxygen in water and the hydrogen-­bonded hydrogen is about twice the bond distance between the oxygen in water and one of its own hydrogen atoms. Hydrogen bonding allows for the formation of very large clusters of water molecules (about 400 molecules per cluster; Luck 1998). The clusters are shifting constantly since the average lifetime of a hydrogen bond is only a few picoseconds (10–12 s). The shifting clusters of water molecules contribute to the solubilization of ions in water, as will be discussed in Section 1.4.2.

Oδ– Hδ+

104.5°

Hδ+

FIGURE 1.3  Charge Distribution among the Atoms in Water

 Hydrogen bonds do not require oxygen atoms. In general, hydrogen bonds may be formed between hydrogen atoms and an electronegative atom bonded to another hydrogen atom. Electronegative atoms have a high affinity for electrons and include O and N.

4

hydrogen bonds: weak bonds formed between an electronegative atom (e.g., oxygen) bonded to an H atom and another H atom

8  

Chapter 1  Getting Started with the Fundamental Concepts

1.4.2  Solubility of Ionic Species Throughout this text, it will be assumed that salts containing sodium, potassium, or chloride at low concentrations dissolve nearly completely in water to form ions. Why are salts so soluble in water? Hydrogen bonding allows for the orientation of a large number of molecules simultaneously. This orientation reduces the field strength of an applied field significantly. More specifically, the orientation reduces the attractive forces between pairs of anions and cations. As the attractive forces are reduced, the salt can be solubilized. For example, you can show that water significantly reduces the attractive forces between Na+ and Cl– (see Worked Example 1.1).

Worked Example 1.1 

Solubilization of Sodium Chloride in Water

Why does NaCl dissolve in water? Solution One reason NaCl dissolves in water is that the potential energy of interaction between Na+ and Cl– is reduced in water. To illustrate this point, calculate the potential energy of interaction between Na+ and Cl– separated by 1 nm in water. The potential energy of two particles of charge q1 and q2 separate by a distance r is given by Coulomb’s law: potential energy

4

q1q2 0

r

where 0 is the permittivity of a vacuum (= 8.854×10–12 J–1C2m–1) and is the relative permittivity or dielectric constant (= 1 for a vacuum and 80 for water). The unit charge on Na+ is +1.602×10–19 C. It is 1 –1.602×10–19 C on Cl–. (The term is Coulomb’s constant, ke, after 4 0 Charles-­Augustin de Coulomb, 1736–1806.) Plugging in the values, the potential energy is –139 kJ per mole in a vacuum and –1.7 kJ per mole in water. Here, the negative sign indicates an energy of attraction. Recall that one mole is 6.022×1023 ions). The attractive potential energy of interaction between Na+ and Cl– in water is very small (only about one-­tenth of the energy of a hydrogen bond). Sodium chloride is soluble in water, mainly because the attractive energy between Na+ and Cl– is reduced by water. (Note: The energy of the system also is altered as the atoms in the salt become more disordered and the water molecules become more ordered around the ions.)

1.4.3  Hydration hydration: the process of forming compounds with water

Once dissolved, the ions become hydrated. In other words, the ions form compounds (called aquo complexes) with water. Water is adept at hydrating both cations

1.6  |  Chapter Summary    9

and anions. Why? Recall from Section  1.4.1 that water has both a partially positive pole on H and partially negative pole on O. Water is said to be very polar. For ­example, when alum (aluminum sulfate) is added to water at low pH, Al3+ forms and is quickly hydrated to Al(H2O)63+. Such structures are frequently abbreviated by omitting the waters of hydration. For example, sometimes Cu2+ is written to ­represent Cu(H2O)42+. Even the proton is hydrated. Free H+ probably does not exist in solution. The free proton is hydrated to form H(H2O)+ = H3O+, H(H2O)2+ = H5O2+, and others. For convenience, the collection of hydrated protons is abbreviated as H+ or H3O+.

1.5  PART I ROADMAP Part I of this book contains the fundamental concepts that must be mastered before calculating the equilibrium concentrations of chemical species. If your chemistry background is limited to freshman chemistry only, you should study each section of each chapter and work the problems assigned. In Chapter 2, concentration units are reviewed. The main take-­home lesson from Chapter 2 is the definition of molar concentration units and why we use molar concentration units for equilibrium calculations. If you have a strong chemistry background, you may wish to focus on Sections 2.3, 2.4, 2.6, and 2.7. Chapter 3 develops the thermodynamic basis of equilibrium. It is critical for you to understand the definitions of the main thermodynamic properties in Sections 3.2 through 3.5. A key thermodynamic property, the equilibrium constant, is developed in Section 3.9. In Chapter 4, the rules for interpreting and manipulating equilibrium expressions are reviewed. The manipulation of equilibrium expressions is a skill that must be honed before attempting equilibrium calculations. Please work through Chapter  4 carefully and practice manipulating equilibria.

1.6  CHAPTER SUMMARY In this chapter, you learned that hydrolysis reactions lead to the formation of a number of chemical species in natural waters. Individual chemical species vary greatly in their importance in natural and engineered systems. Thus, you must learn to calculate the concentrations of individual chemical species, rather than just measuring total elemental compositions in water. Chemical species concentrations may be controlled by rates (kinetics) or thermodynamics (equilibrium). The focus is on systems at equilibrium in this text because many aqueous phase reactions are fast and equilibrium models are useful in many applications. The importance of pH [= –log(H+ activity)] and pe [= –log(e– activity] were discussed. Both pH and pe are called primary variables. In many cases, pH controls H+ transfer (also called proton transfer or acid–base chemistry), and pe controls electron transfer (also called oxidation-­reduction or redox chemistry). Water is a unique chemical species. Because of the charge distribution in its ­constituent atoms, water participates in weak chemical bonds called hydrogen bonds. As a result, salts are very soluble in water and many species become hydrated (i.e., form compounds with water). A roadmap to Part I of the text also was provided in this chapter.

polar: a polar species contains atoms or groups exhibiting different charges (or partial charges)

10  

Chapter 1  Getting Started with the Fundamental Concepts

1.7  PART I CASE STUDY: CAN

METHYLMERCURY BE FORMED CHEMICALLY IN WATER?

1.7.1  Background5 Mercury is another example of a pollutant for which knowledge of chemical speciation is very important. Of particular concern is the chemical species methylmercury, CH3Hg+. Methylmercury can be formed in the aquatic environment by bacterial transformation of mercury. Consumption of methylmercury can lead to severe neurological damage and even death. Infants exposed to methylmercury in utero may be born with devastating disabilities. The primary route of exposure of adults to methylmercury is through the consumption of fish. In environments with significant mercury concentrations, methylmercury can accumulate through the food web, leading to high levels in game fish and shellfish. From the mid-­1950s through the 1970s, symptoms of neurological damage were noticed in the population residing near Minamata Bay in southern Japan. The collection of symptoms was named Minamata disease. Investigations revealed that mercury from a catalyst used in the manufacture of acetaldehyde was migrating to the bay. Another tragic example of methylmercury poisoning occurred in the same time period in Iraq. In this case, people were exposed to methylmercury through bread. The bread was made with grain that had been treated with a fungicide containing methylmercury.

1.7.2  Case Study Question We know that methylmercury can be formed by bacterial transformation of mercury. It is important to know whether other sources of methylmercury are significant. Since this text focuses on chemical reactions, it is of interest to ask whether methylmercury can be synthesized in water by nonmicrobial mechanisms. The question posed in the Part I case study is as follows: Can methylmercury form in significant concentrations from the chemical reaction of methane and mercury in water?

1.7.3  Lessons from Chapter 1 At the conclusion of each chapter in Part 1, important chapter concepts will be applied to the case study. In this chapter, you learned that speciation is important. Thus, you can see that the case study question is poorly posed.

Thoughtful Pause Why is the case study question poorly posed?

 Background material was obtained from NRC (2000) and the National Institute for Minamata Disease website (nimd.go.jp).

5

Historical Note: S.P.L. Sørensen and the p in pH    11

The reactive species have not been specified. The case study question can be refined by identifying the species of interest in the chemical synthesis of methylmercury. Although many species can be considered, this case study will focus on the reaction of Hg2+ and dissolved methane, CH4(aq), to form CH3Hg+. Recall also that methane can exist in the gas phase as methane gas, CH4(g). Methane gas may dissolve into water to form CH4(aq). Thus, from Chapter 1, a refined case study question can be forwarded: Can methylmercury form in significant concentrations from the chemical reaction of dissolved methane, CH4(aq), and Hg2+ in water?

CHAPTER KEY IDEAS • Hydrolysis reactions produce a wealth of dissolved chemical species. • The ability to calculate the concentrations of individual chemical species is critically important in analyzing many environmental problems. • Doses depend on both the required concentration of the target individual species and the chemistry of the water. • The equilibrium state is a useful approximation of many aquatic chemical systems.

CHAPTER GLOSSARY chemical kinetics:  the study of chemical reaction rates hydration:  the process of forming compounds with water hydrogen bonds:  weak bonds formed between an electronegative atom (e.g., oxygen) bonded to an H atom and another H atom hydrolysis reactions:  reactions with water pe:  a primary variable controlling electron transfer and defined by –log(e– activity) pH:  a primary variable controlling H+ (proton) transfer and defined by –log(H+ activity) polar:  a polar species contains atoms or groups exhibiting different charges (or partial charges) primary variable:  a species concentration (or function of a species concentration) that controls the chemistry of a system

HISTORICAL NOTE: S.P.L. SØRENSEN AND THE p IN pH Both of the primary variables introduced in this chapter, pH and pe, use the symbol p. This symbol has come to represent the function: p X = –log X. But what does the p in pH actually stand for? Everyone seems to have an answer to this question. I was taught as a young lad that pH meant “power of hydrogen.” It is common to associate the p to power (or Potenz, German or puissance, French). We now commonly use potentiometric electrodes for the measurement of pH and the p has been associated with potential (or potentia, Latin). You can even find references that relate pH to the Latin pondus hydrogenii, weight of hydrogen. The answer to the p question lies in the original publications of the proposer of the term: Søren Peter Lauritz Sørensen (1868–1939). Sørensen was the director of the Department of Chemistry of the Carlsberg Laboratory in the first third of the twentieth

12  

Chapter 1  Getting Started with the Fundamental Concepts century. (The Carlsberg Laboratory was closely associated with the Carlsberg Brewery and focused on the science of beer making.) In his original paper (Sørensen 1909), Sørensen considered an electrochemical system with the reference electrode side labeled q and the working electrode side labeled p. The hydrogen ion concentration was to be measured on the working side. He denoted the concentration of hydrogen on the working side Cp. Sørensen recognized the usefulness of 10–p and stated: “Für die Zahl p schlage ich den Namen ‘Wasserstoffionenexponent’ und die Schreibweise  pH• vor.” (For the number p, I suggest the name “hydrogen ion exponent” and the notation pH• .) By about 1917, the typography had simplified to the now-­familiar pH (Myers 2010). So why p?6 We do not know exactly what was in Sorensen’s mind in 1909. But it appears that the symbol for the most common primary variables in water chemistry traces back to an arbitrary labeling system by a famous Danish chemist over 100 years ago: p apparently just means p.

Søren Peter Lauritz Sørensen (Polytech Photos/Wikimedia Commons/Public domain)

CHAPTER REFERENCES Hupfeld, H. (1931). “As Time Goes By.” Jackson, G.A. and J.J. Morgan (1978). Trace metal‑chelator interactions and phytoplankton growth in seawater media: Theoretical analysis and comparison with reported observations. Limnol. Oceanogr. 23: 268–282. Luck, W.A.P. (1998). The importance of cooperativity for the properties of liquid water. J. Mol. Struct. 448(2–3): 131–142. Morgan, J.J. and A.T. Stone (1985). Kinetics of chemical processes of importance in lacustrine environments. In: Chemical Processes in Lakes, W. Stumm (ed.), John Wiley and Sons, New York, NY, pp. 389–426. Myers, R.J. (2010). One-­hundred years of pH. J. Chem. Ed. 87(1): 30–32.

Nørby, J.G. (2000). The origin and the meaning of the little p in pH. Trends Biochem. Sci. 25(1): 36–37. NRC (Committee on the Toxicological Effects of Methylmercury, Board on Environmental Studies and Toxicology, National Research Council) (2000). Toxicological Effects of Methylmercury. Washington, DC: National Academy Press. Plato. (c. 375 BCE). The Republic. Part II, 377a. Sillén, L.G. (1959). Graphical presentation of equilibrium data. In: Treatise on Analytical Chemistry. I.M. Kolthoff and P.J. Elving (eds.), New York: Interscience Encyclopedia, pp. 277–317. Sørensen, S.P.L. (1909). Enzymstudien II. Mitteilung. Über die Messungund die Bedeutungder Wasserstoffionen konzentration bei enzymatischen Prozessen. Biochem. Zeit. 21: 131–304.

 For an entertaining and scholarly examination of this issue (and the source of most of the information in this Historical Note), see Nørby (2000). Jens Nørby’s father worked in the Carlsberg Laboratory at the time that Sørensen was the director of the Department of Chemistry.

6

CHAPTER 2

Concentration Units 2.1  INTRODUCTION Use of the correct and consistent units is key to solving problems in the sciences and engineering. Environmental engineers and scientists often are interested in the concentrations of chemical species. Thus, you must decide which of the many possible sets of concentration units makes sense for a given application. Concentration refers to the quantity of the material per volume or mass of the surrounding environment. There are many measures of the quantity of material in the environment and thus many different concentration units. Some concentration units make chemical sense, whereas others persist because they are perceived to be more convenient or have the inertia of common use behind them. In this chapter, the rules for tracking units through calculations will be reviewed. Common units of concentration for chemical species in the environment will be discussed in Section 2.2. The common molar, mass, and dimensionless concentration units for dissolved species are reviewed in Sections 2.3, 2.4, and 2.5, respectively. Equivalents units are introduced in Section 2.6. In Section 2.7, a table for interconverting units is presented. Common units for gas phase and solid phase species are investigated in Sections 2.8 and 2.9, respectively. In Section 2.10, the concept of activity is introduced. The focus of this chapter is on concentration units. Units are the currency by which amounts of the material and volumes of the surroundings are quantified. In addition to units, you also may choose between several concentration scales. Concentration scales have their own standard states and will be discussed more thoroughly in Section 3.7.2. Common concentration scales are molarity, molality, and mole fraction.

concentration: the quantity of the material per volume (or mass) of the surrounding environment

2.2  UNITS ANALYSIS The tracking or accounting system for units is called units analysis. The rules of units analysis are simple: • Addition and subtraction Add or subtract only terms with identical units. • Multiplication, division, and raising to a power When multiplying or dividing terms (or raising terms to a power), also multiply or divide (or raised to a power) the units associated with the terms.

13

14  

Chapter 2  Concentration Units

2.3  MOLAR CONCENTRATION UNITS 2.3.1  Introduction Environmental chemistry often concerns the combinations of chemical species to form other species. Thus, it makes sense to choose concentration units that reflect the proportions in which chemicals combine. One logical choice of units is the number of atoms, molecules, or ions of a given substance per volume (or mass) of the system. After all, if one H+ ion combines with one OH– ion to form one H2O molecule, why not express each concentration in units of the number of ions (or molecules) per liter? This choice of units makes great chemical sense. However, it is unwieldy because the numbers of atoms, molecules, and ions are so large. For example, 1 milliliter of water contains about 3.3×1022 molecules of water. (To put this large number in perspective, 3.3×1022 cans of soft drinks would fill the world’s oceans about nine times over.) If you chose the number per liter concentration units, then the concentrations of the major ions in seawater are [Na+] = 2.82×1023 ions/L and [Cl–] = 3.29×1023 ions/L – not very practical numbers. (Note: As usual, the concentration of A is denoted by [A].)

2.3.2  The Mole mole: (abbr. mol ): one mole is 6.022×1023 atoms, molecules, or ions molar units: concentration units of moles per liter (1 molar = 1 M = 1 mol/L)

You can create a more practical system of concentration units by assigning a name to a large, but arbitrary, number of atoms. The name mole (abbreviation: mol) has been assigned to 6.022×1023 atoms, molecules, or ions. The number 6.022×1023 is called Avogadro’s number (after Amadeo Avogadro, 1776–1856). The value of Avogadro’s number is arbitrary and evolved through a tortuous history (summarized in the Historical Note at the end of this chapter). The concentration units based on the mole are called molar units, typically in moles/L (abbreviated M, with mM = millimoles per liter and μM = micromoles per liter). Thoughtful Pause What are the sodium and chloride ion concentrations in seawater in units of moles per liter? The major ions in seawater in molar units are

Key idea: Molar units are proportional to the number of atoms, molecules, or ions in a given volume

Na

ions L 23 ions 6.022 10 mol

0.468

mol L

0.468 M

Cl

ions L ions 6.022 1023 mol

0.546

mol L

0.546 M

2.82 1023

3.29 1023

It is important to remember that molar units are proportional to the number of atoms, molecules, or ions in a given volume. The proportionality constant relating the number of atoms (or molecules or ions) to the number of

2.3  |  Molar Concentration Units    15

Thoughtful Pause In molar units, what is the constant relating moles to the number of atoms, ­molecules, or ions, and what are the units of this constant?

In molar units, the constant relating moles to numbers is Avogadro’s number (6.022×1023 atoms, molecules, or ions per mole). The molar system is the only set of concentration units with the same constant relating the concentration units to the number of atoms, molecules, or ions for all chemical species. Another way to express this idea is to take the approach of Worked Example 2.1. For any set of concentration units, number . Molar units (where: something = moles) is you can write something X L L the only set of concentration units where X is constant for all chemical species.

Key idea: Molar units are the only set of concentration units with the same constant relating the concentration units to the number of atoms, molecules, or ions for all chemical species

Units Conversion

Worked Example 2.1 

How do you convert the concentration units of milligrams per liter (mg/L) to the concentration units of moles per liter (mol/L)? Solution To answer this question, you do not required values. (In fact, you do not even need to know what the words milligrams and moles mean.) Simply treat the units as terms in an equation: mg X L

mol L

Solving for X: X

mol L L mg

mol mg

Thus, to convert from mg/L to mol/L, multiply by the number of moles per milligram.

To summarize, molar units are the only concentration units with the same constant relating the concentration units to the number of atoms, molecules, or ions for all chemical species. Thus, molar units are especially useful in equilibrium calculations, where combining ratios are critical. You should always use molar units in equilibrium calculations for the concentrations of dissolved species.

2.3.3  Molarity As will be discussed in the Historical Note, Avogadro’s number was set by defining the mass of 1 mole of the 12C isotope of carbon atoms equal to exactly 12 grams. (The 12C isotope of carbon has 6 neutrons in addition to 6 protons and 6 electrons.) With Avogadro’s number fixed, you can use the combining ratios of the elements

Key idea: Unless otherwise stated, always use molar units in equilibrium calculations for dissolved species concentrations

16  

Chapter 2  Concentration Units

relative atomic mass: (atomic weight): the mass of 1 mole of an element molar mass: (molecular weight): the mass of 1 mole of a molecule or ion stoichiometric coefficient: here, the number of occurrences of an atom in a molecule or ion

 orked W example → gram molecular weight: molar mass expressed in grams per mole

the mass of 1 mole of each element. The mass of 1 mole of an element (i.e., the mass of 6.022×1023 atoms of an element) is called its relative atomic mass (previously, atomic weight: tabulated in Appendix A). The molar mass (previously, molecular weight) of a molecule (or ion) is calculated by summing the relative atomic masses of each atom in the molecule (or ion) multiplied by the number of times the atom occurs in the molecule or ion. The number of times the atom occurs is called its stoichiometric coefficient in the molecule or ion. The concept of stoichiometry (from the Greek stoicheion element + metrein to measure) will be used throughout this text. The molar mass of a molecule or ion j containing n different atoms in g/mol is given by: Mj

n

vij mi

i 1

where νij = stoichiometric coefficient of atom i in molecule (or ion) j (in mol i/mol j) and mi = relative atomic mass of atom i (in g/mol). As an example, suppose you wanted to calculate the molar mass of copper ­carbonate, CuCO3. Using the symbols defined above: Mcuco (1)mcu (1)mC (3)mo. Using the 3 r­elative atomic masses tabulated in Appendix A, the molar mass of CuCO3 is (1 mol/ mol)(63.55 g/mol) + (1 mol/mol)(12.01 g/mol) + (3 mol/mol)(16.00 g/mol) = 123.56 g/ mol. Another example of the calculation of molar mass is given in Worked Example 2.2. In the older literature, the term gram molecular weight was used to refer to molar mass expressed in grams (or, more properly, in g/mol). Thus, the statements “The molar mass of water is 18 g/mol,” “The molecular weight of water is 18 g/mol,” and “The gram molecular mass of water is 18 g/mol” are identical. Worked Example 2.2 

Calculation of Molar Mass

What is the molar mass of ferrous ammonium sulfate? Solution The molecular formula is Fe(NH4)2(SO4)2. One mole of ferrous ammonium sulfate contains 1 mole of Fe, 2 moles of N, 2 moles of S, 8 moles of O, and 8 moles of H. The molar mass is M

mFe 2 mN 8 mH 2 mS 8 mO 1 55.85 g / mol 2 14.01 g / mol 8 1.01 g / mol 2 32.06 g / mol 8 16.00 g / mol 284.07 g / mol

The molar mass of Fe(NH4)2(SO4)2 is 284.07 g/mol.

2.3.4  Analytical Concentrations and Formality In some applications, molar units may seem inappropriate. For example, how would you make up a 1 M NaCl solution? When NaCl is added to water, it dissociates nearly completely to Na+ and Cl– ions (see Section  1.4.2). Thus, although you might add 1 mole (about 58.44 g) of NaCl molecules to 1 liter of water, the actual number of moles of NaCl molecules in solution after a short period of time is quite small. The basic question is: Does a 1 M NaCl solution contain 1 M of NaCl molecules before any chemical reactions occur or after chemical reactions occur?

2.3  |  Molar Concentration Units    17

The most common terminology in environmental applications is to label solutions by the number of moles per liter of starting material added, regardless of the final composition of the mixture. For example, a solution made up by diluting 1 mole of HCl to 1 L of water will be referred to as a 1 M HCl solution (even though the actual HCl concentration at equilibrium is only about 1×10–3 M).1 The number of moles per liter of starting material added sometimes is called the analytical concentration. A solution with an analytical concentration of 0.01 M HCl can be made by diluting 0.01 mole of HCl to 1 L of water. The equilibrium concentration of HCl is not 0.01 M. In the older literature, a solution obtained by adding 1 mole of NaCl molecules to 1 L of water was called a 1 formal (or 1 F) NaCl solution. The use of formal concentration units (or formality) is uncommon in environmental applications and will not be used again in this text.

analytical concentration:

number of moles of starting material added per liter of water

2.3.5  Molality Another potential problem with molar concentrations is that they depend on the temperature and pressure of the system. Why? Molar concentrations depend on temperature and pressure because the solution volume depends on temperature and pressure.2 You can create a temperature-­ and pressure-­independent concentration scale by dividing the number of moles of material by the solvent mass rather than the solvent volume. The resulting concentration units are called the molal concentration units: mol . Molal units are related to molar units as follows: kg solvent mol L solution

mol kg solvent

kg solvent

kg solvent

kg solution

L solution

kg solvent mol is the molality (denoted m) and the term is the kg solvent L solution density of the solvent. Thus: The term

molarity

(molality)

kg solvent kg solution

where ρ = density of the solution in kg/L. For aqueous solutions, the solvent is pure water.

Thoughtful Pause Why is the difference between molar and molal units small for dilute aqueous ­solutions at room temperature and pressure?

 After Chapter 7, you will be able to calculate the equilibrium HCl concentration.  By way of justification, assume for a moment that dilute aqueous species behave like ideal gases. The ideal gas law states that (pressure)(volume) = (constant)(number of moles)(temperature); see Section  2.8. In other words, the molar concentration (= number of moles/volume) is proportional to the ratio of the pressure and temperature. Thus, molar concentrations vary with temperature and pressure.

1

2

molal concentration units: concentration units of moles per kg of solvent (1 molal = 1 m = 1 mol/ kg of solvent)

18  

Chapter 2  Concentration Units For dilute aqueous solutions, the mass of the dissolved species is small. Also, 1 kg of solution contains very close to 1 kg of water. In addition, near 25°C and 1 atm of pressure, the density of dilute solutions is near 1 kg/L. Thus for dilute aqueous solutions: molarity

 orked W example →

(molality)

kg solvent

kg solution kg solvent kg solvent (molality) 1 1 kg solution L solution molality (for dilute aqueous solutions)

As an example of the relationship between molarity and molality, suppose you make up a solution of acetic acid (CH3COOH) that is 10% by weight. This means that the solution has 100 g of acetic acid per L of solution. The solution density at 20°C is 1.0138 kg/L. You can verify that the molar mass of acetic acid is 60.06 g/mol. The molar concentrag 100 L solution molarity tion is = 1.67 mol/L solution. The molality is or: g kg solvent 60.06 mol kg solution 1.67 molality 0.9

mol L solution

kg water kg solution

1.0138

kg solution

1.82

mol kg water

1.82 m

L solution

Another example of the conversion between molality and molarity is shown in Worked Example 2.3. Worked Example 2.3 

Conversion of Molal and Molar Units

What is the molarity of a 1 m NaCl solution (at 25°C and 1 atm) if its density is 1.0405 kg/L? Solution From the text: molarity

(molality)

kg solvent kg solution

One mole of NaCl weighs 58.44 g. For a 1 m NaCl solution: mass of solvent = 1 kg water, mass of solution mass of water + mass of solute = 1 kg + 0.05844 kg = 1.05844 kg, and ρ = 1.0405 kg/L. Thus, 1 m NaCl = (1 m)

1 kg 1.05844 kg

(1.0405 kg/L) = 0.98 M.

2.4  |  Mass Concentration Units    19

2.4  MASS CONCENTRATION UNITS 2.4.1  Introduction The amount of material added to the environment typically is quantified by its mass. As a result, mass concentration units are very common. Typical units are milligrams per liter (= mg/L = 10–3 g/L), micrograms per liter (= μg/L = 10–6 g/L), and nanograms per liter (= ng/L = 10–9 g/L). Units analysis reveals that you should multiply molar concentration (in mol/L) by the molar mass (in g/mol) to convert to the mass scale (in g/L) (see also Worked Example 2.1): conc. in g/L = (molar mass in g/mol)(conc. in mol/L) The conversion factor relating mass and molar units is the molar mass. Thus, the conversion factor between mass units and the number of atoms (or moles) is different for every substance having a different molar mass. This observation leads to two problems with mass units. First, mass concentration units do not give you the combining proportions directly. For example, 1 mole of silver ions (Ag+) reacts with 1 mole of chloride ions (Cl–) to form silver chloride. One mole of Ag+ also can react with 1 mole of bromide ions (Br–) to form silver bromide. In molar units, the 1:1 stoichiometry between silver and chloride or silver and bromide is clear. g 0.001 L = 9.27 10 6 M Ag In mass units, a solution containing 1 mg/L of Ag+ g 107.87 mol requires (9.27×10–6 M)(35.45 g/mol) = 0.33 mg/L of Cl or (9.27×10–6 M)(79.90 g/mol) = 0.74 mg/L of Br to satisfy the stoichiometry. Thus, the combining proportion is not obvious when concentrations are expressed in mass units. Another example of conversion between molar stoichiometry and mass units is given in Worked Example 2.4.

Stoichiometry in Mass Units (Agl Example)

Worked Example 2.4 

What mass concentration of iodide ions is required to react with 1 mg/L silver ions if Ag and I react with 1:1 stoichiometry? Solution 1

mg L Ag

g Ag 1 10 3 L

g Ag L g Ag 107.868 mol Ag 1 10

3

9.27 10 3 M Ag

This is equivalent to 9.27 10 16.905

3

gI

gI

L

mol I

1.18 10

3

g or 1.18 mg / L. L

Key idea: The conversion factor relating mass and molar units is the molar mass: concentration in g/L = (molar mass in g/mol) (concentration in mol/L)

←Worked example

20  

Chapter 2  Concentration Units Second, mass concentration units are difficult to compare. For example, you may wish to verify that a 1 mg/L solution of nitrite (NO2–) contains more nitrogen than a 1 mg/L solution of nitrate (NO3–) (see Worked Example 2.5).

Stoichiometry in Mass Units (Nitrogen Example)

Worked Example 2.5 

What is the nitrogen content of a 1 mg/L nitrite solution and a 1 mg/L nitrate solution? Solution 1 10

1 mg / L NO2

46

3

g NO2

L g NO2

2.17 10 5 mol / L NO2 .

mol NO2

Also: 2.17 10 5 mol / L NO2

2.17 10

5

mol NO2

1 mol N

L

mol NO2

2.17 10 mol / L N. 5

Finally: 2.17 10 5 mol/L N 

2.17 10

5

mol N

14 g N

L

mol N

0.30 mg N / L

1 10

mol 1 L NO3

62

Also : 1.61 10

5

1.61 10

5

g NO3 L g NO3 3

1.61 10

5

mol NO3 . L

mol NO3

mol NO3 L mol NO3 L

1 mol N mol NO3

Finally: 1.61 10 5 mol/L N = 1.61 10

5

1.61 10 5 mol / L N

mol N

14 g N

L

mol N

0.23 mg N / L

2.4  |  Mass Concentration Units    21

2.4.2  Mass Concentrations “as” Other Species To allow for comparison of mass concentration units, the concentrations of related species sometimes are calculated by using the molar mass of a common species. If the common species is Y, you say that the concentration of species X is “l mg/L as Y” or “l mg Y/L” or “1 mg X-­Y/L.” This notation means that you are using the molar mass of Y for the molar mass of X. For example, the concentration of the 1 mg/L NO2– solution discussed above can be expressed in terms of mg/L as N (see Worked Example 2.5). You write that the solution concentration is “0.30 mg/L NO2– as N” or “0.30 mg N/L” or “0.30 mg NO2–-­N/L.” In doing so, you are using the molar mass of atomic nitrogen (14 g/mol) for the molar mass of nitrate. When expressing the concentration “as” another species, it is useful to first convert the mass concentration to molar units, then multiply by the molar mass of the “as” species. For example, suppose the personnel at a wastewater treatment plant wish to determine if the measured effluent ammonium (NH4+) concentration of 2.3 mg/L NH4+ violates the discharge limit of 1.6 mg NH4+-­N/L. The ammonium concentration is: 2.3

NH 4

mgNH 4 L 18

10

3

g NH 4

g mg

1.3 10 4 mol / L NH 4

mol NH 4

This is equivalent to: 1.3 10

4

mol NH 4

1 mol N

L

mol NH 4

14

gN mol N

10

3

g mg

or 1.8 mg NH4+-­N/L. Thus, the wastewater is in violation of the discharge limit. See Worked Example 2.6 for another example.

Mass Concentrations as Another Species

Worked Example 2.6 

A rinse water from an electroplating bath has a chromate (CrO42–) concentration of 10 mg/L. What is the chromate concentration expressed as Cr? Solution The bath is 10

10 mg / L CrO 4 2

mg CrO 4 2 L 112

g CrO 4 2

10 3 g mg

8.62 10 5 mol / L CrO 4 2

mol CrO 4 2

Expressed as Cr, the concentration is 8.62 10

5

mol CrO 4 2

l mol Cr

L

mol CrO 4 2

Or : 8.62 10

5

8.62 10 5 mol / L Cr

mol Cr

52 g Cr

103 mg

L

mol Cr

g

4.48 mg / L as Cr

←Worked example

22  

Chapter 2  Concentration Units In performing these calculations, be sure to account for stoichiometry; that is, the number of times the “as” fragment appears in the original substance. In other words:  conc. in mg as Y/L = (conc. in mg X/L)  orked W example →

mol Y

molar mass Y

mol X

molar mass X

For example, suppose 25 mg/L of alum is added at a drinking water treatment plant. Alum is Al2(SO4)3 • 18H2O. What is the alum dose expressed as Al? To do this conversion, you need the molar mass of alum. You can practice your skills by confirming that the molar mass of alum is 666.43 g/mol. The alum dose is:  dose =

25

mg alum g 10 3 L mg g alum 666.43 mol alum

3.75 10

5

mol alum /L

There are 2 moles of Al per mole of alum. Thus, the dose is: dose

2 mol Al mol alum

3.75 10 5 mol alum / L

7.50 10

5

mole Al / L

The mass dose in terms of Al is: dose = 7.50 10

5

mol Al L

26.98

g Al mol Al

2.02 mg/L as Al

Another example is shown in Worked Example 2.7. Worked Example 2.7 

Mass Concentrations as Another Species with Stoichiometry Correction

An industrial waste stream contains 5 mg/L of phenol (C6H6O). The discharge limit for the industry is 3 mg/L total organic carbon (TOC). TOC is the mass concentration of carbon from organic compounds. Does the waste stream meet the discharge limit? Solution The molar mass of phenol is: (6 C)(12 g/mol C) + (6 H)(1 g/mol H) + (1 O)

(16 g/mol O) = 94 g/mol. The phenol concentration is

5

mg phenol 94

10 3 g

L mg g phenol mol phenol

= 5.32×10 mol/L. –5

There are 6 moles of C per mole of phenol. Thus, the TOC concentration is 6 mol C mol phenol

× (5.32×10–5 M phenol) = 3.19×10–4 mol C/L.

This is: (3.19×10–4 mol C/L)×(12 g C/mol)(1000 mg/g) = 3.8 mg/L as C. The TOC is 3.8 mg/L, exceeding the discharge permit.

2.4  |  Mass Concentration Units    23

Why use the “mass as” notation? The advantage of expressing mass concentrations as other species is that species concentrations can be compared easily. If a natural water is said to contain 0.5  mg/L NO3– and 0.2  mg/L NH4+, it is difficult to see at first that ammonium is the larger source of nitrogen. However, when you express the nitrate and ammonium concentrations in the same units (i.e., 0.11 mg/L NO3– as N and 0.16  mg/L NH4+ as N; please verify these values on your own), it becomes clear that ammonium is the larger nitrogen source. In a similar fashion, economists may express currencies in common units (euros or US dollars) to compare prices in ­different countries. The “mass as” notation is particularly useful for compounds that are measured as a group because it allows you to express the concentration of two different species by using the same mass units. Recall from Section  2.2 that you can add or subtract only terms with the same units. Thus, the “mass as” approach allows you to add and subtract species concentrations when using mass units. For example, you may wish to know how much of the total soluble cyanide in a waste stream is present as the species cyanide (CN–) and how much is present as cadmium cyanide (assuming for the moment that cyanide and cadmium cyanide are the only forms of cyanide in the waste). Some common cyanide measurement methods measure the sum of the two species. Expressing the total soluble cyanide as CN eliminates any confusion introduced by the difference in molar masses between cyanide and cadmium cyanide. Thus, if the CN– concentration was 1.1 mg/L as CN and the cadmium cyanide concentration was 4.2 mg/L as CN, the total cyanide concentration can be calculated as 1.1 mg CN/L + 4.2 mg CN/L = 5.3 mg/L as CN. Before leaving the “mass as” notation, please note that the stoichiometry is not always clear from inspection. As an example of the confusion in stoichiometry, consider the case of chlorine. Species containing active chlorine (usually meaning chlorine in the +I oxidation state) often are expressed in units of mg/L as Cl2. The pertinent reactions for several chlorine-­containing species of environmental significance are as follows:

Cl 2

H2 O HOCl

HOCl NH 3

HOCl Cl OCl

H

Key idea: Use the “mass as” notation to convert the mass concentrations of several species to the same set of units to compare or add them

eq. 2.1

H

eq. 2.2

NH 2 Cl H 2 O

eq. 2.3

It is clear from these reactions that 1 mole of HOCl forms from 1 mole of Cl2 (eq. 2.1), 1 mole of OCl– forms from 1 mole of HOCl (eq. 2.2), and 1 mole NH2Cl forms from 1 mole of HOCl (eq. 2.3). Thus, each mole of HOCl, OCl–, and NH2Cl is equivalent to 1 mole of Cl2. As a result: 1 mg/L Cl2 as Cl2 = 1 mg/L HOCI as Cl2 = 1 mg/L OCl– as Cl2 = 1 mg/L NH2Cl as Cl2.

2.4.3  Unusual Mass Concentration Units Having stressed that you can add only terms with the same units, it now must be noted that there are a few common water quality parameters where you add species concentrations expressed in different mass units. One example is total dissolved solids (TDS), operationally defined as the mass of material that passes through a specified filtration operation. The TDS is measured gravimetrically (i.e., by weight). It also can be calculated by adding the masses of the individual components in the water. This approach violates the rules of units analysis because it requires that you add terms with different units. For example, the TDS of a solution containing

total dissolved solids (TDS): sum of the mass concentrations of species passing through a specified filter

24  

Chapter 2  Concentration Units

Key idea: When adding species concentrations in mass units, convert all concentrations to the same units by expressing each value as a common species (exceptions: TDS and salinity)

30 mg/L Na+ and 40 mg/L Cl– is 30 + 40 = 70 mg/L. The sodium ion and chloride ion concentrations have different units (mg/L as Na+ and mg/L as Cl–, respectively). Nonetheless, you add these terms of different units when calculating the TDS from species concentrations. Another example in which species concentrations with different mass units are added is salinity, S(‰). The symbol ‰ is called the per mille symbol, meaning “per thousand.” This makes sense, because salinity is usually reported as g/kg seawater. The salinity of seawater is the sum of the masses of dissolved inorganic species per kg of seawater. For example, a crude model of seawater is 19.35 g/kg Cl–, 10.78 g/kg Na+, 2.71 g/kg sulfate, and 1.28 g/kg Mg2+. The salinity of this mixture is 19.35 + 10.78 + 2.71 + 1.28 = 34.12 g/kg or 34.12‰. It is important to note that TDS and salinity are exceptions to the general rule that masses are not additive when expressed in different units.

2.5  DIMENSIONLESS CONCENTRATION UNITS It is common to use dimensionless units to express the concentrations of chemical species in the environment. Several examples are illustrated below.

2.5.1  Mole Fraction mole fraction: concentration units equal to the number of moles of the substance divided by the total number of moles in the system ­(dimensionless; mole fraction of A = xA)

The mole fraction is the number of moles of the substance divided by the total number of moles in the system. The total number of moles in the system is the sum of the moles of solvent and moles of all dissolved species. In dilute aqueous systems, the solvent is pure water. The mole fraction (denoted x) is used frequently in chemical engineering and is useful in some thermodynamic derivations (see Chapter 3).

2.5.2  Parts per X Pollutant concentrations are frequently expressed in parts per some number of parts. Examples are parts per million (ppm), parts per billion (ppb), and parts per trillion (ppt or pptr). Using the example of ppm, the notation “10 ppm” refers to 10 parts of something per million parts of something else. The word parts can refer to masses or volumes (denoted, for example, ppmm and ppmv, respectively), but typically refers to masses for aqueous or solid systems. In dilute aqueous systems at environmentally important temperatures and pressures, the density of water is very close to 1.0 kg/L. Thus, 1 liter of water weighs approximately 1 kg or 103 g or 106 mg. A concentration of 1 mg/L means 1 mg of material in 1 liter of water = 1 part in 106 mg = 1 part in one million parts = 1 ppm. For dilute aqueous systems near 25°C and 1 atm, the units mg/L and ppm are nearly identical. Similarly, ppb and μg/L are nearly interchangeable in dilute aqueous systems (near 25°C and 1 atm), as are ppt and ng/L. In the more general case: ppm

mg kg of solution

mg L kg

mass concentration in solution densitty in

2.6  |  Equivalents    25

For example, consider a waste stream consisting of 1 mg of total lead (Pb) in 1 L of mg Pb 1 L water (density = 1.0 kg/L at 25°C, 1 atm). The lead concentration is 1 mg/L or = kg 1.0 L 1.0 mg Pb/kg = 1 ppm. If 1 mg of lead was placed in 1 L of a 25% sodium chloride solution (density 1.19 kg/L at 25°C, 1 atm), the solution still would be 1 mg/L. However, in ppm, the total lead concentration is: Pb in ppm

1

mg Pb

L solution kg solution 1.19 L solution

0.84

mg Pb kg solution

←Worked example

0.84 ppm

There are two caveats to remember when using the parts per something units. First, make sure you know whether the parts are mass or volume. Second, remember that ppm and mg/L (or ppb and ug/L or ppt and ng/L) are nearly identical only in dilute solutions when the density of water is very close to 1.0 kg/L.

Key idea: Units of ppm and mg/L are nearly identical in dilute aqueous solutions with solution densities near 1.0 kg/L

2.5.3  Percentage The concentrations of more concentrated solutions are often expressed as a percentage. This dimensionless unit refers to the mass (or volume) of the substance per mass (or volume) of the system. In aqueous solution, the percentage concentration is the mass of the species per mass of solution, expressed as a percentage. Thus: conc. (in % by mass) = 10–4 (conc. in ppm) For dilute aqueous solutions near 25°C and 1 atm: conc. (in % by mass) = 10–4 (conc. in ppm) = 10–4 (conc. in mg/L) In general: conc.(in % by mass) 10 (conc. in ppm ) 10 4

4

mg L kg density in L conc. in

Thus, a 1% sludge stream is very close to 10,000  mg/L solids if the density of the sludge is near 1.0 kg/L. A 1% by mass sodium chloride solution in water has a density of 1.0053 kg/L and therefore a concentration in mg/L of (1% by mass)104 (1.0053 kg/L) = 10,053 mg/L.

Key idea: A 1% solution is nearly 10,000 mg/L if the solution density is near 1.0 kg/L

2.6  EQUIVALENTS Another common set of concentration units in aquatic chemistry is equivalents units. The term equivalents is confusing at first blush because it appears to carry a different meaning with each use. In fact, converting from molar to equivalents units is trivial. Analogous to Example 2.1, units analysis will reveal how to convert from molar to equivalents units: equivalents/L = eq/L = (conversion factor)(mol/L) Thus, the conversion factor that multiplies mol/L to get eq/L must have units of equivalents/mol. A solution containing 1 eq/L is said to be 1 normal (= 1 N). Equivalence concentration units also are called normality.

normality: concentration units in equivalents per liter (1 equivalent/L =

26  

Chapter 2  Concentration Units Now the hard part: How many equivalents are there per mole? The answer to this question depends on the context. In acid–base (proton transfer) reactions, equivalents generally refers to the number of protons transferred (or capable of being transferred). Thus, a 0.5 M HCl solution is (0.5 mol/L)(l eq/mol) = 0.5 eq/L = 0.5 N since HCl has 1 proton (i.e., one H+) available to transfer. The concentration of a 0.5 M H2SO4 solution in eq/L is (0.5 mol/L)(2 eq/mole) = 1 eq/L = 1 N since H2SO4 has two protons available to transfer. Similarly, in redox (electron transfer reactions), equivalents generally refers to the number of electrons transferred. Thus, a 0.4 mM O2 solution is: (0.4 mmol/L)(4 eq/ mol) = 1.6 meq/L = 1.6 mN, since each molecule of O2 can accept four electrons to become two molecules of H2O (formally: O2 + 4e– + 4H+ → 2H2O; more on redox reactions in Chapter 16). You also can use equivalents to express species concentrations in terms of important fragments. For example, a 1.2×10–5 M Cu(CN)3– solution made be said to contain (1.2×10–5 mol/L)(3 eq of cyanide/mol) = 3.6×10–5 eq/L of cyanide. As can been seen by these examples, there are many uses of equivalents units. It is critical to know the basis of comparison. When using equivalents units, always ask, “Equivalent to what?”

Key idea: When using equivalents units, always ask: “Equivalent to what?”

2.7  REVIEW OF UNITS INTERCONVERSION The conversion factors for changing from one set of units to another are listed in Table 2.1. Remember that each conversion factor has units. Table 2.1 Interconversion Factors for Concentration Units in Aqueous Systems [Multiply units in the row by the table entry to get units in the column. Example: To convert mol/L to mg/L, multiply by (1000)(molar mass).]

Molar (mol/L = M)

Molar (mol/L = M)

1

Mole fraction (x)

Mass (mg/L)

“Mass as” (mg/L as Y)

Percentage (mass basis)

1000Mi

1000Mi 10 4 1000MiT 10 4

1 T

1000Mi

T

1

1000MiT

1000MYT

1000MiT

Mass (mg/L)

1 1000Mi

1 1000MiT

1

MY Mi

1

“Mass as” (mg/L as Y)

1 1000MY

1 1000MY T

Mi

1

Mi

Mole fraction (x)

Parts per million (ppmm) Percentage (mass basis)

1000MY

Parts per million (ppmm)

MY

MY

ρ 1000Mi 10 4

1000Mi

MY Mi

1000MiT 10 4

104ρ 1000MiT

Notes: Mi = molar mass of species of interest (g/mol)   MY = molar mass of Y (g/mol) T = total mol/L in solution, including the solvent  The number 1000 has units of mg/g ρ = solution density (kg/L)

10 4

MY Mi

10 10

4

1 Mi

4

MY

1

10

104

1

–4

2.8  |  Common Concentration Units in the Gas Phase    27

2.8  COMMON CONCENTRATION UNITS

IN THE GAS PHASE

Gas phase concentrations usually are expressed in the partial pressure scale. Thus, the concentration of a gas is defined as its partial pressure, Pi, where Pi = (mole fraction of species i)(total pressure). The most common unit in aquatic chemistry for expressing the concentration of a gas is the atmosphere (atm).3 Other concentration units for trace air pollutants are mass concentration (typically μg/m3), parts per X (especially ppmv) = parts by volume per million parts by volume), and percentage units. Gas phase concentration units are dependent on pressure and temperature. Conversion factors for gas phase concentrations are summarized in Table 2.2. An example is given in Worked Example 2.8. The conversion of ppmv to mass concentration units proceeds as follows. The concentration unit ppmv is defined as: concentration (ppmv) = 106

volume of species of interest total volume

n moles of species of interest = 106 = 106 i n total number of moles

ni 106 V n V

where ni = number of moles of species i and n = total number of moles of gas. The ideal gas law states that: PV = nRT, where P = pressure, V = volume, R = ideal gas constant,

Table 2.2 Interconversion Factors for Concentration Units in Gas Phase Systems [Multiply units in the row by the table entry to get units in the column. Example: To convert partial pressure to ppmv, multiply by 106 and divide by the total pressure, P).] Partial pressure (Pi, atm)

Partial pressure (Pi, atm) Mass (μg/m3)

1

10

Volume (ppmv)

6

Mass (μg/m3)

106 RT Mi

10–6P

1 M RT i

Volume (ppmv)

106

1 P

1

RT 1 P Mi

P M RT i

1

Symbols are defined in the text and in the notes to Table 2.1. Units: P (atm), T (K), Mi (molar mass in g/mol), and R = 0.082057×10–3

atm-m3 . mol-K

 The formal (Système International d’Unités, or SI) unit of pressure is the pascal (abbreviated Pa, after Blaise Pascal, 1623–1662), where 1 Pa = 1 N/m2. One atmosphere is 101,325 Pa or 101.325 kPa. One bar of 0.9869 atm. 3

partial pressure: partial pressure = Pi = (mole fraction of species i)(total pressure)

28  

Chapter 2  Concentration Units Worked Example 2.8 

Conversion of Gas Phase Concentration Units

Derive the conversion factor for ppmv to partial pressure in Table 2.2. Use it to calculate the CO2 partial pressure (in atm) if the CO2 concentration in the atmosphere is 420 ppmv. Solution The atmosphere has a total pressure of 1 atm. The units ppmv are a dimensionless concentration and can be thought of as having units of 

L 106 L

mol  = 106 mol

106xi where xi = mole fraction. Thus: xi = (10–6)ppmv. From the definition of partial pressure: PCO2 = xiP = 10–6P(ppmv) = (10–6)(1 atm)(420 ppmv) = 4×10–6 atm Note that 10–6 P is the conversion factor in Table 2.2.

n and T = absolute temperature (K). So: V

P ni . In addition: V RT

Substituting: concentration ppm v

ni 106 V n V

RT 106 P

conc. in Mi

g m3

RT P

g m3 . g molar mass in mol conc.of i in

conc. in Mi

g m3

where Mi = molar mass of species i.

2.9  COMMON CONCENTRATION UNITS

IN THE SOLID PHASE

In solid environments (e.g., soils, sludges, sediments, and other sorbents), common mass concentration units are mg/kg (i.e., milligrams of pollutant per kilogram of solid), parts per X (i.e., parts per million), and percentages. The units of ppm and mg/ kg are equivalent: 1 ppm = 1 mg/kg. For some equilibrium calculations, the concentration of a pure solid phase is given by the mole fraction.

2.10  ACTIVITY In the first section of this chapter, concentration was defined as the quantity of material per volume or mass of the surrounding environment. Now you might ask: Is mass or molar concentration the best measure of how a substance actually behaves in the environment? When seeking to describe the behavior of substances in the environment, mass or molar concentration may not be the most appropriate indicator. To understand the limitations on concentration, consider the problem of determining the behavior of a group of children in a swimming pool. You might start by

2.10  |  Activity    29

simply counting the children. Say there are 20 children in a small pool with a volume of 15 m3. Is the number concentration (also called the bather load, which equals 20 per 15 m3 = 1.3×10–3 per L) the best measure of the sprightliness of the children? The activity of the children may depend on several factors other than their number concentration. For example, the children may be less active if the water temperature is

Thoughtful Pause What factors (besides the number per volume) affect the liveliness of the children?

cooler. Also, the activity of the children may change if adults are present in the pool. In other words, the children may behave differently under different environmental conditions, even at the same number concentration. Similarly, chemical species may behave differently under different environmental conditions, even at the same mass or molar concentration. As with the children in the pool, the behavior of chemical species depends on the water temperature and concentration of other species. In addition, the system pressure may influence the way in which chemical species behave. As an example, consider three glasses of water, each with the same small amount of salt dissolved in the water. One glass is at room temperature, one at elevated temperature, and one at room temperature with a lot of Epsom salts (a form of MgSO4) also dissolved in the water. The mass (and molar) concentrations of sodium and chloride ions will be the same for each glass. However, the ions will be more active in the warmer water and less active with the Epsom salts than in the first glass. The ions are less active in the presence of Epsom salts because the sodium ions interact electrostatically with sulfate ions and the chloride ions interact electrostatically with the magnesium ions. This is analogous to the children in the pool: the children may be less active (i.e., less rambunctious) if they are interacting with the adults in the pool. Scientists have recognized for nearly 100 years that temperature, pressure, and the presence of other chemical species affect the behavior of dissolved chemicals in water. As a result, it has been necessary to develop a new approach in quantifying the amount of material in a system. This approach is an idealized concentration (sometimes called, in the older literature, a thermodynamic concentration), which considers the effects of temperature, pressure, and the presence of other chemical species on the behavior of dissolved chemicals. The idealized concentration is called activity. The activity of species A is denoted as {A}. Activity is the proper way to express the quantity of species in thermodynamic calculations. As you will see in Chapter 3, equilibrium is a thermodynamic concept. Thus, you should use activity (and not concentration) as the sole indicator of species quantity. However, you will find that activity and concentration are nearly identical in dilute solutions (where the concentrations of other chemical species are small) at near standard temperature and pressure. In other words, the conversion factor relating activity and concentration (called the activity coefficient) is nearly 1 in dilute solution near 25°C and 1 atm of pressure. In this text, the effects of other inert chemical species will be ignored for most of the book, and molar concentration units will be used. In Chapter 3, activity will be used in developing thermodynamic relationships. In Chapter 21, a detailed analysis will be made of the effects of other species on activity.

activity: an idealized concentration, used in thermodynamic calculations, which considers the effects of temperature, pressure, and the presence of other chemical species on the behavior of dissolved chemicals

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Chapter 2  Concentration Units

2.11  CHAPTER SUMMARY Units are key to engineering and science calculations, and concentration units are critical when manipulating chemical species concentrations. In equilibrium calculations, molar units (mol/L, or M) are useful because they capture the combining proportions inherent in chemical reactions. In other cases, you may wish to express mass concentrations in the same units so they can be added. Thus, concentrations sometimes are expressed as some other species. In expressing the concentration of species X as Y, use the molar mass of Y as the molar mass of X and consider the number of moles of Y per mole of X. Dimensionless concentration units (mole fraction; parts per million, billion, or trillion; and percentages) also are in common use. For aqueous system with a density near 1 kg/L, 1 part per million (l ppm) is very close to 1 mg/L. Another way to use the same units (or currency) when writing concentrations is to use units of normality (or equivalents/L). In this case, always ask yourself: Equivalent to what? In other words, the basis of the equivalence (i.e., the number of equivalents per mole) must be established clearly. Common equivalences are the number of H+ or electrons accepted or donated. Gas phase and solid phase concentration units also were reviewed in this chapter. Mass or molar concentration units may not be the most appropriate way to describe the behavior of chemical species in solution. In particular, temperature, pressure, and the presence of other chemical species are known to affect the behavior of dissolved chemicals. This observation necessitates a new look at concentration and has led to the development of an idealized concentration called activity. Activity is the most appropriate choice of expressing the quantity of material in thermodynamic calculations. In this text, the effects of other inert chemicals, temperature, and pressure on the behavior of selected chemical species will be ignored until Chapter 21.

2.12  PART I CASE STUDY: CAN

METHYLMERCURY BE FORMED CHEMICALLY IN WATER?

In Chapter l, four species were identified as being important in determining whether methylmercury can be formed at significant concentrations from the reaction of dissolved methane and Hg2+ in water. The four species are CH3Hg+, Hg2+, CH4(aq), and CH4(g). Based on the lessons of this chapter, it should be possible to select units for the four species in the chemical system. Thoughtful Pause What concentration units would you recommend for the species in the Part I case study? In reality, any consistent set of concentration units can be used.4 However, some concentration units make more sense than others.  For thermodynamic calculations, activities make the most sense for the dissolved species (see Section 2.10). However, in the calculations to be considered in this case study, activities and concentrations will be assumed to be interchangeable. Thus, suitable concentration units are sought here.

4

Chapter Glossary    31

In this case study, the concentration units should capture the combining proportions of Hg2+ and CH4(aq). In addition, the water involved is considered to be very dilute. Thus, the dependencies of solution volume on temperature and pressure can be ignored and molal units are not necessary. As a result, molar concentration units appear to be appropriate for the dissolved species. For the gaseous species, concentration units of atmospheres will be used. They are the most commonly used gas phase concentration units.

CHAPTER KEY IDEAS • Molar units are proportional to the number of atoms, molecules, or ions in a given volume. • Molar units are the only set of concentration units with the same constant relating the concentration units to the number of atoms, molecules, or ions for all chemical species. • Unless otherwise stated, always use molar units in equilibrium calculations for dissolved species concentrations. • The conversion factor relating mass and molar units is the molar mass: concentration in g/L = (molar mass in g/mol)(concentration in mol/L). • Use the “mass as” notation to convert the mass concentrations of several species to the same set of units to compare or add them. • When adding species concentrations in mass units, convert all concentrations to the same units by expressing each value as a common species (exceptions: TDS and salinity). • Units of ppm and mg/L are nearly identical in dilute aqueous solutions with solution densities near 1.0 kg/L. • A 1% solution is nearly 10,000 mg/L if the solution density is near 1.0 kg/L. • When using equivalents units, always ask: “Equivalent to what?”

CHAPTER GLOSSARY activity:  an idealized concentration, used in thermodynamic calculations, which considers the effects of temperature, pressure, and the presence of other chemical species on the behavior of dissolved chemicals analytical concentration:  number of moles of starting material added per liter of water concentration:  the quantity of the material per volume (or mass) of the surrounding environment gram molecular weight:  molar mass expressed in grams per mole molal concentration units:  concentration units of moles per kg of solvent (1 molal = 1 m = 1 mol/kg of solvent) molar mass (molecular weight): the mass of 1 mole of a molecule or ion molar units:  concentration units of moles per liter (1 molar = 1 M = 1 mol/L) mole (abbr. mol):  one mole is 6.022×1023 atoms, molecules, or ions mole fraction:  concentration units equal to the number of moles of the substance divided by the total number of moles in the system (dimensionless; mole fraction of A = xA)

32  

Chapter 2  Concentration Units normality:  concentration units in equivalents per liter (1 equivalent/L = 1 eq/L = 1 normal = 1 N) partial pressure:  partial pressure = Pi = (mole fraction of species i)(total pressure) relative atomic mass (atomic weight):  the mass of 1 mole of an element stoichiometric coefficient:  here, the number of occurrences of an atom in a molecule or ion total dissolved solids (TDS):  sum of the mass concentrations of species passing through a specified filter

HISTORICAL NOTE: AMADEA AVOGADRO AND AVOGADRO’S NUMBER

(C. Sentier/Wikimedia Commons/Public domain) The first half of the nineteenth century saw a great deal of progress in the development of atomic theory. The ancients developed the concept that all matter was built from fundamental, indivisible particles (called atoms, from the Greek a not + temnein to cut). The philosophy of atomism was abandoned by the Greeks by 40 bce because the idea of a simple building block was at odds with the apparent complexity of the natural world. A consistent atomic theory was lacking until the work of the influential chemists and physicists of the early 1800s. John Dalton (1766–1844) contributed greatly to the development of modern atomic theory. He struggled to establish a consistent framework for assigning molecular formulas and atomic weights. The experimental work of Joseph Louis Gay-­Lussac (1778–1850) became the basis for determining molecular formulas. His work implied that equal volumes of gases should contain equal numbers of atoms or molecules.5 Enter Lorenzo Romano Amadeo Carlo Avogadro, the count of Quaregna and Ceretto (1776–1856). Avogadro allowed for the acceptance of the equal number for equal volumes idea (later called Avogadro’s law) by proposing in 1811 that some gases were

 Dalton could not accept this result, in part because he believed that all gases were monoatomic. For example, Gay-­ Lussac found that 1 L of CO combined with 0.5 L of oxygen to form 1 L of CO2. To conclude that equal volumes of gas contain equal numbers of molecules from this observation, you must accept that oxygen is O2, not O. Dalton did not think of simple substances as being polyatomic and thus rejected the idea of equal numbers for equal volumes (Pauling 1970). 5

Problems    33

diatomic. Avogadro’s ideas were not appreciated until after his death. Stanislao Cannizzaro (1826–1910) fought for the acceptance of Avogadro’s ideas at the influential Karlruhe Conference of 1860. He used Avogadro’s law to determine atomic weights. Avogadro’s law led to the need to express a large number of atoms or molecules by a convenient unit to describe reasonable gas volumes (say, 1 L). As stated in Section 2.3.2, a large number of atoms or molecules (Avogadro’s number, or NA) is defined to be 1 mole of atoms or molecules. Avogadro’s number can be calculated by several methods.6 For example, consider a crystal containing one mole of NaCl. The crystal contains 2NA atoms (NA atoms of Na and NA atoms of Cl) and has a volume equal to 2NAd3 where d = interatomic distance. The volume also equals the mass divided by the density (or, for molar mass one mole, the molar mass divided by the density). Combining yields: N A . 2 d3 For NaCl, molar mass = 58.448 g/mol, ρ = 2.165 g/cm3, and d = 2.819×10–8 cm. This gives an estimate for NA of 6.026×1023 mol–1. Avogadro’s number is related to the atomic weight scale. In 1961, the mass of 1 mole of 12C (containing six neutrons, six protons, and six electrons) was defined to be exactly 12 g. This definition set the value of Avogadro’s number at 6.022×1023 mol–1.

PROBLEMS Relative atomic mass values may be found in Appendix A. 1. What system of concentration units is most appropriate for following the progress of chemical reactions? 2. Explain the advantages of the “mass as” units (e.g., 1 mg/L as P) over mass units (e.g., mg/L). Illustrate with an example. 3. A drinking water treatment plant adds lime (CaO) at a dose of 200 mg/L as CaCO3. Determine the lime dose in mg/L as Ca units and molar units. 4. What is the phosphate concentration in a lake in mol/L if the measured phosphate concentration is 20 μg/L as P? Assume all the phosphorous is present as HPO42–. 5. In concentrated ferric iron solutions, numerous species may appear as intermediates before iron precipitates as Fe(OH)3(s). What is the total ferric iron concentration in mg/L as Fe if the concentrations of Fe3+, Fe(OH)2+, Fe(OH)2+, and Fe2(OH)24+ are, respectively, 1.0×10–3 M, 8.9×10–4 M, 4.9×10–7 M, and 1.2×10–3 M? 6. Find the total dissolved inorganic nitrogen concentration (nitrate + nitrite + ammonium) in a river in mg/L as N if the concentrations in the river (in mg/L as the compound as written) are: [NO3–] = 2.1 mg/L, [NO2–] = 1.8 mg/L, and [NH4+] = 1.1 mg/L.

7. What is the solids concentration (in mg/L) in a sludge with density 1.1 g/cm3 and 5.6% solids? 8. The main basin of the Great Salt Lake in Utah, USA can be modeled as a 5% NaCl solution (with a density of 1.1 kg/L). The methylmercury concentration in the lake is as high as 25 ng/L. What is the methylmercury concentration in ppt? 9. Seawater typically contains 19.354 g Cl–/kg and 0.0673 g Br–/kg. The chlorinity, of seawater is defined as the number of grams of silver required to precipitate chloride and bromide in 328.5233 g of seawater. Calculate Cl(‰) in typical seawater if 1 mole of silver reacts with 1 mole of chloride or bromide. 10. Calculate the number of grams per liter required to make the following solutions: a. 0.5 M NaCl (recall from Section 2.3.3 that seawater is about 0.47 M Na+ and 0.55 M Cl–) b. 0.1 M K2Cr2O7 (a reagent used in the chemical oxygen demand test) c. A 10 mg/L as C standard made using potassium hydrogen phthalate (KHP, C8H5KO4) d. A 1 mg/L Fe standard made using ferric chloride, FeCl3(s)

6   It is more accurate to say that Avogadro’s number can be back-­calculated by the approach shown. After all, the value of NA = 6.023×1023 mol–1 was used to establish the molar mass of NaCl as 58.448 g/mol. If a different value of Avogadro’s number had been agreed to, then the molar mass of NaCl (and every other substance) would be different.

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Chapter 2  Concentration Units

11. Calculate the molarity and normality of the following solutions (assume density = 1.0 kg/L for all): a. 20 g/L NaOH (assume 1 eq/mol) b. 25 g/L H2SO4 (assume 2 eq/mol) c. 10 g/L HCl (assume 1 eq/mol) d. 15 g/L Ca(OH)2 (assume 2 eq/mol) 12. The major ions in a groundwater sample were found to be as follows: [Ca2+] = 75 mg/L, [Mg2+] = 40 mg/L, [Na+] = 10 mg/L, [HCO3–] = 300 mg/L, [Cl–] = 10 mg/L, [SO42–] = 109 mg/L (values from Benefield and Morgan 1990). a. Calculate the concentration of each ion in molar units and equivalents per liter, where equivalents is defined here as the number of charges per mole. b. If the water is electrically neutral, the sum of the concentration of cations should equal the sum of the concentration of anions. Is the groundwater electrically neutral? c. Calculate the TDS of the water sample. 13. Water quality analysis revealed the following measured concentrations of ions: [Ca2+] = 90 mg/L, [Mg2+] = 30 mg/L, [Na+] = 72 mg/L, [K+] = 6 mg/L, [Cl–] = 100 mg/L, [SO42–] = 225 mg/L, and [HCO3–] = 165 mg/L.

16. The term hardness refers to the sum of the concentrations of the divalent cations (species with charge +2) in water. For many waters, the most important divalent cations are Ca2+ and Mg2+. Hardness is typically expressed in units of mg/L as CaCO3. In this approach, each mole of Ca2+ and Mg2+ is equivalent to 1 mole of CaCO3. a. Calculate the hardness (in mg/L as CaCO3) of a water containing 80 mg/L of Ca2+. b. Calculate the hardness (in mg/L as CaCO3) of the groundwater described in Problem 12. c. Calculate the hardness (in mg/L as CaCO3) of the water described in Problem 13. d. Water with hardness ranging from 150 to 300 mg/L as CaCO3 is considered hard. How much calcium ion (in mg/L as Ca2+) would be required to account for a hardness of 150 mg/L as CaCO3? e. How much magnesium ion (in mg/L as Mg2+) would be required to account for a hardness of 200 mg/L as CaCO3? 17. The average global concentration of carbon dioxide, CO2(g), in the atmosphere is about 420 ppmv currently. Express this concentration in units of μg/m3. Assume 1 atm total pressure and 25°C (= 298 K).

a. Calculate the concentration of each ion in molar units and equivalents per liter, where equivalents is defined here as the number of charges per mole. b. Is the water electrically neutral? (See Problem 12b.) 14. From eq. 2.3, 1 mole of hypochlorous acid (HOCl) reacts with 1 mole of ammonia (NH3) to form 1 mole of monochloramine (NH2Cl). In practice, chlorine concentrations are expressed in mg/L as Cl2 and ammonia concentrations are expressed in mg/L as N. Determine the HOCl concentration (in mg/L as Cl2) and ammonia concentration (in mg/L as N) required to form 1 mg/L NH2Cl as Cl2.

19. A typical limit on arsenic contamination of soil is about 10 mg/kg. What is the limit in ppm? A typical limit on mercury pollution in soil is 100 ppb. What is the limit in mg/kg?

15. Calculate the Fe2+ dose in mg/L required to react with 1 mg/L of dissolved O2 if four moles of Fe2+ are required for each mole of dissolved oxygen.

20. Explain the difference between activity and concentration. Why are both typically expressed in units of moles/L?

18. Two of the National Ambient Air Quality Standards in the U.S. are 0.15 μg/m3 for lead and 0.075 ppmv for ozone. Express both standards in ppmv and ­compare them. Assume 1 atm total pressure and 25°C (= 298 K).

CHAPTER REFERENCES Benefield, L.D. and Morgan, J.S. (1990). Chemical precipitation. In: Water Quality and Treatment. 4th ed., F.W. Pontius (ed.), New York: McGraw-­Hill, Inc., pp. 641–708.

Pauling, L. (1970). General Chemistry. Mineola, NY: Dover Publ.

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CHAPTER 3

Thermodynamic Basis of Equilibrium 3.1  INTRODUCTION 3.1.1  Chapter Overview Thermodynamics (from the Greek therme heat + dynamis power) is the study of thermodynamics: the study transformations of energy. Thermodynamics is a far-­reaching and beautifully self-­ of the transformations of consistent field of study. Thermodynamics seeks to explain phenomena as diverse as energy high jumping (where kinetic energy is converted into potential energy) and thermonuclear weapons. This text is concerned with reactions between chemical species. Thus, the following discussion of thermodynamics will focus on chemical reactions. This is a slice of thermodynamics sometimes called chemical thermodynamics. The intent of this chapter is to show that the concept of equilibrium has thermodynamic underpinnings. To accomplish this task, we must start with some basic thermodynamic principles and definitions. The thermodynamic functions Gibbs free energy and chemical potential will be introduced, along with their dependence on species concentrations. Equilibrium will be defined from these thermodynamic functions. Thus, you will find that the equilibrium state of a chemical system is reflected in the concentrations of the species comprised in the system. The concept of an equilibrium constant will be developed. Further thermodynamic concepts will be introduced in Chapter 21.

3.1.2  Scope As stated in Section 3.1.l, the goal of this chapter is to develop expressions for the equilibrium state of a system based on thermodynamic concepts and the concentrations of chemical species. There are many ways to derive equilibrium expressions from thermodynamic laws. How should you decide which thermodynamic concepts to employ? In this chapter, the discussion will be guided by three principles. First, thermodynamic properties will be developed that are convenient. In some cases, this may mean combining simple properties into groupings to reach the chapter goal in an expedient fashion. Second, thermodynamic properties will be sought that are reusable. This means that we will take advantage of tabulated thermodynamic properties. Third, thermodynamic properties will be developed that describe the system of interest. Thus, the focus will be on the properties of the system, not properties of the matter surrounding the system. 35

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36   Chapter 3  Thermodynamic Basis of Equilibrium

3.1.3  Motivation Before diving into definitions and calculations, it is informative to step back and decide if a discussion of thermodynamics is really necessary. If your goal (from Chapter 1) is to be able to determine the concentrations of individual chemical species at equilibrium, then thermodynamics is absolutely required. Thermodynamics will allow you to determine quickly whether certain chemical reactions will proceed as written. More significantly, thermodynamics will empower you to calculate species concentrations under specified conditions. You will find that thermodynamic properties of chemical reactions will let you determine, for example: • The extent to which ammonia partitions between the atmosphere and a lake • The effect of acid rain on monuments • The dose of soda ash required to neutralize an acid spill • The effect of pH on the disinfection strength of chlorine • The speciation of phosphorous in a drinking water reservoir In short, thermodynamics is the key to many applications of aquatic chemistry.

3.2  THERMODYNAMIC PROPERTIES 3.2.1  Introduction

Key idea: Thermodynamic systems are defined by thermodynamic properties (also called state variables) extensive property: a thermodynamic property dependent on the amount of material in the system Key idea: The values of an extensive thermodynamic property for each portion of a system add up to the value of the extensive thermodynamic property for the whole system.

The state of a system can be defined by a number of thermodynamic properties. Thermodynamic properties sometimes are called state variables because their values depend on the state of the system and not on the manner in which the state was achieved. Thermodynamic properties are divided into two types: extensive and intensive properties. Property types are discussed in Sections 3.2.2 and 3.2.3. Some thermodynamic properties are conserved; that is, they are additive even after transformations of a system or when two systems are combined. Conservation of thermodynamic properties is discussed in Section  3.2.4. Finally, since the state of a system is defined by thermodynamic properties, it is useful to ask: How many thermodynamic properties does it take to define a system? This question is answered in Section 3.2.5.

3.2.2  Extensive Properties As stated in Section  3.2.l, thermodynamic properties can be extensive properties or intensive properties. Extensive properties are dependent on the amount of material in the system. For example, mass is an extensive property because it is dependent on the amount of the substance present. Examples of extensive properties are listed with their common units in Table 3.1. The extensive thermodynamic properties share a unique feature: the sum of the values of an extensive property for each portion of the system equals the value of the extensive property for the system as a whole. For example, if you empty a glass of water one drop at a time, the sum of the volumes (or masses or number of moles of water) of each drop will be equal to the volume (or mass or number of moles) of the original water in the glass.1 1  This definition of extensive properties is valid for systems having constant thermodynamic properties at every point in space.

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3.2  |  Thermodynamic Properties   37

Table 3.1 Extensive and Intensive Properties (common units in parentheses) Type

Extensive Property

Intensive Property

Mass or mole

mass (g or kg) number of moles volume (L) heat capacity (J/K) energy (kJ) enthalpy (kJ) entropy (kJ/K) free energy (kJ)

mass concentration (mg/L) and density (kg/L) molar concentration (mol/L) specific volume (L/kg) and molar volume (L/mol) specific heat (J/g-­K) molar energy (kJ/mol) molar enthalpy (kJ/mol) molar entropy (kJ/mol-­K) molar free energy (kJ/mol) pressure (atm) temperature (K)

Volume Thermal

Other

3.2.3  Intensive Properties Intensive properties are independent of the amount of material. In other words, intensive properties are the same for each packet of material in the system. Temperature is an example of an intensive property. Concentration also is an intensive property since concentration is a mass (or number of moles) per unit volume (or mass). Examples of intensive properties are listed with their common units in Table 3.1. Some intensive properties are formed from extensive properties by normalizing for the amount of material. In such cases, the intensive property is called a specific or molar property. For example, the specific volume is calculated as the volume divided by the mass, and the molar energy is the energy per mole. (Related intensive functions, the partial molar properties, will be discussed in Section 3.6.3.) Recall that intensive properties are the same for each packet of material in the system. Thus, unlike extensive properties, intensive properties are not additive within a system. For example, the temperature of the water in a glass is not equal to the sum of the temperatures of each drop of water in the glass.

intensive property: a thermodynamic property independent of the amount of material in the system Key idea: Intensive thermodynamic properties are not additive within a system

3.2.4  Conservation of Some Thermodynamic Properties Some thermodynamic properties are conserved; that is, they are additive even after transformations of a system or when two systems are combined. The general concept of the conservation laws will be explored in Section 3.4.1. In this section, the thermodynamic properties that are candidates for conservation will be identified.

Thoughtful Pause Which thermodynamic properties are candidates for conservation: extensive or intensive properties? You know that intensive properties are not conserved. Intensive properties are not additive even inside a system, so it does not make sense that they should be additive

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38   Chapter 3  Thermodynamic Basis of Equilibrium when systems are transformed or combined. Only the extensive properties are candidates for conservation. Here you must tread carefully. Although only extensive properties are candidates for conservation, it does not follow that every extensive property is conserved. Hundreds of years of observation have led to the conclusion (see Section 3.4.1) that the extensive properties mass and energy (along with momentum) are conserved. Thus, you can balance mass, energy, and momentum as systems are transformed or combined. It is very important to perform balances only on mass, energy, or momentum and not on intensive properties. One important implication in environmental engineering and science is that we must perform balances on the mass (or mass flux) of chemicals, not on chemical concentrations. This principle is illustrated in Worked Example 3.1. Worked Example 3.1 

Balancing Thermodynamic Properties

What is the sodium ion concentration in a solution formed when one drop (0.05 mL) of seawater ([Na+] = 10.8 g/L) at 25°C is added to 1 L of river water ([Na+] = 6 mg/L) at 25°C? Solution Concentration is not an extensive property and therefore is not additive. Mass is additive and mass = m = VC, where V volume and C = concentration. ms = msw + mrw = VswCsw + VrwCrw (The subscripts s, sw, and rw refer here to solution, seawater, and river water, respectively.) ms = (5×10–5 L)(10.8 g/L) + (1 L)(0.006 g/L) = 0.00654 g Also: Cs =

ms Vs

The density of the final solution is expected to be very close to the density of the river water, so you can approximate: Vs = Vsw + Vrw = 1.00005 L. Thus: Cs

ms Vs

0.00654 g 1.0005 L

6.5 mg / L

3.2.5  How Many Thermodynamic Properties Are Enough? component: a chemical species (or species fragment) that can be varied independently in a system Gibbs Phase Rule: a system with P phases and C components requires C – P + 2 thermodynamic properties

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Thermodynamic properties are used to define the state of a system. You need to know when to stop adding properties to your description of systems. In other words, we would like to know how many thermodynamic properties are required to define a system. The number of thermodynamic properties required to define the state of a system depends on the complexity of the system. More specifically, the number of thermodynamic properties required depends on the number of phases and number of components of the system. The concept of the component is used throughout this text. A component is a chemical species (or fragment of a chemical species) that can be varied independently in a system. A system with P phases and C components requires C – P + 2 thermodynamic properties. This is called the Gibbs Phase Rule (after J. Willard Gibbs, 1839–1903; see the Historical Note at the end of the chapter).

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3.3  |  Why Do We Need Thermodynamics to Calculate Species Concentrations?   39

Here is an example of how to use the Gibbs Phase Rule. The simplest possible aqueous system is a known volume of water. This is a one-­phase, one-­component system (P = C = 1), since it has only a liquid phase and the only component is water. Therefore, two thermodynamic properties are required to define the system, since C – P + 2 = 1 – 1 + 2 = 2. We can use any two thermodynamic properties to describe the system. In aqueous systems, it is convenient to use temperature and pressure as the two thermodynamic properties. Thus, a system of 1 L of pure water is completely and uniquely defined by specifying the temperature and pressure. Now, thermodynamics would be pretty useless if it could describe only systems of only pure water. In fact, the Gibbs Phase Rule is very powerful. It will assist you in describing the equilibrium state of systems. In particular, it will assist in determining the species concentrations at equilibrium. For example, suppose you want to determine the equilibrium concentrations of every species in a system containing P phases and C components. The Gibbs Phase Rule tells you that you need to specify C – P + 2 thermodynamic properties from Table  3.1. In most cases, those properties will be temperature, pressure, and C – P species concentrations (or combinations of species ­concentrations).

←Worked example

3.3  WHY DO WE NEED THERMODYNAMICS

TO CALCULATE SPECIES CONCENTRATIONS?

One of the goals of this text is to help you determine the concentrations of each chemical species involved in a set of chemical reactions. To accomplish this, two features of chemical reactions must be determined. First, you must have a way of deciding whether a chemical reaction will proceed as written. If you can determine that a reaction will not occur under a given set of conditions (e.g., a certain ­temperature, pressure, and set of species concentrations), then clearly it is of little interest to you. Second, you would like to determine if the chemical species concentrations are changing with time. In particular, the goal for most of this text is confined to calculating species concentrations when they do not change with time. The time-­dependent nature of species concentrations will be considered in Chapter 23. How do you know whether a reaction proceeds as written or whether species concentrations are not time-­dependent? We will develop thermodynamic properties specifically for determining whether reactions occur and whether species concentrations change with time. Before developing these thermodynamic properties, it is necessary to examine the concepts of whether reactions occur and the time-­independence of species concentrations in more detail. This examination will require an introduction of three related concepts: spontaneity, equilibrium, and reversibility.

Key idea: Thermodynamics can tell you which chemical reactions are possible (under a given set of conditions) and whether species concentrations are not time-­dependent

3.3.1  Spontaneity A reaction that occurs as written without energy added is called a spontaneous reaction. Spontaneous reactions also are called possible (or favorable) reactions.2 In other words, reactions that are not spontaneous as written are said to be not possible (or unfavorable) under the given set of thermodynamic conditions. Identifying

spontaneous reaction: a reaction occurring as written without energy added

 Formally, spontaneous reactions are said to be exergonic and nonspontaneous reactions are said to be endergonic (from the Greek exo, out or outside; endon, in or within; and ergon work).

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40   Chapter 3  Thermodynamic Basis of Equilibrium spontaneous reactions will allow you to divide of all reactions into two types: possible (i.e., spontaneous) reactions and impossible reactions. It is important to remember that you must specify the set of thermodynamic conditions when determining whether a reaction is possible (spontaneous) or impossible. In other words, you can determine whether a reaction is spontaneous only under a specified temperature, pressure, and set of chemical species concentrations. A reaction could occur spontaneously at, say, one temperature, but not occur spontaneously at another temperature. As an example, the melting of ice to form water is spontaneous at temperatures greater than 0°C (at 1 atm of pressure) but not spontaneous at temperatures less than 0°C (at 1 atm of pressure). In addition, reactions may occur spontaneously at one set of species concentrations, but not spontaneously at another set of species concentrations. For example, the dissolution of sodium chloride to form sodium and chloride ions is spontaneous at 25°C and 1 atm if sodium chloride is the only source of Na+ and Cl–. However, in a saturated brine solution, the dissolution of sodium chloride is not spontaneous.

3.3.2  Equilibrium equilibrium: a state where the thermodynamic properties of the system do not change with time (for a system that does not have a net loss or gain of heat or mass with its surroundings)

Systems in which the species concentrations do not change with time are said to be at equilibrium (from the Latin aequi equal + libra weight). The formal thermodynamic definition of equilibrium is that the thermodynamic properties of the system at equilibrium do not change with time and the system does not have a net loss or gain of heat or mass with its surroundings. Thoughtful Pause Is the formal definition of equilibrium more or less restrictive than the requirement that species concentrations be independent of time? Note that the formal definition of equilibrium is very restrictive. It says that no thermodynamic properties (not just the property of species concentration) change with time at equilibrium. A system at equilibrium also does not have a net loss or gain of heat or mass with its surroundings. Identifying the equilibrium state will allow the division of all spontaneous reactions into two types: reactions at equilibrium and reactions not at equilibrium. The definitions of spontaneity and equilibrium allow you to divide all chemical reactions into groups. One classification scheme is shown in Figure 3.1. Note that all All Chemical Reactions

Spontaneous Reactions (possible reactions)

at equilibrium

Nonspontaneous Reactions (impossible reactions)

not at equilibrium

FIGURE 3.1  Classification of Chemical Reactions (spontaneous reactions are possible and nonspontaneous reactions are impossible under a specified set of thermodynamic conditions)

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3.3  |  Why Do We Need Thermodynamics to Calculate Species Concentrations?   41

reactions at equilibrium are spontaneous, but not all spontaneous reactions are at equilibrium.

3.3.3  Reversibility The concept of equilibrium is closely aligned with the idea of reversibility. A reaction is said to be reversible if the reverse reaction (i.e., conversion of products into reactants) can occur spontaneously with an infinitesimal increase in the product concentration. Consider the reaction: H+ + Cl– → HCl. The reaction is reversible if the reverse reaction (HCl → H+ + Cl–) also occurs spontaneously when the HCl concentration is increased very slightly. In general, a process is said to be reversible if it returns to its initial state when the mass, heat, and energy flows are reversed. What is the relationship between equilibrium and reversibility? Reversible systems are said to be near equilibrium, whereas irreversible systems can be far from the equilibrium state. For a chemical reaction to be at equilibrium, the reaction must be reversible.

reversible reaction: a reaction where the reverse reaction (i.e., conversion of products into reactants) can occur spontaneously with an infinitesimal increase in the product concentration

3.3.4  Summary The connection between spontaneity, equilibrium, and reversibility casts a new light on equilibrium. For a reaction to be at equilibrium, the reaction must be both spontaneous and reversible. In other words, both the reaction and its reverse reaction must be spontaneous. For example, it is known that hydrochloric acid equilibrates quickly with H+ and Cl–. Thus, both the reactions HCl → H+ + Cl– and H+ + Cl– → HCl must be spontaneous. This leads to the expression of equilibrium as reactions that proceed in both directions, denoted HCl ⇄ H+ + Cl–, or more commonly in aquatic chemistry: HCl = H+ + Cl–. At equilibrium, the concentrations of all chemical species do not change with time. This fact sometimes conjures up the image that no reactions are occurring at equilibrium. However, you now know that all chemical reactions and their reverse reactions proceed spontaneously at equilibrium. Overall, the species concentrations are not changing with time. However, reactants are converted to products and products to reactants continuously at equilibrium in such a way that the reactant and product concentrations do not change over time. In the hydrochloric acid example, HCl is being formed continuously from reaction of H+ and Cl–. In addition, HCl is continuously dissociating to form H+ and Cl–. The system is not static at equilibrium, but no net change in species concentration occurs. An analogy may make this clearer. At many college campuses and shopping areas, parking space is at a premium.3 During the day, the number of empty parking spaces remains nearly constant (at some very small number), and the number of filled parking spaces remains nearly constant as well. In spite of this observation, the actual cars parked in the parking spaces change throughout the day. Parking is reversible. The processes of parking and vacating a parking space (analogous to a reaction and its reverse reaction) are spontaneous. The numbers of empty and full spaces (analogous to species concentrations) do not change much over time, even though new empty parking spaces are becoming available and being filled continuously. As with chemical systems at equilibrium, the system is not static, but no net change in the number of empty parking spaces occurs.  This leads to the cynical definition of a university as a place where faculty, students, and staff gather together to complain about parking.

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42   Chapter 3  Thermodynamic Basis of Equilibrium

3.4  THERMODYNAMIC LAWS 3.4.1  Conservation Laws In the context of thermodynamics, laws are concepts held to be true without derivation or proof. The entire codex of thermodynamics is built upon the assumption of three conservation statements: In any process, mass, energy, and momentum are conserved. (An important exception is nuclear reactions, where mass and energy are interconverted according to Einstein’s famous equation: E = mc2.) The conservation laws are logical but unproven. In fact, they are unprovable. You can verify the conservation of mass, energy, and momentum only up to the accuracy of the measurements. The law of conservation of energy is given a special name. It is called the First Law of Thermodynamics and will be explored further in Section 3.4.2.

3.4.2  The First Law of Thermodynamics First Law of Thermodynamics: energy is conserved, so any change in the heat of a system or work done by the system must be accounted for in the internal energy of a system

Key idea: The First Law of Thermodynamics states that energy is conserved and that the internal energy changes only through heat exchange and/or work

The First Law of Thermodynamics states that energy is conserved. Energy may be present in the form of heat or work. Thus, the First Law is written in terms of heat and work. The First Law is a statement that any energy added to the system (i.e., any change in the heat of a system or work done by a system) must be reflected by a change in the internal energy of a system. In other words: dE = dQ + dW where: dE = differential change in the internal energy (E) of the system, dQ = differential change in the heat absorbed (Q) by the system, and dW = differential work done by the system. In this presentation, work done by the system is considered to be negative. Note that the First Law is really two statements in one. First, it is truly a statement that energy is conserved. In other words, the difference between energy added to the system (by heat absorbed and work) and energy released from the system (by heat released and work done) must show up as a change in the internal energy of the system. Second, the First Law states that the only ways that the internal energy changes in a system is through heat exchange (heat absorbed or released) and work. Heat is a fairly common form of energy. The work done by the system is a little more complicated because work can take many forms. For the moment, consider only work done by changes in the pressure or volume of the system. This is called pressure-­ volume work (or P-­V work). Thoughtful Pause How is P-­V work related to pressure and volume? (Hint: Think dimensions.) By dimensional analysis, work (= force×distance) has dimensions of force ML dimensions×length dimensions. Force (= mass×acceleration) has dimensions of , T2 2 ML so work has dimensions of , where M, L, and T are dimensions of mass, length, T2 ML 1 M and time, respectively. Pressure is force per unit area  dimensions: 2 T L2 LT 2

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3.4  |  Thermodynamic Laws   43

and volume has dimensions of L3, so pressure times volume has dimensions of

ML2

. T2 Hence, by dimensional analysis, work has the same dimensions as the product of pressure and volume and is therefore proportional to the product of pressure and volume. In fact: W = PV. If the process is reversible (in other words, if the process can proceed infinitesimally in either direction), then the pressure change must be zero (dP = 0) and: dW = d(PV) = VdP – PdV = –PdV (constant pressure, so dP = 0) The sign convention can be confusing, but be assured that it is consistent. If a gaseous system expands, then it does work and therefore dW must be negative. Expansion means dV > 0, so the negative sign is needed to ensure that dW is negative. Combining the First Law (dE = dQ + dW) and the change in work at constant pressure (dW = –PdV):

dE

dQP

PdV  (constant pressure)

eq. 3.1

where dQP = differential heat absorbed at constant pressure. The subscript P indicates constant pressure.

3.4.3  Enthalpy Equation 3.1 was developed for systems involving P-­V work. As a result, eq. 3.1 is very useful for describing changes in internal energy of gaseous systems. However, eq. 3.1 is not very useful for aqueous systems. Thoughtful Pause Why is eq. 3.1 not very useful for aqueous systems? Equation 3.1 is not very useful for aqueous systems because dV ≈ 0 for most transformations in aqueous systems. As discussed in Section 3.1.2, perhaps we can combine thermodynamic properties to form a new property that is more convenient to use than dE for aqueous systems. A more convenient way to discuss energy changes for chemical reactions in aqueous systems is the extensive thermodynamic function enthalpy (H). Enthalpy is defined as:

enthalpy (H): H = E + PV

H = E + PV What does enthalpy mean? Enthalpy is the sum of internal energy, E, and work (since W = PV). This formal definition probably does not help you understand enthalpy or how to use it. A more meaningful view of enthalpy is found after a little effort. For constant pressure conditions (dP = 0): dH = dE + VdP + PdV = dE + PdV (constant pressure) Rearranging:

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dE = dH – PdV (constant pressure)

eq. 3.2

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44   Chapter 3  Thermodynamic Basis of Equilibrium Comparing eqs. 3.1 and 3.2: dH

Key idea: The change in enthalpy is the heat absorbed at constant pressure

dQP

eq. 3.3



Equation 3.3 gives a better physical feeling for enthalpy. The change in enthalpy is the heat absorbed at constant pressure.

3.4.4  The Second Law of Thermodynamics and Entropy Common experience tells you that the First Law is not adequate to explain all observations of the natural world. In particular, the First Law does not tell you which processes are spontaneous. (Recall that spontaneous processes occur without energy added.) For example, if you place two blocks of identical material at different temperatures in contact with each other, you expect that the temperatures will equilibrate after some time. The First Law does not predict this change in temperature. Thoughtful Pause

Key idea: The First Law of Thermodynamics does not tell you if processes are spontaneous entropy (S): a thermodynamic function related to spontaneous change and a measure of disorder

Second Law of Thermodynamics: entropy increases for spontaneous processes and is related to heat exchange dS total

dS   sys +

dSsurr ≥ 0 and dSsys ≥

dQ T

Two blocks of identical material, each at 20°C, are brought in contact. Does the First Law prevent the temperatures of the blocks from changing spontaneously to 10°C and 30°C? In fact, as demonstrated in the Thoughtful Pause above, the First Law even allows for spontaneous events to occur that you know are impossible! Clearly, a new property is needed to account for spontaneous processes. In 1875, Rudolph Clausius (1822–1888) proposed a thermodynamic function related to spontaneous change called entropy (from the Greek en in + trepein to turn, i.e., to direct). Entropy is denoted S. The entropy of the system, surroundings, and system + surroundings are denoted Ssys, Ssurr, and Stotal, respectively. This means: dStotal = dSsys + dSsurr. Entropy usually is described as a measure of disorder. However, focus for a moment on its original intent as an extensive property required to account for the spontaneity of processes. The Second Law of Thermodynamics describes two properties of entropy. First, entropy increases for spontaneous processes (but does not change for reversible processes). Or:

dStotal



eq. 3.4a

0 only for reversible processes

Second, the entropy of a system is related to the change in the heat absorbed by the system. The relationship between dSsys and dQ is:

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dStotal    0  for spontaneous processes and

dSsys

dQ  for spontaneous processes and T

dSsys =

dQ only for reversible processes T

eq. 3.4b

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3.5  |  Gibbs Free Energy   45

At first glance, these properties of entropy appear to be arbitrary. However, it can be shown that defining dSsys =

dQ for reversible processes leads to the conservation T

statement that dStotal = dSsys + dSsurr = 0 for reversible processes (see, for example, Eisenberg and Crothers 1979). Rearranging eq. 3.4b, note that for reversible processes: dQ    TdSsys  for reversible processes



eq. 3.5

Equation  3.5 is a thermodynamic definition of a reversible process. However, it is not very useful. To form a more useful definition of reversible processes, we must develop a new thermodynamic function.

3.5  GIBBS FREE ENERGY The concept of enthalpy combines internal energy and pressure–volume work. You now know that a more descriptive energy term should include entropy as well to account for reaction spontaneity. The motivation here is to develop a convenient energy term that describes reversibility and spontaneity. Such an extensive property, Gibbs free energy (G), was developed by the ubiquitous J. Willard Gibbs. Gibbs free energy is defined as:

G = E + PV – TS = H – TSsys

Gibbs free energy (G): G H –TS sys

eq. 3.6

Why is this apparently arbitrary collection of thermodynamic properties so useful? Gibbs free energy is useful for four reasons, as discussed in Sections 3.5.1 through 3.5.4.

3.5.1  Gibbs Free Energy Is a Link Between the First and Second Laws Gibbs free energy combines enthalpy and entropy, two important extensive properties of chemical systems. Enthalpy embodies internal energy and work and is related to heat absorbed by the system (Section 3.4.3). Entropy is related to spontaneity and also has a connection to changes in heat (Section 3.4.4). Thus, Gibbs free energy serves as a link between the First and Second Laws.

3.5.2  Gibbs Free Energy Is a Function of Temperature and Pressure For aqueous systems, temperature and pressure are natural intensive properties and G can be expressed in terms of temperature and pressure.4 For reversible processes involving only reversible pressure-­volume work: dG    VdP    Ssys dT  (for reversible  P -V work)





eq. 3.7

Equation 3.7 is derived in Worked Example 3.2.  For systems where temperature and volume are the natural intensive properties, the Helmholtz free energy is used instead of Gibbs free energy. The Helmholtz free energy (after Hermann von Helmholtz, 1821–1894) is denoted A (from the German Arbeit, work).

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46   Chapter 3  Thermodynamic Basis of Equilibrium

Gibbs Free Energy as a Function of T and P

Worked Example 3.2 

Derive: dG = VdP – SsysdT (for reversible P-­V work) Solution From eq. 3.6: dG = dH – TdSsys – dSsysdT From the definition of H: dG = (dE + PdV + VdP) – TdSsys – dSsysdT (*) Now: dE = dQ + dW For reversible P-­V work, dQ = TdSsys (eq. 3.5) and dW = –PdV (since dP = 0 for reversible work). So: dE = TdSsys – PdV (**) Substituting (**) into (*): dG = VdP – SsysdT (for reversible P-­V work) Note: For irreversible systems, dQ < TdSsys. Thus, dG < VdP – SsysdT. Key idea: Gibbs free energy is a measure of reversibility: dG = 0 only for reversible reactions (at constant temperature and pressure) Key idea: Gibbs free energy is a measure of spontaneity: dG ≤ 0 for all spontaneous reactions (at constant temperature and pressure)

3.5.3  Gibbs Free Energy Is a Measure of Reversibility and Spontaneity Gibbs free energy is a convenient way to determine if a system is reversible. From Worked Example 3.2, for any system involving only P-­V work, dG ≤ VdP – SsysdT. For reversible processes involving only P-­V work: dG = VdP – SsysdT. At constant temperature and pressure (dP = dT = 0): dG ≤ 0 for all spontaneous processes and dG = 0 only for reversible processes (constant T and P) Thus, Gibbs free energy tells you whether or not reactions are reversible. The test: dG ≤ 0 for spontaneous reactions but dG = 0 only for reversible processes. The property dG also is a measure of whether a reaction is spontaneous. You can show that for spontaneous reactions at constant temperature and pressure, dG ≤ 0. A derivation of this useful criterion for spontaneity is given in Worked Example 3.3. Worked Example 3.3 

Gibbs Free Energy as a Measure of Spontaneity

How is G a measure of reaction spontaneity? Solution Imagine a system that exchanges only heat with the surroundings at constant pressure i.e., dSsurr

dQP . T

From eq. 3.3: dQP = dH, so dSsurr =

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dH T

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3.5  |  Gibbs Free Energy   47

Now: dStotal

dSsys

dSsurr

dH T

dSsys

Or: TdStotal = TdSsys – dH (*) From the definition of G: dG = dH – TdSsys – SsysdT = dH – TdSsys (**) (T, P constant) Comparing (*) and (**): dG = –TdStotal We know dStotal ≥ 0 for spontaneous reactions. Thus, since T is always > 0, dG ≤ 0 for spontaneous reactions.

3.5.4  Advantages of Gibbs Free Energy Compared to Entropy Why bother relating dG to reversibility and spontaneity when you already know that entropy is related to reversibility and spontaneity? After all, you know already that dQ = TdSsys for reversible processes (eq. 3.5) and dQ ≤ TdSsys for a spontaneous process (from eq. 3.4b). The condition dG = 0 is a much more convenient measure of reversibility for many chemical systems than dQ = TdSsys. Similarly, dG ≤ 0 is a more convenient measure of spontaneity than dQ ≤ TdSsys. Why? The term dSsys includes contributions from the surroundings to the system, not just internal changes to the system entropy from chemical reaction. On the other hand, dG can be calculated by examining the chemical reactions of the system only. As stated in Section 3.1.2, it is preferable to focus on the system of interest rather than having to include an analysis of the surroundings. Remember that both Gibbs free energy and entropy can be related to reversibility and spontaneity. The thermodynamic conditions required for reaction reversibility and spontaneity are reviewed in Table 3.2.

Table 3.2 Conditions of Reversibility and Spontaneity Reaction Type Reversible

Irreversible

Spontaneous

dStotal

0

>0

≥0

  T

dQ ≥  T

dQ 0 for products and < 0 for reactants) enthalpy of reaction: Hrxn v i Hi i

entropy of reaction: Srxn v i Si i

Gibbs free energy of reaction: Grxn v i Gi i

i

vi

i

 Formally: dni = νidξ, where: ξ is the extent of the reaction. Thus, a reactant with νi = –2 loses two moles per mole of reaction and a product with νi = +1 gains one mole per mole of reaction. This formal definition also explains why νi usually is considered negative for reactants and positive for products.

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52   Chapter 3  Thermodynamic Basis of Equilibrium

3.7.2  Calculation of ∆Hrxn, ∆Srxn, and ∆Grxn

 orked W example →

To calculate changes in thermodynamic properties during a reaction, write two new reactions in which reactants are converted to elements at their standard states first and then the elements at their standard states are transformed into products. (Remember that such manipulation is allowed because the values of the thermodynamic properties are path-­independent.) As an example, consider the reaction of aqueous silver ions and aqueous chloride ions to form solid silver chloride: Ag+(aq) + Cl–(aq) → AgCl(s). Using the approach of Section 3.7.l, divide this reaction into three reactions: silver ion → silver in its standard state, chloride ion → chlorine in its standard state, and silver and chlorine in their standard states → solid silver chloride. The standard state of silver is solid silver, and the standard state of chlorine is gaseous chlorine. Thus, you can write the following three reactions: Ag+(aq) → Ag(s)

reaction 1

Cl–(aq) → ½C12(g)

reaction 2

Ag(s) + ½C12(g) → AgCl(s) reaction 3

partial molar enthalpy of formation ( H f ,i ): the

enthalpy of reaction for the formation of a chemical species from elements at their standard states partial molar entropy ( S i ): the entropy of reaction for the formation of a chemical species from elements at their standard states partial molar Gibbs free energy of formation ( G f ,i): the Gibbs free energy of reaction for the formation of a chemical species from elements at their standard states standard enthalpy of reaction ( Hrxn ): the enthalpy of a reaction evaluated under standard-­state conditions and calculated from the standard enthalpies of formation

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What is the advantage of breaking up the reaction into three reactions? Reaction 3 represents the formation of a compound, AgCl(s), from elements in their standard states. Reactions 1 and 2 represent the reverse of the formation of two ions from ­elements in their standard states. Thus, by breaking up the reaction, you have three reactions which relate chemical species to elements in their standard states. Why the focus on elements in their standard state? You know from Section 3.6.3 that the partial molar enthalpy, entropy, and Gibbs free energy are fixed for elements in their standard state at a given temperature and pressure (and equal to zero for Hi and Gi ). The thermodynamic properties for the formation of chemical species from elements at their standard states have been tabulated. These properties are called the partial molar enthalpy of formation ( H f,i ), partial molar entropy ( Si ), and partial molar Gibbs free energy of formation ( G f,i ). The subscript f indicates formation of chemical species from elements in their standard states. Returning to the example of silver chloride formation, you can write the thermodynamic properties of the reaction in terms of the partial molar thermodynamic properties of formation, For enthalpy: ∆Hrxn = ∆Hrxn,1 + ∆Hrxn,2 + ∆Hrxn,3 = H Ag(s) H AgCl(s)

H Ag(s)

1 HCl2 (g) 2

H Ag

H f ,Ag

( aq)

( aq)

1 HCl2 (g) 2

H f ,Cl

( aq)

HCl

( aq)

H f ,AgCl(s)

Evaluating under standard-­state conditions (usually 25°C and 1 atm):

 Hrxn

H f ,AgCl(s)

The thermodynamic function

H f ,Ag

( aq)

H f ,Cl

(aq)

eq. 3.10

Hrxn is called the standard enthalpy of reaction.

Note that it is a partial molar quantity (because the H are partial molar quantities), and thus has units of kJ/mol. In general, for the reaction: aA + bB → cC + dD:

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3.7  |  Changes in Thermodynamic Properties During Chemical Reactions   53  Hrxn

i

vi H f ,i

cH f ,C

dH f ,D

aH f ,A

bH f ,B

A similar definition can be written for Srxn = standard entropy of reaction and Grxn = standard Gibbs free energy of reaction =

i

i G f ,i

i

i i.

i

i Si

3.7.3  Example The standard enthalpy of formation, standard entropy, and standard Gibbs free energy of formation have been tabulated for many chemical species. This greatly facilitates the calculation of the standard enthalpy, entropy, and Gibbs free energy of reaction. Using the language of Section  3.1.2, the standard properties make thermodynamic properties reusable. Consider again the example of the formation of silver chloride from aqueous silver ions and aqueous chloride ions. You can calculate Hrxn , Srxn and Grxn as follows. The pertinent standard partial molar properties of formation are listed in Table 3.3. Using equations analogous to eq. 3.10: Hrxn = –127.1 kJ/mol – (+105.6 kJ/mol –167.2 kJ/mol) = –65.5 kJ/mol Srxn = +96 kJ/mol-­K – (+73.4 kJ/mol-­K + 56.5 kJ/mol-­K) = –33.9 kJ/mol-­K

standard entropy of reaction ( Srxn): the entropy of a reaction evaluated under standard-­state conditions and calculated from the standard entropies standard Gibbs free energy of reaction ( Grxn ): the Gibbs free energy of a reaction evaluated under standard-­state conditions and calculated from the standard Gibbs free energy of formation

←Worked example

Grxn = –109.8 kJ/mol – (+77.12 kJ/mol –131.3 kJ/mol) = –55.62 kJ/mol

Thoughtful Pause How should ∆Hrxn°, ∆Srxn°, and ∆Grxn° be related? From the definition of G, you expect: Grxn = Hrxn – T Srxn . For the formation of silver chloride at 25°C: Hrxn – T Srxn = –65.5 kJ/mol – (298.15 K)(–33.9 k.J/mol-­K) = –55.4 kJ/mol The value of –55.4 kJ/mol is about equal to the calculated value of Grxn (within the limits of accuracy of the thermodynamic data). Another example of calculating standard thermodynamic properties of reaction is given in Worked Example 3.5. Table 3.3 Thermodynamic Data for Ag+(aq), Cl–(aq), and AgCl(s) Species Species

H f (kJ/mol)

S (J/mol-­K)

Ag+(aq)

+105.6

+73.4

+77.12

Cl–(aq)

–167.2

+56.5

–131.3

AgCl(s)

–127.1

+96

–109.8

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G f (kJ/mol)

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54   Chapter 3  Thermodynamic Basis of Equilibrium

Calculation of ΔHrxn, ΔSrxn and ΔGrxn

Worked Example 3.5 

Calculate Hrxn , Srxn and Grxn for the reaction: CO2(g) + H2O(l) → H2CO3(aq) Solution The pertinent thermodynamic data are: Species

Hf   S

   G f

CO2(g)

–393.5  213.6  –394.4

H2O(1)

–285.8  69.9  –237.2

H2CO3(aq) –699.6 187.0 –623.2 (Units: kJ/mol, kJ/mol-­K, and kJ/mol for standard enthalpies of formation, standard entropies, and standard Gibbs free energies of formation, respectively. All values at 25°C = 298 K.) Here, all νi = l. Thus:  Hrxn  Srxn  Grxn

699.6 ( 393.5) ( 285.8) 187.0 213.6 69.9

96.5 kJ / mol-K

623.2 ( 394.4 ) ( 237.2)

 You can verify that:  Grxn

20.3 kJ / mol 8.4 kJ / mol

  Hrxn T Srxn

3.7.4  Interpretation of ΔHrxn, ΔSrxn and ΔGrxn

exothermic reaction: a reaction in which heat is generated endothermic reaction: a reaction in which heat is absorbed

What do Hrxn , Srxn and Grxn mean? For Hrxn , recall that the change in enthalpy is equal to the change in heat at constant pressure (eq. 3.3). Thus: Hrxn = ∆QP under standard-­state conditions. When heat is generated by a reaction (i.e., when ∆QP = Hrxn < 0; see Section  3.4.2 for the sign convention for heat), the reaction is called exothermic. When heat is absorbed during a reaction (i.e., when ∆QP = Hrxn > 0), the reaction is called endothermic. Thoughtful Pause Are the reactions in Section 3.7.3 and Worked Example 3.5 exothermic or endothermic? Note that the reactions in Section 3.7.3 and Example 3.5 are exothermic since Hrxn < 0 for both reactions. The sign of Srxn tells you whether the degree of disorder is increasing, decreasing, or remaining constant for a reaction. When reactions involve transitions from less-­ ordered phases to more highly ordered states, the degree of disorder decreases and Srxn < 0. In predicting the sign of the entropy change, it is important to keep in mind the change in entropy of water during a reaction. For the silver chloride example, Srxn

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3.8  |  Relating Gibbs Free Energy to Species Concentrations   55

is negative since order is increased (i.e., disorder decreased) when the aqueous species form the crystalline product. As discussed in Section 3.5, the change in G at constant temperature and pressure is a measure of spontaneity. You expect Grxn < 0 for spontaneous reactions and Grxn > 0 for reactions that will not occur spontaneously as written under standard-­state conditions. The dissolution of carbon dioxide gas (Worked Example 3.5) is not spontaneous under standard conditions. Note that Grxn is positive in this case in spite of the

fact that heat is given off ( Hrxn < 0) because of the large decrease in entropy under standard conditions ( Srxn 0 for products and < 0 for reactants) reversible reaction:  a reaction where the reverse reaction (i.e., conversion of products into reactants) can occur spontaneously with an infinitesimal increase in the product concentration Second Law of Thermodynamics:  entropy increases for spontaneous processes and is related to heat exchange (dStotal = dSsys + dSsurr ≥ 0 and dSsys ≥

dQ ) T

spontaneous reaction:  a reaction occurring as written without energy added standard enthalpy of reaction ( H rxn):  the enthalpy of a reaction evaluated under standard-­state conditions and calculated from the standard enthalpies of formation standard entropy of reaction ( Srxn ):  the entropy of a reaction evaluated under standard-­state conditions and calculated from the standard entropies standard Gibbs free energy of reaction ( Grxn):  the Gibbs free energy of a reac-

tion evaluated under standard-­state conditions and calculated from the standard Gibbs free energy of formation thermodynamics:  the study of the transformations of energy Third Law of Thermodynamics:  the entropy of a perfectly ordered material approaches zero as the temperature approaches 0 K

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66   Chapter 3  Thermodynamic Basis of Equilibrium

HISTORICAL NOTE: JOSIAH WILLARD GIBBS

J. Willard Gibbs (Popova Olga/Adobe Stock) J. Willard Gibbs was one of the most influential American scientists of the nineteenth century. Gibbs was born in 1839 and named after his father, a professor of sacred literature at Yale Divinity School.9 Gibbs lived most of life in one house in New Haven, Connecticut. Gibbs entered Yale College (now Yale University). in 1854 at age 15. While most people think of him for his transformative ideas about thermodynamics, Gibbs’ PhD dissertation was titled “On the Form of the Teeth of Wheels in Spur Gearing.” In fact, Gibbs was granted the first PhD in engineering in the United States. His graphical approach to gear design would foreshadow his graphical approach to thermodynamics. Gibbs taught Latin and natural philosophy at Yale for three years, followed by a three-­year tour of Europe. Returning to Yale, he was appointed Professor of Mathematical Physics (another US first). He pioneered the use of phase diagrams to visualize thermodynamic relationships. His contributions to thermodynamics were many, but his most influential work was “On the Equilibrium of Heterogeneous Substances,” published in two parts in 1876 and 1878. Gibbs brought theoretical rigor to the mostly experimental field of chemical thermodynamics. He coined the term statistical mechanics and conducted pioneering work in vector analysis and electromagnetics. Although mostly unappreciated in his lifetime, Williard Gibbs is now acknowledged one of the founders of modern physical chemistry. His lasting impact on science is indicated by the range of items bearing his name, among them Gibbs free energy (G), the Wilbraham–Gibbs constant G =

sin 0

d = 1.81937052...), the American Math-

ematical Society Gibbs Lecture, and Crater Gibbs (a 76-­km-­wide crater on the moon, named for Gibbs in 1964). As you continue, your lifelong education, let Gibbs’ words guide you: “One of the principal objects of theoretical research in any department of knowledge is to find the point of view from which the subject appears in its greatest simplicity” (Gibbs 1881).

9  The elder Gibbs has an interesting connection with the abolitionist movement. The rebelled slaves from the Spanish ship La Amistad were jailed in New Haven. Gibbs slowly learned a few words in their native language of Mende. He found a former slave named James Covey to help translate their stories.

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Problems   67

PROBLEMS (Thermodynamic data follow the problems.) 1. Are partial molar enthalpy, entropy, and Gibbs free energy intensive or extensive properties? Why? 2. We sometimes add the volumetric flows of a river and its tributary. Under what conditions can we add volumetric flows? 3. For reactions at constant pressure involving liquids and solids, it is commonly assumed that ∆H ≈ ∆E. Why? 4. To illustrate the relationship in Problem 3, calculate ∆H and ∆E when 100 g of ice melts at 25°C and 1 atm (constant pressure). How much heat is absorbed by the system? The molar volumes are 0.0196 L/mol and 0.0180 L/mol for ice and water, respectively. (See Section 1.4.1 for an explanation of why the molar volumes of ice and water are different and why that is important.). The ∆H values are given at the end of the problems. To interconvert energy from joules to L-­atm, recall that R = 8.314×10–3 kJ/mol-­K = 0.082057 L-­atm/mol-­K. 5. What pressure would be required to convert 1 g of graphite into 1 g of diamond at 25°C? Assume graphite and diamond are incompressible. The densities are 2.25 and 3.51 g/cm3 for graphite and diamond, respectively. Hint: Start with eq. 3.7 and develop a relationship between ΔGrxn and ΔVrxn at constant temperature and pressure. Before you quit your day job to make 1 g (= 5 carats) diamonds, realize that artificial diamonds are made at much higher temperatures and pressures (>100,000 atm) than you calculated. Graphite is dissolved in a molten metal catalyst and diamond precipitates. Why are the temperatures and pressures employed much higher than calculated here? 6. How many thermodynamic properties are needed to define the state of seawater? Use a simplified model of seawater consisting of H2O, Na+, and Cl–. What thermodynamic properties would you choose to define the system? 7. A student who missed a lecture (and avoided the assigned reading) looked at the table of thermodynamic data at the end of the problems and was confused. Taking the tabulated values for O2(aq) as an example, it

Jensen884347_c03.indd 67

G f . Can you explain the miscalculooks like H f TS lation and verify that the data are consistent? 8. What does it mean that G f for Ag+ is positive?

9. Consider the reaction: HSO4– → SO42– + H+ (25°C, 1 atm). Is the reaction endothermic or exothermic if all concentrations are 1 M? Does the reaction proceed spontaneously as written if all concentrations are 1 M?

10. Consider the reaction: NH4+ → NH3 + H+ (25°C, 1 atm). Is the reaction endothermic or exothermic if all concentrations are 0.01 M? Does the reaction proceed spontaneously as written if all concentrations are 1 M? 11. Consider the reaction: H2CO3 → HCO3– + H+ (25°C, 1 atm). What [HCO3–]/[H2CO3] ratio range is required so that the reaction is spontaneous? 12. Calculate K for the reaction in Problem 9 at equilibrium. (At equilibrium, you cannot assume all species concentrations are 1 M.) At what pH are the equilibrium activities of HSO4– and SO42– equal? 13. Calculate K for the reaction in Problem 10 at equilibrium. (At equilibrium, you cannot assume all species concentrations are 0.01 M.) At what range of pH is {NH4+} > {NH3} at equilibrium? 14. Calculate K for the reaction in Problem 11 at equilibrium. (At equilibrium, you cannot make any assumptions about species concentrations.) At what pH are the equilibrium activities of H2CO3 and HCO3– equal? 15. What is the criterion for equilibrium in terms of G? ∆Grxn? Grxn ? What are the criteria for equilibrium in terms of spontaneity and reversibility? 16. From the example discussed in Section 3.8.4, calculate the equilibrium constant for the equilibrium O2(g) = O2(aq) from the equilibrium concentrations of O2(g) and O2(aq). How does your value compare to the accepted value of 1.28×10–3 mol/L-­atm? (Be careful about units.) How does your value compare to that calculated from Grxn ?

17. From the example discussed in Section 3.7.3, calculate the equilibrium constant for AgCl(s) = Ag+ + Cl–. How does your value compare to the accepted value of 2.8×10–10?

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68   Chapter 3  Thermodynamic Basis of Equilibrium Thermodynamic data for chapter problems (25°C, 1 bar ≈ 1 atm):

H2O(l) H2O(s,ice) H+ Ag+ AgCl(s) C(s,graphite) C(s,diamond) H2CO3 HCO3– Cl– NH4+ NH3 O2(g) O2(aq) HSO4– SO42–

 Hf (kJ/mol)

 S (J/mol-­K)

 Gf (kJ/mol)

–285.830 –292.80 0 +105.579 –127.068 0 +1.895 –699.65 –691.99 –167.159 –132.51 –80.29 0 –11.7 –887.34 –909.27

+69.91

–237.129

0 +72.68 +96.2 +5.74 +2.377 187.4 +91.2 +56.5 +113.4 +111.3 +205.138 +11.09 +131.8 +20.1

0 +77.107 –109.789 0 2.900 –623.08 –586.77 –131.228 –79.31 –26.50 0 +16.4 –755.91 –744.53

CHAPTER REFERENCES Eisenberg, D. and D. Crothers (1979). Physical Chemistry with Applications to the Life Sciences. Menlo Park, CA: Benjamin/ Cummings Publ. Co., Inc. Gibbs, J.W. (1881). Letter to the editor. Proc. Am. Acad. Arts Sci. 16 (May, 1880–June, 1881), 420–421. Stumm, W. and J.J. Morgan (1996). Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters. 3rd ed. New  York: John Wiley & Sons, Inc. Wagman, D.D., W.H. Evans, V.B. Parker, R.H. Schumm, I. Halow, S.M. Bailey, K.L. Churney, and R.L. Nuttall (1982). The NBS tables of chemical thermodynamic properties. Selected values for inorganic and C1 and C2 organic substances in SI units. Phys. Chem. Ref. Data 11(Supplement No. 2).

All data except ice from NBS (Wagman et al. 1982)

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CHAPTER 4

Manipulating Equilibrium Expressions 4.1 INTRODUCTION You learned in Chapter 3 that the concept of chemical equilibrium has a thermodynamic basis. At this point, you can write equilibrium expressions and calculate equilibrium constants from Gibbs free energy of formation values. Given enough thermodynamic data, you could calculate the equilibrium constant for any equilibrium. However, there are many instances in environmental engineering and science where you do not have sufficient thermodynamic data to calculate the equilibrium constant for an equilibrium of interest to you. You may know equilibrium constants for related equilibria. But how can information about similar equilibria be used to determine the concentrations of chemical species at equilibrium? To calculate species concentrations efficiently, you must be able to exploit known equilibria. Throughout this text, you will be manipulating systems of equilibrium expressions to calculate species concentrations at equilibrium. In this chapter, some simple rules for the manipulation of equilibrium expressions will be presented. Three points will be emphasized throughout this chapter. First, as discussed in Section 4.2, chemical equilibria can be interpreted as both chemical expressions and mathematical statements. Although the chemical and mathematical forms of equilibria are used for different purposes, they should convey consistent information about the way in which species and species concentrations relate to the equilibrium expression. Second, equilibrium constants can have units associated with them (Section 4.3). Third, knowledge of an equilibrium constant for one reaction allows for the calculation the equilibrium constants for a number of related reactions. To calculate additional equilibrium constants, you must master the skills of reversing (Section 4.4), changing the stoichiometric constants (Section 4.5), and adding equilibria (Section 4.6). With these skills, new equilibria and equilibrium constants can be created, as discussed in Section 4.7.

Key idea: You must be able to exploit known equilibria to calculate species concen­ trations efficiently

4.2  CHEMICAL AND MATHEMATICAL

FORMS OF EQUILIBRIA

4.2.1  Chemical Expressions Chemical equilibria can be written in two ways: chemical expressions and mathematical expressions. Chemical expressions are idealized ways of writing the chemical

69

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70   Chapter 4  Manipulating Equilibrium Expressions Key idea: Equilibria can be written as chemical expressions and mathe­ matical equations Key idea: Chemical expressions show relation­ ships between chemical species: reactants, products, and reaction stoichiometry Key idea: Chemical expressions do not show relationships between species concentrations Key idea: Do not use brackets (e.g., [A]) or braces (e.g., {A}) around species names in chemical expressions Key idea: The mathematical versions of chemical expressions show math­ ematical relationships between species concen­ trations (or activities)

reactions that occur at equilibrium. As an example, consider the dissociation of ammonium (NH+4 ) to ammonia (NH3) and H+. For this chemical equilibrium, you can write the chemical expression: NH 4

NH 3

H

Chemical expressions have several important characteristics. First, chemical expressions describe the relationships between chemical species. In particular, chemical expressions show the reactants, products, and stoichiometry of the chemical reactions. For the ammonium/ammonia example, the chemical expression shows that each mole of ammonium that dissociates produces 1 mole of ammonia and 1 mole of H+. By convention, reaction stoichiometric coefficients equal to 1 are omitted in the chemical expressions. Second, the = symbol in the chemical expression signifies that the reaction is at equilibrium. Thus, the chemical expression tells us that the reaction is reversible and spontaneous. This means that the reactions proceed in each direction. In this way, the symbol = is not a mathematical sign of equality, but rather a shorthand notation for ⇌. In the example, the chemical expression reveals that ammonium dissociates to form 1 mole of ammonia and 1 mole of H+ and also that for each mole of ammonia and H+ that combine, 1 mole of ammonium is formed. Third, chemical expressions have a serious limitation. They do not tell you the relationship between chemical species concentrations. In the ammonium example, the chemical expression does not mean that the concentration of ammonium equals the sum of the concentrations of ammonia and H+ (i.e., the chemical express does not imply that [NH4+] = [NH3] + [H+]). When writing chemical expressions, never place brackets or braces around the names of the chemical species.

4.2.2  Mathematical Equations Equilibria also can be written as mathematical equations. To write an equilibrium as a mathematical equation, set the equilibrium constant equal to the product of the product species concentrations (or activities) raised to their stoichiometric coefficients divided by the product of the reactant species concentrations (or activities) raised to their stoichiometric coefficients (as in Section 3.9. 1).1 For the example above: K

The equation K

NH 3 H NH 4

NH 3 H NH 4

NH 3 H NH 4

is the proper mathematical relationship between

species concentrations. In Chapter  3, the symbol νi for the stoichiometric coefficient of species i in an equilibrium was introduced. Recall that stoichiometric coefficients of reactants are

1  In this chapter, the term stoichiometric coefficient will be used to mean the stoichiometric coefficients in a chemical reaction (i.e., reaction stoichiometric coefficients).

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4.2  |  Chemical and Mathematical Forms of Equilibria   71

considered negative and stoichiometric coefficients of products are considered positive. Thus, you can write in general K

i

i

i

i

i

i

where {i} and [i] are the activity and concentration of species i, respectively.2 The ammonium example demonstrates the relationship between chemical expressions of equilibrium and mathematical statements of equilibrium. In forming the mathematical expressions, you are translating the chemical expressions into mathematical equations. As another example, write the mathematical form of the chemical equilibrium: CO32– + H+ = HCO3–; K. (The species CO32– is carbonate ion and the species HCO3– is bicarbonate ion.) Recall K is written as the product of the product concentrations (activities) divided by the product of the reactant concentrations (activities), each HCO3 HCO3 raise to their stoichiometric coefficients. In this case: K . CO32 H CO32 H Two other illustrations are shown in Worked Example 4.1. It is important to practice so you are able to translate quickly between chemical expressions and mathematical equations.

Interconversion of Chemical Expressions and Mathematical Statements

Worked Example 4.1 

Write the mathematical form of the equilibrium given by: Fe OH 3 s

Fe3

3OH

Write the chemical expression for the equilibrium given by: K

Fe3 PO 4 Fe

2

3

2

PO 4

s 3

2

Solution For the first equilibrium, set K equal to the product of the species activities raised to their signed stoichiometric coefficients: K

Fe 3 OH Fe OH 2 s

3

For the second equilibrium, interpret the numerator term as a product and the denominator terms as reactants. The chemical expression is 3Fe 2

2PO 4 2

Fe 3 PO 4

2

s

 Throughout this chapter, it will be assumed that the solutions are dilute. Thus, activities and concentrations are nearly equal, and you can use concentrations as approximations for activities in equilibrium expressions (see Section 3.9.1 and Chapter 21).

2

←Worked example

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72   Chapter 4  Manipulating Equilibrium Expressions

4.2.3  Example of Chemical Expressions and Mathematical Equations: Le Châtelier’s Principle Key idea: Chemical expressions of equilibria and mathematical equations of equilibria contain the same information and must lead to consistent conclusions Le Châtelier’s principle: a change to a thermody­ namic function controlling the equilibrium will cause the system to adjust to min­ imize the change  orked W ­example →

Chemical expressions and mathematical equations are two ways to write the information contained in chemical equilibria. As you have seen, each approach has its uses. However, the two ways of writing equilibria must lead to consistent conclusions. Throughout this chapter, you will see that chemical expressions and mathematical equations must be consistent in the trends they predict in species concentrations. One way in which predictions from chemical expressions and mathematical equations can be compared is Le Châtelier’s principle (after Henri-­Louis Le ­Châtelier, 1850–1936; see the Historical Note at the end of the chapter). For a system initially at equilibrium, Le Châtelier’s principle states that a change to a thermodynamic function controlling the equilibrium (e.g., a change in the system temperature, system pressure, or species concentration) will cause the system to adjust to minimize the change. For example, an increase in a reactant concentration will “push” the system originally at equilibrium toward products. A decrease in a reactant concentration will “pull” the system originally at equilibrium toward reactants. Le Châtelier’s principle is usually applied to chemical expressions. For the ammonium/ammonia example, assume the system is at equilibrium. Now, imagine that the H+ concentration is increased.

Thoughtful Pause What will happen to the NH3/H+/NH4+ equilibrium if the concentration of H+ is increased?

An increase in the H+ concentration will shift the system away from equilibrium. According to Le Châtelier’s principle, the system will adjust to minimize the impact of the increase in the H+ concentration. To accomplish this, the equilibrium (NH+4 = NH3 + H+) will shift to the left, with a resulting increase in the ammonium concentration and decrease in the ammonia concentration. Is Le Châtelier’s principle consistent with the mathematical version of ­equilibrium NH 3 H , K is constant at conNH 4 stant temperature and pressure. Think of this equation as purely a mathematical conxy struct divorced from the chemistry: K = constant = . If you make y larger but keep z K constant, then one of the following statements must be true: (1) x decreases, (2) z increases, or (3) both x decreases and z increases. Back to the chemistry: To maintain a constant amount of N in the system, you cannot decrease the ammonia concentration without increasing the ammonium concentration by an equal amount. Similarly, you cannot increase [NH+4 ] without decreasing [NH3] by an equal amount. Thus, the ammonia concentration must decrease and the ammonium concentration must increase. In the symbols used previously, x decreases and z increases. The mathematical statement of equilibrium tells you that the ammonium concentration must increase (and the ammonia concentration must decrease) as H+ is increased.

information? In the mathematical expression K

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4.3  |  Units of Equilibrium Constants   73

Thus, Le Châtelier’s principle (usually applied to the chemical expression) is consistent with the mathematical version of equilibrium information. Another instance of Le ­Châtelier’s principle is shown in Worked Example 4.2.3

Le Châtelier’s Principle

Worked Example 4.2 

Using Le Châtelier’s principle, explain why bleach manufacturers bubble chlorine gas into sodium hydroxide solutions (rather than into pure water) when making bleach. The pertinent equilibria are Cl 2 g H2 O

H2 O H

HOCl

Cl

H

OH

Solution From the first equilibrium, the bleach (here, HOCl or hypochlorous acid) concentration is increased when [Cl2(g)] is increased and/or when [H+] is decreased. From the second equilibrium, [H+] is decreased when [OH–] is increased (since the activity of water, a pure liquid, is fixed; see Section 4.3.2). Thus, bleach manufacturers add NaOH to increase the OH– concentration, thus decreasing the H+ concentration and increasing the HOCl concentration. (This saves money by reducing the mass of water shipped per kg of bleach.)

4.3  UNITS OF EQUILIBRIUM CONSTANTS 4.3.1  Introduction What are the units of equilibrium constants? Recall from Section 3.9.2 that equilibrium constants are dimensionless, because each term in the equilibrium constant is normalized to a standard state having the same units as the species activity. You can write mathematical expressions such as logK and Grxn = –RTlnK without concern about the units of K. Although equilibrium constants are inherently dimensionless, it is convenient to associate units with equilibrium constants as a reminder of the concentration units of the species in the equilibrium expression. This is an accounting trick to aid in manipulating equilibria and will be used occasionally throughout this text. In addition, associating units with equilibrium constants gives you another way to check whether the manipulation of the equilibria and equilibrium constants was performed properly: the units associated with the final equilibrium constant should match the final equilibrium expression.

 Le Châtelier’s principle is a powerful tool for understanding the qualitative behavior of simple chemical systems. The great American chemist and double Nobel laureate Linus Pauling wrote (Pauling 1964): “When you have obtained a grasp of Le Châtelier’s principle, you will be able to think about any problem of chemical equilibrium that arises, and, by use of a simple argument, to make a qualitative statement about it.... Some years after you have finished your college work, you may . . . have forgotten all the mathematical equations relating to chemical equilibrium. I hope, however, that you have not forgotten Le Châtelier’s principle.”

3

Key idea: Equilibrium constants are dimensionless, but units are associated with equilibrium constants as a reminder of the units of the species concentra­ tions in the equilibrium

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74   Chapter 4  Manipulating Equilibrium Expressions

4.3.2  Concentration Scales

Worked example →

To associate units with equilibrium constants, you must choose a consistent set of concentration scales. Four types of species are found commonly in equilibria in aqueous systems: dissolved species, pure solids, pure liquids, and gases. As discussed in Chapter  2, dissolved species concentrations in dilute solution are usually expressed using the molarity concentration scale. Thus, dissolved species concentrations commonly are written in units of mol/L = M. For example, the equilibrium constant for NH 3 H M2 ammonium dissociation K may have units of = M associated with NH 4 M it as a reminder that the concentrations of NH3, H+, and NH+4 are in M. Pure solids and pure liquids usually employ the mole fraction concentration scale. Pure solids and pure liquids have a mole fraction of unity. As a result, pure solids and pure liquids (e.g., water) do not affect the units of equilibrium constants. Thus, for the reaction CaCO3(s) = Ca2+ + CO32– you can write Ca 2 CO32 CaCO3 s

K

Ca 2

CO32

The equilibrium constant K in this case has associated units of M2. You will make frequent use of the following reaction: H2O = H+ + OH–. For this reaction, you can write H

K

OH H2 O

H

OH

where K usually has associated units of M2. Gases usually employ the partial pressure concentration scale. Thus, gases commonly have concentrations of atm. For example, for the reaction O3(aq) = O3(g), PO3 O3 g you can write K = with typical associated units of atm/M = O3 aq O3 aq atm-­L/mol. Another illustration of associating units with equilibrium constants may be found in Worked Example 4.3. Worked Example 4.3 

Units Associated with Equilibrium Constants

Using typical concentration units, what units should be associated with the first equilibrium in Example 4.2? The equilibrium is Cl 2 g

H2 O

HOCl

Cl

H

Solution Typical concentration units for gases, pure liquids (such as water), and ­dissolved species are atm, none (mole fraction), and mol/L, respectively. Thus, you would associate the units of mol3/L3-­atm (or M3/atm) with the first ­equilibrium in Example 4.2.

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4.4  |  Reversing Equilibria   75

4.3.3  Interpreting Units Associated with Equilibrium Constants Recall that a general rule of quantitative analysis is to avoid comparing values with different units. For example, it is improper to conclude that a hydrocarbon-­fueled automobile with a gas mileage of 30  miles per gallon is more fuel-­efficient than an automobile with a gas mileage of 12.8 km/L. In fact, the gas mileage is about the same for the two vehicles. Similarly, you cannot compare the values of equilibrium constants with different associated units. For example, K for the equilibrium H3PO4 = PO43– + 3H+ is 3.2×10–22 and K for the equilibrium HCN = CN– + H+ is 6.3×10–11. Can we use this information to compare the ability of phosphoric acid (H3PO4) and hydrocyanic acid (HCN) to produce H+? The answer is: No. The equilibrium constant for the first equilibrium has units of M3 (= mol3/L3) associated with it, whereas the equilibrium constant for the second equilibrium has units of M associated with it. The equilibrium constants have different associated units and cannot be compared. The units associated with equilibrium constants are a reminder that you should not compare values with different (associated) units.

←Worked example

Key idea: Take great care when ­comparing equilibrium constants with different associated units

4.4  REVERSING EQUILIBRIA One of the take-­home lessons from this chapter is that knowledge of one equilibrium constant allows you to easily calculate the equilibrium constants of related equilibria. In this section, you will see that knowing an equilibrium constant allows you to calculate the equilibrium constant for the reaction formed when reactants and products in the original reaction are interchanged. Interchanging reactants and products is called reversing an equilibrium. For the example equilibrium of this chapter (NH4+ = NH3 + H+), the equilibrium constant, K, is about 6.3×10–10 = 10–9.2 M.4 What is the equilibrium constant for the reverse reaction (NH3 + H+ = NH4+)? You can answer this question in two ways: by examining the chemical expression and by examining the mathematical equation for K. From the chemical expression, it is clear that Grxn for the ammonia/H+ association reaction is –1 times Grxn for the ammonium dissociation reaction.

Why are the ∆Grxn ° of reverse reactions equal in absolute value but ­opposite in sign? For any reaction: i

i

G f ,i

products

i

G f ,i

reactants

i

G f ,i

4  Equilibrium constants can take values over many orders of magnitude. In addition, you know that logK values are proportional to Grxn and therefore have special meaning. Thus, we usually express K values as the antilog base 10;

in other words, as 10x.

←Worked example

Key idea:

Thoughtful Pause

Grxn

reversing an equilibrium: an equilibrium is reversed when its reactants and products are interchanged

Grxn values of reverse reactions are equal in absolute value but opposite in sign and K values of reverse reactions are reciprocals of one another.

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76   Chapter 4  Manipulating Equilibrium Expressions Worked example →

Thus, Grxn of reverse reactions are equal in absolute value but opposite in sign, since the reactants and products are simply interchanged for reverse reactions. Now, Grxn

since K e RT , it makes sense that K for the reverse reaction must be equal to the reciprocal of K for the original reaction. For the example, K for the NH3 + H+ association 1 reaction is (K for the ammonium dissociation reaction)–1 = = 10+9.3 M–1. Thus, 9.3 10 M examination of the chemical expression tells you that K for a reaction is the reciprocal of K for its reverse reaction. The mathematical form of the equilibrium also should tell you that K for a reaction is the reciprocal of K for its reverse reaction. This is obvious by inspection. For example, the ammonium dissociation reaction has an equilibrium constant equal to NH 3 H , whereas the equilibrium constant for the ammonia/H+ association NH 4 NH 4 ­reaction is . This shows again that K for the ammonia/H+ association NH 3 H reaction is the reciprocal of K for the ammonium dissociation reaction. You can show that K for a reaction is the reciprocal of K for its reverse reaction in the general case by using the product notation of Section 4.2.2. For any equilibrium (where activity and concentration are nearly interchangeable): K i i . For the i reverse reaction, νi,rev = –νi since reactants and products are reversed in the reverse 1 reaction. Thus: Krev i i ,rev i i . Another example is provided in Worked i i K Example 4.4.

Reversing Equilibria

Worked Example 4.4 

Find the equilibrium constant for the equilibrium HgCl+ = Hg2+ + Cl– if Grxn 38.54 kJ / mol for the equilibrium Hg2+ + Cl– = HgCl+. Solution 38.54 kJ / mol for the equilibrium Hg2+ + Cl– = HgCl+, then Grxn for If Grxn the reverse reaction (HgCl+ = Hg2+ + Cl–) is +38.54 kJ/mol. Thus, K for HgCl+ dissociation reaction is e

Grxn

RT

= 1.76×10–7.

4.5  EFFECTS OF STOICHIOMETRY 4.5.1  Linear Independence Compare the following two reactions: NH+4 = NH3 + H+ with K = K1 and 2NH4+ = 2NH3 + 2H+ with K = K2. The chemical expressions convey the same information: for each mole of ammonium dissociated, 1  mole of ammonia and 1  mole of H+ are

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4.5  |  Effects of Stoichiometry   77

formed. (In addition, for each mole of ammonia associating with 1 mole of H+, 1 mole of ammonium is formed.) Since the chemical expressions NH4+ = NH3 + H+ and 2NH4+ = 2NH3 + 2H+ contain the same information, it would be improper to use both equilibria in describing a chemical system. Why? You cannot use more than one expression that contains the same information in any physical system. For example, if you were solving an algebraic system for the values of x and y, you could not use both the equation x + 2y = 5 and the equation 3x + 6y = 15. Why? The two equations convey the same information. When two equilibria differ only by a constant multiple of their stoichiometric coefficients, we say the equilibria are linearly dependent. This definition of linear dependence is expanded in Section 4.6.2. You will find that you need a set of linearly independent equations to solve chemical systems. Since the mathematical equations representing equilibria come from chemical expressions (see Section 4.2), this means you need a set of linearly independent equilibria to describe a chemical system.

4.5.2  Free Energy and Equilibrium Constants Return to two reactions: NH4+ = NH3 + H+ with K = K1 and 2NH4+ = 2NH3 + 2H+ with K = K2. What is Grxn for each equilibrium? You know Grxn= i G f ,i , so the standard i Gibbs free energy of the second equilibrium should be twice the standard Gibbs free energy of the first equilibrium: doubling the stoichiometric coefficients values doubles Grxn. What about K? You know that K2

exp

Grxn,2 RT

exp

2 Grxn,1 RT

exp

Grxn,1 RT

2

K12 Thus, K for 2NH4+ = 2NH3 + 2H+ is the square of K for NH4+ = NH3 + H+. Is this result consistent with the mathematical form of the equilibrium informaNH 3 H , whereas for the second equilibrium NH 4

tion? For the first equilibrium, K1 K2

NH 3

2

NH 4

H

2

2

= K12. As expected, the mathematical form of the equilibria also

reveals that K for the second equilibrium is the square of K for the first equilibrium. In general, let K be the equilibrium constant for an equilibrium. If you multiply each of the stoichiometric coefficients by a constant c, then the equilibrium constant for the new equilibrium is

i

i

c i i

i

i

c

= Kc.

←Worked example

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78   Chapter 4  Manipulating Equilibrium Expressions Key idea: Multiplying stoichiometric coefficients by a constant multiplies Grxn by the constant and raises K to the power of the constant

The effects of stoichiometry can be summarized as follows. Multiplying each stoichiometric coefficient in an equilibrium expression by a constant does not convey any new information about the equilibrium. However, multiplying stoichiometric coefficients by a constant, c, results in an equilibrium with Grxn equal to c times the Grxn for the original equilibrium and K equal to K for the original equilibrium raised to the c power. You will find that it is sometimes convenient to track logK instead of K. Multiplying stoichiometric coefficients by a constant, c, multiplies Grxn by c and multiplies logK by c (since logKc = clogK). The effects of changing stoichiometric constants also are shown in Worked Example 4.5.

Changing Reaction Stoichiometry

Worked Example 4.5 

Find the equilibrium constant for HCN = H+ + CN– if Grxn for the reaction ½HCN = ½H+ + ½CN– is +27 kJ/mol. Solution If Grxn for the reaction ½HCN = ½H+ + ½CN– is +27 kJ/mol, then Grxnfor the reaction HCN = H+ + CN– is 2(+27 kJ/mol) = +54 kJ/mol. Thus, K for HCN = H+ + CN–at 25°C (298 K) is K Here, Grxn RT

exp

Grxn RT

kJ mol   kJ 8.314 10 3 298 K mol K 54

21.8, so K

e

21.8

3.4 10 10.

You also could calculate the equilibrium constant for ½HCN = ½H+ + ½CN– from Grxn (K = 1.85×10–5) and note that the equilibrium constant for HCN = H+ + CN– is (1.85×10–5)2 = 3.4×10–10.

It should be noted that reversing equilibria (discussed in Section  4.3) is really a special case of changing reaction stoichiometry. Recall that we usually think of stoichiometric coefficients for reactants as negative and stoichiometric coefficients for products as positive. Therefore, reversing an equilibrium (i.e., interchanging reactants and products) is equivalent to multiplying each stoichiometric coefficient by –1. Thus, in reversing an equilibrium, the sign of the Gibbs free energy of reaction is multiplied by –1 and the new equilibrium constant is the reciprocal of the original equilibrium constant (i.e., K is raised to the –1 power).

4.6  ADDING EQUILIBRIA 4.6.1  Free Energy and Equilibrium Constants You can form new equilibria by adding existing equilibria. For example, how would you create a new equilibrium to show the relationship between ammonium, ammonia,

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4.6  |  Adding Equilibria   79

hydroxide ion (OH–), and water? You can accomplish this by adding the following two equilibria: NH 4 Adding: NH 4

NH3

H

reaction 1

H

OH

H2 O

reaction 2

H

OH

NH3

H

←Worked example

H2 O

Eliminating the common species on both sides of the equilibrium (here, only H+ appears on both sides): NH 4

OH

NH3

H 2 O reaction 3

How do Grxn and K for the new equilibrium (reaction 3) compare to Grxn and K for the original equilibria (reactions 1 and 2)? You know from Section 3.2.2 that free energies are additive. Thus, you might expect that Grxn3 = Grxn1 Grxn 2. For the equilibrium constant: K3

exp

exp

exp

Grxn 3 RT Grxn1

Grxn 2

RT Grxn1 RT

exp

Grxn 2 RT

K1 K2 Therefore, adding equilibria results in a new equilibrium with Gibbs free energy equal to the sum of the Gibbs free energies of the individual equilibria and with an equilibrium constant equal to the product of the individual equilibrium constants. Recall that if K3 = K1K2, then logK3 = logK1 + logK2. Thus, when adding equilibrium expressions, add Grxn values and multiply K values or, equivalently, add Grxn values and add logK values. This analysis points to a nice trick involving logK values: logK values are additive when equilibria are added. With this fact, you often can calculate equilibrium constants for new equilibria in your head. The mathematical form of the equilibria should give the same result. Indeed: K1

NH 3 H , K2 NH 4

NH 3 H 2 O NH 4 OH

H2 O , and K3 H OH

From the right-­hand sides of these expressions, it is clear that K3 = K1K2. In the general case, consider equilibrium constants K1

i

i

i ,1

and K 2

i

i

. Then

i, 2

K3, the equilibrium constant formed by adding reactions 1 and 2, can be calculated from

i

i

i ,1

i, 2

= K1K2.

Key idea: When adding equilibria, add Grxn values and multiply K values (or add logK values)

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80   Chapter 4  Manipulating Equilibrium Expressions

4.6.2  Linear Independence Revisited linearly independent ­equilibria: a set of equi­ libria where no equi­ librium expression can be formed by linearly combining some or all of the other equilibria in the set

It is important to note in the example of the previous section that reaction 3 does not convey any information in addition to the information conveyed by reactions 1 and 2 together. In fact, given any two of reactions 1, 2, or 3, it is possible to create the other reaction by judiciously adding and/or reversing the two reactions in hand. It is possible to extend the concept of linearly independent equilibria developed in Section 4.4: a set of equilibria is linearly independent if no equilibrium expression can be formed by linearly combining some or all of the other equilibria in the set. Here, linearly combining means reversing equilibria, multiplying stoichiometric coefficients by a constant, and/ or adding equilibria. As an example, consider the following three equilibria: H 2 CO3    HCO3 2

 orked W example →

HCO3    CO3

2H

H H

2

CO3    H 2 CO3

K1

K2 K3

Thoughtful Pause Are these three equilibria linearly independent? They are not linearly independent. You can form the third equilibria by reversing and adding the first two equilibria: HCO3

H    H 2 CO3

K

  CO32

H    HCO3

K

CO32     H 2 CO3

2H

1 K1 1 K2

K3

1 K1 K2

In expressing the equilibrium relationships between H2CO3, HCO3–, and CO32–, you can use any two of the three equilibria (but not all three). Thoughtful Pause How could you form the first equilibrium from the other two?

 orked W example →

You can form the first equilibria by reversing and adding the other two equilibria: H 2 CO3   2 H   CO32

 CO32

H   HCO3

H 2 CO3    HCO3

K K

H

K1

1 K3 1 K2

1 K 2 K3

Another illustration of linear independence is shown in Worked Example 4.6.

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4.7  |  Creating Equilibria   81

So which equilibria should you use? You will find that certain combinations may make more sense chemically. Mathematically, any linearly independent set will do. This means any three of the carbonate system equilibria listed above, or any three of the aluminum equilibria in Worked Example 4.6.

Linear Independence

Worked Example 4.6 

Is the following set of equilibria linearly independent? Al OH 3 s     Al3 Al

3

OH    AlOH

AlOH 2 Al OH

3OH 2

OH    Al OH

2

OH   Al OH 3 s

2

Solution The fourth equilibrium can be formed by reversing and adding the first three equilibria: Al3

 3OH   Al OH 3 s

AlOH 2   Al3 Al OH

2

Al OH

2

OH

    AlOH 2 OH

OH Al OH 3 s

The set is not linearly independent.

4.7  CREATING EQUILIBRIA In performing equilibrium calculations, you often have to manipulate known equilibria to write a target equilibrium in the form desired. Two skills are required: balancing chemical reactions and manipulating equilibria.

4.7.1  Balancing Chemical Reactions Conservation of mass (Section  3.2.2) requires that chemical reactions are balanced. Balancing reactions is a five-­step process (see also Appendix C, Section C.1.2). The process will be illustrated with the equilibrium between hypochlorous acid (HOCl, a common disinfectant) and chloride. Step 1: Write the known reactants on the left and known products on the right. In the example, write: HOCl = Cl– Step 2: Adjust stoichiometric coefficients to balance all elements except H and O.

←Worked example

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82   Chapter 4  Manipulating Equilibrium Expressions Key idea: To balance a chemical reaction, balance all elements except H and O, add water to balance O, add H+ to balance H, and add e– to balance the charge

In the example, all elements except H and O already are balanced and you still have: HOCl = Cl–. If you were balancing the equilibrium between molecular chlorine (Cl2) and chloride, you would adjust the stoichiometric coefficients to balance the element Cl. You could accomplish this in either one of two equivalent ways: Cl2 = 2Cl– or ½Cl2 = Cl–. Step 3: Add water (H2O) to balance the element O. In the example, oxygen will be balanced if you add water as a product: HOCl = Cl– + H2O. Step 4: Add H+ to balance the element H. In the example, hydrogen will be balanced if you add one H+ as a reactant: HOCl + H+ = Cl– + H2O. At this point in the process, all elements should be balanced. Step 5: Add electrons (e–) to balance the charge. In the example, the left-­hand side has a net charge of +1 and the right-­hand side has a net charge of –1. You must add two e– as reactants: HOCl + H+ + 2e– = Cl– + H2O. The equilibrium is now balanced. Although it may seem slow and unwieldy at first, it is recommended that you follow this process step by step when balancing reactions. Another illustration of balancing reactions is given in Worked Example 4.7. The balancing procedure is extended in Problem 14 at the end of the chapter.

Balancing Chemical Reactions

Worked Example 4.7 

Balance the equilibrium between thiosulfate (S2O32–) and sulfate (SO42–). Solution Follow the procedure in the text: Step 1: Write the known species. S2 O32

SO 4 2

Step 2: Balance all but O, H. S2 O32

2SO 4 2

Step 3: Add H2O to balance O. S2 O32

5H 2 O

2SO 4 2

Step 4: Add H+ to balance H. S2 O32

5H 2 O

2SO 4 2

10 H

Step 5: Add e– to balance the charge. S2 O32

5H 2 O

2SO 4 2

10 H

8e

10H

8e

The balanced reaction is S2 O 3 2

5H 2 O

2SO 4 2

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4.7  |  Creating Equilibria   83

4.7.2  Manipulating Equilibria After balancing the desired reaction, it may be possible to relate it to other equilibria by reversing equilibria, multiplying stoichiometric coefficients by a constant, and/or adding equilibria. It is important to master these skills before you proceed to the mechanics of equilibrium calculations in Part II of this text. Consider the following example. In developing remediation strategies for a landfill leachate, you wish to know the equilibrium constant for the dissolution of cadmium carbonate solid to form divalent cadmium (Cd2+) and carbon dioxide gas. You search tables of equilibrium constants (see Appendix D) and find the following information (with associated units for the K values): CdCO3 s    Cd 2 HCO3    CO32

CO32 H

H 2 CO3

 HCO3

H

H 2 CO3

 CO2 g   H 2 O

K1

10  

11.66

M2

K2

10  

10.3

K3

10  

6.3

K4

10   atm-L / mol

M

M

1.5



You can find the equilibrium constant for the target equilibrium by (1) writing the balanced equilibrium, and (2) manipulating the known equilibria to match the target equilibrium.5 First, balance the reaction by using the five-­step process outlined in Section 4.6. (You may wish to try this first on your own.) Step 1: Write the known reactants on the left and known products on the right. CdCO3 s

Cd 2

CO2 g

Step 2: Adjust stoichiometric coefficients to balance all elements except H and O. All elements except H and O (i.e., Cd and C) are balanced in Step 1. Step 3: Add water (H2O) to balance the element O. CdCO3 s

Cd 2

CO2 g

H2 O

Step 4: Add H+ to balance the element H. CdCO3 s

2H

Cd 2

CO2 g

H2 O

Step 5: Add electrons (e–) to balance the charge. The charges are balanced in Step 4. The final balanced reaction is: CdCO3 s

2H

Cd 2

CO2 g

H 2 O.

 Recall from Section 4.3 that the typical units of solids and liquids are the mole fraction. The mole fractions of pure solids (e.g., CdCO3 (s)) and pure liquids (e.g., H2O) are unity. Thus, the units of pure solids and pure liquids disappear in the equilibrium constants.

5

←Worked example

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84   Chapter 4  Manipulating Equilibrium Expressions Now you need to combine the existing equilibria to obtain CdCO3(s) + 2H+ = Cd2+ + CO2(g) + H2O. Start with the K1 equilibrium. It has CdCO3(s) and Cd2+ in the desired places as a reactant and product, respectively. You now have: CdCO3 s

Cd 2

CO32

K1

10

11.66

M2

You do not want carbonate (CO32–) as a product. Thus, you need to add an equilibrium with carbonate as a reactant to cancel the carbonate as a product in the equilibrium you are building. This can be accomplished by reversing the K2 equilibrium and adding: CdCO3 s    Cd 2   CO32

 CO32

 H    HCO3

CdCO3 s    H    Cd 2

 HCO3

K1

10  

K5

1 K2

K6

11.66

M2

  10.3 M 10

K K1 K 5   1 K2

1

1  0

1.36

M

Notice that you take the reciprocal of K2 when you reverse the K2 equilibrium. Also, you multiply the equilibrium constants when adding reactions. Again, you do not want bicarbonate (HCO3–) as a product. To eliminate HCO3–, reverse the K3 equilibrium and add: CdCO3 s    H    Cd 2  HCO3

 HCO3

 H    H 2 CO3

   CdCO3 s    2 H    Cd 2

 H 2 CO3

   K 6

10  

1.36

M

K 7  10 M 6.3

K8

K6 K7

1

K1 K 2 K3

 10 4.94

Finally, add the K4 equilibrium to eliminate H2CO3 as a product and replace it with CO2(g) as a product: CdCO3 s    2 H

 H 2 CO3

K8

10   4.94

                 H 2 CO3  CO2 g    H 2 O

K4

10   1.5 atm-L / mol

   CdCO3 s    2 H

 Cd 2  Cd 2

 CO2 g    H 2 O K 9

K8 K 4 10  

Worked example →

6.44

K1 K 4 K 2 K3 atm-L / mol

Thus, the target equilibrium (cadmium carbonate solid in equilibrium with ­ ivalent cadmium and carbon dioxide gas) has an equilibrium constant equal to d 106.44 atm-­L/mol. As an aside, this example points out the value of manipulating logK values. You KK showed K9 = 1 4 , so logK9 = logK1 + logK4 – logK2 – logK3. You can do this ­calculation K 2 K3 in your head: –11.66 + 1.5 – (–6.3) – (–10.3) = +6.44, so K9 = 106.44.

4.7.3  Why Learn How to Manipulate Equilibria? In Section  4.7.2, you calculated an equilibrium constant for a target equilibrium by linearly combining other equilibria.

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4.7  |  Creating Equilibria   85

Thoughtful Pause From Chapter 3, is there another way to calculate the equilibrium constant for the target equilibrium? Grxn RT .

You know from Chapter 3 that you can calculate K from the expression K e The standard Gibbs free energy of formation values for CdCO3(s), H+, Cd2+, CO2(g), and H2O are –669.4, 0, –77.58, –394.37, and –237.18 kJ/mol, respectively. These values give Grxn = –39.73 kJ/mol or K = 106.96 at 25°C. This is about three times larger than the value of K = 106.44 atm-­L/mol determined in Section 4.7.2. Such discrepancies are not unusual, given the uncertainty in the thermodynamic data and errors in the equilibrium constants of the constituent equilibria. It is important to become adept at calculating equilibrium constants both from linearly combining other equilibria and from thermodynamics. Why should you learn how to calculate equilibrium constants by manipulating known equilibria when you can always calculate equilibrium constants from thermodynamics? In a reasonably complex aqueous system, it is unlikely you will be able to look up all required equilibria in the form you want. In addition, the G f values may not be available for the species of interest. If G f values are not available, then Grxn cannot be calculated. Thus, it is extremely important to be able to manipulate equilibria to calculate equilibrium constants.

Key idea: Equilibrium constants can be calculated by linearly combining other equilibria or from Grxnvalues Key idea: You should learn to manipulate equilibria since thermodynamic data (K and Gf values) are not always available

4.7.4  A Systematic Approach to Manipulating Equilibria: An Advanced Concept The process described in Section 4.7.2 of linearly combining equilibria to create a target equilibrium may seem somewhat haphazard at first. For simple systems, it is often expedient to play with the equilibria as jigsaw puzzle pieces until the target equilibrium emerges (as was done in the cadmium carbonate example). For more complex systems, you may wish to take a more systematic approach as follows. List the n constituent equilibria that are to be linearly combined to form the target equilibrium. Let m be the total number of chemical species in the n constituent equilibria and in the target equilibrium. Write the stoichiometric coefficients for each species in each equilibrium, using the usual notation that stoichiometric coefficients of reactants are negative and the stoichiometric coefficients of products are positive. Let νi,j be the stoichiometric coefficient of species j in constituent equilibrium i. Now, multiply the stoichiometric coefficients of constituent equilibrium i by a constant ci. For each species j, sum the values of νi,jci over all the equilibria and set the sum equal to the stoichiometric coefficient of species j for the target equilibrium, νtarget,j:

i

ci

i, j

=

νtarget,j. This will yield a linear set of m equations (one for each species) in n unknowns (c1, . . ., cn) that can be solved to find the ci values. From the ci values, you can determine the linear operations to perform on the constituent equilibria to form the target equilibrium and thus the target equilibrium K. As an example, consider the CdCO3(s) problem of Section 4.7.2. The target equilibrium is: CdCO3 s

2H

Cd 2

CO2 g

H 2 O K target

?

←Worked example

ISTUDY

86   Chapter 4  Manipulating Equilibrium Expressions The n = 4 constituent equilibria are: Eq. 1: CdCO3 s

Cd 2

CO32

K1

10

K2

10

10.3 6.3

11.66

Eq. 2: HCO3

CO32

Eq. 3: H 2 CO3

HCO3

H

K3

10

Eq. 4: H 2 CO3

CO2 g

H2 O

K4

101.5 atm-L / mol .

H

M

M

The system has m = 8 chemical species: CdCO3(s), Cd2+, CO2(g), H2CO3, HCO3–, CO32–, H+, and H2O. Writing the matrix of stoichiometric coefficients: Equil.

CdCO3(s)

Cd2+

CO2(g)

H2CO3

HCO3–

CO32–

H+

H2O

1 2 3 4

–1 0 0 0

1 0 0 0

0 0 0 1

0 0 –1 –1

0 –1 1 0

1 1 0 0

0 1 1 0

0 0 0 1

Target

–1

1

1

0

0

0

–2

1

Multiplying the stoichiometric coefficients of constituent equilibrium i by unknowns ci gives: Equil.

CdCO3(s)

Cd2+

CO2(g)

H2CO3

HCO3–

CO32–

H+

H2O

1 2 3 4

–c1 0 0 0

c1 0 0 0

0 0 0 c4

0 0 –c3 – c4

0 –c2 c3 0

c1 c2 0 0

0 c2 c3 0

0 0 0 c4

Target

–1

1

1

0

0

0

–2

1

Summing the columns for each chemical species and setting each sum equal to the stoichiometric coefficient for the target equilibrium yields eight linear equations: c1 c1

1 

1

c4 c3

1  c4

 0

c2 c1

c3 c2

 0 c2  0 c4

c3 1 

2

This system of linear equations can be solved (by inspection in this simple case) to reveal: c1 = 1, c2 = –1, c3 = –1, and c4 = 1. In other words, to obtain the target reaction, add equilibrium 1, the reverse of equilibrium 2 (i.e., multiply the stoichiometric coefficients of the species in equilibrium 2 by –1), the reverse of equilibrium 3, and equilibrium 4. Since free energies are additive: Grxn,target Grxn,1 Grxn,2 Grxn,3 Grxn,4 K1 K 4 and Ktarget . K 2 K3 In general, the n linear and independent equations can be solved by the methods of linear algebra to determine the n values of c1, . . ., cn. In matrix notation: 1,1

1, 2

2 ,1

1, 2





n ,1

n,2

   

1, m 2,m

 n,m

c1 c2  cn

vtarget ,1 vtarget ,2  vtarget ,n

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4.9  |  Part I Case Study: Can Methylmercury Be Formed Chemically in Water?   87

Or: νc = νtarget, where ν, c, and νtarget are matrices, νi,j is the stoichiometric coefficient of species j in constituent equilibrium i, and νtarget,j is the stoichiometric coefficient of species j in the target equilibrium. You can solve the system by linear algebra. In ­general, for this approach: Ktarget =

n i 1

Kici .

4.8  CHAPTER SUMMARY Chemical equilibria can be interpreted as both chemical expressions and mathematical statements. Chemical expressions show relationships between chemical species and mathematical statements show relationships between species concentrations. The chemical and mathematical forms of equilibria convey similar information about the way in which species and species concentrations relate to the equilibrium expression. A useful approach to qualitative analysis is Le Châtelier’s principle. Le Châtelier’s principle states that a change to a thermodynamic function controlling the equilibrium (e.g., a change in the system temperature, system pressure, or species concentration) will cause the system to adjust to minimize the change. Knowing the equilibrium constant of one reaction allows you to calculate the equilibrium constants for a number of related reactions. Reversing an equilibrium results in a new equilibrium with K equal to the reciprocal of the K of the original equilibrium 1 K new . Multiplying the stoichiometric coefficients of an equilibrium by a K old ­constant (say, c) results in a new equilibrium with K equal to the K of the original ­equilibrium raised to the power of the constant (Knew = Koldc). Finally, adding two equilibria results in a new equilibrium with K equal to the product of the equilibrium constants of the original equilibria (Knew = Kold,1Kold,2).

4.9  PART I CASE STUDY: CAN METHYLMERCURY

BE FORMED CHEMICALLY IN WATER?

Recall that the Part I case study involves deciding whether methylmercury, CH3Hg+, can be formed in water from CH4(aq) and Hg2+. To answer this question, you must write a balanced reaction between reactants and products. Try this on your own following the procedures introduced in this chapter. Step 1: Write the known reactants on the left and known products on the right. CH 4 aq

Hg2

CH 3 Hg

Step 2: Adjust stoichiometric coefficients to balance all elements except H and O. All elements except H and O (C and Hg) are balanced in Step 1. Step 3: Add water (H2O) to balance the element O. CH 4 aq

Hg2

CH 3 Hg

Oxygen already is balanced, so no need to add water.

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88   Chapter 4  Manipulating Equilibrium Expressions Step 4: Add H+ to balance the element H. CH 4 aq

Hg2

CH 3 Hg

H

Step 5: Add electrons (e–) to balance the charge. The charges are balanced in Step 4. The final balanced reaction is: CH4(aq) + Hg2+ = CH3Hg+ + H+. From Section  3.11, the values of G f for CH4(aq), Hg2+, and CH3Hg+ are, respectively, –34.4, +164.4, and +112.9  kJ/mol, The G f value for H+ is 0  kJ/mol. Thus Grxn

Grxn = –17.1 kJ/mol and K = e RT = 103.0 at 25°C. If all concentrations were at 1 M (very unlikely), then ∆Grxn = Grxn < 0 and the reaction proceeds spontaneously. To see if methylmercury formation is significant, you will have to do a small equilibrium calculation. If concentrations can be substituted for activities, then: K

CH 3 Hg H CH 4 aq Hg2

Assume for the moment that the only methane-­containing species are CH4(aq) and CH3Hg+ and the only Hg-­containing species are Hg2+ and CH3Hg+. (This is a weak assumption. Other Hg-­containing species likely exist in this system.) If [CH3Hg+] is given the symbol x, then: [CH4(aq)] = total methane – x and [Hg2+] = total Hg – x. Thus: K

x H total methane x total Hg x

For K = 103.0, pH 7, total methane = 10–5 M, and total Hg = 10–7 M, you can calculate that the CH3Hg+ concentration is about 10–7 M. Thus, under these conditions, the CH3Hg+ concentration is small but much larger than the concentration of Hg2+ (which is about 10–12 M). In fact, nearly all the mercury would be in the form of CH3Hg+ if only the species listed are present.

CHAPTER KEY IDEAS • You must be able to exploit known equilibria to calculate species concentrations efficiently. • Equilibria can be written as chemical expressions and mathematical equations. • Chemical expressions show relationships between chemical species: reactants, products, and reaction stoichiometry. • Chemical expressions do not show the relationship between chemical species concentrations. • Do not use brackets (e.g., [A]) or braces (e.g., {A}) around chemical species names when writing chemical expressions. • The mathematical versions of chemical expressions show mathematical relationships between species concentrations (or activities). • Chemical expressions of equilibria and mathematical equations of equilibria contain the same information and must lead to consistent conclusions. • Equilibrium constants are dimensionless, but units are associated with equilibrium constants as a reminder of the units of the species concentrations in the equilibrium.

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Historical Note: Henri-­Louis Le Châtelier and Le Châtelier’s Principle   89

• Take great care when comparing equilibrium constants with different associated units. • Grxn values of reverse reactions are equal in absolute value but opposite in sign, and K values of reverse reactions are reciprocals of one another. • Multiplying stoichiometric coefficients by a constant multiplies Grxn by the constant and raises K to the power of the constant. • When adding equilibria, add Grxn values and multiply K values (or add logK values). • To balance a chemical reaction, balance all elements except H and O, add water to balance O, add H+ to balance H, then add e– to balance the charge. • Equilibrium constants can be calculated by linearly combining other equilibria or from Grxn values. • You should learn to manipulate equilibria since thermodynamic data (equilibrium constants and G f values) are not always available.

CHAPTER GLOSSARY Le Châtelier’s principle:  a change to a thermodynamic function controlling the equilibrium will cause the system to adjust to minimize the change linearly independent equilibria:  a set of equilibria where no equilibrium expression can be formed by linearly combining some or all of the other equilibria in the set reversing an equilibrium:  an equilibrium is reversed when its reactants and products are interchanged

HISTORICAL NOTE: HENRI-­LOUIS LE CHÂTELIER AND LE CHÂTELIER’S PRINCIPLE Who was the person behind Le Châtelier’s principle? Henri-­Louis Le Châtelier was a contemporary of J. Williard Gibbs (see the Historical Note in Chapter 3). Le Châtelier, son of an engineer, was trained as an engineer at Êcole des Mines in Paris. He worked for the French government as a mining engineer in the Corps des Mines for two years. Rather unexpectedly, he was offered a professorship in chemistry at his alma mater. This was the beginning of a career in academia that would span 30 years at Parisian universities, including Êcole des Mines, Êcole Polytechnique, College de France, and the Sorbonne.

Henri-­Louis Le Châtelier (Wikimedia Commons/Public domain)

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90   Chapter 4  Manipulating Equilibrium Expressions In contrast with Gibbs’s theoretical brilliance, Le Châtelier was interested in the applications of chemistry. His training as a problem-­solving engineer was reflected in his accomplishments. Le Châtelier’s work on cements, thermodynamics, combustion, ammonia synthesis, and metallurgy was almost always motivated by practical problems. The principle for which his name is linked was refined by Le Châtelier through a series of papers published in the 1880s. This version of the principle echoes Newton’s Third Law of Motion: “Every change of one of the factors of an equilibrium occasions a rearrangement of the system in such a direction that the factor in question experiences a change in a sense opposite to the original change” (Le Châtelier 1888). Le Châtelier devoted the later years of his career to social issues related to his science and engineering background, including educational reform and the scientific management of industry (called Taylorism, after Frederick W. Taylor, 1856–1915). As a teacher, he changed chemical education by emphasizing general principles over memorization (take note, dear reader). He has been described as “personally the kindliest of men, simple, unassuming, easy to meet, and always anxious to help those, who ask his aid” (Oesper 1931). Imagine sitting in a fin de siécle lecture halls learning about Le Châtelier’s principle from the man himself.

PROBLEMS 1. For each equilibrium, write the mathematical form of the equilibrium. a. Ag

OH

AgOH aq ; K1

b. CuS s

2H

c. Mn OH

2

aq

Cu2

H 2S g ; K 2

Mn 2

2OH ; K3

2. For the three equilibria in Problem 1, state the true units of each equilibrium constant and the units associated with each equilibrium constant. 3. If you add two equilibria (with equilibrium constants K1 and K2) to get a third equilibrium (with equilibrium constant K3), you know from Section 4.6 that K3 = KIK2. Use this information to “prove” that Gibbs free energies are additive. 4. The equilibrium constant for the equilibrium H2O = H+ + OH– is 10–14 M2. What is the equilibrium constant for the equilibrium 3H+ + 3OH– = 3H2O? What units are associated with this equilibrium constant? 5. The equilibrium constant for the equilibrium HAc = H+ + Ac– is 10–4.7 M, where HAc = acetic acid and Ac– = acetate. What is the equilibrium constant for the equilibrium ½H+ + ½Ac– = ½HAc ? 6. Given two equilibria with equilibrium constants K1 and K2, what is the equilibrium constant for the reaction formed by reversing the second equilibrium and adding it to the first? Prove your answer using Gibbs free

energies of reaction and also using the mathematical forms of the equilibria. 7. An environmental engineer is seeking to find a way to recover silver from a metal finishing operation. She finds in a textbook an equilibrium constant of 10+6.3 for the reaction ½Ag2O(s) + H+ = Ag+ + ½H2O. What is the equilibrium constant for the equilibrium containing the same information but with all stoichiometric coefficients integers less than 3? 8. An equilibrium expression is added to its reverse reaction to form a new equilibrium. What is the equilibrium constant of the new equilibrium? What is the Gibbs ­ free energy of reaction for the new equilibrium? Do your answers make sense? 9. Balance the equilibrium between chromate (Cr2O72–) and trivalent chromium (Cr3+). Is there only one answer to this question? 10. Balance the equilibrium between O2(g) and H2O. 11. Balance the reaction for the formation of carbon dioxide from glucose (C6H12O6). 12. Write a balanced equilibrium for the reaction with AlOH2+ as a reactant and Al(OH)+2 as a product. Use the formal method described in Section 4.7. Your answer should include H+. 13. Add the equilibrium from Problem 4 (H2O = H+ + OH–) to your answer from Problem 12 to create an equilibrium that relates AlOH2+ and Al(OH)+2 but does not include H+.

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Chapter References   91 14. In balancing reactions, you sometimes may wish to write a final equilibrium in terms of OH– rather than H+. How would you modify the balancing procedure of Section 4.7 to produce an equilibrium in terms of OH–? Illustrate your procedure with the chlorine example found in Section 4.7. 15. Using the constituent equilibria below, what is the equilibrium constant for the balanced equilibrium between Al(OH)2+ and Al(OH)+2 ? The target equilibrium is the answer to Problem 12.

Al OH 3 s

3H    Al3

Al OH 3 s    AlOH 2 Al3

 2OH    Al OH

H 2 O    H

OH

3H 2 O K1 2OH

2

10   8.3 M 24.7

M3

K2

10  

K3

10   17.9 M

K4

10  

14

2

2

M2

16. Are the four constituent equilibria in Problem 15 linearly independent? Explain your answer.

FeCO3 s Fe

2

Fe 2

CO32

2OH     Fe OH

2

s

K1

10  

10.8

K2

1  0

14.4

M3

K3

  10

10.3

M

1  0

6.3

HCO3     CO32

H

H 2 CO3   HCO3

H

K4

H 2 CO3     CO2 g

H2O

K5

H

OH   H 2 O

K6

M2

M

  1.5 atm-L / mol 10 1  014 M

1

19. Repeat Problem 18 using the systematic method of Section 4.7.3. 20. By trial and error, determine the equilibrium constant for the equilibrium between FeS(s) as a reactant and Fe(OH)2(s) and H2S(g) as products. Use some or all of the constituent equilibria below. Fe 2 Fe 2

 2OH   Fe OH S2     FeS s 2

s

10   14.4 M

3

K2

10   18.0 M

2

K3

10  

14.0

M mol / L-atm

K1

17. Are the five equilibria in Problem 16 (four constituent equilibria plus the target equilibrium) between AlOH2+ and Al(OH)2+ linearly independent? Explain your answer.

HS     S

K4

  10

7.0

K5

10  

1.0

18. By trial and error, determine the equilibrium constant for the equilibrium between FeCO3(s) as a reactant and Fe(OH)2(s) and CO2(g) as products. Use some or all of the constituent equilibria below.

H 2S g   H 2S aq H

K6

H 2S   HS

 H

2

 H

 OH   H 2 O

M

10   14 M

2

21. Repeat Problem 20 using the systematic method of Section 4.7.3.

CHAPTER REFERENCES Le Châtelier, H.L. (1888). Annales des Mines 13(2): 157, as found in: woodrow.org/teachers/chemistry/institutes/1992/LeChatelier.html. Oesper, R.E. (1931). The scientific career of Henry Louis Le Châtelier. J. Chem. Ed. 8(3): 442–461.

Pauling, L. 1964. College Chemistry, 3rd ed. San Francisco, CA: Freeman.

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PA R T I I

Solving Chemical Equilibrium Problems Chapter 5: Getting Started with Solving Chemical Equilibrium Problems Chapter 6: Setting Up Chemical Equilibrium Calculations Chapter 7: Algebraic Solutions to Chemical Equilibrium Problems Chapter 8: Graphical Solutions to Chemical Equilibrium Problems Chapter 9: Computer Solutions to Chemical Equilibrium Problems

The formulation of a problem is often more essential than its solution, which may be merely a matter of mathematical or experimental skill. –Albert Einstein and Leopold Infeld (1938) Part II Haiku Find concentrations With a graph and algebra – Or use a spreadsheet. When I’m working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong. –R. Buckminster Fuller (Darling 2004)

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CHAPTER 5

Getting Started with S­ olving Equilibrium Problems 5.1 INTRODUCTION The second part of this text concerns the techniques used to calculate species concentrations at equilibrium. Part II is the heart of a quantitative investigation of chemical equilibria. It is critical that you master the material in Chapters  5 through  9 before moving on to Part III. In this chapter, the general principles behind the calculation of chemical species concentrations at equilibrium will be presented. In addition, a roadmap for Part II of the text is presented and discussed.

5.2  A FRAMEWORK FOR SOLVING

CHEMICAL EQUILIBRIUM PROBLEMS

5.2.1    Analysis Approaches It is likely that you have some experiences systems analysis from course work in thermodynamics, fluid mechanics, hydraulics, or water quality. These experiences will help you to understand the framework for the calculating species concentrations at equilibrium. In all cases, the goal is the same: to determine the values of state variables in a defined space over a defined time. (See Section 3.2 for a review of state variables.) The space element and time interval that is being studied is called a control volume. For the analysis problems in thermodynamics, fluid mechanics, hydraulics, and water quality, the same approach is used: 1. Define the control volume; enumerate the state variables. 2. Identify the fluxes of mass, energy, and momentum that cross the boundaries of the control volume. 3. Identify processes occurring inside the control volume that affect the state variables.

control volume: the space element and time interval about which inferences are to be made

Expressions for the state variables as functions of the independent variables are then developed. The independent variables usually are the spatial dimensions (x, y, and z) and time. The expressions must consider the control volume geometry (through boundary conditions), control volume history (through initial conditions), fluxes,

95

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96   Chapter 5  Getting Started with ­Solving Equilibrium Problems Key idea: To analyze systems in general, define the control volume and state variables; identify mass, energy, and momentum fluxes crossing the control volume boundaries; and identify processes inside the control volume that affect the state variables

and processes. The resulting expressions constitute a mathematical model of the real world. As engineers and scientists, we attempt to solve the mathematical model to make statements about the state variables in the control volume.

5.2.2  Application to Chemical Equilibrium Modeling For chemical equilibrium modeling, a similar approach is used. However, the characteristics of chemical equilibria allow for simplification of the analysis.

Thoughtful Pause How do the characteristics of chemical equilibria allow us to simplify the analysis process?

Key idea: In chemical equilibrium modeling, define the control volume so that equilibrium occurs between and within phases in the control volume

Knowing that the system is at equilibrium means that the state variables (e.g., temperature, pressure, and species concentrations) do not change with time (Section 3.3.2). The time-­invariant nature of systems at equilibrium eliminates the need to consider changes in species concentrations with time. How do you know that a system is at equilibrium? This can be a difficult question to answer. For now, assume that the decision has been made to use an equilibrium model. Nonequilibrium models are discussed in more detail in Chapter 23. The general analysis approach discussed in Section  5.2 has four steps: (l) define the control volume, (2) enumerate the state variables, (3) identify pertinent fluxes crossing the boundaries of the control volume, and (4) identify processes affecting the state variables that occur inside the control volume. Each step will be discussed in more detail. Control volume and fluxes. The choice of the control volume is influenced by fluxes. A flux means that mass, energy, and/or momentum are transported across the control volume boundary. At equilibrium, no transport of mass occurs across the boundary of the control volume. This means that the control volume must be defined carefully to include material exchange with the phases of interest. Chemical equilibrium modeling involves defining the control volume so that equilibrium occurs between and within phases in the control volume.1 State variables.  The choice of state variables is narrowed in most chemical equilibrium models. In many environmental applications, the temperature and pressure are known.2 Thus, the state variables to be solved for are the equilibrium concentrations of the chemical species. The remaining question is: Which species? In chemical equilibrium calculations, an important step is to identify the species of interest.  There are important differences between a mass flux analysis at steady-­state and an equilibrium model. The steady-­ state and equilibrium models are compared in Appendix B. 2  In this text, it is assumed that the system temperature and pressure are known and constant. Thus, it is assumed that solutions are sufficiently dilute so that heat generated or absorbed by chemical reactions does not affect the temperature and pressure of the system. For some problems (especially gas or nonaqueous phase reactions in low heat capacity media), the effects of reactions on system temperature and pressure must be included. 1

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5.3  |  Introduction to Defining the Chemical System   97

Processes affecting the state variables. Finally, the standard analysis approach includes the identification of processes affecting the state variables. In chemical equilibrium modeling, you know that species concentrations are constrained by thermodynamics. In particular, the equilibrium expressions must be satisfied. Additional constraints on species concentrations will be discussed in Section 5.5 and Chapter 6. The standard analysis approach now can be applied to chemical equilibrium modeling. First, define the system. In particular, you must decide if the system is allowed to equilibrate with gas and/or solid phases. This decision is very important and is discussed in more detail in Section 5.3. Second, identify the chemical species to be considered. Techniques for enumerating chemical species are discussed in Section 5.4. Last, you must identify the equilibrium constraints and other constraints on the species concentrations. This aspect of chemical equilibrium modeling is discussed in Section 5.5.

5.3  INTRODUCTION TO DEFINING

THE CHEMICAL SYSTEM

How do you define the control volume for a chemical system? In many instances, the choice is dictated by (l) the physical system to be modeled, and (2) how closely you wish the chemical model to describe the physical system in question. For example, an equilibrium model for groundwater chemistry would likely include equilibria between dissolved species and minerals. Dissolution/precipitation equilibria should be included since groundwater generally is in intimate contact with the surrounding soil. An equilibrium model for groundwater might exclude gas/liquid exchange with the atmosphere and still be a reasonable approximation of reality. As the groundwater example shows, aqueous systems may (or may not) be in equilibrium with gas or solid phases. Do gas or solid phases exist in a given chemical system? This question will be explored separately for gas and solid phases. Typically, a gas phase exists if we allow it to exist. In other words, we usually define the system to include or exclude a gas phase. What is meant by a gas phase? Gases, at environmentally important temperatures and pressures, exist as uncharged molecules (or atoms). Each molecule (or atom) may exist as two distinct chemical species: a dissolved gas species and a gaseous species. For example, molecular oxygen may be found in the environment as dissolved oxygen, denoted O2(aq), and as gaseous oxygen, O2(g). If the system is defined to include a gas phase, then we allow for the existence of gaseous oxygen. Allowing for gaseous oxygen to exist has two ramifications. First, the system now has an additional chemical species, O2(g). Second, an equilibrium expression that describes the equilibrium partitioning of oxygen between gas and water must be included. In other words, an equilibrium between O2(g) and O2(aq) must be included in the system model. Solid phases are more problematic. First, it is sometimes difficult to know if a solid phase will exist at equilibrium. For example, if you add a small amount of NaCl to water, common experience tells you that the salt likely will dissolve completely. What if you add some calcium carbonate to water? It is difficult to know whether the calcium carbonate will dissolve completely or whether some solid calcium carbonate will remain at equilibrium without performing the calculation (or conducting an experiment). Second, it is possible for more than one solid phase to exist in a system. (In environmental applications, we rarely consider more than one gas phase.) The complexities involved when more than one solid phase exists are discussed in Section 19.4.2.

Key idea: To analyze chemical equilibrium systems, define the system, identify the species of interest, and write the constraints on species concentrations

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98   Chapter 5  Getting Started with ­Solving Equilibrium Problems

5.4  INTRODUCTION TO ENUMERATING

CHEMICAL SPECIES

Key idea: It is important to make an accurate list of the chemical species in the equilibrium model

species list: the list of chemical species to be evaluated in an equilibrium calculation

To determine whether you have enough information to calculate species concentrations at equilibrium, you must first count the number of species in the control volume. Again, the choice of which species to include in the model depends on both the physical system and how accurately you wish your chemical model to describe the physical system. In fact, it is quite easy to miss important species. For example, suppose you tried to predict the effects of equilibration with the atmosphere on the pH (and thus, the H+ activity) of a carbonated cola drink. If your model of the cola drink included only dissolved carbon dioxide and associated chemical species, the effects of opening the bottle on the pH would be greatly overestimated. Why? The pH of the soda is in fact controlled by the large amount of phosphoric acid in the mixture. Without phosphoric acid (and related species) in the list of important species, an equilibrium model cannot predict the pH of the cola drink. On the other hand, carbon dioxide exchange alone can be used to predict accurately the pH of clean rainwater. To calculate the pH of acid rain, other chemical species (such as sulfate, nitrate, and chloride) must be included. Throughout this text, the roster of chemical species to be included in the equilibrium model will be referred to as the species list. Techniques for developing a species list are discussed in Section 6.3.

5.5  INTRODUCTION TO DEFINING

THE CONSTRAINTS ON SPECIES CONCENTRATIONS

5.5.1  Introduction So far, chemical systems have been described from the standpoint of aquatic chemistry. Now it is time to think about the translation of chemical systems into mathematical expressions to be solved. Suppose your species list contains n species.

Thoughtful Pause How many unknowns are there in a chemical system consisting of n species? If the temperature and pressure are known, a system of n chemical species has n unknowns. The unknowns are the n equilibrium concentrations of the n species. How many equations are required to solve a mathematical system of n unknowns? You need n equations.3 Now you must put your chemistry hat on again and come up with n equations to solve the system of n unknowns.

 Please note that to solve a system of n unknowns, n linearly independent equations are required. See Section 4.6.2 for a discussion of the application of linear independence to chemical equilibria.

3

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5.5  |  Introduction to Defining the Constraints on Species Concentrations   99

The process of finding n equations in the n equilibrium concentrations will be described in more detail in Section 6.4. There are four types of equations: mathematical versions of equilibrium expressions, mass balances, a charge balance, and equations from the definition of the standard state of mixtures.

5.5.2  Equilibrium Expressions From Section  4.2.2, equilibria can be written as both chemical expressions and ­mathematical equations. To calculate the concentrations at equilibrium, all pertinent equilibria must be translated into mathematical equations. How many equilibrium expressions will result? The number of equilibrium expressions will vary with the size and nature of the chemical system. However, every aqueous system has at least one equilibrium: the equilibrium between H2O, H+, and OH–.

Key idea: To determine the equilibrium concentrations of n species, n linearly independent equations are required and may be found through mathematical versions of equilibrium expressions, mass balances, a charge balance, and equations from the definition of the standard state of mixtures

5.5.3  Mass Balance Expressions Mass is, of course, conserved. The mass of certain elements is added to the system through the starting materials. The added mass can be redistributed through equilibria into many chemical species. For example (see Section 1.2.2 for details), the addition of copper sulfate to a reservoir results in the formation at equilibrium of at least 10 copper-­containing species. For a closed system of constant volume without solids, conservation of mass means that the sum of the masses of each element at equilibrium must be equal to the added mass of the element.4 As an example, suppose you add enough cyanide to water to generate an initial cyanide concentration of 1×10–3 M. Imagine that, through equilibria, the added cyanide is partitioned into two species, each of which is formed from 1 mole of cyanide per mole of species. The mass balance constraint requires that the sum of the concentrations of the two cyanide-­containing species must be 1×10–3 M. How many mass balance equations are required? Again, the answer depends on the complexity of the system. In general, mass balance equations are written for each element added. (See Section 6.4.2 for a more detailed discussion.) If electron transfer reactions are not considered, you write mass balance equations for each oxidation state of each element added. For all but the simplest system (i.e., pure water), there is at least one mass balance equation.

5.5.4  Charge Balance Expression Water in the environment is electrically neutral. This fact constrains the possible values of species concentrations and will be referred to as the electroneutrality constraint. The electroneutrality constraint means that the sum of the positive charges in solution (contributed by cations) must be equal to the sum of the negative charges in solution (contributed by anions). For each chemical system, you will write one charge balance equation to ensure compliance with the electroneutrality constraint.

4  Remember that mass, not concentration, is conserved. In the case where the volume of the system and the density of the fluid are constant, conservation of mass can be expressed through concentration since mass = concentration×volume. Mass balances usually are expressed as mole balances. Although the total number of moles of material in a system are not necessarily conserved, the number of moles of, say, carbon is conserved. Mass balances sometimes are called material balances.

Key idea: Mass balances usually are written for each element added

Key idea: Species concentrations are constrained by the fact that the sum of the positive charges in solution must be equal to the sum of the negative charges in solution (called the electroneutrality constraint)

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100   Chapter 5  Getting Started with ­Solving Equilibrium Problems Key idea: The activities (concentrations in dilute solution) of pure liquids (such as water) and pure solids are taken to be unity

5.5.5  Other Constraints on Species Concentrations There are two other important constraints on the concentrations of chemical species (and therefore constraints on the unknowns). First, species concentrations cannot be less than zero. Thus, you can reject all negative roots of equations involving species concentrations. Second, the choice of concentration units and standard states of mixtures leads to defined activities of certain species.5 For most systems of environmental interest, the mole fractions of pure liquids (such as water) and pure solids are taken to be unity (see also Section 4.3). This choice of the standard state of mixtures greatly simplifies many equilibrium calculations.

5.6  PART II ROADMAP Part II of this book contains the techniques needed to calculate the equilibrium concentrations of chemical species. The examples provided in Part II are fairly simple. You will encounter more complex systems later in the text. The solution techniques used to solve complex systems are identical to the methods developed in Part II. Thus, it is imperative that you master each chapter of this part of the text. In Chapter  6, the governing equations constraining species concentrations at equilibrium are presented in more detail. By the conclusion of Chapter  6, you will be able to write a complete mathematical description of a chemical system at equilibrium. Chapters  7 through  9 provide techniques to solve the system of ­mathematical equations developed in Chapter 6. Three approaches are presented: algebraic, graphical, and computer calculations. For a given chemical system, one solution technique may be superior to the others. Thus, it is important that you feel comfortable with each solution method. Developing a facility with at least one computer method described in Chapter 9 is especially important. Computer methods allow for the solution of more complex systems comprising many chemical species.

5.7  CHAPTER SUMMARY In this chapter, you learned about a framework for the analysis of chemical systems at equilibrium. To analyze chemical equilibrium systems, three tasks must be accomplished. First, you should define the system as in equilibrium with gas or solid phases or not in equilibrium with gas or solid phases. Second, you should generate a species list to identify the species of interest. Third, you should write the constraints on species concentrations. Constraints stem from equilibrium expressions, mass balances, and electroneutrality. In addition, species concentrations must be nonnegative and comply with the thermodynamic standard states (i.e., pure liquids and solids have activities of unity). Techniques to complete the three tasks are presented in Chapter 6.

 The standard state of a mixture is different from the standard state of an element (described in Section 3.6.2). The standard state of a mixture defines when the activity is equal to the concentration and when the activity (in the chosen concentration scale) is unity.

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 Chapter Glossary   101

5.8  PART II CASE STUDY: HAVE YOU HAD

YOUR ZINC TODAY?

The case study for Part II of this text concerns a dietary supplement containing a metal. Trace metals are necessary for human life. They play vital roles in oxygen transport, enzyme function, and endocrine system activity. If trace metal concentrations in food and water are insufficient, then mineral supplements may be indicated. The chemical form of the supplement and the water chemistry of the mouth and stomach often dictate the speciation of the metal in the digestive system. The form of the metal may be critical in determining whether or not the metal is absorbed by the body and transported to the appropriate site. Formation of the “wrong” species can result in excretion of the metal and loss of its beneficial effects. This case study will explore the chemical speciation of zinc in a zinc acetate supplement. Consider the consumption of a zinc acetate tablet containing 20 mg of zinc as Zn. Assume the tablet is dissolved in 100  mL of water and consumed. The zinc-­ containing water will pass through the mouth (saliva pH 6.5–8.5) and into the stomach (gastric juice pH 1–3). The case study question is: How do the dominant forms of zinc change as the pH of the water is reduced from the pH 6.5–8.5 range to the pH 1–3 range when zinc acetate is consumed?

CHAPTER KEY IDEAS • To analyze systems in general, define the control volume and state variables; ­identify mass, energy, and momentum fluxes crossing the control volume boundaries; and identify processes inside the control volume that affect the state variables. • In chemical equilibrium modeling, define the control volume so that equilibrium occurs between and within phases in the control volume. • To analyze chemical equilibrium systems, define the system, identify the species of interest, and write the constraints on species concentrations. • It is important to make an accurate list of the chemical species in the equilibrium model. • To determine the equilibrium concentrations of n species, n linearly independent equations are required and may be found through mathematical versions of equilibrium expressions, mass balances, a charge balance, and equations from the definition of the standard state of mixtures. • Mass balances usually are written for each element added. • Species concentrations are constrained by the fact that the sum of the positive charges in solution must be equal to the sum of the negative charges in solution (called the electroneutrality constraint). • The activities (concentrations in dilute solution) of pure liquids (such as water) and pure solids are taken to be unity.

CHAPTER GLOSSARY control volume:  the space element and time interval about which inferences are to be made species list:  the list of chemical species to be evaluated in an equilibrium calculation

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102   Chapter 5  Getting Started with ­Solving Equilibrium Problems

HISTORICAL NOTE: “ACTIVE MASS” AND FAMILIAL RELATIONS In Part I of this text, you learned that molar units were the natural units for concentration and activity. Part II is all about establishing systems of n equations as functions of n equilibrium concentrations (or activities). But why concentrations? To our modern eye, this question may seem silly: Of course concentrations! However, early chemists believed that the affinity between chemical species was an inherent property of the species. It took a major conceptual leap to realize that chemical affinity was related to the quantity of material per unit volume: in other words, the concentration. This leap was made by an interesting pair of nineteenth century Norwegians: mathematician turned chemist Cato Maximilian Guldberg (1836–1902) and chemist Peter Waage (1833–1900). Their names are forever linked together in the annals of chemistry because of their publication in 1846 of a principle that came to be known as the law of mass action. (See Ferner and Aronson 2016 for a review). The law of mass action had two parts. First, Guldberg and Waage posited that reaction rates are proportional to the “active masses” of the reactants. They defined “active mass” as the mass per unit volume, or the concentration.6 Second, the law of mass action states that for reversible reactions, the rates of the forward and reverse reactions are equal, resulting in an expression again related to the extensive quantity concentration. (See Section 3.2.2 for a review of extensive properties. Equating the reaction rates of forward and reverse reactions is explored in Section 23.5.2.)

Cato M. Guldberg and Peter Waage (H. Riffart, Berlin/Wikimedia Commons/ Public domain)

 Guldberg and Waage’s original paper was not read widely, since it was published in Norwegian. They republished it in French and German. In the German paper of 1879, they wrote (Guldberg and Waage 1879, p. 71): “Eigentlich verstehen wir unter der activen Masse nur die Menge des Stoffes innerhalb der Actionsphäre; unter sonst gleichen Umständen kann aber die Actionsphäre durch die Volumeneinheit repräsentirt werden.” (By “active mass,” we mean only the quantity of substance inside the sphere of action; all other things being equal, however, the sphere of action can be represented by the unit volume.)

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 Chapter References   103

Guldberg and Waage are linked in more ways than one. They were friends in secondary school in Christiania (now Olso), Norway. They both entered the University of Christiania in 1854. Here is where things get interesting. In 1862, Waage married Johanne Christiane Tandberg Riddervold. The next year, Guldberg married Johanne’s sister, Bodil Mathea Riddervold. (Bodil and Cato were first cousins, since Bodil’s mother, Anna Maria Bull, and Cato’s mother, Hanna Sophie Theresia Bull, were sisters.) This made Guldberg and Waage brothers-­in-­law. . . for the first time. Johanne died in 1869 and Waage remarried in 1870. This time, he married Mathilde Sofie Guldberg, Cato’s sister, making Guldberg and Waage brothers-­in-­law for a second time! (For those of you keeping score at home, Guldberg married his cousin, who also was Waage’s wife’s sister, and later Waage married Guldberg’s sister.) Guldberg and Waage were friends, collaborators, relatives by marriage (twice), and originators of some of the principles behind this text.

CHAPTER REFERENCES Darling, D.J. (2004). The Universal Book of Mathematics. Hoboken, NJ: John Wiley and Sons, p. 34. Einstein, A. and L. Infeld (1938). The Evolution of Physics. London: Cambridge University Press, p. 95.

Ferner, R.E. and J.K. Aronson (2016). Cato Guldberg and Peter Waage, the history of the Law of Mass Action, and its relevance to clinical pharmacology. Br. J. Clin. Pharmacol. 81(1): 52–55. Guldberg, C.M. and P. Waage (1879). Ueber die chemische Affinitat. J. prak. Chem. 19–20: 69–114.

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CHAPTER 6

Setting Up Chemical Equilibrium Calculations 6.1 INTRODUCTION In this chapter, a unified approach to setting up chemical equilibrium problems will be developed. By the end of the chapter, you will be able to generate a species list and write the equations controlling the species concentrations at equilibrium. This chapter expands on the principles of chemical systems at equilibrium that were developed in Chapter 5. The general approach from Chapter 5 is repeated in the Key idea. It may seem a little pedantic to devote an entire chapter to developing the governing equations for species concentrations at equilibrium. The equally important tasks of setting up the equations (Chapters 5 and 6) and solving them (Chapters 7 through 9) have been separated in this text on purpose. It is very important to master the procedure for setting up the equations before you attempt to solve the equations. The steps necessary to build a mathematical model for a chemical system at equilibrium are developed in Sections 6.2 through 6.4 and summarized in Section 6.5.

Key idea: To analyze chemical equilibrium systems, define the system, identify the species of interest, and write the constraints on species concentrations

6.2  DEFINING THE CHEMICAL SYSTEM As discussed in Section  5.2, you must define the system to include or exclude gas phases. A system that includes a gas phase is called an open system. If equilibration with a gas phase is not allowed, the system is closed. The labels “open” and “closed” refer only to the presence of a gas phase and should not be interpreted as indicating the presence or absence of solid phases. Systems may undergo transitions from closed to open states (or from open to closed states). If groundwater is pumped to the surface as a drinking water source, some chemical constituents of the system (dissolved gases) may equilibrate with the atmosphere. As another example, consider an unopened bottle of soda. You could calculate the equilibrium concentrations of key chemical species (e.g., dissolved carbon dioxide) in the soda. Your experience tells you that the concentration of dissolved carbon dioxide will be much smaller after the bottle is opened and the carbon dioxide equilibrates with the atmosphere. At equilibrium (i.e., after the soda goes flat), the total pressure and the dissolved carbon dioxide concentration inside the bottle change from their initial values. It is important to note that the systems (i.e., the control volumes) for the closed-­bottle case and open-­bottle case are different. The control volume is the soda bottle only in the first case and the soda bottle plus the atmosphere in the second case (see Figure 6.1).

open system: an aqueous system in equilibrium with a gas phase closed system: an aqueous system not in equilibrium with a gas phase

105

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106   Chapter 6  Setting Up Chemical Equilibrium Calculations Atmosphere

Atmosphere

closed system

open system

FIGURE 6.1  Control Volumes for Open and Closed Systems (Control volumes are indicated by dashed lines; bottle images: OpenClipart-­Vectors/Pixabay.)

6.3  ENUMERATING CHEMICAL SPECIES Key idea: To enumerate species, identify the starting materials, identify the initial hydrolysis products, enumerate subsequent hydrolysis products, and enumerate other reaction products starting materials: chemicals added to water before equilibration reactions begin

As discussed in Section 5.4, it is necessary to develop a list of the species to be considered in the chemical equilibrium model. Four steps are necessary to generate an accurate species list: identification of the starting materials, identification of the initial hydrolysis products, enumeration of subsequent hydrolysis products, and enumeration of other reaction products. Each step will be discussed in more detail here.

6.3.1  Step 1: Identify the Starting Materials In all environmental applications, you know something about the chemical composition of the system prior to starting the equilibrium calculations. The chemicals added to water initially are called starting materials (also called the recipe; see Chapter 9). Frequently, the total concentrations of some elements are known because the starting materials are known. In the copper sulfate example of Section 1.2.2, the total mass (and, hence, the total concentration at constant system volume) of copper was known. It is important to correctly identify starting materials for two reasons. First, you will use the total concentrations of starting materials as constraints on the concentrations of individual species (see Section 6.4.2). Second, knowledge of the starting materials is critical in generating a list of species in the system. The focus in this section is on this second reason for identifying starting materials.1 The chemical composition of the starting materials often affects the equilibrium composition of the system. For example, equilibrium concentrations may be different if copper is added in the form of copper chloride or copper carbonate instead of in the form of copper sulfate. For the ammonia example in Chapter 4, you need to know if the source of ammonia is ammonia gas, ammonium chloride, or ammonium hydroxide prior to performing equilibrium calculations.

1  As your chemical intuition grows, it will no longer be necessary to be restricted to problems in which the starting materials are known. In the copper example, you can solve for equilibrium concentrations if you know (say, by measurement) the total concentrations of dissolved copper, dissolved chloride, dissolved carbonate, and so on. Or you may know (again, by measurement) the concentrations of individual chemical species (say, Cu2+).

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6.3  |  Enumerating Chemical Species   107

You will find it useful to adopt the convention that water is always a starting material. This makes sense since aquatic systems obviously always contain water. Although it is easy to forget, always include water as a starting material.

6.3.2  Step 2: Identify the Initial Hydrolysis Products As stated in Section 1.2, the primary focus of this text is on chemical species that react with water. To generate a species list, you must ascertain how the added material will dissociate to form the initial hydrolysis products. Dissociation (from the Latin dis not + sociare to join) refers here to the initial bond-­cleaving reactions that form multiple chemical species from the added material. For example, when sodium chloride is added to water, it dissociates at least in part to form sodium ions (Na+) and chloride ions (Cl–). Note that dissociation does not have to be complete. We are not saying at this point that all the NaCl dissociates to form Na+ and Cl– but only that some (as yet unknown) concentrations of Na+ and Cl– are formed by the addition of NaCl to water. The complete dissolution of added materials is discussed in Section 6.4.4. Recall from Section 1.4 that chemicals such as NaCl dissociate in water because water can hydrate the dissolved ions and therefore reduce the attractive forces between the ions. In most cases, the initial dissociation of starting materials is very fast. You can identify the initial hydrolysis products by performing a thought experiment (or Gedankenexperiment) where the first dissociation processes are isolated. For now, you do not care what happens to the initial dissociation products. The job at hand is merely to identify them. The sodium chloride example was very simple. How do you know what chemical species are formed when the starting materials dissociate? How do you even know if the starting material dissociates at all? As you work through this text, you will develop an intuition about dissociation. A few rules will help you get started in developing an intuitive feel for dissociation. Look for the formation of simple ions. To decide whether dissociation occurs, see if the starting materials form simple, common ions. For example, in most aqueous systems of environmental interest, compounds containing sodium or potassium dissociate to form Na+ ions and K+ ions. A list of common ions is provided in Table 6.1. Other examples of the formation of simple ions as initial dissociation products are given in Worked Example 6.1.  Table 6.1 Ions Frequently Formed as Initial Dissociation Products Ion Type

Cations monoatomic1 polyatomic Anions monoatomic polyatomic2

Examples

H+, K+, Na+, Ca2+, Mg2+ NH4+ Cl–, Br–, S2– OH–, SO42–, CO32–, HCO3–, NO3–, PO43–

Notes: 1

 Also any other metal ion of charge +n, Mn+.  Called hydroxide, sulfate, carbonate, bicarbonate, nitrate, and phosphate ion, respectively.

2

material

Key idea: Always include water as a starting

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108   Chapter 6  Setting Up Chemical Equilibrium Calculations Worked Example 6.1 

Initial Dissociation Products

What are the initial dissociation products for the dissolution of Fe(NH4)2(SO4)2 (ferrous ammonium sulfate), K2Cr2O7 (potassium dichromate), and ClO2 (chlorine dioxide) in water? Solution For Fe(NH4)2(SO4)2, common ions that can be formed through initial dissociation are Fe2+, NH4+, and SO42–. Note that the iron is in the +II oxidation state (see Appendix A). The starting material K2Cr2O7 contains potassium and would likely dissociate to form K+. This leaves Cr2O72– initially. For ClO2, no common ions can be formed. The Cl in ClO2 is in the +IV oxidation state, so dissociation to Cl– is not allowed (see text). In all cases, H2O is also a starting material. Thus, H+ and OH– also are initial dissociation products. Thus: Fe(NH4)2(SO4)2 and water dissociate initially to Fe2+, NH4+, SO42–, + H , and OH–. K2Cr2O7 and water dissociate initially to form K+, Cr2O72–, H+, and OH–. Chlorine dioxide likely does not dissociate initially, so only H+ and OH– are formed as initial dissociation products of water.

Key idea: When considering whether dissociation occurs, look for the formation of simple ions (Table 6.1) without a change in the oxidation state

You must be careful about using the list of ions in Table  6.1. It is incorrect to change the oxidation state when dissociation occurs. (See Appendix A for a review of the concept of the oxidation state.) A change in oxidation state means that electron transfer has occurred. Electron transfer (or redox) equilibria will be discussed in Chapter  16. For example, it is proper to allow HCl to dissociate into H+ and Cl–, because both the chlorine in HCl and the chlorine in Cl– are in the –I oxidation state. In addition, the hydrogen in HCl and the hydrogen in H+ are both in the +I oxidation state. However, it is not proper to allow hypochlorous acid (HOCl) to dissociate to Cl–.

Thoughtful Pause Why is it incorrect to allow HOCl to dissociate to Cl–? Key idea: An H+ usually is available for transfer to water if it is bonded to O or N in the starting material

Hypochlorous acid cannot be allowed to dissociate into Cl– because the chlorine in HOCl is in the +I oxidation state and the chlorine in Cl– is in the –I oxidation state. Look for H+ transfer. As you will see in Chapter 11, the transfer of H+ to water (also called proton transfer or acid–base chemistry) is an important part of aquatic chemistry. Always look for H+ transfer to water. In general, an H+ is usually available for transfer to water if it is bonded to O or N in the starting material. Thus, you might think to dissociate HNO3 to H+ and NO3– and you might think to dissociate NH4+ to NH3 and H+, but you would not dissociate CH4 to H+ and CH3–. Why? The H in methane (CH4) is bonded to C in the starting material, not to O or N.2 2  In general, H+ will dissociate from a species (and be available for transfer) if H and the atom it is bonded to have very different affinities for electrons. A difference in electron affinity leads to a highly polar bond (see Section 1.4) called an ionic bond. Examples include the H-­N bond, H-­O bond, and H-­Cl bond. If H and the atom it is bonded to have similar affinities for electrons, the bond is called a covalent bond and H+ dissociation is much less likely.

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6.3  |  Enumerating Chemical Species   109

Always include H2O, H+, and OH– in the species list. Aqueous systems, by definition, always contain water. Recall from Section 6.3.1 that water should be considered a starting material in all aquatic equilibrium systems. As you will see later in this chapter, water dissociates to H+ and OH–. Therefore, H2O, H+, and OH– are always in the species list.

species list

Key idea: Always include H2O, H+, and OH– in the

6.3.3  Step 3: Enumerate Subsequent Hydrolysis Products After the starting materials have been dissociated, expand the species list by adding the chemical species formed by hydrolysis of the initial dissociation products. A useful technique for determining subsequent hydrolysis products is to ask repeatedly whether a given dissociation product reacts with water. If a given initial dissociation product does not react with water, abandon it for the moment and move to the next initial dissociation product. Some chemical species, called nonhydrolyzing species, do not react appreciably with water. Common examples include Na+, K+, and F–. If a given initial dissociation product does react with water, then write the hydrolysis product(s). Question whether the new hydrolysis products react with water. This process continues with each new generation of species until no new hydrolysis products are generated.

6.3.4  Step 4: Enumerate Other Reaction Products In the final step of generating a species list, you must assess whether any species on the list equilibrates with any other species on the list. This process again continues with each new generation of species until no new species are generated. How do you know whether two or more species react? You will develop an intuition for conjuring up equilibria and reaction products as you work through this text. A useful tool is to examine lists of chemical equilibria and look for reactions involving the species on the species list. If the species list contains all the reactants (or all the products) in an equilibrium, then include all the products (or all the reactants) from the equilibrium in the species list. For example, if the species list contains HCO3– and H+, then H2CO3 should be added to the list because of the equilibrium: H2CO3 = HCO3– + H+. This approach also helps generate a list of equilibria to be considered for inclusion in the model. Examples of equilibria in water are found in Appendix D. If the system is open, be sure to include species resulting from equilibria partitioning of mass between gas and aqueous phases.

nonhydrolyzing species: chemical species that do not react appreciably with water Key idea: Na+, K+, and F– are nonhydrolyzing and do not react appreciably with water

Key idea: Use equilibria to determine whether species on the species list react to form other species

6.3.5  An Example of Generating a Species List An example will make the process of generating a species list a bit clearer. The problem at hand is to generate a species list for the addition of ammonium chloride (NH4Cl) to water in a closed system.

Thoughtful Pause Try generating a species list for the addition of NH4Cl to water in a closed system on your own.

←Worked example

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110   Chapter 6  Setting Up Chemical Equilibrium Calculations The starting materials are NH4Cl(s) and H2O. (Don’t forget water as a starting material!) Now allow each starting material to dissociate. Ammonium chloride will dissociate to form the common ions ammonium and chloride. The other starting material (water) will dissociate to form H+ and OH–. After allowing only the starting materials to dissociate, the chemical system is Starting materials: NH4Cl(s), H2O Partial species list from starting materials and initial dissociation products: NH4Cl(s), H2O, NH4+, Cl–, H+, and OH– You now have four species (NH4+, Cl–, H+, and OH–) that came from the dissociation of starting material. You must ask whether any of these species hydrolyze. To answer this question, look for equilibria for which you already have all the reactants or all the products in the species list. Recall the following equilibrium from Chapter 4: NH 4

NH 3

H

The species list contains all the reactants in this equilibrium (only ammonium in this case), so the products must be added to the species list. The species list now is Partial species list: NH4Cl(s), H2O, NH4+, Cl–, H+, OH–, and NH3 What about Cl–? From Section 3.8.2, you are aware of the equilibrium: HCl = H+ + Cl–. All the products in this equilibrium are in the species list, and so the reactant, HCl, must added to the species list. The system becomes Starting materials: NH4Cl(s), H2O Species list: NH4Cl(s), H2O, NH4+, Cl–, H+, OH–, NH3, and HCl species list diagram: a visual representation of species and their origin(s), including starting materials  orked W example →

The process of generating a species list for this example is summarized in Figure 6.2. Species list diagrams such as Figure 6.2 are useful in several ways. They allow you to look for possible interacts between species in a partially completed species list. In addition, they show clearly from which starting material each species comes. For example, in Figure 6.2, it is clear that NH4+ and NH3 both come from ammonium chloride. This information will be helpful in generating mass balances.

Thoughtful Pause Try generating a species list diagram for the system in Worked Example 6.2.

Starting Materials:

NH4+

NH3

H2O

NH4Cl

Cl–

H+

OH+

HCl

FIGURE 6.2  Species List Diagram for the Addition of Ammonium Chloride to Water

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6.3  |  Enumerating Chemical Species   111

Another illustration of generating a species list is given in Worked Example 6.2. Worked Example 6.2 

Generating a Species List

Write a species list for an open system in equilibrium with CO2(g) in the atmosphere. Solution The starting materials are CO2(g) and H2O. The first dissociation products of water, as usual, are H+ and OH– (because of the equilibrium H2O = H+ + OH–). To decide on the hydrolysis products of carbon dioxide, refer to the pertinent equilibria in Problem 18 of Chapter 4 (repeated here): HCO3 H 2 CO3 H 2 CO3

CO32 H HCO3 H CO2 (g) H 2 O

The first hydrolysis product of CO2(g) is H2CO3. To determine subsequent reaction products, note that the species list contains all the reactants (only H2CO3) of the middle equilibrium. Thus, the products (only HCO3– is new) should be added to the species list. Similarly, CO32– is added because all the reactants from the first equilibrium (only HCO3–) are on the species list. Thus, the species list is CO2(g), H2CO3, HCO3–, CO32–, H2O, H+, and OH–. A species list diagram for the dissolution of CO2(g) in water is shown in Figure 6.3. Starting Materials:

CO2(g) H2CO3

H2O H+

OH–

HCO3– CO32– FIGURE 6.3  Species List Diagram for the Addition of CO2(g) to Water

6.3.6  When Is the Species List Complete? It is sometimes difficult to know if you have included all pertinent species in a species list. How do you know if your species list is complete? For many chemical equilibrium calculations, you can use existing lists of equilibria to make sure that all pertinent species and equilibria have been included. Lists of equilibria are provided in Appendix D of this text. Proceed with caution when using existing lists of equilibria. A given species may be easily ignored in some applications but may be important in others. From the copper example in Chapter 1, Cu2+ is important for algal toxicity, even though its concentration in the aquatic environment at equilibrium is very small. On the other hand, we usually ignore the species N2(aq) in environmental systems because it is unreactive, even though its concentration is relatively large (about 5.2×10–4 M in equilibrium with the atmosphere at 25°C). When in doubt, include all species for which valid thermodynamic data are available.

Key idea: When generating a species list, include all species for which valid thermodynamic data are available

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112   Chapter 6  Setting Up Chemical Equilibrium Calculations As an example of when to stop adding species, consider whether the species NH2– should be included in the example of Section 6.3.5. After all, you can write the equilibrium: NH 3

NH 2

H

This equilibrium is not commonly listed as important in aqueous systems. You can show (upon completion of Part II of this text) that the equilibrium concentration of NH2– in a solution formed by adding 1×10–3 M NH4Cl to a closed system of pure water is about 10–23 M, or about six ions per liter of water. Is NH2– important? The answer depends on the application, of course. However, it is pretty unlikely that a species present at several ions per liter will influence the chemistry of the system significantly.

6.4  DEFINING THE CONSTRAINTS

ON SPECIES CONCENTRATIONS

Key idea: Always include the self-­ ionization of water as an equilibrium

The species list leads to a mathematical system with n unknowns; namely, the equilibrium concentrations of the n species. As discussed in Section 5.5, the n equations necessary to solve for the n species concentrations at equilibrium are of four types: equilibrium expressions, mass balances, electroneutrality, and constraints based on the standard state of mixtures. In addition, the species concentrations must be nonnegative. Each of these constraints will be discussed in more detail below.

6.4.1  Equilibrium Constraints

self-­ionization of water: the equilibrium H2O = H+ + OH– ion product of water (KW): the equilibrium constant for the self-­ionization of water (= 1.01×10–14 at 25°C)

Key idea: Make sure the equilibria in the equilibrium list are linearly independent

Every aquatic chemical system has at least one equilibrium. The ubiquitous equilibrium is the expression relating H2O, H+, and OH–. This equilibrium is called the ­self-­ionization of water and is given by: H2 O

H

OH

K

KW

The equilibrium constant for this reaction is called the ion product of water and is given a special symbol, KW. At 25°C, KW = 1.01×10–14. Always include the self-­ionization of water in aqueous equilibrium calculations. What other equilibria should be included in an equilibrium model? You should include all equilibria populated completely by the species in the species list. In fact, you may have compiled this list of equilibria already as you were generating the species list. In this way, the list of pertinent species and list of pertinent equilibria are intertwined. The slate of equilibria developed during the generation of the species list may have to be modified before inclusion in the equilibrium model. Recall that you need n linearly independent equations to solve a system of n unknowns. Thus, the equilibria to be included in the model must be linearly independent. Remember from Section 4.6.2 that a set of equilibria is linearly independent if no equilibrium expression can be formed by linearly combining the other equilibria. In this context, “linearly combining” means reversing equilibria, multiplying stoichiometric coefficients by a constant, and/ or adding equilibria. Always check to see that the equilibria in the equilibrium list are linearly independent.

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6.4  |  Defining the Constraints on Species Concentrations   113

Returning to the ammonium chloride example, a list of potential equilibria is as follows (from Section 4.6.1, 6.3.5, and the self-­ionization of water): NH 4 NH 4 HCl HCl H2 O

NH 3 H OH NH 3 H 2 O H Cl OH Cl H 2 O H OH

K1 K2 K3 K4 KW

(The subscripts on the equilibrium constants in this list, except for KW, are arbitrary.)

Thoughtful Pause Is this a legitimate list of equilibria for the NH4Cl/H2O system? The equilibria in the list are not linearly independent and thus the list cannot be used as is. For example, the K2 equilibrium can be formed by reversing the KW reaction and adding it to the K1 equilibrium. Similarly, the K4 equilibrium can be formed by reversing the KW reaction and adding it to the K3 equilibrium. A proper (and linearly independent) set of equilibria would be KW, either K1 or K2, and either K3 or K4. From the equilibria listed, the four sets of linearly independent equilibria are (KW, K1, K3); (KW, K1, K4); (KW, K2, K3); and (KW, K2, K4).3 For this example, assume you have selected the set KW, K1, and K3. This choice may seem arbitrary now (and it is somewhat arbitrary), but it will appear to be more natural after you learn about acid–base chemistry in Chapter 11. You now need to translate these four equilibrium expressions into mathematical equations. (Please try this on your own before continuing.) The chemical statement of the system is: Starting materials: NH4Cl(s), H2O Species list: NH4Cl(s), H2O, NH4+, Cl–, H+, OH–, NH3, and HCl Equilibria: NH 4 NH 3 H HCl = H Cl H 2 O = H OH

K1 K3 KW

If the solution is dilute enough so that activities and concentrations are interchangeable, the mathematical equivalent of the chemical system becomes: Unknowns: [NH4Cl(s)], [H2O], [NH4+], [Cl–], [H+], [OH–], [NH3], and [HCl]

 Four other sets of equilibria also describe the system: (K1, K2, K3); (K1, K2, K4); (K1, K3, K4); and (K2, K3, K4). These are perfectly reasonable sets. In this text, the approach is emphasized in which the self-­ionization of water always is included in the equilibrium list.

3

Key idea: Translate the equilibrium expressions into mathematical equations (as described in Chapter 4)

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114   Chapter 6  Setting Up Chemical Equilibrium Calculations Equations from equilibria: K1 K3 KW

NH 3 H NH 4 H

Cl HCl

H

OH H2 O

You now have eight unknowns and three equations. Five equations to go.

6.4.2  Mass Balance Constraints

Worked example →

The next equations are provided by mass balances. Recall from Section 5.5.3 that mass balances are written for each element represented in the species list. There are two important exceptions. You do not need to write mass balances on the elements H and O, since water acts as an “infinite reservoir” of H and O. Second, you should write different mass balances for each oxidation state unless electron transfer is allowed. In one-­phase systems (here, a closed system without solids), mass balances are true constraints on species concentrations: the mass added must be equal to the mass in the closed system at equilibrium. The use of mass balances in multiphase systems will be considered in Part IV of this text. As you approach mass balances for the first time, it is probably easiest to write mass balances on each element or oxidation state. For example, if you add sulfuric acid (H2SO4) to water, you may write a mass balance on sulfur: the sum of the mass of sulfur in all species must equal the mass of sulfur added. However, mass balances do not have to be written on elements. Mass balances can be written on any fragments of starting materials. For the example of sulfuric acid, note that sulfuric acid equilibrates with bisulfate (HSO4–) and sulfate (SO42–). Each species contains the fragment SO4. Thus, instead of writing a mass balance on S, you could write a mass balance on the fragment SO4, called total sulfate.4 Here: total sulfate = [H2SO4] + [HSO4–] + [SO42–]. The species list diagram for the H2SO4/H2O system is shown in Figure 6.4. The sum of the concentrations of the boxed species sum to the total sulfate and therefore must be equal to the concentration of H2SO4 added. Starting Materials:

H2SO4

HSO4–

H2O

H+

OH–

SO42– FIGURE 6.4  Species List Diagram for the Addition of Sulfuric Acid to Water (boxed species concentrations sum to the total sulfate concentration)

 By convention, the “total” is named after the “bare” (formally, uncomplexed) form: total sulfate, total carbonate, total ammonia, etc.

4

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6.4  |  Defining the Constraints on Species Concentrations   115

As another example, acetic acid (formally, ethanoic acid, CH3COOH) hydrolyzes to acetate (CH3COO –), but the CH3COO group remains intact.

←Worked example

Thoughtful Pause What is one possible way to write a mass balance equation for an acetic acid/ H2O system? (Hint: Write the species list diagram.) It would make sense to perform a mass balance on total acetate = [CH3COOH] + [CH3COO –]. The species list diagram is shown in Figure 6.5. Formally, any element, oxidation state, or fragment of starting materials about which a mass balance is written is called a component of the system. Four aspects of the mass balances of the system must be ascertained. First, the number of mass balance equations must be determined. One mass balance must be created for each element, oxidation state, or nondissociating fragment in the system.

component: an element, oxidation state, or fragment of starting materials about which a mass balance is written

Thoughtful Pause How many mass balance equations are required for the NH4Cl/H2O system? In the NH4Cl/H2O system, two elements other than H and O (namely, N and Cl) must be balanced. Thus, you need two mass balance equations. In this example, mass balances will be performed on total soluble N(–III) (= NT) and total soluble Cl(–I) (= ClT).5 Second, the total concentration of each element, oxidation state, or nondissociating fragment must be listed. The total concentration can be determined by the starting materials or measured values. For example, you may know the total amount of iron in the +II oxidation state through measurements, but you may not know the concentrations of individual species (Fe2+, Fe(OH)+, etc.) making up the total Fe(+II). In the current example, NT = ClT since all the N(–III) and all the Cl(–I) in the system come from ammonium chloride (see Figure  6.2). For example, if you add 1×10–3 mole of NH4Cl to 1 L of water, then NT = ClT = 1×10–3 M. Note that both NT and ClT are 1×10–3 M since the system contains 1×10–3 M of N(–III) and 1×10–3 M of Cl(–I). Third, almost all the soluble species in the species list must be associated with at least one mass balance. In assigning species to mass balances, ignore the species Starting Materials: CH3COOH

CH3COO–

H2O

H+

OH–

FIGURE 6.5  Species List Diagram for the Addition of Acetic Acid to Water (boxed species concentrations sum to the total acetate concentration)

 In this text, the subscript T will be used to denote total concentrations, as in CI T. Another common approach is to label the total concentration with the prefix TOT, as in TOTCl. A special kind of mass balance on H+, called the proton condition, will be introduced in Section 8.5.2.

5

←Worked example

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116   Chapter 6  Setting Up Chemical Equilibrium Calculations containing only H and O (since mass balances are not written for H and O). Typically, this means you do not assign H+, OH–, or H2O to mass balances. Species list diagrams (such as Figures  6.2 through 6.4) can assist in assigning species to components. Thus, the generation of the species list and mass balance equations are intertwined.

Thoughtful Pause In the chapter example, to which mass balance should each species in the species list be assigned?

Key idea: Remember that each species concentration in a mass balance equation is multiplied by a stoichiometric coefficient equal to the number of equivalents of the total mass per mole of the species

In the NH4Cl/H2O system, the species NH4+ and NH3 would be assigned to NT and the species Cl– and HCl would be assigned to ClT. The species NH4Cl(s) is not assigned to a component because it is a solid and not a soluble species. Fourth, determine the stoichiometry of the element (or oxidation state or fragment) in each species associated with a mass balance. In the example, NH4+ and NH3 each contain one N (the basis of NT) and the species Cl– and HCl each contain one Cl (the basis of ClT). In determining the stoichiometry, you are using the concept of equivalents (see Section  2.6). In other words, NH4+ and NH3 each contain one equivalent relative to NT and Cl– and HCl each contain one equivalent relative to ClT. To perform the mass balance, add the products of the mass balance stoichiometric coefficients and the species concentrations and set the sum equal to the total concentration of the element, oxidation state, or fragment. In the example: NT

1 eq / mo1 NH 4

ClT

1 eq / mol Cl

1 eq / mol NH 3 l eq / mol HCl

The coefficients mean that NH4+ contains 1 mole (or one equivalent) of N per mole of NH4Cl, NH3 contains 1 mole (or one equivalent) of N per mole of NH3, and so on. We commonly leave off the stoichiometric coefficients if they are equal to 1. Do not forget that the stoichiometric coefficients are present. The stoichiometric coefficients are really there for units conversion, to convert the units of mole/L of species to eq/L of total mass. The mathematical equivalent of the chemical system now is: Unknowns: [NH4Cl(s)], [H2O], [NH4+], [Cl–], [H+], [OH–], [NH3], and [HCl] Equations from equilibria: K1 K3 KW

NH 3 H NH 4 H

Cl HCl

H

OH H2 O

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6.4  |  Defining the Constraints on Species Concentrations   117

Equations from mass balances: NT

NH 4

ClT

Cl

NH 3 HCl

Another illustration of the development of mass balance equations is shown in Worked Example 6.3. In the NH4Cl/H2O example system, you now have five equations to describe the eight unknowns: three more equations to go. Worked Example 6.3 

Mass Balance

A metal plating wastewater contains cyanide, iron, nickel, and zinc. Develop a mass balance equation for total cyanide in the wastewater if the species list contains the following cyanide-­bearing species: HCN, CN–, Fe(CN)63–, Ni(CN)42–, Zn(CN)2(aq), Zn(CN)3–, and Zn(CN)42–. Solution It makes sense to conduct a mass balance on the fragment CN (called total cyanide CNT) since each species contains the CN fragment. We must determine how many times the fragment CN appears in each species. This is easy to determine from the chemical formulas. Thus: CNT

(1 eq / mol)[HCN] (1 eq / mol)[CN ] (6 eq / mol) Fe CN 6 3 (4 eq / mol) Ni CN 4 2 (2 eq / mol) [Zn CN

CNT

2

aq ]

(3 eq / mol) Zn CN 3 (4 eq / mol) Zn CN 4 2 [HCN] [CN ] 6 Fe CN 2[Zn CN

2

6

aq ] 3 Zn CN

3

3

4 Ni CN 4 Zn CN

4

2

4

2

(Note that the coefficients have units of eq CN/mol of species = number of CN fragments per molecule or ion of species.)

6.4.3  Electroneutrality (Charge Balance) Constraint The starting materials, including water, almost always are uncharged. If you perform a number balance on electrons, the system at equilibrium also must have a net charge of zero. This is expressed in the electroneutrality constraint or charge balance. The electroneutrality constraint requires that the sum of the equivalents of positive charges is equal to the sum of the equivalents of negative charges. For a system of

electroneutrality constraint (charge balance): the requirement that the sum of the equivalents of positive charges is equal to the sum of the equivalence of negative charges

ISTUDY

118   Chapter 6  Setting Up Chemical Equilibrium Calculations constant water volume, the number of equivalents of positive charge per liter must be equal to the equivalents of negative charge per liter. You calculate the equivalents of positive or negative charge per liter by summing the products of the charge on each cation (or anion) and the cation (anion) concentration.

Thoughtful Pause For the NH4Cl/H2O system, what is the charge balance equation? The charge balance in the ammonium chloride example is: Key idea: Remember that each species concentration in the charge balance equation is multiplied by a stoichiometric coefficient equal to the number of equivalents (charges) per mole of the species

1 eq/mol NH 4     1 eq /mol H     1 eq /mol Cl     1 eq/mol OH Here, “1 eq/mol” means “one equivalent of charge per mole” (i.e., the charge on the species is +1 or –1). We usually write the charge balance without the units on the coefficients: NH 4

H

Cl

OH

Again, do not forget that each species in the charge balance is multiplied by a coefficient. The coefficient represents the number of charges on the species. Please note that the coefficient has units of equivalents of charge/mole (or charges/ion). There are three important features of the charge balance equation. First, each chemical system has only one charge balance equation. Second, neutral (uncharged) species do not appear in the charge balance equation. Third, [H+] always appears on the cation side of the charge balance and [OH–] always appears on the anion side of the charge balance in aqueous systems since H+ and OH– always are in the species list. The mathematical equivalent of the NH4Cl/H2O chemical system becomes: Unknowns: [NH4Cl(s)], [H2O], [NH4+], [Cl–], [H+], [OH–], [NH3], and [HCl] Equations from equilibria: K1 K3 KW

NH 3 H NH 4 H

Cl HCl

H

OH H2 O

Equations from mass balances: NT

NH 4

ClT

Cl

NH 3 HCl

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6.4  |  Defining the Constraints on Species Concentrations   119

Equation from charge balance: NH 4

H

Cl

OH

You now have six equations and eight unknowns and need two more equations.

6.4.4  Other Constraints The final equations in the model come from the activities of pure liquids and solids in the standard state for mixtures adopted in Sections 5.5.4 and 4.3. Water in freshwater systems is considered a pure liquid. Thus, we have a new equation: mole fraction of water = 1. What about NH4Cl(s)? To understand the constraints on [NH4Cl(s)], think about the solubility of solids. You will take up the case of the solubility of solids in Chapter 19. For now, you should recognize that solids either exist or do not exist. A solid does not exist if it dissolves completely when added to water. If the solid does not exist, its concentration is zero. If a pure solid exists, it has unit activity (i.e., mole fraction = 1). It is fairly common to have salts as starting materials. A salt is a compound in which the transferable H+ has been replaced by another cation. If the cation is K+, Na+, or NH4+, then the salt is very soluble in water. For example, the solubility of ammonium chloride in water at 25°C is about 370 g/L (almost 7 M). Thus, a small amount of ammonium chloride added to water is expected to dissolve completely. The assumption of complete dissolution (or complete dissociation) is appropriate for many common salts containing K+, Na+, or NH4+ (e.g., NaOH, NaCl, NaHCO3, and KOH). With the assumption of complete dissolution, you have generated another equation: [NH4Cl(s)] = 0. You now have a complete mathematical model as follows: Unknowns: [NH4Cl(s)], [H2O], [NH4+], [Cl–], [H+], [OH–], [NH3], and [HCl] Equations from equilibria: NH 3 H NH 4

K1

H

K3

Cl HCl

H

KW

OH H2 O

Equations from mass balances: NT

NH 4

NH 3

ClT

Cl

HCl

H

Cl

Equation from charge balance: NH 4

OH

salt: a compound in which the transferable H+ has been replaced by another cation

ISTUDY

120   Chapter 6  Setting Up Chemical Equilibrium Calculations Other equations: mole fraction of H 2 O = 1 (standard state) NH 4 Cl(s) 0 (complete dissolution) You have eight unknowns (eight species concentrations) and eight equations (three from equilibria, two from mass balances, one from the charge balance, and two other). Techniques for solving this mathematical system will be discussed in Chapters  7 through  9. To solve the system, you require numerical values of K1, K3, and KW and numerical values for the total masses of NT and ClT.

6.5  REVIEW OF PROCEDURES FOR SETTING

UP EQUILIBRIUM SYSTEMS

It is imperative that you become very familiar with the process of setting up equilibrium systems described in Sections 6.2–6.4. Several shortcuts are possible and will be discussed later. For now, it is recommended that the procedures presented here be followed closely. For convenience, the procedures are summarized below and in Appendix C.2: Step 1: Define the chemical system. Goal: Decide if an open or closed model is to be used. Tool: Match the chemical model to the system to be modeled. Step 2: Generate a species list. Goal: Create a list of n unknowns (equilibrium concentrations of n species). Tools: Identify the starting materials. Always include H2O as a starting material. Identify the initial hydrolysis products. Look for the formation of simple ions (as shown in Table 6.1). Look for H+ transfer. Always include H2O, H+, and OH–. Enumerate subsequent hydrolysis products. Enumerate other reaction products. Generate a species list diagram (see Figures 6.2–6.5). Step 3: Define the constraints on species concentrations. Goal: Create a list of n equations. Tools: Equilibrium constraints. Always include the self-­ionization of water. Mass balance constraints. Identify elements, oxidation states, or fragments to serve as the bases of mass balances (components). Find the total concentration of mass to be balanced. Associate each soluble species with at least one mass balance (exceptions: H2O, H+, and OH–). Determine the stoichiometry of each soluble species with respect to its component(s).

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6.6  |  Concise Mathematical Form for Equilibrium Systems   121

Electroneutrality constraint (charge balance) Other constraints Activity of pure liquids = 1 (mole fraction) Activity of pure solids = 1 (if solid exists) Activity of solids = 0 (if solid dissolves completely)

6.6  CONCISE MATHEMATICAL FORM

FOR EQUILIBRIUM SYSTEMS

It is possible to write the mathematical form of a closed equilibrium system in a very concise format. Let Ci be the equilibrium concentration of species i. Assume the system has ns species, neq equilibria, and nmb components. For each equilibrium, let ieq , j be the stoichiometric coefficient of species i in equilibrium j.6 For a given equilibrium j, ieq ,j will be zero for any species not involved in the reaction. Recall from Chapters 3 and 4 that ieq , j is negative for reactants and positive for products. For each component, let imb be the stoichiometric coefficient of species i in mass ,j balance j.7 Again, for a given component, imb , j will be zero for any species that is not a part of the mass balance. Let CT,j be the total concentration of total mass j. Finally, let ich be the signed charge on species i (< 0 for anions, > 0 for cations, and equal to zero for uncharged species). The mathematical form of a closed equilibrium system can be written as: Equilibria: K j

ns i 1

ns

log K j

i 1

eq i, j

ns

Mass balances: CT , j

Charge balance:

eq i, j

Ci

i 1

ns i 1

ch i

Ci

, for j = 1 to neq log Ci , for j = 1 to neq mb i, j

Ci for j = 1 to nmb

0

Other constraints: Ci ≥ 0 for all i mole fraction of i = 1 if species i is a pure liquid activity of i = 1 if species i is a pure solid activity of i = 0 if species i is a solid that dissolves completely

 Previously (Sections 3.7.1 and 4.4), the stoichiometric coefficients for chemical reactions were given the symbol νi. At the risk of confusion, symbols here are changed to emphasize that there are several types of stoichiometric coefficients in chemical equilibrium systems. The superscripts indicate the type of stoichiometric coefficients: eq for equilibrium, mb for mass balances, and ch for the charge balance. 7  This is analogous to the stoichiometric coefficient used in Section 3.7.2 to calculate standard enthalpies of formation (where νi = number of occurrences of element j in species i). 6

ISTUDY

122   Chapter 6  Setting Up Chemical Equilibrium Calculations This mathematical form is the basis of the computer calculation methods described ch mb in Chapter 9. Note that ieq , j , i , j , and i can be written as matrices. Thus, this concise form is really a linear algebra representation of the system. The chapter example can be translated into the concise form. For simplicity, stipulate that the activity of NH4Cl(s) is zero and the mole fraction of water is 1 so that those two species can be eliminated from the species list. The compact form becomes: log NH 4 log K1 log K3 log KW

1 0 1 0 1 0 1 1 0 0 0 0 1 1 0

log Cl

0 1 0

log H log OH

log NH3 log HCl

NH 4 NT ClT

1 0 0 0 1 0 0 1 0 0 0 1

Cl H OH NH3 HCl

NH 4 Cl 0

1 1 1

1 0 0

H OH NH 3 HCl

Note that the system is nonlinear in the unknowns (i.e., nonlinear in species concentrations) because the equilibrium equations are expressed in terms of the log of the concentrations.

6.7  CHAPTER SUMMARY A detailed protocol for setting up chemical equilibrium problems was developed in this chapter. The first step is to define the chemical system; that is, to decide if an open or closed model is to be used. The second step is to generate a species list. This can be accomplished by identifying the starting materials (always include H2O as a starting material), identifying the initial hydrolysis products (by looking for the formation of simple ions and H+ transfer), enumerating subsequent hydrolysis products, and enumerating other reaction products. It is useful to draw a species list diagram to identify the source of each hydrolysis species.

ISTUDY

6.8  |  Part II Case Study: Have You Had Your Zinc Today?   123

The third step is to define the constraints on species concentrations. The constraints stem from equilibria (including the self-­ionization of water), mass balances, electroneutrality (charge balance), and other considerations. Mass balance constraints can be determined by identifying components, finding the total concentration of each component, associating each soluble species with a component (except for the species H2O, H+, and OH–), and determining the stoichiometry of each soluble species with respect to its component(s). Only one charge balance equation exists and it insures that the sum of the signed equivalents of charge is zero. Other common constraints are that the mole fraction of a pure liquid is unity, the activity of a pure solid is unity, the activity of a solid is zero if it dissolves completely, and all species concentrations must be nonnegative.

6.8  PART II CASE STUDY: HAVE YOU HAD

YOUR ZINC TODAY?

Recall from Section 5.8 that the Part II case study involves determining the form of zinc when a zinc acetate tablet dissolves in your mouth and is swallowed. Having competed this chapter, you should be able to set up the chemical system for zinc acetate in water. The procedure reviewed in Section 6.5 will be followed. Please try each step on your own before reading the answer. The first step is to define the chemical system. For describing the water chemistry in your mouth and stomach, a closed system appears to be reasonable. Thus, the control volume will not include a gas phase. The second step is to make a species list. First, the starting materials need to be identified. For the problem at hand, the starting materials are zinc acetate and water. Solid zinc acetate, Zn(CH3COO)2(s), will be abbreviated here Zn(Ac)2(s). Second, identify the initial hydrolysis products. It is logical to assume that Zn(Ac)2(s) will dissolve in water to form Zn2+ and Ac–. Water, of course, forms H+ and OH– as initial hydrolysis products. Third, enumerate subsequent hydrolysis products. To generate a list of subsequent hydrolysis products, it is useful to ask whether each of the initial hydrolysis products reacts with water. You know that H+ and OH– do not react further with water, so you need only examine whether Zn2+ and Ac– react with water. By examining the equilibria in Table D.1, it is apparent that Zn2+ reacts with water (or reacts with the water hydrolysis product OH–) to form ZnOH+. Soluble zinc hydrolysis products, formed from subsequent reactions with OH– (from Table D.1), include the species Zn(OH)20, Zn(OH)3–, and Zn(OH)42–. From Table D.3, it appears that the insoluble hydrolysis product Zn(OH)2(s) also should be included in the species list. How about Ac–? It is logical that acetate (Ac–) might accept a proton from water. The hydrolysis product (acetic acid, or HAc) also can be seen in Table D.1. Fourth, enumerate other reaction products. To accomplish this task, examine lists of equilibria to see if all the products or all the reactants of a given equilibrium are on the preliminary species list. Table D.2 lists several soluble species that can be formed by the reaction between Zn and Ac–. The new species include ZnAc+, Zn(Ac)20, and Zn(Ac)3–. The resulting species list is as follows: Species  list:  Zn Ac            Zn OH

2

0

2

s ,  Zn 2 ,  Ac ,  H 2 O,  H ,  OH ,  ZnOH ,

,  Zn OH

           ZnAc ,  Zn Ac

2

0

3

,  Zn OH

,  and  Zn Ac

4 3

2

,  Zn OH

2

s ,  HAc,

←Worked example

ISTUDY

124   Chapter 6  Setting Up Chemical Equilibrium Calculations Fifth, generate a species list diagram. Although this step is not required, it does provide a nice summary of the species formed and their origin. Also, species list diagrams are a good way to check visually whether other species may be formed. A species list diagram for the Zn(Ac)2(s)/H2O system is shown in Figure 6.6. The third step is to define the constraints on the species concentrations. First, write a list of pertinent equilibria. From the tables in Appendix D, a suitable list of equilibria can be found (subscripts on the equilibrium constants, except for KW, are arbitrary): H2 O Zn 2

H OH OH ZnOH

Zn 2

2OH

Zn OH

2

Zn

3OH

Zn OH

3

2

KW K1

0

K2 K3

2

Zn 2 4OH Zn OH 4 Zn 2 2OH Zn OH 2 s HAc Ac H Zn 2 Ac ZnAc

K4 K5 K6 K7

Zn 2

K8

Zn

2Ac

Zn Ac

3Ac

2

Zn Ac

2

0

K9

3

Equilibrium constraints are given by the mathematical forms of the equilibria and will be listed at the end of this section. The fourth step is to write the mass balance constraints. You can verify that two mass balances have to be written. Logical choices are to write mass balances for Zn(+II)T and AcT. The resulting mass balances on soluble species are: Zn

II

T

Zn 2 ZnAc

AcT

ZnOH

Zn(OH)2 0  

Zn(Ac)2 0

Ac     HAc

Zn(OH)3

Zn(OH)4 2

Zn(Ac)3

Zn Ac    2 Zn(Ac)2 0

3 Zn(Ac)3

Make sure that you understand where the mass balance equations come from and can explain the coefficient of each term in the mass balances. Starting Materials:

H2O

Zn(Ac)2(s)

Zn2+

Ac–

ZnOH+ ZnAc+ 0 Zn(OH)2 Zn(Ac)20 – Zn(OH)3 Zn(Ac)3– Zn(OH)42–

H+

OH–

HAc

FIGURE 6.6  Species List Diagram for the Addition of Solid Zinc Acetate to Water

ISTUDY

6.8  |  Part II Case Study: Have You Had Your Zinc Today?   125

The fifth step is to write the electroneutrality constraint. The charge balance for this system is: H

2 Zn 2 OH

ZnOH

ZnAc

Zn(OH)3

2 Zn(OH)4 2

Zn(Ac)3

Again, make sure that you can justify the stoichiometric coefficient of each term in the charge balance equation. The sixth step is to list other constraints in the system. In solving this system, we will assume that Zn(Ac)2(s) dissolves completely and that its activity is zero. In addition, we will assume that Zn(OH)2(s) does not form at all. (You can verify this assumption in the pH range of 2–6.5 after completing Chapter 19.) Finally, the mole fraction of pure water will be assumed to be unity. As a final assumption, we assume in this case study that activities and concentrations are interchangeable. After eliminating Zn(OH)2(s) and translating the equilibria into mathematical constraints, the final mathematical form of the system is as follows: Species list: Zn Ac

2

s ,  Zn 2 ,  Ac ,  H 2 O,  H ,  OH ,  ZnOH ,

     Zn OH      Zn Ac

2

0

2

0

,  Z Zn OH

,  Zn OH

3

,  and  Zn Ac

4

2

,  HAc,  ZnAc ,

3

Equations from equilibria: H

KW

OH , K1 H2 O

Zn(OH)3

K3

Zn

OH

2

3

ZnOH , K2 Zn 2 OH Zn(OH)4 2

, K4

Zn

ZnAc , K8 Zn 2 Ac

K7

2

OH

Zn(Ac)02 Zn

Ac

2

2

4

Zn(OH)2 0 Zn 2

, K6

OH

2

,

H Ac , HAc Zn(Ac)3

, and K 9

Zn 2

Ac

Equations from mass balances: Zn

II

Zn 2     ZnOH

T

Zn(OH)2 0

                    Zn((OH)3

Zn(OH)4 2

                    Zn(Ac)2

Zn(Ac)3

AcT =  Ac

HAc

0

ZnAc

ZnAc

2 Zn(Ac)2 0

Equation from the charge balance: H

2 Zn 2 OH

ZnOH Zn(OH )3

ZnAc

=

2 Zn(OH)4 2

Zn(Ac)3

 

3 Zn(Ac)3

3

,

ISTUDY

126   Chapter 6  Setting Up Chemical Equilibrium Calculations Other equations and assumptions: mole fraction of H2O = 1 (standard state) [Zn(Ac)2(s)] = 0 (complete dissolution) Zn(OH)2(s) does not form Activities and concentrations are interchangeable. The mathematical system consists of 14 unknowns and 14  equations. The 14 equations include 9 equilibria, 2 mass balances, 1 charge balance, and 2 other constraints. You can verify that the equations are linearly independent.

CHAPTER KEY IDEAS • To analyze chemical equilibrium systems, define the system, identify the species of interest, and write the constraints on species concentrations. • To enumerate species, identify the starting materials, identify the initial hydrolysis products, enumerate subsequent hydrolysis products, and enumerate other reaction products. • Always include water as a starting material. • An H usually is available for transfer to water if it is bonded to O or N in the starting material. • Always include H2O, H+, and OH– in the species list. • Na+, K+, and F– are nonhydrolyzing and do not react appreciably with water. • Use equilibria to determine whether species on the species list react to form other species. • When generating a species list, include all species for which valid thermodynamic data are available. • Always include the self-­ionization of water as an equilibrium. • Make sure the equilibria in the equilibrium list are linearly independent. • Translate the equilibrium expressions into mathematical equations (as described in Chapter 4). • Remember that each species concentration in a mass balance equation is multiplied by a stoichiometric coefficient equal to the number of equivalents of the total mass per mole of the species. • Remember that each species concentration in the charge balance equation is multiplied by a stoichiometric coefficient equal to the number of equivalents per mole of the species (charges per ion).

CHAPTER GLOSSARY closed system:  an aqueous system not in equilibrium with a gas phase component:  an element, oxidation state, or fragment of starting materials about which a mass balance is written electroneutrality constraint (charge balance):  the requirement that the sum of the equivalents of positive charges is equal to the sum of the equivalence of negative charges

ISTUDY

Historical Note: Salts of the Ocean   127

ion product of water (KW):  the equilibrium constant for the self-­ionization of water

(= 1.01×10–14 at 25°C) nonhydrolyzing species:  chemical species that do not react appreciably with water open system:  an aqueous system in equilibrium with a gas phase salt:  a compound in which the transferable H+ has been replaced by another cation self-­ionization of water:  the equilibrium H2O  =  H+ + OH– species list diagram:  a visual representation of species and their origin(s), including starting materials starting materials:  chemicals added to water before equilibration reactions begin

HISTORICAL NOTE: SALTS OF THE OCEAN The approach described in this chapter has been used countless times to calculate the concentrations of chemical species at equilibrium. One example is an early equilibrium model for the chemistry of the major ions in seawater. The model was developed by geochemist Robert M. Garrels (1916–1988) and Mary E. Thompson (Garrels and Thompson 1962). Garrels brought the ideas of equilibrium calculations to geochemistry. He and Charles L. Christ (1916–1980) wrote an influential book on geochemistry in 1965 called Solutions, Minerals, and Equilibria. For more on the story of how the model was developed, see Thompson (1992). In developing their model, Garrels and Thompson considered the major cations (Ca2+, Mg2+, Na+, and K+) and anions (Cl–, SO42–, HCO3–, and CO32–) in seawater. They were interested in how much of each cation was found in compounds formed with each anion and how much of each anion was found in compounds formed with each cation. The compounds formed between cations and anions in water with high salt concentrations (such as seawater) are called ion pairs. Their species list was formed from the ions listed above and the ion pair combinations. Chloride does not associate with the cations listed, so Cl– and all ion pairs containing chloride were excluded from the species list. Potassium ion does not associate with bicarbonate or carbonate, so those ion pairs were excluded. Water, H+, and OH– were not of interest in the model, since the pH of seawater is known and none of the species in the species list exhibit important acid–base chemistry at the pH of seawater. The species list was: Ca 2+, Mg2+, Na +, K +, SO 4 2 , HCO3 , and CO32 ; CaSO 4 0 , CaHCO3+, and CaCO30 ; MgSO 4 0 , MgHCO3+, and MgCO30 ; NaSO 4 , NaHCO30 , and NaCO3 ; KSO 4 Seventeen species means 17 unknowns, and therefore 17 equations are needed.

ISTUDY

128   Chapter 6  Setting Up Chemical Equilibrium Calculations Where did the 17 equations come from? Since chloride (the ion in seawater of largest concentration) was excluded from the model, the charge balance was not used. The equations came from equilibria and mass balances. Garrels and Thompson employed seven mass balance equations, one for each of Ca(+II), Mg(+II), Na(+I), K(+I), S(+VI), HCO3, and CO3. They used separate mass balances on HCO3 and CO3 (rather than the usual mass balance on total dissolved carbonate) because the pH was fixed and the total concentrations of bicarbonate and carbonate were known. For example, the mass balances on Ca(+II) and S(+VI) were: Ca

II

T

Ca 2+ + CaSO 4 0 +  CaHCO3+ + CaCO30

S

VI

T

SO 4 2 + CaSO 4 0 + MgSO4 0 +  NaSO4 + KSO 4

The remaining 10 equations were the mathematical forms of equilibria from equilibrium expressions such as: NaHCO30

Na

NaHCO3 ; K

In seawater, activities and concentrations are not identical. Procedures for interchanging activities and concentrations will be discussed in Chapter 21. After corrected for activities, Garrels and Thompson identified the most important species for each component. Their results are summarized in Table 6.2. Note that the cation-­based components are mostly in their free forms, with a moderate contribution of the sulfate ion pairs to calcium and magnesium. The magnesium and sodium ion pairs are significant contributors to the total concentrations of the three anion-­based components.

Table 6.2 Contributions to Major Components in Seawater (Adapted from Garrels and Thompson 1962) CT (m)1

Predominant Species (% of total component)

Ca(+II)

0.0104

Ca2+ (91%), CaSO40 (8%), CaHCO3+ (1%)

Mg(+II)

0.0540

Mg2+ (87%), MgSO40 (11%), MgHCO3+ (1%)

Na(+I)

0.4752

Na+ (99%), NaSO4– (1.2%)

K(+I)

0.100

K+ (99%), KSO4– (1%)

S(+VI)

0.0284

SO42– (54%), MgSO40 (21.5%), NaSO4– (21%)

HCO3

0.00238

HCO3– (69%), MgHCO3+ (19%), NaHCO3+ (8%)

CO3

0.000269

MgCO30 (67%), NaHCO3– (17%), CO32– (9%)

Component

Note: Total component concentration in molal (mole per kg of seawater).

1 

ISTUDY

Problems   129

PROBLEMS 1. Write a complete mathematical model for a closed system consisting of water only at equilibrium. Be sure to include starting material(s), a species list, equilibria (if any), mass balances (if any), a charge balance equation, and other constraints. Make sure you have n independent equations for n unknowns. Assuming the activity and concentration are interchangeable, how do [H+] and [OH–] compare at equilibrium? 2. Given your answer to Problem 1 and that KW  ≈ 10–14 at 25°C, explain why pH 7 is called “neutral pH.” A fluid with [H+] or [OH–] greater than 1 M is no longer called water. What is the approximate pH range of water? (Your answer is approximate because you have ignored the differences between activity and concentration in this problem.) 3. Write a complete mathematical model for a closed system at equilibrium consisting of NaCl(s) added to water. See Problem 1 for a description of what to include. Assume NaCl(s) dissolves completely. Without calculating species concentrations, how do [H+] and [OH–] in this system compare to [H+] and [OH–] in Problem 1? (Again, ignore the differences between activity and concentration.) See Table 6.1 for help with the initial dissociation and Section 6.3.5 for help with the fate of chloride ion. 4. Write a complete mathematical model for a closed system at equilibrium consisting of hydrochloric acid (HCl) added to water. See Problem 1 for a description of what to include. Which is larger: [H+] or [OH–]? From your answer to this question, do you expect the pH to be greater than or less than 7? Given that hydrochloric acid is a very strong acid, does your answer make sense? 5. Write a complete mathematical model for a closed system at equilibrium consisting of sodium hydroxide (NaOH) added to water. See Problem 1 for a description of what to include. Which is larger: [H+] or [OH–]? From your answer to this question, do you expect the pH to be greater than or less than 7? Given that sodium hydroxide is a very strong base, does your answer make sense? 6. Write a complete mathematical model for a closed system at equilibrium consisting of acetic acid (HAc) added to water. See Problem 1 for a description of what to include. The equilibria you need are a subset of the equilibria listed in Section 6.9. Which is larger: [H+] or [OH–]? From your answer to this question, do you expect the pH to be greater than or less than 7? Does your answer make sense?

7. Write the equation for the mass balance on total soluble zinc for the system in Worked Example 6.3. Use only the species in the species list in Worked Example 6.3. 8. How many mass balances are needed to describe a closed system formed by adding Fe(NH4)2(SO4)2 (ferrous ammonium sulfate) to water? Explain your answer. (See Worked Example 6.1 for initial dissociation products.) 9. Using the equilibria and species list diagrams in Section 6.4, write the mass balance equations for total soluble nitrogen and total soluble sulfur in the system described in Problem 8. 10. Using the equilibria from Appendix D, write a mass balance on total soluble copper if Cu2+ is added to water in a closed system. 11. Repeat Problem 10 and write a mass balance on total soluble copper if Cu2+ is added to water in an open system where carbon dioxide dissolves into the water. 12. Write mass balances for total soluble K and total soluble Cr if K2Cr2O7 (potassium dichromate) is added to water in a closed system. See Worked Example 6.1 for the initial dissociation products. Dichromate (Cr2O72–) can accept one or two H+ from water to form two other species. 13. Write the charge balance equation (electroneutrality constraint) for the system in Worked Example 6.2. 14. Write the charge balance equation (electroneutrality constraint) for the system in Worked Example 6.3. Include the appropriate species in Worked Example 6.3 and also H2O, H+, and OH–. 15. How would the mathematical model of the NH4Cl/H2O system differ if one-­half the starting ammonium chloride is replaced with NH4OH? 16. Write a complete mathematical model for a closed system at equilibrium consisting of Zn(OH)2(s) added to water. Do not assume that Zn(OH)2(s) dissolves completely. See Section 6.9 for the appropriate equilibria. 17. In preparing a cake recipe, a baker adds some baking soda (NaHCO3) to water. Generate a species list and species list diagram (such as Figures 6.2–6.5) for this system. You will need to use at least some of the equilibria in Worked Example 6.2. Assume the system is closed. Is this reasonable? 18. Write a complete mathematical model for the system in Problem 17. Assume the system is closed (is this reasonable?) and NaHCO3 dissolves completely. You will need to use at least some of the equilibria in Worked Example 6.2.

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130   Chapter 6  Setting Up Chemical Equilibrium Calculations 19. Would the pH be larger or smaller if the baker had added baking powder (primarily Na2CO3) rather than baking soda? Assume the number of moles per liter of Na2CO3 added here was the same as the number of moles per liter of NaHCO3 added in Problem 17, the system is closed, and Na2CO3 dissolves completely. You should be able to decide which system has the larger [H+] without actually calculating equilibrium concentrations.8 (Hint: You can modify the model in Problem 18

very slightly and use it to represent the system in this problem.) 20. Show how the equations presented in Section 6.6 can be used to model the systems described in Problem 3. 21. Show how the equations presented in Section 6.6 can be used to model the system described in Problems 17 and 18. 22. Show how the equations presented in Section 6.6 can be used to model the system described in Section 6.9.

CHAPTER REFERENCES Garrels, R.M. and M.E. Thompson (1962). A chemical model for sea water at 25 degrees C and one atmosphere total pressure. Am. J. Sci. 260(1): 57–66.

Thompson, M.E. (1992). The history of the development of the chemical model for seawater. Geochim. Cosmochim. Acta 56(8): 2985–2987.

 Commercial baking soda is mainly NaHCO3 → HCO3– + Na+. Applying Le Châtelier’s principle to the equilibria in Worked Example 6.2, you can see that baking soda requires the addition of acid (often vinegar or soured dairy products) to produce CO2(g) and leaven baked items. Commercial baking powder contains Na2CO3 (or sometimes a mixture of Na2CO3 and NaHCO3) and an acid, usually tartaric acid (C4O6H6) or cream of tartar (a potassium salt of tartaric acid). Thus, baking powder does not require an additional acid to produce CO2(g) since carbon dioxide is produced when baking powder mixes with water. As a result, commercial baking powder often contains starch to absorb moisture and prevent CO2(g) formation in the baking powder container. 8

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CHAPTER 7

Algebraic Solutions to Chemical Equilibrium Problems 7.1  INTRODUCTION After working through Chapter 6, you should feel confident that you can translate a chemical equilibrium system into a mathematical system consisting of n equations in n unknowns. The unknowns are the equilibrium concentrations of the n species. The n equations come from equilibria, mass balances, electroneutrality, and other constraints. In this chapter, algebraic solution methods for solving a system of n equations in n unknowns will be discussed. First, you will investigate the nature and linearity of the mathematical equations to be solved. Second, a solution method based on conversion of n equations in n unknowns into one equation in one unknown will be introduced. Third, you will use information about the chemical system to make approximations and simplify the mathematical equation(s) to be solved. A shared example will be used to demonstrate both solution techniques.

Key idea: Solution techniques for chemical equilibrium systems must be capable of solving n equations in n unknowns

7.2  BACKGROUND ON ALGEBRAIC

SOLUTIONS

7.2.1  Nature of the Mathematical System Before discussing solution strategies, it is necessary to examine the nature of the equations. You may be familiar with techniques to solve n linear equations in n unknowns (e.g., 5x + 4y = 13 and x – y = –1). Recall that a system of equations is linear in the unknowns if each term in each equation is a constant or an unknown multiplied by a constant (i.e., each term is an unknown raised to the power of 0 or 1). Linear systems are quite easy to solve. Unfortunately, chemical equilibrium systems do not give rise to linear systems of equations.

Thoughtful Pause What equations make the mathematical form of chemical equilibrium systems n ­ onlinear?

131

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132   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems The equilibrium expressions are nonlinear in the unknowns since they are of the form K Key idea: Chemical equilibrium systems give rise to nonlinear systems of equations because equilibrium expressions contain products of species concentrations Worked example →

C A

c a

D B

d b

(for the equilibrium aA + bB = cC + dD). Thus, you must

use nonlinear solution techniques to calculate species concentrations in chemical ­equilibrium systems.

7.2.2  Chapter Example To illustrate the nonlinear solution techniques presented in this chapter, you will calculate the species concentrations at equilibrium in a 1:100 dilution of vinegar. Vinegar can be modeled as a 0.7 M acetic acid solution. Recall from Chapter 2 that this notation 0.7 M. In other 100 words, the acetic acid concentration is 7×10–3 M before any hydrolysis reactions are allowed to occur. To set up the problem, follow the steps in Chapter 6. You may wish to try this on your own before reading further. Assume in this example that the solution is dilute and activities and concentrations are interchangeable. Acetic acid (CH3COOH) is abbreviated as HAc. Acetate (CH3COO–) is abbreviated as Ac–. means that the total concentration of acetic acid and related species is

Step 1: Define the system. 0.7 The system is closed with no solids and M acetic acid (HAc). The starting 100 materials are H2O and HAc. Step 2: Generate a species list. HAc hydrolyzes in water to form acetate Ac–. Thus, the species list is H2O, H+, OH–, HAc, and Ac–. Step 3: Define the constraints. Equilibrium constraints: H2O = H+ + OH–; KW (self-­ionization of water) HAc = H+ + Ac–; K Mass balance: Total acetate = AcT = 7×10–3 M = [Ac–] + [HAc] Charge balance: [H+] = [OH–] + [Ac–] Other constraints: Mole fraction of H2O = 1 All concentrations ≥ 0 The compete mathematical model is: Unknowns: [H2O], [H+], [OH–], [HAc], and [Ac–]

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7.3  |  Method of Substitution   133

Constraints: KW K AcT H

H H

OH  H2 O

eq. 7.1

Ac HAc

7 10 3 M OH

Ac

HAc

Ac

Mole fraction of H 2 O 1

eq. 7.2

All concentrations 0

7.3  METHOD OF SUBSTITUTION 7.3.1  Introduction The method of substitution is a commonly used approach to solve both linear and nonlinear systems. The goal of the method of substitution is to convert a system of n equations in n unknowns into one equation in one unknown. In the language of equilibrium chemistry, we seek to convert a system of n equations in n species concentrations into one equation in one species concentration. In general, the first step in the method of substitution is to write equations expressing the concentration of each variable as a function of one variable. For the simple example in Section 7.2.1 (5x + 4y = 13 and x – y = –1), you might rewrite the second equation to express y in terms of x: y = x + 1. The second step in the method of substitution is to insert the expressions from the first step into the as yet unused equation to create one equation in one unknown. In the simple example, insert y = x + 1 into 5x + 4y = 13 to get 5x + 4(x + 1) = 13. Now solve the one equation in one unknown (here, x = 1) and back-­substitute to find the values of the other variables (here, y = x + 1 = 2). The key to applying the method of substitution to equilibrium systems is to write equations expressing the concentration of each species as a function of the concentration of one species. Which species should be used as the common currency? In most cases, any species will do. Usually, you select a meaningful species that is expected to be found at a reasonable concentration. If the equilibrium concentration of the selected species is very small, rounding errors may crop up in the calculations. For H+ transfer (acid–base) reactions, it is logical to express the concentration of each species in the system as a function of H+. Later in the text, you will use the method of substitution to express species concentrations in terms of species other than H+.

7.3.2  Method of Substitution Procedure To apply the method of substitution, use the following five-­step approach. First, simplify the system by setting the mole fractions of water equal to unity, activities

Key idea: The method of substitution is a way to convert a system of n equations in n unknowns (i.e., n species concentrations) into one equation in one unknown (i.e., one species concentration; often [H+])

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134   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems Key idea: Use equilibria and mass balance equations to write each species concentration in terms of one species concentration Key idea: Substitute the equations for each species concentration as a function of one species concentration into the charge balance equation to derive one equation in one unknown Key idea: In engineering calculations, manipulate symbols first, then substitute known values for constants (in other words, when you plug and chug, chug first, then plug)

of all solids equal to unity, and activities of all solids that dissolve completely equal to zero. In this way, you reduce the number of unknowns and simplify the KW expression. Second, express each species concentration in terms of one species concentration. This step is the key to the method of substitution. You usually will find the equations for writing every species concentration in terms of one species concentration through the equilibria and mass balances. Third, substitute the equations derived in the previous step into the unused equation. Thoughtful Pause If you have followed the procedure so far, what is the unused equation? If the equilibria and mass balance equations are used to express each species concentration in terms of one species, then the charge balance equation will remain unused. Thus, substitute into the charge balance equation to end up with one equation in one unknown. Fourth, insert known equilibrium constants and total masses and solve the one equation for the one unknown. Notice that to this point, you have been writing equations with no numbers. It is always a good idea in engineering calculations to manipulate symbols first, then substitute known values for constants. Thus, you derive one equation in one unknown first and then substitute in the equilibrium constants and total masses. There are many ways to solve the resulting equation. Several solution methods are outlined in Appendix A. Finally, back-­substitute into the previously developed equations to calculate the  equilibrium concentration of each species. This five-­step solution process is summarized in Appendix C (Section C.3.1). An example using the diluted vinegar system is shown below. Step 1: Simplify the system by setting the mole fraction of water equal to unity, activities of all solids equal to unity, and activities of all solids that dissolve completely equal to zero. In the example, you can use the fact that the mole fraction of water is unity (eq. 7.2) to modify the KW expression (eq. 7.1). The system becomes: Unknowns: [H+], [OH–], [HAc], and [Ac–] Equations:

KW



K



AcT



H

H H

OH

eq. 7.3

Ac HAc

eq. 7.4

7 10 3 M OH

Ac

Ac

All concentrations ≥ 0

HAc

eq. 7.5 eq. 7.6

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7.3  |  Method of Substitution   135

Step 2: Express each species concentration in terms of one species concentration. In the example, express each species concentration in terms of [H+]. This is accomplished most easily by using the equilibria and the mass balance equation: KW H

OH

H

HAc

Ac K

Also : HAc So: AcT Or: Ac



AcT

K

And: HAc

K H AcT

(from eq. 7.4)

Ac H

Ac

(from eq. 7.3)

(from eq. 7.5)

Ac K AcT Ac

K

H

AcT

H

Summarizing:

OH

KW H



Ac

K K H



HAc

K

H

H

eq. 7.7 AcT

eq. 7.8

AcT

eq. 7.9

Step 3: Substitute the equations derived in Step 2 into the unused equation in Step 1. After Step 2, one equation will remain used. In the example, the unused equation is the electroneutrality constraint (charge balance equation). Substituting eqs. 7.7 through 7.9 into eq. 7.6:

H

KW H

K

K H

AcT

eq. 7.10

This gives one equation in the one unknown, [H+]. Step 4: Insert known equilibrium constants and total masses and solve. There are many ways to solve for [H+] in eq. 7.10. For example, you could rewrite eq. 7.10 as: H

KW H

K K H

AcT

0

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136   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems Or, clearing fractions: H

3

K H

2

AcT K

KW H

KKW

0

Using the methods described in Appendix A and for AcT = 7×10–3 M, K = 10–4.7, and KW = 10–14, you can show that [H+] = 3.64×10–4 M or pH 3.44 (if activity and concentration are nearly equal). (For a remarkable exact solution, see the addendum to Chapter 11.) Step 5: Back-­substitute to calculate the equilibrium concentration of each species. From eqs. 7.7–7.9: H

3.64 10 4 M

OH

KW H

Ac

K K H

HAc

K

H

2.75 10

H

11

M

AcT

3.64 10 4 M

AcT

6.64 10 3 M

You may wish to verify that these values satisfy equations 7.3–7.6. Another illustration of the method is shown in Worked Example 7.1.

Worked Example 7.1   

Method of Substitution

What are the species concentrations at equilibrium for a closed system consisting of 1×10–4 M NH4Cl in water? Solution From Chapter 6, the mathematical statement of the system is: Unknowns: NH 4 Cl s , H 2 O , NH 4 , Cl , H , OH , NH 3 , and HCl Equilibria: K1

NH 3 H NH 4

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7.3  |  Method of Substitution   137

K3 KW

H

Cl HCl

H

OH H2 O

Mass balances: NT

NH 4

ClT

NH 3

Cl

HCl

Charge balance: NH 4

H

Cl

OH

Other equations: Mole fraction of H 2 O 1 NH 4 Cl(s)

0 (complete dissolution )

Following the steps for the method of substitution: Step 1: Simplify the system. The mole fraction of H2O = 1 and [NH4Cl(s)] = 0 (complete dissolution). Thus: Unknowns: NH 4 , Cl , H , OH , NH 3 , and HCl Equilibria: K1 K3

NH 3 H NH 4 H

Cl HCl

KW

H

OH

NT

NH 4

ClT

Cl

HCl

NH 4

H

Cl

Mass balances: NH 3

Charge balance: OH

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138   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems Step 2: Express each species concentration in terms of [H+]. KW H

OH

NH 3 H K1

NH 4

H

HCl

NH 4

Cl NH 3 H K1

NH 3 H

Cl

Cl

K3

KW H

So: OH

HCl

NH 3

ClT

So: N T

Cl

K3

NT

HCl

ClT

Cl

K3 K3

K3 ClT H H

H

K1 K1 H

NH 3 NH 4

K1

H

H

ClT NT NT

Step 3: Substitute into the unused charge balance equation. K1

H

H

NT

H

K3 ClT K3 H

KW H

Step 4: Insert numbers and solve. With NT = ClT = 1×10–4 M, K1 = 10–9.3, K3 = 10+3, and KW = 10–14, then [H+] = 2.45×10–7 M or pH 6.61 (assuming that activity and concentration are nearly equal)

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7.4  |  Method of Approximation   139

Step 5: Back-­calculate other species concentrations. H OH Cl HCl NH 3 NH 4

2.45 10 7 M KW = 4.08×10–8 M H K3 K3

K3 ClT = 1.00×10–4 M H H

H

K1 K1 H K1

H

H

ClT = 2.45×10–14 M NT = 2.04×10–7 M NT = 9.98×10–5 M

A final note about the method of substitution. Please think of this method as a process rather than as a way to generate equations to be memorized. Although eq. 7.10 is the general equation for [H+] in systems with one acid that dissociates to form one H+ (such as acetic acid), do not bother to memorize it. While memorization of the end equations is unnecessary, it is important to practice the process so you can derive a single equation in one unknown (e.g., eq. 7.10) quickly.

7.4  METHOD OF APPROXIMATION 7.4.1  Introduction The method of substitution described in Section 7.3 works well for aqueous systems. The substitution approach will be used many times in this text. However, the method has some drawbacks. First, the method of substitution is tedious. Its use may lead to errors due to the number of manipulations required in larger systems. Second, as a brute force method, it does not increase your appreciation for the underlying chemistry. The chances for errors can be reduced by making the system simpler. How can the system be made simpler? The system can be made simpler by reducing the number of unknowns. One way to reduce the number of unknowns is to make approximations. In making approximations, you disregard species that you believe to be significantly smaller in concentration than other species in the same equation.

7.4.2  Where Are Approximations Applied? A key point in making approximations is selecting the appropriate equation (or equations) to which the approximations are applied. The nature of the equilibrium,

Key idea: To make approximations, disregard species that you think are significantly smaller in concentration than other species in the same equation

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140   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems mass balance, electroneutrality, and other constraints dictate where approximations can be used.

Thoughtful Pause To which equations in the mathematical model of equilibrium can approximations be applied? (See Section 7.2.2 for an example of the mathematical form of a chemical equilibrium system.) Key idea: Apply approximations only to equations where the species concentrations are added (i.e., only to mass balance and charge balance equations)

Apply approximations only to the equations where the species concentrations are added. This point is very important. It does not make sense to ignore a species in c d C D . Even small concentrations must be an equilibrium expression e.g., K a b A B included in the equilibrium expressions since the concentrations are multiplied by one another. Since K has a finite value, none of [A], [B], [C], or [D] can be set equal to zero and ignored. However, you can apply approximations to the mass balance and charge balance equations, where the species concentrations are additive. For additive equations (as in the mass balance equation AcT = [Ac–] + [HAc] or the charge balance equation [H+] = [OH–] + [Ac–]), you often can ignore species at low concentrations relative to the other species without introducing significant error. Approximations typically are applied to either mass balance equations or the charge balance equation or both. For simple systems (especially where H+ transfer is one of the main processes), it makes sense to apply approximations to the charge balance equation. Why? You often can guess about the relative importance of [H+] and [OH–]. Also, [H+] and [OH–] are both in the charge balance equation. Recall from Section 1.3 that [H+] is large under acidic conditions (low pH) and [H+] is small under basic conditions (high pH). From the self-­ionization of water (KW = [H+][OH–] if the mole fraction of water is unity), you know that if [H+] is large (low pH, acidic conditions), then [OH–] will be relatively small. Similarly, if [H+] is small (high pH, basic conditions), then [OH–] will be relatively large. This information is summarized in Table 7.1. Table 7.1 Relative Values of pH, [H+], and [OH–] Under Acidic, Neutral pH, and Basic Conditions Condition

pH

[H+]

[OH–]

Acidic Neutral pH Basic

Low Near 7 High

High Moderate Low

Low Moderate High

Note: The exact pH, [H+], and [OH–] and at neutral pH depend on the temperature, pressure, and concentrations of ions in solution.

Thus, a good first approximation is to guess whether the system is acidic or basic.

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7.4  |  Method of Approximation   141

Thoughtful Pause How can you use the guess of acidity or basicity to reduce the number of unknowns in the system?

Key idea: By estimating the equilibrium pH, you usually can ignore either [H+] or [OH–] in equations where they are additive (i.e., the charge balance equation)

If you think the system is acidic, then ignore [OH–] relative to [H+] in any equation where they are additive. Typically, you ignore [OH–] relative to [H+] in the charge balance equation. If you think the system is basic, then you can ignore [H+] relative to [OH–] in the charge balance equation. Approximations also can be applied to mass balance equations. A discussion of approximations in mass balance equations will be postponed until Chapter  11 because it is often difficult to guess which species should be ignored in a mass balance equation. In the acetic acid example, it is not clear which of [HAc] or [Ac–] is larger at equilibrium. After you examine the nature of acid–base equilibria in Chapter  11, it will become easier to make approximations in the mass balances equations.

7.4.3  Checking Approximations It is essential to check the validity of every approximation made in chemical equilibrium calculations. If you guess that the system is basic (and therefore [OH–] >> [H+]), you must check the calculated equilibrium values to confirm that [OH–] is much greater than [H+]. How large should [OH–] be relative to [H+]? The answer to this question depends on how much error you is willing to accept. Error analysis in approximations will be discussed in Section 7.4.5. In addition to checking the assumptions, you should check to see how well the original equilibrium and mass balance equations are satisfied. Students approaching equilibrium calculations for the first time often are hesitant about making approximations. You may worry that you will make the wrong approximation. Do not be afraid to make approximations  – as long as you check them. An incorrect assumption always will be revealed if checked. For example, if you assume the solution is acidic, but the calculated equilibrium [H+] is much smaller than the calculated equilibrium [OH–], then the assumption is wrong. The calculation should be repeated with another assumption (in most cases, with the opposite assumption – here, that the system is basic and [H+] can be ignored relative to [OH–]). Making an incorrect assumption costs a little time and effort but will not result in a permanent error as long as you check the approximation and revise your work if you find the approximation to be inappropriate. (An exception is illustrated in the footnote at the end of Section 7.4.4.)

7.4.4  Method of Approximation Procedure The substitution method can now be revised to include approximations. To apply approximations, use the following six-­step approach. First, simplify the system by setting the mole fractions of all pure liquids, activities of all pure solids equal to unity, and activities of all solids that dissolve completely equal to zero. Second, make

Key idea: After calculating species concentrations, check approximations and the original equations; if the assumption proves incorrect, make a different assumption/approximation and iterate

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142   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems approximations in the additive equations (mass balance and/or charge balance equations). A common approach is to assume acidic conditions and ignore [OH–] relative to [H+] in the charge balance equation or assume basic conditions and ignore [H+] relative to [OH–] in the charge balance equation. Third, manipulate the equations (as modified by the approximations) to end up with one equation in one unknown. Fourth, insert known equilibrium constants and total masses and solve the one equation for the one unknown. Fifth, back-­substitute into the previously developed equations to calculate the equilibrium concentration of each species. Finally, check the assumption and original equations. If the assumption is shown to be invalid (or the original equations are not satisfied to the degree required), make the opposite assumption and iterate. This solution process is summarized in Appendix C, Section C.3.2. Examples using the acetic acid and sodium acetate systems are shown below and in Worked Example 7.2, respectively. Step 1: Simplify the system by setting the mole fraction of water equal to unity, activities of all solids equal to unity, and activities of all solids that dissolve completely equal to zero. In the example, you can use the fact that the mole fraction of water is unity (eq.  7.2) to modify the KW expression (eq.  7.1). The system becomes (equations from Section 7.2.2 are repeated here for convenience): Unknowns: [H+], [OH–], [HAc], and [Ac–] Equations:

KW



K



AcT

H

OH

eq. 7.3

Ac HAc

7 10 3 M H



H

OH

Ac

eq. 7.4 HAc Ac

eq. 7.5 eq. 7.6

All concentrations 0 Step 2: Make approximations in the additive equations. Since acetic acid is being added to water, it makes sense that the resulting solution may be acidic. Thus, assume as a starting point that [H+] >> [OH–]. If [H+] >> [OH–] and [H+] = [OH–] + [Ac–] (charge balance, eq. 7.6), then:

Ac 

H

eq. 7.11

Step 3: Write one equation in one unknown. Insert eq. 7.11 into the K equilibrium (eq. 7.4) and mass balance ­equation (eq. 7.5): K

H

Ac HAc

2

H HAc

(from eqs. 7.4 and 7.11)

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7.4  |  Method of Approximation   143

Also : HAc

AcT

Ac

2

H HAc

So : K Or : H

2

H

AcT

K H

Solving:1 H

AcT



(from eqs. 7.5 and 7.11)

2

H

KAcT K

H

0

K 2 4 KAcT 2

Step 4: Insert known equilibrium constants and total masses and solve. For AcT = 7×10–3 M and K = 10–4.7, you can show that [H+] = 3.64×10–4 M or –3.64×10–4 M. Ignoring the negative root (since [H+] and all other species concentrations must be nonnegative): [H+] = 3.64×10–4 M or pH 3.44 (assuming activity and concentration are nearly equal). Step 5: Back-­substitute to calculate the equilibrium concentration of each species. H OH Ac HAc

3.64 10 4 M KW H K K

2.75 10

K H H

H

11

AcT

3.64 10 4 M

AcT

6.64 10 3 M

Step 6: Check the assumptions and original equations and iterate if necessary. The assumption is reasonable: [H+] (= 3.64×10–4 M) is in fact much greater than [OH–] (= 2.75×10–11 M). Checking the original equations, you will find: H

3.64 10 4 M 3.64 10 4 M

Ac HAc

HAc

6.64 10 3 M Ac

= 2.00×10–3 M = 10–4.7 M ≈ K

6.64 10 3 M 3.64 10 4 M

7.00 10 3 M

AcT

Thus, the assumption and original equations are satisfied very well and no iteration is necessary. Note that [H+] turned out to be about 13 million times greater than [OH–]. Ignoring [OH–] in the charge balance was a very good choice. You know it was a good choice only after checking the assumption.

 Recall that for the equation ax2 + bx + c, x

1

b

b2 4 ac . In the example, a = 1, b = K, and c = –KAcT. 2a

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144   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems Worked Example 7.2 

Method of Approximation

Determine the equilibrium concentrations when 7×10–3 M of sodium acetate (NaAc(s), the sodium salt of acetic acid) is added to water. Use the method of approximation. Solution The system is identical to the chapter example, with the added species (Na+ and NaAc(s)) and an additional mass balance on sodium. Following the steps for the method of approximation: Step 1: Simplify the system. The mole fraction of H2O = 1 and [NaAc(s)] = 0 (complete dissolution of the sodium salt). Thus: Unknowns: Na , Ac , HAc , H , and OH Equilibria: K

H

Ac HAc

KW

H

AcT

Ac

NaT

Na

OH

Mass balances: HAc

Charge balance: Na

H

Ac

OH

Step 2: Make approximations in the additive equations. Looking at the charge balance, it seems logical that [Na+] > [Ac–] (since the salt dissociates completely to Na+, but some Ac– is converted to HAc). Thus, from the charge balance, [H+] < [OH–] and you might guess that the solution is basic. The charge balance becomes: [Na+] = [Ac–] + [OH–]. Step 3: Write one equation in one unknown. From the example in the text: OH Ac

KW H K K H

AcT

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7.4  |  Method of Approximation   145

Substituting into the modified charge balance equation: NaT 2

Or: NaT H

KW H K NaT

K

K H

AcT

AcT KW H

KKW

0

Step 4: Insert numbers and solve. With AcT = NaT = 7×10–3 M, K = 10–4.7, and KW = 10–14 and using the quadratic equation, then [H+] = 5.34×10–9 M or pH 8.27 (assuming activity and concentration are nearly equal). Step 5: Back-­calculate other species concentrations. OH HAc

KW = 1.87×10–6 M H K

H

H

Ac

K K H

Na

NaT

AcT = 1.87×10–6 M AcT

7.00×10 3 M

7.00×10 3 M

Step 6: Check assumptions and original equations. [OH–] is about 350 times [H+]. Thus, there is expected to be little error introduced from the assumption. To summarize: H

5.34 10 9 M pH 8.27

OH

1.87 10 6 M

HAc

1.87 10 6 M

OH

7.00 10 3 M

Na

7.00 10 3 M

In this case, the assumption that the system is acidic was shown to be valid. What if you had made the wrong assumption initially? It is hard to make the wrong assumption in this simple system. For example, it makes no sense to assume that [OH–] >> [H+] since the charge balance equation ([H+] = [Ac–] + [OH–]) shows that [H+] must be equal to or greater than [OH–] to ensure that [Ac–] is non-­negative. To demonstrate what happens when an incorrect assumption is made, rework the problem with AcT = 10–10 M. If you assume the solution is acidic, then [H+] ≈ [Ac–] and eq. 7.11 is regenerated. With AcT = 10–10 M and K = 10–4.7, you can calculate [H+] = 1.00×10–10 M (pH 10) and [OH–] = 1.00×10–4 M. Thus, the assumption that [H+] >> [OH–] is incorrect. As stated earlier, you cannot assume that [OH–] >> [H+] in this case because the charge balance shows that [H+] cannot be smaller than [OH–]. The only other approximation available with the charge balance equation is that [H+] ≈ [OH–]. With this assumption: [H+] ≈ [OH–] = 1.00×10–7 M (pH 7.0), [Ac–] ≈ 10–10 M, and [HAc] ≈ 5×10–13 M. In this case, the new assumption (i.e., [H+] ≈ [OH–]) is valid. The solution makes sense: the pH will not be much different than 7 if a very small amount of the

Key idea: Incorrect approximations almost always will be disclosed if the approximations are checked

ISTUDY

146   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems acid is added to water (AcT = 10–10 M is equivalent to about 10 drops of vinegar in an Olympic-­sized swimming pool). This example points out that incorrect approximations will be disclosed if the approximations are checked. Changing the assumptions, recalculating, and rechecking the results will lead eventually to the correct equilibrium concentrations.2

7.4.5  Error Analysis in Approximations The approximation made in the acetic acid problem with AcT = 7×10–3 M was that [H+] is much greater than [OH–]. In checking the assumption, you discovered that [H+] was over 7 orders of magnitude greater than [OH–]. In this case, it is obvious that the assumption was valid. What if the concentration of H+ was only 100 times greater than the concentration of OH–? Or only 10 times greater? In such cases, a significant error may be introduced if the assumption is accepted. As a general rule of thumb, you can usually ignore a species concentration in an additive equation if it is at least 100 times smaller than other concentrations in the additive equation.3 However, the magnitude of the error depends on the nature of the chemical system. For simple acids such as acetic acid, the equilibrium pH depends on two parameters: the total amount of acid added (C) and K. The error caused by an approximation also depends on C and K. The error in the equilibrium pH at different values of the total acid added and K is shown in Figure 7.1.

Thoughtful Pause From Figure 7.1, what values of C and K result in the greatest error between the exact pH and pH estimated through approximation? Does this make sense?

9

Exact pH pH assuming [H+] >> [OH–]

8 Calculated pH

Key idea: Species concentrations in an additive equation usually can be ignored in an equilibrium calculation if their concentrations are at least 100 times smaller than other concentrations in the additive equation

7

C: 10–6 M

6

10–5 M

5

10–4 M 10–3 M

4 3 10–10

10–9

10–8

10–7

10–6

10–5

10–4

K FIGURE 7.1  Exact pH and pH Assuming [H ] >> [OH ] (see text for details) +

–

 There are exceptions to this statement. Try reworking Example 7.1 by using the reasonable approximations that [Cl–] = ClT and [NH4+] = NT. You will calculate an equilibrium pH of 7.0 and all assumptions will check out. The exact pH is 6.61. Why the difference? As you will see in Chapter 8, the charge balance is not very sensitive in this case. A new kind of equation will be introduced in Chapter 8 to account for this situation. 3  The species concentration may be ignored in terms of its contribution to the linear equation (i.e., in terms of its contribution to the charge balance or mass balance equations). The species may still play an important role in the chemical system even at low concentration. For example, the species may interact with other species or exert toxicity. You ignore the species only in magnitude relative to other species strictly for the purposes of equilibrium calculations. 2

ISTUDY

7.4  |  Method of Approximation   147

Error in Equilibrium pH (pH from approximation – exact pH)

Note that the error (i.e., the difference between the exact pH and the pH calculated with the approximation) is largest for very small C and very small K. This makes sense. The addition of a very small amount of acid (or addition of an acid that does not dissociate very much to “produce” [H+]) will result in a pH close to that of water without the acid. In other words, if C is very small and/or K is very small, you would expect the pH to be near 7. For such values of C and K, the assumption that [H+] >> [OH–] is incorrect and a larger error in the equilibrium pH will result. Remember that an error in the equilibrium pH may translate into sizable errors in the equilibrium concentrations of other species. At first glance, the information in Figure  7.1 is discouraging. From Figure  7.1, it appears that significant errors in the calculated equilibrium pH may result if you make the incorrect assumption. How do you know an error occurs if you evaluate the system only with an approximation and never calculate the exact pH? You know an error will occur if you check the assumption. In this case, check the assumption by comparing the calculated values of [H+] and [OH–]. If you assume [H+] >> [OH–] but the calculated ratio of [H+] to [OH–] is small, then the assumption should be questioned. For the example of an acid in water, the error in the equilibrium pH is plotted as a function of the calculated ratio of [H+] to [OH–] in Figure 7.2. In the calculation, it was assumed that [H+] >> [OH–]. Note that the error is significant only when the H+ concentration is less than about 100 times great than the OH– concentration. Imagine conducting an equilibrium calculation for an acid similar to acetic acid and assuming that [H+] >> [OH–]. If you found that [H+] was only 10 times larger than [OH–], then the assumption was inappropriate and significant errors in the equilibrium pH may arise (see Figure 7.2).4 The system should be reanalyzed with a different assumption. Figure  7.2 only applies to systems with acids that hydrolyze to produce one H+. However, the discussion in this section demonstrates that if you check the assumptions, you will know if errors are creeping into the analysis. The bottom line: do not be afraid to make approximations, as long as you check them. 1.0 0.8 0.6

Likely Likely unacceptable acceptable error error

0.4

0.2 0.0 0.001

0.01 Calculated

0.1 [H+] [OH–]

1 Assuming

10

100

[H+]

[OH–]

>>

FIGURE 7.2  Error in the Calculation of the Equilibrium pH as a Function of  

[H+] >> [OH–])

H+ OH

1000

(calculation assumes

 Figure 7.2 was created by using the data in Figure 7.1. Thus, K was varied from 10–10 to 10–4 and C was varied from H essentially 10–6 to 10–3 M. Over this range of K and C values, the relationship between pHapprox – pHexact and OH is independent of K and C. Caution should be exercised when using the line in Figure 7.2 outside of the indicated ranges of K and C. 4

Key idea: Do not hesitate to make and check ­ ssumptions a

ISTUDY

148   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems

7.5  CHAPTER SUMMARY In this chapter, algebraic methods for the solution of chemical equilibrium problems were presented. Two methods were discussed. In the method of substitution, each species concentration is expressed in terms of the concentration of one species. For acid–base chemistry problems, it is useful to express every species concentration in the species list as functions of [H+]. This usually is accomplished by using the equilibria and mass balance equations. The expressions can be substituted into the charge balance equation to derive one equation in one unknown. The nonlinear equation can be solved numerically by the techniques outlined in Appendix A. In the method of approximation, one or more species concentrations are ignored in the additive equations (mass and charge balance equations). This approach generally makes the resulting one equation in one unknown easier to solve. Be sure to check the assumption and original equations, changing the assumption and iterating if the assumption proves to be incorrect.

7.6  PART II CASE STUDY: HAVE YOU HAD

YOUR ZINC TODAY?

The Part II case study concerns the speciation of zinc in the mouth and stomach upon consumption of a zinc acetate tablet. Recall from Section 5.8 that 20 mg zinc as Zn were dissolved in 100 mL of water and consumed. We are interested in the zinc species formed in the transition from pH 6.5–8.5 (saliva) to pH 1–3 (gastric juices). As a result of your efforts in this chapter, you should be able to determine the zinc species at fixed pH. For the present, the speciation of zinc will be determined at pH 6.5 and 3. Other pH values will be explored in Chapter 8. The chemical system was developed in Section 6.8. It will be revised slightly and restated here. The total soluble zinc concentration, Zn(+II)T , is equal to

0.02 g Zn 65.38

g Zn

=

mol Zn

3.06×10–4 mole in 100 mL, or 3.06×10–3 M. Since zinc acetate is Zn(Ac)2(s), the total soluble acetate, AcT , is 2×Zn(+II)T or 6.12×10–3 M. As stated in Section 6.8, we assume that Zn(Ac)2(s) dissolves completely and we ignore the formation of Zn(OH)2(s). In addition, the mole fraction water is unity, and activities and concentrations will be used interchangeably. In this portion of the case study, we know the pH so the concentration of H+ is fixed. Fixing [H+] adds an equation (i.e., [H+] = known value). Since unknown species are required to fix the pH, the usual approach in systems of fixed pH is to ignore the charge balance equation. Using the equilibrium constants from Appendix D, we have: Unknowns: Zn 2 , Ac , OH , ZnOH , Zn OH Zn OH

4

2

, HAc , ZnAc , Zn Ac

2

0

2

0

, Zn OH

,

3

, and Zn Ac

3

ISTUDY

7.6  |  Part II Case Study: Have You Had Your Zinc Today?   149

Equilibria: KW

H

OH

10

K1

ZnOH Zn 2 OH Zn OH

K2

Zn

2

Zn 2

Zn 2 H

K6

0

OH

10

10.2

3

10

13.9

4

10

15.5

3

4

2

OH

Ac HAc

10

ZnAc Zn 2 Ac

K7

Zn Ac

K8

Zn

2

Zn 2

4.7

10 0

Ac

Zn Ac

K9

2

3

Ac

5

2

OH

Zn OH

K4

10

2

Zn OH

K3

14

1.57

2

10

1.1

3

10

1.57

Mass balances: Zn

II

T

Zn 2 ZnAc

AcT

Ac

HAc

ZnOH Zn Ac ZnAc

Zn OH 2

0

2

0

Zn OH

Zn Ac

2 Zn Ac

2

0

Zn OH

3

3

3 Zn Ac

Other equations and assumptions: mole fraction of H2O = 1 (standard state) [Zn(Ac)2(s)] = 0 (complete dissolution) Zn(OH)2(s) does not form pH is fixed and the charge balance equation is ignored activities and concentrations are interchangeable

3

4

2

ISTUDY

150   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems The system could be solved exactly without approximation. It is a system of 11  equations in 11 unknowns. However, the algebra is much simpler if we assume that some species concentrations are negligible. Of course, any assumptions must be checked and shown to be valid. Logical assumptions are in the number of Zn-­OH species and number of Zn-­Ac species formed. The pH is reasonably low. Therefore, [OH–] is small and it is reasonable to start with the assumption that the formation of higher Zn-­OH species (e.g., Zn(OH)20, Zn(OH)3–, and Zn(OH)42–) can be ignored. In addition, AcT is fairly small, so the formation of [Zn(Ac)20] and [Zn(Ac)3–] will be ignored as a starting point. The system becomes: Unknowns: [Zn2+], [Ac–], [OH–], [ZnOH+], [HAc], and [ZnAc+] Equilibria: KW

H

K1

ZnOH Zn 2 OH

K6 K7

H

OH

10

Ac HAc

14

10 10

ZnAc Zn 2 Ac

5

4.7

10

1.57

Mass balances: Zn AcT

II

T

Ac

Zn 2 HAc

ZnOH

ZnAc

ZnAc

Other equations and assumptions: mole fraction of H2O = 1 (standard state) [Zn(Ac)2(s)] = 0 (complete dissolution) Zn(OH)2(s) does not form pH is fixed and the charge balance equation is ignored activities and concentrations are interchangeable The modified system is 6 equations in 6 unknowns. There are many ways to proceed. As discussed in this chapter, a useful approach is to express the concentration of each species in terms of a small number of species. K Since [H+] is fixed, it is valuable to write [OH–] as a function of [H+]: [OH–] = W . H

ISTUDY

7.6  |  Part II Case Study: Have You Had Your Zinc Today?   151

The species concentrations [ZnOH+], [ZnAc+], and [HAc] also can be written in terms of [Zn2+] and [OH–]: ZnOH

K1 Zn 2

OH H

HAc

K1 KW

Zn 2 H

ZnAc

K 7 Zn 2

Ac

Ac K6

Thus, the mass balances become:

Zn

II

AcT

Zn 2

T

Ac

K1 KW H

1

H K6

1

3.06 10 3 M

K 7 Ac

6.12 10 3 M

K 7 Zn 2

eq. 7.12 eq. 7.13

Equations  7.12 and  7.13 can be solved as two equations in two unknowns. Another approach is to solve for [Ac–] in eq. 7.13 and substitute the new expression for [Ac–] into eq. 7.12. After substituting and rearranging:

a Zn 2

2

a

K7 1

b

1

c

Zn

b Zn 2

c

eq. 7.14

0, where:

K1 KW H

H K6

1

II

T

K1 KW H

K 7 AcT

Zn

II

T

H K6

1

Solving eq. 7.14 with the quadratic equation at pH 6.5: [Zn2+] = 10–2.60 M. You can back-­ substitute and solve for [Ac–] and other zinc-­containing species: Ac

1

H K6

AcT

ZnOH

K1 KW

ZnAc

K 7 Zn 2

K 7 Zn Zn 2 H Ac

2

10

10

2.26

5.10

M

10

3.26

M

M

Checking the approximations, the other zinc-­containing species are all below 10–6 M and can be ignored (at least relative to Zn2+ and ZnAc+). Thus, at pH 6.5, about 83% of the zinc is Zn2+, with about 17% as ZnAc+. About 90% of the total acetate is in the form of Ac– at pH 6.5, with most the remainder as ZnAc+. You may wish to verify that at pH 3, the approximations are even more valid. At pH 3, over 99.5% of the zinc is in the form Zn2+. About 98% of the total acetate is found as HAc at pH 3.

ISTUDY

152   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems

CHAPTER KEY IDEAS • Solution techniques for chemical equilibrium systems must be capable of solving n equations in n unknowns. • Chemical equilibrium systems give rise to nonlinear systems of equations because equilibrium expressions contain products of species concentrations. • The method of substitution is a way to convert a system of n equations in n unknowns (i.e., n species concentrations) into one equation in one unknown (i.e., one species concentration, often [H+]). • Use equilibria and mass balance equations to write each species concentration in terms of one species concentration. • Substitute the equations for each species concentration as a function of one species concentration into the charge balance equation to derive one equation in one unknown. • In engineering calculations, manipulate symbols first, then substitute known values for constants (in other words, when you plug and chug, chug first, then plug). • To make approximations, disregard species that you think are smaller in concentration than other species in the same equation. • Apply approximations only to equations where the species concentrations are added (i.e., only to mass balance and charge balance equations). • By estimating the equilibrium pH, you usually can ignore either [H+] or [OH–] in equations where they are additive (i.e., the charge balance equation). • After calculating species concentrations, check approximations and the original equations; if the assumption proves incorrect, make a different assumption/ approximation and iterate. • Incorrect approximations almost always will be disclosed if the approximations are checked. • Species concentrations in an additive equation usually can be ignored in an equilibrium calculation if their concentrations are at least 100 times smaller than other concentrations in the additive equation. • Do not hesitate to make and check assumptions.

HISTORICAL NOTE: WHAT’S IN A NAME? The common names of the short-­chained organic acids mentioned in this chapter (including in the chapter problems to follow) have interesting origins. Each name is related to the main source of the acid in nature. Acetic acid (formally, ethanoic acid) comes from the Latin acetum, vinegar. Vinegar, of course, is the most common natural source of acetic acid. The related Latin word acere means “to be sour” and is likely also the source of the word acid. Thus, the origins of acetic and acid are related. Formic acid (formally, methanoic acid) stems from the Latin formica, ant. All ants belong to the family Formicidae. The so-­called formicine ants possess sacs that contain relatively high concentrations of formic acid. Citric acid is found in citrus fruits. The word citrus has referred over time to a cedar-­ like conifer (Tetraclinis articulate) and to the citron tree (Citrus medica, so-­named by the Greeks as it grew in Media, a region in modern-­day Iran). The fruit of the citron tree (a citron) was one of the first citrus fruits. Why did citrus refer to both a cedar-­like conifer and a citrus tree? According to the Oxford English Dictionary, it may be because both the fruit and the scented wood were used to discourage moths.

ISTUDY

Problems   153

The word malic comes from the Latin malum, apple. Malic acid is responsible for the tartness in green apples, rhubarb, and wine. It was first isolated from apple juice by Swiss chemist Carl Wilhelm Scheele (1742–1786). Scheele also was the first scientist to record crystallizing citric acid. His source? Lemon juice. As stated in Problem 17, the three-­carbon carboxylic acid is formally called propanoic acid and informally referred to as propionic acid. You may recognize the prefix prop-­ as referring to a three-­carbon fragment: propane is the three-­carbon hydrocarbon with no double bonds, while propene has one double bond. Prop-­ is one of the seemingly frustrating prefixes associated with the numbers of carbon atoms in organic chemistry nomenclature. Once you hit five carbons, the familiar Greek prefixes are used: penta-­, hexa-­, hepta-­, octa-­, etc. But those shorter fragments seem strange: meth-­, eth-­, prop-­, and but-­, for one-­, two-­, three-­, and four-­carbon chains, respectively. Just like the organic acids, these prefixes stem from the common sources of a particular short-­chained organic compound. Methyl ultimately comes from the Greek μέθυ wine + ‛υ'λη wood. The ancients produced methanol (wood alcohol or wood spirits) from the distillation of wood. In 1729, August Sigmund Frobenius (d. 1741) published a paper about the properties of diethyl ether (formally, ethoxyethane) called “An Account of a Spiritus Vin AEthereus, Together with Several Experiments Tried Therewith.” This led to the word ether being associated with the familiar solvent and ultimately to the prefix eth-­ for two-­ carbon moieties. The prefix pro-­ comes from propanoic acid, or to be more precise, from propionic acid. The three-­carbon carboxylic acid is the first acid in the series to behave like the other fatty acids (e.g., to form a potassium salt with a soapy characteristic). It was named propionic acid by the French chemist Jean-­Baptiste Dumas (1800–1884) after the Greek proto first + πιον fat: the first fatty acid. The prefix but-­ stems from the four-­carbon carboxylic acid called butyric acid (formally, butanoic acid). The acid, in turn, was named from the Latin butyrum, butter. When butter goes rancid, it releases butyric acid. The acid is also responsible for the odor of vomit.

PROBLEMS 1. Professor Whoops thinks it is possible to transform the equations representing a chemical system at equilibrium into a linear set of equations. Professor Whoops performs a log transformation of the equilibria; for example: logK = log[Ac–] + log[H+] – log[HAc]. This is now a linear equation in three unknowns: x = log[Ac–], y = log[H+], and z = log[HAc]. Will Professor Whoops’s idea generate a linear system of equations in the new unknowns? Why or why not? 2. At 30°C, KW = 10–13.83. Determine the equilibrium pH of pure water (containing only H2O, H+, and OH–) at 30°C. 3. Suppose you know a water contains only H2O, OH–, H+, and various cations and anions that do not react appreciably with water. If you measure the pH to be less than 7 at 25°C, what can you say about the relative concentrations of the cations and anions? Explain your answer. 4. The mineral acids (e.g., HCl) significantly reduce the pH when added to water. For a 1×10–3 M HCl solution

(i.e., the addition of 1×10–3 moles of HCl per liter of water), find the pH and [HCl] at equilibrium. Solve this problem by guessing whether the solution will be acidic or basic. Is it true that HCl dissociates nearly completely in this system? Use the equilibrium: HCl = H+ + Cl– with K = 10+3. 5. Why does adding a small amount of HCl reduce the pH of pure water, but adding a small amount of NaCl not change the pH of pure water significantly? Use the system descriptions of both systems to explain your answer. 6. From your answer to Problem 4, it appears that HCl dissociates nearly completely. Rework Problem 4 using the assumption that HCl dissociates completely rather than the assumption about the relative concentrations of H+ and OH–. 7. Suppose you add 5×10–4 M of NaOH to the water from Problem 4 (1×10–3 M HCl solution). What is the resulting pH?

ISTUDY

154   Chapter 7  Algebraic Solutions to Chemical Equilibrium Problems 8. Resolve the acetic acid problem in this chapter (with AcT = 7×10–3 M) by expressing each species concentration in terms of [Ac–] rather than in terms of [H+]. Solve the system in three ways: (a) using no assumptions, (b) assuming [Ac–] >> [HAc], and (c) assuming [HAc] >> [Ac–]. 9. Find the equilibrium pH for a 1×10–3 M solution of NaOH. State and check all assumptions. 10. How does the pH of a C M NaAc solution compare to the pH of a C M HAc solution to which C moles per liter of NaOH have been added? Support your answer with a system description. 11. Redo the sodium acetate problem in Worked Example 7.2 by making the assumption that [H+] >> [OH–]. Show that this assumption leads to spurious results. 12. Redo the sodium acetate problem in Worked Example 7.2 by making the assumption that [HAc] >> [Ac–]. Show that this assumption leads to spurious results. 13. One drop (0.05 mL) of 21% by weight ammonium hydroxide (NH4OH) is added to 1 liter of water. What is the equilibrium pH if the system is closed? Assume ammonium hydroxide dissociates completely. The density of 21% by weight NH4OH is 0.957 kg/L. 14. Cyanide (CN–) can bind with an enzyme to cause catastrophic interference with oxygen utilization leading to death. How far below the pK would one have to acidify water so that the cyanide concentration is less than 5% of [CN–] + [HCN]? In other words, find the value of pK – pH so that the cyanide concentration is less than 5% of [CN–] + [HCN]. 15. Verify that a C M HCl solution has an equilibrium pH of about –logC. As an example, the pH of a 10–2 M HCl solution is about –log(10–2) = 2. This relationship is not valid for small values of C. (As discussed in the text, as C goes to zero, the equilibrium pH goes to 7 at 25°C.) Below what values of C does this relationship fail? 16. The final equations in one unknown for the acetic acid system in the chapter example are repeated below:

Exact solution: [H+]3 + K[H+]2 – (AcTK – KW) [H+] – KKW = 0 Assumption that [H+] >> [OH–]: [H+]2 + K[H+] – KAcT = 0

Compare these two equations and discuss whether their differences are consistent with the assumption that [H+] >> [OH–] in the second equation. Hint: Multiply the assumption equation through by [H+]. 17. An industry has a waste stream consisting only of the organic chemical propanoic acid (informally, propionic acid). Propanoic acid, CH3CH2COOH, undergoes

hydrolysis chemistry similar to acetic acid. Its conjugate base, CH3CH2COO–, is called propanoate. a. When a 10–3 M propanoic acid solution is prepared, the equilibrium pH is 3.97. Find the equilibrium constant for the dissociation of propanoic acid to propanoate. b. The waste stream (consisting only of propanoic acid and propanoate in water) is at pH 4.7. Using your answer from part a, find the total propanoic acid concentration (= [CH3CH2COOH] + [CH3CH2COO –]). 18. Formic acid (formally, methanoic acid, CHO2H) donates one H+. How much formic acid would you have to add to water to obtain a pH of 5.0? The pertinent equilibrium is: formic acid = formate + H+; K = 10–3.77. 19. Consider two acids: HA1 = A1– + H+; K1 and HA2 = A2– + H+; K2. If C1 moles/L of HA1 are required to reach a certain equilibrium pH (less than 7), what concentration of HA2 is required to reach the same equilibrium pH? You can assume [H+] >> [OH–] at equilibrium. The focus of this text is building chemical equilibrium models of the environment. We sometimes start with simple models and then revise them to develop more realistic (and more complicated) models. This process is illustrated in Problems 20 through 22. 20. The simplest model of orange juice is a citric acid solution. Citric acid (formally, 2-­hydroxy-­1,2,3-­ propanetricarboxylic acid; C6H8O7) is found in orange juice at about 8 g/L (not mg/L). Find the equilibrium pH of orange juice. Citric acid can be written as H3C. Although the chemistry of citric acid is more complex, for this problem (and Problems 21 through 22) consider only the equilibrium: H3C = H2C– + H+; K = 10–3.14. 21. A slightly less crude model for orange juice is a mixture of 8 g/L of citric acid and 1.6 g/L potassium ion (again, g/L not mg/L). Find the equilibrium pH using this model of orange juice. See Problem 20 for more information about citric acid. 22. A more refined model for orange juice would include the other important cations in orange juice: Na+ (50 mg/L), Mg2+ (100 mg/L), and Ca2+ (100 mg/L). The other important acid is malic acid (formally, hydroxybutanedioic acid; C4H6O5), present in orange juice at about 2 g/L. The pertinent equilibrium for malic acid is: H2M = HM– + H+; K = 10–3.4. What is the revised equilibrium pH of orange juice, including all the cations listed and both citric and malic acid? Assume that none of the cations hydrolyze at reasonably low pH. See Problem 20 for more information about citric acid.

ISTUDY

CHAPTER 8

Graphical Solutions to Chemical Equilibrium Problems 8.1 INTRODUCTION Chapter  7 provided some basic techniques for calculating species concentrations at equilibrium. In theory, you now have all the tools you need to determine equilibrium concentrations. Through equilibria, mass balances, a charge balance, and other thermodynamic constraints, a system of n equations in n unknowns can be generated. The system can be solved by the algebraic methods in Chapter 7. Graphical solutions to chemical systems at equilibrium were popularized by the Swedish geochemist Lars Gunnar Sillén (1916–1970). Sillén’s wonderful review of graphical techniques begins with the same question you may have: “Why should one bother to represent equilibria graphically?” (Sillén  1959). One reason for learning another solution technique is that algebraic methods become very tedious as the number of species (and, hence, the number of unknowns) increases. Computer solutions (Chapter 9) are one way to ease the computational burden. Graphical solutions to aquatic equilibrium systems, the subject of this chapter, are useful to help visualize species concentrations. A second reason for mastering graphical solution methods is as follows. Although algebraic techniques give the species concentrations at equilibrium, they do not tell you how species concentrations change with a change in the primary variable. (See Section 1.3 for a review of the primary variable concept.) For example, you found in Section 7.3.2 that the equilibrium pH of a hundredfold dilution of vinegar was about 3.4, and the equilibrium concentrations of acetic acid and acetate were about 6.6×10–3 and 3.6×10–4 M, respectively. What are the acetic acid and acetate concentrations if the pH is raised to 7? To answer this question with algebraic methods requires significant additional work. Thus, it is desirable to develop solution methods that readily yield information on how species concentrations change with a change in the primary variable. Finally, as you discovered in Chapter 7, algebraic methods become much easier if you can make assumptions about which species dominate the equations and which species can be ignored. However, it is not always easy to guess which species can be ignored. You will find in this chapter that graphical methods allow you to identify easily which species contribute significantly and which can be ignored in the additive equations (i.e., mass balance and charge balance equations).

Key idea: Graphical solutions help to (1) visualize species concentrations; (2) show how species concentrations change with a change in the primary variable; and (3) identify easily which species contribute significantly and which can be ignored in the additive equations

155

ISTUDY

156   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems Key idea: Graphical solutions can show the concentration of each species at each value of the primary variable and the equilibrium concentrations of each species

Sillén (1959) summarized the last two reasons for employing graphical methods succinctly when he wrote: “A graph may tell us at one glance how a complicated equilibrium is shifted on varying conditions, which species predominate, and which are negligible.” In this chapter, one type of graphical solution technique, the log concentration diagram, will be developed. In general, graphical methods are used for two purposes. First, graphs will be created to show how the concentration of each species varies with the primary variable. These graphs will give you the concentration of each species at every value of the primary variable. Second, graphical methods can be used to determine equilibrium concentrations of each species. Equilibrium occurs at one value of the primary variable.

8.2  LOG CONCENTRATION AND

pC-­pH DIAGRAMS

8.2.1  Introduction

Primary Variable Value 0

log Concentration

log concentration diagram: a plot of the log of the species concentrations against the primary variable (usually pH)

As stated in Section  8.1, we are interested in developing graphs to show quantitatively how species concentrations vary with a primary variable. The graphs to be developed will be similar to most other plots in science and engineering; namely, the independent variable (in this case, the primary variable) will be plotted on the abscissa (or x-­axis) and the dependent variables (in this case, species concentrations) will be plotted on the ordinate (or y-­axis).1 Typically, we plot the log of the species concentrations, since the concentrations vary over many orders of magnitude. A plot of the log (base 10) of the species concentrations against the primary variable is called a log concentration diagram. An example of the axes for a log concentration diagram is shown in Figure 8.1. Log concentration diagrams are sometimes confusing at first because the y-­axis typically is numbered “down” from zero. You are probably used to y-­axes being

0

2

4

6

8

10

–2 –4 –6 –8 –10

FIGURE 8.1  Axes for a General Log Concentration Diagram

 For those readers interested in the etymology of scientific terms, the word abscissa comes from the Latin linea absicca (linea line and ab + scindere to cut off), literally a line that “cuts off” the distance from the origin. The word ordinate comes from the Latin linea ordinate applicata, literally a line applied in an orderly manner.

1

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8.2  |  Log Concentration and pC-­pH Diagrams   157

numbered “up” from zero. Do not be alarmed. In the log concentration diagram, species concentrations increase as one moves “northward” up the ordinate, in accordance with your common experiences with plotting data. Remember that the ordinate is a log scale: one unit increase in a log concentration diagram represents a tenfold increase in the species concentration. Also note that the choice of zero for the maximum value of the log of the species concentration (log[C] = 0 or [C] = 1 M) is arbitrary. Species concentrations can be greater than 1 M, and if so, the ordinate must be rescaled.

Key idea: Concentrations increase as one moves “up” the y-­axis in a log concentration diagram

8.2.2  A Taste of Acid–Base Chemistry and pC-­pH Diagrams Log concentration diagrams are used in particular with equilibrium descriptions of acid–base chemistry. Although acid–base chemistry will be presented in detail in Chapter 11, a taste of acid–base chemistry is presented here to provide common examples of log concentration diagrams. Of all possible equilibria, a special name is given to equilibria of the form of the reaction in eq. 8.1: acid

base nH

This is called an acid dissociation equilibrium. One of the reactants is an acid, and the products include a corresponding (or conjugate) base and one or more H+. If n = 1, we say that the acid is a monoprotic acid. Acetic acid, introduced in Chapter 7, is a monoprotic acid: CH3COOH = CH3COO– + + H or HAc = Ac– + H+. The equilibrium constants for equilibria with monoprotic acids written in the form of eq. 8.1 are given a special name: Ka. Thus: monoprotic acid

conjugate base H ; K

acid dissociation ­equilibrium: a reaction of the form: acid = base + nH+

Ka

There are many types of log concentration that differ only in the choice of the ­primary variable. Thoughtful Pause What is the logical primary variable for acid–base chemistry? With acid–base reactions, the logical primary variable is pH. Thus, it makes sense to plot pH values on the abscissa and log concentration values on the ­ordinate. It is common is to plot p(concentration) = –log(concentration) values on the ­ordinate, rather than log(concentration) values. The resulting plots are called pC-­pH diagrams. An example of a pC-­pH diagram is shown in Figure 8.2. Again, the interpretation of the pC-­pH diagram is straightforward. Species concentrations increase as you move up the ordinate. Also, pH increases as you move to the right, again in line with common scientific plots. In other words, more acidic conditions are on the left, and more basic conditions are on the right in pC-­pH diagrams. Note, however, that this means the H+ activity decreases as you move to the right. Also, pC-­pH diagrams are log-­log plots: a one-­unit decrease in pC or pH represents a tenfold increase in the species concentration or H+ activity.

pC-­pH diagram: a log concentration diagram with pH on the abscissa and pC = p(concentration) = –log(concentration) values on the ordinate Key idea: In a pC-­pH diagram, more acidic conditions are on the left (so the H+ activity decreases to the right)

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158   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems pH 0

0

2

4

6

8

10

12

14

2

pC

4 6 8 10 12 14 FIGURE 8.2  Axes for a General pC-­pH Diagram

8.2.3  Plotting Species Concentrations on pC-­pH Diagrams You can draw lines representing the species concentrations as functions of pH on a pC-­pH diagram. We know that aqueous systems always contain H2O, H+, and OH–. Thoughtful Pause What does the line representing [H+] look like on a pC-­pH diagram?

Key idea: On a pC-­pH diagram, the line representing [H+] slopes downward to the right with a slope of 1 and the line representing [OH–] slopes upward to the right with a slope of –1

To determine how the line representing any species appears on a pC-­pH diagram, do what you have done many times in your science and engineering courses: Write an expression for the dependent variable as a function of the independent variable. In this case, write an expression for pC as a function of pH. Writing –log[H+] as a function of pH, you can show that p[H+] ≈ pH (assuming the solution is dilute enough so that concentrations and activities are equal). In other words, the line representing [H+] on a pC-­pH diagram is a straight line with a slope of 1: p[H+] = mpH + b, where m = slope = 1 and b = intercept = 0. How about [OH–]? Recall that all aqueous systems must satisfy the equilibrium constraint given by the self-­ionization of water (KW = [H+][OH–] = 10–14 at 25°C and with the mole fraction of water as unity and concentrations and activities interchangeable). From this equilibrium, you can verify that pKW = pH + pOH or: pOH

pH pK w

Thus, the line representing [OH–] on a pC-­pH diagram is a straight line with a slope of –1 and an intercept of pKW (intercept = 14 if KW = 10–14), assuming dilute conditions.2 It is customary to refrain from plotting the H2O line since the activity of water usually is constant, very large relative to other species, and independent of pH. 2  At first glance, the signs of the slopes of the [H+] and [OH–] lines may seem confusing. You are used to a positive slope meaning that the y values increase with increasing x values and the line slopes upward to the right. In the pC-­ pH diagram, a positive slope (as with the [H+] line) also indicates that the y values, p(concentration) values, increase with increasing pH. This means that the species concentration decreases with increasing pH. A positive slope trends downward to the right.

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8.2  |  Log Concentration and pC-­pH Diagrams   159

The H+ and OH– concentration lines must always have slopes of 1 and –1, respectively, because of the definition of pH and the self-­ionization of water (as long as activity and concentration are equal). Thus, a pC-­pH diagram for a dilute aqueous system will always have the H+ and OH– lines as shown in Figure 8.3 (if dilute at 25°C and 1 atm). As with any well-­constructed scientific plot, you must label the axes and lines. The concentration units in pC-­pH diagrams are assumed to be mol/L, unless otherwise indicated. Before using pC-­pH diagrams to solve equilibrium problems, it is worthwhile to become very familiar with the interpretation of lines on a pC-­pH diagram. Three points are important. First, you read off concentrations on a pC-­pH diagram as you do with any other plot. Read off the concentration corresponding to the point where an imaginary line perpendicular to the x-­axis passes through the pH of interest and intersects the line representing the species concentration. For example, in Figure 8.4, pOH at pH 8 equals 6 or [OH–] = 10–6 M at pH 8. Second, as always, species concentrations are equal at the pH where their lines cross on a pC-­pH diagram. For example, from Figure 8.4, you can see that [H+] = [OH–] = 10–7 M at pH 7: concentrations are equal at pH 7 and the lines cross at pH 7. Third, do not forget that the y-­axis is logarithmic. Two species with pC values one unit apart differ in concentration by tenfold. For example, the concentration of [OH–] at pH 7.5 (=10–6.5 M) is 10 times greater than the concentration of [H+] at pH 0

0

2

4

6

8

10

12

14

2 4 pC

6 8 [H+]

[OH–]

10 12 14

FIGURE 8.3  General pC-­pH Diagram with [H+] and [OH–] Lines

pH 0

0

2

4

6

2

8

10

12

At pH 8…

pC

4 6 8 10 12

…pOH = 6 ([OH–] = 10–6M) [OH–]

concentrations are equal where lines cross

14 FIGURE 8.4  Examples of How to Interpret pC-­pH Diagrams

[H+]

14

Key idea: Always label the axes and concentration lines in pC-­pH diagrams

Key idea: Recall that concentrations on a pC-­pH diagram are logarithmic, and species concentrations are equal at the pH where the lines cross

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160   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems pH 7.5 (=10–7.5 M). For more practice, see Worked Example 8.1. Another example of interpreting pC-­pH diagrams and finding the equilibrium pH value is presented in Section 8.2.4.

Interpreting pC-­pH Diagrams I

Worked Example 8.1 

pC

For the pC-­pH diagram below (for a 10–2 M HAc solution), find (1) the [Ac–] at pH 8, (2) the relative concentrations of OH– and HAc at pH 6, and (3) the pH where [HAc] and [OH–] are equal and their concentrations at this pH.

0 2 4 6 8 10 12 14

pH 0 2 4 6 8 10 12 14 [Ac–] [HAc]

[OH–]

[H+]

Solution 1. At pH 8, the Ac– concentration is a little less than 10–2 M, because the Ac– line is a little below pC = 2 at pH 8. 2. At pH 6, [HAc] is significantly larger than [OH–] since the HAc line is several log units above the OH– line. At pH 6, [HAc] ≈ 10–3.5 M and [OH–] = 10–4.5 M, so [HAc] is about 104.5 ≈ 30,000 times larger than the concentration of OH–. 3. [HAc] and [OH–] are equal where their lines cross. This occurs at about pH 8.3, where their concentrations are about 10–5.7 M.

8.2.4  Solving a Simple Equilibrium System with a pC-­pH Diagram

Worked example →

The pC-­pH diagram would be useful if it could only be used to show species concentrations as a function of pH, as shown in Section  8.2.3. However, pC-­pH diagrams are much more powerful. They also can be used to determine the equilibrium pH and equilibrium concentrations of all plotted species. The use of pC-­pH diagrams for solving equilibrium systems is demonstrated in this section with a very simple example: pure water. A formal process for finding the equilibrium position from a pC-­pH diagram will be developed in more detail in Section 8.3.2. A system of pure water has three species (H2O, H+, and OH–), one equilibrium H OH Kw , no mass balances, one charge balance equation ([H+] = [OH–]), H2 O and the constraint that the mole fraction of water is unity. Using the self-­ionization of water, plot the concentrations of H+ and OH– (see Section 8.2.3). The resulting pC-­pH diagram was shown in Figure 8.3. To get this far, you have used the equilibrium constraint (with the mole fraction of water set equal to unity).

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8.2  |  Log Concentration and pC-­pH Diagrams   161

Thoughtful Pause What other information is available to determine the equilibrium position of the system? You have not yet used the charge balance. The charge balance is: [H+] = [OH–]. Where in Figure 8.3 are the concentrations of H+ and OH– equal? As with any ordinate (y-­axis) values, the concentrations are equal where the lines intersect. This occurs at pH 7. What are the concentrations of each species at pH 7? To determine species concentrations at pH 7, read the values off the graph as you have done many times in the past with x-­y graphs. Locate the equilibrium pH (here, pH 7) on the x-­axis. Draw a dotted line at pH 7 parallel to the y-­axis (Figure 8.5). The species concentrations are given by the y-­axis values at the points where your first dotted line intersects each species line. In this case, both [H+] and [OH–] are equal to 10–7 M at pH 7 (Figure 8.5). The key point here is that the equilibrium position of the system is the one point where the charge balance is satisfied. Now, admittedly, determining the equilibrium pH and concentrations of all species in pure water is not a difficult task. However, using graphical methods, you were able to calculate the equilibrium pH and species concentrations at equilibrium without doing any algebra. You also can find the species concentrations at nonequilibrium pH values. For example, you can find the concentrations of H+ and OH– at pH 9. The algebra involved in finding the species concentrations at pH 9 is not difficult, but the pC-­pH diagram immediately shows that [H+] = 10–9 M and [OH–] = 10–5 M at pH 9, as seen in Figure 8.6. It is clear from Figure 8.5 that [H+] [Ac–]

FIGURE 8.9  Determination of the Equilibrium pH for 0.007 M Acetic Acid

14

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8.3  |  Using pC-­pH Diagrams with More Complex Systems   165

Of course, you need to check this assumption. It is clear from Figure 8.9 that [OH–] is very small compared to [Ac–] at pH 3.4; in fact, [OH–] is seven orders of magnitude smaller than [Ac–] at pH 3.4. To summarize the graphical method (see also Appendix C, Section C.4.1): Step 1: Set up the pC-­pH diagram. Set up the chemical system and prepare a pC-­pH diagram with lines representing the H+ and OH– concentrations. Step 2: Simplify. Simplify the system by setting the mole fraction of water and the activity of all pure solids equal to unity and the activity of all solids that dissolve completely equal to zero. Step 3: Express concentrations as functions of [H+]. Express each species concentration in terms of [H+] by using equilibria and mass balance(s). Step 4: Plot. Plot the functions derived in Step 3 on the pC-­pH diagram. Step 5: Find the equilibrium pH. To find the equilibria pH (and equilibrium concentrations of all plotted species), find the pH where the charge balance is satisfied. Step 6: Check assumptions. If assumptions were made in Step 5 (i.e., if ions were assumed to be of negligible concentration), check all assumptions graphically and iterate if the original assumptions are invalid. Another example of finding the equilibrium pH of a monoprotic acid solution with a pC-­pH diagram is shown in Worked Example 8.3.

Worked Example 8.3 

Finding the Equilibrium pH with a pC-­pH Diagram

Use a pC-­pH diagram to find the equilibrium pH of 1 L of pure water to which 1×10–3 mole of ammonia has been added. Solution From Example 7.1 (which involved adding NH4Cl(s) to water), you can develop the following system: Unknowns: H 2 O , H , OH , NH 3 , and NH 4 Equilibria: K1 KW

NH 3 H NH 4 H

OH H2 O

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166   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems Mass balance: NT

NH 4

NH 3

H

OH

Charge balance: NH 4 Other equations: Mole fraction of H2O = 1 After simplifying and expressing each species concentration as a function of [H+] (see Worked Example 7.1): KW H

OH NH 3 NH 4

K1 H

K1 K1

H

NT NT

H

pC

Plotting the species on a pC-­pH diagram:

0 2 4 6 8 10 12 14

pH 0 2 4 6 8 10 12 14 [NH4+] [NH3] [OH–]

[H+]

Equilibrium occurs where the charge balance is satisfied: NH 4

H

OH

Ignoring [NH4+] relative to [H+] yields pH 7 (where [H+] = [OH–]), but a glance at the pC-­pH diagram reveals that [NH4+] is not negligible compared to [H+] at pH 7. Ignoring [H+] relative to [NH4+] yields pH 10.1 (dashed line, where [NH4+] = [OH–]) and the assumption is valid. The equilibrium pH is about 10.1.

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8.4  |  Special Shortcuts for Monoprotic Acids   167

8.4  SPECIAL SHORTCUTS

FOR MONOPROTIC ACIDS

8.4.1  Introduction As promised in Section 8.1, the log concentration diagram provides information about all species concentrations at all pH values and allows for rapid identification of the equilibrium concentrations. Using spreadsheets, you can draw pC-­pH diagrams fairly rapidly. For acids, shortcuts allow for the creation of approximate pC-­pH diagrams in a few seconds. The shortcuts will be developed here for monoprotic acids. The shortcuts will be extended to other acids in Chapter 11.

8.4.2  Factors Determining the Equilibrium pH with Monoprotic Acids Equations 8.1 and 8.2 show the dependency of acetate and acetic acid concentrations on pH. You can write these equations for a general monoprotic acid, HA. Assume HA dissociates to A– as follows: HA = A– + H+, with K = Ka. In general:

A



HA

Ka Ka H Ka

H

AT

eq. 8.3

AT

eq. 8.4

H

The concentrations of A– and HA clearly depend on pH. To generalize a pC-­pH diagram for any monoprotic acid, you must look at factors other than pH that influence the acid and conjugate base concentrations.

Thoughtful Pause What factors determine the A– and HA concentrations at any pH?

From eqs. 8.3 and 8.4, it appears that AT and Ka will influence the [HA] and [A–] concentrations at any pH. What factors determine the equilibrium pH? Recall that you must apply a charge balance to the pC-­pH diagram to determine the equilibrium pH. The charge balance equation for a monoprotic acid is: [H+] = [A–] + [OH–]. Since [OH–] depends on KW, it makes sense that the equilibrium pH in general will depend on AT, Ka, and KW. (If AT and Ka are sufficiently large, the solution will be acidic, and the appropriate approximation is [H+] ≈ [A–]. In this common case, KW does not affect the equilibrium pH very much.)

Key idea: For a monoprotic acid, the equilibrium pH depends on AT, Ka, and KW

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168   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems pH 0

2

4

pH 6

8

0

0 [A–]

4 [HA]

6

[H+]

[HA]

pH

pC

6

8

0

[H+]

2 4

[A–] [OH–]

6 8

0

2

[HA]

4 [H+]

6

[OH–]

8

(b)

(e) pH

pH 2

0

4

6

0

8

[H+]

Key idea: For a monoprotic acid, the equilibrium pH decreases as AT or Ka increases (unless AT or Ka is very small and the solution pH is near neutral)

Animation: To run an animation showing how the species concentrations and the equilibrium pH vary continuously with AT and Ka, see the example in Appendix E, Section E.2.

4

6

8

4 [HA]

[HA]

2 pC

pC

2

8

2

0

0

6

4

[A–]

2

[HA]

8

(d)

pC

0

4

6

[OH–]

8

pH 2

8

[H+]

(a)

0

6

4 6

[OH–]

8

4 [A–]

2 pC

pC

2

2

0

[A–] [OH+] (c)

4

[A–]

[H+]

6 8

[OH–] (f)

FIGURE 8.10  Effect of AT and Ka on the Equilibrium pH of a Solution Containing a Monoprotic Acid (see Table 8.1 for conditions)

In a qualitative sense, how do AT and Ka affect the equilibrium pH? It makes sense that as you add more acid (increase AT), the equilibrium pH should decrease (the solution becomes more acidic). Also, as Ka increases, the acid dissociation equilibrium favors H+ liberation. Thus, as Ka increases, the solution should become more acidic and the equilibrium pH should decrease. These observations are illustrated in Figure 8.10. Comparing Figures 8.10a, b, and c (right three figures), which differ only in AT, you can see that the equilibrium pH (indicated by the dotted line) decreases as AT increases (from c to b to a). Similarly, comparing Figures  8.10d, e, and f (left three figures), which differ only in Ka, you can see that the equilibrium pH decreases as Ka increases (from f to e to d). The effects of AT and Ka can be shown more dramatically by allowing AT and Ka to vary continuously. You can run an animation to show how the species concentrations and the equilibrium pH vary continuously with AT and Ka using the example in Appendix E (Section E.2).

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8.4  |  Special Shortcuts for Monoprotic Acids   169

Table 8.1 Conditions for the pC-­pH Diagrams in Figure 8.9 Diagram

AT (M)

Ka

Figure 8.9a Figure 8.9b Figure 8.9c Figure 8.9d Figure 8.9e Figure 8.9f

10 10–4 10–6 10–2 10–2 10–2

10–2 10–2 10–2 10–2 10–4 10–6

–2

8.4.3  The System Point and Equilibrium pH This chapter started with the premise that graphical solutions were much easier than algebraic solutions. So far, it does not appear that graphical solutions are all that simple. To determine the equilibrium pH, you must draw the [H+] and [OH–] lines and then plot the [A–] and [HA] lines using eqs. 8.3 and 8.4, respectively. The simplicity of the graphical method comes from a clever shortcut for solving systems involving acids. To use the shortcut, remember that the equilibrium concentrations depend on AT and Ka (and, to a lesser extent, on KW). To follow the shortcut, first locate the system point. The system point is the point on the pC-­pH diagram where pH = pKa and pC = pAT. The system point is located in Figure 8.11a for a monoprotic acid with pKa = 5 and pAT = 3 (i.e., a total concentration of 10–3 M). Why is the system point important? From the pC-­pH diagrams in Figure  8.11b, note that slopes of the [HA] and [A–] lines change at about the system point. In fact, [HA] is about equal to AT when the pH is at least about 1.5 units smaller than the pKa. At pH values above the pKa, the concentration of [HA] decreases by tenfold for each dpHA unit increase in pH i.e., 1 . Similarly, [A–] is about equal to AT when the pH is dpH at least about 1.5 log units larger than the pKa. At pH values below the pKa, the concendpA tration of [A–] decreases by tenfold for each unit decrease in pH i.e., 1 . These dpH observations can be confirmed with eqs. 8.3 and 8.4 (see Problem 4). From this analysis, the second step in the shortcut (after locating the system point) is to sketch the [HA] and [A–] lines to within about 1.5 pH units on either side of the pKa. When sketching the lines, make sure that the lines, if extended, would go through the system point. Lines for [HA] and [A–] to within about 1.5 pH units of the pKa are sketched in Figure 8.10b. Note again that the lines, if extended (dotted lines in Figure 8.10b), would intercept the system point. For the third step in the shortcut, sketch the lines for HA and A– concentrations near the system point.

Thoughtful Pause How do the [HA] and [A–] lines look near the system point?

system point: the point on the pC-­pH diagram where pH = pKa and pC = pAT

Key idea: Acid and conjugate base concentration lines away from the system point should intersect the system point if extended

Key idea: Monoprotic acid and conjugate base concentration lines intersect about 0.3 log units below the system point (at pH = pKa and pC about 0.3 log units below pAT)

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170   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems

0 2 4 6 8 10 12 14

0

2

4

6

8 10 12 14

pC

0 2 4 6 8 10 12 14

pH

[OH–]

[H+] (a)

0

2

4

pH 6 8 10 12 14

0 2 4 6 8 10 12 14

0

[A–] [HA] –]

[OH

2

4

6

8 10 12 14 [A–] [HA]

[OH–]

[H+] (b)

3

4

pH 5

[H+]

6

7

[A–]

4 5

(c)

0

3 pC

pC

pC

pH

6

[HA] [H+] (d)

FIGURE 8.11  Steps Followed in Applying the Shortcut Method (pKa = 5 and pAT = 3; parts (a), (b), (c), and (d) discussed in the text)

From eqs. 8.3 and 8.4, at pH pKa, you can see that [HA] ≈ [A–]. From the mass balance, [HA] = [A–] = ½AT. Thus: pHA = pA = –log(½AT) ≈ pAT + 0.3. Therefore, the [HA] and [A–] lines intersect at pH ≈ pKa and about 0.3 log units “below” pAT. In other words, the [HA] and [A–] lines intersect about 0.3 log units below the system point. In a crude pC-­ pH diagram, you can curve the [HA] and [A–] lines by eye so that they intersect about 0.3 log units below the system point (see Figure 8.11c, shown in detail in Figure 8.11d). Finally, once again find the equilibrium pH by noting where the charge balance is satisfied; that is, where [H+] = [A–] + [OH–]. The equilibrium pH is indicated in ­Figures 8.11c and d. A summary of the shortcut approximation to the graphical method for monoprotic acids follows (see also Appendix C, Section C.4.2): Key idea: The graphical shortcut allows for very rapid estimation of the equilibrium pH of monoprotic acid systems

Step 1: Locate the system point. Prepare a pC-­pH diagram with lines representing the H+ and OH– concentrations and locate the system point (where pH = pKa and pC = pAT). Step 2: Draw species concentration lines away from the system point. Draw the [HA] and [A–] lines at least 1.5 pH units away from the system point by noting that (1) below the pKa, pHA ≈ pAT and pA increases by 1 log unit for every unit decrease in pH below the pKa, and (2) above the pKa, pA ≈ pAT and pHA increases by 1 log unit for every unit increase in pH above the pKa. Make sure that the [HA] and [A–] lines, if extended, would both go through the system point. Step 3: Curve the lines. Curve the [HA] and [A–] lines so that they intersect about 0.3 log units below the system point. Step 4: Find the equilibrium pH. To find the equilibria pH (and equilibrium concentrations of all plotted species), find the pH where the charge balance is satisfied. Check any assumptions made in the charge balance and iterate if necessary.

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8.5  |  When Graphical Methods Fail: The Proton Condition   171

With this shortcut, a pC-­pH diagram for a monoprotic acid can be sketched in literally a few seconds. Armed with only pencil and paper, you can use this approach to estimate the equilibrium pH of a monoprotic acid system to within a few tenths of a log unit in a minute or two. Another illustration is provided in Worked Example 8.4. The graphical shortcut will be extended to other acid systems in Chapter  1l. You can run an animation to show how to create a pC-­pH diagram for a monoprotic acid by using the example in ­Appendix E (Section E.3).

Sketching pC-­pH Diagrams

Worked Example 8.4 

Sketch a pC-­pH diagram and find the equilibrium pH for a 10–3 M HCN solution if the pKa of HCN is 9.2.

pC

Solution The system point is at pC = p(total cyanide) = 3 and pH = pKa = 9.2. The pC-­pH diagram for the system is sketched below:

0 2 4 6 8 10 12 14

pH 0 2 4 6 8 10 12 14 [HCN] [CN–] [OH–]

[H+]

The equilibrium pH occurs at the pH where the charge balance is satisfied. The species list is H2O, H+, OH–, CN–, and HCN. Thus, the charge balance is: H

OH

CN

This occurs at about pH 6.1. Note: At pH 6.1, [H+] ≈ [CN–] and [OH–] is (barely) negligible.

8.5  WHEN GRAPHICAL METHODS FAIL:

THE PROTON CONDITION

8.5.1  Ambiguous Results from the Graphical Method Armed with the graphical method and its handy shortcut, you should feel confident in solving for the equilibrium composition of monoprotic acid systems. To test the general applicability of the graphical method, consider a simple modification of the acetic acid problem solved in Section 8.3.2. We will attempt to find the equilibrium pH of a 0.007 M solution of sodium acetate (abbreviated NaAc). Assume, as usual, that sodium acetate dissociates completely to Na+ and Ac–.

Animation: To run an animation showing how to draw a pC-­pH diagram for a monoprotic acid, see the example in Appendix E, Section E.3.

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172   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems

Thoughtful Pause How will the calculation of the equilibrium pH of a NaAc solution differ from the calculation of the equilibrium pH of an HAc solution by the graphical approach?

Using the shortcut approach, you have the same acid system at the same total acetate concentration. Thus, you have the same AT and Ka and therefore the same system point. You could sketch the pC-­pH diagram in the same way as Figure 8.8 (except that, for completeness, you might add a horizontal line at pC = 10–2.15 M to represent the concentration of sodium ion). The pC-­pH diagram for NaA with the line representing the sodium ion concentration is given in Figure 8.12. In the last step of the shortcut, we finally see a big difference between the HAc and NaAc systems: They have different charge balances. For the NaAc system, the charge balance is: Na

H

Ac

OH

Where is this satisfied? You know that the Na+ concentration is relatively large ([Na+] = 10–2.15 M at all pH), and thus you might guess that [Na+] >> [H+]. So the charge balance becomes: Na Ac OH If you assume that [Ac–] is negligible compared to [OH–], you would look for the pH where [Na+] ≈ [OH–]. This happens at pH ≈ 11.2. At this pH, [Ac–] is clearly not negligible compared to [OH–]. Thus, make the opposite assumption: [OH–] is negligible compared to [Ac–] and thus [Na+] ≈ [Ac–].

Thoughtful Pause At what pH in Figure 8.12 is [Na+] ≈ [Ac–], with [OH–] negligible ­compared to [Ac–]?

pH 0

0

2

4

6

8

10

12

[Na+]

2

pC

4 6

[Ac–]

[HAc]

8 10

[OH–]

[H+]

12 14 FIGURE 8.12  pC-­pH Diagram for the NaAc/H2O System (pAcT = 2.15)

14

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8.5  |  When Graphical Methods Fail: The Proton Condition   173

A big problem looms. The pC-­pH diagram in Figure  8.12 shows that the charge balance apparently is satisfied at every pH value between about pH 6 and pH 10! However, thermodynamics tell you that only one equilibrium pH can exist.

8.5.2  The Proton Condition What is the problem here and how can we resolve it? In fact, there is only one pH where the charge balance is satisfied exactly. The problem is that the graphical method in its present form is simply not sensitive enough to find the one equilibrium pH. You can get around this problem by modifying the charge balance. Recall that the charge balance is: [Na+] + [H+] = [Ac–] + [OH–]. It was noted above that [Na+] = AcT = 10–2.15 M. Substituting:

AcT

H

Ac

OH

eq. 8.5

From the mass balance: AcT = [HAc] + [Ac–]. Substituting into eq.  8.5 and simplifying:

HAc

H

OH

eq. 8.6

Is eq.  8.6 more valuable in finding the equilibrium pH than the original charge balance? Find where eq.  8.6 is satisfied on Figure  8.12. You can see with ease that eq. 8.6 is satisfied at about pH 8.2, where [HAc] ≈ [OH–] (≈ 10–5.8 M) and [H+] ≈ 10–8.2 M is negligible. This new equation (eq. 8.6) is called the proton condition. Another example of using mass and charge balances to derive the proton condition is given in Worked Example 8.5. As you have seen, the proton condition is very useful. It is a good way to eliminate species that have large, constant concentration. The proton condition is a linear combination of the charge balance and the mass balance. Therefore, the set of charge balance/mass balance/proton condition is not linearly independent (see Section 6.4). You can use only two of the charge balance, mass balance, and proton condition. In general, use the proton condition as a replacement for the charge balance. Thus, you can use the proton condition to find the equilibrium pH with graphical methods, and you can use the mass balance and proton condition together in the algebraic approach.

Writing the Proton Condition from Mass and Charge Balances

Worked Example 8.5 

Using the mass and charge balances, write the proton condition for a solution formed by adding sodium carbonate (Na2CO3) to water. Solution The starting material Na2CO3 will dissociate initially to Na+ and CO32– ­(carbonate). Carbonate reacts with water to form HCO3– and H2CO3. The charge balance is: H

Na

HCO3

OH

2 CO32

Key idea: pC-­pH diagrams sometimes appear to point to more than one equilibrium pH value because of poor resolution Key idea: You can eliminate species of large constant concentration in the charge balance (e.g., Na+, K+, and Cl– by using mass balances

proton condition: a mass balance on H+

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174   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems If the sodium carbonate solution is initially C M, then the mass balances are (after complete dissociation of the sodium carbonate): Na So:

2C H 2 CO3

Na

HCO3

2 H 2 CO3

HCO3

CO32

C

CO32

Substituting into the charge balance equation and simplifying, the proton condition becomes: 2 H 2 CO3

HCO3

H

OH

8.5.3  The Proton Condition as a Proton Balance

zero level of protons: the number of protons in each starting material

The proton condition was developed in Section 8.5.2 as an alternative to the charge balance equation. So far, deriving the proton condition seems a bit hit or miss: combine the mass balance and charge balance equations until you eliminate species such as Na+ that are present at high concentrations at all pH values. The proton condition has a much deeper significance than a seemingly random combination of the mass balance and charge balance equations. The proton condition is really a mass balance on protons.4 To do a mass balance on protons, you must define a zero level of protons. With a zero level, you can balance species contributing to an excess of over the zero level and a deficiency of protons relative to the zero level. The choice of the zero level is arbitrary. However, it is convenient to use the starting materials as the zero level of protons. Recall from Section 6.3.1 that the starting materials are water and the chemical species added to water. For the proton condition, it is important to always include water as a starting material. For the addition of NaAc to water, the starting materials are H2O and NaAc. (Remember that NaAc dissociates completely to Na+ and Ac–. Thus, you could list the starting materials as H2O, Na+, and Ac–.) To do a proton balance (i.e., to write the proton condition), you simply identify (1) the species in your species list with more protons than their source in the starting materials and (2) the species in your species list with fewer protons than their source in the starting materials. Multiply each species by the number of excess or deficient protons relative to the starting materials and equate the concentrations of excess and deficient protons. Note that in this approach, you do not include any species with the same number of protons as its starting material. An example will make this clearer. For the NaAc problem, the starting materials and species list are given below: Starting materials: H 2 O, NaAc

Na and Ac

Species list: H 2 O, H , OH , Na , Ac , and HAc What is the source of each species in the species list? Always assume that H2O, H+, and OH– come from water. It is reasonable to write that Na+ comes from Na+ and both  The word proton here is used as a synonym for H+. A proton and H+ represent the hydrogen atom with an electron removed.

4

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8.5  |  When Graphical Methods Fail: The Proton Condition   175

Table 8.2 Summary of Proton Balance Information for the H2O/NaAc System Species

H2O H+ OH– Na+ Ac– HAc

Source

Number of Excess or Deficient Protons Relative to the Source

H2O H2O H2O Na+ Ac– Ac–

0 1 excess 1 deficient 0 0 1 excess

Ac– and HAc come from Ac–. Now, H2O has the same number of protons as its source (namely, water itself), whereas H+ and OH– have, respectively, one more proton and one less proton than water.5 How about Na+, Ac–, and HAc? Sodium ion has the same number of protons as its source (namely, Na+). Acetic acid has one more proton than its source (Ac–), and acetate has the same number of protons as its source (Ac–). This information is summarized in Table 8.2. Now you can balance protons. For each of the species with protons in excess of their source, determine the concentration of excess protons. The concentration of excess protons is given by: (number of excess protons)(species concentration). In a similar fashion, determine the concentration of deficient protons. Since protons are neither created nor destroyed (just exchanged), you can equate the concentration of excess and deficient protons. In the example: (1 excess proton) H

(1 excess proton) HAc

(1 deficient proton) OH

or: H

HAc

OH

Writing proton conditions takes a little practice. There are a several aspects of developing proton conditions that you should keep in mind. First, remember that H+ always is considered to have one excess proton, and OH– is considered to have one deficient proton. Second, it is easier to look for excess and deficient protons if you allow salts to dissociate completely before doing a proton balance (as with the NaAc example). One more example: the HAc/H2O system. Try developing the proton condition for the HAc/H2O system on your own before you read further. Starting materials: H2O, HAc Species list: H2O, H+, OH–, Ac–, and HAc The proton condition is: (1 excess proton)[H+] = (1 deficient proton)[Ac–] + (1 deficient proton)[OH–] or: H

Ac

OH

5  It may seem a little strange to view H+ as having one more proton than water. Recall from Section 1.4.3 that the symbol H+ represents several chemical species, including H3O+. If you think of H+ as H3O+, then it is much easier to accept H+ as having one more proton than water.

Key idea: The proton condition can be formed by balancing protons; that is, by equating the number of excess protons over starting materials with the number of deficient protons over starting materials

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176   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems In this case, the proton condition is identical to the charge balance equation. At least, the charge balance and the proton condition for the HAc/H2O system are nearly identical. Thoughtful Pause What are the differences between the charge balance and the proton condition for the HAc/H2O system? In the charge balance, the coefficients in front of the species concentrations are the number of charges per ion (or number of equivalents per mole, where charge is the currency of equivalence). In the proton balance, the coefficients in front of the species concentrations are the number of excess/deficient protons per mole (or number of equivalents per mole, where protons relative to the starting material is the currency of equivalence). This is, admittedly, a subtle difference, but it points to the importance of the units of the coefficients in such expressions. The proton condition for the Na2CO3/H2O system (examined in Worked Example 8.5) is derived by using the excess/ deficient proton concept in Worked Example 8.6. Note that the excess/deficient proton approach allows proton conditions to be written almost by inspection. Worked Example 8.6 

Writing the Proton Condition from the Species List and Starting Materials

Using the species list and starting materials, write the proton condition for a solution formed by adding sodium carbonate (Na2CO3) to water. Solution The starting materials are H2O and Na2CO3 (which dissociates initially to Na+ and CO32–). The species list contains H2O, H+, OH–, Na+, H2CO3, HCO3–, and CO32–. The number of excess or deficient protons for each species relative to its starting material is as follows: Species

H2O H+ OH– Na+ H2CO3 HCO3– CO32–

Source

Excess

H2O H2O H2O Na+ CO32– CO32– CO32–

0 +1 –1 0 +2 +1 0

Setting the number of excess protons equal to the number of deficient protons: 2 H 2 CO3

HCO3

H

OH

As required, this is the same proton condition as in Worked Example 8.5.

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8.6  |  Chapter Summary   177

8.5.4  Uses of the Proton Condition The proton condition can be used to solve equilibrium systems by either algebraic or graphical methods. What role should the proton condition play in the set of equations to be solved? As stated in Section 8.5.2, the mass balances, charge balance, and proton condition are not linearly independent. You can generate any equation from the other two (see Problems 8.12 and 8.13). As a result, you can use only two of the three equations in your list of equations. Usually, it is easiest simply to replace the charge balance equation with the proton condition. When should you use the charge balance equation and when should you use the proton condition? The simple answer is that the proton condition always will give an acceptable solution.6 The proton condition is very useful for systems with salts as starting materials. When the starting material does not include a salt, the proton condition and the charge balance give the same algebraic equation. In general, use the proton condition to solve chemical equilibrium systems. Although the proton condition may seem awkward at first, you will find that, after some practice, you can write the proton condition as easily as you can write the charge balance equation.

8.6  CHAPTER SUMMARY Graphical solutions to equilibrium systems are used to show how species concentrations change with a change in the primary variable and to determine the equilibrium pH and species concentrations. Graphical solutions also allow you to identify easily which species contribute significantly and which can be ignored in the additive equations. The most common graphical tool is the pC-­pH diagram, a type of log concentration diagram where –1 times the log of each species concentration is plotted as a function of pH. To make a pC-­pH diagram, write each species concentration in terms of [H+]. This process requires the use of the equilibrium expressions and mass balances equations. With monoprotic acids (and, as you will see in Chapter  11, with other acids as well), a shortcut procedure can be employed to draw species concentration lines on a pC-­pH diagram. In the shortcut method, the system point is identified. The system point occurs where pC = p([HA] + [A–]) and pH = pKa. The pHA and pA lines are drawn away from the system point and allowed to intersect 0.3 log units below the system point. To find the equilibrium pH with a pC-­pH diagram, simply find the pH where the charge balance is satisfied. You may wish to use and check assumptions about ignoring certain species relative to other species to find the equilibrium pH. The equilibrium pH also can be found at the pH where the proton condition is satisfied. The proton condition comes from a mass balance on protons and balances protons in excess and deficient to the starting materials. The proton condition works better than the charge balance equation because the proton condition does not contain unreactive ions that limit the resolution of graphical methods for determining equilibrium concentrations. With a little practice, you can write the proton condition as readily as the charge balance equation.  In fact, the charge balance also always gives the correct equilibrium pH. However, the typical resolution of a pC-­pH diagram is not sufficient to use the charge balance to find the equilibrium pH if the charge balance includes ions with relatively large, constant concentrations.

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178   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems

8.7  PART II CASE STUDY: HAVE YOU HAD

YOUR ZINC TODAY?

Worked example →

Recall that the Part II case study was probing the speciation of zinc as a zinc acetatewas swallowed. The Part II case study can be completed by drawing a pC-­pH diagram for the Zn(Ac)2(s)/H2O system. Most of the work in solving this system was done in Section  7.6. The Zn2+ concentration can be determined by solving the equation: a Zn 2

2

b Zn 2

0 , where:

c

K1 KW H

a

K7 1

b

1

c

Zn( II )T 1

H K6

K1 KW H

1

K 7 AcT

Zn( II )T

H K6

The equilibria and mass balance expressions are: KW

H

K1 K6

ZnOH Zn 2 OH

AcT

10

ZnAc Zn 2 Ac Zn 2 Ac

14

10

Ac HAc

H

K7 Zn( II )T

10

OH

5

4.7

10

1.57

ZnOH [HAc]

ZnAc ZnAc

The approximations were shown to be valid in the pH range of 2 to 6.5 in Section 7.8. The quadratic equation can be solved for [Zn2+] at any pH. After calculating [Zn2+], you can determine the concentrations of acetate ion and the other zinc-­containing species by back-­substitution as follows: Ac

1

H K6

AcT K 7 Zn 2 Zn 2 H

ZnOH

K1 KW

ZnAc

K 7 Zn 2

Ac

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Chapter Key Ideas   179

0

0

1

2

pH 3

4

5

6

[Zn2+]

2

pC

4 6 8

[Zn(Ac)20 ]

[ZnAc+] [ZnOH+]

10

[Zn(Ac)3–] [Zn(OH)20]

FIGURE 8.13  pC-­pH Diagram for the Part II Case Study

The calculations are performed most easily with a spreadsheet. The resulting pC-­ pH diagram for the zinc-­containing species is shown in Figure 8.13. Note that [Zn2+] is the predominant zinc-­containing species at low pH. As the pH approaches the pKa of acetic acid (4.7), the [ZnAc+] becomes a greater fraction of the total zinc. Figure 8.13 can also be used to verify the assumptions that Zn(Ac)20 and Zn(Ac)3– can be ignored – relative to ZnAc+ and that Zn(OH)20, Zn(OH)3 , and Zn(OH)42– can be ignored relative to ZnOH+ (see Problem 17). What is the significance of this equilibrium calculation? The significance lies in the relative absorption of the zinc-­containing species. If the most highly absorbed species is Zn2+, then pH would make little difference in the ability of the body to absorb zinc from the zinc acetate tablets. However, if any other zinc-­containing species dominate zinc absorption, then pH may play a major role in the uptake of zinc. Note that the concentrations of all zinc species except Zn2+ are smaller in the low pH of the gastric juices than in the near-­neutral pH of saliva.

CHAPTER KEY IDEAS • Graphical solutions help to (1) visualize species concentrations; (2) show how species concentrations change with a change in the primary variable; and (3) identify easily which species contribute significantly and which can be ignored in the additive equations. • Graphical solutions can show the concentration of each species at each value of the primary variable and the equilibrium concentrations of each species. • Concentrations increase as one moves “up” the y-­axis in a log concentration. • In a pC-­pH diagram, more acidic conditions are on the left (so the H+ activity decreases to the right). • On a pC-­pH diagram, the line representing [H+] slopes downward to the right with a slope of 1 and the line representing [OH–] slopes upward to the right with a slope of –1. • Always label the axes and concentration lines in pC-­pH diagrams. • Recall that concentrations on a pC-­pH diagram are logarithmic, and species concentrations are equal at the pH where the lines cross. • The equilibrium pH on a pC-­pH diagram is the pH where the charge balance is satisfied. • To construct a pC-­pH diagram, write each species as a function of pH and plot the resulting equations.

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180   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems • To find the equilibrium conditions with a pC-­pH diagram, find the pH where the charge balance is satisfied. • To find the pH where the charge balance is satisfied, assume that one or more species concentrations in the charge balance are negligible and check your assumptions. • pC-­pH diagrams allow you to check assumptions very quickly. • For a monoprotic acid, the equilibrium pH depends on AT, Ka, and KW. • For a monoprotic acid, the equilibrium pH generally decreases as AT or Ka increase (unless AT or Ka is very small and the solution pH is near neutral). • Acid and conjugate base concentration lines away from the system point should intersect the system point if extended. • Monoprotic acid and conjugate base concentration lines intersect about 0.3  log units below the system point (at pH = pKa and about 0.3 log units below pAT). • The graphical shortcut allows for very rapid estimation of the equilibrium pH of monoprotic acid systems. • pC-­pH diagrams sometimes appear to point to more than one equilibrium pH value because of poor resolution. • You can eliminate species of large constant concentration in the charge balance (e.g., Na+, K+, and Cl–) by using mass balances. • The proton condition can be formed by balancing protons; that is, by equating the number of excess protons over starting materials with the number of deficient protons over starting materials.

CHAPTER GLOSSARY acid dissociation equilibrium:  a reaction of the form: acid = base + nH+ concentrations against the primary variable (usually pH) log concentration diagram:  a plot of the log of the species pC-­pH diagram:  a log concentration diagram with pH on the abscissa and pC = p(concentration) = –log(concentration) values on the ordinate proton condition:  a mass balance on H+ superpositioning:  the ability to superimpose concentration lines from numerous species on one pC-­pH diagram system point:  the point on the pC-­pH diagram where pH = pKa and pC = pAT zero level of protons:  the number of protons in each starting material

Historical Note: Who Was First? Who drew the first pC-­pH diagram? We probably will never know the answer to that question. The likely candidates come from a group of Scandinavian chemists working in the first half of the twentieth century. There is no doubt that graphical methods for solving and representing chemical systems was popularized (at least in publications in English) by Swedish geochemist Lars Gunnar Sillén. By the time Sillén wrote the review article cited earlier in 1959, the term system point was in common use. Sillén noted that “logarithmic diagrams with a master variable” had been in use for “many decades” in Scandinavia by the late 1950s. He cites a paper by Danish chemist Niels Bjerrum7 (1879–1958) in 1914 as the earliest 7  Bjerrum made important contributions to our understanding of ionic solutions and acid–base chemistry. He measured pH before Sørenson named it. (See the Historical Note in Chapter 1 for more on Sørenson and the naming of pH.)

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Problems   181

example of a diagram plotting the log of something versus a primary variable. In Bjerrum’s case, the titration error was plotted as a function of pH. The work of Sillén and Bjerrum underlies most of this text. Sillén made his reputation in the study of minerals in contact with water. The mineral sillenite (Bi2O3) was named after him. Sillén’s work was a great influence on Robert Garrels (see the Historical Note in Chapter 6) and Swiss chemist Werner Stumm (1924–1999). Stumm’s textbook with his former student James J. Morgan (1942–2020) brought aquatic chemistry to the forefront of environmental engineering and science education. For more on Stumm and Morgan, see the Historical Note in Chapter 24.

PROBLEMS 1. Create a pC-­pH diagram for a 10–3 M HCN solution. The pKa of HCN is 9.2. Sketch the HCN and CN– lines carefully. Write the charge balance equation for the system and find the equilibrium pH. 2. Create a pC-­pH diagram for a 10 M HOCl solution. The pKa of HOCl is 7.54. Sketch the HOCl and OCl– lines carefully. Write the charge balance equation for the system and find the equilibrium pH. –3

3. In Problem 15 in Chapter 7, you verified that a C M HCl solution has an equilibrium pH of about –logC (as long as C was greater than some value). Reverify this relationship (and find the minimum value of C) by using a graphical approach. 4. For a monoprotic acid, verify that: a. At pH values above the pKa, the concentration of [HA] decreases by tenfold for each unit increase in pH i.e.,

dpHA dpH

1 .

b. At pH values below the pKa, the concentration of [A–] decreases by tenfold for each unit decrease in pH i.e.,

dpA dpH

1 .

5. Why does the proton condition eliminate the readily dissociable ions such as Na+, K+, and Cl– from the charge balance? Why does the proton condition for adding an acid of the form HA to water look the same as the charge balance equation? 6. Write the proton condition when the following salts are added to water: a. NH4Cl

c. KH2PO4

b. Na2S

d. (NH4)2CO3

See Appendix D for ideas on the pertinent equilibria to generate species lists. 7. Worked Examples 8.5 and 8.6 were about the proton condition for Na2CO3. Write the proton conditions for NaHCO3 and H2CO3 by inspection. Using the three

proton conditions, determine which system (Na2CO3, NaHCO3, or H2CO3) will have the lowest equilibrium pH and which will have the highest equilibrium pH. 8. Find the equilibrium pH of a 10–3 M NaCN solution by using graphical methods. Try to find the equilibrium pH with the charge balance and separately with the proton condition. 9. An engineer is designing a chlorine contact chamber to disinfect poultry-­processing wastes. (Note: These wastes are often contaminated with the pathogenic bacteria Salmonella.) She can design the system to use chlorine gas (equivalent to using HOCl) or liquid bleach (equivalent to using sodium hypochlorite, NaOCl) as the chlorine source. The total chlorine concentration to be added is 7.1 mg/L as Cl2 (1×10–4 M chlorine). The species HOCl is a much stronger disinfectant than the species OCl–. The pKa for HOCl is 7.54. a. If chlorine chemistry controls the pH, which system provides more disinfection? This problem requires you to calculate equilibrium concentrations of HOCl and OCl– in 10–4 M HOCl and 10–4 M NaOCl systems separately. b. In a complex waste stream such as this, is the assumption that “chlorine chemistry controls the pH” reasonable? Why or why not? 10. Solve Problem 13 in Chapter 7 using graphical methods. 11. It is possible to write general approximations for a monoprotic acid, HA. a. Show graphically that the equilibrium pH for a solution obtained by adding C moles/L of HA to water is pK a pC . What assumptions did you make? about 2 b. Show that the equilibrium pH for a solution obtained by adding C moles/L of NaA to water is about 14 pK a pC . What assumptions did you make? 2 c. What is the equilibrium pH for a solution obtained by adding C moles/L of HA and C moles/L of NaA to water?

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182   Chapter 8  Graphical Solutions to Chemical Equilibrium Problems 12. Show that the proton condition for a NH4Cl solution is a linear combination of the mass balance equations and the charge balance equation.

15. For a monoprotic acid system, how would the lines change if you created a pC-­pOH diagram rather than a pC-­pH diagram?

13. Show that the equations formed by the proton condition, mass balance equations, and charge balance equation for a NaHCO3 solution are not linearly independent.

16. The pKa for formic acid is 3.77. Using graphical methods, find the equilibrium pH for total formate concentrations of pCT = 2, 3, 4, 5, and 6. Plot the equilibrium pH against the pCT value. Does this plot make sense?

14. An acid–base indicator usually is a monoprotic acid where the acid and conjugate base absorb light differently. Suppose the acid (HA) appears colorless and the conjugate base (A–) appears red. You are interested in the error in the indicator (i.e., whether the solution changes color exactly at the pKa). If the pKa is 7.0 and the total concentration is 0.001 M, use a pC-­pH diagram to determine when the solution changes color if the color can be detected when (i) 10% of the total is A– or (ii) 50% of the total is A–.

17. Explain how the results in Figure 8.12 justify the assumptions in the zinc case study. (Any zinc-­containing species in the species list that do not appear in Figure 8.13 have concentrations less than 10–10 M over the pH range in the figure.) 18. Demonstrate the value of the proton condition over the charge balance by using a pC-­pH diagram to find the pH of a 10–4 M NaCl solution.

CHAPTER REFERENCE Sillén, L.G. (1959). Graphical presentation of equilibrium data. In: Treatise on Analytical Chemistry. I.M. Kolthoff and P.J.

Elving (Eds.). New  York: Interscience Encyclopedia, New  York, pp. 277–317.

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CHAPTER 9

Computer Solutions to Chemical Equilibrium Problems 9.1 INTRODUCTION Many equilibrium systems of interest in the environment are sufficiently ­complicated to render algebraic and graphical solution methods very tedious. For systems with many species, computer methods can be used to calculate equilibrium concentrations quickly. As with any computer solution methods, it is imperative that you understand both the chemical system itself and how to interpret the computer output. In this chapter, you will become familiar with several computer solution methods. The ­purpose of this chapter is to increase your comfort level with one or more computer solution ­techniques so that you can solve increasingly complex systems. In Section 9.2, a problem is introduced that will be solved by all computer methods presented in the chapter. The chapter equilibrium problem is simple and familiar to you. It will allow you to try out the software on well-­trodden ground. Section 9.3 introduces you to spreadsheet solutions. You will find that spreadsheets are ideal for solving equilibrium problems. Using a consistent format for your spreadsheets will prove beneficial when tackling large systems. Equilibrium calculation software is discussed in general in Section 9.4. Three software packages are presented in more detail. In Section 9.5, a program called Nanoql SE is introduced. Nanoql SE was developed specifically for use with this text and follows the species/equilibrium/mass balance/charge balance approach you have used so far. In Section 9.6, you will be introduced to MINEQL. MINEQL is a very popular equilibrium chemistry software package. It uses an approach called the Tableau method, which is common to many equilibrium calculation programs. A popular implementation of MINEQL called Visual MINTEQ is discussed in Section 9.7.

Key idea: Many environmental chemistry problems are difficult to solve manually

9.2  CHAPTER PROBLEM To illustrate the solution of equilibrium problems by computer, a sample problem will be set up and solved in several ways in this chapter. The chapter problem is to

183

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184   Chapter 9  Computer Solutions to Chemical Equilibrium Problems determine the species concentrations in a system consisting of 1×10–3 M acetic acid (HAc) in water. The system is: Species list: H2O, H+, OH–, HAc, and Ac– (= acetate) Equilibria: H2 O

H

OH ; K w

HAc

Ac

H ;K

Or: KW K

H

OH H2 O

H Ac [HAc]

Mass balance: AcT

1 10 3 M

Ac

HAc

Charge balance: Other:

H

OH

Ac

Mole fraction of H 2 O  1 All concentrations 0 Excluding water, this system is four equations in four unknowns.

9.3  SPREADSHEET SOLUTIONS 9.3.1  Introduction

Key idea: To solve equilibrium problems with a spreadsheet, enter equations for species (based on equilibria) and vary the –log(conc.) values of components until the mass and charge balances are satisfied

Spreadsheets are invaluable for the solution of small-­to medium-­sized chemical equilibrium problems. The use of spreadsheets for finding roots of equations was discussed in Chapter 7 and Appendix A. In this chapter, spreadsheets will be used to set up and solve equilibrium systems in a more general fashion. The general approach to using spreadsheets for the solution of equilibrium problems is as follows. Equilibrium constants and added total masses are entered. In addition, the equations for the mass balances (sum of species concentrations weighted by their mass balance stoichiometric coefficients) and electroneutrality equation (charge balance or proton condition) are entered. Components are entered into the spreadsheet with guesses for their concentrations. Species are entered into the spreadsheet with equations for their concentrations as functions of the components, other species concentrations, and equilibrium constants. If the species concentration is fixed (e.g., H2O), then its fixed concentration is entered rather than an equation. Values of the component concentrations are calculated by the spreadsheet software to meet the mass balance and electroneutrality constraints. In most

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9.3  |  Spreadsheet Solutions   185

equilibrium calculations by spreadsheet, it is useful to have the spreadsheet program calculate pC [= –log(concentration)] values for each component rather than to have the software calculate concentrations for each component. Calculating pC values minimizes round-­off and underflow errors when component concentrations are very small. You have several constraining functions on the system: one for each mass balance plus electroneutrality. In a general optimization routine (such as found in Microsoft Excel’s Solver), the constraining functions are divided between the objective function and constraints. To meet the objective function, one cell in the spreadsheet (called the target cell) is set equal to a fixed value. To meet the constraints, other cells are set equal to other values. You can use the target cell to satisfy one mass balance or the electroneutrality condition and the constraints to account for the other mass balances and/or electroneutrality condition.

9.3.2  Setting up the Spreadsheet Spreadsheet solutions are much easier if you take the time to set up your spreadsheet carefully. Divide the spreadsheet into six regions for components, species, mass balances, electroneutrality, equilibria, and notes. A nice approach is shown in Figure 9.1. The spreadsheet shown in Figure 9.1 is discussed in Appendix E, Section E.5.1. Note the six regions in the spreadsheet. For each of the two components, guesses of their pC values are entered. In the example, pC guesses are entered in cells D2, D3, E2, and E3. (The spreadsheet program eventually will vary your initial guesses. Entering the values twice allows you to keep a record of your initial guesses.) The formulas in cells C2 and C3 calculate the concentrations from the pC values in cells D2 and D3. The formula in cell C2 is =10^(-­1*D2) and the formula in cell C3 is =10^(-­1*D3). Equations for the species concentrations are found in cells C7 through C11. The equations express the concentration of each species as a function of the components, other species concentrations, and equilibrium constants. (For illustrative purposes, the formulas in cells C7 through C11 are shown in cells D7 through D11 in Figure 9.1.) For KW example, cell D9 contains the formula =I3/C3 to show that OH . For water, H the fixed concentration (mole fraction) of 1 is entered in cell C7. The equation for the total acetate mass balance, expressing the total mass as the sum of the weighted species concentrations, is in cell D15. Cell D15 contains the formula =C10+C11 to indicate that AcT = [HAc] + [Ac–] (shown in cell E15 for illustrative purposes). The numerical value of the added total mass is entered in cell C15. Either the charge balance or proton condition can be used to account for electroneutrality. In this example, the charge balance formula is entered in cell C19. It is useful to enter the electroneutrality constraint as a function that is equal to zero. For example, the charge balance is entered as the sum of the eq/L of positive charges minus the sum of the eq/L of negative charges (see formula in cell D19 in Figure 9.1). The equilibrium information is written in columns H and I. It is recommended that you write out the chemical expressions to help you remember which equilibrium is found in each row (see column H in Figure 9.1). If you wish, pK or log K values could be entered into another column and K values calculated in column I. Finally, write a few notes to yourself about how you set up the problem. It is useful to note what was fitted (component concentrations or pC values of the components) and to identify both the objective function and constraints.

Spreadsheet available: For the spreadsheet for the chapter example, see Appendix E, Section E.5.1 ←Worked example

Key idea: Write out the equilibrium expressions and write notes to help you remember the details of the chemical system and calculation

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186   Chapter 9  Computer Solutions to Chemical Equilibrium Problems

FIGURE 9.1  Spreadsheet for the Chapter Problem

9.3.3  Solving Systems with Spreadsheets The dialog box for Solver is shown in Figure 9.2. As shown, Solver is changing the values of –log[H+] and –log[Ac–] to make the calculated AcT equal to the added AcT, subject to the constraint that the sum of equivalents of cations minus the sum of equivalents of anions is zero. After you press the Solve button, the optimum values of the component concentrations are calculated by varying their pC values. The species concentrations in column C are updated automatically. The final concentrations are shown in Figure 9.3. You can verify that the solution is correct.

FIGURE 9.2  Solver Dialog Box for the Chapter Problem

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9.3  |  Spreadsheet Solutions   187

FIGURE 9.3  Final Results from the Spreadsheet Solution

Whenever you use a generalized optimization routine, it is up to you to decide if the results are accurate enough. To judge accuracy, examine the calculated mass balances and calculated electroneutrality (cells D15 and C19, respectively, in Figure 9.3). If the difference between the calculated mass balance and the added total mass is too large for your purposes (or if the calculated charge balance is very different from zero), the calculation should be repeated. The calculation also must be repeated if Solver does not converge. The accuracy of the results can be changed and convergence problems addressed in two ways. First, the initial component pC guesses can be altered. Second, the Precision setting (accessed through the Option menu on the Solver dialog box) can be adjusted. The results in Figure  9.3 were obtained with a precision setting of 1×10–­10. This approach can be extended to larger systems with ease. For example, a system with 41 species is solved by using the spreadsheet method in Chapter 24. The template spreadsheet can be modified to your purposes. You can add columns to automatically calculate the species concentrations in mass units, calculate pC values for each species, or enter the equilibrium constants as pK values. You can change the objective function and constraints if you wish. Another example is a system comprising the 5 mg/L as Cl2 of hypochlorous acid (HOCl) in water. An example spreadsheet for this system prior to calculation is shown in Figure 9.4. Like acetic acid, HOCl dissociates in water to form hypochlorite (OCl–) with an equilibrium constant K equal to 10–7.54. In the spreadsheet in Figure 9.4, the K values (cells I2 and I3) are calculated from pK values (cells J2 and J3, respectively). The added chlorine in mol/L in cell C15 is calculated from the added chlorine in units of mg/L as Cl2 inputted in cell E15. Formulas were added in cells E10 and E11 to calculate the concentrations of the chlorine-­containing species in units of mg/L as Cl2. The formulas are as follows: I2: =10^(-­1*J2), I3: =10^(-­1*J3), C15: =E15/1000/70.9 (since the molar mass of Cl2 is 70.9 g/mol), E10: =C10*1000*70.9, and E11: =C11*1000*70.9. Note that concentrations are converted to mol/L prior to the calculation, calculated in mol/L, then converted to other units after calculation.

Key idea: Check the optimization results to see if sufficient resolution has been obtained by examining the predicted total masses and charge balance

←Worked example

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188   Chapter 9  Computer Solutions to Chemical Equilibrium Problems

FIGURE 9.4  Spreadsheet for 5 mg/L Cl2 in Water

9.4  EQUILIBRIUM CALCULATION SOFTWARE 9.4.1  General Approaches Chemical equilibrium systems generally are solved by using one of two approaches. In the first approach, the Gibbs free energy of the system is minimized with mass balances as constraints. An example of minimizing Gibbs free energy to determine the equilibrium composition of a system was given in Section 3.8.4. The minimization of free energy approach is very common in chemical engineering, where temperatures and pressures often are very different than standard state conditions. In environmental engineering and science applications, a different approach to solving equilibrium systems usually is used. This approach is the basis of Part II of this text: find the species activities that satisfy the mathematical system created from equilibrium expressions, mass balances, a charge balance (if applicable), and other constraints (such as species with fixed activities). As discussed in Chapter 4, the mathematical system is nonlinear in the species activities because the equilibrium expressions are nonlinear in the species activities. The equivalency of the energy minimization and equilibrium/ mass balances/charge balance approaches is explored in Appendix B, Section B.3.

9.4.2  Equilibrium Calculation Programs A number of software packages have been developed to set up and solve chemical equilibrium systems. Most of the programs are based on MINEQL. Many commercial packages exist, and a short list of available software is provided in Table 9.1. For a more extensive review of available software, see Butler (1998).

9.4.3  Importance of Components Recall from Chapter 3 that components are the minimum set of building blocks from which all species can be formed. Most equilibrium calculation software is based on

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9.4  |  Equilibrium Calculation Software   189

Table 9.1 Chemical Equilibrium Software Packages Used in  Environmental Engineering and Science Applications CHESS Developed by: Jan van der Lee System setup: Tableau method Application: Geochemical Commercial version: chess.ensmp.fr/index.html MINEQL Developed by: F.M.M. Morel System setup: Tableau method Application: General Commercial version: MINEQL+ by Environmental Research Software (mineql.com) Free version: DOS ver 3.01 is free at mineql.com MINTEQA2 Developed by: US EPA Center for Exposure Assessment Modeling System setup: Tableau method Application: General Commercial version: Scientific Software Group (scisoftware.com) Free version: US EPA at epa.gov/ceampubl/minteq.htm Related: Visual MINTEQ, maintained by J.P. Gustafsson at vminteq.lwr.kth.se Nanoql and Nanoql SE Developed by: J.N. Jensen System setup: As per Part II of this text Application: General Commercial version: Included with this text PHREEQC Developed by: US Geological Survey (USGS) System setup: Tableau method Application: Geochemistry Commercial version: PHREEQE at Scientific Software Group (www. scisoftware.com) Free version: wwwbrr.cr.usgs.gov/projects/ GWC_coupled/phreeqc/ Windows version: geo.vu.nl/users/posv/phreeqc.html Related: WATEQ4F, maintained by USGS at wwwbrr.cr.usgs.gov/ projects/ GWC_chemtherm/software.htm

the concept of components to make the calculations more efficient. You have a lot of freedom in choosing the components to describe a system. In fact, the components do not have to be real chemical species (see Morel and Hering 1993). However, there are two common rules used to form components. First, it is common practice to use the set of fully dissociated ions as components. For example, the species CdCl+ can dissociate into Cd2+ and Cl–. Neither Cd2+ nor Cl– can dissociate further. Thus, Cd2+ and

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190   Chapter 9  Computer Solutions to Chemical Equilibrium Problems Key idea: It is common practice to use fully dissociated ions, H2O and H+, as components

Cl– would be used as components of the system. On the other hand, H2CO3 can dissociate into HCO3– and H+. Bicarbonate (HCO3–) can dissociate further into CO32– and H+. Thus, the appropriate components would be CO32– and H+. From CO32– and H+, you can “make” bicarbonate (from the equilibrium CO32– + H+ = HCO3–) and H2CO3 (from the equilibrium CO32– + 2H+ = H2CO3). Second, it is common practice to include H2O and H+ as components in all aqueous chemical systems.

9.5  Nanoql SE 9.5.1  Introduction Key idea: Nanoql SE solves small-­to moderate-­sized aquatic equilibrium problems in a spreadsheet environment

Nanoql is a platform written in Visual Basic and designed to solve small-­to moderate-­ sized aquatic equilibrium systems. It was introduced in the first edition of this text. The name acknowledges the legacy of MINEQL (see the Historical Note at the end of this chapter) and the focus on smaller systems. The motivation behind the development of Nanoql was to produce an equilibrium calculation software package that simulates the manual solution method. For the second edition of this text, Nanoql has been ported to Microsoft Excel. The current version is called Nanoql SE (Nanoql Spreadsheet Edition). It uses (and requires) Excel’s Solver Add-­in. As with the other software packages in Table 9.1, Nanoql SE will provide you with equilibrium concentrations of species so you can create pC-­pH diagrams and titration curves (Chapter 12). Nanoql SE allows you to vary species concentrations, total masses, temperature, and ionic strength (see Chapter 21). Nanoql SE is visually similar to the spreadsheets presented in Section 9.3. It automates the relationships between components and species, thermodynamics, mass balances, electroneutrality, choice of constraints, and Solver control. Nanoql SE is available on the publisher’s website for this text. The best way to learn Nanoql SE is to simply try it out. A brief manual is provided in Appendix F. You may want to read Sections F.2 and F.3 before proceeding to Section 9.5.2. Many Nanoql SE examples will be given in this text and are discussed in Section F.4. Additional hints for using Nanoql SE to solve specific types of chemical equilibrium problems are given in Appendix F.

9.5.2  Using Nanoql SE to Solve the Chapter Problem

Nanoql SE example

To get your feet wet with Nanoql SE, open the spreadsheet. Instructions for entering the chapter example may be found in Table 9.2. The process is straightforward. On the Start tab, type an “x” in the cell next to the component acetate (CH3COO–). (Nanoql SE components are shown in bold through this text.) The components H2O and H+ are preselected. Remember that we usually designate the component as the fully dissociated species in a family (see also Section 9.6.1). In the chapter problem, selecting the component CH3COO– is the way to select the species CH3COO– and CH3COOH. While on the Start tab, enter the total concentration of the component CH3COO– and a guess for component H+. Your Start tab should look like Figure 9.5. You can enter any guess for H+. Move to the System tab. You can change the guesses for the components (shown in red text in column C) if you wish. An example of the System tab for the chapter problem is shown in Figure  9.6. Nanoql SE took the components you entered and generated equilibria from them. For example, look in cell D11 to see that Nanoql SE

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9.5  |  Nanoql SE   191

Table 9.2 Nanoql SE Instructions for the Chapter Example (Steps in parentheses are optional.) Step Instruction

(1) 2 3 (4) 5 (6)

7

Purpose

Start tab: Tap Clear if a previous system Clear previous system from the Start tab. exists. Start tab: Type x in cell C21. Select the acetate component. Start tab: Type 0.001 in cell D21. Enter total mass for the component. [Start tab should look like Figure 9.5]. System tab: Tap Clear Wkst. Clear previous system from the System tab. System tab: Tap Generate. Generate system. System tab: Adjust pC guesses in cells Adjust initial guesses. E2-­E4 if desired. [System tab should look like Figure 9.6]. System tab: Tap Solve. Solve system.

FIGURE 9.5  Start Tab for the Chapter Problem in Nanoql SE

generated the formula =D4^1*D3^1/I3. You can confirm that this formula is­ Ac H , where K is the equilibrium constant for the equilibrium ­stating HAc K HAc = Ac– + H+ (equilibrium in cell H3, with its value in cell I3). Tap Solve to solve. Nanoql SE will calculate the species concentrations and display them on the System tab. The results will look like Figure  9.6. The calculated equilibrium pH of 3.90 is slightly higher than the pH calculated in Section 9.3 (pH 3.88) because the equilibrium constants are slightly different. Nanoql SE calculates all equilibrium constants from a thermodynamic database.

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192   Chapter 9  Computer Solutions to Chemical Equilibrium Problems

FIGURE 9.6  System Tab for the Chapter Problem in Nanoql SE Prior to Solution

Table 9.3 Nanoql SE Instructions for Generating pC-­ pH Data for the ­Chapter Example (Assume the system has been entered and solved as in Table 9.2.) Step Instruction

Key idea: You can make a pC-­pH diagram in Nanoql SE by varying the concentration of H+

Purpose

1

Start tab: Type 0, 14, 0.5, and x in cells G5-­J5, respectively. Enter pH range.

2

System tab: Type x in cell B18.

Vary a species concentration.

(3)

System tab: Adjust pC guesses in cells E2-­E4 if desired.

Adjust initial guesses.

7

System tab: Tap Vary.

Vary pH.

If you want to draw a pC-­pH diagram after solving the system, then enter the pH range on the Start tab (see Table 9.3). Go back to the System tab, type “x” in cell B18 [next to the label: “Total concentration (H+ for pC-­pH diagram)”] and tap Vary. The results will be written to the Vary Results tab. You can use Excel’s graphing capabilities to make a pC-­pH diagram.

9.6  THE TABLEAU METHOD AND OTHER

EQUILIBRIUM CALCULATION APPS

9.6.1  Expressing Equilibrium Information with Components Tableau method: a common way to set up chemical systems by expressing species as functions of components

In the Tableau method, you construct species from the components. One way to determine how to make up each species from your set of components is to combine equilibrium expressions and write the species concentration as a function of component concentrations. The stoichiometric coefficients for the components are given by the exponents in this expression. Thus, from the equilibrium CO32– + 2H+ = H2CO3 it is easy to see that: [H2CO3] = K[CO32–]1[H+]2. Therefore, we say that H2CO3 is “made up” of one part of the component CO32– and two parts of the component H+.

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9.6  |  The Tableau Method and Other Equilibrium Calculation Apps   193

Thoughtful Pause How would you express the species OH– with components H2O and H+? To “make” OH– from the components H2O and H+, manipulate the equilibrium for the self-­ionization of water (H2O = H+ + OH– ; KW): 1

OH   KW H 2 O H

1

Thus, OH– is “made up” of +1 part of the component H2O and –1 part of the component H+. If you have selected the components carefully, then you should be able to write the concentration of each species as a linear combination of the component concentrations (assuming here that activities and concentrations are interchangeable). For example, consider the simplest aqueous system: pure water, with species H2O, H+, and OH–. The logical components are H2O and H+. Making up each species from components: H2O:

1 part component H2O and 0 parts component H+ equilibrium: H2O = H2O; K = 1 or [H2O] = [H2O]1

H+:

0 parts component H2O and 1 part component H+ equilibrium: H+ = H+; K = 1 or [H+] = H+]1

OH–:

1 part component H2O and –1 part component H+ equilibrium: H2O = H+ + OH–; KW or [OH–] = KW[H2O]1[H+]–1

You can write this information compactly in a tabular form, with components as columns and species as rows: Components

Species

H2O

H+

H2O

1

0

H

+

0

1

OH–

1

–1

Adding in the equilibrium constants for the equilibria used to express the species as components: Components

Species

H2O

H+

logK

H2O

1

0

0 (K = 1)

H

+

0

1

0 (K = 1)

OH–

1

–1

–14 (K = 10–14)

The entries in the table help to recreate the information in the system equilibria. The columns with component headings give the stoichiometric coefficients for each species. Consider the component columns as an n by m matrix, where each element, Xij, gives the stoichiometric coefficient of component j in species i (for n species and

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194   Chapter 9  Computer Solutions to Chemical Equilibrium Problems m components). The log of the equilibrium constant is listed in the rightmost column. For the equilibrium H2O = H+ + OH– : Or:

logKW

14

 log OH

log OH

log H 2 O

log H 2 O

log H

log H logKW

In matrix form: log[H 2 O] log[H ] log[OH ]

1 0 1

0 log[H 2 O] 1 log[ H ] 1

0

0 logKW

In general:

S

XC K,  where 

eq. 9.1

S = n×1 matrix of log of the species concentrations S(i) = log of the concentration of species i X = n×m matrix of component coefficients X(i,j) = coefficient for making up species i from component j C = m×1 matrix of log of the component concentrations C( j) = log of the concentration of component j K = n×1 matrix of log of the equilibrium constants for making up species from components K(i) = log of the equilibrium constant for making up species i One of the advantages of using components is that it is possible to solve for the m component concentrations rather than the n species concentrations. Thus, it is necessary to rearrange eq. 9.1 to solve for C as follows: C = (XTX)–1(XT(S – K)) where XT is the transpose of X (that is, the matrix formed when the rows and columns of X are interchanged) and A–1 is the inverse of A (that is, A–1A = I = the identity matrix = a matrix with 1s in the main diagonal and 0s elsewhere). In the pure water exam1 0 2 1 1 0 1 0 1 , XT ple, you can verify that X , XT X , and 1 2 0 1 1 1 1 (XT X)

1

as required.

1 2 1 . Note that (XT X) 1 (XT X) 3 1 2

1 2 1 2 1 2

2 1

1 2

1 0 0 1

I,

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9.6  |  The Tableau Method and Other Equilibrium Calculation Apps   195

In the tableau, it is customary to leave table entries blank when the stoichiometric coefficients or logK values are equal to zero. Thus, the tableau accounting for the equilibrium information in pure water is: Components H2O

Species

H 2O H+ OH–

1 1

H+

logK

1 –1

–14

Now it is time to come up with a set of components for the chapter problem. Recall that the species list is H2O, H+, OH–, HAc, and Ac–. Thoughtful Pause What components can be used to describe the chapter problem? There are many different components that can be used to describe the system in the chapter problem. Using the approach from Section 9.6.1 (i.e., use the most dissociated forms and always include H2O and H+ as components), suitable components are H2O, H+, and Ac–. The species can be formed from the components as follows: H2O:

1 part component H2O, 0 parts component H+, and 0 parts component Ac– equilibrium: H2O = H2O; K = 1 or [H2O] = [H2O]1

H+:

0 parts component H2O, 1 part component H+, and 0 parts component Ac– equilibrium: H+ = H+; K = 1 or [H+] = [H+]1

OH–:

1 part component H2O, –1 part component H+, and 0 parts component Ac– equilibrium: H2O = H+ + OH– ; KW or [OH–] = KW[H2O]1[H+]–1

HAc:

0 parts component H2O, 1 part component H+, and 1 part component Ac– 1 equilibrium: HAc = H+ + Ac– ; logK = 4.7 or [HAc] = [H+]1[Ac–]1 K

Ac–:

0 parts component H2O, 0 parts component H+, and 1 part component Ac– equilibrium: Ac– = Ac–; K = 1 or [Ac–] = [Ac–]1

The partial tableau is then: Components H2O

Species

H2O H+ OH– HAc Ac–

1 1 1

H+

–1 1

Ac–

1 1

logK

–14 4.7

←Worked example

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196   Chapter 9  Computer Solutions to Chemical Equilibrium Problems

9.6.2  Expressing Mass Balance Information with Components recipe: a way in the Tableau method of expressing the starting materials as functions of the components Worked example →

The bottom half of the tableau will include information on the mass balances. As always, the mass balance constraints are determined by the starting materials (often called the recipe in the Tableau method). It is necessary to express each starting material as a function of the components. One approach for doing this is to write a mass balance for the elements in each starting material as a linear function of the components by determining the necessary stoichiometric coefficients for each component. An example will make this process clearer. The starting materials for the chapter example are HAc and H2O. You need to find the stoichiometric coefficients a, b, c, d, e, and f so that the elements balance in the following expressions, where the starting materials are expressed as being made up of the components: HAc

H2 O

H2 O

H2 O

a

H

d

H

b

Ac

c

e

Ac

f

The unique solution is: a = 0, b = 1, c = 1, d = 1, e = 0, and f = 0. Incorporating this information into the tableau, the complete tableau for the chapter example becomes: Components H2O

Species

Recipe

H2O H+ OH– HAc Ac– HAc H2O

H+

1

Ac–

1 –1 1

1

–14 4.7

1 1 1

1

1

logK

The mass balance information in the tableau is interpreted as follows. The column for each component column gives the stoichiometric coefficients in the mass balances for each component. Thus: TOT H 2 O    H 2 O TOT H

H

TOTAc

HAc

OH  

OH

HAc                

Ac

(Total component concentrations typically are expressed with the symbol TOT in the Tableau method.) The rows in the Recipe section yield the expressions for the total mass of each component: TOT H 2 O

H 2 O  added  55.5 M

TOT H

HAc  added

TOTAc

HAc  added  AcT 1 10 3 M

AcT 1 10 3 M                

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9.6  |  The Tableau Method and Other Equilibrium Calculation Apps   197

Combining: TOT H 2 O  TOT H 

H2 O

H

OH

OH

TOTAc    HAc

55.5 M                 HAc

Ac

1 10 3 M                

1 10 3 M

A final comment on the expression of mass balance information in the tableau: Be aware that the stoichiometric coefficients may not always be positive. An example of negative stoichiometric coefficients in a tableau mass balance statement is given in the next Thoughtful Pause. Thoughtful Pause How would you write the elemental composition of NaOH as a function of the components Na+, H2O, and H+? For example, for NaOH you would write: NaOH = (Na+)1(H2O)1(H+)–1.

9.6.3  Expressing the Electroneutrality Equation with Components The electroneutrality equation can be generated by a linear combination of rows of the tableau. From the mass balances on the components: TOT H    H     OH     HAc   TOTAc    HAc     Ac                   Or: H     OH     Ac You can generate the electroneutrality equation more easily if only one component is charged. Since H+ is usually included as a component, the electroneutrality equation will be given by TOT H if H+ is the only charged component. For example, you can rework the chapter example with components H2O, H+, and HAc. The resulting tableau is shown in Table 9.4. Table 9.4 Complete Tableau with H2O, H+, and HAc as Components Components H2O

Species

Recipe

H2O

H+ OH– HAc Ac– HAc H2O

H+

HAc

logK

1 1

1 –1 –1

1

1 1 1

–14 –4.7

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198   Chapter 9  Computer Solutions to Chemical Equilibrium Problems

9.6.4  Solving Chemical Systems with the Tableau Method The tableau summarizes the equations necessary to solve the chemical system. The equations are usually solved by computer rather than by hand; however, an example of manual calculation with a tableau is given here. Computer solutions are illustrated in Section 9.6.7. The tableau for the chapter example with components H2O, H+, and Ac– is repeated here as Table 9.5 for convenience. From the component columns of the tableau (see also Section 9.6.4): TOT H 2 O 

H2 O

TOT H 

H

TOTAc 

HAc

OH

OH

55.5 M                  1 10 3 M                

HAc

Ac

1 10 3 M

From the species rows of the tableau (ignoring rows with logK = 0): OH

1

HAc

1

KW H 2 O H 1 H K

1

Ac

1

KW 1 H K

H2 O H Ac

1

Table 9.5 Complete Tableau for Worked Example with H2O, H+, and Ac– as Components Components

Species

H2O

H2O

H+

1

–1

HAc Ac Recipe

1

HAc

1 1

 Here, K refers to the original acid dissociation constant: K = 10–4.7.

1

–14 1 1

–

H2O

logK

1

H+

OH–

Ac–

1

1

4.7

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9.6  |  The Tableau Method and Other Equilibrium Calculation Apps   199

Because of the large fixed concentration of water, the TOT H2O equation becomes meaningless. Ignoring the TOTH2O equation and recognizing that the mole fraction of H2O is 1 in the [OH–] expression: TOT H

H

OH

TOTAc

HAc

OH

KW H

HAc

1 H K

HAc

Ac

1 10 3 M

1 10 3 M

Ac

Substituting the expressions for [OH–] and [HAc] in the expressions for TOT H and TOTAc yields: TOT H



TOTAc 



H H

KW H

1 H K

Ac

1 10 3 M

eq. 9.2 eq. 9.3

Ac     Ac    1 10 3 M 

Equations 9.2 and 9.3 represent two nonlinear equations in two unknowns, namely [Ac–] and [H+]. As noted above, the TOTH2O equation usually adds little information since the concentration of water is large and constant. Thus, water is sometimes merely assumed to be a component and not listed explicitly in the tableau as a component, species, or starting material. For example, a simplified tableau for the chapter problem is shown in Table 9.6.

Table 9.6 Simplified Tableau for Worked Example with H+ and Ac– as Components Components H

+

Species

H2O H+ OH

logK

1 –

HAc

–1 1

HAc

–14 1 1

Ac

–

Recipe

Ac–

1

1

4.7

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200   Chapter 9  Computer Solutions to Chemical Equilibrium Problems

9.6.5  Summary of the Tableau Method The steps involved in making a tableau can be summarized as follows (also listed in Appendix C, Section C.5): Key idea: To nuke a tableau (1) make a list of species and equilibria, (2) select the components, (3) draw the initial tableau, (4) enter the component stoichiometric coefficients from equilibria, and (5) enter the component stoichiometric coefficients from mass balances

Step 1: Make a list of species and equilibria. Step 2: Select the components. The usual approach is to include H+ (and sometimes H2O) and the most dissociated forms of each species family as components. Step 3: Draw the initial tableau. Make one column for each component and one column labeled logK. Make rows for each species and rows at the bottom of the tableau for each starting material. Step 4: Enter the component stoichiometric coefficients from equilibria. Rearrange the equilibria to express each species as a product of the components raised to a power. For example, the self-­ionization of water can be rearranged to: [OH–] = KW[H2O][H+]–1. Enter the coefficients for each component on the row for that species. Step 5: Enter the component stoichiometric coefficients from mass balances. Express the elemental composition of each starting material as a linear combination of the components. For example, with components Na+, H2O, and H+: NaOH = (Na+)1(H2O)1(H+)–1. Enter the coefficients for each component on the row for that starting material.

9.6.6  Using Software Based on the Tableau Method Software packages based on the Tableau method require the user to select species and enter the tableau information. Tableau information generally is entered as a table, similar to Figure  9.7. After entering the tableau information, the system is solved by the software for the component concentrations. Species concentrations are back-­ calculated by the software and can be examined in tabular output or plotted. The true power and beauty of the Tableau method is shown more clearly in larger examples. The main advantages of the Tableau method are computational efficiency and expandability to larger systems. For example, the final equations in the Tableau method (eqs. 9.2 and 9.3) usually are solved numerically. The system has been simplified to two unknowns, namely, the concentrations of the components, [Ac–] and [H+] (assuming that the component H2O has mole fraction = 1). A species-­ based approach would yield a system of four equations in four unknowns, namely, [H+], [OH–], [Ac–], and [HAc] (assuming that H2O has mole fraction = 1). Since the number of computational steps in nonlinear solution methods generally increases dramatically with the number of unknowns, the Tableau method offers great computational efficiency. The main disadvantage with the Tableau method is that the user must preprocess the system information to generate the tableau. The process of making a tableau will become more familiar to you with practice. In addition, the choice of components sometimes can influence the results of the calculation.

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9.7  |  Visual MINTEQ   201

9.7  VISUAL MINTEQ This chapter concludes with a nice implementation of the Tableau method using a visual interface in Visual MINTEQ. This app was written by Jon Petter Gustafsson (see Table 9.1 for details). To solve the chapter example with Visual MINTEQ, you first must choose how the app will calculate pH. There are three options, labeled “Fixed at. . .,” “Calculated from mass balance,” and “Calc. from mass & charge balance.” Choose to calculate the pH from the mass and charge balance. Now you can add the component acetate. Check the checkbox to “Show organic components” and then select “Acetate-­1” from the list. Set its total concentration to 0.001 molal. Now tap the “Add to list” button to add the acetate component to your component list. The procedure is summarized in Table 9.7. The setup screen will look like Figure 9.7. Table 9.7 Visual MINTEQ Instructions for the Worked Example Step

Instruction

Purpose

1

Select “Calc. from mass & charge balance” next to the pH label. Make sure the “Show organic components” ­checkbox is checked. Select the component “Acetate-­1” from the ­dropdown menu. Type 0.001 in the Total concentration box.

Select method for calculating pH. Add organics.

2 3 4 5

6

Select acetate component.

Enter total mass for the ­component. Tap Add to list. Add component and its total concentration. [Entry screen should look like Figure 9.7.] Tap Run MINTEQ. Solve system.

FIGURE 9.7  Entry Screen for the Chapter Example in Visual MINTEQ

←Worked example, Visual MINTEQ

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202   Chapter 9  Computer Solutions to Chemical Equilibrium Problems

FIGURE 9.8  Output Screen for the Chapter Example in Visual MINTEQ

Tap the “Run MINTEQ” button to solve the system. The output screen is shown in Figure  9.8. The calculated equilibrium pH of 3.91 is slightly higher than that ­calculated with Nanoql SE, probably because of slight differences in the thermodynamic database.

9.8  CHAPTER SUMMARY Computer software can speed up equilibrium calculations significantly and allow for the solution of more complex systems than with manual calculations. Equilibrium calculations lend themselves to solution by spreadsheets. Orderly arrangement of the spreadsheets will help you keep track of calculated results. The computer programs Nanoql and Nanoql SE were written to simulate the solution process taught in Part II of this book. Nanoql SE requires no or very little preprocessing of the chemical system before entering the system into Excel. The Tableau method was developed to be computationally efficient. The main unit in Tableau-­based programs is the component. MINEQL is an example of an equilibrium calculation software package that uses the Tableau method. It requires some (often significant) preprocessing of the chemical system by the user to write it in terms of components and not species.

9.9  PART II CASE STUDY: HAVE YOU HAD Worked example → Spreadsheet available Nanoql SE Example

YOUR ZINC TODAY?

Recall that the Part II case study concerns the speciation of zinc in the mouth and stomach upon consumption of a zinc acetate tablet. The system was set up in Section 7.6. If you use computer software, it is fairly straightforward to calculate the concentrations of all species at any pH of interest. As an example, a spreadsheet for the system is discussed in Section E.5.2. The Nanoql SE chemical system for the Part II case study may be found in Section F.4.2. With either tool, you can regenerate the species concentrations at pH 3 and 6.5 from Section 7.6 and the pC-­pH diagram from Section 8.6. Generating a tableau for the case study system may be difficult now but will be within your abilities by the conclusion of Part IV of this text.

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Historical Note: ALGOL to VBA   203

CHAPTER KEY IDEAS • Many environmental chemistry problems are difficult to solve manually. • To solve equilibrium problems with a spreadsheet, enter equations for species (based on equilibria) and vary the –log(conc.) values of components until the mass and charge balances are satisfied. • Write out the equilibrium expressions and write notes to help you remember the details of the chemical system and calculation. • Check the optimization results to see if sufficient resolution has been obtained by examining the predicted total masses and charge balance. • It is common practice to use fully dissociated ions, H2O, and H+, as components. • Nanoql SE solves small-­ to moderate-­sized aquatic equilibrium problems in a spreadsheet environment. • You can make a pC-­pH diagram in Nanoql SE by varying the concentration of H+. • To make a tableau (1) make a list of species and equilibria, (2) select the components, (3) draw the initial tableau, (4) enter the component stoichiometric coefficients from equilibria, and (5) enter the component stoichiometric coefficients from mass balances.

CHAPTER GLOSSARY recipe:  a way in the Tableau method of expressing the starting materials as functions of the components Tableau method:  a common way to set up chemical systems by expressing species as functions of components

HISTORICAL NOTE: ALGOL TO VBA Computational techniques for aquatic systems have paralleled computing power for over 70 years. Stuart Brinkley laid out the equations necessary for equilibrium calculations in 1947 (Brinkley 1947). (These equations were summarized in Section 6.6.) For comparison, ENIAC (usually regarded as the first large-­scale electronic computer) was announced to the public in 1946. Brinkley recommended a linear algebra approach, with solution of the nonlinear equations through use of the Newton-­Ralphson method. These methods were coded into two software packages: COMICS (Concentrations in mixtures of Metal Ions and Complexing Species; Perrin and Sayce 1967) and HALTAFALL (Ingri et al. 1967).2 HALTAFALL came out of the research group of Lars Gunnar Sillén (see the Historical Note in Chapter 8). Software based on the HALTAFALL algorithm, originally written in ALGOL, is still in use. One of the most influential equilibrium calculation software packages was ­developed by James J. Morgan and his student François M.M. Morel at Caltech

 HALTAFALL expanded a program called HALTA to include solids. The name HALTAFALL comes from the Swedish halt concentration + falla precipitate. The authors added cryptically “. . .a Scandinavian reader may find other associations as well.”

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204   Chapter 9  Computer Solutions to Chemical Equilibrium Problems (Morel and Morgan 1972). Their work became the basis of MINEQL and many of the other packages listed in Table 9.1. Morel and Morgan demonstrated their program with an impressive chemical system consisting of 791 dissolved species, one gas-­ phase component, and 83 possible solid phases. The system converged to a solution after 56 iterations in less than 60 seconds. From Brinkley’s framework at the dawn of electronic computing to programs written in FORTRAN for mainframe computers (MINEQL) to BASIC programs for personal computers (MICROQL) to apps based on Microsoft’s .NET Framework 4.5 (Visual MINTEQ) to educational tools embedded in Excel (Nanoql SE, via Visual Basic for Applications, or VBA), computational tools for aqueous equilibrium calculations have adapted to user needs and computing power.

PROBLEMS 1. List the component(s) you would use for the following starting materials. There is more than one answer for each system. For this problem, follow common practice to include H2O and H+ as components in systems containing water and to otherwise specify the most dissociated species as the component(s). a. Water and HOCl b. Water and H2CO3 c. Water and NaHCO3 2. Modify the template spreadsheet to find the equilibrium pH of a 1×10–3 M HOCl solution. The pK for the dissociation of HOCl is 7.54. 3. Use a spreadsheet to find the equilibrium pH of a 1×10–3 M sodium acetate solution. Assume that sodium acetate dissociates completely to Na+ and Ac–. In solving the problem, set up a spreadsheet similar to Figure 9.1. 4. Use a spreadsheet to find the equilibrium pH of a solution containing 1×10–3 M acetic acid and 1×10–3 M NaOH. Assume that sodium hydroxide dissociates completely to Na+ and OH–. 5. Enter the chapter example into Nanoql SE and solve for the equilibrium pH. Make a pC-­pH diagram using Nanoq SE. Show that the only place on the pC-­pH diagram where charge is balanced is at the equilibrium pH. 6. Use Nanoql SE or Visual MINTEQ to find the equilibrium pH of a 1×10–4 M HOCl solution. 7. Use Nanoql SE or Visual MINTEQ to find the equilibrium pH of a 1×10–3 M sodium acetate solution and plot

its pC-­pH diagram. Hint: You can take the system you created in Problem 5, add the component Na+, and set its total concentration equal to 1×10–3 M. 8. Generate a system with components H2O, H+, and S2– in Nanoql SE to determine the species that form in the system. Confirm that the mass balance and electroneutrality equations produced by Nanoql SE are correct. 9. Explain the meanings of each row and column in the tableau in Figures 9.4 and 9.5. 10. Enter the chapter example in a computer program that employs the Tableau method and solve for the equilibrium pH. Make a pC-­pH diagram. 11. Write the tableau for the sodium acetate system in Problem 3. Use a computer program that employs the Tableau method to solve for the equilibrium pH. Draw a pC-­pH diagram. 12. Using the computer method of your choice, find the equilibrium pH for the following systems: a. 1×10–2 M NH3 b. 1×10–2 M NH4Cl c. 1×10–3 M H2CO3 13. The chapter example can be set up using six different sets of components. Two sets were illustrated in the tableau in Section 9.6.3 and Figure 9.4. List the four other sets of components and the corresponding tableau for each set.

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Chapter References   205

CHAPTER REFERENCES Brinkley, S.R. (1947). Calculation of the equilibrium composition of systems of many constituents. J. Chem. Phys. 15(2): 107–110. Butler, J.N. (1998). Ionic Equilibrium: Solubility and pH Calculations. New York: John Wiley & Sons, Inc. Ingri, N., W. Kakolowicz, L.G. Sillén, and B. Warnqvist (1967). High-­ speed computers as a supplement to graphical methods – V. HALTAFALL, a general program for calculating the composition of equilibrium mixtures. Talanta 14(11): 1261–1286.

Morel, F. and J. Morgan (1972). A numerical method for computing equilibria in aqueous chemical systems. Environ. Sci. Technol. 6(1): 58–67. Morel, F.M.M. and J.G. Hering (1993). Principles and Applications of Aquatic Chemistry. New York: John Wiley & Sons, Inc. Perrin, D.D. and I.G. Sayce (1967). Computer calculation of equilibrium concentrations in mixtures of metal ions and complexing species. Talanta 14(7): 833–842.

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PA R T I I I

Acid–Base Equilibria in Homogenous Aqueous Systems Chapter 10: Getting Started with Acid–Base Equilibrium in Homogeneous ­Aqueous Systems Chapter 11 : Acids and Bases Chapter 12: Acid–Base Titrations Chapter 13: Alkalinity and Acidity

A base is best described as an acceptor of hydrogen nuclei. –Thomas M. Lowry (1923) Part III Haiku Acids and bases, Homogeneous systems: Use the pKa. The general principles of chemical equilibrium can similarly be used in discussion with a weak base... and also of salts formed by weal acids and weak bases. –Linus Pauling (1970)

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C H A P T E R   10

Getting Started with Acid–Base Equilibrium in ­Homogenous Aqueous ­Systems 10.1 INTRODUCTION At this point, you have nearly all the tools you need to solve chemical systems at ­equilibrium. You know how to set up chemical systems by listing species, writing ­pertinent equilibria, devising mass balances, and writing the electroneutrality condition. You know how to solve systems by making approximations to simplify the mathematics (and subsequently checking the assumptions), converting systems to one equation in one unknown, or grinding it out by using algebra or computer software. Given that you now know nearly everything you need to know, you may wonder why there are another dozen or so chapters in this text. The answer is simple: There is a lot more to aquatic chemistry than being able to solve for species concentrations at equilibrium. The richness of the field comes from the variety of reaction types and the myriad applications in natural and engineered systems. Another reason for the rest of the text is that you need more than algebra to understand chemical processes in the environment. To be sure, performing equilibrium calculations is a necessary part of your mastery of aquatic chemistry. However, fluency in aquatic chemistry also requires a feeling for important reactions and species. You will learn to estimate the values of the primary variables and the effects of chemical change on speciation. You will begin this odyssey in Part III of this text, where you will explore the first type of chemical equilibrium: acid–base equilibria in homogeneous systems.

Key idea: Mastery of aquatic chemistry requires a feeling for important reactions and species, as well as fluency in equilibrium calculations

10.2  HOMOGENEOUS SYSTEMS 10.2.1  What Are Homogeneous Systems? A homogeneous system (from the Greek homo-­same + -­genes kind) consists of only one phase. In this text, homogeneous system will refer to an aqueous chemical system with no gas or solid phases. In addition, exclude for the moment other liquid phases,

homogeneous system: a chemical system consisting of only one phase

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210   Chapter 10  Getting Started with Acid–Base Equilibrium in ­Homogenous Aqueous ­Systems

dissolved gases: dissolved, aqueous-­phase species that may equilibrate with a gas phase

dissolved solids: dissolved, aqueous-­phase species that may equilibrate with a solid phase

such as pure benzene or oils or gasoline. To put it another way, we are considering systems in which all the species (except one) are dissolved aqueous species that carry the suffix (aq). The one exception is water itself. Water is the only pure liquid included in the homogeneous systems in this text.1 The definition of a homogeneous system seems fairly straightforward. Unfortunately, several classes of species carry names that are somewhat ambiguous. For example, what are we to make of the dissolved gases? Are they dissolved species? Are they gas-­phase species? Are you supposed to include or exclude them in homogeneous systems? Dissolved gases are dissolved aqueous species that may equilibrate with a gas phase. For example, the species O2(aq) (also called dissolved oxygen, or DO) is an appropriate species for inclusion in a homogeneous system since it is a dissolved, aqueous-­phase species. At sufficiently high concentrations (or appropriate temperature and pressure conditions), the dissolved gases can form bubbles. At this point, you no longer have a homogeneous system and must use the tools of Chapter 18 to calculate species concentrations at equilibrium. Similarly, dissolved solids are considered to be dissolved, aqueous-­phase species. Under the proper conditions, dissolved solids can precipitate, forming one or more solid phases. Again, precipitation creates a nonhomogeneous system. The tools presented in Chapter 19 will be needed to calculate species concentrations.

10.2.2  How Realistic Are Homogeneous Models? In Parts III and IV of this text, we will assume that the systems are homogeneous. By assuming that the system is homogeneous, we clearly are placing large restrictions on the applicability of the model.

Thoughtful Pause Can you think of a truly homogeneous natural or engineered system?

Key idea: Most waters are not in homogeneous systems

In fact, it is difficult to conjure up a system with any environmental relevance that is truly homogeneous. For example, the top layers of lakes and rivers may be in equilibrium with the atmosphere. To model these waters, it is necessary to use a nonhomogeneous system comprising aqueous and gas phases. Similarly, the bottom layers of a lake may be in equilibrium with the solids in the sediments, necessitating a nonhomogeneous system. Even the middle layers may contain suspended solids in chemical equilibrium with the water. Suspended solids are solid phases. Groundwaters usually are in equilibrium with the surrounding soil. Even water flowing full in pipes is not necessarily homogeneous. Important equilibria may occur between water and the walls of the pipe. Turbulent flow may induce cavitation and thus a small gas phase. Reasonable descriptions of each of these waters may require nonhomogeneous systems. So when can you assume that a system is homogeneous? We usually use the homogeneous assumption for three reasons. First, the system of interest may be nearly

 Any single pure liquid could be considered. For example, much of the text could be repeated by using, say, ­ ethanol instead of water. m

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10.3  |  Types of Reactions in Homogeneous Systems   211

homogeneous. For example, waters at mid-­depth in a deep lake may be sufficiently isolated from the atmosphere and sediments. If suspended solids do not dissolve appreciably, then a homogeneous model may be reasonable. Second, the system may not be homogeneous at all, but the equilibria with other phases may not affect the processes of interest. A good example is dissolved nitrogen, N2(aq). Dissolved nitrogen is nearly inert (see Section 6.3.6) and nitrogen exchange between the atmosphere and water usually is ignored in equilibrium calculations. Similarly, if you were interested in the conversion of OCl– to HOCl in an open beaker, you would most likely ignore the exchange of chlorine between the atmosphere and water. How do you know if nonhomogeneity affects the processes of interest? You do not know unless you perform the equilibrium calculations twice, first assuming homogeneous conditions and then repeating the calculation in the presence of other phases. Third, we sometimes impose homogeneous conditions as an idealized case. For example, suppose you wish to understand the relationships between apparently related compounds such as carbonic acid, bicarbonate, and carbonate. It might make sense to study the chemistry first as an isolated homogeneous system without influence from gas or solid phases. After you complete the analysis of the idealized case, more realistic conditions could be introduced.

Key idea: Homogeneity is assumed when the system of interest is nearly homogeneous, not significantly affected by exchange with other phases, or treated on purpose as an ideal case

10.3  TYPES OF REACTIONS

IN HOMOGENEOUS SYSTEMS

Parts III and IV of this text are concerned with reactions that occur in homogeneous systems. The focus is on two chemical species: H+ (the proton) and e– (the electron). Recall from Section 1.3 that these two species control a great deal of aquatic chemistry. As a result, pH (= –log10{H+}) and pe (= –log10{e–}) are logical choices for primary variables. Protons and electrons can undergo two types of reactions. First, the proton or electron can be transferred from one chemical species to another. Proton transfer reactions are called acid–base reactions and are the subject of Part III. Electron-­transfer reactions are called oxidation-­reduction (or redox) reactions. Redox reactions are the focus of Chapter 16 in Part IV. Second, the proton or electron can be shared between chemical species. The sharing of a proton results in a special kind of low-­energy bond called a hydrogen bond (see Section 1.4.1). Hydrogen bond formation allows for very rapid transfer of protons. As a result, acid–base reactions usually are very fast in water. Electron-­sharing reactions are called complexation reactions. Complexation reactions are the subject of Chapter 15 in Part IV. The names for the proton-­ or electron-­transfer reactions and proton-­ or electron-­sharing reactions are summarized in Table 10.1. Table 10.1 Summary of Proton or Electron Transfer and Sharing Reactions Species

Proton (H ) Electron (e–) +

Transfer Reactions

Sharing Reactions

acid–base reactions redox reactions

hydrogen bond formation complexation reactions

Key idea: ­Protons and electrons can be transferred or shared between chemical species

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212   Chapter 10  Getting Started with Acid–Base Equilibrium in ­Homogenous Aqueous ­Systems

10.4  THE WONDERFUL WORLD

OF ACIDS AND BASES

10.4.1  Equilibrium Calculations with Acids and Bases

Worked example →

In Chapter 11, you will learn how to perform and interpret equilibrium calculations with systems containing acids and bases. You have been exposed to equilibrium calculations with simple acids and bases in Chapters 7 and 8. Chapter 11 will prompt you to explore systems of acids and bases provide much greater detail. Before you leap into the details of Chapter 11, a qualitative view of the effects of chemical addition on equilibrium pH is in order. This exercise will help develop your intuition with acids and bases in water. Recall that pure water has an equilibrium pH of 7.0 at 25°C. You saw in Chapter 7 that when some compounds were added to water, the equilibrium pH was less than 7. Other compounds produced an equilibrium pH greater than 7 when added to water.

Thoughtful Pause How can you predict whether the equilibrium pH will be greater than or less than 7?

Key idea: The effects on pH of adding an acid or base to water are reflected in the charge balance

The key to estimating whether the equilibrium pH will be greater than, less than, or about equal to 7 is in the charge balance. Recall that for pure water, the charge balance is:

OH

H

eq. 10.1

Now add an uncharged compound to water and do the thought experiment described in Section  6.3.2: Look for its hydrolysis products. Say that the chemical addition produces only one monovalent (i.e., singly charged) anion as the added material reacts with water (i.e., one anion other than H+ or OH–). Assume for the moment that this initial hydrolysis product does not react further with water or anything else.

Thoughtful Pause What is the new electroneutrality equation? The new electroneutrality equation will be:

H

new

OH

new

A

new



eq. 10.2

where A– is the anionic hydrolysis product. Now, how does the hydrogen ion concentration in eq.  10.2 compare with the hydrogen ion concentration in eq.  10.1? If the terms in eqs. 10.1 and 10.2 were unconstrained variables, then you have no idea

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10.4  |  The Wonderful World of Acids and Bases   213

how [H+]new compares to [H+]! However, [H+] and [OH–] are constrained by the self-­ ionization of water, so: [H+][OH–] = [H+]new[OH–]new = 10–14 at 25°C (assuming the concentrations and activities are interchangeable).

Thoughtful Pause Given that the hydrogen and hydroxide ion concentrations are constrained, how do [H+]new and [H+] compare? Even if [A–]new is small (say, 10–7 M), the hydrogen ion concentration must increase and the hydroxide ion concentration must decrease when the hydrolyzing compound in this example is added. In other words, [H+]new > [H+]. This point is illustrated in Figure 10.1. Note that the electroneutrality condition is satisfied for both conditions in Figure 10.1. When the compound in the example is added, [H+] increases and [OH–] decreases to maintain the charge balance. If [H+]new > [H+], then the pH after addition must be smaller than the pH before addition (assuming that activity and concentration are interchangeable in both systems). Thus, adding a compound that produces anions but not cations will cause the pH to decrease. This little exercise raises two questions. First, how can an added compound produce anions but not cations? After all, if a salt such as NaCl(s) is added to water, both a cation (Na+) and an anion (Cl–) are produced. In fact, an electron balance tells you that it is impossible to add a neutral compound that dissociates to form an anion but not a cation. However, there are compounds that hydrolyze to produce one or more anions and only the cation “produced” is H+. Thus, no new cations are produced compared to pure water, and the situation in Figure 10.1 is valid. Compounds that hydrolyze to form only anions and H+ are part of the class of species called acids.2 The second question raised by Figure  10.1 is as follows: How do we know [A–]new = 10–7 M? In general, we do not know [A–]new without doing a complete equilibrium calculation. The value of [A–]new = 10–7 M was used for illustration. However, as long as [A–]new is greater than zero, then [H+]new > [H+]. Again, Chapter 11 will provide the quantitative tools to allow for the calculation of the concentration of each species at equilibrium. 2E–07

Concentration (M)

Clean Water

[A–] = 10–7 M

2E–07 [A–]

1E–07

5E–08 [H+]

[OH–]

[H+]

[OH–]

0E+00 FIGURE 10.1  Charge Balances with and without Chemical Addition

 The many definitions of acids will be explored in Chapter 11. The concept of an acid is much more general than just “compounds hydrolyzing in water to form anions and H+.” As a simple example, ammonium (NH4+) is an acid. It dissociates to form ammonia (NH3) and H+.

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10.4.2  Mixtures of Acids and Bases Nearly all aquatic systems contain mixtures of acids and bases. In fact, as you will see in Chapter 11, water itself is both an acid and a base. Therefore, the addition of an acid or base to water means that you have a mixture of acids and bases. Thus, to determine the primary variable pH in a real water, it is necessary to be able to handle mixtures of acids and bases. Another important example of acid–base mixtures in engineered systems is the neutralization of acidic (or basic) wastes. A simple example will demonstrate how you can treat mixtures of acids and bases qualitatively. Suppose you take the solution described in Section 10.4.1 and add NaOH to it. Before NaOH is added, the charge balance is: Key idea: Charge balances also change when acids and bases are mixed

H

A

OH

eq. 10.3

(Equation 10.3 is just eq. 10.2 with the subscript “new” removed for clarity.) How will the charge balance change after adding NaOH? Sodium hydroxide is expected to dissociate completely to Na+ and OH–. The new charge balance is:

H

Na

OH

A

eq. 10.4

Note that eq.  10.4 is just eq.  10.3 with an additional term, the concentration of sodium ion, on the left-­hand side. There are only three ways to balance eq.  10.4 after adding the additional term on the left-­hand side: (1) [H+] could decrease; (2) [OH–] could increase; or (3) [A–] could increase. Note that possibilities (1) and (2) are really the same since [H+][OH–] = 10–14. Thus, we really have only two possibilities: either [H+] decreases and [OH–] increases, or [A–] increases. Now [A–] cannot increase without limit. It is constrained by both equilibrium constraints and by at least one mass balance. For certain values of [Na+] (i.e., for certain doses of NaOH), the concentration of A– is very constrained. At these NaOH doses, the pH increases a lot since [OH–] must increase to maintain the charge balance. At other values of [Na+], the concentration of A– is not very constrained. At these NaOH doses, the pH increases very little, since [A–] can increase within its constraints to maintain the charge balance. In Chapter  12, you will learn to translate these qualitative arguments into quantitative tools as solution techniques for mixtures of acids and bases are developed. You will find that mixtures can be analyzed by a small extension of the principles learned in Chapter 11. Why devote an entire chapter to a small extension of Chapter 11? The analy­sis of mixtures of acids and bases is a good example of the importance of developing intuition and approximations for aquatic systems. In many cases, you can sketch the dependence of the equilibrium pH on the added acid or base concentration without doing any calculations at all. You will develop this “feel” for acid–base ­mixtures in Chapter 12. Key idea: Water quality parameters can be developed to quantify the response of a system to the addition of acid or base

10.4.3  Responses to Acidic or Basic Inputs The qualitative analysis of acid–base mixtures in the previous section can be generalized as follows: Aquatic systems vary greatly in their response to acid or base additions depending on their chemical composition. As stated in Section 10.4.2, the tools for quantifying that response will be developed in Chapter 12. In Chapter 13, water-­quality parameters

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10.7  |  Part III Case Study: Acid Rain   215

that influence and help explain the response to acid or base addition will be elucidated. In particular, you will develop a set of related parameters that describe how the system composition influences the response to acidic or basic inputs. One important parameter is alkalinity. The meaning and uses of alkalinity are presented in Chapter 13.

10.5  PART III ROADMAP Part III of this book discusses the applications of equilibrium calculation techniques to systems containing acids and bases. In addition, an intuitive feel for the behavior of acids and bases will be cultivated in Part III. By the conclusion of this part of the text, you should know what acids and bases are and have a qualitative and quantitative understanding of their behavior in water. In Chapter 11, the history of the concepts of acids and bases is reviewed, culminating in their definitions. Calculation techniques are developed for systems of individual acids and bases. Dimensionless parameters are developed to describe the relative concentrations of acids and bases in the same chemical family. By the conclusion of Chapter  11, you should be very comfortable with answering quantitative questions about individual acids and bases in aquatic systems. Chapter 12 provides solution techniques for mixtures of acids and bases. This allows for the analysis of much more complicated and realistic problems in aquatic chemistry. Graphical analysis techniques play an important role in summarizing how the equilibrium pH changes with acid or base addition. In Chapter 12, you will find that the change in equilibrium pH from adding acids or bases depends on the system. In other words, each system responds differently to acid or base addition. In Chapter 13, the tools for quantifying the response of a system to acid or base addition will be developed.

10.6  CHAPTER SUMMARY In this chapter, some qualitative ideas were developed to aid in your understanding of acids and bases in homogeneous systems. A homogeneous system in this text is an aqueous chemical system with no gas or solid phases. Many systems of environmental interest are not homogeneous. However, we use homogeneous models when the system of interest is (1) nearly homogeneous, (2) not significantly affected by exchange with other phases, or (3) treated on purpose as an ideal case. The species H+ (proton) and e– (electron) are related to the primary variables pH and pe, respectively. Both protons and electrons can undergo two types of reactions: transfer and exchange. Part III deals with proton-­transfer reactions. The details of equilibrium calculations with proton-­transfer reactions are considered in Chapters 11 through 13. The charge balance was used in this chapter as a qualitative tool in assessing the effects on pH of acid addition. A more quantitative discussion is the focus of the remainder of this part of the text.

10.7  PART III CASE STUDY: ACID RAIN The application of scientific principles to analyze and solve pollution problems is relatively recent. As with all scientific ventures, the analysis of pollution problems

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216   Chapter 10  Getting Started with Acid–Base Equilibrium in ­Homogenous Aqueous ­Systems has matured. Until the third quarter of the twentieth century, it was common to segregate pollution by the type of media impacted. Air, water, and solid waste pollution were studied separately. Only in the last 40 years or so has the strong relationship among air, water, and solid waste pollution been recognized. The focus now is on intermedia transport of pollutants (e.g., transport from air to water or water to air) and multimedia effects. A prime mover in the development of a multimedia focus is acid rain. It is now well-­established that certain air pollutants from natural and anthropogenic sources can dissolve in rain, snow, and fog to produce highly acidic conditions. For example, sulfur dioxide, produced from the combustion of coal and other fossil fuels, can react with water in the troposphere to produce sulfuric acid, H2SO4(aq). Similarly, oxides of nitrogen [NO(g) and NO2(g)], produced primarily from transportation sources, can be oxidized and can combine with atmospheric water vapor to form nitric acid, HNO3(aq). As a final example, hydrochloric acid, HCl, is produced in the gas phase from chloride formed from the combustion of chlorine-­containing polymers (such as polyvinyl chloride, PVC) and liberated from volcanoes. Thus, fairly acidic rain can be produced. In this case study, the impact of acidic inputs on the pH of rainfall will be examined. In addition, the effects of acid rain on other waters will be explored. In Chapter 11, the pH of acid rain will be determined. The impact of acid rain will be elucidated in Chapter 12, where the mixture of acid rain and a surface water will be analyzed. In Chapter 13, new water quality parameters will be used to simplify the calculations on the impacts of acid rain on surface water pH.

CHAPTER KEY IDEAS • Mastery of aquatic chemistry requires a feeling for important reactions and species, as well as fluency in equilibrium calculations. • Most waters are not in homogeneous systems. • Homogeneity is assumed when the system of interest is nearly homogeneous, not significantly affected by exchange with other phases, or treated on purpose as an ideal case. • Protons and electrons can be transferred or shared between chemical species. • The effects on pH of adding an acid or base to water are reflected in the charge balance. • Charge balances also change when acids and bases are mixed. • Water quality parameters can be developed to quantify the response of a system to the addition of acid or base.

CHAPTER GLOSSARY dissolved gases:  dissolved, aqueous-­phase species that may equilibrate with a gas phase dissolved solids:  dissolved, aqueous-­phase species that may equilibrate with a solid phase homogeneous system:  a chemical system consisting of only one phase

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Historical Note: “An Evil of the Highest Magnitude”   217

HISTORICAL NOTE: “AN EVIL OF THE HIGHEST MAGNITUDE” The Part III case study is about acid rain. The term “acid rain” was coined by Scottish chemist Robert Angus Smith (1817–1884). Smith was born in Glasgow. He attended the University of Glasgow for one year, studying for the ministry. Smith later went to Germany as a tutor, eventually studying chemistry under Justus von Liebig (1803–1873) at the University of Giessen. Smith’s professional life was spent in Manchester, perhaps the world’s first industrial city (Hall 1998). Surrounded by industrial emissions, Smith turned his attention to the effects of pollution on the environment. His neologism was published in 1858: It has often been observed that the stones and bricks of buildings, especially under projecting parts, crumble more readily in large towns, where much coal is burnt, than elsewhere. Although this is not sufficient to prove an evil of the highest magnitude, it is still worthy of observation, first as a fact, and next as affecting the value of property. I was led to attribute this effect to the slow, but constant, action of the acid rain. (Smith 1858)

Hypothesizing a link between coal combustion, rain acidification, and the dissolution of building materials was a pretty impressive feat for mid-­nineteenth century science. (For the chemistry of why acid rain dissolves some building materials, see Section 19.5.2.)

Crystals from the Evaporation of Manchester Rain (Smith  1872/james-hixon/Public domain) A partial list of Smith’s publications reads like a primer on environmental engineering and science: • On Sewage and Sewage Rivers (Smith 1855) • Disinfectants and Disinfection (Smith 1869) • Air and Rain: The Beginnings of a Chemical Climatology (Smith 1872) In spite of this output, Smith’s work was unappreciated in his time.3 One professional obituary read: To keep the air in our towns fresh and wholesome, to restore the water of our streams to its pristine clearness, to preserve the freshness and verdure of the fields and woods, to  For more on Smith’s reputation and legacy, see Eyler (1980).

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218   Chapter 10  Getting Started with Acid–Base Equilibrium in ­Homogenous Aqueous ­Systems sweeten the atmosphere of the crowded dwellings in cities,—­this was the kind of work to which Smith dedicated his life, and at which he laboured to the very last. There have been greater chemists, no doubt; his name is not associated with any fundamental discovery in chemistry, and his attempts at theorising were not always very happy; but in his true vocation, as the chemist of sanitary science, Smith worked alone, and we have yet to find the man on whom his mantle has fallen.4 (Thorpe 1884)

CHAPTER REFERENCES Eyler, J.M. (1980). The conversion of Angus Smith: the changing role of chemistry and biology in sanitary science, 1850–1880. Bull. Hist. Med. 54(2): 216–234. Hall, P. (1998). Cities in Civilization: Culture, Innovation, and Urban Order. London: Weidenfeld & Nicolson. Lowry, T.M. (1923). The uniqueness of hydrogen. J. Soc. Chem. Ind. 42(3): 43–47. Pauling, L. (1970). General Chemistry. Mineola, NY: Dover Publ., p. 482.

Smith, R.A. (1859). XVIII.–On the air of towns. Q. J. Chem. Soc. XI: 196–235. Smith. R.A. (1869). Disinfection and Disinfectants. Edinburgh: Edmonston and Douglas. Smith. R.A. (1872). Air and Rain. The Beginnings of a Chemical Climatology. London: Longmans, Green, and Co. Thorpe, T. (1884). Robert Angus Smith. Nature 30: 104–105.

Smith. R.A. (1855). Sewage and Sewage Rivers. Memoirs of the Literary and Philosophical Society of Manchester, session 1854–5, Vol. XII: 155–175.

 Or woman. . . You no doubt have noticed that the Historical Notes so far seem to dwell on “old white men.” This reflects the dearth of opportunities for women and people of color to contribute to Western science in the early days of environmental science and engineering. For some counterexamples, see the Historical Note in Chapter 23.

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C H A P T E R   11

Acids and Bases 11.1 INTRODUCTION As stated in Section 1.3, pH is a primary variable in aquatic systems. Why the emphasis on pH? We choose pH as a primary variable of interest because H+ has such a strong influence on aquatic chemistry and toxicity endpoints and because pH is easily measured. In this chapter, you will begin to explore the equilibria in which protons, H+, are transferred between chemical species. These equilibria constitute equilibrium acid–base chemistry. The chapter begins by examining the formal definitions of the words acid and base. The definitions of acid and base have changed over time. This history will be followed to arrive at definitions that are both consistent and useful. Acids and bases will be defined by equilibria, as well as words. The thermodynamic concepts of acid and base strength will be quantified by the equilibrium constants for their defining equilibria. Next, the discussion of acids and bases will be extended to polyprotic species, those acids and bases that donate or accept more than one proton. The solution techniques from Part II will be applied to polyprotic acid and base systems. In particular, the shortcut graphical method for monoprotic acids developed in Chapter 8 will be broadened and applied to systems containing polyprotic acids. Finally, compact equations called alpha values will be developed. Alpha values show concisely the effects of pH on the fractional contributions of acidic and basic forms to the total mass.

Key idea: Acid–base chemistry (and pH) are important because H+ is transferred in many reactions that influence the water quality of aquatic systems

11.2  DEFINITIONS OF ACIDS AND BASES 11.2.1  Early History of Acids and Bases You may have noticed that the title of Section 11.2 refers to more than one definition of acids and bases. The concept of acids and bases has evolved over the last 200 years or so.1 In several cases, a new definition of acid or base had to be developed when exceptions were found to old definitions. In the early days of modem chemistry (mid-­eighteenth century), acids were thought to be acids because they contained some common substance. The substance made liquids containing acids taste sour, hence the name acid (from the Latin acere sour; see also the Historical Note to Chapter 7). Antoine-­Laurent du Lavoisier 1   The progression of acid–base concepts presented here was influenced by material at the ChemTeam web site (dbhs. wvusd.k12.ca.us/ChemTeamIndex.htm).

219

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220   Chapter 11  Acids and Bases Key idea: Lavoisier thought, incorrectly, that all acids contained oxygen

(1743–1794) concluded that the common substance was oxygen. His commitment to this idea is shown in the name he selected for oxygen, principe oxygine, meaning “acidifying principle.”

Thoughtful Pause Can you think of a substance that is categorized as an acid today but that does not fit Lavoisier’s definition of an acid?

Lavoisier was wrong, of course. Many acids do not contain oxygen, including common acids (e.g., HCl), less common acids (HCN and HSCN), and some compounds that may not look like acids at first glance (e.g., NH4+). One of the troublesome compounds of the day that did not fit Lavoisier’s definition was chlorine. Chlorine was originally named oxymuriatic acid (after muriatic acid, HCl, named from the Latin muria brine), with the prefix oxy-­ because all acids were supposed to contain oxygen. Humphry Davy (1778–1829) showed that molecular chlorine did not contain oxygen.2 He suggested, based on acids such as HCl and H2S, that perhaps hydrogen was the common substance among acids. To explain why some hydrogen-­ containing substances were not acids, Justus von Liebig proposed that an acid was a hydrogen-­containing substance in which the hydrogen could be replaced by a metal. Thus, according to this definition, HCl is an acid because the hydrogen can be replaced by, say, sodium to form NaCl. On the other hand, methane (CH4) is not an acid because CH3Na does not exist in nature. (See also the Historical Note at the end of this chapter.) All was well for 50 years. Then, in his PhD dissertation of 1884, Svante August Arrhenius (1859–1927) shocked his dissertation committee with the idea that molecules could break apart in a solvent (such as water) to form charged species called ions. Arrhenius received the Nobel Prize in Chemistry in 1903 for this work, later called the ionic theory. Arrhenius extended his ionic theory to acids and bases. The Arrhenius definitions of acids and bases are as follows: Arrhenius acids: a substance that produces H+ in aqueous solution Arrhenius bases: a substance that produces OH– in aqueous solution

Arrhenius acids are substances that produce H+ in aqueous solution, and Arrhenius bases are substances that produce OH– in aqueous solution.

11.2.2  Brønsted-­Lowry Definitions Arrhenius had taken a giant leap forward in linking H+ to acids. He also provided a definition for a base that was consistent with the observations that bases neutralize acids. However, the Arrhenius definitions did not describe the behavior of some common bases, such as ammonia and sodium carbonate. The next step in the evolution of defining acids and bases was provided independently and nearly simultaneously by the Danish chemist Johannes Nicolaus Brønsted (1879–1947) and the English chemist Thomas Martin Lowry (1874–1936). Brønsted postulated that acids donate a proton (H+) to the solvent and bases accept a proton from the solvent.   To see why molecular chlorine was considered an acid, recall that chlorine undergoes the following equilibrium in water: Cl2 + H2O = HOCl + H+ + Cl–. HOCl is hypochlorous acid.

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11.2  |  Definitions of Acids and Bases   221

Lowry used the following formalization for an acid, HA:

HA H 2 O

A

H3O



Together, the Brønsted-­Lowry definitions are as follows: Brønsted-­Lowry acids are substances that donate to water3 (to produce H3O+). Brønsted-­Lowry bases are substances that accept a proton from water (to leave OH–).

These concepts of acids and bases as, respectively, proton donors and proton acceptors are very powerful. The definitions link acids and bases very tightly. They are consistent with Arrhenius’s idea that bases are somehow connected with OH–.

Brønsted-­Lowry acid: a substance that donates a proton to water Brønsted-­Lowry base: a substance that accepts a proton from water

Thoughtful Pause How are the Brønsted-­Lowry definitions consistent with Arrhenius’s ideas about bases? With the Brønsted-­Lowry definitions, you can see that the OH– comes from water, not necessarily from the base itself. We will pause in our romp through history here, although the definition of acids and bases will undergo one more major change in Chapter 14. For the majority of this text, the Brønsted-­Lowry definitions will be used: acids are proton donors and bases are proton acceptors. In other words, acid–base chemistry (in the Brønsted-­Lowry sense) means proton transfer chemistry.

11.2.3  Acid Dissociation Reactions and Base Association Reactions

equilibria

Following the approach of Brønsted and Lowry, you also can define acids and bases by chemical equilibria. (This material was presented briefly in Section 8.2.2, but will be developed in more detail here.) For acids, you can write:

Hn Aa

H2 O

H n 1A a

1

H3O

eq. 11.1

A monoprotic acid is acid that donates one proton: n = 1 in eq. 11.1. If n > 1, the acid is called a polyprotic acid. For example, diprotic acids and triprotic acids can donate two and three protons, respectively. For an uncharged, monoprotic acid (n = 1 and a = 0):

HA H 2 O

A

H3O

eq. 11.2

Equations  11.1 and  11.2 are called acid dissociation reactions. Any substance for which you can write an equilibrium in the form of eq.  11.1 or  11.2 is called an acid (formally, a Brønsted-­Lowry acid). The equilibrium constant for the equilibria in eqs.  11.1 and  11.2 is given a special name. The equilibrium constants are called   The focus here is on water because this is a text on aquatic chemistry. However, one of the strengths of the Brønsted-­ Lowry definitions is that they can be applied to proton-­donating and proton-­accepting solvents other than water (e.g., ethanol).

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Key idea: The Brønsted-­Lowry definitions linked acids and bases by relating them both to proton transfer Key idea: Acid–base equilibria are proton-­transfer

Key idea: Brønsted-­Lowry acids and bases also can be defined by acid dissociation reactions and base association reactions monoprotic acid: an acid that donates one proton (to water) polyprotic acid: an acid that donates more than one proton (to water) acid dissociation reaction: a reaction of the form of eq. 11.1 or 11.2

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222   Chapter 11  Acids and Bases acid dissociation constant (Ka): the equilibrium constant for an acid dissociation reaction

acid dissociation constants and are given the special symbol Ka. Remember: Ka is just an equilibrium constant. It is the equilibrium constant for equilibria of the form of eq. 11.1 or 11.2, in which an acid dissociates. You can develop a similar system for bases. For bases, you can write: B

base association ­reaction: a reaction of the form of eq. 11.3 base association ­constant (Kb ): the equilibrium constant for a base association reaction

ampholyte: a substance that can both donate and accept a proton

H2 O

HB

OH

eq. 11.3

Equation 11.3 is called a base association reaction. Any substance for which you can write an equilibrium in the form of eq.  11.3 is called a base (again, formally, a Brønsted-­Lowry base). The equilibrium constant for the equilibrium in eq. 11.3 is given a special name. The equilibrium constants are called base association constants and are abbreviated Kb. Again, Kb is just an equilibrium constant. It is the equilibrium constant for equilibria of the form of eq. 11.3.

11.2.4  Ampholytes Some species can both donate a proton to water and accept a proton from water. Such compounds are both acids and bases. They are given special names and are called amphoteric species (from the Greek ampho both) or ampholytes (short for amphoteric electrolytes).

Thoughtful Pause Can you name the most common ampholyte in aqueous systems? The most common ampholyte in aquatic systems is water itself! As eq. 11.4 shows, water can both donate a proton to water (and hence is an acid) and accept a proton from water (and thus is a base): Key idea: Water is amphoteric (acts as both an acid and a base)

H2 O H2 O



OH

H3O

eq. 11.4

Other common ampholytes include species with intermediate degrees of protonation (i.e., species that are neither completely protonated nor completely deprotonated). For example, phosphoric acid (H3PO4) is a triprotic acid. It dissociates in water to form three other species: H2PO4–, HPO42–, and PO43– (phosphate).

Thoughtful Pause Are H3PO4, H2PO4 , HPO42–, and PO43– acids, bases, or ampholytes? –

Phosphoric acid is called an acid because it can only donate protons to water. Phosphate a base because it can only accept protons from water. H2PO4– and HPO42– are ampholytes. For example, H2PO4– can donate a proton to water to form HPO42– or accept a proton from water to form H3PO4. For more practice in identifying acids, bases, and ampholytes, see Worked Example 11.1.

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11.3  |  Acid and Base Strength   223 Worked Example 11.1 

Identification of Acids, Bases, and Ampholytes

Identify the following as an acid, base, or ampholyte, using the Brønsted-­Lowry definitions: OH–, HCN, NH3, NH4+, and HCO3–. Solution You can write the following equilibria, analogous to eqs. 11.1 and 11.3: OH

H2O

OH

HCN H 2 O

CN

H3O

NH3

NH 4

OH

NH 4

H2O H2O H2O

HCO3

H2O

HCO3

H2O

NH3

H3O

H 2 CO3 OH

CO32

H 3O

Thus, you would classify the species as follows: OH– is a base, HCN is an acid, NH3 is a base, NH4+ is an acid, and HCO3– is amphoteric. The example of phosphoric acid shows two ampholytes formed from partially protonated species that are part of a polyprotic acid family. Sometimes, ampholytes can be formed when more than one acidic group is found on the same molecule. For example, ethanedioic acid (commonly, oxalic acid: HOOC-­COOH) and 1,2-­benzenedicarboxylic acid (commonly, o-­phthalic acid) each have two acidic functional groups. The partially protonated species (for oxalic acid, the monoanionic species HOOC-­COO–) are ampholytes. A good example of complex systems exhibiting ampholytic behavior are the macromolecules that contribute color to natural waters, called humic and fulvic acids. Another type of ampholyte has both acidic and basic functional groups on the same backbone. Examples include the amino acids (see Problem 23) and some types of polymers.

11.3  ACID AND BASE STRENGTH 11.3.1  Introduction How should you compare the relative strengths of acids and bases? Acids donate protons to water, so it makes sense to develop an acid strength scale that measures the tendency of a compound to donate a proton to water. There are several ways to construct such a scale. The most common approach is to compare the thermodynamic tendency of an acid to donate protons to water.

Thoughtful Pause What is an appropriate quantitative measure of the thermodynamic tendency of an acid to donate protons to water?

properties

Key idea: Acid and base strength are thermodynamic

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224   Chapter 11  Acids and Bases acid strength: thermodynamic tendency of an acid to donate a proton to water base strength: thermodynamic tendency of a base to accept a proton from water

A reasonable quantitative measure is Ka (or pKa). Thus, acid strength is defined as the thermodynamic tendency of an acid to donate a proton to water. Acid strength is quantified by Ka or pKa. Similarly, you can define base strength as the thermodynamic tendency of a base to accept a proton from water and quantify base strength by Kb or pKb.

11.3.2  Properties of Acid and Base Strength The concepts of acid and base strength have a number of important elements. First, since acid and base strength are quantified by an equilibrium constant, acid and base strength are thermodynamic properties. This means that acid and base strength (as with all thermodynamic properties) may change as a function of temperature, pressure, and the concentrations of other species. Second, acid and base strength are defined for one-­proton transfer equilibria (eqs. 11.2 and 11.3). For example, in considering the acid strength of carbonic acid (H2CO3), use the equilibrium constant for the transfer of one proton (i.e., K for H2CO3 + H2O = HCO3– + H3O+), not the equilibrium constant for the transfer of two protons (i.e., not K for H2CO3 + 2H2O = CO32– + 2H3O+). Third, one acid is said to be stronger than another acid if it dissociates more completely to donate a proton to water. Thus, from eqs. 11.1 and 11.2, a stronger acid has a larger Ka. This means that a stronger acid has a smaller pKa. Thus, HOCl (hypochlorous acid), with a pKa of 7.54, is a stronger acid than ammonium (pKa 9.3, both values at 25°C). Very strong acids have very small (and even negative) pKa values: the pKa of HCl is about –3. Similarly, stronger bases have larger Kb values (and thus smaller pKb values). Other examples of ranking acids and bases by their strengths are given in Worked Example 11.2. Worked Example 11.2 

Acid and Base Strength

Rank the following acids by their acid strength: H2S (hydrogen sulfide; pKa = 7.1), HNO3 (nitric acid; Ka = 10+3), and H2O2 (hydrogen peroxide; pKa= 11.7). Rank the following bases by their base strength: NH3 (ammonia; Kb = 10–4.7), CN– (cyanide; pKb = 4.8), and CO32– (carbonate; pKb = 3.7). Solution To compare, express the acid strength for all acids (or bases) in terms of either Ka (Kb for bases) or pKa values (pKb for bases). The choice is arbitrary, but pKa and pKb values are more convenient. Thus: H 2S HNO3 H 2O2

pK a pK a pK a

7.1 3 11.7

NH3

pK b

4.7

CO32

pK b

3.7

CN

pK b

4.8

The acid strength is given by HNO3 >> H2S > H2O2. The base strength is given by CO32– > NH3 ≈ CN–.

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11.3  |  Acid and Base Strength   225

11.3.3  Acid Strength of H+ What is the acid strength of H+? To determine its acid strength, write an equilibrium similar to eq. 11.1 with H+ as a reactant:

H

H2 O

H3O ; K a

?

eq. 11.5

What is Ka for H+? Formally, the standard Gibbs free energy of reaction for this equilibrium is defined to be zero, and thus Ka is defined to be 1. Why was the standard Gibbs free energy of reaction defined to be zero? Recall from Section 1.4.3 that H+ is really an abbreviation for (H2O)nH+. Thus, it makes sense that the free energy required to convert H+ to H3O+ will be zero. There are several important ramifications of the fact that Ka ≡ 1 for the equilibrium in eq. 11.5. First, it means that H+ is a strong acid with pKa ≡ 0. (You can show that the pKa of H3O+ also is zero: see Problem 2.) Second, it means that you can save some ink when you write the Ka expressions in eqs. 11.1 and 11.2. For example, adding eq. 11.2 and the reverse of eq. 11.5 (see Chapter 4 for a review of the rules of adding and reversing equilibria) results in: HA H 2 O H3O

A

H3O   K1

H

H 2 O     K 2

Ka 1   1  1 K1 K 2 K a

HA   A    H       K

This little exercise shows that the equilibrium HA + H2O = A– + H3O+ and the equilibrium HA = A– + H+ have the same equilibrium constant. Thus, you can stop writing acid dissociation reactions in the form of eqs. 11.1 and 11.2 and write them as eqs. 11.6 and 11.7:

Hn Aa



HA

H n 1A a A

1

H

H

eq. 11.6 eq. 11.7

The equilibria in eqs. 11.1 and 11.6 have the same equilibrium constant, as do the equilibria in eqs. 11.2 and 11.7. You can use equilibria in the form of eqs. 11.6 and 11.7 in most instances and use the equilibrium forms in eqs. 11.1 and 11.2 only when you want to emphasize the transfer of protons to water.

Key idea: Since the equilibrium H3O+ = H+ + H2O has K ≡ 1, then the equilibria HA + H2O = A– + H3O+ and HA = A– + H+ have the same equilibrium constant

11.3.4  Conjugate Bases and Acids When an acid donates a proton to water, it produces H3O+ and another species. The “other species” is called the conjugate base of the acid. Thus, from Worked Example 11.1, you can see that CN– is the conjugate base of HCN and ammonia is the conjugate base of the ammonium ion. Similarly, when a base accepts a proton from water, it produces OH– and the conjugate acid of the starting base.

Thoughtful Pause How are the acid strength of an acid and the base strength of its conjugate base related?

conjugate base: the base formed when an acid dissociates conjugate acid: the acid formed when a base accepts a proton

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226   Chapter 11  Acids and Bases You can write the acid dissociation equilibrium and Kb equilibrium for the conjugate base of an uncharged, monoprotic acid as follows: HA Key idea: The pKa of an acid and the pKb of its conjugate base sum to pKW (= 14 at 25°C) Key idea: The conjugate base of a strong acid is a weak base

 orked W example →

A

A

H2 O

H2 O

H

H                    K a

HA OH      Kb  for  conjugate  base

Adding: OH

K

KW

K a Kb

To summarize in words: the product of the Ka of an acid and the Kb of its conjugate base is KW. An equivalent statement is this: the sum of the pKa of an acid and the pKa of its conjugate base is pKW. At 25°C, where pKW = 14:

pK a

pKb (of the conjugate base) 14

eq. 11.8

Equation 11.8 is very useful. It gives you a lot of information from the knowledge of just one equilibrium constant. If you know the pKa of an acid, then you can calculate easily the pKb of its conjugate base (from pKb = 14 – pKa ). Equation 11.8 implies a truism in aquatic chemistry: the conjugate base of a strong acid is a weak base (and, conversely, the conjugate base of a weak acid is a strong base). For example, HC1 is a strong acid. Its pKa is about –3. Thoughtful Pause What can you say about the base strength of chloride (Cl–)? Chloride must be a very weak base since it is the conjugate base of a very strong acid. In fact, pKb for chloride is 14  minus the pKa for HCl or 17. The implication is that chloride will not accept a proton appreciably from water to form HCl. As your common experiences tell you, dumping table salt (NaCl) into a glass of water does not lead to the formation of appreciable quantities of hydrochloric acid. The relationship between conjugate acids and bases is explored further in Worked Example 11.3.

Worked Example 11.3 

Conjugate Acids and Bases

The Ka for phosphoric acid (H3PO4) is 10–2.1. What can you say about the acid strength of phosphoric acid? What can you say about the base and acid strengths of H2PO4–? Solution The pKa of phosphoric acid (= 2.1) is fairly small, so phosphoric acid is a relatively strong acid. The conjugate base of H3PO4 is H2PO4– (since H3PO4 = H2PO4– + H+). Thus, H2PO4– is a relatively weak base, with pKb = 14 – 2.1 = 11.9. With the information provided, you can make no comment about the acid strength of H2PO4–. Its acid strength is quantified by the equilibrium constant of H2PO4– = HPO42– + H+.

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11.3  |  Acid and Base Strength   227

11.3.5  Limits to Acid and Base Strength Are there limits to the strengths of acids and bases? For all practical purposes, you can put limits on the strengths of acids and bases in aqueous systems. Fluids with pH < 0 or pH > 14 generally are considered to be nonaqueous (i.e., the solvent is no longer water). Very strong acids (i.e., acids with pKa < 0) are essentially completely deprotonated (i.e., donate at least one proton completely to water) at all pH values in water. Thus, you should not expect to see appreciable concentrations of the most protonated forms of strong acids such as HCl or H2SO4 in water. Only the conjugate bases of these acids (Cl– and HSO4–, along with its conjugate base, SO42–) are expected to be found in water at reasonable concentrations. In the same vein, you should not expect to see appreciable concentrations of strong bases such as NH2– or S2– in water. Only their conjugate acids will be present (e.g., NH3 and its conjugate acid NH4+, and HS– along with its conjugate base, H2S).4

11.3.6  A Brief Review of Equilibrium Calculations with Monoprotic Acids and Bases You have been exposed to equilibrium calculations with monoprotic acids and bases in Chapters 7 through 9. In this section, the key points in the equilibrium calculations will be reviewed. Recall that to perform algebraic equilibrium calculations with monoprotic acids, you should write the species list, equilibria (one acid dissociation equilibrium for each acid plus the self-­ionization of water), mass balance equation(s) (one mass balance equation for each acid family), and the charge balance or proton condition. As a shortcut for the graphical method, draw the lines for the H+ and OH– concentrations, identify the system point [the point where pC = p(total acid) = pAT and pH = pKa], and draw lines for the acid and conjugate base. The line for the acid (HA) concentration is about equal to AT at pH > pKa. The [HA] and [A–] lines cross at pH = pKa and pC = pAT + 0.3 (i.e., 0.3 log units below the system point). (The [HA] and [A–] lines, if extended, would cross at the system point.) The equilibrium position of the system is the single point on the pC-­pH diagram where the charge balance or proton condition is satisfied. The proton condition is more useful than the charge balance in systems having a large concentration of a species with constant concentration, such as when the salt of an acid (e.g., NaA) is added to water. A typical pC-­pH diagram for a monoprotic acid is shown in Figure 11.1. The charge balance (and proton condition) for the addition of HA to water is: [H+] = [A–] + [OH–]. This is satisfied near point A in Figure  11.1, where [H+] ≈ [A–] and [OH–] is small enough to be ignored. The proton condition for the addition of NaA to water is: [H+] + [HA] = [OH–]. This is satisfied near point B in Figure 11.1, where [HA] ≈ [OH–] and [H+] is small enough to be ignored.

 Acids stronger than about the strength of 100% sulfuric acid are called superacids. An example of a superacid is fluorosulfonic acid, HSO3F. Fluorosulfonic acid is about 40 times stronger than 100% sulfuric acid and, of course, completely deprotonated in water. The acid strength of superacids cannot be measured by their pKa values. In fact, the concept of a pKa is meaningless, since all superacids are completely deprotonated in water. Instead, their acid strengths are quantified by acidity functions. Acidity functions are based on equilibria where the superacids are allowed to protonate bases other than water. 4

ISTUDY

228   Chapter 11  Acids and Bases 0

2

4

6

pH

8

10

12

14

0 system point

2

pC

4

[A–]

A B

6

[HA]

8 10 12

[H+]

[OH–]

14 FIGURE 11.1  pC-­pH Diagram for a Monoprotic Acid with pKa = 6 and pAT = 3 [dashed lines indicate the equilibrium pH for an HA solution (left line, point A) and a NaA solution (right line, point B)]

Two results from the algebraic and graphical solutions are important to remember. First, the HA and A– concentrations depend on AT, pH, and pKa. Specifically, from eqs. 8.3 and 8.4: Ka [A ]   AT  and K a [H ] [HA] 

[H ] K a [H ]

 AT

Thus, the concentration of HA is greater than [A–] at pH < pKa. Conversely, the concentration of A– is greater than [HA] at pH > pKa. Second, at pH = pKa, the HA and A– concentrations are equal. You may wish to verify these conclusions for the monoprotic acid in Figure 11.1.

11.4  POLYPROTIC ACIDS 11.4.1  Definitions

Key idea: Since Ka is defined for a one-­proton transfer, the dissociation of polyprotic acids usually is written as a stepwise process

As stated in Section 11.2.3, a polyprotic acid can donate more than one proton to water. Common examples of diprotic acids (which can donate two protons to water) in the aquatic environment include H2CO3, H2S, and H2SO4. The donation of protons to water from polyprotic acids usually is written as a stepwise process.5 For a general diprotic acid, H2A: H 2 A    HA

HA    A

2

 H        K a,1

 H         K a,2



Because of the stepwise nature of proton transfer, Ka,1 is always greater than Ka,2 and thus pKa,1 < pKa,2. For diprotic acids, the amphoteric form (HA– in the equilibria above) is given the name of the most basic form with the prefix bi-­. Examples include the carbonate family  It is not required to write polyprotic acid dissociation as a stepwise process. Usually, the equilibria are written as one-­proton transfers because Ka is defined for one-­proton transfers. You could write any linearly independent set of n equilibria to express the relationships among the n + 1 species in an n-­protic acid family. For example, you could express the relationships between H2A, HA–, and A2– by: H2A = A2– + 2H+ and HA– = A2– + H+.

5

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11.4  |  Polyprotic Acids   229

H2CO3 (carbonic acid), HCO3– (bicarbonate), and CO32– (carbonate), and the sulfide family H2S (hydrogen sulfide), HS– (bisulfide), and S2– (sulfide).

11.4.2  Solving Equilibrium Systems with Polyprotic Acids You can use the algebraic, graphical, and computer methods discussed in Part II to solve equilibrium systems containing polyprotic acids. The example of a 10–3 M H2CO3 solution (closed system) will be solved by each method in this section. We must pause for a moment and introduce one new symbol. Recall from Section  1.4.3 that H+, (H2O)H+, (H2O)2H+, and so on are almost indistinguishable. Thus, the symbol [H+] is used to represent the sum [H+] + [(H2O)H+] + [(H2O)2H+] + other similar species. Similarly, the species H2CO3(aq) and CO2(aq) are nearly indistinguishable. It is common to represent the sum of their concentrations by one symbol: [H2CO3*]. The system is as follows (if activity and concentration are interchangeable): Unknowns: [H2O], [H+], [OH–], [H2CO3*], [HCO3–], and [CO32–] Starting materials: H2O and H2CO3* Equilibria: OH ;  KW (self - ionization  of  water )

H 2 O    H

H 2 CO3*    H



HCO3

 H

HCO3 ;  K a,1 10

CO32 ;  K a,2

10

6.3

10.3



Equilibrium equations: KW K a,1 K a,2

[H ] [OH ] [H 2 O] [H ] [HCO3 ] [H 2 CO3 ]

[H ] [CO32 ] [HCO3 ]

Mass balance: CT

1 10 3 M [H 2 CO3 ]

[HCO3 ]

[CO32 ]

Electroneutrality: Charge balance: [H+] = [OH–] + [HCO3–] + 2[CO32–] Or proton condition: [H+] = [OH–] + [HCO3–] + 2[CO32–] Other constraints: Mole fraction of H 2 O 1 All concentrations 0

←Worked example [H2CO3*]: Defined as [H2CO3(aq)] + [CO2(aq)]

ISTUDY

230   Chapter 11  Acids and Bases Remember that the mass balance(s), charge balance, and proton condition are not linearly independent. This system can be solved by the algebraic, graphical, and computer methods. Examples of each solution technique will be presented. In the brute force algebraic method, express each species concentration in terms of the concentration of one species (usually [H+]). From the equilibria and mass balances, you can confirm after a little algebra:

[H 2 CO3 ]



[HCO3 ]



[CO32 ]

[H ]2

[H ]2 K a,1[H ] K a,1 K a,2 K a,1[H ]

2

[H ]

K a,1 K a,2

[H ]

K a,1[H ] K a,1 K a,2

eq. 11.9

CT

eq. 11.10

CT

eq. 11.11

K a,1[H ] K a,1 K a,2

2

CT

Substituting into the charge balance (or proton condition) yields:

[H ]

K a,1[H ]CT 2

[H ]

2 K a,1 K a,2CT

K a,1[H ] K a,1 K a,2

KW [H ]

eq. 11.12

Solving (see Appendix A for hints on solving this equation): 2.22 10 5 M   (pH   4.65)

[H ] [OH ]

4.51 10

M

9.78 10 4 M

[H 2 CO3 *]

2.21 10 5 M

[HCO3 ]

10

[CO32 ]

4.99 10

11

M



Using the method of approximation, you might guess that the solution is mildly acidic since a relatively weak acid (pKa of H2CO3* is 6.3) is being added to water. Thus, the right-­hand side of the charge balance should be dominated by the weaker bases.

Thoughtful Pause Why is the charge balance dominated by the weaker bases? The weaker bases generally are higher in concentration because the stronger bases will be more protonated at an acidic pH. Therefore, from the charge balance: [H+] ≈ [HCO3–] or, using eq. 11.9: [H ]

2

[H ]

K a,1[H ]CT K a,1[H ] K a,1 K a,2

ISTUDY

11.4  |  Polyprotic Acids   231

The denominator of the expression for HCO3– (eq. 11.10) is about equal [H+]2 under K a,1CT pK a,1 pCT 6.3 3 acidic conditions.6 Thus: [H ] or [H+]2 ≈ Ka,1CT or pH ≈ 2 2 [H ] = 4.65. You may wish to check that the assumptions are valid. The assumptions were: In the charge balance: [HCO3–] >> 2[CO32–] and [HCO3–] >> [OH–] In the denominator: [H+]2 >> Ka,1[H+] and [H+]2 >> Ka,1Ka,2 This example points out the value of chemical intuition in selecting the approximation. The wise assumption of mildly acidic conditions here converted a fourth-­order 6.3 3 polynomial equation in [H+] to arithmetic you can do in your head: pH ≈ = 2 4.65. Working problems like this is one of the best ways to strengthen your chemical intuition and save you work in the long run. To use the graphical method, merely plot the expressions in eqs.  eq.  11.9–11.11 and  find where the charge balance (or proton condition) is satisfied. A pC-­pH ­diagram for the carbonic acid system is shown in Figure 11.2. The equilibrium pH is  again approximated from the charge balance at the pH where [H+] ≈ [HCO3–] (about where the lines for [H+] and [HCO3–] cross). This occurs at about pH 4.6– 4.7 (dotted line in Figure 11.2). The approximations in the charge balance, [HCO3–] >> 2[CO32–] and [HCO3–] >> [OH–], can be checked easily in the pC-­pH diagram. A shortcut to the graphical method with polyprotic acids will be discussed in Section 11.4.3. Of course, polyprotic acid systems can be solved with computer methods. A spreadsheet for a 10–3 M H2CO3* solution is discussed in Section E.5.3. The Nanoql SE input may be found in Section F.4.3. Try running the system and calculating equilibrium concentrations. A related chemical system is solved in Worked Example 11.4 by algebraic, approximation, graphical, and computer methods.

0

0

2

2

pC

4

4

6

pH

8

10

12

14

[HCO3–] [CO32–]

[H2CO3*]

6 8 10 12

[OH–]

[H+]

14 FIGURE 11.2  pC-­pH Diagram for the Addition of 10–3 M H2CO3* to Water (Dashed line represents the equilibrium pH where the charge balance and proton condition are satisfied.)

 Why is the denominator in eq. 11.10 about equal to [H+]2 under acidic conditions? The term [H+]2 is larger than 16.6 = 8.3. Ka,1[H+] if the pH is less than pKa,1 = 6.3. Also, [H+]2 is larger than Ka,1Ka,2 = 10–16.6 if the pH is less than 2 Thus, [H+]2 is the largest term in the denominator if the pH is less than 6.3. 6

Spreadsheet available: A spreadsheet for a 10–3 M H2CO3* solution is discussed in Appendix E, Section E.5.3. Nanoql SE Example: The Nanoql SE file for a 10–3 M H2CO3* solution is discussed in Appendix F, Section F.4.3.

ISTUDY

232   Chapter 11  Acids and Bases

Salt of a Diprotic Acid

Worked Example 11.4 

Following the exarnple in the text with the H2CO3* system, find the equilibrium concentrations for a 1×10–3 sodium bicarbonate solution. Solution If NaHCO3 → Na+ + HCO3– was added to water instead of carbonic acid, the following changes would be made to the process shown in the text: 1. Add Na+ as a species. 2. Change starting materials to NaHCO3 → Na+ + HCO3– and H2O. 3. Add a new mass balance equation: [Na+] = 10–3 M. 4. Include Na+ as a cation in the charge balance. 5. Change the proton condition to: [H+] + [H2CO3*] = [CO32–] + [OH–]. Algebraic solution: The system has the same equilibria as in the text, so eqs. eq. 11.9–11.11 are still valid. Since Na+ is present, it is more valuable to use the proton condition than the charge balance. Plugging eqs. eq. 11.9–11.11 into the proton condition yields a variation of eq. 11.12: [H ]

[H ]2 CT

K a,1 K a,2CT

KW

[H ]

where: Δ = [H+]2 + Ka,1[H+] + Ka,1Ka,2. Solving: [H ]

5.50 10 9 M   (pH  8.26)

[OH ]

1.82 10 6 M 1.08 10 5 M

[H 2 CO3 ] [HCO3 ] [CO32 ]

9.80 10 4 M 8.94 10 6 M

Approximation method: You expect the pH to be closer to neutral pH than in the case of H2CO3* addition. Thus, Δ is about equal to Ka,1[H+]. You might start by ignoring [OH–] relative to [CO32–] in the proton condition. Thus, the proton condition becomes: [H ]

[H ]CT K a,1

K a,2CT [H ]

or: CT

[H ]2 1 CT

Since =

K a,1

pK a,1 pK a,1 2

>>

1,

[H+]2

6.3 10.3 = 8.3. 2

K a,1 ≈

K a,2CT

Ka,1Ka,2

or

[H ]

K a,1 K a,2

or

pH

ISTUDY

11.4  |  Polyprotic Acids   233

Checking the assumptions, you can show that at pH 8.3, Δ = 2.6×10–15 ≈ K C Ka,1[H+] = 2.5×10–15. Also: [OH–] = 10–5.7 M and [CO32–] = a,2 T = 10–5.0 M, [H ] – so the assumption that [OH ] can be ignored relative to [CO32–] is of questionable validity. Graphical method: Figure 11.2 is still valid. We seek the point where the proton condition is satisfied, i.e., where [H+] + [H2CO3*] = [CO32–] + [OH–]. Near neutral pH, you expect [H+] and [OH–] to be small, so the proton condition becomes [H2CO3*] ≈ [CO32–]. From Figure 11.2, the [H2CO3*] = [CO32–] lines cross at about pH 8.2–8.3. Computer method: In Nanoql SE, add the component Na+ to the carbonate example and set its concentration to 0.001 M. With the default thermodynamic data, you will ­calculate pH 8.29.

11.4.3  A Shortcut to the Graphical Method for Polyprotic Acids You can develop a shortcut method for graphical solutions to polyprotic acid systems similar to the shortcut method developed for monoprotic acids in Section 8.4. For a diprotic acid, consider three pH ranges: pH < pKa,1, pKa,1 < pH < pKa,2, and pH > pKa,2. In the lowest pH range, the denominator in eqs. 11.9–11.11 is about equal to [H+]2. Thus:





[H 2 CO3 ]

[HCO3 ] [CO32 ]

[H ]2

2

[H ]

2

[H ]

[H ]2 K a,1[H ] K a,1 K a,2 K a,1[H ] K a,1[H ] K a,1 K a,2 K a,1 K a,2 K a,1 [H ] K a,1 K a,2

CT

CT CT

CT K a,1 [H ]

CT

K a,1 K a,2 [H ]2

CT

Therefore, where pH < pKa,1, you expect: 1. The [H2CO3*] line should have a slope of zero and be about equal to CT. 2. The [HCO3–] line should be downward sloping to the left with a slope equal K to –1 (i.e., parallel to the [OH–] = W line). [H ] 3. The [CO32–] line should be downward sloping to the left with a slope equal to –2. Also note that [CO32–] is much smaller than either [H2CO3*] or [HCO3–], since [H+] >> Ka,2 (pH pKa,2, you expect:

Key idea: pC-­pH diagrams for diprotic acids have two system points (corresponding to the two pKa values), with the slope of the pC lines changing at the system points

1. The [H2CO3*] line should be downward sloping to the right with a slope equal to +2. Also note that [H2CO3*] is much smaller than either [HCO3–] or [CO32–], since [H+] > pKa,1 = 6.3) in this pH range. 2. The [HCO3–] line should be downward sloping to the right with a slope equal to +1 (i.e., parallel to the [H+] line). 3. The [CO32–] line should have a slope of zero and be about equal to CT. The graphical shortcut method takes advantage of the two system points for a diprotic acid. The first system point occurs at pH = pKa,1 and pC = pCT. The second system point occurs at pH = pKa,2 and pC = pCT. The behavior of the species in any diprotic acid system with respect to pH can be summarized as follows. The p(concentration) of the most protonated form of the diprotic acid (H2CO3* in the example) is about equal to CT at pH values less than pKa,1 (i.e., to the left of the first system point), increases in a 1:1 ratio with increasing pH at

ISTUDY

11.4  |  Polyprotic Acids   235

pH values between the two pKa values (i.e., between the system points), and increases in a 2:1 ratio with increasing pH at pH values greater than pKa,2 values (i.e., to the right of the second system point). Similarly, the p(concentration) of the next most protonated form of the diprotic acid (HCO3– in the example) decreases in a 1:1 ratio with pH at increasing pH values less than pKa,1 (i.e., to the left of the first system point), is about equal to CT at pH values between the two pKa values (i.e., between the system points), and increases in a 1:1 ratio with increasing pH at pH values greater than pKa,2 (i.e., to the right of the second system point). Finally, the p(concentration) of the least protonated form of the diprotic acid (CO32– in the example) decreases in a 2:1 ratio with pH at increasing pH values less than pKa,1 (i.e., to the left of the first system point), decreases in a 1:1 ratio with increasing pH at pH values between the two pKa values (i.e., between the system points), and is about equal to CT at pH values greater than pKa,2 (i.e., to the right of the second system point). As with monoprotic acids, it is quite easy to sketch the pC-­pH diagram for a polyprotic acid. The extension of the shortcut method of Section  8.4 to n-­protic acids is fairly straightforward. Consider an n-­protic acid, HnA, which dissociates in water to form n other species: Hn–1A–, Hn–2A2–, . . . , An–. Each species is of the form HiAi–n. The sum of the concentrations of the n species containing the fragment -­A is AT. To sketch the pC-­ pH diagram (see also Appendix C, Section C.4.3), use the following approach: Step 1: Locate system points. Prepare a pC-­pH diagram with lines representing the H+ and OH– concentrations. Locate the n system points. These occur where pH = pKa,1 and pC = pAT, pH = pKa,2 and pC = pAT, . . . , and pH = pKa,n and pC = pAT. Step 2: Draw species lines. Draw the lines for the p(concentration) of each species at least 1.5 pH units away from each system point. The p(concentration) line of species HiAi–n has the following slopes: i – n at pH values less than the first system point, i – n + j between the jth and ( j + 1)th system point, and i at pH values greater than the final (nth) system point. (Recall that a negative slope means that the pC decreases with increasing pH and thus the line is upward sloping to the right.) Make sure that the lines, if extended, would go through the system points. Step 3: Make species lines intersect below the system points. Curve the lines so that they intersect about 0.3 log units below the system points. Step 4: Find the equilibria pH. To find the equilibria pH (and equilibrium concentrations of all plotted species), find the pH where the charge balance or proton condition is satisfied. Check any assumptions made in the charge balance and iterate if necessary. With this shortcut, a pC-­pH diagram for any acid can be sketched quickly and the equilibrium pH estimated with little fuss.

11.4.4  Power of the Proton Condition The real value of the proton condition shows itself with polyprotic acids. Once you have solved the system formed by adding HnA to water, you can easily calculate the equilibrium conditions for adding various salts to water (generalized as NamHpA where m + p = n). For example, after putting in the work on the H2CO3* system, you can

Key idea: To apply the graphical shortcut to polyprotic acids, locate the system points (at pC = pAT and pH = pKa,i), draw species lines (making sure that species lines intersect below the system points), and find the equilibrium pH where the proton condition is satisfied

ISTUDY

236   Chapter 11  Acids and Bases Key idea: The proton condition is valuable in working with the salts of polyprotic acids Worked example →

find the equilibrium pH for systems where NaHCO3 (sodium bicarbonate) and Na2CO3 (sodium carbonate) are added to water. The three systems (adding H2CO3*, NaHCO3, or Na2CO3 to water) have the same set of equilibria and differ only in their mass balance on the sodium ion and in their proton condition. Worked Example 11.4 demonstrated the ease in extending the H2CO3* system to  NaHCO3 solutions. You also can apply the previous work to Na2CO3 solutions. Begin by writing the proton condition for a Na2CO3 solution (starting materials are Na2CO3 → 2Na+ + CO32– and H2O): [H ] 2[H 2 CO3 ] [HCO3 ] [OH ] Note the factor of 2 in front of the carbonic acid concentration, because H2CO3* has two more protons than its source in the starting materials, namely CO32–. You might expect the pH to be high. Thoughtful Pause Why should the pH be high for a sodium carbonate solution? The right-­hand side of the proton condition contains only [OH–]. Thus, [OH–] must be much larger than [H+].7 At high pH, [H+] and [H2CO3*] are expected to be small and the proton condition becomes: [HCO3–] ≈ [OH–]. From Figure 11.2, this occurs at about pH 10.5–10.6 (exact solution: pH 10.56).

11.5  ALPHA VALUES (DISTRIBUTION FUNCTIONS) 11.5.1  Introduction In developing analytical solutions for monoprotic acids (Section 8.4) and polyprotic acids (Section  11.4), you combined the equilibria and mass balances to express the concentration of acids, bases, and ampholytes in terms of the total concentration multiplied by a function of [H+] and Ka values. For example, eqs. 8.3 and 8.4 stated that, for Ka [H ] AT and [HA] = AT , where AT = [HA] a monoprotic acid: [A–] = K a [H ] K a [H ] + [A–] (see also Section 11.3.6). Note that these expressions are of the form:

where f and g are the functions

[A ] AT

f ([H ], K a )

[HA] AT

g([H ], K a )

Ka

K a [H ]

and

[H ] K a [H ]

, respectively.

  The same conclusion can be reached from the charge balance: [H+] + [Na+] = [HCO3–] + 2[CO32–] + [OH–]. Since [Na+] = 2×10–3 M is large, the species concentrations on the right-­hand side also must be large and [OH–] is expected to be much larger than [H+]. 7

ISTUDY

11.5  |  Alpha Values (Distribution Functions)   237

The function f and g are of the same form for any monoprotic acid. They are called distribution functions (or ionization fractions). The functions have been given many names over the years. Since f and g usually are given the symbols α1 and α0, respectively, they will be referred to as alpha values in this text. The alpha values for a monoprotic acid have a number of useful interpretations and characteristics. First, the alpha values represent the fraction of the total concentration found as either [HA] or [A–]. The alpha values are truly fractions and vary in magnitude between 0 and 1. Second, the alpha values encapsulate all the information on how [HA] and [A–] change with pH and Ka. In fact, alpha values separate the dependence of species concentrations on pH and Ka from the dependency on AT. Third, the alpha values sum to one. You can show with a little algebra that α0 + α1 = 1. It is perhaps more important to see without algebra that α0 plus α1 must equal one. Since the alpha values represent the fraction of the total concentration that is [HA] or [A–], the sum of the alpha values must be unity. Fourth, the alpha values are easily interconvertible. Note that: 1

0

Ka

Ka

K a [H ]

[H ]

[H ]

[H ] Ka

K a [H ]

0

1

For monoprotic acids, these interconversions are pretty obvious. They are more useful with polyprotic acids because they save time in the calculations.

11.5.2  Alpha Values for Diprotic Acids The alpha values for diprotic acids can be generalized from eqs. 11.9–11.11 as follows: [H 2 A]

0 AT ,  where:

0

[HA ]

1 AT ,  where:

1

[H ]2

2

2

[A 2 ]

2 AT ,  where:

[H ]2

[H ]

[H ]2 K a,1[H ] K a,1 K a,2 K a,1[H ] K a,1[H ] K a,1 K a,2 K a,1 K a,2 K a,1[H ] K a,1 K a,2

Note that there are three alpha values to accommodate the three species in the family. Also, note that the alpha values have different mathematical expressions for monoprotic acids and diprotic acids. Importantly, they have the same interpretation as fractions of the total concentration. In other words, α0 for a monoprotic acid and α0 for a diprotic acid are calculated with different formulas, but both are equal to the fraction of total acid made up by the most protonated form of the acid. You also can see the patterns in the nomenclature: the symbol α0 always is used for the most protonated form of the acid (H2A for diprotic acids and HA for monoprotic

alpha values (distribution functions ionization fractions): functions of pH and the Ka values that describe the fractional contribution of each species to the total concentration in its mass balance Key idea: Alpha values range from 0 to 1 and sum to 1

ISTUDY

238   Chapter 11  Acids and Bases acids). You can show (both by logic and algebra) that the alpha values for diprotic acids sum to one. In addition, the alpha values are interconvertible. Comparing α0 and α1: 0

[H ] K a,1

K a,1

1 and

1

and

2

(same as monoprotic acids)

0

[H ]

Comparing α1 and α2: 1

Key idea: The alpha values α0 and α1 have the same meaning for monoprotic and diprotic acids, but different equations for monoprotic and diprotic acids

[H ] K a ,2

2

K a,2

1

[H ]

Finally, comparing α0 and α2: 0

[H ]2 K a,1 K a,2

and

2

2

K a,1 K a,2

0

[H ]2

To summarize, the alpha values for diprotic acids can be confusing when you first meet them. The symbols α0 and α1 have the same meaning for monoprotic and diprotic acids (i.e., the fraction of the total that is the most-­protonated form for α0 and the fraction that is the next-­most-­protonated form for α2). However, the symbols α0 and α1 have different equations for monoprotic and diprotic acid: same meaning, different equations.

11.5.3  Alpha Values for n-­Protic Acids You can extrapolate the formulas above to alpha values for n-­protic acids. For n-­protic acids, the ith alpha value (denoted αi, where i = 0 to n) relates the concentration of the species Hn–iAi– to the total concentration:8 [Hn–iAi–] = αiAT. You can derive the following relationships: [H ]n

i

n j 0 ([ H

n i 0

Key idea: Alpha values can be extended to n-­protic acids

i

i

i n j

]

i k 1

K a,k j k 1 K a,k )

1 [H ] j j k i 1

i

K a,k

j  ,  i

j

You should verify these relationships for n = 1 (monoprotic acids) and n = 2 (diprotic acids).

 This nomenclature assumes that the most protonated form of the acid, HnA is uncharged and that all n protons are transferrable to water. If the most protonated form of the acid has q total H (n of which are transferrable to water) and carries a charge of p, then αi is associated with the species Hq–iAp–i. An example is NH4+/NH3, where q = 4, n = 1 and, p = 1. In this case, α1 (i = 1) is associated with the species Hq–iAp–i = H3N0 = NH3. 8

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11.7  |  Part II Case Study: Acid Rain   239

11.6  CHAPTER SUMMARY Although the definitions of acid and base have changed over time, the Brønsted-­Lowry definitions are very useful in aquatic chemistry: acids donate protons to water and bases accept protons from water. Acids and bases can be defined by equilibria as well. For example, the acid dissociation reaction for an uncharged monoprotic acid is HA + H2O = A– + H3O+ (thermodynamically equivalent to HA = A– + H+). The equilibrium constants for equilibria of this form are called acid dissociation constants and are denoted by the symbol Ka. Similarly, bases can be defined by equilibria of the form A– + H2O = HA + OH–, with K = Kb, the base association constant. Acid strength is measured by Ka or pKa values and base strength by Kb or pKb values. Thus, acid and base strengths are thermodynamic properties and change with thermodynamic conditions. Acids and bases can donate or accept more than one proton. Acids that donate more than one proton to water are called polyprotic acids. We usually think of polyprotic acids as donating protons stepwise in a series of one-­proton reactions, each with its own Ka value. In this chapter, the solution techniques from Part II were applied to polyprotic acid and base systems. A shortcut graphical method for polyprotic acids and their salts was developed. In the shortcut, a diprotic acid was found to have two system points, corresponding to the two pKa values. Compact equations were developed to describe how the fraction of the total concentration contributed by each species depends on pH. These equations are called alpha values or distribution functions. Alpha values can be written for n-­protic acids.

11.7  PART II CASE STUDY: ACID RAIN After completing this chapter, you now have a good understanding of how to perform equilibrium calculations with acids and bases. In this first portion of the Part III case study, you will apply your newly minted skills to the analysis of acid rain. You collected a sample of potentially acidic rainwater and submitted it to an analytical laboratory. Being familiar with the chemistry of acid rain (as reviewed briefly in Section 10.7), you asked the lab to measure the concentrations of common ions in the sample. You received the following results: [SO42–] = 0.5 mg/L as S, [NO3–] = 0.6 mg/L as N, and [Cl–] = 0.7 mg/L as Cl. The lab also reported that the total inorganic carbon concentration, [H2CO3*] + [HCO3–] + [CO32–], is 0.1 mg/L as C. Upon reading this list, you realized that you forgot to ask the lab to measure pH! The sample has been discarded, but you trust the analytical results (including the fact that no cations other than H+ apparently are in the sample in appreciable concentrations). You are a little confused about why the lab measured inorganic carbon but did not report the concentrations of HCO3– or CO32–. Being too embarrassed to tell your boss that you forgot to have pH measured, you are determined to calculate the pH of the rainwater. You know that H2SO4, HNO3, and HCl are very strong acids. Thus, it makes sense that their conjugate bases (SO42–/HSO4–, NO3–, and Cl–) are very weak bases and likely will not accept protons from water. Therefore, it is a good approximation that SO42–, NO3–, and Cl– are the only forms of S(+VI), N(+V), and Cl(–I) in the sample, respectively. You develop a mathematical model of the system: Unknowns: [H2O], [H+], [OH–], [H2CO3*], [HCO3–], [CO32–], [SO42–], [NO3–], and [Cl–] Starting materials: H2O, anions, and inorganic carbon

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240   Chapter 11  Acids and Bases Equilibria: H2 O

H

H 2 CO3 * HCO3

OH ; KW (self - ionization of water )

H

HCO3 ; K a,1 10

CO32 ;  K a,2 10

H

6.3

10.3

Equilibrium equations: [H ] [OH ] [H 2 O]

KW

[H ][HCO3 ]

K a,1

[H 2 CO3 ]

[H ][CO32 ]

K a,2

[HCO3 ]

Mass balances:

CT

[H 2 CO3 *]

ST

2

NT

[HCO3 ]

g 1 10 4   L g 12   mol

2

[CO3 ]

8.3 10 6 M

[SO 4 ]

g 5 10 4  L g 32  mol

1.6 10 5 M

3.1 10 5 eq / L

[NO3 ]

g 6 10 4   L g 14   mol

4.3 10 5 M

4.3 10 5 eq / L

ClT

[Cl ]

g 7 10 4   L g 35.45  mol

2.0 10 5 M

2.0 10 5 eq / L

Charge balance: [H ]

[OH ]

Or: [H ]

[HCO3 ]

[OH ]

2[CO32 ]

[HCO3 ]

Other constraints: Mole fraction of H 2 O 1 All concentrations 0

2

2[SO 4 2 ]

2[CO3 ]

[NO3 ] 5

9.5 10 eq / L

[Cl ]

ISTUDY

Chapter Key Ideas   241

You might guess that the pH is acidic. Thus, you can ignore [CO32–] and [OH–] in the charge balance. The charge balance becomes: [H ] [HCO3 ] 9.5 10 5 eq / L Based on your experience, you know that [HCO3–] = α1CT where α1 = 2

[H ] α1 ≈

K a,1[H ]

K a,1[H ] K a,1 K a,2

K a,1 [H ]

.  At low pH (at least at pH values well below pH 6.3),

. Thus: [H ]

K a,1 [H ]

CT

9.5 10 5 eq / L

Or:

[H ]2 9.5 10 5 [H ] K a,1CT

0

eq. 11.13

Solving: [H+] = 9.5×10–5 M or pH 4.0. Now you realize why the lab failed to report bicarbonate and carbonate concentrations. At pH 4.0 and with 0.1 mg/L as C total inorganic carbon, the concentrations of bicarbonate and carbonate are very, very small. In other words, the inorganic carbon in the sample is primarily H2CO3*. Should you have a great deal of confidence in this calculated pH value? In a word, no. Note that the term Ka,1CT in eq. 11.13 is small compared to the other terms. Thus, [H+] = 9.5×10–5 M = 2[SO42–] + [NO3–] + [Cl–]. Errors in the measurement of sulfate, nitrate, or chloride may have a significant effect on the calculated pH. Also, the presence of any unmeasured cations or anions may skew the calculated pH as well. In the absence of a measured pH value, we will proceed in the case study with the calculated value of the acid rain pH as 4.0.

CHAPTER KEY IDEAS • Acid–base chemistry (and pH) are important because H+ is transferred in many reactions that influence the water quality of aquatic systems. • Lavoisier thought, incorrectly, that all acids contained oxygen. • The Brønsted-­Lowry definitions linked acids and bases by relating them both to proton transfer. • Acid–base equilibria are proton-­transfer equilibria. • Brønsted-­Lowry acids and bases also can be defined by acid dissociation reactions and base association reactions. • Water is amphoteric (acts as both an acid and a base). • Since the equilibrium H3O+ = H+ + H2O has K ≡ 1, then the equilibria HA + H2O = A– + H3O+ and HA = A– + H+ have the same equilibrium constant. • The pKa of an acid and the pKb of its conjugate base sum to pKW (= 14 at 25°C). • The conjugate base of a strong acid is a weak base. • Since Ka is defined for a one-­proton transfer, the dissociation of polyprotic acids usually is written as a stepwise process. • pC-­pH diagrams for diprotic acids have two system points (corresponding to the

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242   Chapter 11  Acids and Bases two pKa values), with the slope of the pC lines changing at the system points. • To apply the graphical shortcut to polyprotic acids, locate system points (at pC = pAT and pH = pKa,i), draw species lines (making sure that species lines intersect below the system points), and find the equilibrium pH where the proton ­condition is satisfied. • The proton condition is valuable in working with the salts of polyprotic acids. • Alpha values range from 0 to 1 and sum to 1. • The alpha values α0 and α1 have the same meaning for monoprotic and diprotic acids, but different equations for monoprotic and diprotic acids • Alpha values can be extended to n-­protic acids.

CHAPTER GLOSSARY acid dissociation constant (Ka):  the equilibrium constant for an acid dissociation reaction acid dissociation reaction:  a reaction of the form of eq. 11.1 or 11.2 acid strength:  thermodynamic tendency of an acid to donate a proton to water alpha values (distribution functions ionization fractions):  functions of pH and the Ka values that describe the fractional contribution of each species to the total concentration in its mass balance ampholyte:  a substance that can both donate and accept a proton Arrhenius acids:  a substance that produces H+ in aqueous solution Arrhenius bases:  a substance that produces OH– in aqueous solution base association constant (Kb):  the equilibrium constant for a base association reaction base association reaction:  a reaction of the form of eq. 11.3 base strength:  thermodynamic tendency of a base to accept a proton from water Brønsted-­Lowry acid:  a substance that donates a proton to water Brønsted-­Lowry base:  a substance that accepts a proton from water conjugate acid:  the acid formed when a base accepts a proton conjugate base:  the base formed when an acid dissociates monoprotic acid:  an acid that donates one proton (to water) polyprotic acid:  an acid that donates more than one proton (to water)

HISTORICAL NOTE: WHY IS A BASE A BASE? The origins of the word acid were explored in the Historical Note to Chapter 7. But why is a base called a base? The first application of the word base to basic compounds was by French chemist Guillaume François Rouelle (1703–1770) in 1754 (Jensen 2006). At the time, salts were thought of as compounds formed from the reaction of acids with “something else.” This should make some sense to you. Most of the strong acids you have encountered so far are compounds of H+ and an anion (a weak conjugate base). An example is HCl. If you combine enough HCl and NaOH, you eventually would precipitate out NaCl(s). In Roulle’s day, most of the common acids were volatile liquids that were purified

ISTUDY

Problems   243

by distillation. An example is acetic acid. To convert a volatile acid (a “spirit” in eighteenth-­century parlance) to a solid salt required reaction with a “base.” In this sense, a base was called a base because it provided the base for the formation of a salt. Roulle is best remembered for his skills in the lecture hall. He has been described as “. . .a dynamic teacher, volatile, unmethodical, impetuous, a kind of impecunious Bohemian of science.  .  .” (Lemay and Oesper  1954).9 Among his students were the great Lavoisier (see Section 11.2.1) and Joseph Louis Proust (1754–1826), “discoverer” of the mass balance in chemical reactions.

PROBLEMS Note: Most of these problems can be solved by either algebraic, graphical, or computer methods. Try using more than one approach with some problems to get a feel for the advantages and disadvantages of the three approaches. 1. How many acids, bases, and ampholytes are there in the species formed when an n-­protic acid is added to water? Recall that an n-­protic acid can donate n protons to water. List the species formed when the tetraprotic acid H4A is added to water. 2. What is the pKa of H3O+? What is the pKb of [OH–]? 3. Water is an ampholyte. Calculate its pKa and pKb. Is water a weak or a strong acid? Is water a weak or a strong base? 4. Show quantitatively that the conjugate base of a strong acid is a weak base. Use the standard measures of acid and base strength. 5. Show quantitatively that the conjugate acid of a strong base is a weak acid. Use the standard measures of acid and base strength. 6. If the pKa of a monoprotic acid, HA, is 4.2, write the base association equilibrium and calculate the Kb value for its conjugate base, A–. 7. The pKb of cyanide (CN–) is 4.8. What is its conjugate acid? What is the pKa of its conjugate acid? 8. A student isolates a monoprotic acid. A 0.01 M solution of the acid has a pH of 3.75. What is the pKa of the acid? 9. A monoprotic acid with pKa = 4.5 has an equilibrium pH of 3.79. What is the total concentration of the acid plus conjugate base in the solution? 10. A 0.05 M solution of the sodium salt of an acid has a pH of 9.35. What is the pKa of the acid and pKb of its ­conjugate base?

11. The potassium salt of a monoprotic acid with pKa = 5.3 has an equilibrium pH of 8.15. What is the total concentration of the acid plus conjugate base in the solution? 12. How does log

[HCO3 ] change with pH? Hint: Use [H 2 CO3 ]

alpha values. 13. For a monoprotic acid, how do the concentrations of the acid and conjugate base compare at pH = pKa? What percentage of the total concentration is made up of the acid and the conjugate base at pH = pKa? 14. For a diprotic acid, how do the concentrations of the acid (H2A) and ampholyte (HA–) compare at pH = pKa,1? What percentage of the total concentration is made up of H2A and HA– at pH = pKa,1? 15. Phthalic acid (benzene-­1,2-­dicarboxylic acid) is a diprotic acid with pKa values of 2.89 and 5.51. It is used in the production of phthalic anhydride (2-­benzofuran-­1,3-­dione) which, in turn, is used for the production of plastics and plasticizers. Sketch a pC-­pH diagram for a 0.01 M phthalic acid solution. Use it to find the equilibrium pH of 0.01 M solutions of phthalic acid (here, H2P), potassium hydrogen phthalate (KHP), and potassium phthalate (K2P). 16. What is the pH of a 0.01 M solution of aniline (phenylamine)? It is an organic base with a pKb of 9.4. Aniline is uncharged and its conjugate acid has a charge of +1. 17. An industry is discharging an acetic acid waste stream. (Recall that acetic acid is CH3COOH.) Its National Pollutant Discharge Elimination System (NPDES) permit requires pH > 4 and total organic carbon (TOC) < 250 mg/L as C. What is the maximum strength of acetic acid waste (as TOC in mg/L) that can be discharged

9   You can only wonder what his teaching evaluations would have looked like (perhaps on EvaluerMesProfesseurs.com?). Many professors would love to be called a Bohemian of science, impecunious or not.

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244   Chapter 11  Acids and Bases without violating either permit requirement? (Recall that TOC, in mg/L, measures the number of milligrams of carbon per liter.) 18. An industry is discharging waste stream that is a dilute ammonia solution. Its NPDES permit requires pH < 10 and total nitrogen < 100 mg/L as N. What is the maximum strength of the ammonia waste (as mg/L N) that can be discharged without violating either permit requirement? 19. Besides baking soda (NaHCO3) and baking powder (mainly Na2CO3), another leavening agent in cooking is ammonium carbonate, (NH4)2CO3, also called baker’s ammonia or hartshorn.10 a. What is the proton condition for the addition of ammonium carbonate to water? b. Using graphical techniques, calculate the equilibrium pH and the equilibrium concentrations of all the dissolved species for a 10–3 M (NH4)2CO3 solution. 20. Three manufacturers are producing carbonated bottled water. One company introduces carbonate by bubbling carbon dioxide into water until CT = 1×10–4 M (equivalent to adding H2CO3* until CT = 1×10–4 M). Another manufacturer adds sodium bicarbonate until CT = 1×10–4 M, and the third company adds sodium carbonate until CT = 1×10–4 M. Using one pC-­pH diagram, estimate the pH of the bottled water from each of the three manufacturers. 21. Chlorendic acid [4,5,6,7,7-­hexachlorobicyclo-­(2.2.1)-­ hept-­5-­ene-­2,3-­dicarboxylic acid] is a highly chlorinated compound used as a flame retardant. It is a diprotic acid with pKa,l = 3.1 and pKa,2 = 4.6. Chlorendic acid inhibits the growth of the alga Chlorella at pH 3.5, but not at pH 5.0 or pH 7.5 (Hendrix et al. 1983). a. Using graphical techniques, which form of chlorendic acid (H2C, HC–, or C2–) is not responsible for the inhibitory effect? You can assume any total ­concentration. b. Using graphical techniques, calculate the distribution (i.e., the fraction) of chlorendic acid as H2C, HC–, and C2– at pH 3.5, 5.0, and 7.5. 22. Professor Whoops attempts to make a 10–2 M NaH2PO4 solution. The good professor accidentally grabs the bottle of Na2HPO4 salt instead of the bottle of NaH2PO4 salt. What pH did Professor Whoops expect the solution to be? What was the actual pH of the solution as made? (One approach: Sketch a pC-­pH diagram to solve this problem. For phosphoric acid: pKa,1 = 2.1, pKa,2 = 7.2, and pKa,3 = 11.)

23. Amino acids have a monoprotic acid group and a monoprotic base in the same molecule. The acid group is a carboxylic acid, with the following acid dissociation reaction: -­COOH = COO – + H+. The basic group is an amine, with the following base association reaction: -­NH2 + H2O = -­NH3+ + OH–. A fully protonated amino acid can be written as HOOC-­R-­NH3+, where R is an organic functional group. Typically, the pKa values for the -­COOH group are about 2 to 3 and the pKb values for the -­NH2 group are about 4 to 5. a. In principle, an amino acid could take four forms, depending on the pH. The four forms are: HOOC-­ R-­NH2, –OOC-­R-­NH2, HOOC-­R-­NH3+, and –OOC-­ R-­NH3+. Explain why the species HOOC-­R-­NH2 is not formed in significant quantities. b. Find the pH range where each of the other three species dominates. The other three species are –OOC-­ R-­NH2, HOOC-­R-­NH3+, and the overall neutral – OOC-­R-­NH3+. (The doubly charged species –OOC-­ R-­NH3+ is called a zwitterion.11 c. The isoelectric point (or isoelectric pH, abbreviated pI) is defined as the pH where the fraction of the acid group dissociated is equal to the fraction of basic group that is protonated. At pH = pI, the amino acid has no overall charge. Develop an expression for pI as a function of pKa and pKb. What is the pI of the amino acid alanine if the pKa of the carboxylic acid group is 2.3 and the pKb of the amino group is 4.3? (The pKa of the protonated amino group is 9.7.) 24. It is possible to write general expressions for the alpha values of a monoprotic acid at any pH relative to the pKa of the acid. a. Verify that α0 = 0.760 and α1 = 0.240 at pH = pKa – 0.5 for any monoprotic acid. b. What are the values of α0 and α1 for any monoprotic acid at pH = pKa – 1? c. Show that

0

pH = pKa ± x.

1

1 10

x

and

1

1

1 10  x

at

25. Show that the equilibrium pH of a NaHA solution is about equal to the average of pKa,1 and pKa,2 if the concentrations of H+ and OH– can be ignored relative to other species. Here, NaHA is the sodium salt of HA– and part of the diprotic acid system: H2A/HA–/A2–. State any assumptions you made. Does this trick apply to Worked Example 11.4?

  Hartshorn originally was made by grinding the horns of harts or reindeer. Modern hartshorn is ammonium carbonate.  Zwitterion comes from the German Zwitter hermaphrodite. Hermaphrodite was the son of Hermes and Aphrodite. According to Greek myth (popularized in Ovid’s Metamorphosis), Hermaphrodite was attracted to the nymph Salmacis. The two became one being, who inherited the worst characteristics of both males and females. 10 11

ISTUDY

Addendum: A Surprising Exact Solution   245 26. Biochemists often use the Henderson-­Hasselbach equation to relate the equilibrium pH to the ratio of the acid and conjugate base concentrations. The Henderson-­ Hasselbach equation for a monoprotic acid is pH = [A ] . pKa + log [HA] a. Derive the Henderson-­Hasselbach equation for a monoprotic acid. b. Derive the following version of the Henderson-­ Hasselbach equation for a diprotic acid (e.g., H2A/

30. Problem 29 can be extended to diprotic acids. Show that the pH of a solution consisting of C mole/L of H2A and C mole/L of NaHA is about equal to pKa,1. Show how you could make a solution by combining NaHA and Na2A to achieve an equilibrium pH of pKa,2. 31. Repeat Problem 21 in Chapter 7 (model for orange juice). This time, treat citric acid as a triprotic acid (H3C) with pKa values of 3.14, 4.8, and 6.4.

[A 2 ] 1 HA /A ): pH = . pK a,1 pK a,2 log [H 2 A] 2 –

that the pH of a solution consisting of C mole/L of the monoprotic acid HA and C mole/L of the conjugate base NaA is about equal to pKa. What restrictions are necessary for the equilibrium pH to be about pKa?

2–

32. Repeat Problem 22 in Chapter 7 (model for orange juice). This time, treat citric acid as a triprotic acid (see Problem 31 for pKa values) and malic acid as a diprotic acid (H2M) with pKa values of 3.4 and 5.2.

27. For a 10–3 M Na2S solution (pKa.1 = 7.1 and pKa.2 = 14) a. Calculate the pH at equilibrium. b. At equilibrium, which is larger, [HS–] or [S2–]? 28. Find the total carbonate concentration (= [H2CO3*] + [HCO3–] + [CO32–]) of a pure Na2CO3 solution so that [HCO3–] = [CO32–] at equilibrium. 29. A truism in acid–base chemistry is that the pH of a solution consisting of equimolar concentrations of an acid and its conjugate base is about equal to the pKa. Show

33. Butler (1964) discussed a way to calculate the Ka of a monoprotic acid from measuring the pH of a mixture of CHA mole/L of HA and CNaA mole/L of NaA. Derive an expression for Ka as a function of CHA, CNaA, and the measured [H+] (assuming activities and concentrations are interchangeable).

ADDENDUM: A SURPRISING EXACT SOLUTION You practiced calculating the equilibrium pH of monoprotic acids a lot in this chapter by making assumptions. The exact solution for [H+] for a monoprotic acid with acid dissociation constant Ka and total concentration AT is, remarkably:

[H ]

1 K 3 a

2 Q cos

2 3



eq. 11.14

where: Q = Ka2 + 3KaAT + 3KW, R = Ka(2Ka2 + 9KaAT  – 18KW), S = θ = tan

1

27S if R > 0 or θ = tan R

1

27S R

R2

4Q3 , and 27

if R < 0.

Cosine? Inverse tan? Where in the world does eq.  11.14 come from? The analysis starts, as usual, with the charge balance: [H+] = [A–] + [OH–]. Using the usual Ka K approach, you can rewrite this as: [H+] = AT + W . Clearing fractions, the K a [H ] [H ] charge balance can be written as a cubic in [H+]: [H ]3 K a [H ]2

( K a AT

KW )[H ] K a KW

0

ISTUDY

246   Chapter 11  Acids and Bases This is of the form ax3 + bx2 + cx + d = 0, with x = [H+], a = 1, b = Ka, c = –(KaAT + KW), and d = –KaKW . Using Cardano’s formula (after Italian mathematician Gerolamo Cardano, 1501–1576), the general cubic has the solution: x =

1 b C 3a

Q C

where

R 2 4Q3 , Q = b2 – 3ac, and R = 2b3 – 9abc + 27a2d. There are three roots, 2 since C (the cubic root of a real or complex number) always has three values. Comparing the cubic in [H+] to the general form: Q = Ka2  +  3KaAT  +  3KW and C = 

3

R

4Q3 , so R 2 4Q3 27S i 27S . 27 The remaining development requires that S > 0 so that  27S  is real.12 Fol-

R = Ka(2Ka2 + 9KaAT – 18KW). Define S = 

R2

lowing the general solution: C  =   3 R i 27S . The radicand is a complex 2 2 number if S > 0. It can be written in polar coordinates as r(cosθ + isinθ), where r=

R 2

2

and θ = tan

27S 2 1

2

1 R2 2

27S R

1 4Q3 2

27S

Q3/ 2 and θ = tan

27S if R > 0 R

1

if R < 0.

From de Moivre’s formula (after French mathematician Abraham de Moivre, 1667– 2 k 3

1754), the three values of C are: C = Q cos and 2. Therefore, the three roots are: [H+] = simplified,13 since C

Q C

1 K 3 a

C

2 k . So [H+] = 3

2 Q cos

2 k 3

i sin

for k = 0, 1,

Q . This expression can be C 1 K 3 a

2 Q cos

2 k . 3

4Q3 = Ka4(AT2 + 4KW) + 4Ka3AT(AT2 + 5KW) + 4Ka2KW 27 (3AT2 – 2KW) + 12KaATKW2 + 4KW3. Therefore, S is minimized with respect to AT when AT = 0. If AT = 0, then Smin =  After a lot of tedious algebra, you can show that S =

12

R2

4Ka4KW – 8Ka2KW2 + 4KW3 = 4KW(Ka2 – KW)2. Therefore, the minimum S is positive, so S > 0 and expression Smin = 4KW(Ka2 – KW)2 suggests that S could be equal to zero if Ka = zero, S is strictly greater than zero as required.  Here: C + 

13

Q  = C

Q cos

2 k 3

i sin

2 k 3

Q Q cos

2 k 3

27S is real. The

KW . However, because AT is never

i sin

2 k 3

. The

Q term can be C

evaluated by multiplying the numerator and denominator by the complex conjugate of the denomiQ 2 k 2 k i sin nator, namely cos . Since cos2α + sin2α = 1, the result is: C  +     =  C 3 3 Q cos

2 k 3

i sin

2 k 3

Q cos

2 k 3

i sin

2 k 3

2 Q cos

2 k . 3

ISTUDY

Addendum: A Surprising Exact Solution   247

As usual, we are interested only in positive roots for [H+], so the term in parentheses must be negative. You know Q = Ka2 + 3KaAT + 3KW > Ka2. This means that

Q >K, a

2 k > –½. If you work through all the 3 possibilities,14 you can show that only k = 1 gives a positive value for [H+]. This is a remarkable result. In the history of mathematics, Cardano’s solution showed the practical value of complex numbers. Similarly, this exact solution shows that you can jump into the complex plane and emerge back with real solutions to equilibrium chemistry problems. Equation 11.14 is easily implemented in a spreadsheet. Enter values for Ka and AT in cells. Then enter the formulas for Q, R, and S. This allows for the calculation of θ and finally, [H+]. Therefore, with a spreadsheet, eq. 11.14 allows for rapid calculation of the equilibrium pH for any monoprotic acid. For example, the equilibrium pH values for monoprotic acids with pKa values between 2 and 10 and pAT values between 2 and 8 are shown in Figure 11.3. Notice how the equilibrium pH values tend toward 7 for very dilute acids (left side of Figure  11.3). Figure  11.3 also is a graphical representation of Problem 25. Notice that the equilibrium pH is about the average of the pKa and pAT for weak acids (see pK a pAT the lines of about slope 1 on the right side of the figure). Formally, pH ≈ 2 for Ka –½ for k = 1, but not for k = 0 or k = 2. If R < 0, then the argument of the inverse 3 tangent function is negative. The inverse tangent of a negative argument lies between  and 0. If R < 0, then 2 2 k 27S θ = tan 1 . Therefore, ≤ θ ≤ π. In this range of θ, cos > –½ for k = 1, but not for k = 0 or 2 3 R  If R > 0, then θ = tan

14

k = 2. Therefore, k = 1.

1

ISTUDY

248   Chapter 11  Acids and Bases

CHAPTER REFERENCES Butler, J.N. (1964). Ionic Equilibrium: A Mathematical Approach.  New York: Addison-­Wesley Publishing Co.

Jensen, W.B. (2006). The origin of the term “base.” J. Chem. Educ. 83(8): 1130.

Hendrix, P.F., J.A. Hamala, C.L. Langner, and H.P. Kollig (1983). Effects of chlorendic acid, a priority toxic substance, on laboratory aquatic ecosystems. Chemosphere 12(7/8), 1083–1099.

Lemay, P. and R.E. Oesper (1954). The lectures of Guillaume Francois Rouelle. J. Chem. Educ. 31(7): 338–343.

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C H A P T E R   12

Acid–Base Titrations 12.1 INTRODUCTION You now have a good understanding of acids and bases and the determination of equilibrium concentrations in systems containing acids and bases. In this chapter, the acid–base concept will be taken one step further by investigating mixtures of acids and bases. It may seem strange to devote an entire chapter to the mixing of acids and bases. However, as you will see, many important environmental processes involve acid–base combinations. The act of combining acids and bases is called an acid–base titration. The word titration comes from the French titre, meaning title or qualification, and refers to the use of titrations for the determination of the purity of gold or silver alloys. You may have been introduced to the concept of adding a base to an acid solution with a burette in the ubiquitous titration laboratory in high school and freshman chemistry courses. In an acid–base titration, small volumes of a strong acid (or base) are added to the solution of interest and pH is monitored. Your experiences may have convinced you that mixing acids and bases is a “pure chemistry” concept best relegated to introductory chemistry courses. In fact, many global and local problems in environmental engineering and science have at their heart the mingling of acids and bases in water. Three examples will show the wide applicability of combining acids and bases in the aquatic environment. The first example concerns the early development of the Earth’s oceans. One prevailing theory purports that the oceans were created when the planet cooled sufficiently so that water vapor could condense into rain. The major anions in the early marine environment are thought to have come from the dissolution of volcanic gases into rainwater. The gases [CO2, SO2, and HCl(g)] combined with rainwater to form the corresponding acids [namely, H2CO3, H2SO4, and HCl(aq)]. (This also was discussed in the Part III case study.) The acidic rain titrated the basic materials in rock to form cations. The anions from the gases (e.g., Cl–, SO42–, HCO3–, and CO32–) and the resulting dissolved cations (such as Ca2+, Na+, and K+) became the mineral soup of the early oceans (Figure 12.1). In essence, the composition of the major chemical species in the early oceans came from the largest, most sustained titration in the 4.5-­billion-­year history of the Earth.1 For the second example of acid–base mixtures, skip ahead a few billion years. In more modern times, anthropogenic contributions have increased the discharge of gases to the atmosphere, which dissolve to form acidic precipitation, acid rain or acid snow. The term acid rain was first coined by the Scottish chemist Robert Angus Smith. (See the Historical Note in Chapter 10 for details.) Smith took note of the damage to buildings in large cities  An early contributor to this theory was Lars Gunnar Sillén, who gave us the pC-­pH diagram discussed in Chapter 8. For a review, see Sillén (1967).

1

acid–base titration: the combining of acids and base

titrations

Key idea: Many environmental problems involve

249

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250   Chapter 12  Acid–Base Titrations

CO2, SO2, HCI(g) H2CO3, H2SO4, HCI(aq)

Ca, Na, K

HCO3–, CO32–, SO42–, CI–, Ca2+, Na+, K+

FIGURE 12.1  A Schematic View of the Formation of the Early Oceans by Titration of Minerals with Acids from Volcanic Gases (rain cloud image: OpenClipart-­Vectors/Pixabay; volcano image: Clker-­Free-­Vector-­ Images/Pixabay)

neutralization: to bring the pH to near-­neutral pH by the addition of acids or bases

from acid rain produced primarily from the combustion of coal. The damage occurred as the acid rain titrated the bases (calcium and magnesium) in the building materials. This kind of defacement has been observed with revered structures throughout the world, from the US Capitol Building and Westminster Abbey to the Parthenon and the Taj Mahal. Again, the equilibrium chemistry of acid rain is explored in the case study. A third example of the intermingling of acids and bases comes from the design of treatment systems. The generation of acidic wastes is very common in industry. In fact, the strong inorganic acids (called mineral acids, e.g., HCl, HNO3, and H2SO4) are among the most commonly reported chemicals released into the environment from industrial sources. In many cases, acidic waste streams are neutralized (i.e., brought to near-­neutral pH) by the addition of a strong base before discharge to municipal wastewater treatment facilities.2 The calculation of the dose of the base requires a quantitative understanding of acid–base interactions.

12.2  PRINCIPLES OF ACID–BASE TITRATIONS 12.2.1  Nomenclature

Titrant: solution added during a titration

Key idea: Titrant concentrations usually are expressed as eq/L (i.e., moles of protons accepted or donated per liter)

In an acid–base titration, an acid (or base) solution of known concentration is added to another solution (or sample) of known volume. The solution added is called the titrant. (The solution being titrated is sometimes called the titrand.) Usually we want to know the resulting pH when a specified volume, v, of the titrant is added to a known volume, V0, of the original solution (or sample). The concentration of the titrant usually is expressed as equivalents per liter. Here, “equivalents” refers to the number of moles of protons being exchanged. For example, sodium hydroxide (NaOH) accepts one proton from water and thus has one equivalent per mole. If a 10–2 M NaOH solution is used as a titrant, then its concentration is (10–2 mol/L)(1 eq/mol) = 10–2 eq/L = 10–2 N. A 0.2 M H2SO4 titrant would be expressed as (0.2 mol/L)(2 eq/mol) = 0.4 eq/L = 0.4 N since sulfuric acid donates two protons (at all but the lowest pH values in water).  Acidic wastes may come from liquid waste streams (as discussed in the text) or from gas scrubbers. In gas scrubbers, basic solutions are used to trap the acidic components in off-­gases. In either case, an acidic aqueous stream is generated and must be treated.

2

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12.2  |  Principles of Acid–Base Titrations   251

In most cases, the titrant strength is assumed to be large, so the titrant volumes are small compared to the volume being titrated. Thus, dilution by the titrant usually is small (but is easily accounted for if desired).

12.2.2  Titrations and Chemical Equilibria In the classic presentation of acid–base titrations, chemistry texts usually go through a standard set of combinations: titration of a strong acid with a strong base, titration of a weak monoprotic acid with a strong base, and so on. It is not necessary to analyze all possible types of titrations if you approach the titration solution at any point in the titration as a chemical equilibrium problem. The key to analyzing mixtures of acids and bases is to use the tools developed in Chapters 5–9. Imagine that you are titrating a solution containing a monoprotic acid (HA, with total concentration AT) with NaOH.

Key idea: Titrations are just chemical equilibrium problems ←Worked example

Thoughtful Pause What species would the system include? The system would contain H2O, H+, OH–, HA, A–, and Na+ (since NaOH dissociates completely to Na+ and OH–).

Thoughtful Pause Does the species list change during the titration?

It is important to note that the species list does not change during the titration. The concentrations of the species at equilibrium change as NaOH is added. Adding NaOH changes [Na+] (since NaOH is assumed to dissociate completely). This will change the concentrations of all the other species to maintain the electroneutrality condition. You can account for the acid–base equilibria and mass balance on HA and A– in several ways. One simple approach is to use alpha values. For the monoprotic acid system: [HA] = α0AT and [A–] = α1AT. As usual, plug these expressions into the charge balance or proton condition.

Thoughtful Pause Is it better to use the charge balance or proton condition in this case? Recall that the proton condition is extremely useful for eliminating the unreactive ions, such as Na+, from consideration. In a titration, you want to include the sodium

Key idea: The species list does not change during a titration

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252   Chapter 12  Acid–Base Titrations Key idea: Use the charge balance to solve chemical equilibrium problems involving titrations

CB: the concentration of strong base cations (in eq/L) in solution at any point during the titration

ion concentration since [Na+] is changing as you titrate. Thus, you are better off here with the charge balance. The charge balance is: H



Na

A

OH

eq. 12.1

Recall that each term in the charge balance is in units of charges per liter or equivalents per liter, where equivalents means the charge on the ion. For titrations with base, replace [Na+] with a general symbol for the equivalents of base added (or, more specifically, the equivalents of cations added by titrating with a base that dissociated completely). The general symbol is CB, the concentration of strong base cations in eq/L based on charge. Why use the term strong base? We are assuming that the base in the titrant dissociates completely. This means that the titrant is a strong base (see Section 11.3). Substituting the alpha values and CB into eq. 12.1: or:



H

+ CB =

1

AT

OH

CB



1

AT + OH H

eq. 12.2

Recall that CB here is CB “in the beaker.” If the titrant volume (v) is expected to be large compared to the original HA volume (V0), you can write eq. 12.2 in terms of the CB of the titrant and correct for dilution:

CB of titrant

v v V0

1

AT

V0 v V0

OH

H

eq. 12.3

The volume correction on AT is needed where very accurate results are imperative or when v is greater than about 5% of V0.

12.2.3  Titration Curves How do you use equations such as eqs. 12.2 or 12.3? Recall from Section 12.2.1 that you want to find the pH for every value of CB. For a monoprotic acid, this is a relatively easy task. The simple solution is to set up a spreadsheet with a column of pH (or Ka [H+]) values and a column with the formula α1AT + [OH–] – [H+], or AT + Ka H KW  – [H+]. The latter column will be the CB values. The pair of columns gives pH-­CB H pairs. An example spreadsheet for using this approach to calculate CB from pH for the titration of 10–2 M acetic acid with strong base is shown in Figure 12.2. The CB values are in column F. For clarity, the formulas used are shown in the column to the right of the calculated values.

FIGURE 12.2  Example Spreadsheet for Back-­Calculating CB from pH for the Titration of 10–2 M Acetic

Acid with Base

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12.2  |  Principles of Acid–Base Titrations   253

Alternatively, you could use a fitting routine such as Solver to find the equilibrium pH for each value of CB. In this text, you are encouraged to use Nanoql SE to perform equilibrium calculations. With Nanoql SE, you can enter water, acetate, and Na+ as components. The Nanoql SE system for the Na+/acetic acid system is discussed in Section F.4.4. Vary the concentration of Na+ (see Appendix F for instructions on how to vary a component concentration) and plot pH against CB (= [Na+]). As an example of the titration of a monoprotic acid with a strong base, consider the titration of 10–2 M acetic acid (pKa 4.7) with NaOH. Thoughtful Pause Take a moment and use one of the approaches described in the previous ­paragraph to generate (pH, CB) pairs. Here is a sample calculation. The expanded version of eq. 12.2 is: CB



Ka

Ka H

KW H

AT

H



You can calculate the CB required to achieved pH 5 by letting [H+] = 10–5 M: CB

10 M eq 10 2 M 1  mol M 10 5 M 4.7

10

4.7

eq 10 14 M 1  5 mol 10 M

eq mol

10 5 M 1 

 6.65 10 3 eq /L





Note the use of associated units with equilibrium constants and the explicit conversion of mol/L to eq/L. Pairs of CB and pH for this example are plotted in Figure 12.3. Figure 12.3 is called a titration curve. A titration curve is a plot of pH versus CB. Note that titration curves are log-­linear plots: the y-­axis (pH) is logarithmic, but the x-­axis (CB) is linear. To visualize how the equilibrium pH varies with total acid, pKa, and CB for the titration of a monoprotic acid, run the animation in Appendix E, Section E.4. The titration curve contains a large amount of information. Even a cursory glance at Figure  12.3 tells you that the same amount of NaOH has a different effect on pH,

Nanoql SE Example: The Nanoql SE system for the Na+/acetic acid system is discussed in Section F.4.4 Key idea: The same amount of strong base or strong acid added has a different effect on pH depending on how much base or acid has been added previously

← Worked example titration curve: a plot of pH against a measure of the amount of titrant added (usually CB or CA)

14 12

pH

10 8 6 4 2 0 0.000

0.005

0.010

0.015

CB (eq/L) FIGURE 12.3  Titration Curve for the Titration of 10–2 M Acetic Acid with NaOH

0.020

Animation: To run an animation on the titration of a monoprotic acid with a strong base, see the example in Appendix E, Section E.4

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254   Chapter 12  Acid–Base Titrations depending on how much base has been added previously. (This is a roundabout way of saying that the slope in Figure 12.3 is not constant.) For example, adding the first 5×10–3 eq/L (= 5 meq/L) of base raises the pH from about 3.4 to about 4.7. The second 5 meq/L addition raises the pH to about 8.8 and the third addition causes the pH to increase to about 11.7. The pH increase for the first 0.9  meq/L added is only about one pH unit, whereas a 4.3 pH unit increase is observed for the addition of the next 0.2 meq/L aliquot. The titration curve tells you how much base you must add to achieve a certain pH. This information also can be calculated directly from the charge balance. The use of the charge balance to determine the quantities of materials needed to neutralize an acid or base can be found in Worked Example 12.1. Worked Example 12.1 

Neutralization of an Acidic Waste

The Sourpuss Vinegar Company produces spiced and herbed vinegar products for the domestic market. The factory discharges between 50,000 and 200,000 gallons per day (131 to 525 L/min) of a liquid waste stream that contains only acetic acid at a concentration of 80 mg/L as acetic acid. Size a metering pump to dose 50% sodium hydroxide at a rate to meet an effluent pH standard of 7.2. Solution

g as CH 2 COOH L The waste stream is or 1.33×10–3 M acetic acid. Performing g 60 CH 2 COOH mol a charge balance after NaOH addition yields eq. 12.2: 0.08

CB

1

AT

OH

H

where AcT = [HAc] + [Ac–] = 1.33×10–3 M. At pH 7.2, substitute: H   10  

7.2

OH   10   1

M

6.8

Ka Ka H

M 10

10

4.7

4.7

10

7.2

   0.997

Thus: CB = (0.997)(1.33×10–3 M) + 10–6.8 M – 10–7.2 M = 1.33×10–3 M = 1.33×10–3 eq/L. We must add enough NaOH to achieve a sodium ion concentration of 1.33×10–3 M. Fifty percent NaOH has a density of 1.53 kg/L. Thus, its concentration is (0.5 NaOH/kg solution)(1.53 kg solution/L solution) = 0.765 kg/L, or kg g 0.765 1000 L kg = 19.1 M. g 40 mol

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12.3  |  Equivalence Points   255

The rate of acetic acid released is (131 L/min)(1.33×10–3 M) = 0.174 mole/ min at the low flow rate and (525 L/min)(1.33×10–3 M) = 0.698 mole/min at the mol 0.174 min larger flow rate. The NaOH flow rate required is = 9.11 mL/min to mol 19 1 . mol L 0.698 min = 36.5 mL/min. Thus, the recommended flows are 9.11 to 36.5 mL/ mol 19.1 L min (or 3.47 to 14.0 gpd). Figure 12.3, a plot of pH versus CB, is a titration curve for the titration with a strong base like NaOH. Recall that the assumption is that NaOH dissociates completely in water. Also, CB is defined as the Na+ concentration from NaOH in eq/L. You could instead titrate with a strong acid (say, HCl). You know that HCl dissociates completely at any reasonable pH and concentration in water. Assign the symbol CA to the concentration of Cl– from HCl. In general, CA is the concentration of anions in eq/L dissociated from the strong acid. By analogy with Figure 12.3, the titration curve for titration with a strong acid is a plot of pH versus CA.

CA: the concentration of strong acid anions in solution at any point during the titration (in eq/L)

12.3  EQUIVALENCE POINTS 12.3.1  Introduction The approach discussed in Section 12.2 is all you need to do equilibrium calculations with acid and base mixtures. However, the brute force method discussed above is not very satisfactory. It does not give you much insight into titration chemistry. For example, the grind-­it-­out approach does not help you understand why the pH is so constant at moderate and large base additions. It also does not tell you why the pH is so sensitive to the amount of base added at CB = 1×10–2 N (see Figure 12.3). Why worry about such insights when you can just calculate the pH from eqs. 12.2 or 12.3? Without chemical intuition, we are doomed to make mistakes in environmental engineering and science. Take the case discussed in Worked Example 12.1. A small error in the base addition rate may result in a significantly different pH than expected. This question is explored quantitatively in Worked Example 12.2. Worked Example 12.2 

Sensitivity of pH to CB

Based on Worked Example 12.1, a pump with a nominal discharge of 3.5 gpd was selected. The actual discharge is 3.5 gpd ± 10%. What is the expected pH range of the treated waste at 50,000 gpd? Solution The actual discharge is 3.15 to 3.85 gpd. Therefore, the actual CB is 19.1 M 3.15 gpd = 1.20×10–3 M at 3.15 gpd. Similarly, CB is 1.34×10–3 M 50, 000 gpd and 1.47×10–3 M at 3.5 and 3.85 gpd, respectively. Inserting into the charge

Key idea: Although titration curves are relatively easy to generate, an appreciation of the features of the titration curve requires chemical insight

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256   Chapter 12  Acid–Base Titrations balance and solving for pH, the actual pH values are 5.7, 9.0, and 10.1 at 3.15, 3.5, and 3.85 gpd, respectively. Note: To achieve a pH range of 7.15 to 7.25 (i.e., the target pH), a discharge rate between 3.4684 and 3.4707 gpd is required (3.4697 gpd for pH 7.2). This is a tolerance of about ±0.001 gpd, or about ±2.6 μL/min!

12.3.2  Equivalence Points and the Proton Condition

f function: a normalized measure of the amount of base titrant added and equal to the eq/L of base added divided by the moles/L of initial material (units: eq/mol)

Titration curves such as Figure 12.3 often are presented by normalizing the x-­axis. This is accomplished by dividing the titrant concentration (in eq/L) by the concentration of the initial material being titrated (in moles/L). For the titration of acids with a base, this function is called f : f

eq  base added L

  mol  of initial total material L CB AT  

OH

AT

H

The function f has units of eq/mol. You can redraw Figure  12.3 by rescaling the x-­axis using the f function, as shown in Figure 12.4. Figure 12.4 allows for further interpretation of the titration curve. The slope of the titration curve is steepest at f = 1. The slope appears to be increasing at f = 0. It is small at f = 0.5 and at the higher pH values ( f greater than about 1.25 or so in this example). What is so special about f = 0 and f = 1? These points will be referred to, respectively, as the 0th equivalence point and the 1st equivalence point. The word equivalence here should lead you to ask, as always: Equivalent to what? To answer this question, 14 12 10 pH

equivalence point: the point in a titration when the titrated solution has a charge balance identical with the proton condition of a solution formed using an acid (or base) or conjugate base (or conjugate acid) as a starting material

1

8 6 4 2 0 0.0

0.5

1.0

1.5

2.0

f FIGURE 12.4  Titration Curve Using the f Function (10 M acetic acid titrated with NaOH) –2

ISTUDY

12.3  |  Equivalence Points   257

examine the charge balance at f = 1, where CB = AT . You know: CB = [Na+] and AT = AcT = [Ac–] + [HAc]. Substituting into eq. 12.1, the charge balance becomes:

H

HAc

OH (charge balance at f

1)

eq. 12.4

Thoughtful Pause What solution has the same proton condition as eq. 12.4?

Equation 12.4 is the proton condition for a NaAc solution! In other words, a solution of acetic acid titrated to f = 1 is identical to (or equivalent to) an equimolar solution of sodium acetate. You cannot tell the difference between a 10–2 M acetic acid solution titrate to f = 1 with NaOH and a 10–2 M sodium acetate solution (if the titrant does not dilute the solution being titrated appreciably). What about f = 0? This point in the titration curve is even easier to interpret. It is obvious that an untitrated 10–2 M acetic acid solution is identical to (or equivalent to) a 10–2 M acetic acid solution.3 Thus, the 0th and 1st equivalence points correspond to pure solutions of acetic acid and sodium acetate, respectively. What about f = ½? The point f = ½ is called the half-­equivalence point (or, formally, the first half-­equivalence point). At f = ½, CB = ½AT = ½([HAc] + [Ac–]). Since CB = [Na+], the charge balance becomes: H

1

Or : H

2

HAc 1

2

HAc

Ac

Ac 1

2

Ac

OH OH



Now, if [H+] and [OH–] are small relative to the acetic acid and acetate concentrations, the charge balance becomes: [HAc] ≈ [Ac–]. Using the general symbols, the charge balance becomes: [HA] ≈ [A–].

Thoughtful Pause At what pH does [HA] = [A–]?

An acid and its conjugate base are equal in concentration at pH = pKa. (Recall that [HA] = [A–] means α0 = α1, so [H+] = Ka or pH = pKa.) Thus, f = ½ occurs at about pH = pKa as long as [H+] and [OH–] are small compared to the acid and conjugate base concentrations. You can verify that pH ≈ pKa = 4.7 at f = ½ for the acetic acid titration curve in Figure 12.4.

Formally, at f = 0, CB = [Na+] = 0 and the charge balance (eq. 12.1) becomes: [H+] = [Ac–] + [OH–]. This is the charge balance and proton condition for an acetic acid solution.

3

Key idea: The solution formed during the titration of a monoprotic acid to the first equivalence point is identical to a solution of the conjugate base Key idea: The pH of a monoprotic acid titrated to f = ½ is about equal to the pKa as long as [H+] and [OH–] are small compared to the acid and conjugate base concentrations

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258   Chapter 12  Acid–Base Titrations

12.3.3  Shortcut Method for Titration Curves of Monoprotic Acids The information in Section  12.3.2 can be used to sketch a titration curve for any uncharged monoprotic acid (say, HA at a total concentration of C M). The pH values at f = 0, ½, and 1 can be estimated easily. The pH values at the 0th and 1st equivalence points are the equilibrium pH values of a C M HA and C M NaA solution, respectively. Recall that you can use one pC-­pH diagram to estimate these pH values: simply apply the proton conditions for HA and NaA solutions. The proton condition for HA as the starting material is [H+] = [A–] + [OH–], and the proton condition for NaA as the starting material is [H+] + [HA] = [OH–]. The pH at f = ½ is estimated as pH = pKa. Be sure to check graphically whether you can assume that [H+] and [OH–] can be ignored at this pH. With three points estimated, the rest of the titration curve can be sketched by hand. The curve should be steepest at f = 1. The procedure can be summarized as follows (see also Appendix C, Section C.6.1): Step 1: Prepare a pC-­pH diagram for a C M HA solution. Step 2: Find the pH at the zeroth and first equivalence points. Apply the proton conditions for HA and NaA solutions to the pC-­pH diagram to find the pH at f = 0 and f = 1, respectively. Step 3: Find the pH at the half-­equivalence point. The half equivalence point occurs at about pH = pKa. Using the pC-­pH diagram, check to make sure that [H+] and [OH–] are much smaller than [HA] and [A–] at pH = pKa. Step 4: Connect the dots. Draw a smooth curve through the three points determined in the previous steps. Make sure the slope is steepest at f = 1. An example will make the use of this procedure clearer. Hemoglobin (Hb) is an important protein in blood. It helps transport oxygen and serves a major role in regulating the pH and acid–base chemistry in red blood cells. Protonated hemoglobin (HHb+) has a pKa of about 7.2. Since the oxygen-­carrying capacity of Hb varies with pH, you may want to know how the pH in the red blood cell is affected by acid or base addition. To sketch the titration curve, first find the pH values at f = 0, 1, and ½. Blood contains about 15 g of total hemoglobin per 100 mL or about 2.2×10–3 M since the average molecular weight of Hb is about 68,000 g/mole. Thus, the total hemoglobin concentrations is HbT = [HHb+] + [Hb] = 2.2×10–3 M. The pC-­pH diagram for the HHb+/ Hb system is shown in Figure 12.5. pH 0 2

0

2

4

6

8

10

[HHb+]

[Hb]

[OH–]

[H+]

12

14

4 pC

Key idea: To sketch the titration curve for a monoprotic acid: (1) Prepare a pC-­pH diagram for a C M HA solution; (2) Apply the proton conditions for an HA and NaA solution to the pC-­pH diagram to find the pH at f = 0 and f = 1, respectively; (3) Locate the half-­equivalence point at about pH = pKa (use the pC-­pH diagram to check and see if [H+] and [OH–] are much smaller than [HA] and [A–] at pH = pKa); and (4) Draw a smooth curve through the three points determined in the previous steps, making sure that the slope is steepest at f = 1

6 8 10 12 14

FIGURE 12.5  pC-­pH Diagram for 2.2×10–3 M Hemoglobin (Added as the chloride salt; dashed lines (left to

right) are f = 0, ½, and 1.)

ISTUDY

pH

12.3  |  Equivalence Points   259 12 11 10 9 8 7 6 5 4

0.0

0.5

1.0 f

1.5

2.0

FIGURE 12.6  Titration Curve for 2.2×10–3 M Hemoglobin (Hemoglobin is added as its chloride salt.)

Imagine that a neutral salt, say HHbCl → HHb+ + Cl–, is added to water at 2.2×10–3 M as a chemical model of a red blood cell and titrated with base. At f = 0, the untitrated solution is equivalent to an HHb+ solution. It has the following proton condition: [H+] + [HHb+] = [OH–]. Imposing this proton condition onto Figure 12.5, the equilibrium pH at f = 0 is about 4.9. At f = 1, the titrated solution is equivalent to an Hb solution. Thoughtful Pause What is the proton condition for an Hb solution? The proton condition for an Hb solution is [H+] + [HHb+] = [OH–]. From Figure 12.5, this is satisfied at about pH 9.3. At f = ½, the pH is about equal to the pKa 7.2, since [H+] and [OH–] are negligible at pH 7.2 (which you can check easily on Figure 12.5). The pH values at f = 0, 1, and ½ are shown in Figure 12.6. You can sketch the titration curve by connecting the dots with a steeper slope at f = 1. For greater accuracy, you must develop the titration curve equation from the charge balance. At any point in the titration: H



HHb

Na

OH

Cl

From mass balances on chloride and hemoglobin: HbT



HHb

Hb

ClT

Cl

Thus, the charge balance becomes (with CB = [Na+]):

CB

OH

Hb

H

OH

0

HbT

H

This equation is plotted in Figure  12.6. Another illustration of the titration of a monoprotic acid is provided in Worked Example 12.3. Worked Example 12.3 

Titration of a Monoprotic Acid

Both benzoic acid (formally, benzenecarboxylic acid) and the salt of its conjugate base (sodium benzoate) are used as food preservatives. A food producer wishes to transition from the manufacture of 10–­2 M stock solutions of benzoic acid to 10–­2 M stock solutions of sodium benzoate by adding NaOH. Can you estimate

ISTUDY

260   Chapter 12  Acid–Base Titrations the pH of the mixture at the beginning, middle, and end of the t­ ransition? How does the pH of the stock solutions change during the transition?

pC

Solution Let benzoic acid be represented by HA and benzoate by A–, with [HA] + [A–] = 10–­2 M. The pKa of HA is 4.2. A pC-­pH diagram for the benzoic acid system is shown below:

0 2 4 6 8 10 12 14

pH 0 2 4 6 8 10 12 14 [A–] [HA] [OH–]

[H+]

The beginning, middle, and end of the transition correspond to f = 0, ½, 2 and (dashed lines). At f = 0, the mixture is equivalent to benzoic acid (proton condition: [H+] = [A–] + [OH–]) and thus the pH is 3.1. At f = 1, the mixture is equivalent to benzoate (proton condition is: [H+] + [HA] = [OH–]) and thus the pH is 8.1. At f = ½, the pH is about equal to the pKa if [H+] and [OH–] small (confirmed in the pC-­pH diagram), so: pH = pKa = 4.2. At any point in the titration curve, the charge balance is: H

Na

A

OH

CB

A

OH

H

Or:

pH

This titration curve is plotted below with the equivalence points and half-­ equivalence point highlighted. 14 12 10 8 6 4 2 0

0.0 0.5

1.0

1.5

2.0

f

To summarize, the pH is about 3.1, 4.2, and 8.1 at f = 0, ½, and 1, respectively.

ISTUDY

12.3  |  Equivalence Points   261

12.3.4  Titration of a Base with Strong Acid The discussion of the titration of an acid with base in Sections 12.3.2 and 12.3.3 can be repeated for titration of a base with acid. For a monoprotic base, NaA, you might titrate with a strong acid (such as HCl) that dissociates completely in water. A typical charge balance would be: H



Na

OH

Cl

A

From mass balances: AT



HA

A , NaT

Na , and AT

NaT

Substituting CA for [Cl–] (see Section 12.2.3): CA



HA

H

OH

0

AT

H

OH

Once again, you can normalize the x-­axis of the titration curve. For the titration of bases with an acid, the function is called g: g



mol  of initial total material L CA AT

12 10

pH

g function: a normalized measure of the amount of acid titrant added and equal to the eq/L of acid added divided by the moles/L of initial material (units: eq/mol)

eq  acid added L

As with f, the function g has units of eq/mol. As an example, you could titrate a solution of ammonia with a strong acid (say, HCl). The resulting titration curve for a 10–3 M ammonia solution is shown in Figure 12.7. The interpretation of the titration curve in Figure 12.7 is similar to the interpretation of titration curves with base as the titrant. At g = 0 (the zeroth equivalence point), the solution is equivalent to a 10–3 M NH3 solution (pH 10.1). At g = 1 (the first equivalence point), the solution is equivalent to a 10–3 M solution of the conjugate acid of

8 6 4 2 0.0

Key idea: The titration of a monoprotic base is analogous to the titration of a monoprotic acid: the charge balance describes the titration curve

0.5

1.0

1.5

g FIGURE 12.7  Titration Curve for the Titration of 10 M Ammonia with Strong Acid –3

2.0

←Worked example Key idea: The solution formed by the titration of a monoprotic base to the first equivalence point is identical to a solution of the conjugate acid

ISTUDY

262   Chapter 12  Acid–Base Titrations ammonia, namely, ammonium (i.e., the solution is equivalent to, say, a 10–3 M NH4Cl solution: equilibrium pH ≈ 6.1). At g = ½, the concentration of base and conjugate acid are nearly equal ([NH3] ≈ [NH4+]) if the concentrations of H+ and OH– can be ignored. Thus, the pH at g = ½ is about equal to the pKa of the conjugate acid (NH4+ in this case, pKa = 9.3). Another instance of the titration of a monoprotic base with strong acid is shown in Worked Example 12.4.

Titration of a Monoprotic Base

Worked Example 12.4 

A dye manufacturer uses aniline (benzenamine, C6H8NH2) as a feedstock for synthesizing dyes. The various synthesis reactions require different pH values for optimum yield. The manufacturer wishes to know how the pH of a 10,000-­gallon tank containing aniline at one-­half its aqueous solubility changes as a function of the number of gallons of 40% (w/w) HCl added. Solution At its heart, this is a titration problem. Let B = aniline, HB+ = conjugate acid (C6H5NH3+), and BT = [B] + [HB+]. To find the pH values at the equivalence points, sketch the pC-­pH diagram (see below). The pKa of HB+ is 14 – pKb (aniline) = 4.6. The aqueous solubility is 36.07 g/L g 36.07 L or = 0.388 M, so BT = ½ (0.388 M) = 0.194 M. g 93 mol pH

pC

Key idea: For a monoprotic acid or base: f (titration of the acid) + g (titration of the conjugate base) = 1

0 2 4 6 8 10 12 14

0 2 4 6 8 10 12 14 [B] [HB+] [OH–] [H+]

At g = 0 (CA = 0), the mixture is equivalent to benzamine (proton condition: [H+] + [HB+] = [OH–]), and thus the pH is 8.95. At g = 1 (CA = 0.194 eq/L), the mixture is equivalent to the conjugate acid (proton condition: [H+] = [HB] + [OH–]), and thus the pH is 2.66. At g = ½ (CA = 0.097 eq/L), the pH is about equal to the pKa if [H+] and [OH–] are small (confirmed in the pC-­pH diagram). The charge balance at any point in the titration curve is: H

HB

CA

H

Cl

OH

Or: HB

OH

ISTUDY

12.3  |  Equivalence Points   263

The titration curve is: 10

pH

8 6 4 2 0 0.0

0.1

0.2

0.3

0.4

CA (eq/L)

You must translate the units of CA from eq/L to gallons of HCl/10,000 gallons. The density of 40% HCI is 1.1977 g/cm3 = 1197.7 g/L. Thus, its concentrag 1197.7 L tion is = 13.14 M = 13.14 N. So: 1 gal of 40% HCl g HCl g HCl 0.4 36.45 g 40% HCl mole eq 1 gal L per 10,000 gal represents a CA of = 1.314×10–3 eq/L. To convert 10, 000 gal CA from eq/L to gallons of 40% HCl per 10,000 gallons, divide by 1.314×10–3 as shown below: 13.14

10

pH

8 6 4 2 0

0

100

200

300

Acid Added (gal)

The foregoing analysis shows the similar nature of the titration of an acid with base and the titration of the conjugate base of the acid with acid. For example, the titration of an acid, HA, with NaOH is related to the titration of NaA with HCl. You can show (see Problem 3) that f + g = 1 at any point in the titration. An example with the titration of HOCl is shown in Figure 12.8. Using the lower x-­axis, the titration curve (read left to right) shows the titration of HOCl with strong base. Using the upper

10

1.0

g 0.5

0.0

9 pH

8 7 6 5 4

0.0

0.5 f

FIGURE 12.8  Titration Curve for the HOCl/OCl– System ([HOCl] + [OCl–] = 10–3 M)

1.0

Key idea: The titration of a monoprotic acid with base and the titration of the conjugate base of the acid with acid trace out the same titration curve

ISTUDY

264   Chapter 12  Acid–Base Titrations x-­axis, the titration curve (read right to left) shows the titration of the conjugate base with strong acid. Common salts of the conjugate base are NaOCl and Ca(OCl)2. The titration curve in Figure 12.8 provides a great deal of information about the chemistry of HOCl and OCl–. Thoughtful Pause Can you estimate the pKa of HOCl from the titration curve in Figure 12.8? At f = g = ½, you expect the pH to be about equal to the pKa if [H+] and [OH–] are negligible in the charge balance (here, if they are small compared to [OCl–]). Checking to see if [H+] and [OH–] are negligible, the pH at f = g = ½ is about 7.5, so [H+] = 10–7.5 M and [OH–] = 10–6.5 M. Thus, [H+] and [OH–] are negligible and the pKa is about 7.5. The actual pKa is 7.54.

12.3.5  Titration Curves at Extreme pH Values As discussed, titration curves can be sketched over a reasonable pH range from a pC-­ pH diagram and knowledge of the pKa of the acid in the system being titrated. The titration curves tend to level off at large positive or negative values of f and g.4 In other words, titration curves level off at extreme pH values. Why is this so? To answer this question, return once more to the charge balance. For the titration of a monoprotic acid (e.g., a C M HA solution) with a strong base, NaOH, the charge balance at any point in the titration is: [H+] + [Na+] = [OH–] + [A–]. At very high pH, OH KW the charge balance becomes: [Na+] ≈ [OH–] or CB ≈ [OH–] or f ≈ . TakC C H C ing logs and then solving for pH: pH = log + log( f ). Go to your favorite spreadsheet KW program and plot the function: y = constant + log(x). You will find that this function flattens out at large values of x. A similar argument can be made to show that the titration of a base with an acid should show a fairly flat titration curve at large g values (see Worked Example 12.5). Worked Example 12.5 

Titration of a Monoprotic Base at Low pH

Why does the titration curve for the titration of a base flatten out at larger g values (low pH)? Solution Consider the titration of a monoprotic base (e.g., a C M B solution, where B + H+ = HB+) with a strong acid, say HCl. The charge balance at any point in the titration is: [H+] + [HB+] = [Cl–] + [OH–]. At very low pH, this becomes: [H+] H 10 pH ≈ [Cl–] or CA ≈ [H+] or g ≈ . Rearranging: pH ≈ –pC – log(g). The C C function: y = constant – log(x) flattens out at large values of x. 4   At first blush, it may seem strange that f and g can take on negative values, since they appear to be the ratio of two positive numbers. It is possible, of course, to titrate an acid (say, C M HA) with an acid (say, HCl). In this case, it is C CA C CB CA . More general definitions of f and g are: f = B and g = A . Thus, f is negcommon to define: f = C C C th ative at pH values below the 0 equivalence point (acid added to acid) and g is negative at pH values above the 0th equivalence point (base added to base).

ISTUDY

12.4  |  Titration of Polyprotic Acids   265 12

pH

10 8 6 4 2 0.0

0.5

1.0

1.5

2.0

Key idea: Titration curves for the titration of an acid with base flatten out at high pH because K CB ≈ [OH–] = W and the H titration curve is a plot of CB pH versus CB or f C

f FIGURE 12.9  Titration Curve for the Titration of 10–3 M Phenol with a Strong Base

Return to the question: Why does the titration curve flatten out at extreme pH values? There are two factors contributing to this behavior. First, at extreme pH, nearly all the acid (or base) being titrated has been converted to its conjugate base (or conjugate acid) form. Thus, the pH is linear with log( f ) [or –log(g); see Worked Example 12.5]. Second, the titration curve is a log-­linear plot: a plot of –log{H+} versus f (or g). These two factors combine to cause the flattening of the titration curve at extreme pH values. One ramification of this flattening behavior is that features of the titration curve can be hidden. Consider the titration of a 10–3 M solution of phenol (pKa = 9.9) with base. The titration curve is shown in Figure 12.9. Note that the titration curve is nearly featureless. The slope of the titration curve at f = 1 is flat, not steep. Thoughtful Pause Why is the titration curve for 10–3 M phenol featureless? The phenol titration curve does not show the usual features because phenol is a very weak acid. By the point in the titration when f = 1, the pH is already high. By this point, [Na+] is already nearly equal to [OH–]. (You may want to confirm this point for yourself; see Problem 5) Thus, the flattening out of the titration curve at high pH is present in the titration curve for phenol even at relatively low f values. Note also that the pH at f = ½ is 9.79, not pKa = 9.9, since [OH–] is not negligible.

12.4  TITRATION OF POLYPROTIC ACIDS 12.4.1  Introduction The approach discussed in Section 12.3 also can be applied to diprotic acids. For the titration of a diprotic acid (say, a C M solution of H2A), the titration curve once again is described by the charge balance equation:

Na

H

HA

2 A2

OH

At f = 0, 1, and 2, CB = [Na+] is equal to 0, C (= [H2A] + [HA–] + [A2–]), and 2C (= 2[H2A] + 2[HA–] + 2[A2–]), respectively. You can show (Problem 4) that the titration solution is equivalent to H2A, NaHA, and Na2A solutions at f = 0, 1, and 2, respectively. The important equivalence points in the titration curve are summarized in Table 12.1.

Key idea: Titration curves for the titration of a base with acid flatten out at low pH because CA ≈ [H+] and the titration curve is a plot of pH CA versus CA or g C Key idea: Titration features can be lost at very low and high pH where the titration curve flattens out

ISTUDY

266   Chapter 12  Acid–Base Titrations Key idea: By using only one pC-­pH diagram, you can calculate the pH at f = 0, 1, and 2 easily for a diprotic acid (H2A) by imposing the proton conditions for H2A, NaHA, and Na2A systems. Key idea: The pH is about equal to pKa,1 at f = ½ (if [H+] and [OH–] can be ignored and if [A2–] HCO3–. For saline lakes (e.g., the Dead Sea), Na+ and Cl– usually dominate, although the dominant anion can vary (Wetzel, 1983).

Key idea: The major ions in the majority of fresh waters are Na+, K+, Ca2+, Mg2+, SO42–, Cl–, and HCO3–

ISTUDY

292   Chapter 13  Alkalinity and Acidity acid anions and the symbol CB to represent the sum of the eq/L of strong base cations. Equation 13.5 becomes:

H

CB

OH

CA

HCO3

2 CO32

eq. 13.6

H

eq. 13.7

Or:

CB C A

HCO3

2 CO32

OH

Equations 13.6 and 13.7 include all “other” ions from strong bases and acids in CB and CA respectively, and ignore “other” ions from weak acids and bases (but see Section 13.6). One more simplification: Express the bicarbonate and carbonate ions in terms of their alpha values. Thus, the charge balance becomes:

CB C A

1

2

2

CT

OH

H

eq. 13.8

In eq. 13.8, CT again is the total dissolved carbonate concentration: CT = [H2CO3*] + [HCO3–] + [CO32–]. How does all this relate to alkalinity? Comparing eqs. 13.3 and 13.8: alkalinity equation: Alk = CB – CA = (α1 + 2α2)CT + [OH–] – [H+]

Alk

CB C A

1

2

2

CT

OH

H

eq. 13.9

Equation  13.9 is called the alkalinity equation. It is one of the few equations in this text that you should memorize (although you can derive it quickly from the charge balance).

13.4  CHARACTERISTICS OF ALKALINITY

AND ACIDITY

13.4.1  What Does Alkalinity Mean?

Key idea: Alkalinity commonly is used as an indicator of buffering, even though it is not a measure of buffer intensity Worked example →

Remember that alkalinity is the acid neutralizing capacity for waters in which the acid–base chemistry is dominated by the carbonate family. Alkalinity tells you how much acid the water can accept to make it equivalent to an H2CO3* solution ( f = 0). Sometimes, alkalinity is thought of as a measure of buffering. In this sense, high alkalinity waters are thought of as being well buffered. Although natural waters with high alkalinities usually are well buffered, you must be careful not to confuse alkalinity with buffer intensity. Alkalinity is a capacity. It measures the total amount of acid a water can accept (to a specified equivalence point). Buffer intensity measures the response of a system to acid (or base) input.3 As an example, consider a 1×10–3 N NaOH solution and a 1×10–3 M NaHCO3 solution. You can show (see Section 13.5.3) that both solutions have the same alkalinity. In other words, it takes the same amount of acid to make each solution equivalent to

  Alkalinity and the buffer intensity (β) are not identical, but they are related. Recall that alkalinity is equal to the

3

pH f

ANC if the carbonic acid system dominates. One way to define ANC is: ANC = pHi is the initial pH of the sample and pHf = 0 is the pH at f = 0.

pHi

0

CB (dpH) pH

pH f

0

(dpH), where pHi

ISTUDY

13.4  |  Characteristics of Alkalinity and Acidity   293 12

1×10–3 M NaOH

f=0

pH

10 8 6

1×10–3 M NaHCO3

4 0.0000 0.0002 0.0004 0.0006 0.0008 0.0010 0.0012 Acid Added (eq/L) FIGURE 13.3  Comparison of the pH During Titration for 1×10–3 N NaOH and 1×10–3 M NaHCO3

Solutions

an H2CO3* solution containing the same CT as the starting solution.4 However, the pH values for the two solutions along the path from the starting solutions to f = 0 are very different (see Figure 13.3). The bicarbonate solution clearly provides better pH buffering near neutral pH. Although alkalinity is not a direct measure of the buffer intensity, alkalinity is still commonly used as an indicator of buffering. Why? Alkalinity is a great measure of buffering if the lower pH of interest is near f = 0 for the CT in your system. As you will see in Section 13.4.2, this is a pH range of about 4.3–4.7 for most natural waters. Not coincidently, this lower pH of 4.3–4.7 corresponds to the pH at which aquatic biota begin to be adversely affected in natural water bodies. If you are more interested in a different pH range, however, alkalinity may be misleading. Say, for example, that you have a water with pH 7.4, CT = 1.08×10–3 M, and Alk = 1×10–3 eq/L. (You should verify that these values are consistent with eq. 13.3.) Suppose you are concerned about the pH dropping below 6.8 as a result of an acidic input. Although the alkalinity is 1×10–3 eq/L, it takes only about 1.8×10–4 eq/L of strong acid (about 18% of the alkalinity) to reduce the pH to 6.8. In this case, alkalinity is not so valuable. Use caution when accepting alkalinity as an indicator of buffering if the lower pH range of interest is different than 4.3–4.7. Alternative types of alkalinity for different pH ranges of interest will be developed in Section 13.4.3. Another example of the relationship between alkalinity and the desired pH range is shown in Worked Example 13.1. Worked Example 13.1 

Alkalinity and the pH Range of Interest

A river water has a CT value of 7.8×10–4 M and pH 7.9. An acidic input is anticipated. If the trout population in the river is to be maintained, the river pH cannot drop below 7.5. If the trout are relocated, the river pH can drop to as low as 5.8. Explain quantitatively the value of alkalinity in determining the vulnerability of the river pH to the acidic input. Assume CT remains unchanged after the acidic input. Solution First, calculate the alkalinity.

  Note that the CT values of the starting solutions are different: 1×10–3 N NaOH has CT = 0, and 1×10–3 M NaHCO3 has CT = 1×10–3 M. 4

Key idea: Use caution when accepting alkalinity as an indicator of buffering if the lower pH range of interest is different from 4.3–4.7

ISTUDY

294   Chapter 13  Alkalinity and Acidity At pH 7.9, α1 = 0.972 and α2 = 0.00387. From eq. 13.3, Alk = 7.8×10–4 eq/L. After the acidic input, the charge balance is: H

original CB

1

2

2

CT

OH

original C A added C A

You know: original Alk = original CB – original CA. So: H

original Alk

1

2

2

CT

OH

added C A

Solving for the added CA, an acidic input of 3.0×10–5 eq/L, or about 4% of the alkalinity, is required to get to pH 7.5. To get pH 5.8 requires an acidic input of 5.8×10–4 eq/L, or about 76% of the alkalinity. Thus, alkalinity is a reasonable estimate of the acidic input required to reach pH 5.8, but greatly overestimates the acidic input required to get to pH 7.5.

13.4.2  Measuring Alkalinity and Acidity by Titration Alkalinity is the amount of strong acid required to make a water equivalent to an H2CO3* solution ( f = 0 for H2CO3*), and acidity is the amount of strong base required to make water equivalent to a Na2CO3 solution (g = 0 for Na2CO3). This means that alkalinity and acidity are tied intimately to the H2CO3* titration curve. It also means that titration with strong acid is an excellent way to measure alkalinity. Similarly, titration with strong base is used to measure acidity. Alkalinity is measured by titrating the water sample to the zeroth equivalence point for H2CO3* for that sample. Thoughtful Pause How do you know the pH at the zeroth equivalence point for H2CO3* for a given water sample? Key idea: The pH values at the equivalence points for alkalinity and acidity depend on the total carbonate concentration (CT) of the sample

endpoint: a key point in a titration, usually selected to be an estimate of a true equivalence point in a titration

The pH of the H2CO3* equivalence point depends on CT. Waters with higher CT have a lower pH at f = 0 (more weak acid present at f = 0). Similarly, waters with higher CT have a higher pH at g = 0 (more weak base present at g = 0). This is shown graphically in Figures 13.4 and 13.5. In Figure 13.4, note that for natural waters (CT ≈ 1 mM), the pH at f = 0 is about 4.7 and the pH at g = 0 ( f = 2) is about 10.5. As shown in Figures 13.4 and 13.5, the exact pH at the f = 0 and g = 0 equivalence points varies with CT. To be precise, you would have to measure CT with every water sample, calculate the pH at the equivalence points, and titrate to those pH values. However, because the range of CT values in most natural waters is small, we use one set of equivalence point pH values for most waters. For routine analysis, alkalinity is determined by titration to pH 4.5. Since this pH value does not necessary correspond to exactly f = 0, we refer to it as an endpoint, rather than an equivalence point. In general, the endpoint pH depends on the alkalinity (see Problem 8). Endpoints generally are set in three ways: fixed pH (e.g., pH 4.5 as an indicator of f = 0), inflection points (where the entire titration curve is generated in the laboratory, and the strong acid or base required to reach an inflection point is recorded), and pH

ISTUDY

13.4  |  Characteristics of Alkalinity and Acidity   295 11 10

pH

9 8 7

CT = 1×10–3 M CT = 5×10–4 M CT = 1×10–4 M

6 5 4 0.0

0.5

1.0

1.5

2.0

f

pH

FIGURE 13.4  Carbonic Acid Titration Curves for Different CT Values

12 11 10 9 8 7 6 5 4 0.0001

g = 0 (f = 2)

f = 1 (g = 1) f = 0 (g = 2) 0.001

0.01

CT (M, log scale) FIGURE 13.5  pH Values at Integral Equivalence Points as a Function of CT

indicators. A pH indicator is an acid–base pair where the acid is one color or colorless in solution and the conjugate base is a different color in solution. Only a small amount of the indicator is added to the sample to avoid changing the sample pH. If the pKa of the acid is near the pH of the endpoint, then a color change is observed at pH values near the endpoint. Characteristics of several acid–base indicators are listed in Table 13.2. In the older literature, the endpoint approximating the f = 0 equivalence point is called the methyl orange endpoint because the pH indicator methyl orange was used to visualize the endpoint. The preferred pH indicator currently is bromcresol green (or the bromcresol Table 13.2 Common Endpoint Indicators Indicator

Acid Color

Methyl orange Bromcresol green

Red (HB+) Yellow (HA)

Phenolphthalein Metacresol purple

Clear (H2A) Yellow (HA)

Base Color

Near f = 0 (g = 2): Orange (B) Blue (A–) Near f = 2 (g = 0): Pink (A–) Purple (A–)

pKa

Transition pH Range

3.8 4.8

4.5–3.2 5.4–3.8

9.7 8.3

8.2–10.0 7.6–9.2

methyl orange endpoint: a pH endpoint of about pH 4.5 to 3.2, previously used to estimate the alkalinity equivalence point (f = 0 for carbonic acid)

ISTUDY

296   Chapter 13  Alkalinity and Acidity green-­methyl red mixed indicator). Phenolphthalein is used for endpoints near neutral pH (e.g., the titration of a strong acid with a strong base; see also Section 13.4.3).

13.4.3  Types of Alkalinity

phenolphthalein ­alkalinity: the amount of strong acid required to bring the sample to the f = 1 (NaHCO3) equivalence point (an endpoint of about pH 8.2–8.3)

0th equiv. point 11

1st equiv. point

2nd equiv. point

10 9 pH

Key idea: Different types of alkalinity are defined for different endpoints

As discussed in Section 13.4.1, alkalinity is not always the best measure of buffering, especially if you are more interested in the ability to neutralize acids in the neutral pH range and are not interested in letting the pH drop to about 4.5. As a result, you can define several types of alkalinity, depending on the sample and pH range of interest (see also Sawyer and McCarty 1978). Each type of alkalinity is related to the titration to a specific equivalence point (or, in practice, a specific endpoint). Each type will be discussed below and illustrated with example titration curves for three hypothetical samples. The samples all have CT = 1 mM, where the equivalence points at f = 0, 1, and 2 are about pH 4.7, 8.2, and 10.5, respectively. The samples have different pH and alkalinity values: Sample A (Figure 13.6) has a pH value of 7.0 (between the zeroth and first equivalence points), sample B (Figure 13.7) has a pH value of 9.5 (between the first and second equivalence points), and sample C (Figure 13.8) has a pH value of 10.8 (greater than the second equivalence point). Alkalinity (also total alkalinity, or Alk) is defined as the amount of strong acid required to bring the sample to the f = 0 (H2CO3*) equivalence point, an endpoint at about pH 4.5. The hypothetical titration curve for a sample at pH 7.0 is shown in Figure 13.6, along with the alkalinity. Alkalinity describes the total acid neutralizing capacity of a water, within the pH range of biological interest. It can be calculated from the proton condition for H2CO3*. The proton condition is [H+] = [HCO3–] + 2[CO32–] + [OH–] and Alk = (α1 + 2α2)CT + [OH–] – [H+]. The main contributor to alkalinity near pH 7 is HCO3–. Phenolphthalein alkalinity (also p-­alkalinity, or p-­Alk).5 Phenolphthalein alkalinity is defined as the amount of strong acid required to bring the sample to the

8

(total) Alk

7 6 5 4 acid added

base added

FIGURE 13.6  Example Alkalinity Titration for a Sample Originally at pH 7.0 (Sample A in the text)

  The name stems from the pH indicator phenolphthalein, which is commonly used to indicate the endpoint (see Table 13.2). However, the name “phenolphthalein alkalinity” is used regardless of the manner in which the endpoint is determined. For example, the endpoint could be determined by pH, inflection point, or another pH indicator (e.g., metacresol purple). The word phenolphthalein comes from phenol + phthal (from naphtha) + ein (from -­ene, meaning a double bond). The word naphtha comes from the Persian neft, perhaps related to the Greek nephos, meaning a cloud or mist. Nephos is the source of the word nephelometric, found in the common units of turbidity: nephelometric turbidity units, or NTU.

5

ISTUDY

13.4  |  Characteristics of Alkalinity and Acidity   297

11 10

0 th equiv. point

1st equiv. point

2nd equiv. point

(total) Alk

pH

9 8 7

bicarb. alk. + ½(carb. alk.)

p-Alk = ½(carb. alk.)

6 5 4 acid added

base added

FIGURE 13.7  Example Alkalinity Titration for a Sample Originally at pH 9.5 (Sample B in the text)

11

0th equiv. point

1st equiv. point

2nd equiv. point

(total) Alk

10

OH-Alk

pH

9 8 7

bicarb. alk. + ½(carb. alk.)

p-Alk = ½(carb. alk.)

6 5 4 acid added FIGURE 13.8  Example Alkalinity Titration for a Sample Originally at pH 10.8 (Sample C in the text)

f = 1 (NaHCO3) equivalence point, an endpoint at about pH 8.3. (Note from Figures 13.4 and 13.5 that the f = 1 equivalence point is fairly independent of CT.) If the pH of the sample is less than 8.3, then the phenolphthalein alkalinity is less than zero, but usually reported as zero (i.e., no acid is needed to reach pH 8.3, as in Figure 13.6). The hypothetical titration curve for a sample at pH 9.5 is shown in Figure 13.7, along with the phenolphthalein alkalinity. Phenolphthalein alkalinity describes the acid neutralizing capacity of a water near neutral pH. Phenolphthalein alkalinity can be calculated from the proton condition for NaHCO3. The proton condition is [H+] + [H2CO3] = [CO32–] + [OH–] and p-­Alk = (α2 – α0)CT + [OH–] – [H+]. Except at very high pH, the main contributor to p-­Alk is CO32–. Caustic alkalinity (also OH-­Alk, or hydroxide alkalinity). Caustic (from the Greek kaiein to burn) alkalinity is defined as the amount of strong acid required to bring the sample to the f = 2 (Na2CO3) equivalence point, an endpoint of about pH 9.5–11. If the pH of the sample is less than the endpoint pH, then the caustic alkalinity is less than zero (but usually reported as zero). The hypothetical titration curve for a sample at pH 10.8 is shown in Figure 13.8, along with the caustic alkalinity. Caustic alkalinity can be calculated from the proton condition for Na2CO3. The proton condition is [H+] + 2[H2CO3] + [HCO3–] = [OH–] and OH-­Alk = [OH–] – (α1 + 2α0)CT – [H+]. The main contributor to caustic alkalinity is OH–.

caustic alkalinity: the amount of strong acid required to bring the sample to the f = 2 (Na2CO3) equivalence point

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298   Chapter 13  Alkalinity and Acidity

carbonate alkalinity: the portion of the total alkalinity contributed by CO32– (= 2(p-­Alk) – OH-­Alk)

bicarbonate alkalinity: the portion of the total alkalinity contributed by HCO3– (= total alkalinity – carbonate alkalinity = total alkalinity – 2(p-­Alk) + OH-­Alk)

 orked W example →

Carbonate alkalinity. As long as the pH of the sample is less than about 10.5, the total alkalinity is primarily from HCO3– and CO32–. The total alkalinity can be divided into two parts: carbonate alkalinity (the portion of the total alkalinity contributed by CO32–) and bicarbonate alkalinity (the portion of the total alkalinity ­contributed by HCO3–). After a sample has been titrated to the phenolphthalein endpoint, almost all of the carbonate has been converted to bicarbonate (since pH 8.3 < pKa,2 = 10.3), but almost none of the bicarbonate has been converted to carbonic acid (since pH 8.3 > pKa,1 = 6.3). The carbonate alkalinity is defined to be twice the p-­Alk.6 The relationship between the titration curve and the carbonate alkalinity is shown in Figures 13.7 and 13.8. Bicarbonate alkalinity. The bicarbonate alkalinity is the difference between the total alkalinity and the carbonate alkalinity. The amount of strong acid required to move from the phenolphthalein endpoint to the pH 4.5 endpoint is the bicarbonate alkalinity plus the remaining one-­half of the carbonate alkalinity. Thus: bicarbonate alkalinity

total alkalinity carbonate alkalinity total alkalinity 2( p - Alk ) OH Alk total alkalinity 2( p -A Alk)

The relationship between the titration curve and the bicarbonate alkalinity is shown in Figures 13.7 and 13.8. You can calculate the types of alkalinity in a sample from a titration. Suppose you have a 200 mL water sample and titrate with 0.05 N sulfuric acid. The starting pH is 8.6. It takes 0.3 mL to titrate to pH 8.3 and 19.8 mL to titrate to pH 4.5.

Thoughtful Pause What are the Alk, p-­Alk, caustic Alk, carbonate Alk, and bicarbonate Alk of the water sample?

The caustic alkalinity (OH-­Alk) is zero, since the pH of the sample is less than the caustic alkalinity endpoint of about pH 9.5. The total Alk is the eq/L to bring the sample to the methyl orange (pH 4.5) endpoint. Therefore, the total Alk is

(19.8 mL) 0.05 200 mL

eq L

= 5.0×10 eq/L. The p-­Alk is the eq/L to bring the sample to the phenolphthalein –3

(pH 8.3) endpoint. Therefore, the p-­Alk is

(0.3 mL) 0.05 200 mL

eq L = 7.5×10–5 eq/L. The car-

bonate Alk is 2(p-­Alk) – OH-­Alk = 2(7.5×10 eq/L) – 0 = 1.5×10–4 eq/L. The bicarbonate –5

  This definition is strictly true only if you are between the first and second equivalence points. In general, you must correct for OH-­Alk when calculating the carbonate and bicarbonate alkalinity. Thus: carbonate alkalinity = 2(p-­Alk) – OH-­Alk, and bicarbonate alkalinity = total alkalinity – 2(p-­Alk) + OH-­Alk.

6

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13.4  |  Characteristics of Alkalinity and Acidity   299

Alk is the Alk minus the carbonate Alk or 5.0×10–3  – 1.5×10–4 = 4.9×10–3 eq/L. You can see that most of the alkalinity is provided by bicarbonate (bicarbonate Alk > carbonate Alk >> OH-­Alk). This makes sense, since HCO3– is the predominant carbonate species at pH 8.6. Other illustrations of the different types of alkalinity are presented in Worked Example 13.2. For another take on carbonate and bicarbonate alkalinity, see Problem 31.

Worked Example 13.2 

Alkalinity Types from Titration Curves

pH

A 250 mL lake water sample was titrated with 0.1 N HCl, and the titration curve shown below was obtained. Find the alkalinity, p-­Alk, OH-­Alk, carbonate alkalinity, and bicarbonate alkalinity. 8 7 6 5 4 3 0 2 4 6 8 10 12 14 0.1 N HCl Added (mL)

Solution The initial pH of the sample is well below the caustic alkalinity endpoint (pH 9.5), so OH-­Alk = 0. The initial pH also is below the p-­Alk endpoint (about pH 8.3: note that the inflection point around the NaHCO3 equivalence point is not fully developed). Thus, p-­Alk = 0. Since the carbonate alkalinity is 2(p-­Alk), then the carbonate alkalinity = 0. The total alkalinity is the acid added to reach about pH 4.5 or about eq (12 mL) 0.1 L = 4.8×10–3 eq/L or 240 mg/L as CaCO (see 12 mL. The Alk is 3 250 mL Section 13.4.6 for alkalinity units). The bicarbonate alkalinity is equal to the total Alk since OH-­Alk and carbonate alkalinity are zero. To summarize: Alk = 4.8×10–3 eq/L (240 mg/L as CaCO3) p-­Alk = 0 OH-­Alk = 0 carbonate alkalinity = 0 bicarbonate alkalinity = 4.8×10–3 eq/L (240 mg/L as CaCO3)

13.4.4  Types of Acidity Different types of acidity also can be defined, depending on the endpoint. The three types of acidity are total acidity, CO2 acidity, and mineral acidity.

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300   Chapter 13  Alkalinity and Acidity

CO2-­Acy: the amount of strong base required to bring the sample to the g = 1 (NaHCO3) ­equivalence point mineral acidity (H-­Acy): the amount of strong base required to bring the sample to the g = 2 (H2CO3*) equivalence point

Acidity (also total acidity, or Acy). Acidity is analogous to alkalinity. Acidity is defined as the amount of strong base required to bring the sample to the g = 0 (Na2CO3) equivalence point. Acidity describes the total base neutralizing capacity of a water, within the pH range of biological interest. It can be calculated from the proton condition for Na2CO3. The proton condition is [H+] + 2[H2CO3] + [HCO3–] = [OH–] and Acy = [H+] – (2α0 + α1)CT – [OH–]. CO2 acidity (also CO2-­Acy). CO2-­Acy is analogous to p-­Alk. It is defined as the amount of strong base required to bring the sample to the g = 1 (NaHCO3) equivalence point, an endpoint of about pH 8.3. CO2 acidity describes the base neutralizing capacity of a water near neutral pH. It can be calculated from the proton condition for NaHCO3. The proton condition is [H+] + [H2CO3] = [CO32–] + [OH–] and CO2-­Acy = [H+] + (α0 – α2)CT – [OH–]. Mineral acidity (also H-­acidity, or H-­Acy). Mineral acidity is analogous to caustic alkalinity. Mineral acidity is defined as the amount of strong base required to bring the sample to the g = 2 (i.e., H2CO3*) equivalence point, an endpoint of about pH 4.5. For all but the most acidic waters, the mineral acidity is less than zero. The main contributor to H-­Acy is H+ (i.e., strong acids). Mineral acidity can be calculated from the proton condition for H2CO3*. The proton condition is [H+] = [HCO3–] + 2[CO32–] + [OH–] and H-­Acy = [H+] – (α1 + 2α2)CT – [OH–].

13.4.5  Comparison of Alkalinity and Acidity Types Characteristics of the types of alkalinity and acidity are summarized in Table 13.3. The types of alkalinity and acidity are related as follows (see also Problem 17): Acy OH - Alk 0 CO2 Acy p - Alk H - Acy Alk 0



0

Table 13.3 Summary of the Characteristics of Alkalinity Name

Alkalinity (total alk., Alk) Phenolphthalein alkalinity (p-­Alk) Caustic alkalinity (OH-­Alk) Carbonate alkalinity Bicarbonate alkalinity Acidity (total acidity, Acy) CO2 acidity (CO2-­Acy) Mineral acidity (H-­Acy)

Equiv. Point

f=0 (H2CO3*) f=1 (NaHCO3) f=2 (Na2CO3) N/A N/A g=0 (Na2CO3) g=1 (NaHCO3) g=2 (H2CO3*)

Formula

pH

Main ­Contributor

(α1 + 2α2)CT + [OH–] – [H+] (α2 – α0)CT + [OH–] – [H+] [OH–] – (α1 + 2α0) CT – [H+] 2(p-­Alk) – OH-­Alk Alk – carbonate alk. [H+] + (2α0 + α1) CT – [OH–] [H+] + (α0 – α2) CT – [OH–] [H+] – (α1 + 2α2) CT – [OH–]

4.5

HCO3–

8.3

CO32–, OH–

9.5–­11

OH–

N/A

CO32–

N/A 9.5–­11 8.3

HCO3– H2CO3*, HCO3– H2CO3*

4.5

H+

ISTUDY

13.4  |  Characteristics of Alkalinity and Acidity   301

It is possible for the types of alkalinity and acidity to take negative values. It is common practice not to report negative alkalinity or acidity values, but rather to report them as zero and use the positive corresponding acidity or alkalinity values, respectively. Thoughtful Pause How would you report the p-­Alk and CO2-­Acy for the sample shown in Figure 13.6? The p-­Alk is the eq/L of acid needed to get to the bicarbonate equivalence point. The CO2-­Acy is the eq/L of base needed to get to the same equivalence point (about pH 8.3). The pH in this sample is less than 8.3, so a negative amount of acid is needed. You would report the p-­Alk as zero. The CO2-­Acy can be determined in Figure 13.6 by the horizontal distance between the sample and the f = g = 1 equivalence point. It has a positive value.

13.4.6  Units of Alkalinity One common set of units for alkalinity, from eqs. 13.3 and 13.9, is equivalents per liter (or meq/L). Clearly, from the alkalinity equation (eq. 13.9), alkalinity will be calculated in units of eq/L if CB and CA are in units of eq/L. Thoughtful Pause What are the units of alkalinity if it is calculated from Alk = (α1 + 2α2)CT + [OH–] – [H+]? Recall that eq. 13.9 was derived from a charge balance. Therefore, each term has units of eq/L moles or charges/L. You could rewrite eqs. 13.3 and 13.9 as:

Alk

1 eq/mol

1

2 eq/mol

2

CT

1 eq / mol OH

1 eq/mol H



(Remember that α1CT = [HCO3–] and α2CT = [CO32–], so the terms have one and two equivalents of charge per mole, respectively.) The units of eq/mol on the coefficients usually are not written out but cannot be ignored. Recall that CT, [OH–], and [H+] have units of mol/L. Thus, using either eq. 13.3 or 13.9, you will calculate the alkalinity in units of eq/L.7 There is another common set of units for alkalinity based on the “mass as” nomenclature (see Section 2.4.2). Alkalinity commonly is expressed in units of mg/L as CaCO3. Recall this means that you use the molar mass of CaCO3 for the molar mass of alkalinity. Why use these units? As you will see in Chapter 19, calcium carbonate (in the form of the mineral calcite) is a common source of alkalinity in natural waters. Also, 7  It may not be immediately apparent that alkalinity will be in units of eq/L if calculated from eq. 13.3. Recall that eq. 13.3 was derived from a proton condition. Thus, the units of the coefficients in eq. 13.3 are “excess or deficient protons over the zero proton level.” Since the zero proton level is set here for uncharged species (H2CO3* and H2O), x excess protons and x deficient protons correspond to x eq of charge per mole. Thus, alkalinity will be in units of eq/L if calculated from eq. 13.3.

Key idea: Alkalinity will be calculated in units of eq/L if CB and CA are in units of eq/L

ISTUDY

302   Chapter 13  Alkalinity and Acidity calcium carbonate is a convenient choice for alkalinity units because the molar mass of CaCO3 is about 100 g/mol. How do you convert units of eq/L to units of mg/L as CaCO3? As calcium carbonate dissolves, it produces Ca2+, which has 2 eq of charge per ion. Thus: 100 g 1 eq/L Alk = 1 Key idea: 1 eq/L Alk is equivalent to 50,000 mg/L Alk as CaCO3 Worked example →

eq L

mol CaCO3 2 eq

= 50 g/L Alk as CaCO3.

mol CaCO3 To summarize:

1 eq / L Alk 50 g / L Alk as CaCO3 50, 000 mg / L Alk as CaCO3 1 meq / L Alk 50 mg / L Alk as CaCO3 As examples of units conversion, 2.1×10–3 eq/L Alk is equivalent to:



2.1 10 3 eq / L

(2.1 10 3 eq / L)(50,000 mg / L as CaCO3 per eq / L) 105 mgg / L as CaCO3

An alkalinity of 175 mg/L as CaCO3 corresponds to:

mg 175   as CaCO3 L meq mg 50    as CaCO3 per  L L  3.5 meq /L

175 mg/L as CaCO3

You should practice converting the units of alkalinity between eq/L (or meq/L) and mg/L as CaCO3 until the conversion becomes second nature to you.

13.5  USING THE DEFINITIONS OF ALKALINITY

TO SOLVE PROBLEMS

13.5.1  Alkalinity Definitions It is very important to see that eq. 13.9 is really three equations in one. First, eq. 13.9 tells you that: Key idea: The alkalinity equation is really three equations in one: Alk = (α1+ 2α2)CT + [OH–] – [H+], CB – CA = (α1+ 2α2)CT + [OH–] – [H+], and Alk = CB – CA



Alk

(

2

1

2

)CT [OH ]

H

This is identical to eq. 13.3. Second, eq. 13.9 tells you that:

CB C A

(

1

2

2

)CT [OH ]

H

as in eq. 13.8. Finally, eq. 13.9 tells you that:

Alk

C B C A

This is a new expression and a very powerful way to calculate alkalinity. The secret to solving alkalinity problems is to select the most appropriate definition from among the three equations in eq. 13.9. A few examples will illustrate that much effort can be saved by selecting the proper definition.

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13.5  |  Using the Definitions of Alkalinity to Solve Problems   303

13.5.2  Alkalinity as (α1 + 2α2)CT + [OH–] – [H+] According to this definition, alkalinity is related to two other parameters: pH and CT. If you know two out of the three parameters (i.e., two of Alk, pH, and CT), then you can calculate the other parameter easily. As an example, consider a water with CT = 2×10–3 M and pH 7.5. For Ka,1 = 10–6.3 and Ka,2 = 10–10.3, you can calculate that the alkalinity is 1.9×10–3 eq/L or 94 mg/L as CaCO3. What is CT if the alkalinity of a water is 140 mg/L as CaCO3 (2.8 meq/L) and the pH is 7.8? You can show that: CT

Alk [OH ] [H ] 1 2 2

or CT = 2.9×10–3 M. The more challenging calculation is to find pH by knowing CT and alkalinity. You can set up a spreadsheet and determine the pH by iteration or use a nonlinear solver. For example, for a water with CT = 1.5×10–3 M and Alk = 1.2×10–3 eq/L, you should be able to confirm that the pH is about 6.9. Other examples are worked out in Worked Example 13.3.

Alkalinity Calculations: Alk = (α1+ 2α2)CT + [OH–] – [H+]

Worked Example 13.3 

Complete the water quality calculations for the following three water samples: Sample #1: Alk = ?, CT = 2.1 mM, pH 6.8 Sample #2: Alk = 157 mg/L as CaCO3, CT = ?, pH 7.2 Sample #3: Alk = 1.5 meq/L, CT = 20 mg/L as C, pH = ? Solution Sample #1 is at pH 6.8, so you can show α1 = 0.760 and α2 = 2.40×10–4. Therefore: Alk = (0.760 + 2×2.40×10–4)(0.0021 M) + 10–7.2 – 10–7.2 = 1.60×10–3 eq/L or 80 mg/L as CaCO3. For Sample #2, Alk =

CT

157

mg as CaCO3 L = 3.14×10–3 eq/L. Rearranging: mg 50 meq

Alk [OH ] [H ] 1 2 2 3.14 10

3

10

6.8

10

7.2

0.887 2(8.877 10 4 )

3.53 10 3 M For Sample #3, Alk = 1.5 meq/L and CT = ­Iterating: pH 7.24.

20

mg

12

as C L = 1.67×10–3 M. mg mmol

Key idea: If you know two of Alk, pH, and CT, then you can calculate the other parameter easily ←Worked example

ISTUDY

304   Chapter 13  Alkalinity and Acidity The expression for alkalinity in the title of this section emphasizes that alkalinity is a function of pH and CT. You could create a three-­dimensional plot showing the relationship among Alk, pH, and CT, but such plots are hard to read. A common approach is to plot lines of equal pH for pairs of Alk and CT. This approach was introduced by American geologist Kenneth S. Deffeyes8 (1931–2017) (see Deffeyes 1965), so these plots sometimes are called Deffeyes diagrams. An example is shown in Figure  13.9. Deffeyes diagrams are easy to generate since a plot of Alk versus CT clearly has a slope of (α1 + 2α2) and an intercept of [OH–] – [H+]. Note that this intercept is nearly zero, except at extreme pH values (see also Figure 13.9). The Deffeyes diagram is difficult to read near neutral pH. In many cases, it is more valuable to plot pH on the x-­axis and show lines for constant Alk (Figure  13.10) or constant CT (Figure 13.11). Although the lines in Figures 13.10 and 13.11 are curved, the diagrams are much easier to read than the Deffeyes diagram near neutral pH. You can verify the Alk, pH, and CT values in the examples above and in Worked Example 13.3 using Figures 13.9, 13.10, and 13.11.

4.0

11

3.5

10.5

10

8.5

8

9.5 9

7.5 7

Alk (meq/L)

3.0 2.5

6.5

2.0 6

1.5 1.0

5.5

0.5

4.55 4 3.5

0.0 –0.5 0.0

0.5

1.0

1.5

2.0 2.5 CT (mM)

3.0

3.5

4.0

FIGURE 13.9  Deffeyes Diagram for Alkalinity (lines are constant pH)

4.0

3.8 4.0 3.6 3.4 3.2 3.0 2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2

3.5

CT (mM)

3.0 2.5 2.0 1.5 1.0 0.5

0.0

0.0 3.0

4.0

5.0

6.0

7.0 8.0 pH

9.0 10.0 11.0 12.0

FIGURE 13.10  CT-­pH Diagram Showing the Relationship Among Alkalinity, CT, and pH (lines are constant Alk in meq/L)

  Deffeyes evidently was quite a character. He was profiled in John McFee’s 1981 book Basin and Range.

8

ISTUDY

13.5  |  Using the Definitions of Alkalinity to Solve Problems   305 4.0

4.0 3.8 3.6 3.4 3.2 3.0 2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0

3.5

Alk (meq/L)

3.0 2.5 2.0 1.5 1.0 0.5 0.0 3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0 11.0 12.0

pH FIGURE 13.11  Alk-­pH Diagram Showing the Relationship Among Alkalinity, CT, and pH (lines are ­constant CT in mM)

A final lesson from Alk = (α1 + 2α2)CT + [OH–] – [H+] concerns the alkalinity near neutral pH. For reasonable values of CT and pH conditions near neutral pH, [HCO3–] is much greater than [CO32–], [OH–], and [H+]. In other words: α1CT >> α2CT, [OH–], and [H+]. Thus, for reasonable values of CT and near neutral pH: Alk

HCO3

CT (for reasonable values of CT and near neutral pH) 1

This approximation is extremely useful. It allows you to estimate the alkalinity of many natural waters with ease. For example, from Table 13.1, you may estimate that the average alkalinity in the Great Lakes is about [HCO3–] = 1.65 meq/L = 83 mg/L as CaCO3. The approximation also allows you to estimate CT. For example, the CT of a mg as CaCO3 125 L water with pH 7.7 and alkalinity = 125 mg/L as CaCO3 = = 2.5 meq/L mg 50 eq meq 2.5 10 3 Alk L –3 is about = = 2.6×10 M. The exact solution for CT is 2.59×10–3 M. eq 1 0.959 mol A CT value of about 2.6×10–3 M also can be estimated from Figures 13.9, 13.10, and 13.11. This is a good example of how the diagrams in Figures 13.10 and 13.11 are much easier to use than the Deffeyes diagram near neutral pH.

13.5.3  Alkalinity as CB – CA This form of the alkalinity definition is very valuable when individual ion concentrations are known. As an example with a pure solution, what is the alkalinity of a 1×10–3 N NaOH solution? You could calculate the alkalinity the hard way: 1. Find the pH. The charge balance, [Na+] + [H+] = [OH–], and KW expression allow you to calculate pH 11.

Key idea: Near neutral pH and for reasonable values of CT: Alk ≈ [HCO3–] = α1CT

←Worked example

ISTUDY

306   Chapter 13  Alkalinity and Acidity 2. Find CT. It is clear that CT = 0 from starting materials. 3. Use eq. 13.3 to calculate the alkalinity. Alk = 1×10–3 eq/L (from eq. 13.3 or Figures 13.9–13.11)

 orked W example →  orked W example →

Key idea: When H+ is released from a process, alkalinity is consumed

 orked W example →

You can calculate the alkalinity much more easily in this case using the CB  – CA expression: Alk = CB – CA = [Na+] – 0 = 1×10–3 eq/L. Again, selecting the most appropriate definition of alkalinity saves a lot of computational time and effort. As another example, find the alkalinity of a 1×10–3 M NaHCO3 solution. Hard way: Find the pH (equilibrium calculation yields pH 8.3), find CT (CT = 1×10–3 M from starting materials), and use eq. 13.3 (or Figure 13.9, 13.10, or 13.11) to calculate the alkalinity (Alk = 1×10–3 eq/L). Easy way: Alk = CB – CA = [Na+] – 0 = 1×10–3 eq/L. The definition of alkalinity as CB – CA also is useful with natural waters. From the values in Table 13.1, you could estimate the average alkalinity in the Great Lakes as CB – CA ≈ [Na+] + [K+] + [Mg2+] + [Ca+2+] – [SO42–]– [Cl–] = 1.64 meq/L,9 very close to the 1.65 meq/L value estimated in Section 13.5.2. Another use of the CB – CA definition of alkalinity is in calculating the consumption of alkalinity from natural and engineered processes. When H+ is released from a process (i.e., when acidity is released), we say that alkalinity is consumed. Why is alkalinity consumed? The addition of strong acid moves you closer to the H2CO3* equivalence point. After the acid is added, the alkalinity is smaller than before the acid is added. In a true sense, alkalinity is consumed.10 For example, a commonly used rule of thumb in environmental engineering is that adding alum to raw drinking water for turbidity removal consumes about 0.5  mg of alkalinity as CaCO3 for every mg of alum added. Where does this rule of thumb come from? Alum dissolves initially in water to produce Al3+ and sulfate: Al2(SO4)3∙18H2O → 2Al3++ 3SO42– + 18H2O. The aquo aluminum ion, Al3+, hydrolyzes to produce aluminum hydroxide solid (see Chapter 19): Al3++ 3H2O → Al(H2O)3(s) + 3H+. Overall, 1 mole of alum (666 g) produces six equivalents of acidity (H+) and thus consumes six equivalents of alkalinity, or (6 eq)(50 g Alk as CaCO3 per eq) = 300 g Alk as CaCO3. This means 300 g Alk as CaCO3 = 0.45 mg ≈ 0.5 mg of alkalinity as CaCO3 are consumed per mg 666 g alum alum. Another illustration of alkalinity consumption is given in Worked Example 13.4. Worked Example 13.4 

Alkalinity Consumption

One way to remove ammonium from water is through the process of breakpoint chlorination. The overall reaction is: 2NH4+ + 3Cl2 → N2(g) + 8H+ + 6Cl–. How much alkalinity must be added per mg of NH4+-­N removed to maintain the pH during breakpoint chlorination? Solution Breakpoint chlorination generates four equivalents of acidity (H+) per mole of ammonium consumed or 9    This equation is written for the data in Table 13.1, where the ion concentrations are in meq/L. If the ion concentrations were in mol/L, the correct equation would be: CB – CA ≈ [Na+] + [K+] + 2[Mg2+] + 2[Ca+2+] – 2[SO42–] – [Cl–]. 10   Here is another interpretation of Alk = CB – CA that may resonant with you. In a real sense, Alk is the amount of strong acid anion (i.e., negative Alk) that you have to add to your sample to make Alk = 0. (See Section 13.2.2 for a similar interpretation of ANC.) You can apply this to any measurement. Your height is the number of cm that you have to remove from you to make your height equal to zero. Don’t try this at home.

ISTUDY

13.5  |  Using the Definitions of Alkalinity to Solve Problems   307

meq Acy 4  mmol NH 4 - N mg N

14   mmol NH 4 - N

0.286  meq Acy/mg N.

Thus, you need 0.286 meq of Alk per mg of NH4+-­N removed (about 14 mg Alk as CaCO3 per mg of NH4+-­N removed) to maintain the pH.

13.5.4  Alkalinity and Acidity as Conservative Properties As discussed in this chapter, alkalinity can be expressed as CB – CA. Since strong base cations and strong acid anions are relatively unreactive in water, the alkalinities are additive when two water samples of equal volume are combined. There is an important caveat to this rule. The rule assumes that no solids precipitate or dissolve. If solids precipitate or dissolve, the situation becomes a bit more complex (see Chapter 19). The conservation of alkalinity (and acidity) aids in the solution of water chemistry problems. Since alkalinity is conserved, you can write a mass balance on it. If two rivers (with flows Q1 and Q2 and alkalinities Alk1 and Alk2) mix to form a river with flow Q and alkalinity Alk, then (assuming no precipitation or dissolution):

eq of Alk per time in river 1 eq of Alk per time in river                          eq of Alk per time in new river, or:             Q1Alk1 Q2 Alk 2 QAlk

As an example, you can find the alkalinity, CT, and pH of the river formed when two tributaries with the following characteristics combine:



Tributary A: flow 15 cfs                     Alk 200 mg / L as CaCO3 , and:                     pH 8.1 Tributary B: flow                     Alk                     CT

7 cfs



100 mg / L as CaCO3 , and:

2.07 10 3 M

First, you must calculate Alk, CT, and pH for both tributaries. Using the approach in Section 13.5.2, you can find that CT for Tributary A is 4.01×10–3 M and the pH of Tributary B is 7.7. Alkalinity is conserved, so:

QA Alk A QB Alk B QAlk

Here, Q = QA + QB if the densities of the waters are similar. Solving: Alk = 168 mg/L as CaCO3 (3.36 meq/L). If no inorganic carbon is lost to the atmosphere or precipitates, then carbon is conserved and:

QA CT ,A QBCT ,B

QA QB CT

Thus: CT = 3.39×10–3 M. Using the alkalinity equation or Figures 13.9–13.11, you can show that the pH of the combined rivers is 8.1. The utility of the conservation of alkalinity is shown again in Worked Example 13.5.

Key idea: Alkalinity and acidity are conserved ←Worked example

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308   Chapter 13  Alkalinity and Acidity Worked Example 13.5 

Conservation of Alkalinity

Acid rain falls on a lake. How much rain (d) would have to fall to reduce the pH of the lake to 6.7? The lake characteristics are pH 7.0, Alk = 30 mg/L as CaCO3, and mean depth = D = 10 ft. The characteristics of the acid rain are pH 4 and [Cl–] = 1×10–4 M. Solution Assume that CT is unchanged after the acid rain addition, no minerals dissolve, and only rain falling on the lake surface is important (i.e., the alkalinity of the soil neutralizes acid rain falling in the drainage basin). Also assume no inputs to or outputs from the lake. mg as CaCO3 30 L The lake alkalinity is = 6×10–4 eq/L. The lake CT mg eq 50, 000 as CaCO3 per L L is

Alk [OH ] [H ] = 7.19×10–4 M at pH 7.0. 2 1 2

After the acid rain addition, the pH is 6.7 and CT is 7.19×10–4 M (assumed constant). Thus, the lake Alk after acid rain addition is (α1 + 2α2)CT + [OH–] – [H+] = 5.14×10–4 eq/L. Alk is a conservative property. Thus: (Alk)(V) = (AlkL)(VL) + (AlkR) (VR), where V = volume and the subscripts L and R represent the lake before and the rain, respectively. Note that AlkR = CB,R – CA,R = 0 – 1×10–4 = –1×10–4 eq/L. If the lake surface area is A, then (assuming the rainfall does not change the lake volume significantly): V = VL = DA and VR = dA. (Alk)( DA) (Alk L )( DA) Alk R meq meq (10 ft) 0.6   0.514   (10 ft) L L   2.075  ft meq 0.1  L VR

So: d

dA

24.9  inches

63.3  cm.

13.6  EFFECTS OF OTHER WEAK ACIDS AND

BASES ON ALKALINITY

13.6.1  Introduction For most natural waters, the members of the carbonate family are the dominant weak acids and weak bases. However, on some occasions, other weak acids or bases may be important. Other weak acids or bases can influence the alkalinity by increasing (for weak bases) or decreasing (for weak acids) the acid-­neutralizing capacity of the water. In such cases, the weak acids and bases must be included in the alkalinity calculation. How should weak acids and bases be included in the alkalinity equation? The easiest way to include the other species is to return to the derivation of alkalinity as a

ISTUDY

13.6  |  Effects of Other Weak Acids and Bases on Alkalinity   309

Table 13.4 Zero Proton Levels for Common Species at pH 4.5 Family

pKa Value(s)

Zero Level (pH 4.5)

2.1, 7.2, 12 9.3 9.3 >4.5 4.5 Monoprotic acid with pKa < 4.5

proton condition. You know that the endpoint of the alkalinity titration is about pH 4.5. Now, define the zero level of protons as the dominate species in the acid–base family at pH 4.5. The proton condition becomes:



(concentration of species with excess protons over the zero level)(no. of excess protons)

(concentration of species with deficient protons over the zero level)(no. of deficient protons)

Note that this definition works even for the carbonate species, OH–, and H+. For the carbonate family, H2CO3* is the dominant species at pH 4.5 (since pKa,1 = 6.3). For the “water family,” H2O is the dominant species at pH 4.5. Other examples are shown in Table 13.4.

13.6.2  Example As an example of the effects of other weak acids and bases on alkalinity, consider the impact of naturally occurring organic acids on the alkalinity of natural waters. These acids are divided into several fractions, primarily fulvic and humic acids. A highly colored water body may have 10 mg/L as C of dissolved organic carbon (DOC). Humic and fulvic acids have many acidic functional groups. At neutral pH, the acidic groups contributing to alkalinity are carboxylic acid groups with concentrations of up to 10 meq per gram of carbon and a typical pKa between 1.5 and 6. This yields a total acid concentration, AT, of (10 meq/g C)(10 mg C/L)(10–3 g/mg) = 0.1 meq/L. In this example, a typical pKa of 5 will be used. Since the pKa of the carboxylic acids is greater than the endpoint pH (4.5), use HA as the zero proton level for the carboxylate species. The proton condition is:

H

OH

HCO3

2 CO32

A

A

H

The alkalinity equation becomes: Alk



1

2

2

CT

OH

The concentration of A– at the endpoint pH is:

1, HA

AT

H

Ka

Ka

AT

10

10

4.5

5

10

5

0.1

meq L

ISTUDY

310   Chapter 13  Alkalinity and Acidity or 0.024 meq/L of ANC. This is a contribution of only about 1 mg/L of Alk as CaCO3. Since most natural waters have much higher alkalinities from carbonate species, fulvic and humic acids usually do not contribute very much to the alkalinity.

13.7  CHAPTER SUMMARY In this chapter, water quality parameters were developed to measure the amount of acid or base a water requires to reach specified endpoints. Natural and process waters rarely are pure acid or base solutions. They usually are acid or base solutions that have been partially titrated to some unknown point in their titration curves. For general acids and base, the acid neutralizing capacity (ANC) is defined to be the amount strong acid required to reach the zeroth equivalence point. Similarly, base neutralizing capacity (BNC) is the amount of strong base required to reach the  nth equivalence point ( f = 1 for monoprotic acids). If the carbonate system ­provides the dominant weak acids and bases, then the ANC is called alkalinity (titration to f = 0 for H2CO3*) and the BNC is called acidity (titration with base to g = 0 for Na2CO3). Expressions for alkalinity and acidity can be developed from proton conditions or charge balances. For alkalinity, the resulting expression is called the alkalinity equation: Alk = CB – CA = (α1 + 2α2)CT + [OH–] – [H+]. Alkalinity can be measured by titration to the H2CO3* endpoint, about pH 4.5 for most waters. Although alkalinity measures the ANC of a water to this endpoint, other types of alkalinity (e.g., phenolphthalein alkalinity) may be more important for other pH ranges of interest. Common units of alkalinity are eq/L and mg/L as CaCO3: 1 meq/L = 50 mg/L as CaCO3. In solving problems involving alkalinity (Alk), it is important to remember that knowing two of Alk, CT, or pH allows you to calculate the third parameter. For the problem at hand, carefully decide whether the information you know allows you to more easily calculate alkalinity from Alk = CB – CA or from Alk = (α1+ 2α2)CT + [OH–] – [H+]. In addition, remember that alkalinity is conserved when waters are mixed (if no solids precipitate or dissolve). Weak acids and bases can influence alkalinity and acidity in certain waters. To include the effects of weak acids and bases, define their zero proton level as the species in highest concentration at the endpoint (about pH 4.5 for alkalinity). Then adjust the alkalinity equation to include species with excess or deficient protons compared to the zero proton level.

13.8  PART III CASE STUDY: ACID RAIN

 orked W example →

From what you have learned in this chapter, it is apparent that the influence of acid rain on cistern water pH can be determined much more easily by using the concept of alkalinity. Remember that alkalinity is conserved if no solids precipitate or dissolve. From Section 12.8, you found that the rainwater has a pH of 4.0, a total carbonate concentration, CT,rain, of 8.3×10–6 M, and a total strong acid anion concentration, CA,rain, of 9.5×10–5 eq/L. This means that the alkalinity is Alkrain = –9.5×10–5 eq/L (or  –4.8  mg/L as CaCO3, or, better, H-­Acy = 9.5×10–5 eq/L). You may wish to confirm that these values are consistent with the alkalinity equation. The river water has a pH of 7.5 and a total carbonate concentration, CT,river, of 1.2×10–3 M. Using the alkalinity equation, you can verify that the alkalinity of the river, Alkriver, is equal to 1.13×10–3 eq/L (or about 56 mg/L as CaCO3, fairly low). When a volume v (= 1500 L) of

pH

8 7.5 7 6.5 6 5.5 5 4.5 4 3.5 3

60 50

pH

40

Alk

30 20 10 0

Alk (mg/L as CaCO3)

ISTUDY

Chapter Key Ideas   311

–10 0

5

10

15 20 25 Time (days)

30

35

40

FIGURE 13.12  Cistern Water pH and Alkalinity in the Part III Case Study

rainwater is mixed with a volume V (= 15,000 L – 1500 L = 13,500 L) of water in the cistern, the resulting concentrations are:

CT Alk

vCT ,rain VCT ,cistern

, and: v V vAlk rain VAlk cistern v V

where the subscript “cistern” represents the water in the cistern prior to the rain falling that day. The mass balance on CT assumes that no inorganic carbon is exchanged with the atmosphere or precipitates. To calculate the pH, simply apply the alkalinity equation each day. The resulting pH and Alk of the cistern water each day are shown in Figure 13.12 (pH values also were shown in Section 12.8). Note that the alkalinity of the cistern water transitions from the alkalinity of the river water (56 mg/L as CaCO3) to approaching the alkalinity of the rainwater (–4.8 mg/L as CaCO3). As a result of this analysis, you now know why the pH in the cistern water dropped so rapidly: the river water had a very low alkalinity. Once the alkalinity of the water in the cistern became sufficiently small, the acidity in the rain caused the pH to drop quickly. The case study shows that the power of alkalinity is in solving water quality problems.

CHAPTER KEY IDEAS • It is possible to probe a system and determine its ability to buffer against changes in pH. • Weak acids (or bases) partially neutralize strong bases (or acids) added to them. • Adding strong acid to a partially titrated weak acid will retrace the titration curve of the weak acid. • ANC and BNC tell you how much added acid or base can be added without significantly changing the pH of the system. • For a monoprotic acid, ANC is the amount of strong acid needed to change α1AT + [OH–] – [H+] from its current value to zero, so ANC = α1AT + [OH–] – [H+]. • For a monoprotic base, BNC is the amount of strong base needed to change α0AT + [H+] – [OH–] from its current value to zero, so BNC = α0AT + [H+] – [OH–]. • The major ions in the majority of fresh waters are Na+, K+, Ca2+, Mg2+, SO42–, Cl–, and HCO3–.

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312   Chapter 13  Alkalinity and Acidity • Alkalinity commonly is used as an indicator of buffering, even though it is not a measure of buffer intensity. • Use caution when accepting alkalinity as an indicator of buffering if the lower pH range of interest is different from 4.3–4.7. • The pH values at the equivalence points for alkalinity and acidity depend on the total carbonate concentration (CT) of the sample. • Different types of alkalinity are defined for different endpoints. • Alkalinity will be calculated in units of eq/L if CB and CA are in units of eq/L. • 1 eq/L Alk is equivalent to 50,000 mg/L Alk as CaCO3. • The alkalinity equation is really three equations in one: Alk = (α1 + 2α2)CT + [OH–] – [H+], CB – CA = (α1 + 2α2)CT + [OH–] – [H+], and Alk = CB – CA. • If you know two of Alk, pH, and CT, then you can calculate the other parameter easily. • Near neutral pH and for reasonable values of CT: Alk ≈ [HCO3–] = α1CT. • When H+ is released from a process, alkalinity is consumed. • Alkalinity and acidity are conserved.

CHAPTER GLOSSARY acid neutralizing capacity (ANC):  equivalents per liter of strong acid required to reach the zeroth equivalence point of a solution ( f = 0) acidity (Acy):  BNC in a water where bicarbonate and carbonate are the dominant weak bases and are used to define the ending point in the titration (i.e., titration to g = 0 for carbonate) alkalinity (Alk):  ANC in a water where the carbonic acid family is the dominant weak acid system and is used to define the ending point in the titration (i.e., titration to f = 0 for carbonic acid) alkalinity equation:  Alk = CB – CA = (α1 + 2α2)CT + [OH–] – [H+] base neutralizing capacity (BNC):  equivalents per liter of strong base required to reach the nth equivalence point of a solution (g = 0) bicarbonate alkalinity:  the portion of the total alkalinity contributed by HCO3– (= total alkalinity  – carbonate alkalinity = total alkalinity  – 2(p-­Alk) + OH-­Alk) carbonate alkalinity:  the portion of the total alkalinity contributed by CO32– (= 2(p-­Alk) – OH-­Alk) caustic alkalinity:  the amount of strong acid required to bring the sample to the f = 2 (Na2CO3) equivalence point CO2-­Acy:  the amount of strong base required to bring the sample to the g = 1 (NaHCO3) equivalence point endpoint:  a key point in a titration, usually selected to be an estimate of a true ­equivalence point in a titration methyl orange endpoint:  a pH endpoint of about pH 4.5–3.2, previously used to estimate the alkalinity equivalence point ( f = 0 for carbonic acid) mineral acidity (H-­Acy):  the amount of strong base required to bring the sample to the g = 2 (H2CO3*) equivalence point

ISTUDY

Historical Note: Can You Pass the Litmus Test?   313

phenolphthalein alkalinity:  the amount of strong acid required to bring the ­sample to the f = 1 (NaHCO3) equivalence point (an endpoint of about pH 8.2–8.3)

HISTORICAL NOTE: CAN YOU PASS THE LITMUS TEST? This chapter discussed some common pH indicators used for alkalinity titrations. As with most organic chemistry, natural products were used as pH indicators before isolated compounds. Many budding water chemists’ first measurement of pH employed the juice of the red cabbage. The class of compounds responsible for the pH-­based color change in red cabbage and many other colorful plants is the anthocyanins (from the Greek for “blue flower”). Michael Faraday (1791–1867) was so enamored with red cabbage that he wrote: “Those very important chemical substances, acids and alkalies, in a free state, possess the power, even in very small quantity, of effecting certain general and regular changes in the tints of some vegetable colours. ...The only substance of the kind perhaps worth keeping in solution is an acid infusion of red cabbage” (Faraday 1830, p. 266). Anthocyanins also are responsible for the blue color of violets (and the red in roses). Over 60 years before Faraday’s report, Robert Boyle (1627–1691) titled Experiment XX in his collection of experiments about colors “Of turning the Blew of Violets into a Red by Acid Salts. . .” (Boyle 1664, p. 246). Cooks of Boyle’s day evidently knew that adding lemon juice (weak citric acid solutions) to “syrup of violets” resulted in a color/opacity change (Turner and Laroche 2011). Probably the first person to notice that the light absorption properties of natural products depended on the addition of acids and bases was the Catalan physician Arnaldus de Villa Nova (or Arnaud de Vilanova, c. 1240–1311; Coleman, 2008). He is thought to have used lichen as an acid–base indicator. Lichen are symbiotic organisms composed of algae and fungi. So what does all of this have to do with the litmus test? You may be familiar with litmus paper. The word litmus originally referred to the blue material derived from lichen. We now use the term to mean a mixture of dyes, mostly derivatives of 7-­hydroxyphenoxazone. Traditional litmus paper is a pretty crude estimator of the pH. It is red below about pH 4.5, turns purple between about pH 4.5 and 8.3, and transitions to blue above about pH 8.3. Despite the poor resolution of litmus paper, a “litmus test” now means a question that sharply determines a politician’s opinion.

Arnaldus de Villa Nova (Michel Wolgemut/Wikimedia Commons/Public domain)

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314   Chapter 13  Alkalinity and Acidity

PROBLEMS 1. Using equations, what is ANC + BNC for a monoprotic acid? Does your answer make sense from the definitions of ANC and BNC in words?

9. Plot the endpoint pH for the alkalinity titration against the alkalinity from 50 to 300 mg/L as CaCO3. Assume the pH of the sample is near neutral.

2. Draw the titration curve for the titration of a 10–2 M acetic acid solution (pKa = 4.7) with strong base. Indicate the ANC and BNC graphically on the titration curve for the solution at pH 5. Calculate ANC and BNC at pH 5. Do your estimated values from the titration curve match your calculated values?

10. Lake Mendota, located in Madison, Wisconsin, USA, is one of the most thoroughly studied lakes in the world. Average water quality conditions are pH 8.5, Alk = 3.4 meq/L, [Ca2+] = 32 mg/L, [Mg2+] = 32 mg/L, [SO42–] = 22 mg/L, [Cl–] = 27 mg/L, [Na+] = 1 mg/L, [K+] = 3.2 mg/L, and total dissolved solids (see Section 2.4.3) = 260 mg/L.

3. Using the definitions of alkalinity and acidity, explain why 1 eq of acidity consumes 1 eq of alkalinity. (Hint: Write an expression for Alk + Acy.) 4. An industry wishes to discharge an acidic waste continuously to a river. Describe briefly the information you would require to determine the impact of the waste on the river pH. 5. The pH of a lake water is measured in the field to be pH 8.0. A 100 mL sample of the water required 5.0 mL of 0.10 N HCl to titrate it to the H2CO3* equivalence point. a. What is the alkalinity of the water? b. What are the concentrations of CT, H2CO3*, HCO3–, and CO32– in the lake? 6. The pH of a lake water is measured to be pH 7.6. The alkalinity is 150 mg/L as CaCO3. a. A standard titration is carried out with a 200 mL sample and 0.02 N H2SO4. (Note: 0.02 normal, not molar.) How many milliliters of titrant will be used in the alkalinity titration? b. What are the concentrations of CT, H2CO3*, HCO3–, and CO32– in the lake? c. What are the pH and concentrations of CT, H2CO3*, HCO3–, and CO32– in the beaker at the end of the titration? Assume the alkalinity is zero at the end of the titration. 7. A 10–3 M NaOH solution and a 10–3 M NaHCO3 solution have the same alkalinity. Sketch the titration curves for each with strong acid. If the goal is to titrate to g = 1, for which solution would it be easiest to overshoot the titration (i.e., accidentally miss the endpoint)? 8. Standard Methods lists the following endpoints for alkalinity titrations, based on the expected alkalinity: pH 4.9, 4.6, and 4.3 for expected alkalinity values of 30, 150, and 500 mg/L as CaCO3, respectively. a. Estimate the corresponding CT for each alkalinity assuming that the sample pH is near neutral. b. Verify that these endpoint pH values are reasonable for the anticipated alkalinity values.

a. Estimate the average CT in Lake Mendota. b. Do charges balance? Include all ions you expect in the lake. c. How does the alkalinity compare to CB – CA for the species listed? d. How does the calculated TDS compare to the measured TDS? 11. Lago De Amatitlan (Lake Amatitlan) is a moderate-­ sized lake in southwestern Guatemala. Typical water quality is as follows (all data are from the International Lake Environment Committee’s World Lake Database, wldb.ilec.or.jp): [Ca2+] = 88.3 mg/L, bicarbonate = 410.4 mg/L, [K+] = 7.4, mg/L, and [Na+] = 219.2 mg/L. a. If the only important unmeasured ion is chloride, estimate the chloride concentration in mg/L. b. Estimate the alkalinity in mg/L as CaCO3. Is it possible to calculate the pH and CT of this lake with the information given? 12. Seneca Lake is one of the Finger Lakes in New York State, USA. Typical water quality values are: [Ca2+] = 41.6 mg/L, [Mg2+] = 9.8 mg/L, [Na+] = 85.3 mg/L, [Cl–] = 129.5 mg/L, Alk = 129 mg/L as CaCO3, and pH 8.4. a. Calculate the alkalinity from the ion concentrations and compare it to the measured value. Explain any discrepancies. b. From the measured alkalinity and pH, calculate the CT and bicarbonate ion concentration. Is the approximation that the alkalinity is about equal to the bicarbonate ion concentration valid? 13. Use the Deffeyes diagram (Figure 13.9) or Figure 13.10 or Figure 13.11 to estimate the new pH if 0.5 meq/L of alkalinity is added to a water initially at pH 7.2 with an initial alkalinity of 110 mg/L as CaCO3. Assume CT does not change and precipitation/dissolution does not occur when alkalinity is added. What is the CT? 14. Use the Deffeyes diagram (Figure 13.9) or Figure 13.10 or Figure 13.11 to estimate the new alkalinity if water

ISTUDY

Problems   315 initially at pH 7.0 with an alkalinity of 150 mg/L as CaCO3 is changed to pH 7.5. Assume CT does not change and precipitation/dissolution does not occur when the pH is changed. What is the CT? 15. Use the Deffeyes diagram (Figure 13.9), Figure 13.10, or Figure 13.11 to estimate the new alkalinity if water initially at pH 8.0 with an alkalinity of 100 mg/L as CaCO3 has its CT increased by 50% at constant pH. Assume precipitation/dissolution does not occur. 16. Using a spreadsheet, recreate one of Figures 13.9, 13.10, or 13.11. 17. Section 13.4.5 presented the following three equations: Acy + OH-­Alk = 0, CO2-­Acy + p-­Alk = 0, and H-­Acy + Alk = 0. Derive all three two ways: (1) using the equations, and (2) using the definitions in terms of ­titration endpoints. 18. Acid Corp. is requesting to discharge 200 m3 of a 0.1 N HCl acidic waste into a lake. The lake has an alkalinity of 125 mg/L as CaCO3, is at pH 7.5, and has a volume of 25,000 m3. Acid Corp. claims that the pH of the lake water after the addition of the acidic waste will not drop below pH 6.8. You have been hired as a consulting engineer to assess the situation. What will be the pH of the lake water after the addition of the acidic waste? Assume that the lake has no other inputs or outputs and that CT stays constant. 19. PollutiCon Inc., a large multinational widget manufacturer, wishes to discharge 50,000 gallons of an acidic waste into Lake Fishkill. The waste is 0.1 M sulfuric acid. Lake Fishkill has the following characteristics: alkalinity 100 mg/L as CaCO3, pH 7.5, mean depth = 7 feet, and surface area = 3 acres. PollutiCon claims that the pH of the lake water after the addition of the acidic waste will not drop below pH 7. You have been hired as a consulting engineer to assess the situation. What will be the pH of the lake water after the addition of the acidic waste? Assume: (1) closed system (no CO2 lost to or gained from the atmosphere), (2) lake volume (mean depth)(surface area), (3) the lake has no other inputs or outputs. 20. Calculate the alkalinity (in units of both eq/L and mg/L as CaCO3) of solutions containing: a. b. c. d.

2×10–3 M NaHCO3 2×10–3 M Na2CO3 + 1×10–3 M HCl 2×10–3 M NaCl 2×10–3 M H2SO4

21. If no solids form or dissolve, does the alkalinity of a 1 L sample of a typical natural water increase, decrease, or stay the same if you:

a. Add a small amount of NaCl to it? b. Add 5 mL of 0.1 M NaOH and 5 mL of 0.1 M H2SO4 to it? c. Titrate it to the H2CO3* endpoint? d. Titrate it to the second equivalence point ( f = 2)? 22. In wastewater treatment plants, ammonium can be ­oxidized biologically to nitrate. The overall reaction is: NH4+ + 2O2 = NO3– + 2H+ + H2O. Based on this reaction, justify the statement that “for each g of ammonia nitrogen (as N) converted, 7.14 g of alkalinity as CaCO3 will be required” (Metcalf and Eddy 2003). 23. The denitrification process (conversion of nitrate to nitrogen gas in the presence of a carbon source such as methanol) proceeds as follows: 6NO3– + 6H+ + 5CH3OH = 3N2 + 5CO2(g) + 13H2O. How much alkalinity as CaCO3 is produced per mg of nitrate-­N converted to nitrogen gas? 24. Ammonia also can be removed from wastewater by ammonia stripping. In this process, ammonium is converted to NH3 by raising the pH to 11 with lime, and ammonia is volatilized into a moving airstream. Create a plot showing the lime dose required (in mg/L) to raise the wastewater pH to 11 as a function of the initial alkalinity of the water (in mg/L as CaCO3) over the normal alkalinity range of wastewaters (50 to 200 mg/L as CaCO3). Lime is Ca(OH)2(s). Assume all the alkalinity is from carbonate species and the added lime. Also assume that the lime dissolves completely and that the initial pH of the wastewater is 7.5. 25. In an anaerobic digester, organics in wastewater sludge are converted into volatile acids (short-­chain organic acids), which are subsequently converted into methane and carbon dioxide by a group of organisms called methogens. In a typical anaerobic digester (pH 7, Alk = 3000 mg/L as CaCO3, volatile acids concentration = 250 mg/L), how much of the alkalinity is due to the weak volatile acids? Assume that the volatile acids are acetic acid with pKa = 4.7. 26. In Figure 13.5, a plot of the pH at the H2CO3* equivalence point versus log(CT) is a straight line. Find the slope of the line from the alkalinity equation. 27. Draw a titration curve like Figures 13.6–13.8 to show the fractions of acidity. 28. A so-­called “universal pH indicator” is made up of thymol blue, methyl red, bromothymol blue, and phenolphthalein. Given the information in the table, under approximately what pH conditions will the color of the indicator be red? Orange/yellow? Green? Blue? ­Violet/purple?

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316   Chapter 13  Alkalinity and Acidity

Indicator

Acid Color

Base Color

Thymol blue Yellow (HA ) Red (H2A) (pKa,1) Methyl red Red (HA) Yellow (A–) Bromothymol Yellow (HA) Blue (A–) blue Thymol blue Yellow (HA–) Blue (A2–) (pKa,2) Phenolphtha- Clear (H2A) Pink (A–) lein –

same density, no precipitation or dissolution of solids occurs, and CT is conserved.

pKa

Transition pH Range

1.7

1.2–­2.8

5.1 7.0

4.4–­6.2 6.0–­7.6

31. Standard Methods gives a version of the following table. Justify the entries in the table using the definitions of the types of alkalinity. It may be helpful to use ­Figures 13.6–13.8.

8.9

8.0–9.6

Titration Result

OH-­Alk

9.7

8.2–10.0

p-­Alk = 0 p-­Alk < ½Alk p-­Alk = ½Alk p-­Alk > ½Alk

0 0 0 2(p-­Alk) – Alk Alk

29. A very simple model for the oceans is a solution at pH 8.3 with CT = 2×10–3 M and total dissolved borate of 4×10–4 M. (This problem ignores the important effects of the saltiness of the water: see Chapter 21.) What is the alkalinity from carbonate species only? What is the contribution from borate? (See Table 13.4 for information on borate chemistry.) 30. Two rivers mix. One river has a flow of 3 m3/s, Alk = 150 mg/L as CaCO3, and pH 7.8. The other river has a flow of 2 m3/s, CT = 1.5×10–3 M, and pH 8.2. Find the alkalinity, CT, and pH of the river formed when the two rivers mix. Assume the water in the two rivers has the

p-­Alk = Alk

Carbonate Alkalinity

Bicarbonate Alkalinity

0 2(p-­Alk) 2(p-­Alk) 2(Alk – p-­Alk)

Alk Alk – 2(p-­Alk) 0 0

0

0

Adapted from Standard Methods. Alk = total alkalinity.

Also justify the following statements (paraphrased from Standard Methods): 1. If p-­Alk < Alk – p-­Alk, then the carbonate alkalinity is 2(p-­Alk) and the bicarbonate alkalinity is Alk – 2(p-­Alk). 2. If p-­Alk > Alk – p-­Alk, then the carbonate alkalinity is 2(Alk – p-­Alk) and OH-­Alk is 2(p-­Alk) – Alk.

CHAPTER REFERENCES Boyle, R. (1664). Experiments and Considerations Touching Colours. First occasionally Written, among some other Essays, to a Friend; and now suffer’d to come abroad as the Beginning of an Experimental History of Colours. London: Henry Herringman. Coleman, W.F. (ed.). (2008). Molecular models of indicators. J. Chem. Educ. 85(8): 1152. Deffeyes, K.S. (1965). Carbonate equilibria: A graphic and algebraic approach. Limnol. Oceanogr. 10(3): 412–426. Faraday, M. (1830). Chemical Manipulation: Being Instructions to Students in Chemistry, on the Methods of Performing

Experiments of Demonstration or of Research, with Accuracy and Success. London: John Murray. Metcalf and Eddy (2003). Wastewater Engineering: Treatment and Reuse. 4th ed. New York: McGraw-­Hill Book Co. Sawyer, C.N. and P.L. McCarty (1978). Chemistry for Environmental Engineers. 3rd ed. New York: McGraw-­Hill Book Co. Turner, S. and R. Laroche (2011). Robert Boyle, Hannah Woolley, and Syrup of Violets. Notes & Queries 58(3): 390–391. Wetzel, R. (1983). Limnology. Philadelphia, PA: Saunders Publ.

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PA R T I V

Other Equilibria in Homogenous ­Aqueous Systems Chapter 14: Getting Started with Other Equilibria in Homogeneous ­Aqueous Systems Chapter 15: Complexation Chapter 16: Oxidation and Reduction

I shall now attempt to show how, by a single type of chemical combination, we may explain the widely varying phenomena of chemical change. –Gilbert N. Lewis Part IV Haiku Electron transfer? Oxidation-­reduction. Shared? Complexation. Here’s metal more attractive. –William Shakespeare

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C H A P T E R   14

Getting Started with Other Equilibria in Homogeneous Aqueous Systems 14.1 INTRODUCTION The acid–base equilibria in Part III introduced you to the concept of the proton donor. Recall that a Brønsted-­Lowery acid donates a proton to water. Since acid-­base reactions are so important in aquatic chemistry, pH is a logical primary variable. However, protons are not the only chemical species that can be donated or accepted between species. The other major “currency” in aquatic chemistry is the movement of electrons between species. As discussed in Chapter 10, electrons can be shared or transferred. The sharing of electrons creates products that are composites of the reactants. This process is shown schematically in Figure 14.1. The formation of composite products from the sharing of electrons is called complexation. In some cases, the composite products are not very stable. The free energy of the system may be minimized if the product reverts back to the reactants. For some reactants, free energy will be minimized if electrons are completely transferred. Complete transfer results in the formation of new species with different oxidation states than the reactants (see Figure 14.1). The transfer of electrons is oxidation-­reduction (or redox) chemistry. The equilibrium tools you mastered in Part II and honed in Part III will be applied to electron-­sharing and electron-­transfer equilibria in Part IV of this text. By the conclusion of Part IV, you will be able to describe and calculate species concentrations in realistic models of homogeneous aqueous systems.

Key idea: In addition to proton donation, electrons can be shared or exchanged

14.2  ELECTRON-­SHARING REACTIONS 14.2.1  Alfred Werner and Molecular Compounds In the early part of the twentieth century, chemists struggled to understand a particular class of compounds. These so-­called “molecular compounds” consisted of a metal and several other groups. One example of interest at the time was Pt(NH3)2Cl2.

319

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320   Chapter 14  Getting Started with Other Equilibria in Homogeneous Aqueous Systems Electron Sharing (complexation) + Stable

Electron Transfer (oxidation-reduction) +

+

FIGURE 14.1  Electron Sharing and Electron Transfer

Key idea: In complexes, certain substituents coordinate with a central metal ion (or ions)

The structure of this compound was not known. Did it form a long chain, such as Cl-­NH3-­Pt-­NH3-­Cl? Were the chlorine atoms associated with ammonia or were they associated with platinum? Alfred Werner (1866–1916) realized that nearly all the molecular compounds could be described by placing the metal ion in a central location and having it associate with four or six substituents. The realization that the substituents were associated with the central metal and not with each other was a major breakthrough. It explained many of the observations concerning molecular compounds. For more on Werner, see the Historical Note at the end of this chapter. Werner used the term coordination number for the number of substituents associated with the central metal, a term still used today. We now call molecular compounds complexes or coordination complexes. The formation of complexes is called complexation. Complexation is the subject of Chapter 15.

14.2.2  Sharing of Electrons During Complexation Why do certain substituents associate with metals? To answer this question, examine a list of common central metal species (e.g., Cu2+, Hg2+, Zn2+, Fe2+, and Fe3+) and substituents (e.g., Cl–, OH–, CN–, and S2–) that coordinate with a central metal ion (or ions).

Thoughtful Pause Based on these lists, what do you think is the nature of the interaction between the metal and substituents?

It appears that the metal ions have a deficiency of electrons relative to their atoms (and hence are positively charged). The substituents have an excess of electrons relative to their atoms (and hence are negatively charged). Thus, it makes sense that electrons may be shared during complex formation.

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14.3  |  Electron Transfer   321

We can now extend Werner’s ideas: Complexes form when a central metal ion and other species share electrons. This model will go a long way toward explaining why certain complexes form in the aquatic environment. However, the model seems to have a few problems. For example, if electron sharing is so important, why do complexes form with uncharged species such as water, ammonia, and organic acids? The answer is that nitrogen and/or oxygen atoms in these species have unshared electrons available to share with the central metal ion. As a result, you may expect that water and OH– will be important contributors to complex formation in water. Another potential problem with the electron-­sharing idea is a bit more subtle. Remember from freshman chemistry that all chemical bonds involve some degree of electron sharing. A covalent bond is one in which the electrons are shared fairly equally (that is, the electrons are associated with both atoms in the bond). In an ionic bond, the electrons are much more closely associated with one atom (e.g., the Cl in NaCl). If all bonds involve electron sharing, then shouldn’t all bond formation be called complexation? More specifically, should the association of H+ with a conjugate base be called complexation, not acid–base chemistry? In answer to the first question, the word complexation often is limited to describing electron sharing with certain types of electrons. Thus, the covalent bonds common in organic chemistry (e.g., in methane, CH4) are not referred to as complex formation. Addressing the second question, you will find it insightful to include H+ as a metal. In this way, acid–base chemistry is a sort of complexation. This concept is explored further in Chapter 15.

14.2.3  Examples of Complexation in the Environment By broadening your view of aquatic chemistry to include electron-­sharing reactions, the richness of homogeneous chemistry will become apparent to you in Chapter 15. For example, you will learn about the chemistry of dissolved metals. In natural systems, numerous species react with metals, including the ubiquitous OH– and naturally occurring acids. Complexation is at the heart of the biochemistry of many vitamins and important enzymes. Complexes also can be formed by reaction between metals and species of anthropogenic origin, such as detergents and cyanide. Complexation is very useful chemistry. For example, the principles of complexation are used to control metal solubility in industrial processes (e.g., metal plating) and in the laboratory (in metal buffers). In addition, the determination of metal concentrations at very low levels with simple instruments is made possible by highly specific complexation reactions.

14.3  ELECTRON TRANSFER 14.3.1  When Electron Sharing Becomes Electron Transfer When the sharing of electrons creates a stable compound, we say that complexation has occurred, and the stable compound is called a complex. In many instances, the complex formed from the initial sharing of electrons is not stable. In this case, the electrons may be completely transferred from one species to another. Consider the reaction of ferric ion with iodide ion. Initially, a complex is formed:

Fe3

I

FeI 2

eq. 14.1

Key idea: Electrons are shared between metals and ­substituents in complexation

Key idea: Acid–base chemistry is a form of complexation

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322   Chapter 14  Getting Started with Other Equilibria in Homogeneous Aqueous Systems The reaction in eq. 14.1 is complexation equilibrium. It is reasonable to assume that the unshared electrons originally in the iodide ion are being shared with the ferric ion in FeI2+. The oxidation state of iron remains at +III and the oxidation state of iodine remains at –I since electrons are merely being shared. The complex reacts with more iodide: FeI 2



I

FeI 2

Again, this is a simple complexation reaction. However, the intermediate FeI2+ is unstable. The free energy of the system is minimized if the intermediate rearranges: FeI 2



Fe 2

I2

eq. 14.2

Note that the oxidation state of the iron has been reduced from +III to +II, and the average oxidation state of the iodines has been increased from –I to –½. Thus, in eq. 14.2, electrons are transferred from one initial chemical species to another. A series of reactions, which started with electrons being shared, ended with electrons being transferred. To complete the reaction scheme, the iodine anion reacts with more ferric ion:

I2

Fe3

I2

Fe 2

In this final reaction, the oxidation state of the iron has been reduced to +II again, and the average oxidation state of the iodines has been increased further to 0. The overall reaction is: Key idea: Electron sharing may result in electron transfer from one species to another to minimize the free energy of the system



2I

2 Fe3

I2

2 Fe 2

eq. 14.3

Note that in the overall reaction, no sign of electron sharing is evident. Recall, though, that electron sharing made electron transfer possible. Since the FeI2+ complex is unstable, free energy is minimized by transferring the electrons from iodine to iron. Reactions such as eq. 14.3 are called overall redox reactions. Eq. 14.3 is an example of oxidation-­reduction chemistry. (The term redox is an abbreviation of reduction-­ oxidation.) In eq. 14.3, the ferric ion is reduced to ferrous ion (that is, its oxidation state is reduced), and iodide ion is oxidized to iodine (i.e., its oxidation state is increased). In Chapter 16, the equilibrium aspects of redox reactions will be considered.

14.3.2  Examples of Redox Reactions in the Environment In many ways, redox chemistry is even more textured than acid–base chemistry. In acid–base chemistry, the Brønsted-­Lowery definition allows for only one acceptor of protons, H2O. In redox chemistry, many different species act as electron acceptors. Thus, many species participate in electron-­transfer reactions. In natural aquatic systems, the cycling of elements between different oxidation states is an important part of chemical ecology. The focus in this text is on chemical processes. However, the importance of redox chemistry in natural systems is best seen by considering the role of electron transfer in biota. Nearly all microorganisms (and all organisms that respire, including humans) require a terminal electron acceptor as the final step in the process of deriving energy from chemical bonds. Organisms undergoing aerobic respiration use oxygen as the terminal electron acceptor. Oxygen accepts electrons and protons to form water.

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14.6  |  Part IV Case Study: Which Form of Copper Plating Should You Use?   323

Redox reactions occur in engineered systems as well. Corrosion is the process by which electrons are transferred from a metal in its zero oxidation state to an electron acceptor. Most chemical disinfectants (such as chlorine and ozone) are strong electron acceptors. They disrupt cellular function by accepting electrons from electron-­rich sites such as sulfhydryl groups (-­SH groups) and double bonds.

14.4  PART IV ROADMAP Part IV of this book discusses systems with electron-­sharing (complexation) and electron-­transfer (redox) equilibria. In Part IV, both equilibrium calculation techniques and qualitative tools will be applied to complexation and redox chemistry. By the conclusion of this part of the text, you will be able to quantify the effects of electron-­sharing and electron-­transfer equilibria on homogeneous aquatic systems. In Chapter  15, the nomenclature of complexation is introduced. The presence of many different elements in complexation equilibria generally means that multiple mass balances are required. Several different classes of species that react with metals are discussed. By the conclusion of Chapter 15, you will be able to determine species concentrations in fairly complicated homogeneous aquatic systems. Chapter  16 provides solution techniques for systems involving electron transfer. The importance of the primary variable pe will be emphasized. Graphical solution techniques showing the effects of both pe and pH on speciation will be developed. By the conclusion of Chapter 16, you will be able to answer questions about the dominant chemical species in homogeneous aquatic systems undergoing acid–base and redox chemistry.

14.5  CHAPTER SUMMARY In this chapter, the fate of electrons in homogeneous aquatic systems was introduced. You found that electrons could be shared or transferred. Electron-­sharing reactions with metals results in the formation of complexes. Complexes have one or more central metal ions and one or more species sharing electrons with the metal(s). Many different types of species can form complexes with metals, including common species in water (such as water itself and OH–). If electron sharing results in the formation of unstable species, then the free energy of the system may be minimized by complete transfer of electrons from one chemical species to another. This is called oxidation-­reduction (or redox) chemistry. The net result is the change in the oxidation state of elements in the species accepting electrons and in the species donating electrons.

14.6  PART IV CASE STUDY: WHICH FORM

OF COPPER PLATING SHOULD YOU USE?

A quick glance around your apartment or office should convince you that hundreds of everyday items are covered with a very thin coat of metal. From doorknobs to razor handles, many metal and plastic objects benefit from metal finishing. In most metal finishing operations, metal in an oxidation state greater than zero is reduced electrochemically to the zero oxidation state on the object to be plated (called the piece or work). This produces a stable metal coating to increase the strength or appearance of the piece.

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324   Chapter 14  Getting Started with Other Equilibria in Homogeneous Aqueous Systems One of the most common types of metal plating operations is copper plating. Copper can be used to provide a number of finishes, from a useful undercoat for additional plating operations to a bright, final finish. Two types of copper plating processes are common. In copper cyanide plating, the source of copper is CuCN(s). Note that the copper source in copper cyanide plating is Cu(+I). In acid copper plating, the source of copper is CuSO4(s); copper in the +II oxidation state. In this case study, the relative merits of copper cyanide and acid copper plating will be examined. The advantages of one plating type over another will be quantified through aqueous equilibrium calculations.

CHAPTER KEY IDEAS • In addition to proton donation, electrons can be shared or exchanged. • In complexes, certain substituents coordinate with a central metal ion (or ions). • Electrons are shared between metals and substituents in complexation. • Acid–base chemistry is a form of complexation. • Electron sharing may result in electron transfer from one species to another to minimize the free energy of the system.

HISTORICAL NOTE: HAUPTVALENZ AND NEBENVALENZ The scientific method is a brutal form of natural selection. Untenable ideas are discarded unceremoniously. Useful ideas are retained as “traits” for the next generation of scientists. Take the compound ferricyanide, Fe(CN)63–. You know CN– is the conjugate base of HCN and so has a charge of –1. This means Fe is in the +III oxidation state. And based on the definition in Section 14.2.1, you would say that Fe has a coordination number of 6 (for the six cyanide ions bonded to the iron). But before Alfred Werner, the structure of ferricyanide made no sense: Fe(+III) had a valance of 3 and should bind three cyanide ions. How did Werner work around the tyranny of the valance bond and establish a useful trait that is the basis of Part IV of this text? Werner’s new idea was that there were two kinds of valance bonds: primary valence (Hauptvalenz) bonds and secondary valance (Nebenvalenz) bonds. We now refer to these two types of valences as the oxidation state and coordination number, respectively. In Werner’s words: Even when they are saturated in the sense of the older theory of valence [i.e., have their oxidation state requirements satisfied], the elementary atoms still possess sufficient chemical affinity to bind other seemingly also saturated atoms and groups of atoms, under generation of clearly defined atomic bonds [i.e., complexation]. (Werner 1913)

This breakthrough idea apparently came to Werner in a dream. Despite lacking a background in inorganic chemistry, Werner “.  .  . awoke one night in 1892 at 2:00 am with the solution to the puzzle  .  .  .  [and] wrote his most important theoretical paper by 5:00 pm.” (Kauffman undated). Now, this may seem like every professor’s fantasy: Have a world-­changing idea in the middle of the night, write a paper about it in 15 hours, and win the Nobel for it. But at the time, Werner was working as an

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 Chapter References   325

unsalaried lecturer (Privatdozent) at the Eidgenössisches Polytechnikum1 in Zurich and being paid individually by students (a practice frowned upon today).

Chemistry lab at ETH Zurich around the Time of Alfred Werner (1886–1889) (Unbekannt/ETH Library/Public domain) His work was so significant that Werner was awarded the Nobel Prize in Chemistry in 1913. His was the first laureate in inorganic chemistry. (Underappreciated inorganic chemists have some justification for their slights. It would be another 58 years before inorganic chemists again would receive the Nobel.) Through his work distinguishing the ideas of oxidation state and degree of complexation, Alfred Werner is the ideal subject for an Historical Note for Part IV of this text.

CHAPTER REFERENCES Kauffman, G.B. (undated). Alfred Werner. Encyclopedia Britannica. britannica.com/biography/Alfred-­Werner. Lewis, G.N. (1916). The atom and the molecule. J. Am. Chem. Soc. 38(4): 762–785.

Shakespeare, W. Hamlet, Act 3, Scene 2. Werner, A. (1913). On the constitution and configuration of higher-­ order compounds. Nobel Lecture, December 11. From the English translation at nobelprize.org/uploads/2018/06/werner-­lecture.pdf.

1   The Polytechnikum now is called the Eidgenössische Technische Hochschule (ETH) or the Swiss Federal Institute of Technology. Among many other accolades, ETH Zurich birthed the Eidgenössische Anstalt für Wasserversorgung, Abwasserreinigung, und Gewässerschutz (Eawag or the Swiss Federal Institute of Aquatic Science and Technology), one of the finest institutions of aquatic chemistry in the world.

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C H A P T E R   15

Complexation 15.1 INTRODUCTION The world of environmental chemistry is much broader than acids and bases. An important component of the aquatic environment is metals. Metals are worthy of study because of their toxicity to humans and aquatic biota. In addition, metals equilibrate with many common aqueous species (called ligands) to form compounds called complexes. Complexes may control the chemistry of the ligands. In this chapter, you will learn about the formation of complexes. In Sections 15.2 and  15.3, complex formation will be discussed on a qualitative basis. You will learn about the roles of both the metal and the ligands. Equilibrium calculation techniques for systems exhibiting complexation will be presented in Section 15.4. A fairly involved system containing acids, bases, metals, and ligands is discussed in Section 15.5. Applications of complexation chemistry are presented in Section 15.6.

15.2 METALS

Key idea: Metals are important because (1) they are toxic to humans and aquatic biota and (2) they form complexes that may control the chemistry of the ligands

15.2.1  Introduction You now have a great deal of experience with acid–base equilibria. So far in this text, acid–base equilibria have been analyzed from the point of view of the acid and conjugate base. For example, the calculation techniques (Chapter  7), alpha values (Chapter 11), and approaches to titration (Chapter 12) have emphasized the role of HA and A–. For the moment, focus on H+ rather than HA or A–. Usually, you write: HA

A

H

or: HA H 2 O

A

H3O

K

Ka

K

Ka

(Recall that the equilibrium constant for the second equilibrium is equal to the equilibrium constant for the first equilibrium because we define K for the equilibrium H+ + H2O = H3O+ to be 1.) To emphasize the fate of H+, rewrite these equilibria as:

H

A

or: H3O

 HA      

K

A

K

HA H 2 O

1 Ka 1 Ka

eq. 15.1

327

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328   Chapter 15  Complexation coordination: the association of two chemical species to form a third species metal: in aquatic chemistry, the positively charged partner in a coordination reaction ligand: a species that coordinates with a metal complex: a compound formed by the coordination of a metal and a ligand (also called coordination complexes or metal-­ligand complexes)

Key idea: Metals and ligands coordinate to form complexes Key idea: Acid–base chemistry is a subset of complexation chemistry

Lewis acid: a substance that accepts an electron pair Lewis base: a substance that donates an electron pair

You can rewrite eq. 15.1 slightly as follows:

H H2 O

A

HA H 2 O

K

1 Ka

eq. 15.2

Equation 15.2 shows the fate of H+ in the HA/A–/H2O system: H+ equilibrates between being associated with water and being associated with A–. We say that H+ coordinates with H2O and with A–. Coordination is the association of two chemical species to form a third species.1 Coordination involves two partners. The positively charged partner is called a metal. The other partner, usually negatively charged or uncharged, is called the ligand. Compounds formed by the coordination of a metal and a ligand are called coordination complexes or metal-­ligand complexes or just complexes. The formation of coordination complexes is called complexation. Thoughtful Pause Name the metal(s) and ligand(s) in eq. 15.2. In eq. 15.2, the only metal is H+ and the ligands are H2O and A–. Because of its high concentration, water is a very important and common ligand in aquatic systems.

15.2.2  Lewis Acids and Bases If you wanted to, you could now completely redo Part III of this text and eliminate the words acid and base. You could say that Part III of this text was devoted to one metal, namely H+. A variety of ligands was considered, including H2O, OH–, SO42–, HSO4–, CH3COO –, HCO3–, and CO32–. The ligands were Brønsted-­Lowry bases, and the metal-­ligand complexes (e.g., CH3COOH) were Brønsted-­Lowry acids. It is completely acceptable to think of acid–base chemistry as a subset of metal-­ligand chemistry.2 1 In fact, in some compilations of equilibrium constants, the Ka values (actually, Ka values) are listed for equilibria involving the metal H+ and the pertinent ligand (conjugate base) as reactants. Another illustration of the similarities between acid–base and complexation chemistry is given in Figure 15.1. Rather than redefining acids as metal-­ligand complexes, you also could redefine metal-­ligand complexes as a sort of acid. The brilliant American chemist Gilbert Newton Lewis (1875–1946) formalized this idea through his notation that electrons are shared between atoms in chemical bonds. Lewis extended the Brønsted-­Lowry definitions of acids and bases to what are now called Lewis acids and bases.3 A Lewis acid is a substance that accepts an electron pair. A Lewis base is a substance that donates 1  The association can be strong or weak. Strong association results in chemical bond formation and is called inner-­ sphere coordination. Weak association may occur if water molecules exist between the coordinating species and is called ion pair coordination. 2  Hartshorn et al. (2015) point out “. . . all the oxoacids [e.g., H2SO4] themselves and their derivatives may be viewed as coordination entities. . ..” 3  Lewis’s words still resonant: “It seems to me that with complete generality we may say that a basic substance is one which has a lone pair of electrons which may be used to complete the stable group of another atom, and that an acid substance is one which can employ a lone pair from another molecule in completing the stable group of one of its own atoms. In other words, the basic substance furnishes a pair of electrons for a chemical bond, the acid substance accepts such a pair.” (Lewis 1923; emphasis his, p. 142).

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15.2  |  Metals   329 As acid-base chemistry… acetic acid donates a proton to water: CH3COOH + H2O = CH3COO– + H3O+

K = Ka

As complexation chemistry… the metal H+ equilibrates between the ligand water and the ligand acetate: CH3COO– + H(H2O)+ = CH3COOH + H2O K =

1 Ka

FIGURE 15.1  Similarities between Acid–Base and Complexation Chemistry

an electron pair. Note that Brønsted-­Lowry acids and bases also are Lewis acids and bases.4 For example, the Brønsted-­Lowry/Lewis base PO43– both accepts protons and donates (or shares) one or more electron pairs. It is important to see that the Lewis acid definition is much broader than the Brønsted-­Lowry definition. For example, all the transition metals are Lewis acids. The environmentally important transition metals are chromium, manganese, iron, cobalt, nickel, copper, zinc, silver, cadmium, gold, and mercury. Other important Lewis acids in the environment come from magnesium, aluminum, and calcium.

Key idea: Brønsted-­Lowry acids and bases also are Lewis acids and bases, but the Lewis acid concept is much broader than the Brønsted-­Lowry acid concept

15.2.3  Coordination Number The number of ligands with which a metal coordinates is called the coordination number. Thoughtful Pause

Key idea: The transition metals are Lewis acids coordination number: the number of ligands with which a metal coordinates

What is the coordination number for H+? H+ has a coordination number of 1, as it generally coordinates with one ligand, as in H+(H2O) ≡ H3O+. Most other metals of environmental interest have coordination numbers of 2, 4, or 6. Common metals in the aquatic environment and their coordination numbers are listed in Table 15.1. The coordination numbers are satisfied at all times in the aquatic environment. This means that uncomplexed metal ions such as Cu2+ and Al3+ do not exist in the environment. However, a shorthand notation is used frequently where the ligand water is not written explicitly. Thus, it is common to write Cu2+ as an abbreviation for the species Cu(H2O)62+ and Al3+ as an abbreviation for the species Al(H2O)63+. If the number of ligands is less than the coordination number, then you generally can assume that the unwritten ligands are water. Thus, we write ZnOH+ as shorthand for the species Zn(OH)(H2O)5+. Complexes that contain more than one coordinating metal are called polynuclear (or multinuclear) complexes. Complexes with two coordinating metals are called dimers, and complexes with three coordinating metals are called trimers. One chemical

 It is easy to see that H+ is both a Brønsted-­Lowry acid and a Lewis acid. It is a bit more difficult to see how other Brønsted-­Lowry acids, say HCl, are Lewis acids. In the Lewis acid–base view, HCl is a considered a Lewis acid because it donates a Lewis acid, namely, H+ (March 1985). Similarly, Cl2 is a Lewis acid because it donates “Cl+”.

4

explicitly

Key idea: The ligand water usually is not written

polynuclear (multinuclear) complexes: complexes with more than one coordinating metal

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330   Chapter 15  Complexation Table 15.1 Coordination Numbers of Common Metals in the Environment1 Coordination Number

Example Metals2

1 2 43 6

H(+I) Ag(+I), Au(+I) Au(+III), Cu(+I), Hg(+II), Mn(+VII) Al(+III), Cd(+II), Cr(+III), Cu(+II), Fe(+II), Fe(+III), Ni(+II), Zn(+II)

Notes:  The coordination number changes with the ligand and environment. Tabulated values are for the most common coordination numbers in aquatic environments (Cotton et al. 1999). 2  H(+) represents H+, Al(+III) represents Al3+, etc. 3  The coordination number of four has two geometric configurations: square planar (ligands in the same plane as the metal and at the corners of a square) and tetrahedral (ligands at the corners of a tetrahedron with the metal in the center. Source: Republished with permission of John Wiley & Sons, from Advanced Inorganic Chemistry, Cotton et al. (1999) permission conveyed through Copyright Clearance Center, Inc. 1

used as a coagulant in drinking water treatment, polyaluminum chloride, is a mixture of polynuclear complexes. One polynuclear complex in polyaluminum chloride is the tridecamer Al13O4(OH)24(H2O)127+.

15.3 LIGANDS 15.3.1  Ligand Characteristics and Nomenclature

Key idea: Ligands donate electron pairs to metals and commonly contain N, O, and/or S monodentate ligand: a ligand capable of forming one bond with a metal chelates: complexes formed with a multidentate ligand

Ligands are Lewis bases and thus donate electron pairs. Thus, ligands must contain atoms with electron pairs to donate. In the aquatic environment, many ligands contain N, O, or S since these atoms have unshared electron pairs in their common oxidation states in the environment. Thus, carboxylic acids (containing the functional group -­COOH or -­COO –) are common ligands in the aquatic environment. Ammonia can be an important ligand as well. Recall that the number of ligands with which a metal can form coordination complexes is called the coordination number of the metal. Ligands also are labeled according to how many bonds they can form with metals. A ligand capable of forming one bond (e.g., -­COO – and chloride) is called a monodentate ligand (from the Latin dentatus tooth). Similarly, bidentate ligands can form two bonds. Amino acids contain both -­NH2 and -­COOH groups and can be bidentate ligands. The copper(+II)-­diglycine complex is shown in Figure 15.2. (For clarity, the water ligands are not shown.) Complexes formed with multidentate ligands are called chelates or chelate compounds (from the Greek chele claw). The formation of chelates is called chelation. Most chelates have a ring structure. For example, a copper(+II)-­diglycine complex shown in Figure 15.2 is a chelate. Note the presence of a ring (cyclic) structure. Sections 15.3.2 through 15.3.4 discuss several ligands and corresponding complexes important in the environment. Recently, IUPAC changed the names of many ligands. For example, the ligand H2O, previously called the aquo ligand, is now called the aqua ligand. This edition of the text retains the older nomenclature while acknowledging the new names.

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15.3  |  Ligands   331 H

H H

C C

O

H O

N

O C

Cu O N H H

H

C H

FIGURE 15.2  Copper(+II)-­Diglycine Chelate

15.3.2  Aquo Complexes (Ligand = H2O) As stated in Section 15.2.1, water is an important ligand because water is present at such high concentrations in aqueous systems. The ligand H2O is given the name aquo (formally, aqua). If water is the only ligand, then the metal-­water complex is called a metal aquo complex. For example, the formal name of Cu(H2O)62+ is hexaaquocopper(+II) ion. We also refer to Cu(H2O)62+ ≡ Cu2+ as free copper. In general, free metal refers to the aquo-­only complex of a metal. The water ligands help explain why some Lewis acids act as Brønsted-­Lowry acids; that is, why some Lewis acids donate protons to water. As an example, consider Al(H2O)63+. This ion can donate a proton to water:

Al H 2 O

6

3

H2 O

Al H 2 O 5 OH 2

H 3 O ; K1

free metal: the aquo-­only complex of a metal

eq. 15.3a

The complexes in equilibrium expressions such as eq.  15.3a usually are written without the water ligands on any metal. The metals here are Al3+ and H+. Without the water ligands: Al3



H2 O

AlOH 2

H ; K1

eq. 15.3b

Note that the equilibria in eq. 15.3a and 15.3b give the same information and therefore have the same equilibrium constant (arbitrarily labeled K1 here). The form in eq. 15.3b is common, but it can be confusing: Where does the H+ in the products come from? In eq. 15.3a, the answer is clear. One of the water ligands in the hexaaquoaluminum ion donates a proton to water. You can see this in eq. 15.3c, where one of the waters of hydration is pulled out and one of its H is bolded.

Al H 2 O

5

H OH

3

H2 O

Al H 2 O 5 OH 2

H H2 O

eq. 15.3c

You can simplify eq. 15.3b even further. “Subtract” (i.e., reverse and add) the self-­ ionization of water from the equilibrium in eq. 15.3b to obtain: Al3

H2 O

AlOH 2

 H

OH

H2 O

K

Al3

OH

AlOH 2       

K

H

K1

1 KW

eq. 15.4

K1

KW

Although eq. 15.3b contains the same information as eq. 15.4 (plus the self-­ionization of water), the two reactions give a very different feeling for the origin of AlOH+. From eq.  15.4, it appears that AlOH2+ comes from the formation of a complex between

Key idea: The acidity of some complexes stems from the ability of aquo ligands to donate protons to water

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332   Chapter 15  Complexation Al3+ and the hydroxide ion. However, from eq.  15.3, it appears that AlOH2+ (or Al(H2O)5OH2+) comes from the fact that the ligand waters are acidic and can donate protons to water. The acidity of the ligand waters can be estimated. Recall that water itself is a pretty weak acid. Its first pKa is 14. In contrast, the water molecules coordinated with metal ions are fairly acidic. For example, the pKa for Al(H2O)5OH2+ (pK1 for the equilibria in eq. 15.3a) is 5. Thoughtful Pause Why are the waters around a metal ion so much more acidic than water itself? It is likely that the protons in the water ligands are electrostatically repelled by the metal ion, reducing the attraction to the ligand water molecule and making the protons more likely to protonate solvent water molecules.

Thoughtful Pause If the ligand water molecules are acidic, what is the expected pH range where aquo complexes are favored?

In a sense, the aquo complexes are analogous to the most protonated form of a moderately strong acid. In other words, the aquo complexes are expected to be the dominant form of the metal only at reasonably acidic pH values.

15.3.3  Hydroxo Complexes (Ligand = OH–) Hydroxo complexes (formally, the hydroxido ligand) have OH– as the only type of ligand other than H2O. As in the Brønsted-­Lowry analogy discussed in Section 15.3.2, hydroxo complexes are analogous to the conjugate bases of moderately strong acids. In this case, the moderately strong acids are the bound aquo (H2O) ligands. Mechanistically, hydroxo complexes are formed when bound water (aquo ligand) donates a proton to water. In practice, we often think of OH– as a ligand. Recall that at high pH, moderately strong acids are increasingly deprotonated as they donate protons to water. Similarly, at high pH, bound aquo ligands are increasingly deprotonated as they donate protons to solvent water. This forms complexes with deprotonated aquo ligands (i.e., complexes with OH–). We often think of this phenomenon as the formation of hydroxo complexes at high pH where the concentration of the ligand OH– is large. The water ligands will donate protons to water in a stepwise fashion to form complexes with increasingly larger numbers of OH– ligands per mole as the pH increases. As an example, copper(+II) goes from being primarily in the form of Cu(H2O)62+ at low pH to the stepwise predominance of Cu(H2O)5OH+, Cu(H2O)4(OH)20, Cu(H2O)3(OH)3–, and Cu(H2O)2(OH)42– as the pH increases. These species are usually written as, respectively, Cu2+, CuOH+, Cu(OH)20, Cu(OH)3–, and Cu(OH)42–. The complexes with five and six OH– ligands do not form appreciably in water.

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15.3  |  Ligands   333

A word about the uncharged soluble species such as Cu(H2O)4(OH)20 ≡ Cu(OH)20. The uncharged soluble hydroxo complexes do not form with all metals. When writing these species, it is important to differentiate them from the hydroxide solids. We usually use a superscript zero to indicate that the species are uncharged. You also can label the species as an aqueous (i.e., dissolved) species. Thus, the symbols Cu(OH)20, Cu(OH)20(aq), and Cu(OH)2(aq) all are used to denote the soluble hydroxo-­copper(+II) complex with two hydroxy groups.

Key idea: Label soluble, uncharged complexes with (aq) or the superscript 0 (for zero charge) to differentiate them from solids

15.3.4  Other Inorganic Ligands In addition to aquo and hydroxo complexes, other inorganic ligands also can be important in natural waters. Several important inorganic ligands are listed in Table  15.2, along with examples of their complexes that are important in the environment and in engineered systems. Oxo complexes (formally, the oxido ligand in complex names) are important with several metals. For example, Cr(+VI) complexes with four O2– ligands to form chromate in water. Chromate is Cr(+VI)O(–II)42– ≡ CrO42–. Another common oxo complex is the familiar carbonate: C(+IV)O(–II)32– ≡ CO32–. Bicarbonate is a mixed-­ligand complex; that is, a complex consisting of two or more different ligands. Bicarbonate is an oxohydroxo complex: C(+IV)O(–II)2OH(–I)– ≡ HCO3–. The formal name of sulfuric acid, H2SO4, treats S(+VI) as a metal: dihydroxidodioxidosulfur (Hartshorn et al. 2015). Complexes with bicarbonate and carbonate also can be important in the environment. They usually are found in lower concentrations than hydroxo complexes in fresh water. The chloro complexes (formally, the chlorido ligand in complex names) are very important in estuarine systems and in seawater. Cyano complexes in the environment typically are found only in waters with high cyanide concentrations (e.g., in metal finishing wastewaters).

15.3.5  Organic Ligands Metal-­organic complexes often control the chemistry of many metals in the environment. Although there are many types of organic ligands, three classes of organic ligands play especially important roles in the environment and in methods for the analysis of metals in the environment.

Table 15.2 Common Inorganic Ligands and Complexes Complex Name (Common)

Ligand

Examples

Oxo Bicarbonato/carbonato Chloro

O HCO3–, CO32– Cl–

Cyano

CN–

CrO42–, CO32– CoCO30, NiCO30, CuCO30, ZnCO30, CdCO30 NiCl+, ZnCl+, AgCl0, AgCl2–, CdCl20, HgCl42– many, e.g., Ni(CN)42–

(Source: Republished with permission of John Wiley & Sons, from Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters, 3rd Edition, Stumm and Morgan 1996: permission conveyed through Copyright Clearance Center, Inc.)

mixed-­ligand complex: a complex consisting of two or more different types of ligands

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334   Chapter 15  Complexation The first class is the simple organic acids. Common organic acids in the environment and in engineered systems include the short-­chained organic acids (acetic and propionic acids found in anaerobic digesters), the amino acids (see Figure 15.2 for an example), and anionic surfactants such as nitrilotriacetic acid (NTA, formally N,N-­ bis(carboxymethoxy)glycine; see Figure 15.3). NTA is a tridentate ligand. Another organic acid, EDTA (ethylenediaminetetraacetic acid or ethylenedinitrilotetraacetic acid or, formally, N,N’-­1,2-­ethane-­diylbis(N-­(carboxymethyl)glycine), forms very strong complexes with many metals of environmental interest. The structure of EDTA is shown in Figure 15.3. EDTA coordinates with all four acidic oxygen atoms and both nitrogen atoms. Thus, it is a hexadentate ligand and can completely coordinate a metal with a coordination number of six. The ligand EDTA forms exceptionally strong complexes. It is used in pollutant analysis methods to bind unwanted metals and is administered in acute metal poisoning cases (called chelation therapy). A second important class of ligands comprises naturally occurring organics in the environment. Litter from plant material is biodegraded by microorganisms into complex, poorly characterized organic matter. In the aquatic environment, the two important organic fractions are called fulvic and humic acids. Fulvic and humic acids are large molecular weight materials with many acidic functional groups. For example, fulvic acid is typically about 40% oxygen and 0.5% nitrogen by weight with a carboxyl acid content of about 6 mmol/g (Thurman and Malcolm 1983). Thus, fulvic and humic acids are expected to be both Brønsted and Lewis acids. The third class of organic ligands comprises those important in biochemistry. The classic example is the porphyrin ring, a modification of porphin (see Figure  15.4). One important metal-­porphyrin complex is hemoglobin, an iron-­porphyrin complex. Hemoglobin is responsible for oxygen transport in the blood of vertebrates. (Related compounds include myoglobin, which transports oxygen in muscle tissue, and hemocyanin, which is responsible for oxygen transport in the blood of some invertebrates.) Recall that iron has a coordination number of six (Table 15.1). Four electron pairs come from the four nitrogen atoms in the porphyrin ring. The iron-­porphyrin system lies in a plane. The fifth electron pair comes from the nitrogen in an amino acid (histidine) in hemoglobin that lies outside the iron-­porphyrin plane. The sixth electron pair comes from reversible binding with oxygen.

HO

OH

O

O

O O N

N OH

HO O

OH

FIGURE 15.3  NTA (left) and EDTA (right)

FIGURE 15.4  Structure of Porphin

N

O

N

NH

NH

N

O

OH OH

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15.4  |  Equilibrium Calculations with Complexes   335

Another important metal-­porphyrin complex is chlorophyll. Chlorophyll, the principal photoreceptor in green plants, is based on porphyrin complexed with magnesium. Vitamin B12 (cobalamin) is based on the complexation of cobalt with a corrin ring (similar to the porphyrin ring). Cobalt is complexed with two other substituents perpendicular to the ring to satisfy its coordination number of six. As a final example, the cytochrome family comprises iron porphyrins used for electron transfer reactions. In the cytochromes, the central iron atom undergoes oxidation-­reduction reactions (see Chapter 16).

15.3.6  Effects of pH With substances exhibiting both Brønsted and Lewis acidity, protons (H+) and other metals (Mn+) compete for binding sites. Thus, the ability of fulvic acid humic acids (along with the simple organic acids) to bind metals depends on the pH. Thoughtful Pause How does pH affect metal binding? At low pH, the activity of H+ is high and H+ tends to displace metals from the binding sites. For example, acidic rainfall will leach aluminum from soils. Why? The functional groups in soils that act as ligands and bind aluminum ions are also Brønsted acids. At low pH, H+ replaces Al3+ and aluminum is released. It is important to remember that pH affects the chemistry of almost every metal and many ligands. Consider a qualitative description of the complexation of cadmium by acetic acid. At very low pH, acetic acid is protonated and is a weaker complexing agent. Thus, at very low pH, cadmium is mostly in the aquo form. As the pH increases, the cadmium-­acetate complex will form. At higher pH values, the hydroxo complexes will dominate. Why? Due to the fixed total dissolved acetate concentration, the Ac– concentration stays fairly constant at pH > pKa of acetic acid, while the concentration of OH– increases exponentially with increasing pH.

Key idea: Metals and H+ often compete for binding sites with ligands that are Brønsted acids

15.4  EQUILIBRIUM CALCULATIONS

WITH COMPLEXES

15.4.1  Introduction Equilibrium calculations for homogeneous systems involving complexation are set up exactly as with acid–base systems. To set up the system, write the species list, pertinent equilibria, mass balance equations, and in some cases the electroneutrality equation. After the equilibria are translated into mathematical equations, the n equations in n unknowns (i.e, n species concentrations) can be solved for. Recall from Chapter  6 that mass balances are needed to account for all species except H+, OH–, and H2O. As a result, systems involving complexation usually have one mass balance equation for each metal and one mass balance equation for each ligand. Mass balances are not needed for the ligands OH– and H2O. Often in aquatic chemistry problems, the total dissolved metal and total dissolved ligand concentrations are known.

Key idea: The equilibrium calculations for homogeneous systems involving complexation are set up exactly as with acid–base systems

ISTUDY

336   Chapter 15  Complexation Key idea: Metals that form hydroxo complexes will precipitate under certain pH and total metal ­concentrations

Before going into the details of equilibrium calculations with complexes, please note: The techniques presented in this chapter are appropriate for homogeneous systems. When metals hydrolyze to form aquo and hydroxo complexes, it is very common for solids to form under certain total metal and pH conditions. For example, Fe(+III) can precipitate as Fe(OH)3(s) if the pH and Fe(+III)T are in certain ranges. Equilibrium calculations in the presence of solids will be covered in Chapter 19. For the moment, consider homogeneous systems only. Thus, many of the examples in this chapter are limited to low total metal concentrations and/or low pH where metal precipitation typically does not occur.

15.4.2  Equilibrium Constants for Complexation formation (stability) constant: the equilibrium constant for a complexation equilibrium, written with the complex as a product

The equilibrium constants for complex formation reactions are called formation constants or stability constants. Remember: A formation constant is just an equilibrium constant. It is an equilibrium constant for a reaction of the form:

aMn

bLm

Ma L b an

bm

K

formation constant

Formation constants for many complexes are listed in Appendix D. Table D.1  lists formation constants for metals with inorganic ligands and Table D.2 lists formation constants for metals with organic ligands. Most formation constants are large. Therefore, it is common to write and compare logK values. Recall that acid dissociation reactions are written as the reverse of the com1 plexation equilibrium shown above (with Mn+ = H+ and K = ). Thus, acid dissociaKa tion reactions have small K values and it is common to write and compare pKa values. Tables D.1 and D.2 list logK values for many complexes. It is good practice to find a formation constant for an equilibrium of interest. Several formation constants from Table D.1 are shown in Table 15.3. Try to use Table 15.3 to find K for: Worked example →

Equilibrium 1: Al3+ + OH– = AlOH2+ Equilibrium 2: Al(OH)42– = Al3+ + 4OH– Equilibrium 3: Al3+ + HPO42– = AlHPO4+ Table 15.3 Excerpt from Table D.1 (tabulated values are logK) Inorganic Ligand (L) Metal (M)

Equilibrium

OH–

PO43–

Al

M + L = ML M + HL = MHL M + 2L = ML2 M + 3L = ML3 M + 4L = ML4

9.00   17.9 25.2 33.3

  6.12

3+

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15.4  |  Equilibrium Calculations with Complexes   337

For Equilibrium 1, M = Al3+ and L = OH–. The metal-­ligand stoichiometry is 1:1, so use the first entry in the table: logK = 9.00 so K = 10+9.00. Equilibrium 2 is not in the right form, but for the reverse reaction (Al3+ + 4OH– = Al(OH)42–): M = Al3+, L = OH–, and the M:L stoichiometry is 1:4. Therefore, K for the reverse of Equilibrium 2 is 10+33.3 and K for Equilibrium 2 is 10–33.3. For Equilibrium 3, M = Al3+ and L = PO43–. So HL = HPO42–. The metal-­ligand stoichiometry is 1:1, so use the second entry in Table 15.2: logK = 6.12 so K = 10+6.12 It is common to say that the complex is strong if the formation constant is large. However, you must be careful when deciding whether a formation constant is large or when comparing formation constants. The magnitude of the formation constant depends on the stoichiometry of the reaction. For example, try to determine whether the tetrachloro-­Hg(+II) complex is strong. The equilibrium constant for Hg2+ + 4Cl– = HgCl42– is K = 10+15.2. This seems like a very large equilibrium constant, but recall that it has associated units of M–4. If the Hg2+ and Cl– concentrations could be held magically at 10–5 M, then the concentration of the complex would be only 10–9.8 M.5 An example of comparing formation constants is given in Worked Example 15.1. Given the potential problems with comparing formation constants, it is more precise to say that a complex is strong in a given system if it is the dominant form of the metal in that system. Worked Example 15.1 

Comparing Formation Constants

Which metal forms a stronger complex with cyanide ion: Fe3+ or Ni2+? Solution In Table D.1 (Appendix D), the only complex formation equilibria for ferric iron and nickel with CN– are: 3

log K1

43.6

2

log K 2

30.2

Fe3

6CN

Fe CN

6

Ni 2

4 CN

Ni CN

4

It is not appropriate to simply compare the values of K1 and K2, since they have different associated units. To determine which complex is stronger, it is necessary to perform an equilibrium calculation to see whether more Fe(CN)63– or more Ni(CN)42– is formed. (For further analysis, see Worked Example 15.3.)

Recall that acid dissociation constants are given a special symbol, namely, Ka. Similarly, some formation constants also are given special symbols. A common naming convention for formation constants is as follows: 1. Use Ki for stepwise formation equilibria (i.e., when the stoichiometric coefficient of the ligand in the equilibrium is 1), resulting in the formation of MLi from MLi–1. Example: For Cu(OH)20 + OH– = Cu(OH)3–, use K3.  The phrase “only 10–9.8 M” is a bit misleading since a concentration of 10–9.8 M may be significant if the complex ­participates in important biological reactions.

5

Key idea: A complex is strong in a given system if it is the dominant form of the metal in that system Key idea: Make sure that the stoichiometry of the equilibria are the same before comparing formation constants

ISTUDY

338   Chapter 15  Complexation 2. Use βi for cumulative formation equilibria (i.e., when the stoichiometric coefficient of the ligand in the equilibrium is i > 1). Example: For Cu2+ + 3OH– = Cu(OH)3–, use β3. 3. Use βmn for formation of the polynuclear complex MmLn.

Key idea: Ki, βi, βmn, *Ki, *βi, and *βmn are just equilibrium constants for formation equilibria written in a specific form

Example: For 2Cu2+ + 2OH– = Cu2OH22+, use β22. 4. Use *Ki, *βi, or *βmn when the ligand is protonated (including reactions with water, where H2O is the protonated form of OH–). Example: For Fe3+ + 2H2O = Fe(OH)2+ + 2H+, use *β2. For simplicity, only the symbol Ki will be used in this text, with i having no significance and chosen simply to assign a unique symbol to each formation constant. However, you should be aware of the naming convention, as formation constants frequently are compiled by using this system.6 Remember: Ki, βi, βmn, *Ki, *βi, and *βmn are just equilibrium constants.

15.4.3  Equilibrium Calculations with Hydroxo Complexes Recall from Section  15.3.2 that the formation of hydroxo complexes stems from the acidity of the aquo ligands. Thus, equilibrium calculations containing only soluble species are similar to acid–base equilibrium calculations. In general, pC-­pH diagrams are useful for displaying the results of calculations. As an example, consider the speciation of dissolved aluminum in treated drinking water. Assume that no solid phases exist (i.e., you sample after filtration) and that the total soluble aluminum concentration is about 0.5 μg/L as Al. Find the major dissolved aluminum species in tap water (pH 7.5–8.0) and determine whether the predominant soluble aluminum species changes when consumed tap water hits the stomach (at about pH 2). For the dissolved aluminum system, the pertinent equilibria are: Al3

 AlOH 2

OH

AlOH 2 Al(OH)2

OH    Al(OH)2 OH

Al(OH)30 OH

 Al(OH)30

 Al(OH)4  

K1

10

9

10

8.8

K3

10

7.7

10

7.9

K2 K4

Thus: [AlOH 2 ] Al OH

2

K1 [Al3 ][OH ] K1 K 2 [Al3 ][ OH ]2 0

Al OH

3

Al OH

4

K1 K 2 K3 [Al3 ] [OH ]3 K1 K 2 K3 K 4 [Al3 ][OH ]4

 The naming convention has certain advantages. You could write β2 = 10+12.8 for Fe2+ and NTA and the reader would know you are referring to the formation equilibrium: Fe2+ + 2L3– = FeL24– (where L3– = unprotonated NTA). Thus, you could tabulate formation constants without writing the equilibria explicitly if you use the naming convention. 6

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15.4  |  Equilibrium Calculations with Complexes   339

A mass balance on soluble aluminum reveals: Al

III

total soluble aluminum

T

Al3 Al3

AlOH 2

Al OH

1 K1 OH

Al OH

2 2

K1 K 2 OH

3

0

Al OH

4

3

K1 K 2 K3 OH

4

K1 K 2 K3 K 4 OH Or: Al

III

Al3

T

1

K1 KW H

K1 K 2 KW2

K1 K 2 K3 KW3

2

H

H

3

K1 K 2 K3 K 4 KW4 H

4

So Al3

K1 KW 1 H 5 10

Here: Al(+III)T =

7

K1 K 2 KW2 H

Al

2

III T K1 K 2 K3 KW3 H

K1 K 2 K3 K 4 KW4

3

H

4

g L = 1.85×10–8 M; pAl(+III) = 7.7. You can calculate T

g mol [Al3+] at any pH, then back-­substitute to calculate the concentrations of all the aluminum-­hydroxo complexes. A pC-­pH diagram for the system is shown in Figure 15.5. Clearly, Al3+ dominates at low pH and Al(OH)4– dominates at high pH. The pC-­pH diagram is a little crowded at intermediate pH values. To look for the dominant species, it is convenient to replot the calculated concentrations on a percentage basis, as shown in Figure 15.6. From Figure  15.6, it is easy to see that Al(OH)4– will predominate in tap water. As the pH decreases when tap water mixes with gastric fluids, the predominant soluble 27

pH 0

0

2

4

6

8

10

12

14

2

pC

4 6

[AI(OH)2+] [AI3+]

[AI(OH)4–]

8 10 12

[AIOH+]

[AI(OH)30]

14 FIGURE 15.5  pC-­pH Diagram for Aluminum (no solid phases, Al(+III)T = 1.85×10–­8 M)

ISTUDY

Concentration (as % of total)

340   Chapter 15  Complexation 100% [AI3+]

[AI(OH)4–]

80% [AI(OH)2+]

60% 40%

[AI(OH)30]

[AIOH+]

20% 0% 5

4

6

7

8

pH FIGURE 15.6  Predominance Diagram for Aluminum (no solid phase)

aluminum species will change to the lower hydroxo complexes (i.e., the complexes with fewer hydroxo ligands). Below, about pH 4.5, Al3+ is present at the highest ­concentration.

15.4.4  Equilibrium Calculations with Other Complexes Complexes with ligands other than hydroxo and oxo ligands require an additional mass balance (or mass balances) to account for the material in the ligand. As an example, consider the speciation of mercury and chloride in a solution where 10–3 mole of mercuric chloride, HgCl2(s), is added to 1 L of water and dissolves completely. Assume the solution has been acidified and the pH is fixed and low enough so that hydroxo complexes are not formed. Because Hg(+II) has a coordination number of 4 (from Table 15.1), you might expect that the mercury-­containing species will include Hg2+, HgCl+, HgCl20, HgCl3–, and HgCl42–.

Thoughtful Pause What other species should be in the species list?

The species list also should include Cl–, H+, OH–, and H2O (assuming that the pH is not extremely low and so HCl can be ignored). However, since the pH is assumed to be fixed, [H+] and [OH–] are not unknowns. As usual, assume that the mole fraction of water is unity. The species list becomes: Cl–, Hg2+, HgCl+, HgCl20, HgCl3–, and HgCl42–. The pertinent equilibria are (see Appendix D, Table D.1, rewritten here as stepwise complexation reactions): Hg2 HgCl

Cl Cl

HgCl2 0 Cl HgCl3

Cl

HgCl

HgCl2 0 HgCl3

HgCl 4 2

K1

10

K3

10

K2 K4

10 10

7.3 6.7 1.0 0.6

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15.4  |  Equilibrium Calculations with Complexes   341

Thus:

HgCl

K1 Hg2

Cl

eq. 15.5



HgCl 2 0

K 2 HgCl

Cl

eq. 15.6



HgCl3

K3 HgCl 2 0 Cl

eq. 15.7



HgCl 4 2

K 4 HgCl3

Cl

eq. 15.8

Thoughtful Pause What are the appropriate mass balances for this system? Mass balances must be written for total dissolved mercury(+II) and total dissolved chloride. The pertinent mass balances are:

Hg CIT

II

T

Cl

Hg2 HgCl

HgCl

HgCl 2 0

2 HgCl 2 0

HgCl3

HgCl 4 2 4 HgCl 4 2

3 HgCl3

eq. 15.9 eq. 15.10

Note the stoichiometric coefficients in the chloride mass balance. For example, 1 mole of HgCl3– contains 3 moles of chloride. Also note that the total chloride concentration is 2×10–3 M, and the total mercury(+II) concentration is 1×10–3 M when 10–3 mole of HgCl2(s) is added to 1 L of water and dissociates completely. What about electroneutrality? Through equations 15.5–15.10, you have 6 equations in 6 unknowns. Another equation is not needed. Does this make sense? It is not possible to write the electroneutrality equation, since you do not know how much strong acid was added to reduce the pH. A useful solution technique with complexation systems is as follows (see also Appendix C, Section C.7). Step 1: Solve for each species concentration as a function of the uncomplexed metal (i.e, aquo complex) and uncomplexed ligand. In this case, write each species concentration in terms of [Hg2+] and [Cl–]. For the mercuric chloride system:

HgCl

K1 Hg2



HgCl 2 0

K1 K 2 Hg2



HgCl3

K1 K 2 K3 Hg2+ Cl

3



HgCl 4 2

K1 K 2 K3 K3 Hg2

Cl

Cl

Cl

eq. 15.11 2



eq. 15.12

eq. 15.13 2



eq. 15.14

Key idea: Make sure that the stoichio­ metry of the equilibria are the same before comparing formation constants Key idea: To solve systems involving complexation: (1) Solve for each species concentration as a function of the uncomplexed metals (i.e., aquo complex) and uncomplexed ligands; (2) Substitute these expressions into the mass balances; (3) Rework to form one equation in one unknown and solve for the uncomplexed metal or ligand concentration; and (4) Back-­calculate the concentrations of the complexed species

ISTUDY

342   Chapter 15  Complexation Step 2: Substitute the expressions from Step 1 into the mass balances. Hg

II

T

Hg2

1 K1 Cl

Cl

1 K1 Hg2

3K1 K 2 K3 Hg

3

K1 K 2 K3 Cl

4

K1 K 2 K3 K 4 Cl ClT

2

K1 K 2 Cl

2 K1 K 2 Hg2 Cl

2

2

Cl

4K1 K 2 K3 K 4 Hg2

Cl

3

Note that you now have two equations in two unknowns, [Hg2+] and [Cl–]. Step 3: Rework the mass balances to obtain one equation in one species concentration. Solve for mercury(+II) mass balance for [Hg2+] and then substitute into the chloride mass balance:

CIT

where: a

K1 2 K1 K 2 Cl 1 K1 Cl

Cl

1 aHg

3K1 K 2 K3 Cl

K1 K 2 Cll

2

II

T

2

K1 K 2 K3 Cl



eq. 15.15

4 K1 K 2 K3 Cl 3

3

K1 K 2 K3 Cl

4

Although messy, eq. 15.15 is one equation in one unknown, namely [Cl–]. ­Solving by iteration, [Cl–] = 1.40×10–5 M, and from the Hg(+II)T expression, [Hg2+] = 7.93×10–8 M. Spreadsheet available: A spreadsheet for a 10–3 moles of HgCl2(s) in 1 L water is discussed in Section E.5.4

Nanoql SE file available: The Nanoql SE input for a 10–3 moles of HgCl2(s) in 1 L water is discussed in Section F.4.5

Step 4: Back-­calculate the concentrations of the complexed species. Thus: [HgCl+] = 1.40×10–5 M, [HgCl20] = 9.86×10–4 M, [HgCl3–] = 1.38×10–7 M, and [HgCl42–] = 7.73×10–12 M. To solve systems involving complexation (l) Solve for each species concentration as a function of the uncomplexed (i.e., aquo complexed) metals and uncomplexed ligands; (2) Substitute these expressions into the mass balances; (3) Rework to form one equation in one unknown and solve for the uncomplexed metal or ligand concentration; and (4) Back-­calculate the concentrations of the complexed species. You can verify that the mass balances are satisfied. Note that HgCl20 is the dominant species, accounting for over 99% of the total dissolved mercury and over 99% of the total chloride. The remaining mercury is mostly HgCl+ and the remaining total chloride is mostly HgCl+ and Cl–. A spreadsheet for this system is discussed in Section E.5.4, The Nanoql SE inputs are discussed in Section F.4.5. In this example, the ratio of Hg(+II)T:ClT was fixed because HgCl2(s) was added to water. If you were to dissolve 10–3 M of Hg(+II) into water and add chloride (by, say, adding NaCl(s)), the changes in species concentrations with ClT would be as shown in Figure  15.7. (To avoid too many lines in Figure  15.7, the concentrations of Hg2+ and Cl– are shown in the top panel and the other species shown in the ­bottom panel.)

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15.4  |  Equilibrium Calculations with Complexes   343

Thoughtful Pause Is complexation important at low total chloride concentrations?

log(CIT) –4.0 0 2

–3.5

–3.0

–2.5

–2.0

–2.5

–2.0

[Hg2+]

pC

4 6 8

[CI–]

10 log(CIT) –4.0 0 2

pC

4

–3.5

[HgCI+]

–3.0

[HgCI20] [HgCI3–]

6 8

[HgCI42–]

10 FIGURE 15.7  pC-­log(ClT) Diagram for the Mercury(+II)-­Chloride System (No solid phases or hydroxo complexes, Hg(+II)T = 1×10–3 M, [HgCl+] plotted with a dashed line for clarity.)

As with so many of the Thoughtful Pauses in this text, the answer to this question is: It depends. The importance of complexation depends on whether you are interested in mercury or chloride. At ClT = 1×10–4 M, most of the mercury is uncomplexed (at low pH) and Hg2+ is the dominant mercury species. However, most of the chloride is complexed and found in the form of HgCl+. At ClT = 1×10–3 M, HgCl+ and HgCl20 are becoming more important in the mercury mass balance. In addition, HgCl+ and HgCl20 are the dominant chloride-­containing species. Thus, most of the mercury is complexed and most of the chloride is complexed. Note also that the concentrations of the complexes are very sensitive to ClT near ClT = 1×10–3 M. At ClT = 1×10–2  M, HgCl20 is the dominant mercury-­containing species and chloride is the chloride-­ containing species at highest concentration. At this ClT, most of the mercury is complexed, but most of the chloride is not complexed since chloride is present in great excess over mercury. Another worked example of complexation is shown in Worked Example 15.2.

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344   Chapter 15  Complexation

Inorganic Complex Formation

Worked Example 15.2 

(The conditions and all formation constants in this example were taken from Butler 1998.) Solid waste from mining operations (mine tailings) frequently contain various FeS(s) minerals. In the presence of bacteria, water will leach through the tailings to form a very acidic liquid stream containing Fe3+ and SO42–. The liquid stream is called acid mine drainage. Determine the speciation of ferric iron and sulfur in acid mine drainage if the water is at pH 2 with Fe(+III)T = 5×10–3 M and S(+VI)T = 10–2 M. Solution The species list includes Fe3+, FeOH2+, Fe(OH)2+, Fe(OH)30, Fe(OH)4–, Fe2(OH)24+, Fe3(OH)45+, FeSO4+, SO42–, and HSO4–. In addition, H+, OH–, and H2O are all at fixed activity at pH 2. The pertinent equilibria are: Fe3

OH   FeOH 2  

Fe3

2OH   Fe OH

2

Fe 3

3OH   Fe OH

3

Fe3

4 OH   Fe OH

4

0

logK1

11.81

logK 2

23.3

logK3

28.8

logK 4

34.4

4

logK 5

25.1

5

logK 6

49.7

logK 7

4.04

logK8

5.38

logK 9

1.99

Fe OH

3

 

2 Fe3

2OH   Fe 2 OH

2

3Fe3

4 OH   Fe3 OH

4

Fe3

SO 4 2   FeSO 4

Fe3

2SO 4 2   Fe SO 4

SO 4 2

H

2

HSO 4 2  

The mass balances are: Fe

III

Fe3

T

FeOH 2

2 Fe 2 OH S

VI

SO 4 2

T

2

Fe OH

4

2

3 Fe3 OH

HSO 4 2

4

FeSO 4

5

0

FeSO 4 2 Fe SO 4

Fe OH

4

Fe SO 4

2

2

Writing each term in the ferric iron mass balance in terms of [Fe3+] and each term in the sulfate mass balance in terms of [SO42–]: Fe

III

T

Fe3

1 K1 OH

K 4 OH K 7 SO 4 2 S

VI

T

SO 4 2

4

K 5 Fe 3

OH

K8 SO 4 2

1 K9 H

2

K 2 OH 2

K 7 Fe3

2

3

K3 OH K 6 Fe3 2 K8 Fe3

2

OH

SO 4 2

4

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15.5  |  Systems with Several Metals and Ligands   345

The system has been reduced to two equations in two unknowns. Solving with Fe(+III)T = 5×10–3 M and S(+VI)T = 10–2 M, [Fe3+] = 1.59×10–4 M and [SO42–] = 2.49×10–3 M. Back-­calculating and summarizing: Fe 3

1.59 10 4 M

FeOH 2

1.03 10 4 M

Fe OH

2

Fe OH

3

Fe OH

4

3.18 10 5 M 0

Fe 2 OH

2

Fe 3 OH

4

FeSO 4

1.00 10

11

M

4.00 10

18

M

4

3.19 10 7 M

5

2.03 10

M

4.46 10 3 M 2.49 10 4 M

Fe SO 4

2

SO 4 2

2.55 10 3 M

HSO 4

10

2.49 10 3 M

Note that over 94% of the ferric iron is complexed with sulfate, while only about 50% of the sulfate is complexed with iron.

15.5  SYSTEMS WITH SEVERAL METALS

AND LIGANDS

15.5.1  Introduction Natural waters contain many metals and ligands. In principle, equilibrium calculations with very complicated natural or process waters can be conducted with the same steps as outlined in Section 15.4.2: 1. Solve for each species concentration as a function of the uncomplexed metals (i.e, aquo complex) and uncomplexed ligands. 2. Substitute these expressions into the mass balances. 3. Rework to form one equation in one unknown and solve for the uncomplexed metal or ligand concentration. 4. Back-­calculate the concentrations of the complexed species.

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346   Chapter 15  Complexation

15.5.2  Example System As an example, consider the addition of nitrilotriacetic acid (NTA) to a natural water. NTA (shown in Figure  15.3) is used as a replacement for phosphate in detergents and for treating boiler water (to complex the calcium and magnesium that make up hardness). The World Health Organization has recommended that drinking water contain no more than 200 μg/L of NTA. Your task is to determine the speciation of metals and ligands in a lake at pH 8 if the total soluble calcium concentration (= Ca(+II)T) is 1×10–3 M, the total soluble copper concentration (= Cu(+II)T) is 2×10–6 M, and the total soluble ferrous iron concentration (= Fe(+II)T) is 2×10–6 M. The calculations will be performed here with two total NTA concentrations (= NT): a low NTA concentration (NT = 10 μg/L = 5×10–8 M) and a very high NTA concentration (2 mg/L = 1×10–5 M). To simplify the symbols, the symbol L3– will be used to represent the uncomplexed (i.e., completely deprotonated) NTA. This system is very complicated (although slightly simplified by the assumption that Fe(+III) species can be ignored). Thoughtful Pause What are the uncomplexed Lewis acids and Lewis bases in the system? The system contains four uncomplexed Lewis acids (H+, Ca2+, Cu2+, and Fe2+) and three uncomplexed Lewis bases (OH–, L3–, and H2O). The pertinent equilibria are (see Appendix D, Table D.2, with three mixed-­ligand complexes added): Complexation with H+: H H H H

L3   HL2

K1

HL2   H 2 L

K2

H 2 L   H3 L

10

10.3 2.9

10 1 KW

2.0

10

6.3

K5

10

8.8

K6

10

1.3

K7

10

12.7

K3

OH   H 2 O

10

K

10  

Complexation with Ca2+: Ca 2

L3

CaL                

Ca 2

2 L3

Ca 2

OH

CaL 2 4          

CaOH          

K4

Complexation with Cu2+: Cu2

L3

CuL  

Cu2

2 L3

CuL 2 4

K8

10

17.4

CuL

OH

CuOHL2

K9

10

4.4

Cu2

OH

CuOH

K10

10

6.5

Cu2

2OH

K11

10

11.8

Cu OH

2

0

 

14

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15.5  |  Systems with Several Metals and Ligands   347

Cu2

3OH

Cu OH

3

Cu2

4 OH

Cu OH

4

2Cu2

2OH

2

Cu2 OH

2

2

K12

10

14.5

K13

10

15.66

K14

10

17.4

Complexation with Fe2+:



Fe 2

L3   FeL

K15

10

8.9

Fe 2

2 L3   FeL 2 4

K16

10

12.0

FeL

H   FeHL0  

K17

10

1.9

Fe 2

OH

K18

10

11.7

Fe 2

OH   FeOH

K19

10

4.6

Fe 2

2OH   Fe OH

2

K 20

10

7.5

Fe 2

3OH   Fe OH

3

K 21

10

13.0

Fe 2

4 OH   Fe OH

4

K 22

10

10.0

L3   FeOHL2 0

2

The pertinent mass balances are: NTAT

L3

HL2 CuL

           

H2 L 2 CuL 2 4

                2 FeL 2 4



CaL

CuOHL2

FeHL0

2 CaL 2 4

T

Ca 2

CaL

CaL 2 4

CaOH

Cu

II

T

Cu2

CuL

CuL 2 4

CuOHL2

CuOH

               

Cu OH

II

T

Fe 2

Cu OH 4

2

FeL

           

FeOHL2

               

Fe OH

2

0

Cu OH

2 Cu2 OH FeL 2 4

Fe OH

2

eq. 15.17



3

eq. 15.18

2

FeHL0

FeOH 3

eq. 15.16

FeOHL2

II

           



FeL

Ca

Fe

H3 L

Fe OH 4

2

2

0



eq. 15.19

           

Note that not all the stoichiometric coefficients in the mass balances are equal to one. This system has 29 unknowns (i.e., 29 species concentrations) and 29 equations (23 equilibria, 4 mass balances, and two species with fixed activities: mole fraction of H2O = 1 and {H+} = 10–8 M (pH 8). The Nanoql SE inputs are discussed in Section F.4.6.

Nanoql SE file available: The Nanoql SE input for the NTA-­Ca(+II)-­ Cu(+II)-­Fe(+II) system is discussed in Section F.4.6

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348   Chapter 15  Complexation

15.5.3  Manipulating the System Although this is a big system, approach it just like other complexation equilibria. You must solve for each species concentration as a function of the uncomplexed metals and ligands, then substitute the resulting expressions into the mass balances (eqs. 15.16–15.19). The mass balances then become: NTAT

L3

1 K1 H

2 K 5 Ca 2

Ca

II

Cu

II

T

T

K 7 K 9 Cu2

OH

K15 K17 Fe 2

H

Ca 2

1 K 4 L3

K 5 L3

Cu2

1 K 7 L3

K8 L3

II

T

Fe 2

K18 L

OH

K 21 OH

3

L3

OH 2

K 6 OH

2

K 7 K 9 L3

K12 OH

3

OH

K13 OH

4

2

OH

1 K15 L3 3

2

L3

2 K16 Fee 2

K18 Fe 2

2 K14 Cu2

K 4 Ca 2

2 K8 Cu2

K15 Fe 2

K11 OH

3

K3 H

K 7 Cu2

L3

K10 OH

Fe

2

K2 H

K16 L3 K19 OH

K 22 OH

2

K15 K17 L3

H

2

K 20 OH

4

The system has now been reduced to four equations (the four mass balance equations) in four unknowns: [L3–], [Ca2+], [Cu2+], and [Fe2+]. At pH 8 and assuming a dilute solution: [H+] = 10–8 M and [OH–] = 10–6 M. Solving and back-­substituting, you can obtain the concentrations shown in Table 15.4. Another equilibrium calculation with more than one metal is shown in Worked Example 15.3.

Key idea: Spreadsheet hints: (1) optimize by changing the p(concentration) or log(concentration) values, not the concentration values themselves, and (2) use relative errors in the objective function, not absolute errors

Table 15.4 Solutions to the NTA Problem (All concentrations are in M.) NTAT = 5×10–8 M Species

NTAT = 1×10–5 M

Exact

Approx.

Exact

[L3–] [HL2–] [H2L–] [H3L]

2.55×10–14 2.02×10–19 1.61×10–24 1.61×10–30

2.45×10–14 1.94×10–19 1.54×10–24 1.54×10–30

3.31×10–9 2.63×10–14 2.09×10–19 2.09×10–25

3.01×10–9 2.39×10–14 1.90×10–19 1.90×10–25

Approx.

[Ca2+] [CaL–] [CaL24–] [CaOH+]

1.00×10–3 5.08×10–11 4.10×10–22 2.00×10–8

1.00×10–3 4.88×10–11 3.78×10–22 2.00×10–8

9.93×10–4 6.56×10–6 6.87×10–12 1.98×10–8

1.00×10–3 6.01×10–6 5.72×10–12 2.00×10–8

ISTUDY

15.5  |  Systems with Several Metals and Ligands   349 NTAT = 5×10–8 M Species

NTAT = 1×10–5 M

Exact

Approx.

Exact

[Cu2+] [CuL–] [CuL24–] [CuOHL2–] [CuOH+] [Cu(OH)20] [Cu(OH)3–] [Cu(OH)42–] [Cu2(OH)22+]

3.91×10–7 4.99×10–8 6.37×10–17 2.50×10–22 1.24×10–6 2.47×10–7 1.24×10–10 1.56×10–15 3.84×10–8

4.07×10–7 4.99×10–8 6.12×10–17 2.50×10–22 1.29×10–6 2.57×10–7 1.29×10–10 1.62×10–15 4.16×10–8

1.20×10–10 2.00×10–6 3.32×10–10 1.00×10–20 3.81×10–10 7.60×10–11 3.81×10–14 4.80×10–19 3.64×10–15

1.33×10–10 2.00×10–6 3.02×10–10 1.00×10–20 4.19×10–10 8.36×10–11 4.19×10–14 5.38×10–19 4.41×10–15

Approx.

[Fe2+] [FeL–] [FeL24–] [FeHL0] [FeOHL2–] [FeOH+] [Fe(OH)20] [Fe(OH)3–] [Fe(OH)42–]

1.92×10–6 3.89×10–11 1.25×10–21 3.89×10–26 2.46×10–14 7.66×10–8 6.08×10–11 1.92×10–11 1.92×10–20

1.92×10–6 3.74×10–11 1.15×10–21 3.74×10–26 2.36×10–14 7.66×10–8 6.08×10–11 1.92×10–11 1.92×10–20

5.45×10–7 1.43×10–6 5.97×10–12 1.43×10–21 9.04×10–10 2.17×10–8 1.72×10–11 5.45×10–12 5.45×10–21

5.83×10–7 1.39×10–6 5.28×10–12 1.39×10–21 8.79×10–10 2.32×10–8 1.84×10–11 5.83×10–12 5.83×10–21

Complex Formation with More than One Metal

Worked Example 15.3 

Return to Example 15.1, where the challenge was to determine whether Fe3+ or Ni2+ forms the stronger complex with cyanide ion. One way to address this question is to add the same amount of total Fe(+II) and Ni(+II) to a solution containing some total cyanide and calculating the concentrations of the complexes formed. For a water at pH 2 with FeT = NiT = 1×10–5 M, and CNT = 1×10–3 M, determine which cyanide complex is formed at the highest concentration. Solution The pertinent equilibria are: CN Fe

3+

Ni 2

H

log K

9.2

3

log K1

43.6

2

log K 2

30.2

HCN

6CN 4 CN

Fe CN Ni CN

6 4

(Note: At pH 2, the hydroxo complexes of iron and nickel can be ignored safely.) Writing mass balances: 3

FeT

Fe3

Fe CN

6

NiT

Ni 2

Ni CN

4

CNT

CN

HCN

6 Fe CN

2

6

3

4 Ni CN

4

2

ISTUDY

350   Chapter 15  Complexation (Note: We cannot write the charge balance, since we do not know what materials were added to set the pH at 2.) Thus: FeT

Fe3

1 K1 CN

6

NiT

Ni 2

1 K 2 CN

4

CNT

CN

1 K H

 6 K1 Fe3

5

CN

4 K1 Ni 2

CN

3

Write expressions for [Fe3+] and [Ni2+] from the metal mass balances: FeT

Fe3

1 K1 CN

6

NiT

Ni 2

1 K 2 CN

4

Substituting into the cyanide mass balance results in one equation in one unknown (i.e., [CN–]). Solving at pH 2 with the total masses specified gives [CN–] = 6.31×10–11 M. Back-­calculating: Fe CN

6

Ni CN

6

3

2.51 10

23

M

2

2.51 10

16

M

Thus, Ni(CN)62– is the stronger complex. (Note: Most of the total cyanide is HCN. At pH 2, H+ outcompetes both Fe3+ and Ni2+ for the cyanide ion.)

15.5.4  Spreadsheet Solution

Spreadsheet available: A spreadsheet for the NTA-­Ca(+II)-­ Cu(+II)-­Fe(+II) system is discussed in Section E.5.5

How can you solve such a complicated system? An easy way to calculate equilibrium concentrations is with a spreadsheet. An example spreadsheet is shown in Figure 15.8 and is available. The equilibria and equilibrium constants are listed in columns F and G, respectively. For ease of entry, logK values were entered (column H) and K values calculated by Excel as: K = 10–logK. The four unknowns are shown in cells A3 through A6. The guesses of their –log(concentration) and the calculated concentrations (= 10–pC) are in columns B and C, respectively. The added total masses (given in the problem statement) are listed in cells G28-­G31. The equations for the concentrations of each species are entered in cells B11 through B38. For example, [CaL–] at equilibrium is equal to K4[Ca2+][L3–], so the formula entered in cell C18 is: =G7*B17*B13. The formulas in column B for [L3–], [Ca2+], [Cu2+], and [Fe2+] refer back to cells B3 to B6, respectively. The calculated total masses are calculated by summing the species concentrations with the appropriate stoichiometric coefficients (see eqs. 15.16–15.19). The calculated total masses are in cells H28–H31.

ISTUDY

15.5  |  Systems with Several Metals and Ligands   351

Key idea: To simplify chemical systems, ignore the species in each mass that are expected to be at low concentrations relative to the other species in the mass balance

FIGURE 15.8  Spreadsheet Solution for NTA Problem

To find the equilibrium concentrations, use a nonlinear tool (such as Solver) to adjust the values of [L3–], [Ca2+], [Cu2+], and [Fe2+] until the calculated total masses (cells H28–H31) match the actual total masses (cells G28–G31). In this example, the values of –log(concentration) of the four species are varied. To adjust the concentrations, you must devise an objective function. Here, calculate the sum of the squares of the relative errors for each total mass. The square of the relative error is:

2

added total mass calculated total mass added total mass

Using Solver, minimize (or set equal to zero) the sum of the squares of the relative errors (cell I32) by changing the pC values of [L3–], [Ca2+], [Cu2+], and [Fe2+] (cells C3 through C6). Initial guesses for the pC values of [L3–], [Ca2+], [Cu2+], and [Fe2+] are required, but can be crude. Two useful tricks aid in the spreadsheet solution. First, optimize by changing the p(concentration) or log(concentration) values, not the concentration values themselves. This increases the likelihood of converging, since the equilibrium concentration values may vary over many orders of magnitude.

ISTUDY

352   Chapter 15  Complexation Second, use relative errors in the objective function, not absolute errors. In this problem, the absolute error for the calcium mass balance will overwhelm the other mass balances, since Ca(+II)T is so much larger than the other total masses. Using relative errors results in more equitable weighting of the errors in calculating the total masses.

15.5.5  Interpreting the Equilibrium Calculations

Key idea: The species at the largest concentration in each mass balance usually depends on the total metal and ligand concentration

The species at the largest concentration in each mass balance usually depend on the total metal and ligand concentrations. Remember that a species may dominate one mass balance, but be a relatively small contributor to another mass balance. This point is illustrated in the example problem. At the lower NTA concentration, nearly all (99.8%) of the NTA is complexed with copper. The copper-­NTA complexes are strong and dominate the NTA mass balance. Nearly all the calcium is free calcium (Ca2+). About 78% of the dissolved copper is in hydroxo complexes, with about 20% in the form Cu2+ and only 2.5% complexed with NTA. Soluble ferrous iron is primarily Fe2+ (96%), with the remainder mostly as hydroxo complexes. At the higher NTA level, the speciation of the organic ligand changes. Although the copper-­NTA complex is strong, the total soluble copper concentration is small relative to the total NTA concentration. This means that the copper becomes saturated and NTA is available to complex with other metals. As a result, about 66% of the organic ligand is complexed with calcium, with about 20% in copper-­NTA complexes and about 14% in ferrous iron-­NTA complexes. Since the total NTA concentration is still so much smaller than the total calcium concentration, calcium is still found almost completely as Ca2+. Copper, on the other hand, is now nearly completely complexed with NTA. Ferrous iron now is present mainly in ferrous iron-­NTA complexes (72%), with 27% free, and 1% in hydroxo complexes.

15.5.6  Approximate Solution The available software allows for the solution of complex systems with relative ease. However, your chemical intuition is sharpened by simplifying the system through approximations. Approximations should be approached in a systematic way. A good approach is to ignore the species in each mass balance that are expected to be at low concentrations relative to the other species in the mass balance. For example, in the calcium mass balance, you know that: CaL



K 4 Ca 2

L3

10 2

CaL 2 4

K 5 Ca 2

L3

CaOH

K 6 Ca 2

OH

Ca 2

6.3

10

8.8

10

4.7

L3

Ca 2 Ca 2

L3

2

at pH 8



Thoughtful Pause Which species in the calcium mass balance can be ignored?

ISTUDY

15.5  |  Systems with Several Metals and Ligands   353

Clearly, [CaOH+] MT. You can MT use the same approach as the previous example, with [M] = , [M*] = a KM L LT MT , and [L] = where a* encapsulates M*-­OH ­complex b K M M K M* M* a* K M* L

formation analogous to a. As an example, revisit the Pb(+II)-­EDTA system. The Pb2+ concentrations as a function of pH for PbT = 10–3 M and LT = 10–2 M are shown in Figure 15.11 with no Mg(+II) and when 0.1 M Mg(+II) is added to the system. Here, a* ≈ 1. Note that the metal buffer system maintains a very constant Pb2+ concentration above about pH 5. K * MT* LT Perrin and Dempsey (1974) give an approximate solution: [M] ≈ M MT K M LT MT (see also Problem 23). The approximate solution gives pPb2+ = 11.21 for the ­conditions in this example. This is within 0.1 of the exact solution in Figure 15.11 for pH > 5.

15.7  CHAPTER SUMMARY Metals are important parts of the aquatic environment because of their toxicity to humans and aquatic biota. Metals form complexes with ligands. Complex formation often controls the chemistry of both the metals and ligands. In this chapter, you learned that the concept of acids and bases can be extended to include complex formation. Lewis acids and bases are defined to be substances which accept or donate an electron pair. All Brønsted-­Lowry acids are Lewis acids. However, the Lewis acid concept is much broader and includes metals as well. Equilibrium calculations for systems exhibiting complexation are similar to acid– base equilibrium calculations. In general, mass balances are required for each metal (other than H+) and each ligand (other than OH–). In most applied problems, the pH is known and thus the electroneutrality equation is not needed. Complexation chemistry can be applied to many areas, including the complexation of metals by naturally occurring organics, biochemistry, metal plating, analytical chemistry, and metal buffering.

←Worked example

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362   Chapter 15  Complexation

15.8  PART IV CASE STUDY: WHICH FORM

OF COPPER PLATING SHOULD YOU USE?

As presented in Section  14.6, the Part IV case study concerns the relative merits of copper cyanide plating and acid copper plating. In this section, the effects of complexation with several metals and ligands on the plating process will be explored. Metal finishers strive to reduce discharges of potentially harmful chemicals into the environment. As a result, the use of cyanide in metal finishing operations is being minimized. In spite of this trend, copper cyanide plating is fairly common. Why? In general, cyanide is used in plating processes to increase the solubility of the metal. The chemistry behind this statement will be explored in Chapter 19. However, cyanide is used in copper plating for other reasons as well. Two reasons for the use of cyanide in copper plating will be explored: economics and preservation of the integrity of the piece. In the discussion of the Part IV case study at the end of Chapter 16, an economic advantage of copper cyanide over acid copper will be quantified. However, the economic advantage disappears if the Cu+ from copper cyanide is converted into Cu2+. The interconversion of Cu+ and Cu2+ can occur according to the equilibrium: Cu(s) Cu2+



2Cu

log K1

6.1

(The equilibrium constant for this equilibrium will be developed in Section 16.11.) 2 Cu Note that since the activity of a pure solid phase such as Cu(s) is 1, then: = 10–6.1 Cu2 or [Cu2+] = 10+6.1[Cu2+]2. Since Cu2+ and Cu+ are the only forms of Cu(+II) and Cu(+I), respectively, in the absence of cyanide, you can write:

Cu

II

10

T

6.1

Cu

I

2

T



This equation indicates significant conversion of Cu2+ to Cu+. To prevent the interconversion of copper oxidation states, cyanide is added. How does cyanide help? Cyanide forms complexes with Cu+ (see equilibria below), but not with Cu2+ (see Appendix D). Cu



2CN

Cu CN

Cu

3CN

Cu CN

Cu

4 CN

Cu CN

logK 2 16.3

2 3

2

4

3

logK3

21.6

logK 4

23.1



Performing a mass balance on Cu(+I): Cu



I

T

Cu

Cu CN

Cu

1 K 2 CN

Cu CN

2 2

K3 CN

3

2

3

Cu CN 4

K 4 CN

Thus:

Cu

1 K 2 CN

2

Cu

I

T

K3 CN

3

K 4 CN

4



4

3



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15.8  |  Part IV Case Study: Which Form of Copper Plating Should You Use?   363

Or:

Cu where: αCu =

1 K 2 CN

Cu

1

2

K3 CN

3

I T

Cu

4

K 4 CN

Note that αCu is like an alpha value for complexation. It represents the fraction of the total Cu(+I) that is Cu+. Thus: αCu ≤ 1. Recall that Cu2+ is still the only form of Cu(+II) in the system, so [Cu2+] = Cu(+II)T. Thus, the equilibrium expression [Cu2+] = 10+6.1[Cu2+]2 becomes: Cu

II

10

T

6.1

Cu

Cu

I

T

2

Cu

1

Thus, by adding cyanide (and therefore reducing the value of αCu), the concentration of Cu(+II) is reduced and the interconversion of Cu+ and Cu2+ is minimized. A second reason for adding cyanide in the copper cyanide process is to avoid the formation of immersion deposits. Immersion deposits occur when the oxidized form of the metal to be plated oxidizes the metal in the piece. This process can be described by the following equilibrium: Mn

P(s)

M(s) P n

where M is the metal to be plated and P is the metal in the piece. Of particular concern with copper is immersion deposits formed during electroplating on zinc. With copper cyanide plating, the pertinent equilibrium is: 2Cu

Zn(s)

2Cu(s) Zn 2+

logK

43.4

(Again, the equilibrium constant for this equilibrium will be developed in Section 16.11.) This equilibrium shows that copper(+I) in the plating bath will dissolve some zinc from the piece and cause pitting in the piece. At equilibrium in the absence Zn 2 of cyanide: = 1043.4 or [Zn2+] = 1043.4[Cu+]2 or: 2 Cu Zn



II

T

10 43.4 Cu

I

T

no cyanide

eq. 15.20

Clearly, the potential for the oxidation of metallic zinc by Cu+ is real and an appreciable quantity of Zn2+ may be formed at equilibrium. Again, the picture changes in the presence of cyanide. With cyanide, [Cu+] = αCuCu(+I)T. Zinc complexes with cyanide as well according to the equilibria:

Zn 2

2CN

Zn CN

2

Zn 2

3CN

Zn CN

3

Zn

2

4 CN

Zn CN

4

0

2

logK 5 logK 6

11.1 16.1

logK 7

19.6

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364   Chapter 15  Complexation From a mass balance on Zn(+II): Zn



II

T

Zn 2 Zn

Zn CN

2

0 2

1 K 5 CN

2

Thus: Zn 2

Zn CN K 6 CN Zn

2

1 K 5 CN

II

T

K 6 CN

3

Zn CN

3

4

3

K 7 CN

K 7 CN

4

2



4

Or: [Zn2+] = αZnZn(+II)T where: αZn =

2

1 K 5 CN

1 K 6 CN

3

K 7 CN

4

Again, αZn is like an alpha value for complexation. It represents the fraction of the total Zn(+II) that is Zn2+. Thus: αZn ≤ 1. At equilibrium in the presence of cyanide:

Zn 2

10 43.4 Cu

2

Zn

Zn

II T

Or, since [Cu+] = αCuCu(+I)T: Zn



II

T

10 43.4 Cu

I

T

2

2 Cu Zn

with cyanide

eq. 15.21

Note that zinc complexes with cyanide less strongly than Cu(+I). You can see this because stability constants K2, K3, and K4 are larger than stability constants K5, K6, and K7, respectively. (The stoichiometries are the same in the equilibria, so it is acceptable to compare the values of the stability constants.) Therefore, αZn > αCu and the term 2 Cu

must be less than one. Thus, comparing eqs. 15.20 and 15.21 with

Zn

2 Cu

< 1, you can

Zn

conclude that cyanide helps to minimize the formation of immersion deposits. Further discussion of this case study and the calculation of the pertinent equilibrium constants may be found in Section 16.11. The equilibrium chemistry of copper cyanide baths is discussed in a much more quantitative fashion in Section 24.2, where more realistic plating conditions are explored.

CHAPTER KEY IDEAS • Metals are important because (1) they are toxic to humans and aquatic biota and (2) they form complexes which may control the chemistry of the ligands. • Metals and ligands coordinate to form complexes. • Acid–base chemistry is a subset of complexation chemistry. • Brønsted-­Lowry acids and bases also are Lewis acids and bases, but the Lewis acid concept is much broader than the Brønsted-­Lowry acid concept. • The transition metals are Lewis acids. • The ligand water usually is not written explicitly. • Ligands donate electron pairs to metals and commonly contain N, O, and/or S.

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Chapter Glossary   365

• The acidity of some complexes stems from the ability of aquo ligands to donate protons to water. • Label soluble, uncharged complexes with (aq) or the superscript 0 (for zero charge) to differentiate them from solids. • Metals and H+ often compete for binding sites with ligands that are Brønsted acids. • The equilibrium calculations for homogeneous systems involving complexation are set up exactly as with acid–base systems. • Metals which form hydroxo complexes will precipitate under the proper pH and total metal concentrations. • A complex is strong in a given system if it is the dominant form of the metal in that system. • Make sure that the stoichiometry of the equilibria are the same before comparing formation constants. • Ki, βi, βmn, *Ki, *βi, and *βmn are just equilibrium constants for formation equilibria written in a specific form. • To solve systems involving complexation (1) Solve for each species concentration as a function of the uncomplexed (i.e., aquo complexed) metals and uncomplexed ligands; (2) Substitute these expressions into the mass balances; (3) Rework to form one equation in one unknown and solve for the uncomplexed metal or ligand concentration; and (4) Back-­calculate the concentrations of the complexed species. • Spreadsheet hints: (1) optimize by changing the p(concentration) or log(concentration) values, not the concentration values themselves, and (2) use relative errors in the objective function, not absolute errors. • To simplify chemical systems, ignore the species in each mass that are expected to be at low concentrations relative to the other species in the mass balance. • The species at the largest concentration in each mass balance usually depends on the total metal and ligand concentration.

CHAPTER GLOSSARY chelates:  complexes formed with a multidentate ligand complex:  a compound formed by the coordination of a metal and a ligand (also called coordination complexes or metal-­ligand complexes) coordination:  the association of two chemical species to form a third species coordination number:  the number of ligands with which a metal coordinates formation (stability) constant:  the equilibrium constant for a complexation equilibrium, written with the complex as a product free metal:  the aqua-­only or aquo complex of a metal Lewis acid:  a substance that accepts an electron pair Lewis base:  a substance that donates an electron pair ligand:  a species that coordinates with a metal metal:  in aquatic chemistry, the positively charged partner in a coordination reaction mixed-­ligand complex:  a complex consisting of two or more different types of ligands monodentate ligand:  a ligand capable of forming one bond with a metal polynuclear (multinuclear) complexes:  complexes with more than one coordinating metal

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366   Chapter 15  Complexation

HISTORICAL NOTE: BRITISH ANTI-­LEWISITE – A WMD-­INSPIRED LIGAND The history of science is woven with threads linking seemingly disconnected topics.8 This Historical Note will take you from attempts to make synthetic rubber in Washington, DC to the battlefields of World Wars I and II to the development of a chelate with medical applications. At the turn of the twentieth century, demand for rubber was skyrocketing. Many scientists were seeking a way to synthesize artificial rubber. The key to synthetic rubber was polymerizing short-­chained hydrocarbons like acetylene (formally, ethyne). Julius Nieuwland (1878–1936) was conducting his doctoral dissertation work at Catholic University of America (CUA) on the reactions of acetylene with just about everything to understand its potential for making artificial rubber. In one experiment, he reacted acetylene with arsenic trichloride (AsCl3) in the presence of aluminum chloride as a catalyst. As Nieuwland wrote in his dissertation: “Inhalation of the [resulting] fumes, even in small quantity caused nervous depression. . .Owing to the poisonous nature of the compounds formed, their thorough investigation was postponed.” (Nieuwland 1904). In fact, inhalation of the fumes sent Nieuwland to the hospital for a few days. Nieuwland had accidentally synthesized dichloro-­(2-­chlorovinyl)arsine a compound we now call lewisite (Figure  15.12). Although Nieuwland postponed a more thorough investigation of the compound, lewisite was not forgotten. So why is the compound called lewisite and not nieuwlandite? We need to expand the dramatis personae of our story to include Winford Lee Lewis. Lewis had a PhD in chemistry from the University of Chicago and was teaching as an associate professor at Northwestern University. As the United States entered World War I, Lewis (later Captain Lewis) took a leave of absence to work for the organization that became the Chemical Warfare Service (CWS), located at CUA and neighboring American University. World War I saw the introduction of chlorine and mustard gas as agents of war. Lewis sought an allied response. He discussed the matter with a chemistry professor at CUA, John Griffin. Griffin remembered that his former doctoral advisee, Julius Nieuwland, had landed in the hospital after experimenting with arsenic trichloride, acetylene, and ammonium chloride. Thus, a potential weapon of mass destruction, lewisite, was rediscovered. Lewisite turned out to be a relatively ineffective chemical warfare agent. Its characteristic smell (see the World War II poster) and rapid hydrolysis in humid conditions lead to it not being used in the field. Nevertheless, the United Sates produced about 20,000 pounds of the chemical agent. It was enough of a threat during World War II CI CI

H

As C H

C CI

FIGURE 15.12  Structure of Lewisite

 For an excellent book on lewisite and British anti-­lewisite, see Vilensky (2005).

8

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Historical Note: British Anti-­Lewisite – A WMD-­Inspired Ligand   367

Lewisite Warning Poster from World War II (National Museum of Health and ­Medicine/Wikimedia Commons/CC BY 2.0) that an antidote was sought. (The threats to rain down lewisite on unsuspecting populations earned it the name “Dew of Death.”) A group at Oxford University hypothesized that compounds with two sulfhydryl groups9 might bind arsenic. This led to the development of dimercaprol [2,3-­bis(sulfanyl)propan-­1-­ol], or British anti-­lewisite (BAL), illustrated in Figure 15.13. Dimercaprol, although itself toxic at high doses, is used in chelation therapy for arsenic and other toxic metals, including mercury and lead. And what of Julius Nieuwland, the man who started it all? After receiving the first PhD in chemistry awarded at CUA, Father Nieuwland became a professor at the University of Notre Dame. He eventually found a way to polymerize acetylene, eventually leading to the development of the synthetic rubber neoprene by scientists at DuPont. For this work, he was inducted into the National Inventors Hall of Fame in 1996. DuPont assisted in the funding of the Nieuwland Science Hall at Notre Dame in 1952. Father Nieuwland also was a friend and master’s advisor of famed Notre Dame football coach Knute Rockne.

HO

FIGURE 15.13  Structure of Dimercaprol

SH SH

9  Due to wartime shortages, the Oxford researchers extracted sulfur-­containing proteins from human hair collected by local barbers.

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368   Chapter 15  Complexation

PROBLEMS 1. Is CN– a Brønsted acid? Lewis acid? Both? Neither? Explain your reasoning. 2. Explain why HCO3– is a Brønsted acid, Brønsted base, and Lewis base, but not a Lewis acid. 3. Many metals (with free metal species M ) form complexes with sulfate. From Appendix D, list the metals that form MSO4(n–2) complexes. Which MSO4(n–2) complex is the strongest and why? n+

4. List all the complexes of mercury from Appendix D that are formed from a 3:1 ratio of ligand to metal. If the ligand is denoted Lm–, which HgL3(2–3m) complex is the strongest and why? 5. Show that:

i

a. Which oxidation state of iron forms the strongest complex with cyanide? b. If [Mn+] is fixed, then determine the order in concentration of the complexes as [CN–] changes.

Fe

2

Cd 2

N O CH3

O NH OH N O NH O

i j 1K j.

6. The following questions refer to the formation equilibria below.

Fe 3

HO

6CN 6CN 2CN

 Fe CN  Fe CN  Cd CN

3 6 4 6 0 2

  logK

43.6

  logK

35.4

  logK

11.1

7. Using the formation constants in Section 15.5.2, find the concentrations of the hydroxo complexes of copper (including the dimer) in a water at pH 6.0 with total soluble copper equal to 1.0×10–6 M. 8. A plot the fraction of total copper present as the dimer versus Cu(II)T at pH 6 is pretty linear. Explain why this is true. You will need the formation constants in Section 15.5.2. 9. Why are dimers formed at very low concentrations when the total metal concentration is small? 10. Desferrioxamine (structure below) is used to treat iron overload in patients receiving frequent blood transfusions. a. Looking at the structure of desferrioxamine and with your knowledge of electron sharing, how many moles of desferrioxamine are required to complex one mole of iron? (Recall that rotation around most single bonds is allowed.) b. Desferrioxamine binds ferrous iron preferentially over calcium (stability constant K = 1031 M for ferrous iron and 109 M for calcium). Determine

O N

OH

NH2

the total desferrioxamine concentration in the blood needed to reduce the iron levels to normal values if the total soluble calcium concentration is 1880 mg/L as Ca and the total soluble iron concentration is 3 mg/L as Fe (three times the normal value of 1 mg/L). The pH of blood is 7.3. Use the equilibrium constants from Section 15.5.2 and assume all the iron is ferrous iron. Assume the complexes are 1:1. 11. Redo Worked Example 15.2 and calculate the equilibrium concentrations using a spreadsheet and the method of approximation. 12. In the acidic gold cyanide plating process described in Worked Example 15.4, 60 g/L of citric acid also is added. Do you expect the citric acid to change the metal and cyanide speciation? Explain qualitatively how the speciation will change. 13. Find the speciation of cadmium in a water containing 1×10–4 M total cadmium and 1×10–4 M total citric acid at pH 7. Use the stability constants in Appendix D. Ignore any polymers of cadmium.

ISTUDY

Chapter References   369 14. A cyanide bath at pH 7 can contain dangerously high levels of HCN. Calculate the change in the HCN ­concentration in a cyanide bath at pH 7 and CNT = 1×10–4 M and if 1.0×10–5 M total Ni(+II) is added. Use the equilibrium constants from Example 15.4. To solve this problem, make an assumption about the predominant form of nickel in the system (by looking at the nickel-­cyanide stability constant and the total Ni(+II)T:CNT). Be sure to check your assumptions. 15. Repeat the calculation in Problem 12 using no approximations. 16. Calculate the fraction of total aluminum that is present as polymer under four conditions: pH 3 and 4 and Al(+III)T = 10–2 M and 10–3 M. Explain why the fraction of Al(+III)T as polymer changes as the total soluble aluminum increases and also as the pH increases. 17. Sometimes the addition of chelating agents has unintended consequences. Jensen and Johnson (1990) discuss the reagents added to measure chlorine by the DPD method. Mercuric chloride is added to the pH buffer to complex trace iodide (I–), since iodide causes problems in the chlorine analysis. EDTA is added to the pH buffer to complex trace metals that interfere in the chlorine analysis. As a result of EDTA addition, some mercury(+II) is complexed inadvertently and more free iodide is available to cause problems. a. Given the equilibria in Appendix D and the equilibrium for the formation of HgI+ below, calculate the equilibrium [I–] if the test solution is at pH 6.2 and contains 1.1×10–8 M total iodide, 1 mg/L total mercury(+II) as HgC12, and 40 mg/L EDTA as its disodium salt. (This is the solution composition if chlorine is measured by the standard method. Iodide is a contaminant of the phosphate buffer.) b. What is the equilibrium [I–] of the test solution in Part A if EDTA is omitted? Hg2+ I

HgI

log K 12.9

18. Why is the leaching of trace metals from soils considered one of the impacts of acid rain? 19. A community uses a groundwater supply for drinking water. The water is fairly hard, with a hardness of 300 mg/L as CaCO3. To minimize soap scum, the residents wish to add a detergent containing nitrilotriacetate (L3–) to their washing machines. The detergent is 3% L3– by weight. If the average washing machine has a capacity of 40 L, how much detergent should be added to reduce the hardness to 25 mg/L as CaCO3? Assume the hardness is from calcium. The molecular weight of L3– is 188 g/mole. 20. The Part IV Case Study in Section 15.8 introduced the idea of an alpha value for complexation. For a metal Mn+ that forms hydroxo complexes up to M(OH)in–i (i = 0 to 3), write the formulas for the four alpha values (α0 through α3) as functions of Ki (see Section 15.4.2 for 2 i

M OH i . Calculate the MT values of the four values for Ni(OH)i2–i (i = 0 to 3) at pH 9, assuming no precipitation. Stability constants may be found in Appendix D.

nomenclature), where αi =

21. Repeat Problem 20 by developing formulas for alpha values using βi values instead of Ki values (see Section 15.4.2 for nomenclature). Confirm that you get the same numerical values of the alpha values for the nickel system at pH 9. 22. Repeat Problem 20 by developing formulas for alpha values using *Ki values instead of Ki values. (See Section 15.4.2 for the nomenclature.) Confirm that you get the same numerical values of the alpha values for the nickel system at pH 9. 23. Derive the approximation for [M] in a metal buffer given in Section 15.6.3. Use the definitions of KM and KM* . Assume that most of the metal M is in the form ML and that [L] is very small.

CHAPTER REFERENCES Butler, J.N. (1998). Ionic Equilibrium: Solubility and pH Calculations. New York: John Wiley & Sons, Inc. Cotton, F.A., G. Wilkinson, C.A. Murillo, and M. Bochmann (1999). Advanced Inorganic Chemistry, 6th edition. New  York: John Wiley & Sons. Perrin, D.D. and B. Dempsey, B. (1974). Buffers for pH and Metal Ion Control. London: Chapman & Hall Ltd.

Hartshorn, R.M., K.-­H. Hellwich, A. Yerin, T. Damhus, and A.T. ­Hutton (2015). Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 87(9–10): 1039–1049. Lewis, G.N. (1923). Valence and the Structure of Atoms and Molecules. New York: Chemical Catalog Co. March, J. (1985). Advanced Organic Chemistry: Reactions, Mechanisms, and Structure. 3rd ed. New York: John Wiley & Sons, Inc.

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370   Chapter 15  Complexation Nieuwland, J. (1904). Some Reactions of Acetylene. PhD dissertation, Catholic University of America.

Terrestrial Humic Materials. R.F. Christman and E.T. Gjessing (Eds.), Ann Arbor, MI: Ann Arbor Science, pp. 1–23.

Stumm, W. and J.J. Morgan (1996). Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters. 3rd ed. New  York: John Wiley & Sons, Inc.

Vilensky, J.A. (2005). Dew of Death: The Story of Lewisite, America’s World War I Weapon of Mass Destruction. Indiana University Press.

Thurman, E.M. and R.L Malcolm (1983). Structural Study of Humic Substances: New Approaches and Methods. In: Aquatic and

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C H A P T E R   16

Oxidation and Reduction 16.1 INTRODUCTION Acid–base and complexation chemistry describe only some of the interactions that occur between chemical species. In addition to exchanging protons and sharing electrons, aquatic species can exchange electrons. The equilibrium calculation tools developed so far in this text can be applied to electron transfer reactions in a straightforward manner. However, some of the thermodynamic principles and graphical calculation tools presented earlier in the text must be extended to account for the new interactions. Electron exchange brings with it many new terms. Important concepts in electron exchange chemistry are reviewed in Section 16.2. The techniques you learned to balance chemical reactions will be broadened to electron transfer reactions in Section  16.3. Correctly balanced electron exchange reactions are essential to determining the dose of some chemicals used in treatment processes. In Sections 16.4 and 16.5, a parameter is developed to quantify the thermodynamic tendency to transfer electrons, just as pKa and pKb were developed to quantify the thermodynamic tendency to exchange protons. Equilibrium calculation techniques are presented in Sections 16.6 and 16.7. New graphical presentation methods are developed in Section 16.8. In Section 16.9, limitations and strengths of redox equilibrium calculations are discussed.

16.2  A FEW DEFINITIONS 16.2.1  Introduction Chemical reactions in which electrons are consumed are called reduction reactions. In reduction reactions, the free electron is a reactant and the oxidation state of at least one reactant is decreased. (The procedure for calculating the oxidation states of elements in a chemical species is reviewed in Appendix C, Section C.1.1.) As a mnemonic, remember that the oxidation states of at least one reactant is reduced in reduction reactions. For example, the following reaction is a reduction:

Cl2 (g) 2e

2Cl

reduction reaction: a reaction in which electrons are consumed (and the oxidation state of one or more reactants is decreased)

eq. 16.1

Note that one of the reactants is the free electron. The overall oxidation state of chlorine gas, Cl(0), is reduced to the oxidation state of the chloride ion, Cl(–I). Note that both material (i.e., chlorine) and electrons (i.e., charge) are balanced in eq. 16.1. 371

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372   Chapter 16  Oxidation and Reduction oxidation reaction: a reaction in which electrons are produced (and the oxidation state of one or more reactants is increased)

redox reaction: a reaction in which reduction or oxidation or both take place (i.e., a reaction in which electrons are transferred) half reaction: a reaction having one or more electrons as either reactants or products

overall redox reaction: a reaction in which electrons are transferred but free electrons do not appear as reactants or products Key idea: Overall reactions are formed by adding together half reactions in such a way that the free electrons cancel out

Chemical reactions in which electrons are produced are called oxidation reactions. In oxidation reactions, the free electron is a product and the oxidation state of at least one reactant is increased. For example, the following reaction is an oxidation: Fe 2



Fe3

e

eq. 16.2

Note that one of the products is the free electron. In addition, the oxidation state of ferrous iron, Fe(+II), is increase to the oxidation state of ferric iron, Fe(+III). Any reaction in which reduction or oxidation or both take place is called a redox reaction (for reduction-­oxidation reaction). Redox reactions are electron transfer reactions.

16.2.2  Half Reactions and Overall Redox Reactions Equations 16.1 and 16.2 have electrons as either reactants or products. Any reaction having one or more electrons as either reactants or products is called a half reaction. In aqueous systems under environmental conditions, free electrons do not exist. In other words, all electrons are associated with other chemical species in chemical systems of environmental interest. Thus, redox reactions such as eqs. 16.1 and 16.2 cannot occur in isolation. Since half reactions cannot exist in isolation, it is sometimes desirable to write the overall redox reaction. In an overall redox reaction, electrons are transferred, but free electrons do not appear as reactants or products. The half reactions are added together in such a way that the electrons cancel out. For example, suppose that your groundwater has a high ferrous iron concentration. Ferrous iron can be oxidized to ferric iron by chlorine and subsequently precipitated. The overall reaction, obtained by adding eqs. 16.1 and 16.2 so that the electrons cancel, is shown in eq. 16.3: Cl2 (g) 2e (Fe



Fe

2

2Fe

2

2Cl 3

Cl2 (g)

e ) 2 2 Fe

3

eq. 16.3

2Cl

Note that all elements and charge balance in eq. 16.3. The methods for balancing half reactions and overall redox reactions and the reasons why you might want to oxidize ferrous iron in drinking water treatment are discussed in Section 16.3. At first blush, eq. 16.3 might not seem like a redox reaction.

Thoughtful Pause Key idea: Half reactions have electrons as reactants or products, but overall redox reactions do not have electrons as reactants or products

How can you tell that eq. 16.3 is a redox reaction?

Return to the definition of a redox reaction in Section 16.2.1. Eq. 16.3 is a redox reaction because the oxidation states of iron and chlorine change. Do not let the absence of the electron as a reactant or product fool you. Half reactions have the electron as a reactant (reduction half reactions) or product (oxidation half reactions), but overall redox reactions do not include the electron as a reactant or product.

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16.2  |  A Few Definitions   373

In overall reactions, certain species are given special names. The compound that oxidizes another species is called the oxidant. The compound that reduces another species is called the reductant. These labels can be confusing at first. In overall reactions, oxidants accept electrons. They oxidize other species and are, in turn, reduced. Reductants donate electrons. They reduce other species and are, in turn, oxidized.

oxidant: a compound that oxidizes another species reductant: a compound that reduces another species

16.2.3  pe Revisited There are many similarities between acid–base chemistry and redox chemistry (see also Sections 16.2.4 and 16.5.2). One similarity concerns the water quality parameters pe and pH. In Section 1.3, pe was introduced as a primary variable in aquatic chemistry. Recall that pe = –log{e–}. As you saw in Section 16.2.1, at equilibrium {e–} is very small in environmental systems. How can pe be useful if the free electron essentially does not exist? To see the usefulness of pe, imagine freezing the chemistry of an aquatic system. Although no free electrons would exist, chemical species would “feel” whether electrons were available to be accepted or donated. For example, if electrons are available (i.e., if a significant concentration of strong electron donors – reductants – was present), an oxidant such as dissolved oxygen may be reduced. If electrons are much less available (i.e., if a significant concentration of strong electron acceptors – oxidants – was present), a reductant such as hydrogen sulfide may be oxidized. Thus, pe is a useful concept as a measure of the availability of free electrons. If electrons are readily available, then the system behaves as if the activity of the free electron is high; thus, pe is small or even negative. These are called reducing conditions. Under reducing conditions, many oxidants will be reduced. In other words, the reduced forms of chemical species will be favored. If electrons are not readily available, the system behaves as if the activity of the free electron is low; thus, pe is large and positive. These are called oxidizing conditions. Under oxidizing conditions, many reductants will be oxidized and the oxidized forms of chemical species will be favored. Oxidizing and reducing conditions are summarized in Table 16.1.

16.2.4  Other Similarities between Acid–Base and Redox Chemistry If you found the discussion in Section 16.2.2 confusing, it might be helpful to consider the analogies with acid–base chemistry. In water, the proton (H+), like the Table 16.1 Characteristics of Reducing and Oxidizing Conditions Characteristic

Reducing Conditions

Oxidizing Conditions

pe Reactions Favored species Environmental examples

low or 0) reductants are oxidized oxidized forms aerated surface waters

Key idea: In an overall reaction, oxidants accept electrons and are reduced, whereas reductants donate electrons and are oxidized pe: a measure of the availability of electrons; pe = –log{e–}

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374   Chapter 16  Oxidation and Reduction electron, is associated with other species (i.e., with Brønsted bases, including H2O), and not “free.” To mimic the language in Section 16.2.3, if a significant concentration of strong proton donors (acids) was present, a Brønsted base such as acetate might be protonated (to form acetic acid). If a significant concentration of strong proton acceptors (bases) was present, a Brønsted acid such as HOCl might become deprotonated. Table  16.1 has an analogy with acid–base chemistry. Acidic conditions correspond to lower pH, where protonated species are favored. Basic conditions correspond to higher pH, where deprotonated species are favored. The analogy between redox chemistry and acid–base chemistry will be extended in Section 16.5.2 and in Table 16.2.

16.3  BALANCING REDOX REACTIONS 16.3.1  Balancing Half Reactions The general procedure for balancing any chemical reaction is given in Appendix C, Section C.1.2. This procedure will be repeated shortly with the example of balancing the reduction of sulfate ion (SO42–) to elemental sulfur S(s) at high pH.

Thoughtful Pause How do you know that the conversion of sulfate ion to elemental sulfur is a reduction?  orked W example →

Key idea: To balance a half reaction: (1) write the known reactants on the left and known products on the right; (2) adjust stoichiometric coefficients to balance all elements except H and O; (3) add water (H2O) to balance the element O; (4) add H+ to balance the element H; and (5) add electrons to balance the charge

The oxidation state of S in the sulfate ion is +VI, and the oxidation state of elemental S is 0. Thus, the oxidation state of sulfur is reduced in converting sulfate to elemental sulfur, and the reaction is a reduction. In the half reaction, electrons should appear as reactants. The half reaction balancing procedure is as follows (see also Section 4.7.1 and Appendix C, Section C.1.2): Step 1: Write the known reactants on the left and known products on the right. In the example: SO 4 2 S(s). Step 2: Adjust stoichiometric coefficients to balance all elements except H and O. In this case, you need only to balance S, and S is already balanced in the first step. Step 3: Add water (H2O) to balance the element O. Thus: SO 4 2 S(s) 4 H 2 O . Step 4: Add H+ to balance the element H. In the example: SO 4 2 8H S(s) 4 H 2 O.

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16.3  |  Balancing Redox Reactions   375

Step 5: Add electrons to balance the charge. The left side has an overall charge of +6, and the right side has an overall charge of 0. Thus, you must add 6 electrons to the left side to balance the charge: SO42– + 8H+ + 6e– → S(s) + 4H2O. It is common practice to express all reactions in terms of species at the highest concentration. For basic conditions, add the reaction 4H2O = 4OH– + 4H+ to eliminate H+. The resulting expression is:

SO 4 2

S(s) 4 OH

4 H 2 O 6e

eq. 16.4

Note that elements and charge balance. Also, electrons are reactants, as anticipated. Another illustration of balancing a half reaction is given in Worked Example 16.1. WORKED EXAMPLE 16.1 

 

Balancing Half Reactions

Balance the half reactions for the reduction of hypochlorous acid (HOCl) to chloride ion at low pH. Balance the half reaction for the oxidation of hydrogen sulfide to sulfate at low pH. Solution Following each step in the balancing procedure: Step 1: Write known participants. HOCl

Cl SO 4 2

H2S

Step 2: Balance all elements except H and O. Cl and S are balanced in Step 1 Step 3: Add water to balance O. HOCl

Cl

H2 O

SO 4 2

H2S 4H2 O Step 4: Add H+ to balance H. HOCl H

Cl

H2S 4H2 O

H2 O

2

10 H

Cl

H2 O

SO 4

Step 5: Add e– to balance charge. HOCl H H 2 S 4H 2 O

The balanced half reactions are: HOCl H+

H 2 S 4H 2 O

2e

SO 4 2 2e

10 H Cl

SO 4 2

8e

H2O 10H

8e

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376   Chapter 16  Oxidation and Reduction

Key idea: Check the change in oxidation states in balanced half reactions

If elements and charges balance in the final half reaction, then the reaction is balanced. However, a useful check on the charge balance is to make sure that the change in oxidation state of the pertinent species makes sense. In eq. 16.4, H and O are in the +I and –II oxidation state as usual (see Section A.2.4) and do not undergo redox chemistry in this reaction. As stated, the oxidation state of S in sulfate ion is +VI, and the oxidation state of S in elemental sulfur is 0. Since 1 mole of sulfate ion is converted to 1 mole of sulfur, you expect eq. 16.4 to require six electrons. This check on the change in oxidation state should become second nature to you. It is a useful check to catch errors in balancing half reactions.

16.3.2  Balancing Overall Reactions To balance overall reactions, two steps are required. First, the reduction half reaction and oxidation half reaction should be balanced separately. Second, add the half reactions in such a way that the electrons cancel. As an example, consider the oxidation of hydrogen sulfide to elemental sulfur, S(s), by hypochlorous acid. The hypochlorous acid will be reduced to chloride ion. Following the protocol for balancing overall reactions (compiled in Appendix C, Section C.1.3): Step 1: Balance the half reactions separately. The half reactions are balanced in Worked Example 16.1. The resulting half reactions are:

Key idea: To balance overall reactions (1) balance the half reactions separately, and (2) add the half reactions to eliminate free electrons

HOCl H

2e

Cl

H 2 O and H 2 S 4 H 2 O

SO 4 2

10 H

8e .

Step 2: Add the half reactions to eliminate free electrons. As written, the reduction half reaction consumes two electrons and the oxidation half reaction produces eight electrons. Thus, to eliminate free electrons, multiply the reduction half reaction by four and add:

 orked W example →

HOCl H 2e Cl H 2 O 4 H 2 S  4H 2 O SO 4 2 10 H 8e 4HOCl H 2 S



4Cl

SO 4 2

eq. 16.5

6H

Another example of balancing overall reactions is given in Worked Example 16.2. Worked Example 16.2   

Balancing Overall Reactions

One measure of the organic content of water is chemical oxygen demand (COD). In the COD test, organic matter is oxidized with dichromate (Cr2O72–) at low pH to CO2, and the dichromate is reduced to Cr3+. Write the balanced overall reaction for the oxidation of ethanol (CH3CH2OH) in the COD test. Solution You must first balance the two half reactions. For chromium species: Starting point: Cr2 O 72

Cr 3

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16.3  |  Balancing Redox Reactions   377

Balancing Cr: Cr2 O 72

2Cr 3

Balancing O: Cr2 O 72

2Cr 3

7H 2 O

Balancing H: Cr2 O 72

2Cr 3

14 H

7H 2 O

Balancing charge: Cr2 O 72

14 H

2Cr 3

6e

7H 2 O

For ethanol: Starting point: CH3CH 2 OH

CO2

CH3CH 2 OH

2CO2

Balancing C:

Balancing O: CH3CH 2 OH 3H 2 O

2CO2

Balancing H: CH3CH 2 OH 3H 2 O

2CO2 12 H

Balancing charge: CH3CH 2 OH 3H 2 O

2CO2 12 H

12e

Adding: Cr2 O 7 2 14H 6e  CH 3 CH 2 OH  3H 2 O CH 3 CH 2 OH  2Cr2 O 7 2

2Cr 3 7H 2 O 2 2CO2 12H 12e 16H

2CO2 4Cr 3

11H 2 O

The balanced overall reaction is: CH 3 CH 2 OH    2Cr2 O 7 2

16H

2CO 2

4Cr 3

 11H 2 O

ISTUDY

378   Chapter 16  Oxidation and Reduction You may be tempted to shortcut the balancing process and balance the overall reaction in one step. For example, you may say to yourself: Why not just balance the  reaction aHOC1 bH 2 S cH 2 O dCl eSO 4 2 f H ? In fact, you can convert the balancing process into an algebra problem. The mass and charge ­balances become: Cl balance: a = d

S balance: b = e

O balance: c = 4e

H balance: a + 2b + 2c = f

Charge balance: 0 = f – d – 2e (net reactant charge = net product charge) Note that this system is five linearly independent equations in six unknowns and therefore has an infinite number of solutions. (All reactions can be balanced in an infinite number of ways, since you can always multiply each stoichiometric coefficient a a , c = 0, d = a, e , by a constant.) Solving for each unknown in terms of a: b 4 4 3a and f . The solution with no fractions and the smallest stoichiometric coeffi2 3a a a cients is: a = 4, b 1, and f 6. This reproduces 1, c = 0, d = a = 1, e 4 2 4 eq.  16.5. Although the algebraic approach works, the balancing process shown here (balance each half reaction separately and add to eliminate free electrons) is recommended because it increases your chemical intuition (and also provides the balanced half reactions). When two balanced half reactions are added, the resulting overall redox reaction should be balanced. It is always wise to check the resulting reaction to make sure: (1) electrons cancel out and do not appear in the final expression; and (2) species are written in terms of their predominant forms under a given set of conditions. For example, eq. 16.5 makes sense below pH 7, where HOCl, H2S, Cl–, and SO42– are larger in concentration than their conjugate acid or basic forms (since the pKa values of HOCl and H2S are 7.54 and 7, respectively). Also, at pH less than 7, it makes sense to write the reaction in terms of H+ rather than OH–.

Thoughtful Pause What would be the appropriate reactions to express the chemistry in eq. 16.5 at higher pH?

You could find the appropriate expression by starting over (i.e., writing the half reactions at higher pH) or by modifying eq. 16.5. Since eq. 16.5 already has been developed, start from it. At higher pH, Cl(+I) is predominantly OCl–, H2S is predominantly HS– (bisulfide ion, pKa,2 = 14), and it is preferable to write the reaction in terms of OH– rather than H+. Adding the appropriate reactions:

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16.3  |  Balancing Redox Reactions   379

4HOCl H 2 S   OCl HS

H

  H

H OH

4OCl

HS

4 Cl

SO 4 2

HOCl

6H

4



H2S

eq. 16.6

H2 O OH

4Cl

SO 4 2

H2 O

In the final expression in eq. 16.6, elements, charges, and oxidation states all balance. All species are the dominant forms at higher pH values. As in Section 16.3.1, you can double-­check the changes in the oxidation state. The Key idea: Check the changes in the oxidation states of all elements must balance in an overall reaction. In change in eqs. 16.5 and 16.6, the oxidation states of H and O do not change. Chlorine changes – oxidation states from +I in HOCl and OCl to –I in the chloride ion. This represents an eight-­electron in balanced overall – change since four moles of HOCl (or OCl ) are reduced. Sulfur is in the –II oxidation reactions – state in H2S or HS and is oxidized to the +VI oxidation state in the sulfate ion. Thus, the oxidation of the S in hydrogen sulfide (or bisulfide) to S(0) also represents an eight-­ electron change. The changes in the oxidation states balance, as required. Although this process takes many words to describe, you will soon be able to perform the oxidation state check in your head in less time than it takes to reread this paragraph.

16.3.3  Disproportionation and Comproportionation Reactions There are several important examples in water chemistry in which a single element is present in more than two (usually three) oxidation states in the same overall reaction. In most cases, the element is both oxidized and reduced in the same reaction. If the element is in one oxidation state in the reactants and two or more oxidation states in the products, then the reaction is called a disproportionation reaction. As an example, consider the dissolution of chlorine gas into water. Molecular chlorine (Cl2, oxidation state 0) forms both hypochlorous acid (HOCl, oxidation state +I) and chloride ion (Cl–, oxidation state –I) at pH 7. Using the usual approach for balancing overall reactions, you can show that:

Cl2 (g) H 2 O

HOCl Cl

H

eq. 16.7

Equation 16.7 is a disproportionation reaction: Cl(0) → Cl(+I) + Cl(–I). You may wish to verify that elements, charges, and oxidation states balance in eq. 16.7. In a comproportionation reaction, chemical species containing an element in two or more oxidation states react to form chemical species in one oxidation state of the element. The reverse of eq. 16.7 [HOCl + Cl– + H+ → Cl2(g) + H2O] is a comproportionation reaction. Another example of a comproportionation reaction is presented in Section 16.3.4.

disproportionation reaction: a reaction in which an element is in one oxidation state in the reactants and two or more oxidation states in the products Key idea: Exercise care when dealing with oxidation states of O other than–II and of H other than +I comproportionation reaction: a reaction in which an element is in two or more oxidation states in the reactants and one oxidation state in the products

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380   Chapter 16  Oxidation and Reduction

16.3.4  Practical Applications of Balancing Overall Reactions

 orked W example →

A balanced overall reaction gives you a great deal of information. In an engineered setting, the stoichiometry of the balanced overall reaction gives chemical doses. For example, in the removal of hydrogen sulfide by chemical oxidation for odor control, a common rule of thumb is that about 9 lb of chlorine (as Cl2) are required per lb of hydrogen sulfide removed (as S). This value stems directly from the stoichiometry in eqs. 16.5 and 16.7. From eq. 16.7, 1 mole of Cl2 forms 1 mole of HOCl. From eq. 16.5, 4 moles of HOCl are required to oxidize 1 mole of H2S. Thus, the dose requirement is: dose

4

moles HOCl mole H 2 S 4



moles HOCl mole H 2 S

 mole Cl 2 mole HOCl mole S 1 mole H2 S

1

g Cl 2 mole Cl 2 g S 32 mole S

70.9

8.9  g HOCl as Cl 2 per g of H 2 S as S 8.9  lb HOCl as Cl 2 per lb of H 2 S as S



This is the source of the rule of thumb is that about 9 lb of chlorine (as Cl2) are required per lb of hydrogen sulfide removed (as S). Here is a slightly more complicated example. Permanganate (sold in the form of potassium permanganate, KMnO4) is a strong chemical oxidant. It is used to oxidize ferrous iron and reduced manganese, Mn(+II), in groundwater. Iron and manganese can cause problems in treated drinking water. They may be oxidized post-­treatment and precipitate as Fe(OH)3(s) and MnO2(s), causing red or black spots on clothing in washing machines. Permanganate is used to oxidize Fe2+ to Fe3+ and Mn2+ to MnO2(s) in the drinking water treatment plant. The ferric iron precipitates rapidly so that removal of both iron and manganese occurs in the plant. (You may have noticed that we add manganese to remove manganese.) Optimization of the permanganate dose is critical because permanganate is relatively expensive and excess permanganate can give the treated water a pinkish hue (a condition called pink water, never popular among drinking water consumers). What is the stoichiometric dose of permanganate required to oxidize reduced iron and manganese? To answer this question, you must write the balanced overall reactions for iron oxidation, manganese oxidation, and manganese reduction. You may wish to do this on your own. First, the balanced half reactions are:

Fe 2

Fe3



Mn 2

2H2 O



MnO 4

4H

e MnO2 (s) 4 H 3e

2e

MnO2 (s) 2 H 2 O

ISTUDY

16.3  |  Balancing Redox Reactions   381

Adding the ferrous iron oxidation and manganese reduction reactions: Fe 2 MnO 4 3Fe



2

Fe3 4H

e 3e

MnO 4

3

4H

MnO2 (s) 2 H 2 O 3Fe3

MnO2 (s) 2 H 2 O

Adding the manganese oxidation and manganese reduction reactions: Mn 2

2H2 O

MnO 4 3Mn 2



4H

MnO2 (s) 4 H  3e

MnO2 s

 2 H 2 O 2 MnO 4

2e

3

2H2 O

2

eq. 16.8

5MnO2 (s) 4 H

You may wish to verify that the elements, charges, and oxidation states balance in the overall reactions. Note that eq. 16.8 is a comproportionation reaction: Mn(+II) reacts with Mn(+VII) to form the intermediate oxidation state Mn(+IV). The required dose for iron oxidation is:

dose

1 mole MnO 4

1 mole KMnO 4

3 moles Fe

1 mole MnO 4

158

g KMnO 4 mole KMnO 4

55.8

g Fe mole Fe

or 0.94 g KMnO4 per g of Fe or 0.94 lb KMnO4 per lb of iron. The required dose for manganese oxidation is:

dose

2 mole MnO 4

1 mole KMnO 4

3 moles Mn

1 mole MnO 4

158

g KMnO 4 mole KMnO 4

54.9

g Mn mole Mn

or 1.9 g KMnO4 per g of Mn or 1.9 1b KMnO4 per 1b of manganese. Another use of stoichiometry to determine dosing is given in Worked Example 16.3.

Worked Example 16.3 

Determining Dose from Stoichiometry

Wastewater treatment plants discharging to sensitive watersheds may be required to dechlorinate their effluent. In most cases, a chemical reducing agent is added. Chlorine in most wastewater effluents is in the form of monochloramine, NH2Cl. Monochloramine is reduced to chloride ion, and the nitrogen is released as ammonium ion. Determine the dose (in lb/d) of sodium bisulfite (NaHSO3) required to dechlorinate 8 mgd of a wastewater containing 2.5 mg/L NH2Cl as Cl2 to a final chlorine residual of 0.2 mg/L NH2Cl as Cl2. Bisulfite ion is oxidized to sulfate ion.

ISTUDY

382   Chapter 16  Oxidation and Reduction Solution To write the overall reaction, you need the two half reactions. The balanced half reactions are: NH 2 C1 2 H HSO3

2e

H2 O

C1

SO 4 2

NH 4

3H

2e

Adding to eliminate free electrons: NH 2 Cl 2H HSO3

2e

H2 O

Cl

SO 4

NH 2 Cl HSO3

2

H2 O

NH 4 3H Cl

2e NH 4

SO 4 2

H

Thus, 1 mole of HSO3– is required to reduce 1 mole of NH2Cl. On a mass basis, the dose is:

1

mole HSO3 mole NH 2 Cl

1

mole NaHSO3

1

mole HSO3 mole Cl 2 mole NH 2 Cl

103

g NaHSO3 mole NaHSO3

70.9

g Cl 2 mole Cl 2

or 1.45 mg sodium bisulfite per mg monochloramine as Cl2. To dechlorinate, it would require (2.5 – 0.2 mg/L monochloramine as Cl2)(1.45 mg sodium ­bisulfite per mg monochloramine as Cl2) = 3.3 mg/L NaHSO3. For loading calculations: kg / d

1 kg / L Q m 3 / d C m 3 / L

Or: 1b / d

8.341 b / gal Q mgd C mg / L

where: Q = flow, C = concentration, and the density of the fluid (ρ) is near 1 g/cm3. Thus, the sodium bisulfite requirement is 8.34 (8 mgd)(3.3 mg/L) = 220 lb/d NaHSO3 (or 100 kg/d).

16.3.5  How Can You Tell if Electrons Are Transferred? In some cases, it is difficult to tell if electrons are transferred in a chemical reaction. In other words, it is sometimes difficult to tell if a reaction is a redox reaction or is not a redox reaction. Usually, you can identify redox reactions by checking to see if the oxidation states of any element has changed. For example, are electrons transferred in the reaction: Cl2(aq) + 2I– → I2(aq) + 2Cl–? Yes: both chlorine and iodine change oxidation states. On other occasions, determining whether electron transfer occurs can be more difficult. Consider the chemistry of aqueous ozone, O3(aq). During reactions in water, ozone is converted to oxygen.

ISTUDY

16.4  |  Which Redox Reactions Occur?   383

Thoughtful Pause Write the balanced reaction for the conversion of ozone to oxygen. If no electrons are transferred, you could write:

2O3(aq )

3O2 (aq )

eq. 16.9

Both species appear to represent oxygen in the zero oxidation state. In fact, ozone is a very strong chemical oxidant and is reduced in water to form oxygen (and other species). Knowing that a reduction occurs, you can derive the balanced half reaction for the reduction of ozone to molecular oxygen and H2O at low pH (please try this before continuing to read):

O3(aq) 2 H

2e

O2 (aq) H 2 O

eq. 16.10

So which reaction is correct: eq. 16.9 or eq. 16.10? In fact, both equations are correct. The half reaction for the reduction of O3 to O2 and H2O is given by eq. 16.10. If you combine eq.  16.10 with the half reaction for the oxidation of water to O2, then you obtain eq. 16.9: O3(aq) 2 H 2H 2 O

2O3(aq)

2e

O2 (aq) 4 H 3O O2 (aq)

O2 aq

H2 O

2

4e

In other words, eq. 16.9 is an overall reaction. The bottom line: When dealing with oxidation states of O and H other than –II and +I, respectively, exercise great care. The ozone–oxygen example shows the misinterpretations that can occur if you treat reaction balancing as an algebra problem. If you simply tried to balance aO3(aq) → bO2(aq) by performing a mass balance on O, then you would get eq. 16.9 quickly. However, this approach misses the true chemistry: aqueous ozone oxidizes water to dissolved oxygen and in turn is reduced to dissolved oxygen and water.1

16.4  WHICH REDOX REACTIONS OCCUR? 16.4.1  Introduction As the examples presented so far in this chapter show, chemical oxidants have a number of important applications in drinking water and wastewater treatment. In selecting an oxidant to achieve a treatment objective, you want to know if the oxidant 1  The most stable resonance structure of ozone reveals that one oxygen atom is in the +I oxidation state and the other two oxygen atoms are in the –½ oxidation state. If we think of both oxygen atoms in O2 as being in the 0 oxidation state, then eq. 16.10 shows one oxygen atom in ozone undergoing a one-­electron reduction [O(+I) to O(0) in O2(aq)], one oxygen atom in ozone undergoing a ½-­electron oxidation [O(–½) to O(0) in O2(aq)], and one oxygen atom in ozone undergoing a 1½-­electron reduction [O(–½) to O(–II) in H2O]. This is a net two-­electron reduction, with a net reduction to form O2 and a reduction to form water.

ISTUDY

384   Chapter 16  Oxidation and Reduction can oxidize the pollutant to a given set of oxidation products. In this section, you will develop tools to decide whether one chemical species can oxidize (or reduce) another chemical species to a given oxidation state.

16.4.2  Redox Reactions and Spontaneity How can you decide whether one chemical species can oxidize (or reduce) another chemical species? In fact, how do you decide whether any reaction proceeds as written? The answer was discussed in Section 3.5. A reaction proceeds as written (in other words, a reaction is spontaneous) if ΔGrxn < 0. Recall from eq. 3.16 (repeated here for convenience):

Grxn

o Grxn

RT ln i i

eq. 16.11

where νi are the stoichiometric coefficients (0 for products). o Recall that Grxn is calculated from the partial molar Gibbs free energy of formation o values (G f ,i ). An example will clarify the use of eq. 16.11. Suppose you and a fellow student are trying to explain why brominated compounds are found in some treated drinking waters. Engineers usually do not add bromine to water. Perhaps the brominating agent is hypobromous acid (HOBr), potentially formed from the oxidation of bromide (Br–) by HOCl. For this hypothesis to be plausible, HOCl must be capable of oxidizing bromide to HOBr. You can test this idea by performing a thermodynamic calculation. First, balance the overall reaction. The balanced half reactions are (try this on your own):

HOCl H

2e

Br

HOBr H

H2 O

Cl

H2 O 2e



Adding: HOCl + Br– → HOBr + Cl–. The pertinent partial molar Gibbs free energy of formation values are –79.9, –104.0, –82.2, and –131.3 kJ/mol for HOCl, Br–, HOBr, and Cl–, respectively. Thus: o Grxn

(1)( 82.2 KJ/mol) (1)( 131.1 KJ/mol) (1)( 79.9 KJ/mol) (1)( 104.0 KJ/mol) 29.6 KJ/mol

To calculate the last term in eq. 16.11, you need the concentrations of each species. A common approach in pure chemistry is to assume that each species is present at a concentration of 1 M. This makes the math easier (since the last term in eq. 16.11 becomes zero), but is unrealistic under environmental conditions. If each species was at 1 M, then ΔGrxn = –29.6 kJ/mol and the reaction is spontaneous as written. Unfortunately, you do not know the concentrations. Reasonable estimates might be: HOCl Br HOBr

Cl

2 mg / L as Cl 2 0.1 mg / L

2.8 10 5 M

5.6 10 7 M

1 10 7 M, and: 20 mg / L

5.6 10 4 M



ISTUDY

16.4  |  Which Redox Reactions Occur?   385

With R = 8.314×10–3 kJ/mol-­K and at 25°C = 298 K, ΔGrxn = –26.4 kJ/mol. Since ΔGrxn < 0, hypochlorous acid is thermodynamically capable of oxidizing bromide to HOBr under the conditions tested. You do not know if this process is slow or fast. You only know that the oxidation is possible. Another determination of whether oxidations or reductions can occur spontaneously is given in Worked Example 16.4.

Spontaneity

Worked Example 16.4 

You are working for a startup environmental technology firm. Your boss is knowledgeable about the use of hydrazine (NH2NH2) to reduce trace levels of dissolved oxygen (DO). Hydrazine is oxidized to nitrogen gas, N2(g), and DO is reduced to H2O. Your boss wants you to investigate the possibility of developing a new synthesis method for hydrazine using the reverse reaction: N2(g) combining with H2O to form NH2NH2 and O2(aq). Is this synthesis reaction possible under 1 atm of nitrogen gas at room temperature? Solution The half reactions are: N 2 (g ) 4 H 2H2 O

4e

NH 2 NH 2

O2 (aq) 4 H

4e

The overall reaction is: N 2 (g) 2 H 2 O

NH 2 NH 2

O2 (aq)

The pertinent partial molar Gibbs free energy of formation values are +158.64, +16.32, 0, and –237.18 kJ/mol for NH2NH2, O2(aq), N2(g), and H2O, o respectively. Thus, Grxn = +649.32 kJ/mol. Under the conditions of the problem, {N2(g)} = 1 atm and the mole fraction of H2O is unity. From the stoichiometry, {NH2NH2} = {O2(aq)} = x. Thus: Grxn

o Grxn

RT ln

N2 H2

O2(aq)

N 2(g) H 2 O

2

Substituting at 25°C: Grxn

649.32 kJ / mol (2.48 kJ / mol) ln ( x 2 )

For ΔGrxn to be negative, x must be less than about 10–57 M. This is less than one molecule of hydrazine in the volume occupied by the Sun. Tell your boss gently that the synthesis idea is thermodynamically unjustified at room temperature.

←Worked example

ISTUDY

386   Chapter 16  Oxidation and Reduction

16.5  REDOX THERMODYNAMICS AND

OXIDANT AND REDUCTANT STRENGTH

16.5.1  Introduction Several oxidants have been referred to in this chapter as “strong oxidants.” What does it mean to be a strong oxidant or a strong reductant? Can we develop an approach or scale to rank oxidants (or reductants) by the compounds they can oxidize (or reduce)? Quantitative tools for environmental redox chemistry will be developed in this section to answer such questions.

16.5.2  Electron Affinity  How do you judge the strength of an acid? Remember that you quantify the strength of an acid by the value of its Ka or pKa. Recall that the Ka is the equilibrium constant for a specific type of equilibrium. You write:

HA H 2 O

H3O

A

eq. 16.12

You could also write: HA = A– + H+; K = Ka. Why? You can use the simplified form of the equilibrium because the equilibrium constant of H+ + H2O = H3O+ is defined to be 1. In essence, water is selected as a standard proton acceptor (for Brønsted acids). In fact, you can think of eq. 16.12 as a proton transfer equilibria consisting of two proton half reactions: a proton-­donating half reaction, HA = A– + H+, and a proton-­accepting half reaction, H+ + H2O = H3O+. In the overall reaction (HA + H2O = A– + H3O+), protons do not appear. By analogy with redox chemistry, the scheme is:



HA

A

H

H 2 O H3O

HA H 2 O

H A

H3O

K

K a (proton donating reaction )

K

1 (proton accepting reaction )

K

( K a )(1)

K a (overall reaction)

Applying this approach to redox equilibria, you could judge oxidant or reductant strength by using a standard form of an equilibrium in which a chemical species acts as an oxidant or reductant. It is common practice to write redox reactions as reductions (i.e., as electron-­accepting reactions). Therefore, you need a standard electron donor, analogous to the use of water as a standard proton acceptor. The usual standard electron donor is hydrogen gas, which is oxidized to H+. Thus, the standard form of the equilibrium for determining reductant strength is:

Xm

1

2

H 2 (g)

Xm

H

1

K

K1



This comprises two half reactions: Xm 1

2

e

H 2 (g)

Xm H

K

1

e

K2 K K3



By thermodynamic convention, K3 is defined to be 1 (just as K for H+ + H2O = H3O+ K is defined to be 1). Thus: K2 = 1 = K1. K3

ISTUDY

16.5  |  Redox Thermodynamics and Oxidant and Reductant Strength   387

What is the best measure of reductant strength? You could use the equilibrium constant for an equilibrium such as Xm + e– = X(m–1) as a measure of reductant strength (e.g., Fe3+ + e– = Fe2+). However, some reductants accept more than one electron. For example, suppose you write: 2X m



2e

2X m

K

1

K4

(As an example: 2Fe3+ + 2e– = 2Fe2+.) Now, K4 = K22 but the oxidant is the same for both equilibria. If you used K2 as a measure of oxidant strength and your friend used K4 as a measure of oxidant strength, you would get different values for the oxidant strength of Xm. 1 To avoid this problem, we usually use logK as our measure of reductant strength, n where n = number of electrons transferred and K equilibrium constant for an equilibrium of the form Xm + ne– = X(m–n). This is usually written: ox ne



red

where: ox = oxidized form, red = reduced form, and the charges are assumed to 1 balance. The expression logK sometimes is called pe° (pronounced: pee ee naught). n In many ways, pe° is analogous to pKa. Remember that pKa quantifies the thermodynamic tendency of an acid to donate a proton. Thus, pKa also quantifies the thermodynamic tendency of the conjugate base to accept a proton. Similarly, pe° quantifies the thermodynamic tendency of an oxidant to accept an electron. This thermodynamic tendency is called electron affinity. Strong oxidants have strong electron affinities and therefore large K and large pe° values.2 The symbol pe° may seem confusing because pe° seems at first to have little to do with pe = –log{e–}. There is a connection between pe° and pe. For a general redox equilibrium, Xm + ne– = X(m–n) with equilibrium constant K, you can show:

pe°: a measure of reductant strength and equal to 1logK, where n = number n of electrons transferred and K = equilibrium constant for an equilibrium of the form Xm + ne– = X(m–n) (or ox + ne– = red)

Xm n Xm

log K

npe log

pe

Xm n 1 log n Xm

So: pe



1 log K n

pe

1 red log ox n

This leads to another way to look at pe°: pe is equal to pe° when the oxidized and reduced forms have the same activity. This is analogous to the idea that pH = pKa when the basic and acidic forms have equal activity. This analysis extends the relationship between pe° and pKa:

pe pe and



pH pK a

1 oxidized form log n reduced form log

unprotonated form protonated form

The analogy between acid–base and redox chemistry is laid out in Table 16.2. 2  The relationship between oxidant or reductant strength and Ka emphasizes that oxidant and reductant are thermodynamic concepts. A strong oxidant accepts electrons to a large extent, but not necessarily quickly. The kinetic analogies to oxidant and reductant are electrophile and nucleophile, respectively. Electrophiles accept electrons rapidly, but not necessarily to a large extent.

Key idea: pe is equal to pe° when the oxidized and reduced forms have the same activity

ISTUDY

388   Chapter 16  Oxidation and Reduction Table 16.2 Detailed Comparison of Acid–Base and Redox Chemistry Acid–Base

Governing equilibrium Mathematical form Scale Condition name Favored species Order

Example

Oxidation-­Reduction

HA = A– + H+; Ka pH = pKa + log low pH high {H+} acidic

ox + ne– = red; K

A A  pH – pKa = log HA HA high pH low {H+} basic

protonated unprotonated (acids) (bases) every base every acid protonated deprotonated stronger bases (larger conjugate acid pKa) are protonated at higher pH than weaker bases (i.e., protonated first during titration with an acid) Starting at high pH (basic conditions) and titrating with an acid, NH3 (conjugate acid pKa 9.3) is protonated before acetate (conjugate acid pKa 4.7) (“before” means “at higher pH”) NH4+ = NH3 + H+; Ka = 10–9.3 HAc = Ac– + H+; Ka = 10–4.7

1 1 1 pe = logK + log ox  pe – pe° = log ox ; pe° = logK n n n red red low pe high {e–} reducing

high pe low {e–} oxidizing

reduced oxidized (reductants) (oxidants) every oxidant every reductant reduced oxidized stronger oxidants (larger pe°) are reduced at higher pe than weaker oxidants (i.e., reduced first during titration with a reductant) Starting at high pe (oxidizing conditions) and titrating with 83.1 a reductant, O2 (pe° = = 20.78) is reduced to H2O before 4 36.2 2– = 6.03) SO4 is reduced to S(s) (pe° = 6 (“before” means “at higher pe”) O2(g) + 4H+ + 4e– = 2H2O; K = 10+83.1 SO42– + 8H+ + 6e– = S(s) + 4H2O; K = 10+36.2

16.5.3  Oxidant Strength and Cell Potentials There is one problem with the use of pe° as a measure of oxidant strength: The K values for redox equilibrium frequently are not known. They are not known, in part, because the oxidation half reaction cannot occur without a corresponding half reaction in which the free electrons are accepted. Not knowing the K values is a big problem. 1 How can you calculate pe° = logK if you do not know K? One approach is to find n G Grxn. If you know Grxn, then (from Section 3.9.2) you can calculate K by: K

e

rxn

RT

.

Unfortunately, the Grxn values also are not known in general. Fortunately, you can measure the energy of the overall reaction by measuring the voltage in an electrochemical cell. How can voltage help? Both voltage and Gibbs free energy are related to the work done by the system. Recall from Section 3.6.5 that electrical work is: electrical work

(potential difference)dq (potential difference)(total charge transferred) (potential difference)(moles of electrons transferred ) (charge per mole of electrons)

The potential difference here is the cell potential, E. We have been using the symbol n to represent the number of moles of electrons transferred. The charge per mole of

ISTUDY

16.5  |  Redox Thermodynamics and Oxidant and Reductant Strength   389

electrons is a constant: 96,485 coulombs of charge per mole of electrons (96,485 C/ mol) = 1 faraday = 1 F (after Michael Faraday, 1791–1867).3 Based on this analysis, the electric work in an electrochemical cell is related to nFE. The cell potential is defined so that the electric work in an electrochemical cell is equal to –nFE. Since the chemical work done is ΔGrxn, then: Grxn



nFE

Note that, by convention, ΔGrxn and E are opposite in sign. Recall that ΔGrxn < 0 for spontaneous reactions. Thus, E > 0 for spontaneous reactions. If all species are at their standard states, then ΔGrxn = Grxn = –nFE°. Here E° is the cell potential with all species at their standard states. Now you have a way of ­calculating pe°: 1. Measure the cell potential in an electrochemical cell with all species at their standard states. This is E°. 2. Calculate Grxn = –nFE°. 3. Calculate K

e

Grxn RT

Grxn

nFE 1 96, 485 J / mol-V 74.29 kJ / mol

Pt wire



Pt wire

salt bridge {H+} = 1 M

{Fe3+} = {Fe2+} =1M

FIGURE 16.1  Electrochemical Cell Showing the Standard Hydrogen Electrode (The salt bridge completes the circuit and allows electrons to flow.)

 A volt is defined so that the energy required to move 1 coulomb of charge through 1 volt is 1 joule. Thus, F also has units of joules per mol-­volt (J/mol-­V). 4  More precisely, the normal hydrogen electrode (NHE) measures the potential of a platinum electrode in a 1 N acid solution where {H2(g)} = 1 atm. The standard hydrogen electrode (SHE) measures the potential of a platinum electrode in a theoretical solution where {H2(g)} = 1 atm and {H+} = 1 M. The equilibrium constant for the half-­reaction in the SHE is defined to be 1 (i.e., its E° is defined to be zero). 3

standard (or normal ) hydrogen electrode: a part of an electrochemical cell where {H2(g)} = 1 atm and {H+} = 1 M ←Worked example

0.77 V

voltmeter H2 gas

Key idea: The Gibbs free energy of reaction and the cell potential are related by ΔGrxn = –nFE (note negative sign)

.

1 4. Finally, calculate pe° = logK. n This procedure relies on measuring the cell potential. To measure the cell potential, an electrochemical cell must be devised in which a standard reductant is oxidized and the oxidant of interest is reduced. Take the example of Fe3+ + e– → Fe2+. It makes sense to link ferric iron reduction to an oxidation with a known K value. We have already defined the equilibrium constant for ½H2(g) = H+ + e– to be 1. A possible electrochemical cell is shown in Figure 16.1. The hydrogen chamber has {H2(g)} = 1 atm and {H+} = 1 M (i.e., very acidic) and is called the standard (or normal) hydrogen electrode.4 If {Fe3+} = {Fe2+} = 1 M, then the voltmeter will measure +0.77 V. Thus, E° = +0.77 V, and:



faraday (F ): one faraday means 96,485 coulombs per mole of electrons transferred (F also can be expressed as 96,485 joules per mol-­volt = 96,485 J/mol-­V)

ISTUDY

390   Chapter 16  Oxidation and Reduction So: K

e

Grxn RT

kJ mol exp kJ 8.314 10 3 298 K mol-K 74.29



1.0 1013

1 Thus: pe° = logK = 13.0. n Typically, E° values for redox equilibria (written as reductions) are tabulated rather than pe° values. Using the equations presented above, it is easy to convert redox thermodynamic data into any form you wish. Interconversion formulas between redox thermodynamic data are listed in Table  16.3. Another example of interconversions between data types is given in Worked Example 16.5. Table 16.3 Interconversions Between Redox Thermodynamic Data (Tabulated formulas show the row parameter as a function of the column parameter.) E°

logK

Grxn

pe°

Grxn

2.303RT logK nF

2.303RT pe F

–nFE°

Grxn

–2.303RTlogK

–2.303RTnpe°

logK =

nFE 2.303RT

Grxn

logK

npe°

pe° =

FE 2.303RT

1 logK n

pe°

E° =

Grxn =

E°

nF

2.303RT Grxn

2.303nRT

Note: Formulas are for the reaction: ox + ne– = red. Also note that ΔGrxn = –nFE and

2.303RT 0.059 59 = V = mV at 25°C. nF n n

Worked Example 16.5 

Redox Thermodynamic Data

Will chlorine gas cause iron to rust spontaneously if all pertinent species are at their standard state at 25°C? Solution For iron to rust, Fe(s) must be oxidized to Fe2+. The relevant equilibria are: Fe

2e E

Fe(s) 0.409 V

Cl2 (g) 2e 2Cl logK 46.0

ISTUDY

16.5  |  Redox Thermodynamics and Oxidant and Reductant Strength   391

Chlorine will rust iron if the overall reaction has E > 0 or E° > 0 with all species at their standard states. Thus, we must convert logK to E° and then find E° for the overall reaction. To convert logK to E°: E From the note to Table 16.3,

2.303RT logK nF

2.303RT = 0.059 V at 25°C. With n = 2 and F

logK = 46.0, E° for chlorine reduction is +1.36 V. Rearranging the equilibria and adding (E° in V in parentheses; see Section 16.6 for the rules of adding E° values): Fe 2

2e ( 0.409)

Cl 2 (g) 2e

2Cl ( 1.36)

Fe(s)

Cl 2 (g) Fe(s)

Fe 2

2Cl

For the overall reaction, E° = 0.409 V + 1.36 V = +1.77 V > 0, so chlorine gas will cause iron to rust spontaneously if Cl2(g), Fe(s), Fe2+, and Cl– are at their standard state. A few environmentally important redox reactions, listed as usual as reductions, are given in Table 16.4 along with pertinent thermodynamic data. You may wish to verify a few of the interconversions between thermodynamic parameters. Table 16.4 is ordered by pe° values. Strong oxidants are the reactants in reactions near the bottom of the table (e.g., oxygen and ozone) and have large pe° values. Strong reductants, for example Zn(s), are products in reactions at the top of the table and have small or negative pe° values.

Table 16.4 Common Environmental Redox Reactions1 Reaction

E° (V)

logK

pe°

Zn + 2e = Zn(s) Fe2+ + 2e– = Fe(s) 2H+ + 2e– = H2(g) S(s) + 2H+ + 2e– = H2S Cu2+ + e– = Cu+ Cu2+ + 2e– = Cu(s) SO42– + 8H+ + 6e– = S(s) + 4H2O Fe3+ + e– = Fe2+ NO3– + 2H+ + 2e– = NO2– + H2O O2(g) + 4H+ + 4e– = 2H2O MnO2(s) + 4H+ + 2e– = Mn2+ + 2H2O Cl2(g) + 2e– = 2Cl– O3(g) + 2H+ + 2e– = O2(g) + H2O

–0.763 –0.409 ≡0 +0.142 +0.158 +0.340 +0.362 +0.770 +0.842 +1.229 +1.21 +1.36 +2.09

–25.8 –13.8 0.0 4.8 2.7 11.5 36.2 13.0 28.3 83.1 40.9 46.0 70.7

–12.9 –6.92 0.0 2.4 2.7 5.7 6.03 13.0 14.15 20.78 20.5 23.0 35.3

2+

–

Notes: 1

 All at 25°C, E° data are from Bard and Faulkner (1980), with logK and pe° values calculated.  The pe° value is from Stumm and Morgan (1996), with E° and logK calculated.

2

Key idea: With reactions listed as reductions, strong oxidants appear as reactants in reactions with large positive pe° values and strong reductants appear as products in reactions with large negative pe° values

ISTUDY

392   Chapter 16  Oxidation and Reduction

16.5.4  How Is the Cell Potential Related to Species Activities? From Section 16.5.3, you know that ΔGrxn is related to the cell potential, E, by: ΔGrxn = –nFE or E =

Grxn

nF

. Since ΔGrxn is a function of species activities, it makes sense that

the E also should depend on the species concentrations. As an example, consider a general reduction in which an oxidized species (ox) accepts n electrons to be reduced to a reduced form: ox + ne– = red. Using hydrogen gas as the electron donor, the overall reaction is: ox ne red 1 H g H 2 2 n H2 g 2

ox



e

n

red nH



(Note: It appears that charges do not balance here because the charge on red is n less than the charge on ox.) The Gibbs free energy of reaction for the overall equilibrium is (see Section 3.8.3):

Grxn

Grxn

RT ln

red H ox H 2 g



n

eq. 16.13

n 2

For the standard hydrogen electrode: {H+} = 1 M and {H2(g)} = 1 atm. Substituting these values and dividing both sides of eq. 16.13 by –nF:

E

E

RT ox ln nF red

eq. 16.14

Or: E

where Nernst equation: an equation relating the cell potential to the activities of the electron-­transferring species

2.303RT nF

0.059 V n

E

2.303RT ox log nF red

59 mV at 25°C. n

Equation 16.14 is called the Nernst equation (after Walther Hermann Nernst, 1864– 1941; see the Historical Note at the end of the chapter). The Nernst equation shows the dependence of the cell potential on the activities of the species undergoing redox chemistry. For example, the Nernst equation for the reduction of ferric iron to ferrous iron (i.e., Fe3+ + e– = Fe2+) is:

E

E

Fe3 RT ln F Fe 2

n 1



Although the Nernst equation usually is written as shown in eq. 16.14, do not forget that it comes from the expression for the Gibbs free energy of reaction. Thus, the numerator in the logarithm argument is really the product of the activities of the reactants in a reduction reaction (excluding e–), each raised to their stoichiometric

ISTUDY

16.6  |  Manipulating Half Reactions   393

coefficients, and the denominator is the product of the activities of the products in a reduction reaction, each raised to their stoichiometric coefficients (where the Nernst equation is written with a plus sign and the standard hydrogen electrode is used). By convention, the reaction must be written as a reduction. For example, the Nernst equation for the reduction of dissolved oxygen to water [O2(aq) + 4H+ + 4e– = 2H2O] is:



E

E

4

O aq H RT ln 2 2 4F H2 O



16.6  MANIPULATING HALF REACTIONS 16.6.1  Introduction Your experiences so far in aquatic chemistry have convinced you of the need to reverse equilibria, multiply equilibria by positive constants, and add equilibria to get them in the form you need. Tracking the thermodynamic parameters as you manipulate redox equilibria can be challenging. The most common form of thermodynamic data listed for redox reactions is E°. Thus, it is important to practice manipulating E° and E values as you manipulate equilibria. Three skills are particularly important: tracking E as you reverse a redox equilibrium, tracking E as you multiply redox equilibria by a positive constant, and tracking E as you add redox equilibria. You should become comfortable analyzing transformations (such as reversal, multiplying by a constant, and adding) from both chemical and mathematical perspectives. In all cases, it is valuable to think about the effects of these transformations on the Gibbs free energy of reaction first. Recall from Section 3.7.1 that (repeated here for convenience):

Grxn

i products

Gi

i

Gi

eq. 16.15

reactants

16.6.2  Reversing a Half Reaction We usually tabulate E° values for half reactions written as reductions (as in Table 16.4), just as we usually tabulate Ka values for acids. Sometimes it is handy to know E or E° for the corresponding oxidation reaction.

Thoughtful Pause How should you calculate E or E° for a reaction where products and reactants are reversed?

From eq.  16.15, swapping reactants and products results in multiplying ΔGrxn by –1. Since ΔGrxn = –nFE and Grxn = –nFE°, reversing a reaction will change the sign of E and E°.

Key idea: Reversing a reaction will change the sign of E and E°

ISTUDY

394   Chapter 16  Oxidation and Reduction

16.6.3  Multiplying a Half Reaction by a Positive Constant As an example of tracking E as you multiply redox equilibria by a positive constant, compare the following two half reactions: Equilibrium 1: Fe3 Equilibrium 2: 2 Fe



Fe 2

e 3

2e

E

2 Fe

2

E1

E

E2

Thoughtful Pause How does E2 compare to E1? From a chemistry perspective, you would expect that E2 and E1 are equal: The cell potential should not change because the chemistry has not changed. Mathematically, you can verify this intuition with the Nernst equation: E1

E1o

Fe3 RT ln F Fe 2

E2

E2

Fe3 RT ln 2F Fe 2

o

2 2

E2

RT ln 2F

o

Fe3 Fe 2

2

We know that E2 o = E1o , since the E° values are the cell potentials when all species are at their standard states (1 M for dissolved species). So: Key idea: The cell potential does not change if you multiply the half reaction by a positive constant



E2

E1o

RT ln 2F

Fe3 Fe 2

2

Fe3 RT ln F Fe 2

E1o

E1



Thus, the cell potential does not change if you multiply the half reaction by a positive constant. You can reach this conclusion by converting the cell potentials to Gibbs free energy values (see Problem 12).

16.6.4  Adding Half Reactions Key idea: Since Gibbs free energy values are additive, when adding redox equilibria convert E° values to Grxn add the ° values, and then Grxn reconvert to E°

When adding redox equilibria, always come back to the truism that Gibbs free energy values are additive. E° values generally are not additive. They are additive only if the number of electrons transferred does not change upon addition. Thus, the most error-­ o free way to calculate E° when adding redox equilibria is to convert E° values to Grxn , add o the Grxnvalues, and then reconvert to E°. Two examples will make this process clearer. Suppose you wish to add two half reactions where the number of electrons transferred changes upon addition. An example is: Zn 2

 orked W example →

Zn

e

2

2 Zn

Zn(s)

e 2

Zn(s) 2e

E1o E2

o

2 Zn(s) E3

0.763 V

o

0.763 V ?



ISTUDY

16.6  |  Manipulating Half Reactions   395

You know from Section  16.6.3 that E3o = E1o = E2 o = –0.763 V. It is clear here that E3o ≠ E1o + E2 o . o How can we derive this value of E3o ? The answer is to convert E° to Grxn . You o o o o o o o o Grxn,1 Grxn,2 = 2 FE1o . know: Grxn,1 FE1 , Grxn,2 FE2 FE1 , and Grxn,3

←Worked example

o Grxn ,3 E1o, as required. 2F o Applying this E-­to-­ Grxn -­then-­add-­then-­back-­to-­E approach to the general case:

So: E3o

ox1 ne ox 2

red1

me

o E1o , Grxn ,1 o E2 o , Grxn ,2

red 2

n m e

ox1 ox 2

nFE1o

E3 o

red1 red 2

o o Energy is additive, so: Grxn Grxn ,3 ,1 o G n m rxn , 3 o o E3o E FE ( n m )F n m 1 n m 2 .

mFE2 o

o ?, Grxn ,3

o Grxn ,2

nFE1o

?



mFE2 o . Therefore:

Now suppose in the general case that we want to reverse the second reaction and add so that electrons cancel (e.g., to create an overall reaction): ox1 ne red 2

red1

m

me

n

ox 2

mox1 nred 2

o Now: Grxn ,3

o Grxn ,1

o E1o , Grxn ,1 o E2 o , Grxn ,2

mred1 nox 2

o Grxn ,2

nmF E1o

E3 o

o ?, Grxn ,3

nmFE1o nmFE2 o ?



E2o . If we consider the overall reaction o Grxn ,3

to be a change of nm electrons, then E1o E2o . This shows that you can nmF add E° values if the electrons cancel. As an example, using the data given below, decide whether molecular chlorine is o capable of oxidizing ferrous iron to ferric iron (if all species are at 1 M) and find Grxn , E°, and K for the overall reaction. The thermodynamic data are: E3o

Fe3

e

Cl2(g) 2e



Fe 2 2Cl

E1o E2

o

0.77 V 1.36 V

o Calculating the standard Gibbs free energies of reaction: Grxn,1 FE1o and o o Grxn,2 2 FE2. You know the iron reaction should be reversed (written as an oxidation) and multiplied by 2. Reversing the reaction will multiply E1o by –1. Multiplying o the stoichiometric coefficients by 2  will have no effect on E°, but will double Grxn o (since Grxn = –nFE°). Thus:

2 Fe 2

2 Fe3

Cl2 (g) 2e

2e 2Cl

E3o

E2 o

o 0.77 V, Grxn ,3

o 1.36 V, Grxn ,2

2 FE3o

2 FE2 o

←Worked example

ISTUDY

396   Chapter 16  Oxidation and Reduction Adding the half reactions: 2Fe2+ + Cl2(g) = 2Fe3+ + 2Cl–. Adding the standard Gibbs free energies of reaction: o Grxn

o Grxn ,3

o Grxn ,2

2 F E3 o E 2 o 2 96.485 kJ / mol-V

0..77 1.36 V

113.9 kJ / mol

Thus: E

o Grxn = +0.59 V and K = e 2F

o

o Grxn RT

= 9.2×1019.

Thoughtful Pause Is the reaction as written favorable if all species are at 1 M? Since ΔGrxn is less than zero (or, equivalently, E > 0), the oxidation of ferrous iron by chlorine gas is thermodynamically favorable if all species are present at 1 M. Note in this case that E° E3o + E2 o since the half reactions were added so that the electrons cancel.

16.7  ALGEBRAIC EQUILIBRIUM

CALCULATIONS IN SYSTEMS UNDERGOING ELECTRON TRANSFER

16.7.1  Calculating Activities as a Function of pe and pH The primary variables in most aquatic systems are pH and pe: pH controls the acid– base chemistry and pe controls the redox chemistry. It is useful to manipulate equilibria to show the chemical speciation at any combination of pe and pH value. As an example, consider the equilibrium representing the oxidation of ammonium ion to nitrate ion. This is an important environmental process called nitrification. Writing the equilibrium as usual as a reduction: Key idea: To calculate activities as functions of pe and pH, rearrange the mathematical form of the equilibrium to obtain an equation of the form: ratio of species activities = a + bpH + cpe

NO3



10 H

8e

For this equilibrium: log K =

logK

3 log 3H 2 O

NH 4

3H 2 O; E

0.88 V

8FE = 119.3 at 25°C, so K = 10119.3. Also: 2.303RT

log{NH 4 } log{NO3 } 10 log{H } 8 log{e }

Using the definitions of pH and pe and assuming that the mole fraction of water is unity:



log K 119.3 log

{NH 4 } 10 pH 8pe, or : {NO3 }

ISTUDY

16.7  |  Algebraic Equilibrium Calculations in Systems Undergoing Electron Transfer   397



log

{NH 4 } 119.3 10 pH 8pe {NO3 }

eq. 16.16

Equation 16.16 can be used to determine how the speciation of nitrogen varies with pH and pe at equilibrium. For example, consider water in an oxidizing environment {NH 4 } (pe = +10) at neutral pH (pH 7). Under these conditions: log = 119.3 – 10(7) – { NO } 3 {NH 4 } 8(10) = –30.7 or = 10–30.7. Therefore, in an oxidizing environment at neutral {NO3 } pH, NO3– is greatly favored over NH4+ at equilibrium. This makes sense: The more ­oxidized form is favored in an oxidizing environment.

Thoughtful Pause What is the ratio of species activities at pe = –5 and pH 7? {NH 4 } At pe = –5 and pH 7, log {NH 4 } = 119.3 – 10(7) – 8(–5) = +89.3 or = 10+89.3. {NO3 } {NO3 } In other words, in a reducing environment at neutral pH, the reduced form, NH4+, is greatly favored over the oxidized form, NO3– at equilibrium. Equation 16.16 also can be used to show how pe and pH are related at a given ratio of the reduced and oxidized forms of chemical species. We often examine the relationship between pe and pH when the activities of the reduced and oxidized forms are 5 119.3 10 pH equal. For the nitrification example, pe = = 14.9 ‒  pH when {NH4+} = 4 8 {NO3–}. This approach allows you to calculate the pe. For a water at pH 6.5 where nitro5 gen chemistry controls the pe and {NH4+} = {NO3–}, pe = 14.9 ‒   (6.5) = +6.8. Another 4 illustration of quantifying the relationship among pe, pH, and species activities is given in Worked Example 16.6. Worked Example 16.6

pe, pH, and Species Activities

What form of sulfur predominates in a bog water at pH 4 and pe = –3? Solution As a start, compare the lowest two oxidation states of sulfur: S(–II) and S(0). The equilibrium is: S(s) 2 H

2e

H 2 S logK

4.8

Thus: log

H2S S(s)

4.8 2 pH 2 pe

At pH 4 and pe = –3: log H 2 S = 4.8 – (2)(4) – (2)(–3) = +2.8. S(s)

ISTUDY

398   Chapter 16  Oxidation and Reduction Thus, H2S predominates. If H2S predominates over S(0), then higher oxidation states such as sulfate are not expected to be important. You can use the equilibria in Table 16.4 to confirm this point. Adding the S equilibria in Table 16.4: SO 4 2

10 H

8e

H 2 S 4 H 2 O; logK

41

Thus: log

At pH 4 and pe = –3: log

{H 2S} {SO 4 2 }

{H 2S} {SO 4 2 }

41 10 pH 8pe

25.

Thus, H2S predominates, as expected. In a bog water at pH 4 and pe = –3, H2S predominates.

16.7.2  Application: Potentiometric Electrodes and the Measurement of pH The Nernst equation shows that the cell potential is related to species activities. Cell potentials caused by species activities are called Nernst potentials. One example of a Nernst potential is the cell potential generated when two solutions of differing compositions are separated by a thin membrane. If one solution is of known composition, the activity of the species of interest can be determined in the other solution through measurement of the Nernst potential. In general:

E

E

X RT ln nF X

side 1 side 2



A common example of the use of Nernst potentials in chemical analysis is the pH probe. The working electrode (i.e., the {H+}-­sensing electrode) is a thin glass membrane shaped like a bulb and developed by Arnold O. Beckman in the 1920s.5 The inside of the bulb is filled with a strong acid of known {H+}. The sample is the other (or outside) solution. A salt bridge (see Figure 16.1) and reference electrode complete the circuit. Thus:



E

E

RT {H }inside ln nF {H }outside

E

RT ln{H }inside nF

RT ln{H }outside nF

5   The glass membrane is thin enough that the H+ in the sample can communicate with the H+ inside the glass bulb (through the activity of Na+ in the glass) without actually being transported across the glass. By analogy, suppose a classroom was equipped with a rubber membrane for a door. People inside the classroom would be aware of the activity of people in the hallway without the transport of people through the membrane.

ISTUDY

16.8  |  Graphical Representations of Systems Undergoing Electron Transfer   399

Or, since n = 1: E

where E

E

E

RT ln{H }inside . F

2.303

RT pHsample F

eq. 16.17

The pH probe must be calibrated. Calibration is the process of determining the values of the adjustable variables so that the response (E here) can be related to the quantity of interest (sample pH here). To calibrate the probe, the pH of a solution of known {H+}, called a standard, is measured.

Thoughtful Pause What is the minimum number of standards that must be measured to calibrate the response of the pH probe? As eq. 16.17 states, the cell potential is linearly related with the pH of the standard or sample. Thus, you must measure the pH of at least two standards to determine RT 2.303RT the slope and intercept E ln{H }inside .6 The Nernst equation F F gives you a way to measure pH and tells you the minimum number of standards that must be measured to calibrate a pH probe.

16.8  GRAPHICAL REPRESENTATIONS OF

SYSTEMS UNDERGOING ELECTRON TRANSFER

16.8.1  Introduction Since pe and pH are primary variables, it is desirable to be able to show how speciation changes with both pe and pH. To show species concentrations as a function of both pe and pH would require a three-­dimensional plot. This is not impossible, of course, but it can be a bit confusing. To simplify the plots, two approaches have been taken to show the effects of redox conditions on speciation. In the first approach, the pH is fixed and pe is used as the primary variable. The resulting plots are called pC-­pe diagrams, with –log(concentration) on the y-­axis and pe on the x-­axis. The pC-­pe diagrams are exactly analogous to the pC-­ pH diagrams of Chapter 8. The second approach is called a pe-­pH diagram. In a pe-­pH diagram, pe is plotted on the y-­axis and pH on the x-­axis. Lines are drawn on the plot to indicate conditions where two species have the same activity. Before discussing each diagram in more detail, the redox limits of water must be established.

 You might wonder why standards are even necessary since it appears that the slope and intercept both can be calculated from temperature, constants, and E°. In fact, the intercept also comprises other terms (including the liquid-­junction potential, a potential difference caused by the different rates of ion migration across an interface). F In practice, pH is operationally defined from pH measurements by the equation: pHsample = pHstandard + 2.303RT (E  – E ). 6

sample

standard

pC-­pe diagram: a graphical representation of redox equilibria at constant pH with –log(concentration) on the y-­axis and pe on the x-­axis pe-­pH diagram: a graphical representation of redox equilibria with pe on the y-­axis and pH on the x-­axis, where lines indicate conditions where two species have the same activity

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400   Chapter 16  Oxidation and Reduction

16.8.2  The Redox Limits of Water What happens to water when the oxidizing or reducing conditions are too extreme? Under highly oxidizing conditions, water is oxidized to oxygen gas. Under highly reducing conditions, water is reduced to hydrogen gas. How high or low does the pe have to be before water is oxidized or reduced? Not surprisingly, the answer depends on the pH. For the oxidation of water:

O2 (g) 4 H

4e

83.1

2 H 2 O; logK

With the mole fraction of water at unity and the activity of oxygen gas expressed by its partial pressure, you can show that pe = 20.78 – pH + ¼PO . If the total pressure of 2 the system is 1 atm, then the largest possible value of PO is 1 atm. Thus: 2



pe

20.78 pH for a pure O2 atmosphere

A similar equation can be developed for the reduction of water to hydrogen gas. The redox equilibrium is:7

2 H 2 O 2e

H 2 (g) 2OH ; logK

28

Again, for the mole fraction of H2O = 1, {H2(g)} = PH  = 1 atm, and recognizing that 2 log{OH–} = logKW + pH = –14 + pH at 25°C: Key idea: The redox and pH ranges of water are intertwined and given by the equations: pe = 20.78 – pH and pe = –pH

pe

pH for a pureH 2 atmosphere

We now have two equations that define the redox stability of water: pe = 20.78 – pH (in equilibrium with a pure O2 atmosphere) and: pe = –pH (in equilibrium with a pure H2 atmosphere) We are only interested in the region between these two extremes (at 1  atm total pressure and 25°C). Note that the redox stability of water depends on the pH. What does this mean? First, it means that we must revisit our pC-­pH diagrams. The pC-­pH diagrams are valid only for a specific pe (or range of pe values). In a perfect world, pC-­pH diagrams should be labeled with the pe range for which they are valid. In addition, the valid pH range in a pC-­pH diagram depends on pe. Previously, we have considered the pH range of water to be 0 to 14. Now you know that the pH range of water depends on the pe. For pH 0 to 14, the pe must be in the range of 0 to +20.78 at pH 0 and in the range of –14 to +6.78 at pH 14 to satisfy the redox stability of water. Thus, all our previous pC-­pH diagrams with x-­axes extending from 0 to 14 should be amended to say that they are valid only at certain pe values. Similar statements hold true for other pH ranges. For example, the pH range of 4 to 10 corresponds to pe = –4 to +16.78 at pH 4 to pe = –10 to +10.78 at pH 10. Second, the dependence of the redox stability range of water on pH means that pC-­pe and pe-­pH diagrams are valid only for specific ranges of pe and pH. For pC-­pe diagrams (which are drawn for a specific pH), this means that you must choose a pe range (x-­axis range) corresponding to the stability of water at that pH value. For pe-­pH diagrams, the standard approach is to draw the lines for the stability of water on the diagram.

  The redox equilibrium for the reduction of water can be obtained by adding the following two reactions: 2H+ + 2e– = H2(g) (K = 1 or logK = 0) and 2H2O = 2H+ + 2OH– (logK = 2logKW = –28). 7

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16.8  |  Graphical Representations of Systems Undergoing Electron Transfer   401

Both pC-­pe and pe-­pH diagrams will be discussed in more detail below. For each type of diagram, three systems will be presented: (1) the H2(g)/H2O/O2(g) system, (2) the Fe3+/Fe2+ system, and (3) the NO3–/NH4+ system from Section 16.7.1.

16.8.3  pC-­pe Diagrams To draw a pC-­pe diagram, the pH is fixed and pC for each species is plotted as a function of pe. For the H2(g)/H2O/O2(g) system, the pertinent equilibria are: O2 (g) 4 H

4e

2 H 2 O 2e



2H2 O

H 2 (g) 2OH

logK

83.1

logK

28

Expressing each species concentration in terms of pe:

pPO

2



pPH

2

83.1 4 pH 4 pe



2 pH 2 pe

The pC-­pe diagram for the system is shown at pH 7 in Figure 16.2. (At pH 7, PO

2

= 55.1 – 4pe and PH = 14 + 2pe.) The y-­intercepts of the lines for H2(g) and O2(g) are 2

both dependent on the pH. The pC-­pe diagram for the H2(g)/H2O/O2(g) system is not very useful, but it does point out an important fact about some redox systems. How would you determine the pe of water at pH 7 in equilibrium with the atmosphere? In the atmosphere, the partial pressures of H2(g) and O2(g) are 5×10–7 atm and 0.209 atm, respectively. Applying these data to Figure 16.2, you can obtain two equilibrium pe values (indicated by the solid circles in Figure 16.2): pe = 13.6 in equilibrium with atmospheric oxygen at pH 7 and pe = –3.8 in equilibrium with atmospheric hydrogen at pH 7.8 Thoughtful Pause How can there be two equilibrium positions of the system? pe –16 0

–12

–8

–4

0

4

8

12

16

pC

10 PH2

PO2

20 30 40

FIGURE 16.2  pC-­pe Diagram for the H2(g)/H2O/O2(g) System at pH 7

  The equilibrium pe value of 13.6 (calculated from the partial pressure of oxygen in the atmosphere) corresponds to a hydrogen partial pressure of about 10–41 atm. This is about 1000 molecules of hydrogen gas in the entire atmosphere – not a very reasonable result. The equilibrium pe value of –3.8 (calculated from the partial pressure of hydrogen in the atmosphere) corresponds to an oxygen partial pressure of about 1070 atm!

8

ISTUDY

402   Chapter 16  Oxidation and Reduction Clearly, water at pH 7 is not in equilibrium with the atmosphere. In fact, many redox systems are not in equilibrium. This idea will be explored further in Section 16.9.1. A more meaningful pC-­pe diagram can be obtained with the Fe3+/Fe2+ system. The pertinent equations for constructing the pC-­pe diagram are: Equilibrium: Fe3+ + e– = Fe2+; logK = 13.0 (= pe°, since n = 1) {Fe 2 } {Fe3 }{e }

K



Mass balance: FeT = [Fe2+] + [Fe3+] Solving each species activity in terms of (assuming activities and concentrations are equal): Fe

2

{e }

1 {e } K

FeT and {Fe } 3

1 K

1 K {e }

FeT

These equations are analogous to alpha values with acid–base chemistry. The pC-­pe diagram at FeT = 1×10–4 M is plotted in Figure 16.3 (assuming that the pH is low enough that hydroxo complexes can be ignored). Note that {Fe2+} is favored in reducing conditions (lower pe) and {Fe3+} is favored in oxidizing conditions (higher pe). Also note that {Fe2+} = {Fe3+} when pe = pe° (dotted line in Figure 16.3), showing the analogy between pe° and pKa, discussed in Section 16.5.2. As a final example of pC-­pe diagrams, consider the NO3–/NH4+ system discussed in Section  16.6.2. At pH 7, the predominate acid–base forms of N(+V) and N(–III) are NO3– and NH4+, respectively. The pertinent equations for constructing the pC-­pe diagram are: Equilibrium: NO3– + 10H+ + 8e– = NH4+ + 3H2O;  logK = 119.2 Mass balance: NT = [NO3–] + [NH4+] pe 0

0

5

10

pe = pe°

{Fe2+}

pC

5

10 {e–} 15

15

{Fe3+}

20 FIGURE 16.3  pC-­pe Diagram for the Fe3+/Fe2+ System at Low pH

20

ISTUDY

16.8  |  Graphical Representations of Systems Undergoing Electron Transfer   403

Solving each species activity in terms of (assuming activities and concentrations are equal): {e }8

{NH 4 }

1 K{H }10



{e }8

1 K{H }10 NT 1 8 } { e K{H }10

N T and {NO3 }

Again, these fractions look like alpha values. At pH 7:



1 = 10–49.2. K{H }10

The pC-­pe diagram at NT = 1×10–3 M is plotted in Figure 16.4. Note that {NH4+} = 1 {NO3–} at pe = log K{H }10 = 6.15 at pH 7. Thus, pe° = 6.15 at pH 7 (dotted line in 8 Figure 16.4). You can see that ammonium ion dominates at lower pe and nitrate ion dominates at higher pe, as expected. The conditions calculated in Section 16.7.1 at pH 7 can be evaluated using Figure 16.4. It is clear from the pC-­pe diagram that the nitrate ion dominates at pe = +10 and pH 7 and that the ammonium ion dominates at pe = –5 and pH 7.

16.8.4  pe-­pH Diagrams Recall that in pe-­pH diagrams, lines are drawn to indicate conditions where two species have the same activity. Returning to the O2(g)/H2O/H2(g) example, the pertinent equations were developed in Section 16.8.2 and are repeated here for convenience: pe = 20.78 – pH (water in equilibrium with a pure O2 atmosphere) pe = –pH (water in equilibrium a pure H2 atmosphere) You can plot the water stability conditions as lines on a pe-­pH diagram to indicate the possible oxidation states of H and O under environmental conditions, as in Figure 16.5. pe 0 2

0

2

4

6

{NH4+}

8

10

{NO3–}

pC

4 6 8 10 FIGURE 16.4  pC-­pe Diagram for the NO3–/NH4+ System at pH 7

{e–}

ISTUDY

404   Chapter 16  Oxidation and Reduction

FIGURE 16.5  pe-­pH Diagram for Water at 25°C (top line: PO = 1 atm; bottom line: PH = 1 atm) 2

2

The top line in Figure  16.5 indicates where {O2(g)} = {H2O}, and the bottom line indicates where {H2(g)} = {H2O}. In Figure 16.5: {O2(g)} = {H2O} = {H2(g)} = 1 (mole fraction for water). It is customary to label each line to show the pe region where a given species predominates. Thus, for example, we put the label for O2(g) at higher pe values than the line indicating where {O2(g)} = {H2O}. Since a fluid with pH < 0 or pH > 14 is not called water, water is defined thermodynamically (at 1 atm and 25°C) by the parallelogram in Figure 16.5. By the principle of superpositioning, you can simply plot other pe-­pH relationships on top of the pe-­pH diagram for water (just as you plotted other pC-­pH relationships on top of the pC-­pH diagram for water in Chapter 8). The general approach to constructing simple pe-­pH diagrams is to write equations of the form pe = apH + b for every equilibrium involving the species of interest and setting the ratio of oxidized and reduced species equal to 1. For electron transfer reactions with no proton transfer (i.e., no acid–­base chemistry), a = 0. In this case, the line of equal activities on a pe-­pH 1 diagram is a line of slope = 0 at pe = pe° = logK. n As an example, consider the reduction of ferric iron: Fe3+ + e– = Fe2+, logK = 13.0. Writing the expression for K in terms of species activities and taking logs:

pe 13.0 log

{Fe 2 } {Fe3 }

For {Fe3+} = {Fe2+}, pe = pe° = 13.0. This line is plotted in Figure 16.6. (For systems containing species other than just water, it is common practice to write the water stability lines as dashed lines. For the remaining pe-­pH plots in this chapter, the dashed lines will not be labeled.) Clearly, this pe-­pH diagram is valid at low pH only, since the hydroxo complexes of ferrous and ferric iron must be included at higher pH values. Note that ferric iron predominates at higher pe values (more oxidizing conditions) and ferrous iron dominates at lower pe values (more reducing conditions). 5 For NH4+/NO3– chemistry, you know from Section 16.7.1 that pe = 14.9 –  pH when 4 {NH 4 } = 1. The resulting pe-­pH diagram is shown in Figure 16.7. The conditions cal{NO3 } culated in Section  16.7.1 also are shown in Figure  16.7. Point A is at pe = +10 and

ISTUDY

16.8  |  Graphical Representations of Systems Undergoing Electron Transfer   405

FIGURE 16.6  pe-­pH Diagram for the Fe3+/Fe2+ System (valid only at low pH)

FIGURE 16.7  pe-­pH Diagram for the NH4+/NO3– System

pH 7, where nitrate ion dominates. Point B is at pe = –5 and pH 7, where ammonium ion dominates. Point C is at pe +6.78 and pH 6.5, where the nitrate and ammonium ion activities are equal. Note that point C lies on the line of equal activity. The pe-­pH diagram gives a qualitative sense of which species dominates. For example, as pe increases, the oxidized forms become more dominant. However, be careful not to overinterpret pe-­pH diagrams. It is not possible at a glance to see that the ammonium ion activity is over 89 orders of magnitude larger than the nitrate ion activity at point B in Figure 16.7 under equilibrium conditions, as calculated in Section 16.7.1. The magnitude of the ratio of activities depends on the number of electrons transferred. You can show that the log of the ratio of activities is equal to the number of electrons transferred multiplied by the vertical distance between the condition of interest and the line of equal activity (see Problem 16).

16.8.5  More Complex pe-­pH Diagrams: The Example of Inorganic Carbon Chemistry When several species are involved, pe-­pH diagrams can become fairly complex. As an example, consider the chemistry of inorganic carbon. For this example, consider only the +4 and –4 oxidation states of carbon. (In doing so, we neglect the rich redox

Key idea: To create a pe-­pH diagram: (1) Make a species list, including species from all reasonable oxidation states; (2) Carefully consider which chemical processes should be included; (3) List the equilibria describing the chemical processes listed in Step 2; (4) Write the expressions for the K values in terms of species activities, then set the mole fraction of water to unity and set the ratio of activities of redox pairs equal to one; (5) Plot the lines generated in Step 4 on a pe-­pH diagram; and (6) Erase line segments in the pe-­pH diagram that are thermodynamically impossible

ISTUDY

406   Chapter 16  Oxidation and Reduction chemistry of organic carbon.) The steps in constructing a more complex pC-­pH diagram will be presented here, using dissolved inorganic carbon as an example (see also Appendix C, Section C.4.4). Step 1: Make a species list, including species from all reasonable oxidation states. For this system, consider only carbon in its highest oxidation state, i.e., C(+IV), and carbon in its lowest oxidation state, i.e., C(–IV). The species list in a closed system would include H2CO3*, HCO3–, CO32–, and aqueous CH4 (methane). Step 2: Carefully consider which chemical processes should be included. It is useful to make a list of how each species in the species list could be related to every other species. You know that H2CO3*, HCO3–, and CO32– are related by proton transfer (acid–base chemistry). Each of these species also can be reduced to methane. A reasonable list of chemical processes is: 1. 2. 3. 4. 5.

H2CO3* deprotonation to HCO3– HCO3– deprotonation CO32– H2CO3* reduction to CH4(aq) HCO3– reduction to CH4(aq) CO32– reduction to CH4(aq)

You also could express this in an extended version of the species list diagrams from Section 6.3.5. An example is shown below. Numbers are the processes listed above. In this species list diagram, the extent of protonation of the C(+IV) species decreases as you move down the diagram and the oxidation state of carbon decreases as you move to the left. H2CO3* 1 4 HCO – 3 + 8e– 2

CH4(aq)

H+

CO 32–

Step 3: List the equilibria describing the chemical processes listed in Step 2. Tables of redox and acid–base equilibria (see Table 16.4 and Appendix D) provide the following possible equilibria for inclusion in the chemical model: H 2 CO3 HCO3

1 H CO 8 2 3

HCO3 CO32 H

H H e

logK1

6.3

logK 2

10.3

3 1 CH 4 (aq) H O logK3 8 8 2

2.71



In this list, equilibrium constants are numbered so that Ki corresponds to process i. By combining the K1 and K3 equilibria, we can describe process 4. Process 5 can be described by combining the K1, K2, and K3 equilibria:

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16.8  |  Graphical Representations of Systems Undergoing Electron Transfer   407



1 HCO3 8 1 CO32 8

9 H 8 5 H 4

1 CH 4 (aq) 8 1 CH 4 (aq) 8

e e

3 H 2 O logK 4 8 3 H 2 O logK 5 8

3.50 4.79



Step 4: Write the expressions for the K values in terms of species activities. Set the mole fraction of water to unity and set the ratio of activities of redox pairs equal to 1. For the equilibria above:



pH

6.3 log

{H 2 CO3 } {HCO3 }

pH 10.3 log pe

2.71 pH

6.3

{HCO3 } 10.3 {CO32 } 1 {H CO } log 2 3 8 {CH 4(aq )}

2.71 pH

pe 3.50

9 1 {HCO3 } pH log 8 8 {CH 4(aq)}

3.50

9 pH 8

pe

5 1 {CO32 } pH log 4 8 {CH 4(aq )}

4.79

5 pH 4

4.79

Step 5: Plot the lines generated in Step 4 on a pe-­pH diagram. The equations are plotted in Figure 16.8. It is important to terminate each line where it is no long applicable. The process 1 and process 2 lines terminate at the O2(g)/H2O dashed line and the H2CO3*/CH4(aq) line, since the processes are valid only for water and for where C(+IV) predominates. Similarly, the process 3, 4, and 5 lines are valid only for specific forms of C(+IV) and terminate at the H2CO3*/HCO3–/ CO32– boundaries.

FIGURE 16.8  pe-­pH Diagram for C(–IV)-­C(+IV)

ISTUDY

408   Chapter 16  Oxidation and Reduction Figure 16.8 shows that bicarbonate ion dominates in a lake at pH 7 and pe = +10, while methane dominates at equilibrium in a swamp at pH 7 and pe = –5.

16.8.6  More Complex pe-­pH Diagrams: The Example of Nitrogen Chemistry As another example of a complex pe-­pH diagram, reconsider the nitrate ion/ammonium ion pe-­pH diagram in Figure 16.7. This pe-­pH diagram is incomplete since several acid–base and redox forms of nitrogen are missing. You can construct a more realistic pe-­pH diagram for the nitrogen system. Species. The common oxidation states of inorganic nitrogen-­containing species in natural waters include N(+VII), N(+V), N(0), and N(–III). The species list would include NO3–, NO2–, NH4+, and NH3. You can ignore dissolved nitrogen gas, N2(g), because it is inert. Also, you can ignore HNO3, because its pKa is about –1. To simplify the system a bit, you can ignore nitrous acid, HNO2 (pKa 3.3). Nitrous acid decomposes in water at low pH. Processes. In the nitrogen system, NO3– clearly could be reduced to NO2–. At this point, we do not know if NO3– could be reduced directly to NH4+ or NH3, so we better include both transformations. Nitrite ion might be reduced to ammonium ion at lower pH (where ammonium ion may be expected to predominate) or reduced to ammonia at higher pH (where ammonia may be expected to predominate). Ammonium ion and ammonia are interconverted by proton transfer. Thus, a reasonable list of chemical processes is: 1. 2. 3. 4. 5. 6.

NO3– reduction to NO2– NO3– reduction directly to NH4+ at lower pH NO3– reduction directly to NH3 at higher pH NO2– reduction directly to NH4+ at lower pH NO2– reduction directly to NH3 at higher pH NH4+ deprotonation to NH3

The species and processes can be summarized in a species list diagram. In this species list diagram, the extent of protonation of the N(–III) species decreases as you move down the diagram, the oxidation state of nitrogen decreases as you move to the left, and numbers represent the processes. 2 + 8e– 4 + 6e–

NH4+

NO 2–

1 + 2e–

NO 3–

6

H+

NH3

Equilibria. The next step is the generation of a list of equilibria: 1

2

NO3

1 NO3 8 1 NO2 6 NH 4

e 5 H 4 4 H 3 NH 3

H

1

e e H

2

NO2

1 NH 4 8 1 NH 4 6

1

2

H2 O

logK1

3 H 2 O logK 2 8 1 H 2 O logK 4 3 logK 6

14.15 14.90 15.14 9.3

ISTUDY

16.8  |  Graphical Representations of Systems Undergoing Electron Transfer   409

The equilibria are numbered so that the Ki equilibrium corresponds to process i. 1 You can generate an equilibrium for process 3 by adding the K1 and times the K6 8 1 equilibria, and you can generate an equilibrium for process 5 by adding the K4 and 6 times the K6 equilibria. The final list of equilibria is: 1





2

NO3

e

1 NO3 8 1 NO3 8

5 H 4 9 H 8

1 NO2 6 1 NO2 6

4 H 3 7 H 6

NH 4

NH3

H

1

e e e e

2

NO2

1

2

H2 O

1 3 H2 O NH 4 8 8 1 3 NH 3 H2 O 8 8

logK1

14.15

logK 2

14.90

logK3

13.74

1 1 NH 4 H O logK 4 6 3 2 1 1 NH3 H O logK 4 6 3 2

H

logK6



15.14 13.59 9.3



The equilibria must be translated into pe-­pH equations. Recall that this is accomplished by using the equilibria and setting the ratio of activities of species undergoing redox chemistry equal to 1. The resulting pe-­pH equations are: pe 14.15 pH (process1)

9 pH (process3) 8 7 pe 13.59 pH (process 5) pH 6

pe 13.74

5 pH (process 2) 4 4 pe 15.4 pH (process 4 ) 3 pe 14.90

9.3 (process 6)



Lines are plotted in Figure  16.9. Again, lines are terminated when transitions to another predominant species occurs.

FIGURE 16.9  pe-­pH Diagram for the N(–III)/N(+III)/N(+V) System Including Ambiguous Processes (dotted lines, see text)

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410   Chapter 16  Oxidation and Reduction

FIGURE 16.10  Final pe-­pH Diagram for the N(–III)/N(+III)/N(+V) System (ignores N2(g), HNO3, and HNO2)

There remain some ambiguous sections of the pe-­pH diagram. Consider the reduction of nitrate ion at pH less than 3.0, the pH at which the lines for processes 1 and 2 intersect. If nitrate ion is reduced at a constant pH of, say 2.5, you would “hit” the process 2 line (reduction to ammonium ion) before you “hit” the process 1 line (reduction to nitrite ion). This means at pH < 3.0, nitrate ion is reduced directly to ammonium ion and the dotted line representing process 1 at pH < 3.0 can be erased. Now consider the reduction of nitrate ion at pH between 3.0 and 9.3. If nitrate ion is reduced at constant pH in this pH range, you would “hit” the process 1 line (reduction to nitrite ion) before you “hit” the process 2 line (reduction to ammonium ion). This means from pH 3.0 to 9.3, nitrate ion is reduced to nitrite ion and the dotted line representing process 2 in this pH range can be erased. Finally, consider the reduction of nitrate ion at pH > 9.3. If nitrate ion is reduced at a constant pH in this pH range, you would “hit” the process 1 line (reduction to nitrite ion) before you “hit” the process 3 line (reduction to ammonia). This means at pH > 9.3, nitrate ion is reduced to nitrite ion and the dotted line representing process 3 in this pH range can be erased. A “cleaned-­up” version of the pe-­pH diagram with the dotted lines removed is shown in Figure 16.10.

16.8.7  More Complex pe-­pH Diagrams: The Example of Chlorine Chemistry Chlorine exists in water in a large number of oxidation states (see Problem 1). Consider the construction of a pe-­pH diagram for chlorine in the +I, 0, and –I oxidation states at a total chlorine concentration of 1×10–3 M. Species. The species list includes species in the +I oxidation state (HOCl and OCl–), species in the zero oxidation state [Cl2(aq)], and species in the –I oxidation state (Cl–). Processes. It is possible that both HOCl and OCl– could be reduced to Cl2(aq) or directly to chloride ion. In addition, Cl2(aq) could be reduced to Cl–. As always, HOCl can donate a proton to water to form hypochlorite ion. Thus, a reasonable list of chemical processes is:

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16.8  |  Graphical Representations of Systems Undergoing Electron Transfer   411

1. 2. 3. 4. 5. 6.

HOCl reduction to Cl2(aq) HOCl reduction directly to Cl– OCl– reduction to Cl2(aq) OCl– reduction directly to Cl– Cl2(aq) reduction to Cl– HOCl deprotonation to OCl–

The species and processes can be summarized in a species list diagram. In this species list diagram, the extent of protonation of the Cl(+I) species decreases as you move down the list, the oxidation state of chlorine decreases as you move to the left, and numbers represent the processes. 2 + 2e– 5 Cl (aq) + 6e– 2

Cl –

1 + e–

HOCl H+

6 OCl –

+ 2e – 4

Equilibria. The next step is the generation of a list of equilibria: HOCl e



1

2

1

Cl2 (aq) e

HOCl



H

2

Cl2 (aq) H 2 O

Cl

OCl

H

logK1

26.6

logK5

23.0

logK6

7.54

The equilibria are numbered so that the Ki equilibrium corresponds to process i. You can generate equilibria for the other processes as follows: process 2 = ½(process 1 + process 5) process 3 = process 1 – process 6 process 4 = ½(process 1 + process 5 – process 6) The final list of equilibria is: HOCl e 1

2

HOCl e

OCl 1 1



H  

2 OCl

e

2

2H    e

2 Cl 2 (aq)

HOCl

1

1

2 Cl 2 (aq)

H2 O

1

1

H   1

H   e

OCl

2 Cl

2 Cl 2 (aq) 1

2 Cl

Cl           H

2

2

26.6

logK3

19.03

logK5

23.0

H 2 O logK2

H2 O 1

logK1

H2 O

logK 4 logK6

24.8 21.03 7.54

The equilibria must be translated into pe-­pH equations. Recall that this is accomplished by using the equilibria and setting the ratio of activities of species undergoing redox chemistry equal to 1. Stoichiometry must be taken in account in the mass balance

ISTUDY

412   Chapter 16  Oxidation and Reduction on chlorine in the system. For example, to account for stoichiometry, when {HOCl} or {OCl–} or {Cl–} is equal to {Cl2(aq)}, then {HOCl} or {OCl–} or {Cl–} is equal to ½ClT and {Cl2(aq)} is equal to ¼ClT.9 As an example, for process 1:

logK1

26.6

pe pH log

Cl 2 (aq) HOCl

pe pH

1

2

log ClT



In this equation, {HOCl} = ½ClT and {Cl2(aq)} = ¼ClT. Performing similar manipulations with ClT = 1×10–3 M yields the following pe-­pH equations: pe = 25.1 – pH (process 1)

pe = 24.8 – ½pH (process 2)

pe = 32.64 – 2pH (process 3)

pe = 21.03 – pH (process 4)

pe = 24.5 (process 5)

pH = 7.54 (process 6)

All of these pe-­pH equations are valid only for ClT = 1×10–3 M. Lines are plotted in Figure 16.11. Again, lines are terminated when transitions to another predominant species occurs. There remain some ambiguous sections of the pe-­pH diagram. It is clear that hypochlorite ion is reduced directly to chloride ion. Therefore, process 3 does not occur and the dotted process 3 line in Figure 16.11 can be erased. Figure 16.12 shows a close-­up view of the stability region for Cl2(aq). You can see that for pH > 0.6, HOCl is reduced directly to chloride ion, while for pH < 0.6, HOCl is reduced to Cl2(aq). Therefore, the lines for process 1 at pH > 0.6 and for process 2 at pH < 0.6 can be erased. The final version of the pe-­pH diagram is shown in Figure 16.13, which will be interpreted in Section 16.9.1.

FIGURE 16.11  Preliminary pe-­pH Diagram for the Cl(–I)/Cl(0)/Cl(+I) System (ClT = 1×10–3 M)

  This may seem a bit confusing. The idea is that under pe and pH conditions at, say, the HOCl/Cl2(aq) boundary, those two species dominate and together make up the total chorine. Therefore: ClT ≈ {HOCl} + 2{Cl2(aq)}. The traditional way of dividing chlorine is to have equal numbers of chlorine atoms as HOCl and as Cl2(aq)}. Thus: {HOCl} = ½ClT and {Cl2(aq)} = ¼ClT. 9

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16.9  |  Applying Redox Equilibrium Calculations to the Real World   413

FIGURE 16.12  Expanded View of the pe-­pH Diagram for the Cl(–I)/Cl(0)/Cl(+I) System (ClT = 1×10–3 M)

FIGURE 16.13  Final pe-­pH Diagram for the Cl(–I)/Cl(0)/Cl(+I) System (ClT = 1×10–3 M)

16.9  APPLYING REDOX EQUILIBRIUM

CALCULATIONS TO THE REAL WORLD

16.9.1  Are Redox Systems Really at Equilibrium? Systems in which electrons are transferred are not as easy to deal with as systems in which protons are transferred. One difficulty in applying redox equilibrium calculations to the real world is that redox reactions can be slow. As a result, redox systems may not be in equilibrium. You saw an example of the nonequilibrium nature of some redox reactions in Section 16.8.3. In that section, it was shown that water cannot be in redox equilibrium with both the oxygen gas and hydrogen gas in the atmosphere at pH 7. Another example of nonequilibrium conditions is chlorine chemistry. The pe-­pH diagram for the +I, 0, and –I oxidation states of chlorine at a total chlorine concentration of 1×10–3 M was shown in Figure 16.13. Thoughtful Pause From Figure 16.13, what form of chlorine is thermodynamically stable?

Key idea: Many redox reactions are slow and may not be at equilibrium

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414   Chapter 16  Oxidation and Reduction From Figure 16.13, you can see that the only form of these three oxidation states of chlorine that is thermodynamically stable is Cl(–I), primarily the chloride ion. Is this true? Yes: At equilibrium, chloride ion predominates. As shown in Figure 16.13, the other oxidation states of chlorine shown oxidize water to oxygen gas at equilibrium and therefore are thermodynamically unstable in water. However, the redox reactions of the 0 and +I oxidation states of chlorine are fairly slow. Aqueous Cl(+I) solutions are reasonably stable in the dark. If the reduction reactions were not slow, chlorine would be a poor choice for a disinfectant and chlorine bleach (which is a concentrated OCl– solution) would be unavailable in the stores.

16.9.2  Measuring pe Key idea: It is difficult to measure pe

Another challenge to applying redox equilibrium calculations to the real world is that it is very difficult to measure pe. In proton transfer (acid–base) chemistry, the primary variable, pH, is easy to measure. In electron transfer (redox) chemistry, the primary variable, pe, is not easy to measure. A full discussion of the reasons behind the difficulties in measuring pe are beyond the scope of this text. However, problems with impurities and nonequilibrium conditions greatly limit our ability to measure a ­thermodynamically reliable pe value in natural waters.

16.9.3  The Value of Redox Equilibrium Calculations

Key idea: Redox equilibrium calculations play an important role in chemical modeling, especially in determining the predominant species in a system

Although pe cannot be measured accurately and easily, redox equilibrium calculations still play an important role in the modeling of aquatic chemistry. Why? In acid–base systems, acid/conjugate base pairs are usually related by the exchange of only one or two protons. Thus, the slopes of the lines in pC-­pH diagrams are small, and small changes in pH result in only small changes in species concentrations. In redox systems, redox pairs often are related by the exchange of many electrons. Thus, the slopes of the lines in pC-­pe diagrams can be quite large (see Figure 16.4 for an example). As a result, species concentrations (and which species predominates) often change significantly only over a small pe range. In other words, once you are at a pe different than pe°, certain species can be ignored with little error. Thus, redox equilibrium calculations play an important role in chemical modeling, especially in determining the predominant species in a system.

16.10  CHAPTER SUMMARY With the introduction of electron transfer in this chapter, you now have been introduced to the major types of equilibria in homogeneous systems. In this chapter, you learned about oxidants (species that accept electrons to become reduced) and reductants (species that donate electrons to become oxidized). Reactions in which electrons are transferred are called redox reactions and may be half reactions (if electrons appear in the reaction) or overall redox reactions (if one or more oxidation states change but electrons do not appear as reactants or products). Redox (electron transfer) chemistry has many similarities with acid-­base (proton transfer) chemistry. Chemical reactions and chemical equilibria must be balanced to be useful. In Section 16.3, you honed your skills on balancing half reactions and overall reactions. To balance an overall reaction, add the half reactions in such a way that electrons cancel. Correctly balanced electron exchange reactions allow for the determination of oxidant and reductant doses in treatment processes.

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16.11  |  Part IV Case Study: Which Form of Copper Plating Should You Use?   415

In Sections 16.4 and 16.5, a system was developed to quantify the thermodynamic tendency to transfer electrons. You learned about several thermodynamic parameters, including the cell potential (linked to species activities through the Nernst equation) and pe° (equal to the equilibrium constant for a reduction reaction divided by the number of electrons transferred). In fact, pe° is analogous with pKa, one of many similarities between acid–base and redox chemistry. The thermodynamic parameters enabled the calculation of species activities at equilibrium through combinations of redox equilibria and mass balances. New graphical presentation methods were developed to display the results of redox equilibrium calculations. A key to both graphical methods is that the redox stability of water (with respect to oxygen and hydrogen gas) depends on the pH. Two plot types are (1) pC-­pe diagrams, plotted with –log(concentration) on the y-­axis and pe on the x-­axis (drawn at a fixed pH), and (2) pe-­pH diagrams, with pe plotted on the y-­axis and pH on the x-­axis. In pe-­pH diagrams, lines are drawn to indicate conditions under which two species have the same activity. Finally, the limitations and value of redox equilibrium calculations were reviewed. Limitations include the nonequilibrium state of some redox systems (e.g., H2(g) in the atmosphere in nonequilibrium with water and the aqueous chlorine system) and the inability to measure pe reliably and easily. In the end, though, redox equilibrium calculations are valuable because often only one oxidation state predominates under a given set of pe and pH conditions. Equilibrium calculations can identify the dominant species.

16.11  PART IV CASE STUDY: WHICH FORM

OF COPPER PLATING SHOULD YOU USE?

Recall that the Part IV case study concerns the relative merits of copper cyanide plating, where CuCN(s) is the copper source, and acid copper plating, where CuSO4(s) is the copper source. After completing Chapter 16, you should be thinking about how the redox chemistry of copper may influence the choice of plating process. It is of interest to ask whether copper cyanide offers a savings in electrical usage over acid copper. Electrical usage stems directly from redox stoichiometry. For example, how much electricity does it take to reduce Cu(+I) to Cu(0)? The electric requirement is quantified in electroplating by the electrochemical equivalent, the number of grams of a metal reduced per amp-­hour of reduction. How do you calculate the electrochemical equivalent? Recall that Faraday’s constant, F, is equal to 96,485 coulombs of charge per mole. One coulomb is 1 amp-­second. Thus, F = 96,485 A-­s/mol. A one-­ electron process would require 1F of charge, and a two-­electron process requires 2F of charge. Thus, the electrochemical equivalent is given by:



electrochemical equivalent g / A-hr

M 3600 nF

s hr



where M = molar mass (in g/mol) = 63.55 g/mol for copper, n = number of electrons transferred, and F = 96,485 A-­s/mole. For copper cyanide plating, n = 1 [Cu+ + e– Cu(s)], while n = 2 for acid copper plating [Cu2+ + 2e– = Cu(s)]. Thus, the electrochemical equivalent is 2.4 g/A-­hr for copper cyanide plating and 1.2 g/A-­hr for acid copper plating. Therefore, one of the advantages of copper cyanide plating is that it uses less electrical power than acid copper plating.

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416   Chapter 16  Oxidation and Reduction Another use of redox chemistry in analyzing copper plating is in determining the pertinent equilibrium constants. For examining the “loss” of Cu(+I) to Cu(+II) (see Section 15.8), the appropriate equilibrium is: Cu(s) Cu2



eq. 16.18

2Cu

(Note that the forward reaction in eq. 16.18 is a comproportionation reaction and the reverse reaction is a disproportionation reaction.) For the avoidance of immersion deposits with copper cyanide plating (see Section 15.8), the pertinent equilibrium is:

2Cu

Zn(s)

Zn 2

eq. 16.19

To determine the equilibrium constants, it is necessary to start with the reactions for which we have thermodynamic data. For many redox reactions, this means starting with the half reactions. The important half reactions are: Cu

e   Cu(s)

E1o

0.52 V

2

e   Cu

E2

o

0.16 V

2

2e   Zn(s) E3

o

Cu Zn



0.76 V



To find the equilibrium constant for the equilibrium in eq. 16.18, it is necessary to add the following two half reactions: Cu(s)

Cu

Cu

e

2

Cu

Cu(s) Cu



e

2

2Cu

E4 o

?

E2

o

0.16 V

E3

o

?



Clearly, the E4 o equilibrium is the reverse of the E1o equilibrium. Remember that reversing reactions changes the sign of E°. Therefore, E4 o = –0.52 V. Because the E4 o o and E2 o equilibria have the same number of electrons, you can calculate E5 = E4 o + o E2 = –0.52 V + 0.16 V = –0.36 V. (Remember in general that the safest way to manipuo o late redox equilibria is to convert the E° values to Grxn values, add the Grxn values, o and then reconvert to E°.) Now: Grxn,5 = –nF E5o = (–1)(–0.36 V)F = (0.36 V)F and:



logK 5

o Grxn ,5

2.303RT

( 0.36 V)

0.36 V

F 2.303RT

0.059 V 2

6.1

Cu Thus, K for the equilibrium in eq. 16.18 is K = = 10–6.1. This value of the equi2 Cu librium constant was used in Section 15.8. To find the equilibrium constant for the equilibrium in eq. 16.19, add the following two half reactions:



2Cu

2e

2Cu(s)

Zn(s)

Zn 2

2e

2Cu

Zn(s)

Zn 2

E6 o

?

E7o

?

2Cu(s) E8o

?



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Chapter Key Ideas   417

Recall that multiplying a reaction by a positive constant does not affect E°, so E6 o = E1o = +0.52 V. The E7 o equilibrium is the reverse of the E3o equilibrium, so: E7 o   = E3o = +0.76 V. Again here E8o = E6 o + E7 o because the number of electrons transferred is the same in the E6 o and E7 o equilibria. So E8o = +0.52 V + 0.76 V = 1.28 V, o Grxn,8 = –nF E8o = –(2)(1.28 V)F and:



logK8

o Grxn ,8

2.303RT

(2)(1.28 V)

2.56 V

F 2.303RT

Thus, K for the equilibrium in eq.  16.19 is K =

0.059 V Zn

2 2

43.4



= 10+43.4, since the activ-

Cu ities of the pure solids are unity. This value of the equilibrium constant was used in Section 15.8.

CHAPTER KEY IDEAS • Overall reactions are formed by adding together half reactions in such a way that the free electrons cancel out. • Half reactions have electrons as reactants or products, but overall redox reactions do not have electrons as reactants or products. • In an overall reaction, oxidants accept electrons and are reduced, whereas reductants donate electrons and are oxidized. • To balance a half reaction (1) write the known reactants on the left and known products on the right, (2) adjust stoichiometric coefficients to balance all elements except H and O, (3) add water (H2O) to balance the element O, (4) add H+ to balance the element H, and (5) add electrons (e–) to balance the charge • Check the change in oxidation states in balanced half reactions. • To balance overall reactions (1) balance the half reactions separately, and (2) add the half reactions to eliminate free electrons. • Check the change in oxidation states in balanced overall reactions. • Exercise care when dealing with oxidation states of O other than –II and of H other than +I. • pe is equal to pe° when the oxidized and reduced forms have the same activity. • The Gibbs free energy of reaction and the cell potential are related by ΔGrxn = –nFE (note negative sign). • With reactions listed as reductions, strong oxidants appear as reactants in reactions with large positive pe° values and strong reductants appear as products in reactions with large negative pe° values. • Reversing a reaction will change the sign of E and E°. • The cell potential does not change if you multiply the half reaction by a positive constant. • Since Gibbs free energy values are additive, when adding redox equilibria convert E° values to Grxn, add the Grxn values, and then reconvert to E°. • To calculate activities as a function of pe and pH, rearrange the mathematical form of the equilibrium to obtain an equation of the form: ratio of species activities = a + bpH + cpe. • Measured cell potentials and the Nernst equation can be used to develop methods to measure species activities.

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418   Chapter 16  Oxidation and Reduction • The redox and pH ranges of water are intertwined and given by the equations pe = 20.78 – pH and pe = –pH. • To create a pe-­pH diagram: (1) Make a species list, including species from all reasonable oxidation states; (2) Carefully consider which chemical processes should be included; (3) List the equilibria describing the chemical processes listed in Step 2; (4) Write the expressions for the K values in terms of species activities, then set the mole fraction of water to unity and set the ratio of activities of redox pairs equal to one; (5) Plot the lines generated in Step  4 on a pe-­pH diagram; and (6) Erase line segments in the pe-­pH diagram that are thermodynamically impossible. • Many redox reactions are slow and may not be in equilibrium. • It is difficult to measure pe. • Redox equilibrium calculations play an important role in chemical modeling, especially in determining the predominant species in a system.

CHAPTER GLOSSARY comproportionation reaction:  a reaction in which an element is in two or more oxidation states in the reactants and one oxidation state in the products disproportionation reaction:  a reaction in which an element is in one oxidation state in the reactants and two or more oxidation states in the products faraday (F):  one faraday means 96,485 coulombs per mole of electrons transferred (F also can be expressed as 96,485 joules per mol-­volt = 96,485 J/mol-­V) half reaction:  a reaction having one or more electrons as either reactants or products Nernst equation:  an equation relating the cell potential to the activities of the electron-­transferring species overall redox reaction:  a reaction in which electrons are transferred but free electrons do not appear as reactants or products oxidant:  a compound that oxidizes another species oxidation reaction:  a reaction in which electrons are produced (and the oxidation state of one or more reactants is increased) pC-­pe diagram:  a graphical representation of redox equilibria at constant pH with –log(concentration) on the y-­axis and pe on the x-­axis pe:  a measure of the availability of electrons; pe = –log{e–} 1 pe°:  a measure of reductant strength and equal to logK, where n = number of n electrons transferred and K = equilibrium constant for an equilibrium of the form Xm + ne–  = X(m–n) (or ox + ne–  =  red) pe-­pH diagram:  a graphical representation of redox equilibria with pe on the y-­axis and pH on the x-­axis, where lines indicate conditions where two species have the same activity redox reaction:  a reaction in which reduction or oxidation or both take place (i.e., a reaction in which electrons are transferred) reductant:  a compound that reduces another species reduction reaction:  a reaction in which electrons are consumed (and the oxidation state of one or more reactants is decreased) standard (or normal) hydrogen electrode:  a part of an electrochemical cell where {H2(g)} = 1 atm and {H+} = 1 M

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Historical Note: Walther Hermann Nernst   419

HISTORICAL NOTE: WALTHER HERMANN NERNST Many students of chemistry learn the name Nernst upon their introduction to electrochemistry. Yet few appreciate Nernst’s great contributions to physical chemistry, thermodynamics, photochemistry, and industrial chemistry. Born in West Prussia in 1864, Nernst studied at several of the great universities of central Europe, including the Universities of Zurich, Berlin, Graz, and Würzburg. His first academic appointment, at Leipzig University, created a sort of late nineteenth-­century Dream Team of  physical chemists: Nernst, Ostwald, van’t Hoff (see Chapter  23), and Arrhenius (see  Chapter  23). In one of the great “academic genealogies” of the late nineteenth century, Nernst studied with Ludwig Boltzmann, and in turn Irving Langmuir (Chapter 22) and Gilbert N. Lewis (Chapter 15) studied with Nernst. Although his early work was in electrochemistry, Nernst’s contributions span physical chemistry and beyond. In 1906, he developed what would come to be called the Third Law of Thermodynamics (see Chapter 3), opening up a new degree of precision in thermodynamic calculations. He collaborated with Otto Haber on the important industrial syntheses of the day. Outside of chemistry, Nernst contributed practical designs of many items, including a replacement for the carbon filament in the light bulb, the use of nitrous oxide to enhance the performance of the internal combustion engine, and an electric piano. The Nernst lamp made him so wealthy that his students referred to him jokingly as the Kommerzienrat (“successful businessman”) (Stone 2014). Nernst played an interesting role in the fame of Albert Einstein. In 1907, Einstein, then an associate professor at the University of Zurich and relatively unknown (despite his annus mirabilis two years prior), published a paper that supported Nernst’s work on what would become the Third Law of Thermodynamics. In early March of 1910, Nernst made a special trip from Berlin to Zurich to meet with Einstein. He was very impressed with the young physicist. Nernst’s reputation

Advertisement for the Nernst Lamp (1903) (Ulrich Schmitt/ Museum of Chemistry, Göttingen)

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420   Chapter 16  Oxidation and Reduction proceeded him. According to George Hevesy (as found in Stone 2014), “Einstein in 1909  was unknown in Zurich. Then Nernst came and people in Zurich said ‘that Einstein must be a clever fellow if the great Nernst comes all the way from Berlin to Zurich to talk to him.’” A prolific writer of textbooks, Nernst was awarded the Nobel Prize in Chemistry in 1920. He died in 1941.

PROBLEMS 1. Chlorine exists in seven oxidation states. Examples include (in alphabetical order): chlorate ion ClO3–

chlorine dioxide ClO2

chloride ion Cl–

chlorite ion ClO2–

chlorine Cl2(aq)

hypochlorous acid HOCl

perchlorate ion ClO4–

a. Determine the oxidation state of each compound and order them from the most negative oxidation state to the most positive oxidation state. b. Write a balanced half reaction for the conversion of each oxidation state to the next highest oxidation state at low pH. (Assume each species listed above is the dominant acid–base species for its oxidation state at low pH.) 2. Phosphorus is found in four oxidation states. The most protonated form of each is listed below in alphabetical order, along with the approximate pKa values for the acid–base system in parentheses. hypophosphorous acid H3PO2 (1.2) phosphine PH3 (no acid–base chemistry) phosphoric acid H3PO4 (2.2, 7.2, 12.4) phosphorous acid H3PO3 (1.3, 6.7) a. Determine the oxidation state of each compound and order them from the most negative oxidation state to the most positive oxidation state. b. Identify the species at highest concentration at pH 10 for each oxidation state. For example, for the oxidation state represented by phosphoric acid, identify whether H3PO4, H2PO4–, HPO42–, or PO43– is at the highest concentration at pH 10. You can do this by inspection without performing an equilibrium calculation. c. Write a balanced half reaction for the conversion of each species identified in Part B to the species with the next highest oxidation state at pH 10. 3. Balance the following overall reactions: a. CH3OH + O2 → CO2 + H2O (at high pH) (oxidation of methanol with oxygen)

b. HS– + MnO4– → S(s) + MnO2(s) (at high pH) (oxidation of bisulfide with permanganate) c. IO3– (iodate) + I– (iodide) → I2 (iodine) (at low pH) (chemistry used to standardize iodate solutions) 4. Calculate the oxidation states of I for each species in Problem 3 Part C and show that the reaction is a comproportionation reaction. 5. Every few years, there is renewed interest in the use of ferrate (FeO42–) for drinking water treatment. Ferrate is a strong chemical oxidant and therefore serves as a disinfectant. It is reduced to Fe(+III) which precipitates as Fe(OH)3(s) and therefore serves as a coagulant. Ferrate can be synthesized from the oxidation of Fe(OH)3(aq) at high pH by OCl–, which is, in turn, reduced to chloride ion. a. Determine the oxidation state of iron in ferrate. b. Write the balanced half reactions and balanced overall reaction for the synthesis of ferrate using the reactions described in the problem statement. c. Determine the dose of hypochlorite required to synthesize ferrate (in g OCl– as Cl2 per g of ferrate as Fe). 6. Dissolved oxygen, O2(aq) or DO, can be measured by a wet-­chemical technique called the Winkler method. The Winkler method consists of three steps outlined below. For each step, write the balanced overall reaction. a. In the first step, NaOH and a great deal of Mn2+ are added to the sample. The Mn2+ is oxidized to MnO(OH)2(s) by O2(aq) at high pH, and O2(aq) is reduced to H2O. b. In the second step, acid and a great deal of I– are added. The MnO(OH)2(s) oxidizes I– to I2 and is reduced to Mn2+ at low pH. c. In the third step, I2 is titrated with thiosulfate (S2O32–). During the titration, I2 is reduced to I– and S2O32– is oxidized to S4O62– (tetrathionate). d. Add together the overall reactions in parts A through C to develop the total overall reaction for the Winkler method. In the standard method, it is stated that 1 mL of 0.025 M Na2S2O3 titrant corresponds to 1 mg/L DO in a 200 mL sample. Is this information consistent with your total ­overall reaction?

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Problems   421 7. Rusting is the process of the oxidation of zero-­valent iron by dissolved oxygen to ferric iron (with subsequent reduction of dissolved oxygen to water). a. Write the half reactions and balanced overall reaction for the rusting of iron. b. Calculate the logK for the overall reaction (see Table 16.4). c. Find the pe for water at pH 7 with a dissolved oxygen Fe 2 concentration of 10 mg/L. What is the Fe(s) at this pe? 8. Hydrogen peroxide (H2O2) is a good chemical oxidant. You may have used hydrogen peroxide as a topical disinfectant and noticed the evolution of oxygen bubbles. a. If the oxidation state of H in hydrogen peroxide is +I, what is the oxidation state of O? b. Write the half reactions and overall reaction for the conversion of hydroxide peroxide to molecular oxygen and water. c. Is the overall reaction a disproportionation reaction, a comproportionation reaction, or neither a disproportionation nor a comproportionation reaction? 9. A drinking water treatment plant is using ozone for disinfection. Plant personnel want to use ozone also to oxidize reduced manganese, Mn2+, to MnO2(s). a. Find the required ozone dose (in g O3 per g Mn2+ oxidized) if ozone is reduced to oxygen (see Section 16.3.5 for the ozone reduction reaction). b. Plant personnel added ozone to their water and noticed a pink water condition. What was the minimum ratio of O3 to Mn2+ that they were using? (See Section 16.3.4 for a discussion of the chemical species responsible for pink water.) 10. Chlorine sometimes is used to oxidize H2S to sulfate ion in the treatment of groundwater. Chorine (here, HOCl) is reduced to chloride ion. The chemistry is discussed in Section 16.3.4. If the chlorine dose is too low, then sometimes the water becomes turbid. This occurs because too little chlorine has been added and the H2S is oxidized to S(0), which subsequently polymerizes to S8(s). Using balanced redox reactions, show how an improper dose can form S(s) and cause the turbidity problem. 11. The following questions concern the COD test, introduced in Worked Example 16.2. a. In the COD test, excess dichromate is added. The dichromate remaining after oxidizing the organics is titrated with a ferrous iron salt. Write the balanced overall reaction for the reduction of dichromate to Cr3+ (while Fe2+ is oxidized to Fe3+). b. Comparing the redox chemistry of dichromate and oxygen, verify the statement that “. . .each milliliter

of a 0.25 N solution of dichromate is equivalent to 2 mg of oxygen” (Sawyer and McCarty 1978). Note: a 1 N solution here means one mole/L of electrons available. 12. Following the example in Section 16.6.1, convert cell potentials to free energies to show that the cell potential does not change if you multiply a half reaction by a positive constant. 13. Using the data in Table 16.4, rank the following from the weakest to strongest oxidant strength: Fe3+, SO42–, Fe2+, O3, and O2. Justify your answer. Assume each species is reduced to its nearest oxidation state using the data in Table 16.4. 14. Heterotrophic bacteria use organic carbon as an electron source. The organics are oxidized to CO2. An example is: ¼CH2O + ¼H2O = ¼CO2 + H+ + e–. The organisms require a terminal electron acceptor (TEA) to accept electrons in the paired reduction reaction. Some bacteria (called facultative bacteria) can use either oxygen or sulfate ion as the TEA. From an energy standpoint, explain why facultative organisms use ­oxygen ­preferentially. 15. Construct a pC-­pe diagram for the N(–III)/N(+III)/ N(+V) system at pH 7. Use the equilibria: NO3– + 2H+ + 2e– = NO2– + H2O; logK = 28.3 and NO3– + 10H+ + 8e– = NH4+ + 3H2O; logK = 119.3. Ignore NH3 at pH 7 and assume a total dissolved nitrogen concentration of 1 mg/L as N. Plot over the pe range of 0 to +10 and pC range of 0 to 15. Justify the pe values where {NH4+} = {NO2–} and {NO2–} = {NO3–}. 16. Show that at any point on a pe-­pH diagram, the log of the ratio of activities of the oxidized and reduced forms is equal to the number of electrons transferred multiplied by the signed vertical distance to the line of equal activity. 17. Construct a pe-­pH diagram for the O3(aq)/H2O system. The pertinent equilibria are in Tables 16.4 and D.4. Assume {H2O} = 1. What can you say about the thermodynamic stability of aqueous ozone? 18. Construct a pe-­pH diagram for the aqueous chlorine system similar to Figure 16.15 for ClT = 5×10–3 M. The pertinent equilibria are HOCl H 1

2

e

1

2

Cl2 (aq) H 2O  ;  logK

Cl2 (aq) e   Cl ; logK

HOCl

OCl

H ;  logK

26.6

23.0 7.54

Note: Because of mass balance considerations, [HOCl] = ½ClT and [Cl2(aq)] = ¼ClT at the Cl2(aq)-­HOCl boundary and [Cl–] = ½ClT and [Cl2{aq)] = ¼ClT at the Cl2(aq)-­Cl– boundary. 19. In Section 16.7.3, it was suggested that alpha values for redox reactions could be developed by analogy with the

ISTUDY

422   Chapter 16  Oxidation and Reduction acid-­base alpha values. Develop redox alpha values for the following system: ox ne mH red C total concentration

ox

red

The alpha values should satisfy the equations: {ox} = αoxC and {red} = αredC. Check your answer with the ammonium ion/nitrate ion example in Section 16.7.1. In that example, αox = αred at pH 6.5 when pe = +6.78.

CHAPTER REFERENCES Bard, A.J. and L.R. Faulkner (1980). Electrochemical Methods: Fundamentals and Applications. New York: John Wiley & Sons, Inc. Sawyer, C.N. and P.L. McCarty (1978). Chemistry for Environmental Engineers. 3rd ed. New York: McGraw-Hill Book Co. Stone, A.D. (2014). Einstein and the Quantum: The Quest of the Valiant Swabian. Princeton, NJ: Princeton University Press.

Stumm, W. and J.J. Morgan (1996). Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters. 3rd ed. New  York: John Wiley & Sons, Inc.

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PA R T V

Heterogeneous Systems Chapter 17: Getting Started with Heterogeneous Systems Chapter 18: Gas–Liquid Equilibria Chapter 19: Solid–Liquid Equilibria

Johnny, finding life a bore, drank some H2SO4; Johnny’s father, an MD, gave him CaCO3. Now he’s neutralized, it’s true, but he’s full of CO2. –Thomas Stephenson Part V Haiku Gas equilibrates. And what becomes of solids? Precipitation. A field of water betrays the spirit that is in the air. It is continually receiving new life and motion from above. It is intermediate in its nature between land and sky. –Henry David Thoreau

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C H A P T E R   17

Getting Started with ­Heterogeneous Systems 17.1 INTRODUCTION In the first four parts of this text, you were introduced to the major types of ­chemical equilibria in homogeneous aquatic systems. Homogenous systems have only one phase. Thus, exchange of mass with gas and/or solid phases was not allowed. The restriction to one phase greatly limits the applicability of the tools you have learned. A quick glance at any surface water will tell you that exchange with the atmosphere may be possible. Many small ponds and shallow rivers exhibit reasonably large surface area-­to-­volume ratios, thus aiding mass transfer. The equilibration of chemical species between the gas and liquid phases, called gas–liquid equilibrium, often sets limits on aqueous phase concentrations. Gas–liquid equilibria are the focus of Chapter 18. Chemical species also can move between solid and aqueous phases. Recall from Section 12.1 that the composition of the early oceans was determined by the leaching of minerals into acidic runoff. This is an example of dissolution. The chemical process of forming solid phases from dissolved species is called precipitation. At equilibrium, the rates of dissolution and precipitation are equal. The equilibrium state is called solid–liquid equilibrium and is the topic of Chapter 19. Dissolved species also can partition onto and from surfaces. Although you will be learning about new types of equilibria in this part of the text, it is important to recognize that no new solution techniques are needed. Systems containing gas–liquid and solid–liquid equilibria can be solved by the same approaches that you already have mastered. You still will be developing species list, writing equilibrium expressions, and conjuring up mass balances. In Part V, you simply will be adding new chemical species and new equilibria to the mix. In fact, in some cases, equilibrium calculations with systems containing gas–liquid and solid–liquid equilibria are easier than with homogenous systems. Why? As you will see, many heterogenous systems have species with fixed concentrations. Surface equilibrium calculations are a bit more complex. Many surfaces in the environment are charged. We must account for the electrostatic interactions between the charged surface and charged species interacting with the surface. Surface equilibrium calculations require additional tools and are discussed in Chapter 22.

Key idea: Systems containing gas–liquid and solid–liquid equilibria can be solved using the methods of Part II

425

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426   Chapter 17  Getting Started with ­Heterogeneous Systems

17.2  EQUILIBRIUM EXCHANGE BETWEEN

GAS AND AQUEOUS PHASES

17.2.1  Introduction Some dissolved species are volatile (from the Latin volare to fly). Volatile compounds may partition into a gas phase that is in contact with the water. The gas phase may be the atmosphere or gas bubbles. You will discover in Chapter 18 that at equilibrium, the gas phase activity and aqueous phase activity of the equilibrating species are related through an equilibrium constant. This is analogous to the way that the activities of an acid and its conjugate base are related through an equilibrium constant. Dissolved species that partition between water and a contacting gas phase are called dissolved gases. Dissolved gases play a critical role in the function of ecosystems. One of the most important dissolved gases in natural waters is dissolved oxygen. Dissolved oxygen is used as a terminal electron acceptor by a number of aquatic organisms, including fish and aerobic bacteria. Dissolved carbon dioxide also is important. With increasing carbon dioxide concentrations in the atmosphere, the oceans serve as important reservoirs for dissolved carbon dioxide. A full understanding of global climate change requires an understanding of how carbon dioxide partitions between the atmosphere and the oceans. In addition, as you will see, increasing atmospheric CO2 concentrations affect the pH of large water bodies. Oxygen and carbon dioxide dissolved in blood play similar roles in mammals and other terrestrial organisms.

17.2.2  Open Systems

Key idea: Gas phase concentrations sometimes can be considered constant, as long as the ratio of the water volume to air volume in the system is small enough

Some environmental systems have gas phases that serve as extremely large reservoirs of chemical species. The classic example is equilibration with the atmosphere. However, even gas phases with smaller volumes than the atmosphere can contain large amounts of gaseous species. The mass of gaseous species is large enough that we often assume that the gas phase concentration is constant even when a gas species partitions into water. Here is an example. Air contains about 20.9% oxygen by volume or about L O2 g O2 g O2 0.209 1.331 0.278 at 20°C. Water in equilibrium with the atmoL air L O2 L air g O2 sphere has a dissolved oxygen concentration at 20°C of about 9 mg / L 0.009 . L water Thus, as long as the ratio of the water volume to air volume in the system is less than about 0.1:1, the mass of oxygen in the water at equilibrium will represent a small ( 12.69, use the approach of Section 19.3.2 [no solid phase, Zn(+II)T = 10–3 M]. The resulting pC-­pH diagram is shown in Figure 19.4.

← Worked example

Key idea: Many metals are soluble at both low and high pH, with minimum solubility at an intermediate pH value Key idea: If the solid phase is assumed to exist under certain conditions, but the calculated total soluble metal concentration exceeds the available metal concentration, then the solid phase does not exist under those ­conditions

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474   Chapter 19  Solid–Liquid Equilibria

FIGURE 19.4  Zinc Speciation with 10–3 M Total Zinc [Zn(OH)2(s) exists at 7.74 < pH < 12.69]

19.4  FACTORS AFFECTING METAL SOLUBILITY 19.4.1  Introduction

Key idea: Metal solubility is affected by the available metal concentration, pH, types of solid phases potentially formed, and ligand concentrations

The only factors affecting metal solubility discussed so far in this chapter are the total available metal concentration and pH. You know pH plays an important role with hydroxide solids such as Zn(OH)2(s). Other kinds of solids can be important in natural and engineered systems, including solids formed with sulfide, phosphate, and carbonate. Thus, it is possible that more than one solid phase can form. In Section 19.4.2, you will explore the effects of multiple solid phases on metal solubility. In Section 19.3.3, it was pointed out that some hydroxide solids dissolve at high pH because of the large [OH–]. This concept can be expanded. In fact, the presence of any additional ligand may affect metal solubility. In Section  19.4.3, you will explore the effects of multiple ligands on metal solubility.

19.4.2  Multiple Solid Phases Key idea: The controlling solid phase is the solid phase producing the smallest equilibrium total metal concentration: the least soluble solid phase controls solubility

Is it possible for two or more solid phases to exist in equilibrium? How do you decide which solid phases form and which solid phase controls the solubility of a metal? To answer these questions, consider the example of ferrous iron and its two common solid forms, Fe(OH)2(s) and FeCO3(s). Suppose that, under a given set of conditions (i.e., pH and total carbonate concentration), Fe(OH)2(s) alone produces a total soluble Fe(+II) concentration, Fe(+II)T, of 4.8×10–5 M, and FeCO3(s) alone produces a Fe(+II)T of 8.0×10–2 M. Under these conditions, Fe(OH)2(s) is much less soluble than FeCO3(s). Why? Fe(OH)2(s) produces much less Fe(+II)T. Thus, Fe(OH)2(s) will form and FeCO3(s) will not form under these conditions. In general, the controlling solid phase is the solid phase producing the smallest equilibrium total metal concentration. In other words, the least soluble solid phase controls solubility. Now you can decide whether it is possible for two or more solid phases to exist in equilibrium. Two or more solid phases exist in equilibrium under the same set of conditions if they produce the same equilibrium total metal concentration. Why? If one

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19.4  |  Factors Affecting Metal Solubility   475

solid produces a smaller equilibrium total metal concentration than a second solid, then the first solid will exist and the second solid will not. Only if they produce the same equilibrium total metal concentration will more than one solid phase exist under one set of conditions. To determine which solid phase controls the solubility of a metal, construct lines of total soluble metal, assuming that only one solid phase exists at a time. For the ferrous iron example, the pertinent equilibria are: H2O = H+ + OH–

KW = 10–14

Fe2+ + OH– = FeOH+

K1 = 10+4.5

Fe2+ + 2OH– = Fe(OH)20

K2 = 10+7.4

Fe2+ + 3OH– = Fe(OH)3–

K3 = 10+11

← Worked example

In this case: Fe(+II)T = [Fe2+] + [FeOH+] + [Fe(OH)20] + Fe(OH)3–] = [Fe2+](1 + K1[OH–] + K2[OH–]2 + K3[OH–]3) If you pretend for the moment that only Fe(OH)2(s) exists, then include the equilibrium: Fe OH

Thus: [Fe2+] =

2

s

Fe 2

 2OH        Ks 0,1

10

15.1



eq. 19.4

K s0,1 , and: [OH ]2

Fe(+II)T =

K s0,1 (1 + K1[OH–] + K2[OH–]2 + K3[OH–]3) [OH ]2

The resulting Fe(+II)T is shown in Figure 19.5A. If you pretend for the moment that only FeCO3(s) exists, then use the KW, K1, K2, and K3 equilibria as before plus the equilibrium: FeCO3 s

Thus: [Fe2+] =

K s 0,2 [CO32 ]

Fe(+II)T =

Fe 2

 CO32 Ks 0,2

10  

10.7

Key idea: Two or more solid phases exist in equilibrium under the same set of conditions if they produce the same equilibrium total metal concentration



eq. 19.5

K s 0,2 , and: 2 CT K s 0,2 (1 + K1[OH–] + K2[OH–]2 + K3[OH–]3) C 2 T

Here, CT is the total carbonate concentration. The resulting Fe(+II)T is shown in Figure 19.5B for CT = 10–4 M. At what pH are both solids present? The key pH is the pH where both solids give the K s 0,1 K s 0,2 same Fe(+II)T. This occurs where or pH 9.26 for CT = 10–4 M. Another 2 [OH ] 2 CT

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476   Chapter 19  Solid–Liquid Equilibria way to determine the pH where both solids exist is explored in Worked Example 19.3. The overall Fe(+II)T is shown in Figure 19.5C. Note that FeCO3(s) controls at pH < 9.26 since it yields the smallest Fe(+II)T at pH < 9.26. Worked Example 19.3 

Multiple Solid Phases

For the Fe(+II) example in the text, find the pH where the transition between the controlling solids occurs. Solution A simple way to find the transition point is to construct an equilibrium between the two solid phases. Reversing eq. 19.5 and adding it to eq. 19.4: Fe(OH)2(s) + CO32– = FeCO3(s) + 2OH– K=

K s 0,1 = 10–4.7 K s 0,2

Since the activities of the two solids are both 1: K=

[OH ]2 [CO32 ]

KW2 2 2 CT [ H ]

You can solve this by trial and error or simplify. At high pH, α2 ≈ K 2 = 10–10.3. Substituting and expanding:

K2 , [H ] K2

KK2'CT[H+]2 – KW2[H+] – K2'KW2 = 0 Solving with the quadratic equation and CT = 1×10–4 M gives [H ] = 5.47×10–10 or pH 9.26. +

Figures such as Figure 19.5 sometimes are called stability diagrams. The dark lines show the stability regions (i.e., the conditions under which each solid controls the solubility).

19.4.3  Effects of Other Ligands on Metal Solubility Suppose that the concentration of a free metal ion is controlled by one solid. Does the soluble metal concentration increase or decrease if a different ligand is added to the system? As an example, suppose that the solubility of Cu(+II) is controlled by Cu(OH)2(s). The question before you is whether the total soluble Cu(+II)T concentration will increase or decrease if, say, chloride is added to the system. In the most general case, you must solve the new equilibrium system to assess whether Cu(+II)T increases [that is, Cu(OH)2(s) dissolves] or decreases [that is, Cu(+II) precipitates] upon addition of chloride. If we narrow the problem a bit, then you can answer the question at hand without doing any calculations. Suppose that only a small

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19.4  |  Factors Affecting Metal Solubility   477

FIGURE 19.5  Total Soluble Fe(+II) in Equilibrium with (A) Fe(OH)2(s), (B) FeCO3(s), and (C) Both Solids

amount of chloride is added so that the pH and total ion concentration do not change appreciably. Also, assume that so little chloride is added that no new coppercontaining solid species form. In the absence of chloride, the total soluble copper is likely made up of four species: Cu(+II)T = [Cu2+] + [CuOH+] + [CuOH)20] + [Cu(OH)42–] [For simplicity, the dimer Cu2(OH)40 and bicarbonate/carbonate complexes are ignored here.] You can rewrite this sum by expressing each species concentration in

← Worked example

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478   Chapter 19  Solid–Liquid Equilibria terms of [Cu2+] and remembering that [Cu2+] = Cu(+II) solubility: Cu(+II)T =

K s0 if Cu(OH)2(s) controls the [OH ]2

K s0 (1 + K1[OH–] + . . .) [OH ]2

If chloride is added, then a new soluble Cu(+II) species is introduced: CuCl+. This species equilibrates with Cu2+ as follows: Cu2+ + Cl– = CuCl+   KCl = 10+0.5 The expression for total soluble copper becomes: Cu(+II)T = [Cu2+] + [CuOH+] + [CuOH)20] + Cu(OH)42–] + [CuCl+] =

Worked example → Key idea: Adding a new ligand usually increases the solubility of metals, assuming that the same solid phase controls before and after ligand addition and the pH and other water quality characteristics (ligand concentrations) do not change ­appreciably

K s0 (1 + K1[OH–] + . . .+ KCl[Cl–]) [OH ]2

Thus, if Cu(OH)2(s) still controls the copper solubility and if the pH does not change, then the total soluble copper concentration in the presence of chloride must be larger than the total soluble copper concentration in the absence of chloride. Why? A new species has been added and the concentrations of the hydroxide complexes are unchanged. In general, adding a new ligand increases the solubility of metals (assuming that the same solid phase controls before and after ligand addition and pH and other water quality characteristics do not change appreciably).4 Of course, you can show quantitatively that ligands increase metal solubility. As an example, consider the use of cyanide to increase the solubility of metals in metal plating operations. What is the effect on copper solubility at pH 8 if 1×10–4 M of total cyanide is added? In the absence of cyanide, the pertinent equilibria are: H2O = H+ + OH–

KW = 10–14

Cu2+ + OH– = CuOH+

K1 = 10+6.2

Cu2+ + 2OH– = Cu(OH)20

K2 = 10+11.8

Cu2+ + 4OH– = Cu(OH)42–

K4 = 10+16.5

Cu(OH)2(s) = Cu2+ + 2OH–

Ks0 = 10–19.3

In metal plating, it is desirable to maintain the highest possible dissolved metal concentration. This means operating at the solubility limit. The total soluble copper concentration is: Cu(+II)T = [Cu2+] + [CuOH+] + [CuOH)20] + Cu(OH)42–]

4   We can be more precise here. The assumption is that the same solid phase controls before and after ligand addition and all other ligand concentrations stay the same. The pH is important because OH– is such an important ligand.

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19.4  |  Factors Affecting Metal Solubility   479

If Cu(OH)2 (s) controls the copper solubility, then: Cu(+II)T = [Cu2+](1 + K1[OH–] + K2[OH–]2 + K4[OH–]4) =

K s0 (1 + K1[OH–] + K2[OH–]2 + K4[OH–]4) [OH ]2

= 1.8×10–7 M at pH 8 Thus, only about 1.8×10–7 M = 12 μg/L of total copper can be maintained at pH 8. If cyanide is added, the following equilibria also must be considered: Cu2+ + 2CN– = Cu(CN)30

K2,CN = 10+16

Cu2+ + 3CN– = Cu(CN)3–

K3,CN = 10+21.6

Cu2+ + 4CN– = CuCN)42–

K4,CN = 10+23.1

HCN = H+ + CN–

Ka = 10–9.2

In the presence of cyanide, the total soluble copper concentration is: Cu(+II)T = [Cu2+] + [CuOH+] + [CuOH)20] + Cu(OH)42–] + [CuCN)20] + [CuCN)3–] + [CuCN)42–] Again, if Cu(OH)2(s) controls the copper solubility, then: Cu(+II)T = [Cu2+](1 + K1[OH–] + K2[OH–]2 + K4[OH–]4 + K2,CN[CN–]2 + K3,CN[CN–]3 + K4,CN[CN–]4) =

K s0 (1 + K1[OH–] + K2[OH–]2 + K4[OH–]4 [OH ]2 + K2,CN[CN–]2 + K3,CN[CN–]3 + K4,CN[CN–]4)

Writing a mass balance for cyanide: CNT = [CN–] + [HCN] + 2[CuCN)20] + 3[CuCN)3–] + 4[CuCN)42–] = [CN–](1 + [H ] + 2K2,CN[Cu2+][CN–] + 3K3,CN[Cu2+][CN–]2 Ka + 4K4,CN[Cu2+][CN–]3) If no cyanide-­containing species precipitate, then CNT = 1×10–4 M. At pH 8, [OH–] K s0 = 10–6 M and [Cu2+] = = 10–7.3 M. Solving the cyanide mass balance for [CN–] [OH ]2

ISTUDY

480   Chapter 19  Solid–Liquid Equilibria (by using a spreadsheet and guesses or Solver): [CN–] = 2.84×10–7 M. Thus, Cu(+II)T = 4.53×10–5 M or 2.9 mg/L of dissolved copper. Note that copper is about 250 times more soluble in the presence of 1×10–4 M of total cyanide at pH 8. In the presence of 1×10–3 M of total cyanide at pH 8, copper is about 2400 times more soluble than in the absence of cyanide. The addition of another ligand can indeed make a metal more soluble. The addition of cyanide also has changed the speciation of dissolved copper. In the absence of cyanide, the dissolved copper is primarily CuOH+, Cu2+, and Cu(OH)20. In the presence of cyanide, the dissolved copper is primarily Cu(CN)3– and Cu(CN)20.

19.5  SOLUBILITY OF CALCIUM CARBONATE 19.5.1  Introduction Key idea: Calcite (pKs0 = 8.48) is the most common mineral form of calcium carbonate, CaCO3(s)

Calcium carbonate, CaCO3, is an important solid phase in natural waters. Calcium carbonate has two common mineral forms: calcite (from the Latin calx lime) and aragonite (named after the Spanish province, Aragon, where it was first found). Calcium carbonate is the main constituent of several rocks, including limestone (at least 50% CaCO3), marble (recrystallized calcium carbonate), and chalk. The pKs0 values for the mineral forms are 8.48 for calcite and 8.34 for aragonite. Calcite is the most common mineral form. Calcium carbonate dissolves to form calcium and carbonate:

CaCO3 s

Ca 2

 CO32 K s 0

10  

8.48

calcite

eq. 19.6

Calcium carbonate is important in aquatic chemistry because it influences several common water quality parameters.

Thoughtful Pause What common water quality parameters are influenced by Ca2+ and CO32–?

hardness (or total hardness): the sum of the concentrations of the divalent cations in water (usually about equal to the sum of the calcium and magnesium ion concentrations) carbonate (or temporary) hardness: hardness from calcium carbonate (called temporary because it can be removed by merely raising the pH)

From eq. 19.6, you can see that calcium carbonate serves as a source of both alkalinity (from the cation Ca2+) and CT (from CO32–). How does Ca2+ affect alkalinity? Recall from Section 13.3 that alkalinity comes from a charge balance. Thus, any cations or anions (that do not cancel each other out) will influence alkalinity. Remember also that pH, alkalinity, and CT are interrelated (see Section 13.5). Thus, you might expect that the dissolution or precipitation of calcium carbonate also may affect the primary variable pH. The calcium ion also affects hardness. Hardness (also called total hardness) is the sum of the concentrations of the divalent cations in water, principally calcium ion and magnesium ion. Thus, in most waters: hardness ≈ [ Ca2+] + [Mg2+]. Hardness typically is expressed in units of mg/L as CaCO3. Here, 1 mmol of Ca2+ and 1 mmol of Mg2+ are each equivalent to 1 mmol of CaCO3 or 100 mg/L as CaCO3. The hardness from calcium is called calcium hardness, and the hardness from magnesium is called magnesium hardness. Hardness from calcium carbonate is called carbonate or temporary hardness. It is temporary because it can be removed by merely raising the pH.

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19.5  |  Solubility of Calcium Carbonate   481

Thoughtful Pause Why can temporary hardness be removed by raising the pH? If Ca2+ comes from calcium carbonate, then the CT must be at least as large as the calcium ion concentration. You can see this from the equilibrium in eq.  19.6. Thus, raising the pH will convert the CT to carbonate and allow for precipitation of most of the calcium as calcium carbonate. Hardness also can come from calcium sulfate, CaSO4(s), often found in the form of the mineral gypsum. Hardness from calcium sulfate is called noncarbonate or ­permanent hardness. It is called permanent hardness because it cannot be removed by merely raising the pH. Why? If Ca2+ comes from calcium sulfate, then the CT may not be at least as large as the calcium ion concentration. Thus, raising the pH, although it will still convert the CT to carbonate, may not provide enough carbonate to precipitate most of the calcium as calcium carbonate. There is a natural relationship between alkalinity and hardness. Consider a water that has no CA and no CB other than Ca2+.

noncarbonate (or permanent) hardness: hardness from calcium sulfate (called permanent because it cannot be removed by merely raising the pH)

Thoughtful Pause What can you say about the relationship between Alk and hardness in this water?

In this water, all the Alk and all the hardness comes from Ca2+. There are no other divalent cations (no other CB), so the total hardness, carbonate hardness, and Alk are all equal.5

19.5.2  Equilibrium Calculations with Calcium Carbonate In many natural systems, calcium carbonate solid exists. If calcium carbonate exists, then its solubility and the resulting dissolved calcium concentration are pH dependent. You can show the pH dependence easily. From eq. 19.6: [Ca2+][CO32–] = Ks0 = 10–8.48, so: [Ca2+] =

Ks0 [CO32 ]

Ks0 2 CT

  The units here can be confusing at first. For this special water, total hardness = carbonate hardness = [Ca2+] and

5

Alk = 2[Ca ]. Suppose [Ca ] = 1 mM. The carbonate hardness is (1 mM) 2+

meq The Alk is 2 mmol

2+

mmol 1 L

50

mg as CaCO3 L = 100 mg/L as CaCO3. meq L

100

mg as CaCO3 L = 100 mg/L as CaCO3. mM Ca

Key idea: If calcium carbonate exists, then its solubility and the resulting dissolved calcium concentration are pH dependent

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482   Chapter 19  Solid–Liquid Equilibria

FIGURE 19.6  Total Soluble Calcium in Equilibrium with CaCO3(s) (CT = 1×10–3 M)

low pH

Key idea: Calcium carbonate is very soluble at

Worked example →

If CT is fixed, you can calculate [Ca2+] as a function of pH. (Remember that for the carbonate system, pKa,1 = 6.3 and pKa,2= 10.3). The calcium solubility is shown in Figure 19.6 for CT fixed at 10–3 M. The solid line is the total soluble calcium concentration.6 Note the changes in slope in the total soluble calcium line at pH 6.3 and 10.3 (caused by the changes in the speciation of dissolved carbonate). You can interpret Figure 19.6 as you have previously. “Inside” the line (i.e., at larger pH and smaller pC values than points on the line), calcium carbonate precipitates. For example, you can see that calcium carbonate would precipitate at pH 8 and [Ca2+] = 10–2 M (point A, Figure 19.6), but it would not precipitate at pH 7 and [Ca2+] = 10–4 M (point B, Figure 19.6). Figure 19.6 tells you that calcium carbonate is very soluble at low pH. Why? At low pH, the carbonate ion concentration is small. Therefore, CaCO3(s) dissolves to  increase the concentration of Ca2+ and maintain the product [Ca2+][CO32–] at Ks0 = 10–8.48. This phenomenon explains a number of observations in the environment. For example, acid rain is known to dissolve limestone monuments in a sort of titration (see Section 12.1 and the Historical Note in Chapter 10 for examples). The low pH of the acid rain solubilizes CaCO3(s). On the other hand, CaCO3(s) precipitates at higher pH values. An example can be found in groundwater remediation. Environmental engineers sometimes use reactive barriers to treat groundwater passively. In some reactive barriers, zero-­valent iron, Fe(0), is used as an electron source (i.e., a reducing agent) to reduce chlorinated organics. As a result of the redox chemistry, the pH increases. Since some groundwaters have appreciable calcium concentrations from equilibrium with CaCO3(s), an increase in the pH may result in calcium precipitation and may clog the reactive barrier. Another equilibrium calculation with calcium carbonate is given in Worked Example 19.4.

6   In this calculation, the species CaOH+ has been ignored. CaOH+ is present in significant concentrations relative to [Ca2+] only above about pH 12.

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19.5  |  Solubility of Calcium Carbonate   483 Worked Example 19.4 

Langlier Index

One way to determine whether a water is undersaturated, saturated, or supersaturated with respect to calcium carbonate is to calculate the Langlier index (also called the LI, the saturation index, and the SI). The Langlier index is the difference between the observed pH and the pH at saturation for a closed system at the observed alkalinity and calcium ion concentration: LI = pHobs – pHsat. Calculate the Langlier index for a water with Alk = 250 mg/L as CaCO3, [Ca2+] = 50 mg/L as Ca, and pH 7.15. Will CaCO3(s) tend to dissolve or ­precipitate in this water? Solution For this water, Alk = 5 meq/L, and [Ca2+] = 1.25 mM. At pH = pHsat (in Ks0 Ks0 equilibrium with calcium carbonate): [Ca2+] = , so: α2 = [CO32 ] 2 CT Ks0 . You can find an expression for CT from the Alk equation: Alk = [Ca 2 ]CT KW Alk [H ] [ H] (α1 + 2α2)CT + [OH–] – [H+]. So: CT = . Substituting, α2 = 2 2 1 [H ] Ks0 2 Ks0 1 2 2 K2 . Dividing by α2: 1 = . RearKW KW 2 2 [Ca ] Alk [H ] [Ca ] Alk [H ] [H ] [H ] ranging, pHsat is the solution to the quadratic: a[H+]2 + b[H+] + c = 0, where: a = K2[Ca2+] – Ks0 b = K2([Ca2+]Alk – 2K s0) c = –K2[Ca2+]KW For the water in question, pHsat is calculated to be 7.02. Thus: LI = 7.15 – 7.02 = +0.13. The water is supersaturated with calcium carbonate (which raises the pH above pHsat), and thus calcium carbonate will precipitate. In general, it is desirable to have a slightly positive LI in a drinking water distribution system so that a protective layer of calcium carbonate will form on the pipes and corrosion will be inhibited. To summarize: LI = +0.13 and calcium carbonate will precipitate.

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484   Chapter 19  Solid–Liquid Equilibria

19.6  MODELS FOR THE ACID–BASE

CHEMISTRY OF NATURAL WATERS

19.6.1  Introduction

CB': the sum in eq/L of all strong base cations except for calcium ion: CB' = CB – 2[Ca2+]

With the addition of calcium carbonate chemistry, you now can create more realistic models of the aquatic environment. In this section, all the pertinent chemistries of aquatic systems will be examined, including open and closed systems, systems with and without calcium carbonate, and systems with and without strong acid anions and strong base cations other than the calcium ion. This analysis will take you from the simplest system (closed to the atmosphere, no solids, and no ions other than H+ and OH–, i.e., pure water) to a general model for surface waters. The cases are outlined in Table 19.1. You know that Ca2+ will dissolve from CaCO3(s). This, of course, contributes to the alkalinity. It will be convenient to think about the alkalinity other than that from calcium ion. Recall the CB is the sum of all the strong base cations in equivalence per liter (concentration×charge). Define CB' as the sum in eq/L of all strong base cations except for calcium ion: CB' = CB – 2[Ca2+]. Waters for which CB' and CA are 0 will be called “clean.” (The required condition is not so strict. It is required only that CB' – CA = 0.) These waters are clean in the sense that other strong base cation and strong acid anions either do not exist or cancel out. Waters where CB' – CA ≠ 0 will be labeled “contaminated.” One final comment before jumping into the calculations: The key to the calculations is the charge balance. For each case, try to write the charge balance on your own. The goal of the calculations is to find the chemical conditions where the charge balances are satisfied.

19.6.2  Pure Water (Closed) The pure water model is closed to the atmosphere with no solids and no CB or CA. The charge balance is: Worked example →

[H+] = [OH–] Table 19.1 Cases for the Acid–Base Chemistry of Natural Waters Description

Pure water (closed) Contaminated water (closed) Clean rainwater Contaminated rainwater Clean groundwater Contaminated groundwater Clean surface water Contaminated surface water

Open?

CaCO3(s)?

No No Yes Yes No No Yes Yes

No No No No Yes Yes Yes Yes

CB' – CA = 0?

Yes No Yes No Yes No Yes No

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19.6  |  Models for the Acid–Base Chemistry of Natural Waters   485

Table 19.2 Results from the Clean Cases in Table 19.1 (PCO = 10–3.37 atm) 2

  Description

  pH

CT (M)

Alk (eq/L)

Pure water Clean rainwater Clean groundwater Clean surface water

7.00 5.58 9.89 8.15

0 1.61×10–5 1.09×10–4 9.78×10–4

0 0 2.17×10–4 8.83×10–4

[Ca2+] (M)

0 0 1.09×10–4 4.87×10–4

Solving with the KW expression gives pH 7. Of course, CT = 0, Alk = [OH–] – [H+] = 0, and [Ca2+] = 0. Results from this and the other cases with unique solutions are summarized in Table 19.2.

19.6.3  Contaminated Water (Closed) For this system, the charge balance can be rearranged to: CB – CA = (α1 + 2α2)CT + [OH–] – [H+] where Alk = CB  – CA. Numerical answers can be obtained if the alkalinity is  known,  as discussed in Chapter  13. For the general case in a closed system, see ­Figures 13.8 and 13.9.

19.6.4  Clean Rainwater The clean rainwater case (water in equilibrium with the atmosphere) was covered in Section 18.4.2. The charge balance can be rearranged to: 0 = (α1 + 2α2)CT + [OH–] – [H+]. From the work you did in Chapter  18: [H2CO3*] = α0CT = KHPCO2 . Thus, CT = K H PCO2 . The charge balance is: 0

0 = (α1 + 2α2)

K H PCO2

+ [OH–] – [H+]

0

This is one equation in one unknown, [H+]. The results are (for PCO2 = 10–3.37 atm): pH 5.58, CT = 1.61×10–5 M, Alk = 0, and [Ca2+] = 0. It is good practice to be able to explain the changes when the system is opened.

Thoughtful Pause Explain the changes (if any) in pH, CT, and Alk as you open up the system. When you open up the water, CO2(g) dissolves from the atmosphere. Dissolved carbon dioxide reacts with water to form carbonic acid, lowering the pH. You also can

← Worked example

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486   Chapter 19  Solid–Liquid Equilibria think about this as follows: Adding CO2(g) generates new anions (bicarbonate and carbonate), so [H+] (the only cation) must increase to maintain the charge balance, thereby decreasing the pH. The alkalinity stays at zero since only H2CO3* (the zero level of alkalinity) is exchanged.

19.6.5  Contaminated Rainwater In this case, the rainwater is in equilibrium with the atmosphere and now contains strong base cations and strong acid anions. The charge balance becomes:

CB C A

1

 2

2

K H PCO2 0

   OH [ ] [H +]

eq. 19.7

In this case, the pH (and therefore CT) depends on Alk = CB – CA. As an example, the average rainwater pH in the United States was reported as 5.10 (Keresztesi et al. 2020). Plugging this value into eq. 19.7 with PCO2 = 10–3.37 atm gives an alkalinity of –7.09×10–3 eq/L (–355 mg/L as CaCO3). The corresponding CT is

K H PCO2

= 1.43×10–5 M.

0

The relationships between pH, CT, and Alk in contaminated rainwater are shown in Figure 19.7 for PCO2 = 10–3.37 atm. Note that Alk < 0 is required for pH values below 5.58 (Section 19.6.4). For more on the mathematics and graphical approaches to the equations in this section, see the addendum at the end of this chapter.

19.6.6  Clean Groundwater

Worked example →

Clean groundwater is in equilibrium with CaCO3(s), not in contact with the atmosphere, and has CB' – CA = 0. In other words, this system is pure water equilibrated with CaCO3(s). The charge balance becomes: 2[Ca2+] = (α1 + 2α2)CT + [OH–] – [H+]

FIGURE 19.7  Alkalinity and Total Dissolved Carbonate for Contaminated Rainwater

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19.6  |  Models for the Acid–Base Chemistry of Natural Waters   487

Ks0 Ks0 . In this special case, [Ca2+] 2 [CO3 ] 2 CT = CT , since all the calcium and all the carbonate comes from CaCO3(s). Why? The system is closed to the atmosphere and contains no solids other than CaCO3(s). Thus: K K s0 [Ca2+] = CT = s 0 . Solving for CT: CT = [Ca2+] = . The charge balance becomes: 2 CT 2 From the solubility constraint: [Ca2+] =

2 

K s0 2

= (α1 + 2α2)

K s0

+ [OH–] – [H+]

2

This equation looks messy, but it is one equation in one unknown, [H+]. Solving: pH 9.89, CT = [Ca2+] = 1.09×10–4 M, and Alk = 2[Ca2+] = 2.17×10–4 M eq/L. Recall that for pure water, the pH is 7.0 and CT = [Ca2+] = Alk = 0.

Thoughtful Pause Explain the changes in pH, CT , and Alk as pure water equilibrates with CaCO3(s). Equilibrating with CaCO3(s) will cause Ca2+ to dissolve, increasing the alkalinity. The charge balance must shift since a cation is being added. Therefore, [H+] must decrease and the pH increases. Carbonate dissolves from calcium carbonate, so CT increases.

19.6.7  Contaminated Groundwater As you know, most groundwaters do not have pH values near 9.9. This means that most groundwaters do have some additional acid inputs and thus some CA in excess over CB'. Consider a general groundwater in equilibrium with CaCO3(s), not in contact with the atmosphere, and where CB' – CA ≠ 0. The charge balance becomes: CB' – CA + 2[Ca2+] = (α1 + 2α2)CT + [OH–] – [H+] In this case, [Ca2+] may or may not be equal to CT , since there may be sources of dissolved carbonate and/or calcium other than CaCO3(s). In any event, the solubility constraint still holds if CaCO3(s) controls the solubility of calcium: Ks0 Ks0 [Ca2+] = . Thus: [CO32 ] 2 CT Alk = CB' – CA + 2

Ks0 = (α1 + 2α2)CT + [OH–] – [H+] C 2 T

This is as far as you can go without knowing CB' – CA and CT . In general, pH, CT , [Ca2+], and Alk are functions of CB'  – CA. In the presence of CaCO3(s), specifying two of Alk, CB'  – CA, pH, CT , and [Ca2+] will allow you to calculate the other three parameters (see the chapter addendum for details). Here is an

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488   Chapter 19  Solid–Liquid Equilibria example. Suppose you analyze a groundwater sample and find that it is at pH 7 and has a calcium ion concentration of 80 mg/L (2×10–3 M). In other words, you know pH and K K s0 [Ca2+]. You can find CT by rearranging [Ca2+] = s 0 so: CT = = 3.96×10–3 M. 2 2 [Ca ] 2 CT You can calculate the alkalinity from pH and CT in the usual way: Alk = 3.31×10–3 eq/L (165 mg/L as CaCO3). And finally, you can calculate CB' – CA from Alk – 2[Ca 2 ] = –6.93×10–4 eq/L (–35 mg/L as CaCO3).

Thoughtful Pause What does the negative value of CB’ – CA mean? The negative value of CB' – CA means that the sum of the strong acid anions, CA, exceeds the sum of the strong base cations other than the calcium ion, CB'. In other words, the groundwater has an excess of strong acid anions over strong base cations other than the calcium ion. Negative values of CB' – CA are not unusual.

19.6.8  Clean Surface Water Worked example →

Clean surface water is in equilibrium with both the atmosphere and CaCO3(s) and has CB' – CA = 0. For this system, the charge balance becomes: 2[Ca2+] = (α1 + 2α2)CT + [OH–] – [H+]. The atmospheric carbon dioxide fixes the concentration of H2CO3* and K K P CT = H CO2 . From the solubility constraint: [Ca2+] = s 0 . Combining these two 0 2 CT Ks0 0 expressions: [Ca2+] = . The charge balance becomes: 2 K H PCO2 2

K P Ks0 0 = (α1 + 2α2) H CO2 + [OH–] – [H+] K P 2 H CO2 0

With PCO2 = 10–3.37 atm, you can calculate: pH 8.15, CT = 9.78×10–4 M, [Ca2+] = 4.87×10–4 M, and Alk = 2[Ca2+] = 8.83×10–4 eq/L. You can think of clean surface water as clean groundwater opened up to the atmosphere. Recall that clean groundwater had pH 9.89, CT = [Ca2+] = 1.09×10–4 M, and Alk = 2.17×10–4 M eq/L.

Thoughtful Pause Explain the changes in pH, CT, and Alk as clean groundwater is exposed to the atmosphere.

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19.6  |  Models for the Acid–Base Chemistry of Natural Waters   489

Equilibrating with the atmosphere causes CO2(g) to dissolve, increasing CT and lowering the pH (with the addition of H2CO3*). At the lower pH, CaCO3(s) is more soluble, so [Ca2+] increase and therefore the alkalinity increases. Before leaving this case, note that you can get a very good answer with some reasonable approximations. If you are anywhere near neutral pH (where [H+], [CO32–], and [OH–] can be ignored), the charge balance is approximated by 2[Ca2+] ≈ [HCO3–]. You K P K a,1 know [HCO3–] = α1CT = 1 H CO2 . Since 1 for any diprotic acid: [HCO3–] = [ H ] 0 0 K1 K H PCO2 . Substituting in the expressions for [Ca2+] and [HCO3–]: [H ]

2

K1 K H PCO2

K s 0 [H]2 K1 K 2 K H PCO2

[H ]

Simplifying: 2 K12 K 2 K H2 PCO 2

[H ]3



2Ks0



At PCO2 = 10–3.37 atm:

[H ] ≈ + 3

≈

10

10

6.3

24.16

2

M

2

M3

10

10.3

10

mol L-atm 10 8.48 M2

M 10

24.46

1.5

2

10

3.37

atm

2

M3

24.46

So [H+] ≈ 10 3 M or 10–8.15 M or about pH 8.15. If you remembered that log(2) is about 0.30, you can do this calculation without a calculator: pH ≈ 2 6.3 10.3 2 1.5 2 3.37 0.30 8.48 8.15. This is the exact answer to two 3 ­decimal places because the assumptions are so good (i.e., [H+], [CO32–], and [OH–] are small relative to [HCO3–].)

19.6.9  Contaminated Surface Water This is the most general case: in equilibrium with both the atmosphere and CaCO3(s), with CB' – CA ≠ 0. The charge balance is: Alk = CB' – CA + 2

K P Ks0 0 = (α1 + 2α2) H CO2 + [OH–] – [H+] 2 K H PCO2 0

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490   Chapter 19  Solid–Liquid Equilibria

FIGURE 19.8  Alkalinity, CB' – CA, Calcium Ion Concentration, and Total Dissolved Carbonate for Contaminated Surface Water

In this case, pH, CT , [Ca2+], and Alk are functions of CB' – CA. If you know one of CB' – CA, pH, CT, and [Ca2+], then you can calculate the other four parameters plus the alkalinity. For example, suppose you measure the pH of a lake to be 7.8 and you suspect that the water is in equilibrium with both the atmosphere and calcium carbonate. You can K P Ks0 0 calculate CT = H CO2 = 0.44 mM, [Ca2+] = = 2.45 mM, CB' – CA = (α1 + 2α2) 2 K H PCO2 0 Ks0 0 K H PCO2 + [OH–] – [H+] – = –4.48 meq/L, and Alk = CB' – CA + 2[Ca2+] = K P 2 H CO2 0 0.43 meq/L. The general results are plotted in Figure 19.8. You can double-­check the example at pH 7.8 using Figure 19.8.

19.6.10  Importance of the Clean Cases

Worked example →

The calculated values in Table 19.2 are useful in making assessments of water quality. Consider the groundwater from Section 19.6.7: pH 7, [Ca2+] = 2 mM, CT = 3.96 mM, and Alk = 3.31 meq/L. Is this a “clean” groundwater? To answer this question quickly, compare its water quality parameters to those of clean groundwater in Table 19.2. It is clear that the groundwater in question has a lower pH, larger calcium ion concentration, larger alkalinity, and larger CT than groundwater with CB' – CA = 0. Since the pH is lower than the clean model system, you know that CB' – CA < 0. The groundwater is not “clean.” The excess strong acid anions would lower the pH, dissolving more CaCO3(s) and therefore increasing [Ca2+], Alk, and CT . As another example, consider the surface water from Section 19.6.9: pH 7.8, [Ca2+] = 2.45 mM, CT = 0.44 mM, and Alk = 0.43 meq/L. Is this a “clean” surface water? Again, Table 19.2 helps. Compared to the clean surface water, the lake in question has lower pH, slightly lower calcium ion concentration, lower CT , and smaller alkalinity. Since the pH is lower than the clean model system, you know that CB' – CA  1000 bars = 987 atm). Such conditions may be found in the ocean. However, in the ocean (even at 1000 bars), the effects of ionic strength and temperature on equilibrium constants can exceed the effects of pressure on equilibrium constants. To show the effects of pressure on K, reconsider eq. 21.8: dG = VdP – SsysdT. This dG d ln K relationship suggests that and therefore should be related to the change in dP dP volume in a system at constant temperature. A more detailed analysis reveals the following relationship at constant T: d ln K



dP

 Vrxn

RT



eq. 21.11

Here, Vrxn is the change in the molar volume of reaction. It can be calculated from the standard partial molar volumes (V°) of the reactants and products. The standard molar volume is the contribution of a species (per mole) to the volume of solution under standard conditions. At low pressures, Vrxn is relatively constant. The integrated form of eq. 21.11 is:

KP

K P 1 exp

 Vrxn P 1 RT

eq. 21.12

In eq. 21.12, KP is the equilibrium constant at pressure P and KP=1 is the equilibrium constant at 1 atm.

21.4.2  Examples of the Effects of Pressure on K  A few examples will demonstrate the small influence of pressure on most equilibrium constants. Consider the self-­ionization of water: H2O = H+ + OH–. The V° values for H2O, H+, and OH– are 18.07, 0, and –3.98 cm3/mol, respectively (Byrne and Laurie 1999).

 The calculations summarized in Figure 21.2 used a pKs0 value for amorphous Al(OH)3(s) rather than the pKs0value for α-­Al(OH)3(s) from Appendix D. Using a pKs0 for the amphorous solid and adjusting for experimental data, Edzwald (2020) found minimum Al(+III)T values at pH 6.7 at 5°C and 6.3 at 20°C. Van Benschoten et al. (1994) found fitted values from field data at drinking water treatment plants for pKs0 and Hrxn of 32.3 and 15.9 kJ/mol, respectively. 2

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21.5  |  Chapter Summary   529

Table 21.5 ΔV°rxnValues for Selected Weak Acids Equilibrium

ΔV°rxn (cm3/mol)

H2O = H+ + OH– H2CO3 = H+ + HCO3– HCO3– = H+ + CO32– H2S = H+ + HS– NH4+ = H+ + NH3

–22.05 –27.4 –28.1 –15.0 +6.9

Key idea: The change in lnK with pressure is related to the change in the molar volume of reaction, Vrxn

(Source: Adapted from Byrne and Laurie 1999; 25°C, 1 atm)

So: Vrxn = VH

VOH

VH

2O

= 0 + (–3.98) – 18.07 = –22.0 cm3/mol = –0.022 L/mol.

You can evaluate KW at 2 atm:

KP KP

2 1

L (2 atm 1 atm ) mol exp L-atm (0.082057 (298.115 K) mol-K ( 0.022

1.000

Clearly, pressure has little effect on the KW equilibria. Assume for a moment that the V° values are still valid at higher pressures. At the mean depth of Lake Superior (about 15  atm), KW is about 1.01 times larger than at the surface (all at 25°C). The average depth of the sea floor is about 4000 m or about 400 atm. At 400 atm and at 25°C, KW is about 1.4 times its value at 1 atm. Owen and Brinkley (1941) reported that KW at 2000 bar was 2.36 times its value at 1 bar. As you can see, pressure corrections are important only for the oceans. Table 21.5 lists several Vrxn values for some weak acids. Solids in general are more soluble at higher pressures. Bryne and Laurie (1999) noted a linear increase in the lnKs0 of solids up to almost 1,000 atm in experimental work. Millero (1982) found that the Ks0 values for calcite and aragonite increased about 1.7-­to 1.9-­fold from 1 bar to 1000 bar at 2°C at the ionic strength of seawater.

21.5  CHAPTER SUMMARY In this chapter, the effects of the medium composition on the equilibrium position of a system were determined. An important measure of the medium composition is the ionic strength: I =

1 2

n

i 1

zi2Ci , where zi and Ci are the charge on species i and concentra-

tion of species i, respectively. The concentration (Ci) and activity (ai) of a species are related through an activity coefficient, γi: ai = γiCi. The single-­ion activity coefficient can be estimated through several approximations. In general, increasing I decreases the activity of ions. Increasing I generally increases the activity of uncharged species slightly. The latter process is called salting out. Once the single-­species activity coefficients are calculated, the system can be solved at equilibrium. The usual process is followed, except activity coefficients are used to express all equilibrium constants in terms of concentrations rather than activities.

←Worked example

Key idea: The effects of pressure on equilibrium constants usually are small

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530   Chapter 21  Thermodynamics Revisited: The Effects of Ionic Strength, Temperature, and Pressure Thermodynamic arguments were used to derive relationships between equilibrium constants and temperature. The dependency of K on temperature is described by the van’t Hoff equation. The integrated form of the van’t Hoff equation is K Hrxn 1 1 , where K and K are the equilibrium constants at temperaln 2 2 1 K1 R T1 T2 tures T2 and T1, respectively. This form of the equation is valid only if Hrxn is fairly independent of temperature. According to the van’t Hoff equation, endothermic reactions will show increasing equilibrium constants with temperature and exothermic reactions will show decreasing equilibrium constants with temperature. With the van’t Hoff equation, equilibrium constant values can be calculated at temperatures other than 25°C. Some solids, including CaCO3(s), show higher solubility at lower temperatures. The effects of pressure on equilibrium constants also was investigated in this chapter. Equilibrium constants were found to vary with pressure as follows:  Vrxn P 1 , where KP is the equilibrium constant at pressure P and K P K P 1 exp RT KP=1 is the equilibrium constant at 1 atm. This equation is valid only if Vrxn is independent of pressure over the pressure range of interest. In general, pressure exerts only a small effect on equilibrium constants. Solids are more soluble at high pressures.

CHAPTER KEY IDEAS • Chemical species may behave differently under different environmental conditions, even at the same mass or molar concentration. • Even if two species do not react with one another, the behavior (i.e., the activity) of each species will be influenced by the other species. • Concentration (Ci) and activity (ai) of a species are related through an activity coefficient, γi: ai = γiCi. • At higher ionic strength, the single-­ion activity coefficient de­ creases (i.e., the activity of the ions decreases as the ions of interest are constrained by other ions). • Always use single-­ion activity coefficient approximations only over the ionic strength range for which they are valid. • At higher ionic strength, the single-­species activity coefficients for uncharged species usually increase. • Equilibrium expressions are written in terms of activities, but charge and mass balances are written in terms of concentrations. • Endothermic reactions will show increasing equilibrium constants with temperature and exothermic reactions will show decreasing equilibrium constants with temperature. • For many equilibria of environmental interest and over temperatures of 0°C to 30°C or so, Hrxn is fairly independent of temperature. • Over a temperature range in which Hrxn is independent of temperature,  K Hrxn 1 1 . ln 2 K1 R T1 T2 • Never assume that the solubility of a solid will increase with increasing temperature. • The change in lnK with pressure is related to the change in the molar volume of reaction, Vrxn. • The effects of pressure on an equilibrium constant is given by  Vrxn P 1 K P K P 1 exp , if Vrxn is independent of pressure. RT

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Historical Note: Jacobus Henricus van’t Hoff   531

CHAPTER GLOSSARY K:  an equilibrium constant based on concentrations, not activities Debye-­Hückel limiting law:  an approximation for the single-­ion activity coefficient as a function of ionic strength: log10 i Azi 2 I ,

c

1

where A

0.5

L2

1 mol 2

at 25°C

infinite dilution standard state:  γi → 1 as I → 0 ion size parameter (a):  a parameter used in a common extension of the Debye-­ Hückel limiting law that accounts for the finite size of ions ionic strength (I):  a measure of the concentration of charged species and given by I =

1 2

n

i 1

zi2Ci

salting out:  the tendency of most uncharged species to partition out of water to a larger extent at higher ionic strengths salting-­out coefficient:  the parameter k in the common model for the single-­species activity coefficient for uncharged species: log10 i kI van’t Hoff equation:  the relationship showing how equilibrium constants change Hrxn d ln K with temperature: , where Hrxn is the standard enthalpy dT RT 2 of reaction

HISTORICAL NOTE: JACOBUS HENRICUS VAN’T HOFF Jacobus Henricus van’t Hoff 3 was one of the most creative chemists of his time, leaving his imprint on nearly all branches of chemistry. In organic chemistry, he created the idea of asymmetrical carbon atoms (stereochemistry, in modern terms). The idea evidently came to him during a walk. He published the revolutionary concept in a small pamphlet several months before submitting his doctoral dissertation on cyanoacetic and malonic acids. In physical chemistry, his work on the effects of temperature on the equilibrium position of a system was first published in 1884. The concept was to be expanded by Le Chatelier (see Chapter 4). The next year, van’t Hoff was to make another breakthrough: using osmotic pressure to demonstrate that the ideal gas law can be translated to solutions. This idea, coupled with Arrhenius’s dissociation theory, took chemistry in a new direction. Stereochemistry, solution chemistry: We take these ideas for granted. But chemistry was in a very different place in van’t Hoff’s day. In secondary school (Hogere Burgerschool), he was taught that the chemical formula for water was HO, although later in school he was told that “. . . a new formulation with H2O was beginning to make headway” (Walker 1913).

3   This text has adopted the common English spelling of the name van’t Hoff. However, ‘t is an abbreviation of het (the), so the family name is spelled more properly as “van ‘t Hoff.” It means “from the homestead.”

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532   Chapter 21  Thermodynamics Revisited: The Effects of Ionic Strength, Temperature, and Pressure

Jacobus van’t Hoff (Nicola Perscheid/Wikimedia Commons/Public domain)

It could be argued that van’t Hoff was one of the first environmental chemist. He used laboratory studies to understand an important aquatic chemistry phenomenon: the equilibrium precipitation of salts in the ocean. His work was performed to better understand the salt deposits underlying Stassfurt, Germany. His work was published in 1905 and 1909, the last significant work before his death in 1911. Although van’t Hoff received many honors over the years, his greatest recognition was as recipient of the first Nobel Prize in Chemistry in 1901. Imagine that first Nobel ceremony at the Musical Academy in Stockholm, with honorees van’t Hoff, X-­ray discoverer Wilhelm Conrad von Röntgen, Emil Adolf von Behring (developer of therapies for diphtheria and tuberculosis), and French poet Rene François Armand (Sully) Prudhomme. Evidently, van’t Hoff enjoyed himself in Stockholm. At the banquet following the ceremony, picture this indelible image: “There were many toasts and a splendid ambience. And in the small hours, two marshals carried the little van’t Hoff on a gold chair around the room” (NobelPrize.org n.d.). For one person to have developed such far-­reaching and innovative ideas seems almost beyond belief. During the Nobel Prize ceremony, van’t Hoff’s work was described as the most important work in theoretical chemistry in 100 years. Many people now consider van’t Hoff, along with fellow laureates Arrhenius and Wilhelm Ostwald, to be the founders of physical chemistry.

PROBLEMS Note: Thermodynamic data are listed at the end of the problems. 1. What is the ionic strength of a water of the following chemical composition? 30 mg/L Ca2+, 183 mg/L HCO3–, 35 mg/L Na+, 27 mg/L Cl–, 9 mg/L Fe2+, pH 7.0, 9 mg/L Mg2+, and 15 mg/L SO42– 2. Experimentalists often add sodium perchlorate, NaClO4(s), to control the ionic strength. If sodium perchlorate dissociates completely to sodium and perchlorate, how many grams per liter of the salt would

you have to add to 1 L of water to set the ionic strength at 0.02 M? 3. What is the relationship between I and TDS in a NaCl solution? 4. Show that, for a C M solution of a salt MxYz (which dissolves to form xC moles/L of Mz+ and zC moles/L of Yx–), the ionic strength is ½xz(x + z)C. 5. What is –log[H+] of water in an estuary at pH 8.2 and I = 0.05 M? 6. A chloride-­specific electrode reports the activity of Cl– in a sample as 1.2 mM. Another analytical method

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PROBLEMS   533 gives a chloride ion concentration of 1.1 mM. What is the approximate ionic strength of the sample? Use the Güntelberg approximation and check to see if it is an appropriate model. 7. The following data were collected on the Henry’s law constant for CO2(g) at 25°C (modified from Onda et al. 1970). For I values in NaCl of 0.5, 1.0, 2.0, and 3.0 M, the corresponding values of KH were 0.02964, 0.02650, 0.02102, and 0.01685 mol/L-­atm. Estimate the salting-­out coefficient. 8. Compare the Ks0 values of MnCO3(s) in fresh water (I → 0) and in estuarine waters (I = 0.08 M) at 25°C. If each of the waters contained a concentration of CO32– equal to 10–6 M and MnCO3(s) controls the Mn solubility, what is the Mn2+ concentration in each water? What is the Mn2+ activity in each water?

a. Will CaCO3(s) precipitate in the distribution system? b. Will CaCO3(s) precipitate in a home water heater (at about 140°F = 60°C)? c. Find the lowest temperature at which CaCO3(s) precipitates at equilibrium in the water described above. 14. This problem extends the analysis of the water described in Problem 13. a. Will CaCO3(s) precipitate in a home water heater (60°C) if you consider the effects of temperature on all pertinent equilibria in the system? b. Find the lowest temperature at which CaCO3(s) precipitates you consider the effects of temperature on all pertinent equilibria in the system.

10. At what temperature does the pKa for bicarbonate become 6.4?

15. A chemical reactor containing water is maintained at 50 atm and 25°C. What is KW in the reactor? What is neuL-atm tral pH in the reactor? Use R = 0.082057 mol-K

11. How much stronger or weaker as an acid is H2CO3* at 5°C compared to 25°C?

16. The value of pKa for the equilibrium H2S = HS– + H+ is 7.0 as I → 0, T = 25°C, and P = l atm.

9. What is “neutral” pH at 10°C?

12. Compute the equilibrium constant at 10°C for: MnS(s,pink) + H+ = Mn2+ + HS–.

a. Find the ionic strength (at 25°C and P = 1 atm) needed to decrease the pKa (expressed in concentration) to 6.9. b. Find the temperature (as I → 0 and P = l atm) needed to decrease the pKa (expressed in concentra-

(Hint: First calculate the equilibrium constant at 25°C given that the pKa,1 for H2S at 25°C is 7.0.) Is MnS(s,pink) more or less soluble at higher temperatures? 13. A treated drinking water has the following characteristics: pH 7.5, alkalinity = 75 mg/Las CaCO3, [Ca2+] = 50 mg/L as Ca, and temperature of the distribution system = 77°F (= 25°C). For this problem, ignore ionic strength effects and assume that the only equilibrium affected by temperature is the precipitation of CaCO3(s).

tion) to 6.9. Assume Hrxn at this temperature is

equal to Hrxn at 25°C. c. Find the pressure (as I → 0 and 25°C) needed to decrease the pKa (expressed in ­concentration) to 6.9. Assume Vrxn at this pressure is equal to Vrxn at 1 atm (= –15.0 cm3/mol).

H f values in kJ/mol (from Stumm and Morgan 1996): See Table 21.5 for Vrxn values. Ca2+

–542.83

CO32–

–677.1

H2S

–39.75

CaCO3(s) –1207.4

HCO3–

–692.0

H+

0

HS–

–17.6

MnS(s)

–213.8

H2CO3*

–699.6

H2O(l)

–285.83

OH–

–230.0

Mn2+

–220.7

OH–

–3.98

V° values in cm3/mol (from Bryne and Laurie 1999): Ca2+

–16.83

CO32–

–3.78

H+

0

H2O(l)

18.07

HCO3–

24.29

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534   Chapter 21  Thermodynamics Revisited: The Effects of Ionic Strength, Temperature, and Pressure

CHAPTER REFERENCES Butler, J.N. (1982). Carbon Dioxide Equilibria and Their Applications. Reading, MA: Addison-­Wesley Publ. Co. Byrne, R.H. and S.H. Laurie (1999). Influence of pressure on chemical equilibria in aqueous systems – with particular reference to seawater (Technical Report). Pure Appl. Chem. 71(5): 871–890. Edzwald, J.K. (2020). Aluminum in drinking water: Occurrence, effects, and control. J. Am. Water Works Assoc. 112(5): 34–41. Kielland, J. (1937). Individual activity coefficient of ions in aqueous solutions. J. Amer. Chem. Soc. 59: 1675–1678. Langelier, W.F. (1936). The analytical control of anti–corrosion water treatment. J. Amer. Water Works Assoc. 28(10): 1500–1521. Metcalf and Eddy (2003). Wastewater Engineering: Treatment and Reuse, 4th ed. New York: McGraw-Hill Book Co. Millero, F.J. (1982). The effect of pressure on the solubility of minerals in water and seawater. Geochim. Cosmochim. Acta 46(11): 11–22. NobelPrize.org. (n.d.). From the first Nobel Prize award ceremony, 1901. nobelprize.org/ceremonies/from-­the-­first-­nobel-­prize-­award-­ ceremony-­1901.

Onda, K., E. Aada, T. Kobayashi, S. Kito, and K. Ito. (1970). Salting-­out parameters of gas solubility in aqueous salt solutions. J. Chem. Eng. Japan 3(1): 18–24. Owen, B.B. and S.R. Brinkley (1941). Calculation of the effect of pressure upon ionic equilibria in pure water and in salt solutions. Chem. Rev. 29(3): 461–474. Snoeyink, V.L. and D. Jenkins (1980). Water Chemistry. New York: John Wiley & Sons, Inc. Stumm, W. and J.J. Morgan (1996). Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters, 3rd ed. New  York: John Wiley & Sons, Inc. Sun, M.S., D.K. Harriss, and V.R. Magnuson (1980). Activity corrections for ionic equilibriums in aqueous solutions. Can. J. Chem. 58(12): 1253–1257. van Benschoten, J.E., J.N. Jensen, and M.A. Rahman (1994). Effects of temperature and pH on residual aluminum in alkaline-­treated waters. J. Environ. Eng. 120(3): 543–559. Walker, J. (1913). van’t Hoff Memorial Lecture. J. Chem. Soc. Trans. 103: 1127–1143.

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C H A P T E R   22

Aquatic Chemistry of ­Surfaces 22.1 INTRODUCTION Solids produce surfaces. In Chapter 19, you learned about the equilibrium partitioning between solids and the chemical species that precipitate to form solids. In this chapter, you will learn about the partitioning of dissolved species between water and surfaces. Surfaces control the chemistry of some dissolved species. The focus of this chapter is equilibrium models for the interactions between dissolved species and surfaces important in natural and engineered systems. A number of new terms to describe surfaces and surface–solute interactions are introduced in Section 22.2. Simple equilibrium partitioning models are discussed in Section  22.3, including isotherms and ion exchange. The concept of surface complexation modeling is introduced in Section 22.4 in a simplified form. Section 22.5 provides a deeper look into surface complexation modeling and the effects of ­surface charge.

22.2 NOMENCLATURE 22.2.1  Introduction The production of solids creates particles, and with particles come surfaces. A surface is the interface between a solid phase and a fluid such as water or air. Surfaces provide sites for chemical interactions with dissolved species. Dissolved chemical species can interact with surfaces in a number of ways. For example, pollutants may bind irreversibly to surfaces. Some pollutants decompose at high-­energy surface sites. Cations and anions also can precipitate onto surfaces through a process called surface precipitation. The focus in this chapter is reversible interactions between dissolved species and surfaces. Such interactions can be analyzed by using equilibrium tools. By way of introduction, surface chemistry comes with its own terms. The accumulation of mass on a surface is called adsorption.1 The dissolved species being adsorbed is called the adsorbate. The solid phase onto which the adsorbate is adsorbed is called the adsorbent.  The words adsorption and absorption have different meanings. Absorption refers to a substance coming into a space, as water in a sponge. It comes from ab + sorbēre, to take in. The prefix ad-­in adsorption has the sense of “toward.” Remember: A pie in the face is adsorption. Eating pie is absorption.

surface: the interface between a solid phase and a fluid (such as water or air) Key idea: Surfaces provide sites for chemical interactions with dissolved species adsorption: the accumulation of mass on a surface adsorbate: the substance being adsorbed adsorbent: the solid phase onto which the adsorbate is adsorbed

1

535

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536   Chapter 22  Aquatic Chemistry of ­Surfaces

22.2.2  Environmental Surfaces

hydrous ferric oxides (HFO): Fe(+III)-­containing minerals with large surface areas hydrous manganese oxides (HMO): precipitated manganese-­ containing floc with large surface areas Key idea: Some porous materials have a great deal of internal surface area

Many chemical processes other than precipitation generate particles and surfaces. The most important process is erosion. Erosion transmits enormous quantities of solids to the oceans through streams. The total suspended sediment flux to the oceans is about 1.4 × 1016 g/year (Fyfe 1987), a mass of material each year equivalent to a cube of soil about 19 kilometers on a side. Other sources of surfaces include algae, bacteria, and colloidal material such as aquatic humic and fulvic acids. Colloids are particles with characteristic lengths less than about 1 μm. The most studied natural surfaces in the environment are the most common minerals. Examples include gibbsite, α-­Al(OH)3(s) (see Section 19.2.2) and calcite, CaCO3(s) (see Section  19.5.1). Another important natural surface in the environment come from a class of minerals called hydrous ferric oxides, or HFOs. HFOs are Fe(+III)-­ containing minerals with large surface areas. An example of an HFO is the iron oxyhydroxide goethite, α-­FeO(OH)(s). A related surface comes from floc formed by the precipitation of manganese.2 These surfaces are called hydrous manganese oxides, or HMOs. HMOs also have large surface areas available for adsorption. Clays (aluminosilicates) also can be important adsorption sites in natural systems. An example is the clay mineral illite.3 Some engineered solids with small diameters are porous and can exhibit very large internal surface areas. For example, one gram of nonporous colloidal particles with a diameter of 1 μm and a density of 2.5 g/cm3 would contain about 2.4 m2 of surface area. Activated carbon, a highly porous adsorbent, can exhibit greater than 1,000 m2 per gram of material. A mass of this material equal to the mass of a euro coin (7.5 g, about the mass of 1½ US nickels) has more surface area than a soccer pitch (or almost 1.7 American football fields). The surface area is almost entirely associated with the internal pores.

22.2.3  Surface Characteristics The focus of this chapter is on modeling the equilibrium partitioning of chemical species between the dissolved phase and a surface. An important characteristic is the total site concentration, NT , usually expressed in mol/L. You can estimate NT by experiment. For example, the total concentration of acidic sites can be determined by acid– base titration. The total site concentration also can be estimated from other surface characteristics, specifically the surface area, site density, and solids concentration. The surface area usually is normalized to the mass of the solid phase and expressed as the specific surface area, A, usually in units of m2/g. The site density on the surface, Ns, is expressed as sites per area (sites/nm2). The site density also can be normalized to the metal content of the surface and expressed as moles of sites per mole of metal (mol/mol). The solids

 Geothite was known since antiquity. It was named in 1806 in honor of Johann Wolfgang von Goethe (1749–1832). In addition to his famed literary output, Goethe studied mineralogy. 3  Illite is found throughout the world. It was named by scientists at the University of Illinois Urbana-­Champaign after the State of Illinois, origin of a number of representative samples. Illite has many forms, but is usually represented by the chemical formula K0.65Al2.0[Al0.65Si3.35O10](OH)2. 2

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22.2  |  Nomenclature   537

concentration, Cs, can be written in terms of mass (g/L) or moles (moles of metal per liter). The relationships between NT and A, Ns, and Cs are given in eqs. 22.1 and 22.2.





mol

NT

1018

L

NT

mol L

Ns

A

nm m2

2

m2 sites g Ns Cs g L nm 2 sites NA mol

mol metal mol Cs mol metal L

eq. 22.1

eq. 22.2

In eq.  22.1, NA is Avogadro’s number (see Section  2.3.3): 6.022 × 1023 sites per mole of sites. An example of surface characteristics calculations is given in Worked Example 22.1.

Worked Example 22.1 

Surface Characteristics

Karamalidis and Dzombak (2010) give the following surface characteristics of gibbsite: specific surface area = 32 m2/g and site density = 8 sites/nm2. Find the total site concentration for a 2 g/L suspension of gibbsite. Solution Using eq. 22.1 with A = 32 m2/g, Ns = 8 sites/nm2, and Cs = 1 g/L:

NT

mol L

1018

nm m2

2

sites g 2 2 L nm sites 6.022 1023 mol

32

m2 g

8

Or 8.50 × 10–4 mol/L.

22.2.4  Characterization of Solute–Surface Interactions There are several types of reversible interactions between dissolved species and surfaces in water. The types are characterized by the strength of the adsorption, the role of water in adsorption, or the type of solute. The terms chemisorption and physisorption are used to describe the type of adsorption by the adsorption strength. In chemisorption, chemical bonds are formed between the adsorbate and adsorbent. Due to bond formation, chemisorption represents a stronger form of adsorption. Physisorption does not involve the formation of chemical bonds and therefore is a weaker interaction. The primary form of attraction in physisorption is due to weak interactions between dipoles called van der Waals forces (after Dutch physicist Johannes Diderik van der Waals, 1837–1923). The van der Waals forces create interaction energies of about 0.4 to 4 kJ/mol, much

chemisorption: a stronger form of adsorption that involves the formation of chemical bonds physisorption: a weaker form of adsorption that involves van der Waals forces but no chemical bonds

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538   Chapter 22  Aquatic Chemistry of ­Surfaces surface complexation: interactions between species in solution and surfaces to form surface-­ bound complexes (analogous to homogeneous complexation) ion exchange: surface complexation with ions

weaker than even hydrogen bonds (about 20 kJ/mol between water molecules, see Section 1.4.1). Another way to characterize adsorption is by the role of water in the adsorption mechanism. In an inner sphere surface complex, the adsorbate and adsorbent bond through one or more water molecules. If water molecules do not act as bridging species during adsorption, then the interaction is called an outer sphere surface complex. Adsorption also can be characterized by the type of adsorbate. The term surface complexation is a general term. Ion exchange refers to the reversible partitioning of ions between water and an adsorbent. If the mechanism is unknown, the partitioning process sometimes is just called sorption.

22.3  ISOTHERMS AND ION EXCHANGE isotherm: a quantitative relationship between the mass adsorbed per mass of adsorbate and the dissolved species concentration at constant temperature

22.3.1  Introduction Several simple models have been developed to describe the equilibrium partitioning of chemical species between the dissolved and adsorbed phases. A relationship between the mass adsorbed per mass of adsorbate and the dissolved species concentration is called an isotherm. The name comes from the idea that the relationship is defined for a constant temperature.

22.3.2  Langmuir Isotherm The simple equilibrium approach to surface complexation can be applied to the adsorption of chemical species onto solid surfaces. Consider the adsorption of an uncharged pollutant P onto a surface ≡S. (In general, the symbol ≡ is used to indicate a surface.) An example is the adsorption of chloroform onto activated carbon. For simplicity, assume that the pH is fixed so that the acid–base chemistry of the surface does not change and ignore surface charge effects. Applying the general equilibrium calculation approach: Species list: S, P, SP

Equilibrium:

S P



SP; K

Or: K



SP S P



Mass balance: Key idea: The symbol ≡ indicates that the chemical species is a surface

Mass balances are required on P and on the sites. For the pollutant: PT = [P] + [≡SP] For sites: ST = [≡S] + [≡SP] Note that ST is just the total site concentration NT .

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22.3  |  Isotherms and Ion Exchange   539

In adsorption, we are interested in the relationship between the adsorbed species concentration and the dissolved species concentration. Thus, we seek a relationship between [≡SP] and [P]. From the equilibrium expression: SP



K P

S

From the mass balance on sites: S



ST

SP

Combining: [≡SP] = K[P](ST – [≡SP]) or:

SP

K P ST 1 K P

eq. 22.6

The term in brackets in eq. 22.6 is a sort of alpha value for adsorption, with ≡SP analogous to HA, P analogous to H+, and K analogous to 1 . Ka

If the molar mass of P is MP and the mass concentration of the adsorbate is M (usuSP M P ally in g/L), then the mass adsorbed per mass of adsorbate is . Multiplying M MP both sides of eq. 22.6 by M , the isotherm becomes: SP M P



K P m ST 1 K Pm

M

eq. 22.7

K , S ′ = ST M P = mass of sites as P per mass of adsorbate, and MP T M [P]m = mass concentration of P = [P]MP . Equations  22.6 and  22.7 are called the ­Langmuir isotherm (after American chemist Irving Langmuir, 1881–1957). Many forms of the Langmuir isotherm are in use. It is common to introduce the following symbols:

where K′ =

qe = mass adsorbed per mass of adsorbate

SP M P

M Ce = pollutant concentration (usually in mass units) = [P]m qmax = mass of sites (as P) per mass of adsorbate = ST′ b = a measure of the strength of adsorption K Thus, the Langmuir isotherm becomes:

qe

qmax bCe 1 bCe



K MP eq. 22.8

Eq. 22.8 may look far removed from the surface characteristics and equilibrium discussed earlier in this chapter. It is important to note that qmax is just a normalized form of NT and b is just a normalized form of the equilibrium constant relating ≡S, P, and ≡SP. (See also Problem 2.) The bottom line is that the Langmuir isotherm comes from an equilibrium calculation approach utilizing surface characteristics.

Langmuir isotherm: an isotherm given by: q bC qe = max e (terms 1 bCe defined in the text)

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540   Chapter 22  Aquatic Chemistry of ­Surfaces The key to solving problems with isotherms is to perform a mass balance on P: initial P mass = mass of P adsorbed + mass of P in solution at equilibrium, or C0V = qeMV + CeV, where C0 is the initial concentration of the pollutant before adsorption and V is the system volume.4 Therefore: qe =

C0 Ce M

and so: qmax bCe 1 bCe

C0 Ce M

eq. 22.9

A common challenge in environmental engineering is to solve for the adsorbent dose (M) required to achieve a specified equilibrium concentration (Ce). Suppose that b = 26.2 L/mg and qmax = 8.1 mg/g for the adsorption of phenol onto a particular type of activated carbon. You want to find the activated carbon dose that will reduce the 1 bCe phenol concentration from 100 mg/L to 5 mg/L. Solving for M : M C0 Ce qmax bCe mg L 1 26.2 5 mg L 11.8 mg / L activated carbon. L mg mg 8.1 26.2 5 g mg L Another common problem is to solve for Ce if you know C0, the mass of the adsorbent (M), and isotherm parameters. An example is given in Worked Example 22.2.

Langmuir Isotherm

Worked Example 22.2 

For the adsorption of phenol onto activated carbon described in the text, find the aqueous equilibrium concentration of phenol if the initial phenol concentration is 100 mg/L and the activated carbon dose is 10 mg/L. Solution Recall:

qmax bCe 1 bCe

bCe 2 Key idea: Ion exchange can be characterized by surface properties and adsorption equilibria

So: Ce

C0 Ce M

. Solving for Ce:

qmax bMCe C0 qmax bM

0

qmax bM 2b

2

4 bC0

With b = 26.2 L/mg, qmax = 8.1 mg/g, M = 0.01 g/L, and C0 = 100 mg/L, Ce = 1.9 mg/L.

4  This is the appropriate mass balance for a batch reactor when the absorbent is added. For a well-­mixed flow-­through basin, you can replace V with Q (flow) and the equation is still valid. The common application here is the addition of powdered activate carbon (PAC) to a well-­mixed basin. For a fixed-­bed reactor (packed, say, with granular activated carbon or GAC), the mass balance at the point of breakthrough is: QC0t = qem, where t is the breakthrough time and m is the mass of absorbent in the fixed bed.

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22.3  |  Isotherms and Ion Exchange   541

22.3.3  Linear Isotherm In some environmental applications, pollutant concentrations are very small. For example, in relatively clean groundwater, PT is small and therefore [P] and [P]m are small. In this case, the denominator in eq. 22.8 is about 1 and qe and Ce are linearly related: qe



K d Ce

where Kd = qmaxb. This is called a linear isotherm. Linear isotherms are used frequently in contaminant transport in groundwater.

22.3.4  Freundlich Isotherm The Freundlich isotherm (after German chemist Herbert Freundlich, 1880–1941) is a relationship between qe and Ce of the form: qe



1 K f Cen

adsorbent with K f

is 20  mg/L. Plugging in: qe

L mg 3.2

1 n

1

and 1/n = 0.7 if the equilibrium concentration

mg g

L mg

1 n

20

mg L

0.7

26.1 mg / g. You can solve 1

for Ce or M as described with the Langmuir isotherm by setting qe = Kf Cen =

Another example is given in Worked Example 22.3. In a way, the Freundlich isotherm makes no sense. Worked Example 22.3 

C0 Ce M

Freundlich Isotherm

For the Freundlich isotherm example described in the text, find the aqueous equilibrium concentration of the pollutant if its initial concentration is 10 mg/L and the adsorbent dose is 10 mg/L. Solution

1

Recall: K f Cen Kf

mg 3.2 g

C0 Ce M

Freundlich isotherm: an isotherm of the form: qe

where Kf (the Freundlich constant) and 1/n usually are considered empirical constants. Note that the units of Kf depend on the value of 1/n. As an example of the use of the Freundlich isotherm, calculate the amount adsorbed of a pollutant on a certain mg 8.2 g

linear isotherm: a linear relationship between the mass adsorbed per mass of absorbent and the equilibrium concentration in air or water

. Solving for Ce requires the use of a nonlinear solver. For 

1

L n , 1/n = 0.7, M = 0.01 g/L, and C0 = 10 mg/L, Ce = 8.6 mg/L. mg

Thoughtful Pause Why does the mathematical form of the Freundlich isotherm make no sense?

.

K f Cen (terms defined

in the text)

←Worked example

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542   Chapter 22  Aquatic Chemistry of ­Surfaces The Freundlich isotherm predicts that qe will increase without limit as Ce increases. This suggests a limitless number of adsorption sites. In spite of this logical inconsistency, the Freundlich isotherm often fits environmental sorption data well. Although the Freundlich isotherm is often treated as an empirical equation, it can be derived from adsorption equilibria (see the Addendum to this chapter).

22.3.5  Ion Exchange Ion exchange reactions occur in both natural and engineered systems. Equilibrium models are valuable in analyzing ion exchange. In engineering applications, ion exchange is used to remove ions from water. The adsorbents are called ion exchangers. (Ion exchangers sometimes are called resins because they are polymeric resins.) They are characterized by their total exchange capacity, analogous to NT. A typical value is about 5 meq/g, with equivalents determined by the charge on the ion. Second, exchangers are characterized by their capacity for removing specific ions. This is called selectivity. In general, multivalent ions (such as Ra2+, Ba2+, CrO42–, and SO42–) are preferentially removed over monovalent ions (such as Na+, H+, Cl–, and OH–). To assess selectivity, consider the following equilibrium: n RH A n



Rn A nH ; K HA

(This equilibrium is written for cation exchange. You could write an analogous equilibrium for anion exchange.) The species ≡RH represents a cation exchanger with H+ bonded to the exchange site. We say that the ion exchanger in the H form This equilibrium expresses the idea that n sites on a resin in the hydrogen form exchange one cation with a charge of n+ and release n H+, all with an equilibrium constant K HA . In the language of ion exchange, the equilibrium constant K HA sometimes is called a selectivity coefficient. It can be used to determine the relative affinities of ions for specific ion exchangers.5 Note that K HA

{H }n { R n A} { RH}n {A n }

. (Ionic strength effects likely

are important here.) Using C for dissolved species (in units of eq/L) and q for sorbed CHn qA species (in units of eq/g): K HA . qHn CA Just as acid dissociation constants quantify the relative affinity of bases for H+, K HA values quantify the relative affinity of a surface for different ions (or the relative affinity of one ion for different exchangers). In summary, ion exchange in natural and engineered systems is characterized by surface properties (e.g., total exchange capacity, analogous to NT) and adsorption equilibrium measures, such as K HA.

5

 While K HA is a perfectly suitable measure of selectivity, a more common measure of affinity is the separation

factor, α, defined as:

A H

C H qA qH C A

K HA

qH CH

n 1

.

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22.4  |  Introduction to Surface Complexation Modeling   543

22.4  INTRODUCTION TO SURFACE

COMPLEXATION MODELING

22.4.1  Introduction When species in solution interact with surfaces to form surface­bound complexes, the form of interaction is called surface complexation. The term surface complexation modeling or SCM refers to a specific way to model surface complexation where the charge of the surface affects the interaction. In this section, the general idea of surface complexation is presented, with no attempt to “correct” for surface charge. In general, surface charge is important. It is neglected in this section only as a teaching tool to get you used to surface complexation modeling one concept at a time. Surface charge models will be introduced in Section 22.5.

22.4.2  Analogy with Heterogeneous Complexation Surface complexation is analogous to homogeneous complexation discussed in ­Chapter 15. For example, you know that copper complexes with the hydroxide ion in part as follows: Cu2



OH

CuOH ;

By analogy, aquo cupric ion can complex with a surface site, as shown in eq. 22.10:

Cu2

SOH

SOCu

H ; K

eq. 22.10

In eq. 22.10, ≡S represents a surface and ≡SOH is a surface oxide site. The surface also can be written as ≡M, where M is a generic surface metal. If the nature of the surface is known, then the surface metal can be specified. For example, for the surface complexation of aquo cupric ion onto goethite is written:

Cu2

FeOOH

FeOOCu

H



As with the total site concentration NT, you can express the concentrations of surface species in units of moles of surface species per liter of water. Thus, the statement that “[≡SOCu+] = 10–4 M” means that 10–4 mole of the surface species ≡SOCu+ are found per liter of water. These concentration units also are analogous to the homogeneous case.

22.4.3  Equilibrium Model for Complexation at a Surface On the simplest level, you might apply the normal procedure for equilibrium calculations to systems containing complexing surfaces. For example, suppose that you expose the surface ≡SOH to water containing a metal M. To make this example more straightforward, assume that M does not precipitate or form complexes other than the aquo complex and a metal-­surface complex. Write the aquo form of the metal as M2+.

Key idea: Surface charge is important and affects surface–solute interactions

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544   Chapter 22  Aquatic Chemistry of ­Surfaces Worked example →

Many surfaces of interest in environmental engineering and science exhibit acid– base chemistry. Suppose the site in this example also donates a proton to water: SOH



SO

H

The equilibrium system becomes (as usual, you may wish to try this on your own before reading further): Species list: H 2 O, H , OH , M2 , SOH, SOM , and SO

Equilibria:

H2 O

M2

SOH SOH

H

OH SOM SO

KW H

H

10

14

K



Ka

Or: KW

K

Ka

[H ][OH ] H2 O [ SOM ][H ] [M2 ][ SOH] [ SO ][H ] [ SOH ]

Mass balance: Mass balances are required on the metal and on the sites. For the metal: MT = [M2+] + [≡SOM+] For sites: ST = [≡SOH] + [≡SOM+] + [≡SO–] Charge balance:

[H ] 2[M2 ] [ SOM ] [OH ] [ SO ] Other: Dilute solution, so that concentrations and activities are interchangeable. Mole fraction of water is unity.

Thoughtful Pause Is this system solvable?

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22.4  |  Introduction to Surface Complexation Modeling   545

The system has six unknowns ([H+], [OH–], [≡SOM+], [M2+], [≡SOH], and [=SO–]) and six linearly independent equations (three from equilibria, two mass balances, and one charge balance). Thus, a solution should be possible. The solution method is similar to that employed in homogeneous complexation (see Chapter  15): Express both mass balances in terms of the concentration of two species. In the example, you can show that:

MT

ST

K

[M2 ] [ SOM ] [M2 ] 1

SOH [H ]



eq. 22.11

[ SOH] [ SOM ] [ SO ] K [ M2 ] [H ]

[ SOH] 1

Ka [H ]

eq. 22.12

Solve for [≡SOH] in eq. 22.12 and substitute into eq. 22.11: MT

K [M ] 1 [H ] 2

ST K [ M2 ] 1 [H ]

Ka [H ]

Solving for [M2+]:

K [M2 ]2 ([H ] K a

K ST

MT )[Cu2 ] MT ([H ] K a )

0

You can prepare a pC-­pH diagram if you know the total concentrations and equilibrium constants. Assume K = 10–1.5, Ka = 10–8, MT = 10–6 M, and ST = 10–4 M. You may want to generate the pC-­pH diagram on your own. The resulting pC-­pH diagram is shown in Figure 22.1.

Thoughtful Pause Does the pC-­pH diagram make qualitative sense?

You can think of the system as comprising two metals, H+ and M2+, and one ligand, ≡SO – (since all other ligands were ignored). At low pH, [H+] is large and H+ outcompetes M2+ for ≡SO –. Therefore, at low pH, the surface is mainly ≡SOH and the metal is mainly M2+. As the pH increases (i.e., as [H+] gets smaller), M2+ begins to compete with H+ for surface sites. Since the total available M is only 10–6 M, [≡SOM+] levels out at 1 × 10–6 M. It is common to plot only the surface-­bound metal concentration as a function of pH. Usually, the surface-­bound metal concentration is expressed as a

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546   Chapter 22  Aquatic Chemistry of ­Surfaces pH 0

0

2

4

6

8

10

12

14

2 [≡SOH]

pC

4

[≡SO–]

[M2+]

6

[≡SOM–]

8 10 12

[OH–]

[H+]

14

Metal Adsorbed (% total M)

FIGURE 22.1  pC-­pH Diagram for a Metal M in the Presence of a Complexing Surface (surface charge ignored)

100 80 60 40 20 0

0

2

4

6

pH

8

10

21

14

FIGURE 22.2  Adsorption Edge for the Metal in the Example

percentage of the total metal concentration adsorption edge plot: a plot of the surface-­bound metal concentration (as a percentage of the total metal) as a function of pH

[ SOM ]

100% . Such a plot, MT called an adsorption edge plot, is shown in Figure 22.2. Note that the bound M concentration is very sensitive to pH over a narrow pH range. The bound M increases from about 10% to about 90% over two pH units (about pH 4.5–6.5). This is a relatively sharp adsorption edge.

22.5  SURFACE COMPLEXATION MODELING 22.5.1  Introduction As stated earlier, the simple equilibrium approach discussed in Section  22.4 fails to consider the effects of the surface charge on complexation. Many surfaces in natural waters are negatively charged. The charge often stems from ionized weak acids at the surface. In clays, the surface charge comes from the replacement of one metal (e.g., Si4+) with a metal containing a lower positive charge (e.g., Al3+). This results in an excess of anionic counterions and an overall negative charge.

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22.5  |  Surface Complexation Modeling   547

You might expect that the affinity of ions in aqueous solution for a charged surface would be influenced by the surface charge. For example, a metal cation such as Cu2+ may have an attractive electrostatic interaction with a surface exhibiting a high negative charge. In addition, the complexation of H+ or other metals onto the surface will affect the surface charge. In this way, surface complexation itself changes the affinity of ions for the surface. In other words, the surface complexation equilibrium constants are not constant but depend on the surface charge. The surface complexation equilibrium constants in Section 22.4 ignored the effects of surface charge. These are sometimes called intrinsic equilibrium constants (or Kint) because they represent the intrinsic chemical interaction that is independent of electrostatics. In other words, Kint can be calculated strictly from the intrinsic Gibbs free energy, Gint . With a charged surface, the Gibbs free energy now is the sum of two terms: the intrinsic Gibbs free energy and the free energy associated with electrostatics:

G

Gint

Gelectro

Gelectro

ZF



eq. 22.14

where ΔZ = change in the surface charge for a given equilibrium and F = the Faraday constant = 96,485 C/mol. Combining eqs. 20.13 and 20.14:

K app

e

G RT

e

Gint RT

e

Gelectro RT

K int e

intrinsic equilibrium constants: equilibrium constants uncorrected for electrostatic effects (symbol: K int)

eq. 22.13

The charged surface generates an electric potential difference at the surface (i.e., a voltage difference), Ψ. The electrostatic portion of the Gibbs free energy is the work required to move the ions across the potential difference Ψ and can be calculated by:

Key idea: Surface complexation equilibrium constants are not constant but depend on the surface charge (which, in tum, depends on the degree of surface complexation)

ZF RT



where Kapp is the apparent equilibrium constant. The apparent equilibrium constant describes how the species concentrations are related when a charged surface is involved. It is a function of the surface charge.

apparent equilibrium constant: an observed equilibrium constant dependent on the surface charge (symbol: Kapp)

Key idea: The surface charge density, calculated from the excess of charges at the surface, can be used to calculate the surface potential

22.5.2  Charge Distribution Models To properly account for surface charge, you need to use Kapp values in equilibrium calculations. This means you need to know the surface potential Ψ. Unfortunately, you cannot measure Ψ. You can, however, calculate Ψ from the surface charge. Several models of surface charge exist, including the double layer, triple layer, and constant capacitance models. A common model in environmental engineering and science is the double-­layer model (Dzombak and Morel  1990). In this model, the charge distribution near a surface is treated as two layers called the electric double layer (EDL).6 The EDL is envisioned as a fixed charge on the surface with a diffuse charge penetrating into the water surrounding the surface. The diffuse layer commonly is modeled by the Guoy-­Chapman diffuse layer model, developed independently by Georges Guoy (1854–1926) in 1910 and David Leonard Chapman (1869–1958) in 1913. 6  Electric double layer theory underlies the Debye-­Hückel limiting law for single-­ion activity coefficients discussed in Section 21.2.3.

electric double layer (EDL): a conceptualization of the charge distribution around a charged surface as consisting of two charged layers Guoy-­Chapman diffuse layer model: a model for the charge distribution at and near a surface, with a fixed charge on the surface and a diffuse charge penetrating into the water surrounding the surface

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548   Chapter 22  Aquatic Chemistry of ­Surfaces According to the Guoy-­Chapman diffuse layer model, the surface charge density, σ, at 25°C in the presence of c mol/L of a 1:1 monovalent electrolyte such as NaCl can be calculated by:

C / m2

1

0.1174 c 2 sinh 19.46



eq. 22.15

where Ψ is in volts and sinh(x) = hyperbolic sine of x = ½(ex – e–x).7 How does this equation help? The surface charge density can be calculated by the excess of positive charges over negative charges at the surface. In general, the excess of positive charge is calculated by summing the concentrations of the positively charged surface species multiplied by their charge and subtracting the concentrations of the positively charged surface species multiplied by their charge. This is a charge balance on surface species only. The excess positive charge will have units of moles of charge (or equivalents of charge) per liter.

Thoughtful Pause How can you convert units of eq/L to a surface charge?

surface concentration (s): s = ACs and usually has units of m2/L

Recall from Sections 16.5.3 and 22.5.1 that the Faraday constant is the number of coulombs of charge per mole. So F×(excess positive charge) will have units of coulombs per liter. To get the surface charge density, divide by the surface concentration, s. Using the symbols from Section 22.2.3:

s

ACs

where A is the specific surface area in m2/g and Cs is the solids concentration in g/L (see Section 22.2.3). For example, the surface in the example in Section 22.4.3 had two charged surface eq species, ≡SOCu+ and ≡SO–. The excess positive surface charge is 1  [≡SOCu+] – mol eq 1  [≡SO–] = [≡SOCu+] – [≡SO–]. The corresponding surface charge density is: mol

F [ SOCu ] [ SO q ] s

eq. 22.16

Thus, Ψ can be determined by eq. 22.15 after calculating the surface charge density from eq. 22.16. Note that surface complexation changes the surface charge density by changing the concentrations of ≡SOCu+ and ≡SO–.

 Formally, the Guoy-­Chapman diffuse layer model for the surface charge density is: σ = (8RTεε0c × 103)½sinh

7

z F , 2 RT

where ε = relative permittivity of water (see Section 1.4.1), ε0 = permittivity of a vacuum, and z = valance of the 1:1 electrolyte (= 1 for NaCl). Using values at 25°C and z = 1 gives eq. 22.14. For small x, sinh(x) ≈ x. So at small Ψ, sinh(19.46Ψ) ≈ 19.46Ψ, and σ (C/m2) = 2.28c½Ψ. For a swamping electrolyte, c is about equal to the ionic strength I, so σ (C/m2) ≈ 2.28I½Ψ at small Ψ.

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22.5  |  Surface Complexation Modeling   549

22.5.3  Equilibrium Calculations with Charge Correction To illustrate the effects of surface charge, the problem in Section 22.4.3 will be solved again. Everything stays the same as in Section  22.4.3, except for the equilibrium constants involving surfaces. For these equilibria: Cu2



SOH SOH

SOCu SO

K app

H

K aapp

H

F RT

K int e

K aint e

F RT





The signs in the exponents stem from the fact that the surface gains one positive charge per mole in the first reaction (ΔZ = +1) and the surface loses one charge per mole in the second reaction (ΔZ = –1). You can calculate Ψ from eqs. 22.15 and 22.16. As an example, assume s = 10 m2/L and medium of 0.1 M NaCl (c = 0.1 M), with Kint = 10–1.5, and K aint = 10–8 as before. As an illustration of the solution method, species concentrations will be calculated only at pH 6. To accomplish this equilibrium calculation, you must iterate since the surface charge is a function of surface species concentrations. Assume a starting surface charge density. A surface charge density of zero is a convenient starting point. If the surface charge density is zero, then Ψ = 0 from eq. 22.15. This means that Kapp = Kint for the two equilibria involving surfaces. You can calculate all species concentrations in the normal fashion as in Section 22.4.3. You will find that [≡SOCu+] = 7.57 × 10–7 M and [≡SO–] = 9.82 × 10–7 M. For the next iteration, calculate the surface charge density from these calculated concentrations: F [ SOCu ] [ SO ] s



96, 485

C mol

7.57 10 7 M 9.82 10 7 M

m2 10 L

0.00218 C / m 2 Now you can recalculate Ψ = –0.00302 V.8 Since Ψ ≠ 0, then Kapp ≠ Kint, so you have to calculate the new Kapp values:

K app

K int e

F RT

10

1.5

exp

96, 485

C mol

0.00302 V

J 298 K 8.31446 K mol

0.0294

 This value was calculated using the inverse hyperbolic sin function. The approximation in the footnote to

8

Section 22.5.2 is valid here and Ψ ≈

0.0218 2.28 I

C

m 2 = –0.00302 V. 2.28 0.1

←Worked example

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550   Chapter 22  Aquatic Chemistry of ­Surfaces



K

app a

K e int a

F RT

1.08 10

10

1.5

exp

96, 485

C mol

0.00302 V

J 298 K 8.31446 K mol



8

(Note that 1 joule is 1 coulomb-­volt, so the units work out.) Now recalculate the [≡SOCu+] and [≡SO–] values with the new equilibrium constant values. This allows you to calculate another round of surface charge density, surface potential, and surface concentrations until the surface concentrations converge. After a number of iterations, the system converges: σ = –0.00816 C/m2, Ψ = –0.0112 V, [≡SOCu+] = 6.67 × 10–7 M, and [≡SO–] = 1.51 × 10–6 M. Since the surface charge density and surface potential are small, the solution without surface charge correction is a reasonable approximation to the correct solution. Copper was 76% surface-­complexed at pH 6 with no charge correction (Section 22.4.2) and 67% surface-­complexed with charge correction. The iterative approach lends itself nicely to a spreadsheet solution. Set up the spreadsheet as usual, but add columns for σ, Ψ, and the Kapp values. Set σ = 0 in one row. In subsequent rows, calculate σ from the surface species concentrations in the previous row. The spreadsheet for the chapter example is shown in Figure 22.3. Note that σ is set to zero initially in cell A8. The formula in cell A9 is =$H$2/$H$1*(J8-­K8). In this way, the surface concentrations from the previous iteration are used to calculate σ in the current iteration.

22.5.4  Summary of the Surface Complexation Method The steps involved in using the surface complexation method can be summarized as follows (also listed in Appendix C, Section C.9): Step l: Set up the equilibrium in the usual fashion. Step 2: Assume the surface charge density is zero (σ = 0). If σ = 0, then Kapp = Kint.

FIGURE 22.3  Spreadsheet for the Chapter Example

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22.5  |  Surface Complexation Modeling   551

Step 3: Calculate the species concentrations as usual. F Step 4: Calculate σ =  (excess surface charge concentration). s Step 5: Calculate the surface potential, Ψ, from σ (C/m2) = ­ 0.1174c½sinh(19.46Ψ). For small Ψ, Ψ ≈

2.28 I

.

Step 6: Calculate Kapp = K int e

ZF RT

for all surface equilibria.

Step 7: Iterate (back to Step 3) until the concentrations converge.

22.5.5  Point of Zero Charge The point of zero charge, PZC, is the pH at which the net charge on the surface is zero. Using the language from Section 22.5.2, the PZC is the pH where the excess positive surface charge is zero. The PZC is easy to calculate for surfaces with simple acid–base surface chemistry in the absence of other cations or anions. (In other words, “clean” systems as defined in Chapter 19.) As an example, the pertinent acid–base equilibria for gibbsite are:

AlOH 2

AlOH 0

AlOH 0

AlO

H H

K aapp ,1

K aint,1 e

K aapp ,2

K aint,2 e

F RT



F RT

Note the similarities to a dissolved diprotic acid. In this example, the surface charge F density is σ = s ({≡AlOH2+} – {≡AlO–}). The surface is uncharged when {≡AlOH2+} = { AlOH 0}K aapp { AlOH 0}{H } ,2 – {≡AlO –}. Note that {≡AlOH2+} = and {≡AlO } = . app {H } K a ,1 Therefore, the surface is uncharged when

pH =

pK aint,1

2

pK aint,2

{ AlOH 0}{H }

{ AlOH 0}K aint,2

K aint,1

{H }

, or

. (Remember that Kapp = Kint when σ = 0.) For gibbsite, pK a,int1 = 7.17

and pK a,int2 = 11.18, giving a PZC of 9.2. Titration data show a PZC of about 9.0 (Karamalidis and Dzombak 2010). As another example, you can estimate the PZC in the ≡SOH-­Cu2+ system from the uncorrected approach in Section 22.2.4.

point of zero charge (PZC): the pH at which the net charge on the surface is zero ←Worked example

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Surface Charge (C/mol)

552   Chapter 22  Aquatic Chemistry of ­Surfaces 0.10 0.08 0.06 0.04 0.02 0.00 –0.02 –0.04 –0.06 –0.08 –0.10 5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 6.0 pH

FIGURE 22.4  Surface Charge for the Chapter Example

Thoughtful Pause From Figure 22.1, what is the approximate PZC for the ≡SOH-­Cu2+ example? In this example, the surface charge is F([≡SOCu+] – [≡SO–]). By looking at [≡SOCu+] and [≡SO–] in the pC-­pH diagram in Figure 22.1, you can see that the surface is positively charged at pH < 5.8 ([≡SOCu+] > [≡SO–]) and negatively charged at pH > 5.8 ([≡SOCu+] < [≡SO–]). Therefore, the PZC is about 5.8. A more careful calculation is shown in Figure  22.4 using surface charge corrections. The PZC in this example is about pH 5.84. The PZC is a useful parameter. It tells you whether the electrostatic interactions will be stronger with cations or with anions. In drinking water treatment, one goal of coagulation is to reduce the charge on naturally occurring particles to near zero so that they will collide and stick together in the flocculation step. The PZC concept is useful in determining the optimum coagulant dose.

22.5.6  Surface Complexation Intrinsic Equilibrium Constants In a way, calculations involving surface complexation are similar to calculations involving ionic strength corrections. In both cases, the basic setup follows the familiar pattern established in this text. The only changes are that the values of the equilibria constants change as the solution converges. (For an example of iteration in ionic strength calculations, see the Integrated Case Study on phosphate buffers in Section 24.5.) The equilibrium calculations, as always, require thermodynamic data. Both cations and anions can sorb to surfaces. Representative intrinsic equilibrium constants for the complexation of cations and anions by common mineral surfaces are listed in Appendix D in Tables D.6 and D.7, respectively. Always use caution with surface complexation equilibrium constants because the equilibria can take many forms.

22.6  CHAPTER SUMMARY This chapter focused on the equilibrium partitioning of dissolved species between water and surfaces. Important surfaces in the environment include gibbsite, calcite, hydrous ferric oxides (HFOs) such as goethite, hydrous manganese oxides (HMOs), and

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Chapter Glossary   553

clays. Reversible interactions between dissolved species and surfaces are characterized by the strength of the adsorption, the role of water in adsorption, or the type of solute. An isotherm is a relationship between the mass adsorbed per mass of adsorbate and the dissolved species concentration. Common isotherms in environmental engineering and science are the Langmuir, linear, and Freundlich isotherms. Each isotherm can be related back to governing equilibria and surface characteristics. Ion exchange also can be modeled by an equilibrium approach. Surface complexation modeling (SCM) is an approach to modeling the equilibration of species between dissolved and surface-­associated forms that considers surface charge. The common approach is model the surface with a fixed charge on a diffuse layer in solution (electric double layer), with the diffuse layer charge described by the Guoy-­Chapman diffuse layer model. The surface charge density (σ) can be calculated from the concentrations of surface species and used to calculate a surface potential (Ψ). The surface potential is used to calculate an electrostatic Gibbs free energy term, which in turn can be used with an intrinsic equilibrium constant (K int) to calculate an apparent equilibrium constant (K app). The apparent equilibrium constants are used in equilibrium calculations. Typically, iterative solutions are required until the surface concentrations converge.

CHAPTER KEY IDEAS • Surfaces provide sites for chemical interactions with dissolved species. • Some porous materials have a great deal of internal surface area. • The symbol ≡ indicates that the chemical species is a surface. • Ion exchange is characterized by surface properties and adsorption equilibria. • Surface charge is important and affects surface–solute interactions. • Surface complexation equilibrium constants are not constant but depend on the surface charge (which, in turn, depends on the degree of surface complexation). • The surface charge, calculated from the excess of charges at the surface, can be used to calculate the surface potential.

CHAPTER GLOSSARY adsorbate:  the substance being adsorbed adsorbent:  the solid phase onto which the adsorbate is adsorbed adsorption edge plot:  a plot of the surface-­bound metal concentration (as a percentage of the total metal) as a function of pH adsorption:  the accumulation of mass on a surface apparent equilibrium constant:  an observed equilibrium constant dependent on the surface charge (symbol: Kapp) chemisorption:  a stronger form of adsorption that involves the formation of chemical bonds electric double layer (EDL):  a conceptualization of the charge distribution around a charged surface as consisting of two charged layers 1

Freundlich isotherm:  an isotherm of the form: qe = Kf Cen (terms defined in the text)

Guoy-­Chapman diffuse layer model:  a model for the charge distribution at and near a surface, with a fixed charge on the surface and a diffuse charge penetrating into the water surrounding the surface hydrous ferric oxides (HFO):  Fe(+III)-­containing minerals with large surface areas hydrous manganese oxides (HMO):  precipitated manganese-­containing floc with large surface areas

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554   Chapter 22  Aquatic Chemistry of ­Surfaces intrinsic equilibrium constants:  equilibrium constants uncorrected for electrostatic effects (symbol: K int) ion exchange:  surface complexation with ions isotherm:  a quantitative relationship between the mass adsorbed per mass of adsorbate and the dissolved species concentration at constant temperature q bC Langmuir isotherm:  an isotherm given by: qe = max e (terms defined in the text) 1 bCe linear isotherm:  a linear relationship between the mass adsorbed per mass of absorbent and the equilibrium concentration in air or water physisorption:  a weaker form of adsorption that involves van der Waals forces but no chemical bonds point of zero charge (PZC):  the pH at which the net charge on the surface is zero surface complexation:  interactions between species in solution and surfaces to form surface-­bound complexes (analogous to homogeneous complexation) surface concentration (s):  s = ACs and usually has units of m2/L surface:  the interface between a solid phase and a fluid (such as water or air)

HISTORICAL NOTE: FROM “CAT’S CRADLE” TO THE “SWISS MODEL” TO SURFACE COMPLEXATION MODELING Our current understanding of adsorption started with Irving Langmuir. He established the modern view of surfaces when he wrote: The surface of a crystal must then consist of an arrangement of atoms as definite as that existing in the interior of the crystal, although slightly different from the latter. The surface must thus be looked upon as a sort of checkerboard containing a definite number of atoms, of definite kinds arranged in a plane lattice formation. The space between and immediately above (away from the interior) these atoms is surrounded by a field of electromagnetic force more intense than that between the atoms inside the crystal (Langmuir 1916).

Langmuir even has an unusual connection to one of the most popular novelists of the twentieth century. H.G. Wells (1866–1946) visited the General Electric labs where Langmuir spent most of his long career. Langmuir tried to convince Wells to write a story about ice stable at room temperature. Wells evidently didn’t bite on the idea, but one of Langmuir’s chemistry colleagues at General Electric later mentioned the idea to his brother (who also worked at GE as a publicist). The colleague was Bernard Vonnegut (1914–1997), known for his work on cloud seeding and brother of erstwhile publicist Kurt Vonnegut (1922–2007). The stable ice idea showed up in Vonnegut’s 1963 novel Cat’s Cradle as “ice-­nine” (Ball 2018). Langmuir may have been the model for the character Felix Hoenikker in the novel. Our understanding of environmental surfaces was extended by the work of Paul Schindler and Werner Stumm and their students in the 1970s. Their work was summarized in the so-­called “Swiss model.” See Figures 20.2 and 22.5 for sketches and Stumm (1993) and Brown and Calas (2012) for reviews. More recently, David Dzombak and François Morel standardized the field onto the EDL/Guoy-­Chapman model for the modeling of surface charge. Dzombak’s doctoral dissertation (published as Dzombak and Morel 1990) was a defining paper in the field.

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Problems   555

FIGURE 22.5  The “Swiss Model” for Surface Complexation. Source: Stumm (1993)/with permission of Springer Nature.

PROBLEMS 1. Calculate the solids concentration required to produce 1 mM of sites with gibbsite. See Worked Example 22.1 for the surface characteristics. 2. For the Langmuir isotherm, develop the relationship between qmax and NT . Develop the relationship between b and the equilibrium constant relating ≡S, P, and ≡SP. 3. For the adsorption of DDT onto activated carbon, the following Langmuir isotherm constants were found: b = 303 L/mg and qmax = 41.3 mg/g. Find the aqueous equilibrium concentration of DDT if 100 mg of activated carbon and 1 μg of DDT are combined in 1 L of water. 4. A river to be used for a drinking water supply contains the pesticide parathion at 50 μg/L. Powdered activated carbon is to be added to reduce the concentration to 10 μg/L. Find the PAC dose in mg/L. For parathion: qmax = 530 mg/g and b = 4.16 L/mg. 5. Suppose the river in Problem 4 water also contains phenol at 200 μg/L (in addition to the parathion at 50 μg/L). Using the PAC dose you calculated for parathion alone, find the equilibrium concentrations of parathion and phenol. For phenol: qmax = 103 mg/g and b = 1.15 L/mg. For this problem, you will need to use the competitive Langmuir model. For n adsorbates: qmax ,i biCe,i C0,i Ce,i qe,i , where: qe,i = mass n M 1 biCe,i i 1

of species i per mass adsorbent, C0,i = initial concentration of species i, Ce,i = equilibrium concentration of species i, M = adsorbent concentration, and qmax,i and bi are Langmuir constants for species i. (The competitive Langmuir model is valid only if the

qmax,i values are nearly equal, which is not true in this problem.) 6. Determine the percent removal of a compound from water if its initial concentration is 20 mg/L, the adsorbent concentration is 50 mg/L, and the appropriate Freundlich isotherm constants

mg are Kf = 5.4 g

L mg

1 n

and 1/ n

0.9.

7. In Freundlich’s original paper (Freundlich 1907), he obtained the following data for the adsorption of propionic acid onto charcoal: Ce (mmol/cm3)

qe (mmol/g)

0.0201 0.0516

0.785 1.22

0.1298

1.71

0.2466

2.11

0.6707

2.94

1.58

3.78

Calculate Kf and 1/n for these data. Freundlich calculated Kf = 3.463 and 1/n = 0.354 (although he used the symbols β and 1/p, for Kf and 1/n, respectively.) What units did Freundlich intend? 8. In Section 22.3.5, the following cation exchange ­equilibrium was given: n≡RH + A n+ = ≡RnA + nH+; K HA. Write the corresponding anion exchange equilibrium for an anion exchange resin in the OH form (≡ROH) equilibrating with an anion of the form Bn–. B , Express the resulting equilibrium constant, K OH in terms of species activities and the C-­q notation.

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556   Chapter 22  Aquatic Chemistry of ­Surfaces 9. Estimate the PZC for HFO. The equilibrium constants below are from Dzombak and Morel (1990). Is the charge on HFO positive or negative at pH 7? At pH 9.5? FeOH 2 FeOH

int FeOH H K a, 1

FeO

int H K a, 2

10 10

7.29

8.93

10. For the system in Problem 9, what sign would you use in F

the e RT term for each of the two equilibria? Explain your answer. 11. For the system in Problem 9, calculate the surface charge density at pH 7 and at pH 9.5. Assume a medium of 0.1 M NaCl, total site concentration of 10–4 M, and a surface concentration of 10 m2/L. 12. Redo the example in Section 22.5.3 at pH 4.0, 4.5, 5.0, 5.5, and 6.0 and plot the adsorption edge plot. ­(Concentrations at pH 6.0 are given in the text.)

ADDENDUM: THE FREUNDLICH ISOTHERM AND ADSORPTION EQUILIBRIA While the Freundlich isotherm generally is thought of as a purely empirical equation, it does have a connection with the principles in this chapter. The following derivation follows the classical approaches of Halsey and Taylor (1947) and Sips (1948). Recall from Section 22.3.4 that the Langmuir isotherm is based on an adsorption equilibrium: ≡S + P = ≡SP, with equilibrium constant K. You know from Chapter 3 that you can write K as: Grxn

K = e RT , where Grxn is the free energy of reaction for adsorption. To simplify the notation, let E = Grxn. The usual assumption is that the adsorption sites are identical and have the same adsorption energy. In other words, you can use one value of E to characterize the adsorption. In this derivation, assume that the sites on the adsorbent are not identical, but rather have a distribution of adsorption energies. Let N(E) be the number concentration of sites with adsorption energy between E and E + dE. Also let [≡SP](E) be the concentration of filled sites with adsorption energy between E and E + dE. Equation 22.6 becomes: [ SP](E )dE

K [P]N ( E ) dE 1 K [P]

e

E RT [P]N ( E )

1 e

E RT [P]

dE

Now consider an exponential distribution of adsorption energies: N(E) = E0e

[ SP]

cE RT

. Substitute and integrate: E0 e

[ SP]( E )dE

( c 1) E RT [P]

1 e To evaluate the integral, let x = 1 e [P] [ SP]

RTE0 [P]c

0

E RT [P] E RT

dE

. Then:

x (c 1) dx 1 x

The integral evaluates to a constant, so [≡SP] is proportional to [P]c at constant temperature. Using the definitions from Section 22.3.2: qmax = (constant) Cec. This is of the form of the Freundlich isotherm. This derivation shows that the Freundlich parameter 1 can be related to the exponential distribution constant n 1 c. In other words, can be related to the distribution of n adsorption energies.

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Chapter References   557

CHAPTER REFERENCES Ball, P. (2018). Putting the science in fiction. www.chemistryworld. com/opinion/the-­c hemistry-­t hat-­i nspired-­h g-­w ells/3008994.­ article, May 13. Brown, G.E., Jr. and G. Calas (2012). Section 4. Historical perspectives on surface chemistry and environmental interface ­chemistry  – The contributions of Langmuir, Krauskopf, and Stumm. Geochem. Perspectives 1(4–5): 520–525. Dzombak, D.A. and F.M.M. Morel (1990). Surface Complexation Modeling: Gibbsite. New York: John Wiley & Sons. Freundlich, H. (1907). Über die Adsorption in Lösungen. Z. physik. Chemie, Stöchiometrie und Verwandtschaftslehre 57(4): 385–470. Fyfe, W.S. (1987). From molecules to planetary environments: Understanding global change. In: Aquatic Surface Chemistry:

Chemical Processes at the Particle-­Water ­Interface. W. Stumm (Ed.), pp. 495–508. Halsey, G. and H.S. Taylor (1947). The adsorption of hydrogen on tungsten powders. J. Chem. Phys. 15(9): 624–630. Karamalidis, A. and D.A. Dzombak (2010). Surface Complexation Modeling: Gibbsite. New York: John Wiley & Sons. Langmuir, I. (1916). The constitution and fundamental properties of solids and liquids. Part I. Solids. J. Am. Chem. Soc. 38(11): 2221–2295. Sips, R. (1948). On the structure of a catalyst surface. J. Chem. Phys. 16(5): 490–495. Stumm, W. (1993). From surface acidity to surface reactivity: Inhibition of oxide dissolution. Aquatic Sci. 55(4): 273–280.

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C H A P T E R   23

Chemical Kinetics of Aquatic Systems 23.1 INTRODUCTION So far in this text, you have had great success in using equilibrium models to describe aqueous systems. At equilibrium in an isolated system, species concentrations do not change with time. However, not all chemical systems are at equilibrium. In this chapter, your collection of modeling tools will be extended to allow the investigation of nonequilibrium systems. In Section 23.2, you will see examples of species that are not found at their equilibrium concentrations. You will learn that nonequilibrium conditions may exist for species that react slowly. You will be able to relate the rate of a reaction qualitatively to the energy profile as the reaction proceeds from reactants to products. A more quantitative discussion of reaction rates may be found in Section  23.3. In that section, you will learn about the relationship between reaction stoichiometry and the number of species participating in the reaction. You will discover the relationships among the reaction rate, rates of change of each species with respect to time, and reaction stoichiometry. An experimental measure, the kinetic order, also will be introduced in this section. Finally, you will learn an approach to developing expressions for the rates of change of species concentrations with several reaction types. In Section  23.4, you will learn about several simple rate expressions. Each rate expression will be integrated to yield equations showing species concentrations over time. The units of the rate constants and time for half completion of the reaction will be presented for the most common rate expressions. The kinetics of reactions on surfaces also will be explored. You will explore more complex systems, including reversible and sequential reactions, in Section 23.5. For all rate expressions, environmental examples will be provided. The focus of Section 23.6 is the effects of temperature and ionic strength on rate constants. It is interesting that the common models describing the effects of temperature and ionic strength on rate constants were developed from the models describing the effects of temperature and ionic strength on equilibrium constants. You will see the quantitative relationship between the energy profile of a reaction and its rate constant in this section.

559

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560   Chapter 23  Chemical Kinetics of Aquatic Systems

23.2  THE NEED FOR CHEMICAL KINETICS 23.2.1  Not All Reactions Are at Equilibrium

Key idea: Thermodynamics tells you what reactions are possible but not how fast the reactions occur

Chemical equilibrium is a thermodynamic concept. Recall from Chapter 3 that thermodynamics tells you what reactions are possible. However, not all possible reactions actually proceed to equilibrium. As an example, consider the following innocent question: Do glucose solutions spontaneously combust? To answer this question, write the balanced redox reaction for the oxidation of glucose with oxygen



oxidation half-reaction: 1        C 6 H12 O6 + H 2 O = CO2 g  + 4H + 4e 6 reduction half-reaction:         O2 g  + 4H + 4e  = 2H 2 O overall redox reaction:

1         C 6 H12 O6 + O2 g  = CO2 g  + H 2 O 6

eq. 23.1

The standard Gibbs free energy of reaction for eq. 23.1 is negative. Based on your experience, you would say that the reaction in eq. 23.1 proceeds spontaneously under standard conditions and that glucose is expected to bum in air. In reality, you are unlikely to observe spontaneous combustion of glucose. Why? In fact, the reaction in eq. 23.1 proceeds very slowly. Thermodynamics tells you what reactions are possible, but not how fast the reactions occur. There is a need for chemical kinetics because some reactions of environmental interest in water are slow and not at equilibrium over the time scale of interest.

23.2.2  Why Are Some Reactions Not at Equilibrium?

equilibrium control (thermodynamic control): the situation in which it takes little or no energy to start the process of converting reactants to products and therefore species concentrations are determined by thermodynamics

To understand why some chemical systems are not at equilibrium, it is necessary to review why systems equilibrate. Recall from Chapter  3 that the driving force in approaching equilibrium is the minimization of the Gibbs free energy of the system. The example of oxygen partitioning between air and water was discussed quantitatively in Section  3.8.4. Recall that the energy profile was similar to that shown in Figure 23.1. The equilibrium position corresponds to the minimum in free energy. The equilibrium composition is determined by the equilibrium constant, which is in turn calculated from the standard Gibbs free energy of reaction. Not all reactions have energy profiles similar to Figure  23.1. One important feature of Figure 23.1 is that all system compositions have free energies less than the free energy of the reactants. In other words, it takes little or no energy to start the process of converting reactants to the equilibrium position. A system with this characteristic is said to be under equilibrium (or thermodynamic) control. The species concentrations in a system under equilibrium control are at their equilibrium values. In the case of the energy profile in Figure 23.1, it also takes little or no energy to start the process of converting products to the equilibrium position. The energy profile in Figure  23.1 is similar to the energy path taken by a skateboarder standing at the lip of a concrete trough. With little energy, the skateboarder can drop into the trough and travel back and forth until reaching the equilibrium

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23.3  |  Reaction Rates   561

Gibbs Free Energy

∆Grxn

Pure reactants

Equilibrium position

Pure products

FIGURE 23.1  Energy Profile for a System under Equilibrium Control

Gibbs Free Energy

∆G‡

Pure reactants

Pure products

FIGURE 23.2  Energy Profile for a System under Kinetic Control

position at the bottom of the trough. At the equilibrium position, the potential energy of the skateboarder is minimized. Another common energy profile is shown in Figure 23.2. It is important to see that the systems in Figures  23.1 and  23.2 have the same ΔGrxn. However, for the system depicted in Figure 23.2, energy must be added before reactants can proceed to the equilibrium position. The system in Figure 23.2 is said to be under kinetic control. In this system, the energy barrier to be overcome (ΔG‡) is more important than the difference in energy between reactants and products (ΔGrxn). This is analogous to a skateboarder who must go over a small hill before entering the trough. The size of the hill (analogous to ΔG‡) will determine the progress of the skateboarder, not the depth of the trough (analogous to ΔGrxn). To return to the question at hand, why are all reactions not at equilibrium? Some reactions require energy input to overcome an energy barrier. This means that significant time may be required to reach the equilibrium position, where the energy of the system is minimized.

23.3  REACTION RATES 23.3.1  Mechanisms and Molecularity The field of kinetics is concerned with the change in species concentrations over time. Thus, the rates of chemical reactions become important. In Section  23.3.2, you will

kinetic control: the situation in which it takes energy to start the process of converting reactants to products and therefore species concentrations are determined by kinetics Key idea: Some reactions require energy input to overcome an energy barrier and therefore are under kinetic control and not at equilibrium

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562   Chapter 23  Chemical Kinetics of Aquatic Systems

chemical mechanism: the nature of the interaction between chemical species in a chemical reaction

molecularity: the total number of reactant molecules (or ions) participating in the reaction

learn how the rates of change of chemical species are related and how to determine the rates for some reactions by looking at reaction stoichiometry. Before diving into the details, take a moment to reflect on how chemical reactions occur. Chemical reactions in homogeneous systems take place with chemical species that are close enough to interact. The interaction can take many forms, such as proton transfer, complexation, and electron exchange (redox reactions). If you had a viewing device with sufficient resolution, you could observe the exact ways that chemical species interact. The nature of the interaction is called the chemical mechanism. Kinetics and mechanisms are related. Chemical kinetics provides important information for the elucidation of chemical mechanisms. Also, predicted mechanisms can be used as a starting point to study chemical kinetics. When the time-­dependent nature of chemical reactions is to be emphasized, the reactions are written with a small arrow pointing from reactants to products. An example is the chemical reaction: A + 2B → 3C. Just as with equilibria, the symbols A, B, and C here represent species, not species concentrations. The numbers in front of the species names are stoichiometric coefficients. They represent the number of molecules (or ions) participating in the reaction. Thus, writing the reaction as A + 2B → 3C symbolizes that 1 mole of A and 2 moles of B react to form 3  moles of C. The total number of reactant molecules (or atoms or ions) participating in the reaction is called the molecularity of the reaction. The reaction A + 2B → 3C has a molecularity of 3 (1 from A plus 2 from B). The terms unimolecular, bimolecular, and termolecular describe reactions with molecularities of 1, 2, and 3, respectively. It is common practice to ignore the role of water in determining the molecularity of reactions in the aqueous phase. For example, the dissociation of acetic acid (HAc) to acetate (Ac–) and H+ usually is written as a unimolecular reaction (HAc → H+ + Ac–), even though you know that the actual mechanism is proton transfer to water (i.e., the bimolecular reaction HAc + H2O → H3O+ + Ac–).

23.3.2  Rates and Stoichiometry What does the “rate of a chemical reaction” mean? To understand the idea of the rate of a reaction, it is easier to start with the rate of change of the concentrations of indid A vidual species. Thus, we are seeking expressions for, say, in a system containdt ing species A. Determining the rates of change of individual species can be difficult. However, it is fairly easy with simple systems to calculate how the rate of change of the concentration of one species relates to the rate of change of the concentration of another species. As an example, suppose you are interested in a chemical system described only by the following chemical reaction: A + B → C. Without doing experiments, it is not d A d B d C possible to determine , , or . However, you can determine by inspection dt dt dt d C d A d B how , , and are related. Suppose you know by experiment that after 5 dt dt dt d A seconds of reaction, is equal to –3×10–3 mole/s. dt

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23.3  |  Reaction Rates   563

Thoughtful Paue What is d [B] after 5 seconds of reaction? dt   d B also is equal to –3×10–3 mole/s after 5 seconds dt of reaction, since the reaction stoichiometry tells you that 1 mole of B is consumed for every mole of A consumed. What about d C ? Again, by inspection, you know that dt d C 3 10 3 mole/s after 5 seconds of reaction, since the reaction stoichiometry dt dictates that 1 mole of C is produced for every mole of A or B consumed. Now consider the chemical reaction: A + 2B → C. Again, assume you know by d A d B experiment that after 5 seconds of reaction, 3 10 3 mole/s. Now what is dt dt after 5 seconds of reaction? You can see by the reaction stoichiometry that B is consumed twice as fast as A. For each mole of A that is consumed, 2 moles of B are cond B sumed. Therefore, 6 10 3 mole/s after 5 seconds of reaction. dt You know by inspection that

Thoughtful Paue d [ C ] What is   after 5 seconds of reaction in this case? dt d C 3 10 3 mole/s after 5 seconds of dt reaction, since the reaction stoichiometry dictates that 1  mole of C is produced for every mole of A consumed or for every 2 moles of B consumed. The relationships between the rates of changes of species concentrations can be generalized. To generalize, note that it is useful in chemical kinetics to treat the stoichiometric coefficients of reactants as negative and the stoichiometric coefficients of products as positive. This trick also was used with equilibria. With this sign convention in mind, consider this slightly more general reaction: A A B B CC D D, where νi is the stoichiometric coefficient for species i. The rates of changes of the individual species are related by: Again, by inspection, you know that



1 d A dt A

1 d B dt B

1 d C dt C

1 d D dt D

eq. 23.2

d A are negdt 1 d D 1 d A d D ative so is positive and both νD and are positive so also dt dt dt D A Note that the signs work out in eq. 23.2. For example, both νA and

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564   Chapter 23  Chemical Kinetics of Aquatic Systems Key idea: In a reaction, the rate of change of each species concentration divided by the species stoichiometry is the same for each species

rate of a chemical reaction: the rate of change of any species in the reaction over time divided by the signed stoichiometric coefficient of the species

is positive. For the most general reaction involving n species A1, A2, ... , An, you can n

write the reaction itself as

i

A i and the relationships between the rates of change of

i 1

species concentrations as

1 d Ai dt i

1 d Aj . dt j

How can you use these observations? First, relationships between the rates of change of species concentrations will save you time. Once you find an expression for the rate of change of one species concentration with time, you can write immediately the rate of change of all the other species in the reaction. Second, you can use these observations to write an expression for the rate of a chemical reaction. Equation 23.2 defines the rate of a chemical reaction as the rate of change of any species in the reaction over time divided by the signed stoichiometric coefficient of the species: rate

1 d Ai for any species A i in the reaction dt Ai

Another calculation of reaction rates is given in Worked Example 23.1. Worked Example 23.1 

Reaction Rates

Methylene chloride (formally, dichloromethane: CH2Cl2) reacts with bisulfide to form dithiomethane: CH 2 Cl 2

2 HS

CH 2 SH

2

2Cl

The reaction is monitored in the lab by measuring the chloride concentration. What is the reaction rate and rate of methylene chloride disappearance after 2 minutes of reaction if chloride appears at a rate of 10 mmol/min after 2 minutes? Solution 1 d Ai for any species Ai. If Ai is chloride, then the reacdt i tion rate after two minutes is (½)(10 mmol/min) = 5 mmol/min. 1 d Ai If Ai is methylene chloride, then = reaction rate = 5 mmol/min. 1 dt Thus, the rate of change of methylene chloride after two minutes is –5 mmol/min. The reaction rate is

23.3.3  Kinetic Order and Elementary Reactions At first appearance, it seems that the concepts of molecularity and stoichiometry could be combined to determine reaction rates. For example, suppose you know that

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23.3  |  Reaction Rates   565

species A and B are the only reactants and react with 1:1 stoichiometry. In this case, the molecularity is 2. If A and B react with 1:1 stoichiometry, then you might expect d A d B that the reaction rate is proportional to the concentrations of both dt dt A and B. If the reaction rate is proportional to the concentrations of both A and B, then it makes sense that the reaction rate is proportional to the product of the reactant concentrations: rate ~ [A][B]. Thus, you might be willing to accept reaction rates d B such as: d A k A B . In this expression, k is some proportionality condt dt stant and the negative sign is used to indicate that A and B are disappearing (so that d A and d A are both less than zero). Equations such as d B k A B are dt dt dt called rate expressions or rate laws. Proportionality constants such as k are called rate constants.1 The problem with this approach is that the reaction mechanism must be known before both the molecularity and stoichiometry of the reaction can be determined. In many cases, reaction rates are determined experimentally. If reaction rates are determined experimentally, then you do not really know how many molecules (or atoms or ions) react. In other words, you do not know the true molecularity. All you know is apparently how many molecules (or atoms or ions) react. This apparent molecularity is called the reaction order. If the experimentally deterd A k A B , then we dt say that the reaction is first order in A, first order in B, and second order overall. The reaction order with respect to each species is the exponent on the concentration of that species in the rate expression. The overall reaction order is the sum of the exponents on the species concentrations in the rate expression. The difference between the (true) molecularity and apparent molecularity (i.e., the reaction order) is important. The (true) molecularity is determined by the true mechanism. The reaction order comes from the hypothesized mechanism through experimental data or good guessing. For some (but certainly not all) chemical reactions, the reaction order and molecularity are identical. Reactions for which the reaction order and molecularity are identical are called elementary reactions. Another way to think of elementary reactions is that in an elementary reaction, the reaction order (and thus the rate expression) can be determined from the stoichiometry. As an example, suppose that you measure the rate at which the concentrations of species A and B change over time. You find that A disappears twice as fast as B appears. mined rate of change of species A with time is described by

Thoughtful Pause What is your guess of the reaction stoichiometry?

1  Rate constants have a deeper meaning than just a proportionality constant. The meaning of rate constants is explored more fully in Section 23.6.

rate expression (rate law): a mathematical expression for the rate of change of a species concentration rate constant: the proportionality constant in the rate expression that relates the rate of change of a species concentration with a function of species concentration reaction order: the apparent molecularity, calculated as the sum of the exponents on the species concentrations in the rate expression

elementary reactions: reactions for which the reaction order and molecularity are identical and thus the reaction order (and the rate expression) can be determined from the reaction stoichiometry

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566   Chapter 23  Chemical Kinetics of Aquatic Systems You might guess that the reaction stoichiometry is: 2A → B. If the reaction is not elementary, then you cannot write the rate expression. However, if the reaction is elementary, then the rate expression is given by the stoichiometry.

Thoughtful Pause What is the rate expression if the reaction is elementary?

d B dt expression for

If the reaction is elementary, then the rate expression for B is

2

k A .

Using the approach of Section  23.3.2, the rate A is: d A d A 2 2 12 k A or 2k A . dt dt Similarly, if you know the reaction is elementary, then you can deduce the reaction stoichiometry from the rate expression. Suppose in another reaction that B is produced and the rate of change of the concentration of B is found to be first order in the concentrations of both A and C.

Thoughtful Pause Key idea: Reaction rates are written as functions of reactant concentrations only

What is the reaction stoichiometry if the reaction is elementary?

The reaction stoichiometry is likely to be that 1 mole of A reacts with 1 mole of C to produce 1 mole of B, or A + C → B. Without being stated explicitly, an important convention has been used to write rate expressions in the examples shown in this section. The convention is that all the reaction rates are written as a function of reactant concentrations only. In other words, a reaction such as A + C → B, if elementary, means that A and C react in some fashion to form B. Thus, the rates of change of A, B, and C all depend only on reactant concentrations (i.e., only on [A] and [C]) and not on [B]. If the rate of change of the concentration of B is found to depend on [B], then a new reaction must be written.

23.3.4  Systems of Elementary Reactions In environmental settings, it is not unusual to have many chemical reactions involving the same species occurring simultaneously. Thus, a given species may be found in several or even dozens or hundreds of reactions. Regardless of the number of reactions, it is fairly simple to write rate expressions for each species if the reactions are elementary. Three basic principles are used to write rate expressions with a system of reactions. First, as discussed in Section 23.3.3, the reaction rates for individual reactions are determined from reactant concentrations only.

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23.3  |  Reaction Rates   567

Second, the reaction rates for individual reactions can be determined directly from the reaction stoichiometry for elementary reactions. Third, rates are additive. Thus, if the rate of change of the concentration of A with respect to time is –k1[A][B] from one reaction and +k2[C]2 from another reaction, then the overall rate of change of [A] with time is the sum of the rates: d A dt



k1 A B

Key idea: Rates are additive because mass is conserved

2 k2 C

Before reviewing the calculation procedure, make sure it makes sense to you that rates are additive. Recall that mass is conserved. In a constant volume system, you are used to writing mass balances in terms of concentration. Imagine you had a closed system to which you added some H2CO3* and Na2CO3 (which dissociates completely to Na+ and CO32–). You could write:

[HCO3 ] [HCO3 ] from H 2 CO3* [HCO3 ] from CO32

Now differentiate both sides with respect to time: d[HCO3 ] dt

rate of [HCO3 ] formation from H 2 CO3 rate of [HCO3 ] formation from CO32



As this simple example shows, rates are additive because mass is additive. And mass is additive because mass is conserved. The process of writing rate expressions with more complex systems can be broken down into a series of steps. The steps are shown here and summarized in Appendix C (Section C.10). The process will be illustrated with the following set of elementary reactions:



Reaction 1: A B C; Reaction 2: C A B; Reaction 3: 2C A;

k1 k2 k3

We seek to write the rate expressions for the concentrations of A, B, and C. Step 1: Identify the elementary reaction(s) that include the species of interest as either a reactant or product. For A, the pertinent reactions containing A are reactions 1, 2, and 3. Species B appears only in Reactions 1 and 2. Species C appears in all three reactions. Step 2: Write the rate expressions for each species of interest, using the elementary reactions identified in Step 1. Each rate expression should be the rate constant multiplied by the reactant concentrations, each raised to their stoichiometric powers. The rate is negative (multiply the rate by –1) with when the species of interest is a reactant. The rate is positive when the species of interest is a product. Remember to correct for the stoichiometry of the species of interest, if the stoichiometry is not one.

Key idea: To write rate expressions with a system of elementary reactions, remember that (1) the reaction rates for individual reactions are determined from reactant concentrations only, (2) the reaction rates for individual reactions can be determined directly from the reaction stoichiometry for elementary reactions, and (3) rates are additive

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568   Chapter 23  Chemical Kinetics of Aquatic Systems For A: rate  from  reaction 1    k1 A B rate    0  since  A  is  consumed rate  from  reaction 2     k2 A C rate    0  since  A  is  consumed 2

rate  from  reaction 3    k3 C   rate    0  since  A  is  formed For B: rate  from  reaction  1 

k1 A B

             rate    0  since  B is  consumed   rate  from  reaction  2   k2 A C              rate    0  since  B is  formed                For C: rate  from  reaction  1  k1 A B rate    0  since  C  is  formed rate  from  reaction  2   k2 A C rate    0  since  C  is  consumed 2

rate  from  reaction  3  2 k3 C rate    0  since  C  is  consumed                 Note that the rate expressions use only reactant concentrations. Also note the factor of 2 in the rate of [C] over time with reaction 3. For that reaction: d C 2 2 1 k3 C , so the rate from reaction 3 is d C 2k 3 C . 2 dt dt Step 3: Add up the rates from each elementary reaction to find the overall rate of change of each species concentration. From Step 2:

d A dt d B dt



k1 A B

d C dt

k2 A C

k1 A B k1 A B

2 k3 C

k2 A C

k2 A C

2 2 k3 C

eq. 23.3 eq. 23.4 eq. 23.5

If you have experience with solving differential equations, you will recognize that eqs. 23.3–23.5 constitute a set of three coupled differential equations in three unknowns. This set of equations can be solved numerically to find [A], [B], and [C] over time if

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23.4  |  Common Rate Expressions   569

the rate constants and initial concentrations of A, B, and C are known. For another example of writing rate expressions, see Worked Example 23.2. Worked Example 23.2 

Systems of Reactions

When ozone gas dissolves in water, it decomposes fairly rapidly. One set of pertinent reactions in “pristine” water is: O3 OH   HO2 O2 HO2 O3   HO2 O3 HO2    H O2 O 2 O3   O 2 O3   O3 H 2 O OH   OH OH O3   HO2 O2

O2

k1 k2 k3 k4 k5 k6

Write an expression for the rate of change of the ozone concentration with time. Solution Although the reaction sequence is complex, simply apply the steps in the text. The reactions having ozone as a reactant or product are reactions 1, 2, 4, and 6. d O3 The rate terms for from reactions 1, 2, 4, and 6 are, respectively, –k1[O3] dt [OH–], –k2[HO2–][O3], –k4[O2–][O3], k5[O3][H2O], and –k6[OH][O3]. Note that the rate expressions use reactant concentrations only. Adding the rate expressions for the individual reactions: d O3 dt

k1 [OH ] k2 [HO 2 ] k4 [O 2 ] k5 [O3 ][H 2 O] k6 [OH ] [O 3 ]

23.4  COMMON RATE EXPRESSIONS 23.4.1  Introduction A number of simple rate expressions are seen repeatedly in aquatic chemistry. Several common forms will be discussed in this section. For each form, the rate law will be presented. As indicated in Section 23.3, the rate laws are differential equations and are sometimes called the differential form of the rate law. To determine the species concentration over time, the rate expression will be integrated. This sometimes is called the integrated form of the rate law. In addition, environmental examples of each rate law will be presented. The important features of each rate expression are summarized in Table  23.1. In addition, examples of species concentrations over time with each rate law are shown in Figure 23.3.

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570   Chapter 23  Chemical Kinetics of Aquatic Systems Table 23.1 Summary of the Characteristics of the Simple Kinetic Orders Order

Elementary Reaction

Differential Form2

Zero

None

First

A → products

d A dt

k1 A

Second

2A → products

d A dt

k2 A

d A dt

A + B → products

nA → products

nth

d A dt

k0

kn A

Units of k1

[A] = [A]0 – k0t

mol L-time

A0 2k0

1 time

ln 2 k1

L mol-time

1 A 0 k2

L mol-time

note 3

Ln 1 mol -time

note 5

A 2

A

k2 A B

d A dt

Integrated Form

A

A 0e

k1t

A0 1 k2 A 0 t B B0 e A0

A

0

B0 A

0

0 k2t

1

note 4

n

t½

n 1

Notes: 1

 Assumes the concentration units are mole/L.  It is traditional to define the rate constants here so that the rate laws are written as

2

and

d A dt

d A n kn A , not dt

n nkn A .

d A dt

d A 2 k2 A , not dt

A0 3  If [A]0 ≥ 2[B]0, then [A] will never be as small as and the half-­time is infinite. If [A]0 < 2[B]0, then t1 2 2k0 A

 t1 2

5

A A

kn t 1 n

1 1 n

ln 2 B0

A0 B0 . A 0 k2

1 n 0

2n 1 1 for n 1 n 1 kn

100

Concentration (mM)

 

4

1 n 0

2 2 k2 A ,

Zero order: k0 = 2 mM/min

80 60

First order: k1 = 1 min–1 40

Second order: k2 = 1 mM–1min–1

20 0 0

5

10 Time (min)

FIGURE 23.3  Examples of Zero-­, First-­, and Second-­order Kinetics

15

20

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23.4  |  Common Rate Expressions   571

23.4.2  Zero-­Order Kinetics Recall that the kinetic order is the sum of the exponents on species concentrations in the rate law. The zero-­order rate expression refers to a rate law in which the exponent for each species is zero. For zero-­order disappearance: d A dt



k0 A

0

k0

(The subscript 0 is used here simply as a reminder that the rate constant is a zero-­ order rate constant.) Note that a zero-­order reaction has a constant reaction rate; that d A is, is constant. In addition, zero-­order rate constants have units of mol/L-­time. dt The rate expression can be integrated to show: [A] = [A]0 – k0t, where [A]0 is the concentration of A at t = 0. Thus, a substance exhibiting zero-­order disappearance will show a linear decrease in concentration over time (see Figure 23.3). One way of expressing the rate of a reaction is to determine the time required to reduce a species concentration to one-­half is initial value. This is called the half-­time of the reaction (or sometimes, the half-­life of species A) and is given the symbol t½. From the integrated form of the rate expression, the half-­time of a zero-­order reaction A0 is . Note that the half-­time increases with increasing [A]0. 2k0 Recall that elementary reactions have rate laws that come directly from reaction stoichiometry. Thus, no elementary reaction by itself can exhibit zero-­order kinetics. In spite of this conclusion, zero-­order rate expressions are observed fairly frequently in environmental systems. Why? Zero-­order kinetics can result as the limiting form of more complicated rate laws. One example of this phenomenon is the chemical reaction of adsorbed species (called heterogeneous catalysis). Using the symbols from Section 22.3.2, consider a species P equilibrating with a surface ≡S to form a surface complex ≡S-­P with an equilibrium constant K. From eq.  22.6, the adsorbed species concentration is given by:

SP

K P ST , where ST is the total site concentration. 1 K P

Now imagine that the adsorbed species reacts on the surface to form a product, P´:

SP

P ; rate constant

k

Therefore, the rate of change of [P′] is:

d P dt

k

SP

k

K P ST 1 K P

This is called the Langmuir-­Hinshelwood equation (after Irving Langmuir and d P Cyril Norman Hinshelwood, 1897–1967). If K[P] >> 1, then kST . Thus, at the dt

Key idea: In the zero-­order rate law, the exponent for each species is zero and the species will show a linear change with time half-­time of the reaction (t½): the time required to reduce a species concentration to one­half its initial value

Key idea: No elementary reaction can exhibit zero-­order kinetics

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572   Chapter 23  Chemical Kinetics of Aquatic Systems beginning of the adsorption process, when [P] is large (and K[P] may be >> 1), the formation of P´ will exhibit zero-­order kinetics with this type of catalysis, since

d P dt

constant

kST

Another example comes from the transfer of gases between the aqueous phase (where they exist as dissolved gases) and gaseous phase. A common kinetic model takes the following form:

d A(aq ) dt

kL a

PA KH

A(aq )

where kL is a mass-­transfer coefficient, a is the ratio of the interfacial surface area to the liquid volume, PA is the partial pressure of A in the gas phase, and KH is the Henry’s law constant. Under conditions in which [A] is small and PA is constant (e.g., in the early stages of mass transfer from the atmosphere to a small liquid volume):

d A(aq ) dt

k L aPA KH

Thus, zero-­order kinetics is observed.

23.4.3  First-­Order Kinetics For first-­order disappearance, the rate law is:

Key idea: In the first-­order rate law, the exponent for the species is 1 and the species will show an exponential change with time

d A dt

k1 A

1 Thus, first-­order rate constants have units of . The rate expression can be time integrated to show:

A

A 0e

k1t



Thus, a substance exhibiting first-­order disappearance will show an exponential decrease in concentration over time (see Figure 23.3). Similarly, exponential growth in dN a population stems from a first-­order rate law: rN , where N is the population dt at time t and r is a first-­order growth rate constant. The half-­time for a first-­order process is

ln 2 . First-­order kinetics is the only kinetic order where t½ is independent of the k1

initial concentration. Many processes in the environment exhibit first-­order kinetics. Examples include radioactive decay and exponential growth. A good example of first-­order kinetics is Chick’s law of disinfection, a common model for disinfection (after British microbiologist Harriette Chick, 1875–1977; see the Historical Note at the end of this chapter). Chick’s law states that the rate of disinfection is proportional to the number concendN tration of surviving organisms. Thus: k1 N , where N is the number of surviving dt

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23.4  |  Common Rate Expressions   573

organisms per liter. If Chick’s law holds, then you would expect an exponential decrease over time in surviving organisms. Another illustration is shown in Worked Example 23.3.

Worked Example 23.3 

First-­Order Kinetics

The pesticide ethylene dibromide (EDB, formally 1,2-­dibromoethane: BrH2C-­ CH2Br) can hydrolyze (react with water) to form the corresponding alcohol. The first-­order hydrolysis rate constant is 6×10–9 s–1 at pH 7 and 25°C (Haag and Mill 1988). What is the half-time of EDB pH 7 and 25°C? Most uses of EDB were banned in 1984. If the EDB concentration in a water was 2 μg/L in 1984, when did the EDB concentration fall below the drinking water standard of 0.05 μg/L at pH 7 and 25°C? Solution The reaction is first order in EDB. If the hydrolysis reaction is elementary, dC then the appropriate rate expression is: kC. where C is the EDB condt centration. Integrating: C C0 e k1t, where C0 is the initial EDB concentration. 1 C Rearranging: t ln k C0 ln 2 ln 2 The half-­time is = 3.7 years. The time required to reduce k 6 10 9 s 1 g 0.05 1 C 1 L = the concentration from 2 to 0.05 μg/L is: ln ln g k C0 6 10 9 s 1 2 L 6.15×108 seconds, or 19.5 years. This would have occurred in about 2004. The hydrolysis reaction is slow.

Even more common than first-­order elementary reactions is a kinetic form that appears as first-­order kinetics. Consider the reaction of two species as follows: A + B → products. If this is an elementary reaction, then the reaction is second order and the rate expression is given by:

d A dt

k A B

Now imagine that [B] is present initially at a very high concentration relative to the initial concentration of A. In this case, [B] will stay pretty constant at its initial value [B]0. If [B] is pretty constant, then the rate expression can be written as d A k A, dt where k′ = k[B]0. This form of first-­order kinetics is called pseudo-­first-­order kinetics. The pseudo-­first-­order rate constant, k′, is the product of a second-­order rate constant and (usually) the concentration of another species. Pseudo ­first-­order kinetics is very common in environmental systems. For example, reactions with H+ at constant pH

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574   Chapter 23  Chemical Kinetics of Aquatic Systems (and ionic strength) generally result in pseudo-­first-­order kinetics with k´ = k[H+]. Pseudo-­first-­order kinetics also is shown in Worked Example 23.4 Worked Example 23.4 

Pseudo-­First-­Order Kinetics

The compound ethyl acetate (formally, ethyl ethanoate) is used as a solvent, as an insecticide, and to impart a fruity aroma. It hydrolyzes by three mechanisms: reaction with water (considered first order with rate constant kW = 5×10–10 s–1), reaction with H+ (second order, kH = 1.1×10–4 M–1s–1), and reaction with OH– (second order, kOH = 1.1×10–1 M–1s–1). Show how ethyl acetate hydrolysis depends on pH. Solution Assume the reactions are elementary. Since rates are additive, the overall rate expression is: dC dt

(kW

kH [H ] kOH [OH ]C

where C is the ethyl acetate concentration.

dC If the pH is constant, then this becomes: kC , where: k = kW + kH[H+] dt ln 2 + kOH[OH–]. The half-­time is and is plotted as a function of pH below. Note k that the slowest rate (largest half-­time) occurs at about pH 5.5. Half‐time (years)

20 15 10 5 0

5

6

7

8

pH

23.4.4  Second-­Order Kinetics Second-­order kinetics can take two very different forms. In the first form, a chemical species reacts with itself to form products: 2A → products. The resulting rate expression is d A dt



2 k2 A

Be a little careful here. At times, the stoichiometric coefficient is incorporated into the rate constant, as in Table 23.1. Sometimes you see the stoichiometric coefficient 2 written explicitly: d A 2 k2 A . dt

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23.4  |  Common Rate Expressions   575

A0 1 and the half-­time is . Note that the A 0 k2 A 0 k2 t half-­time decreases with increasing [A]0. The units of the second-­order rate constant L are . This “second order in A” form of second-­order kinetics is relatively rare mol-time in environmental systems. It is sometimes used to model particle-­particle collisions, where [A] is interpreted as the number of particles per unit volume. A more common form of second-­order kinetics involves the interaction of two chemical species: A + B → products. This might be called the “first order in both A and B” case. In this case: The integrated form is A

d A dt



k2 A B

eq. 23.6

L . Clearly, mol-time the differential equation in eq. 23.6 cannot be solved because it is one equation in two unknowns (namely, [A] and [B]). From the stoichiometry of the reaction, it makes sense that the amount of A consumed in the reaction must be equal to the amount of B consumed. The amount of A consumed is [A]0 – [A]. The amount of B consumed is [B]0 – [B]. So: [A]0 – [A] = [B]0 – [B], or: Again, note that the units of the second-­order rate constant are



B

B

A

0

0

A

Substituting into eq. 23.6:

d A dt

k2 A

B

0

A

0

A

Or:

d A B0 A

A

0

A

k2 dt

eq. 23.7

Solving the differential equation in eq. 23.7:

A

B0 A0 0 B 0 B 0 A 0 k2 t e 1 A0

eq. 23.8

Does eq. 23.8 make sense? One way to decide if expressions such as eq. 23.8 are reasonable is to evaluate them at the extreme values of t. As you can see from eq. 23.8, [A] = [A]0 at t = 0, as required. As t → ∞, [A] → 0 if [B]0 > [A]0. This makes sense: In the presence of excess B, all A will be consumed eventually. However, if [A]0 > [B]0, then [A] → [A]0 – [B]0 as t → ∞. Again, this makes sense: If A is in excess, B will be consumed completely given enough time. If [B]0 moles/L of B are consumed, then [B]0 moles/L of A will be consumed and [A] → [A]0 – [B]0. Equation 23.8 is not valid if [A]0 = [B]0. If [A]0 = [B]0, then eq.  23.7 is equivalent to d A = –k2[A]2 and the dt ­“second order in A” solution holds.

Key idea: Second-­order kinetics typically come from reactions that are second order in one reactant or from reactions that are first order in two reactants

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576   Chapter 23  Chemical Kinetics of Aquatic Systems 100 80 [A]0 > [B]0 ([B]0 = 50 mM)

[A] (mM)

60 40

[A]0 = [B]0 ([B]0 = 100 mM)

20 0

[A]0 < [B] ([B] = 150 mM) 0 0

0

2

4

6

8

10

12

14

16

18

20

Time (min) FIGURE 23.4  Effects of lnitial Concentrations on the Concentration Profiles in Second-­Order Reactions (k2 = 0.005 L/mmol-­min, [A]0 = 100 mM)

Figure 23.4 shows three cases of second-­order kinetics. In all cases, [A]0 = 100 mM. The case of [B]0 > [A]0 is illustrated with [B]0 = 150 mM. Note that [A] decreases to zero, as expected. The case of [A]0 > [B]0 is illustrated with [B]0 = 50 mM. Note that [A] appears to be leveling out near [A]0 – [B]0 = 50 mM. Finally, the case of [A]0 = [B]0 is shown with [B]0 = 100 mM. Another illustration of second-­order kinetics is shown in Worked Example 23.5. Worked Example 23.5 

Second-­Order Kinetics

Dissolved ozone reacts with toluene with a second-­order rate constant of 14 M–1 s–1 (Staehelin and Hoigné 1983). If the initial dissolved ozone and toluene concentrations are 2 mg/L and 10 mg/L, respectively, what will the toluene and ozone concentrations be after 20 minutes? Solution The initial ozone and toluene concentrations are 1 10 92

2

g L = 1.09×10–4 M, respectively.

2 10 48

3

g L = 4.17×10–5 M and

g mol

g mol

Plugging these values into eq. 23.8 (species A corresponding to toluene) with k = 14 M–1 s–1 and t = 1200 s yields a toluene concentration of 8.57×10–5 M or 7.9 mg/L. Assigning A to ozone yields a dissolved ozone concentration of 1.87×10–5 M or 0.9 mg/L. Note: By 4400 seconds, the ozone has decreased to about 1% of its initial value and the toluene concentration is approaching its limiting value of [toluene]0 – [ozone]0 = 6.7×10–5 M or about 6.2 mg/L.

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23.5  |  More Complex Kinetic Forms   577

23.5  MORE COMPLEX KINETIC FORMS 23.5.1  Introduction To analyze more complex kinetic systems of elementary reactions, write the set of differential equations and solve them with the appropriate initial conditions. The required initial conditions typically are the initial concentrations of all species. Writing the set of differential equations is fairly easy, using the approach shown in Section 23.3.4. Solving the differential equations can be tedious. Numerical solution methods frequently are needed. In this section, some representative examples of more complex systems are presented and solved.

23.5.2  Kinetics of Reversible Reactions Most of this book has focused on reversible reactions at equilibrium. Reversible reactions also can be analyzed to determine how the species change over time as equilibrium is approached. Without doing the mathematics, you know there must be a link between kinetics and equilibrium.

Thoughtful Pause What is the link between the kinetic and equilibrium approaches to a system with reversible reactions?

The kinetic equations for a reversible system must give the equilibrium concentrations as t → ∞. To demonstrate this point, consider a simple reversible system consisting of two reactions: A B



B; k f A; kr

This system is equivalent to: A = B. Both kf and kr are first-­order rate constants. The subscripts used in this example are a reminder that the rate constants refer to the forward (reactant-­to-­product) and reverse (product-­to-­reactant) reactions that make up the reversible system. Following the procedure in Section  23.3.4, you can write the rate expressions for going on.)

d A d B and . (You may wish to try this on your own before dt dt



d A dt

kf A

kr B



d B dt

kf A

kr B

eq. 23.9

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578   Chapter 23  Chemical Kinetics of Aquatic Systems

Key idea: The kinetic equations for a reversible system must give the equilibrium concentrations as t → ∞

A mass balance reveals that [A] + [B] must equal the total initial concentrations added. For this example, consider the addition of A only ([B]0 = 0), so: [A] + [B] = [A]0 or: [B] = [A]0 – [A]. Substituting into eq. 23.9: d A dt



kf A

kr A

0

A

Solving with [A] = [A]0 at t = 0:

Similarly: B

A

0

A

A

0

A

A

0

kf

kf

kr

kf

kr

kr

1

kf e kr

1

kf e kr

k f kr t

k f kr t



eq. 23.10

.

There are several interesting features of the time-­dependent solution to the ­reversible system (eq. 23.10). First, you can compare the time-­dependent solution as B t → ∞ with the equilibrium solution. At equilibrium, K, where K is the equilibA kf rium constant for the equilibrium A = B. As t → ∞ in eq. 23.10, A A0 k f kr kr kr B kr A0 and B . Thus: . This implies that K . The kinetic k f kr kf A kf c­ alculations yield a remarkable result: an equilibrium constant can be expressed as the ratio of the forward and reverse rate constants. k In fact, you can show that r K much more easily. At equilibrium, the species kf d A concentrations do not change with time. From eq. 23.9: = – kf [A] + kr[B] = 0 dt B kr B at equilibrium. Thus: at equilibrium. Since K at equilibrium, then, as A kf A kr before, K .2 kf A second interesting feature of eq. 23.10 is the kinetics of the approach to equilibrium.

Thoughtful Pause What is the apparent kinetic order and rate constant for the approach to equilibrium? Based on eq.  23.10, the approach to equilibrium appears to be first order, with a first-­order rate constant equal to kf + kr. Thus, the sum of the forward and reverse rate constants determines the apparent reaction rate constant.  This analysis assumes that the forward and reverse reactions go through the same transition state. This statement is not true for all equilibria.

2

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23.5  |  More Complex Kinetic Forms   579

You can use the discussion in this section to decide if elementary reactions can be considered at equilibrium over the time frame of interest. We almost always think of acid–base reactions as being at equilibrium. Why? The usual explanation is that proton-­transfer reactions are fast. In fact, the transfer of a proton from H3O+ to a conjugate base of the form A– is exceedingly fast. As an example, the second-­order rate constant for OH– + H3O+ → 2 H2O is about 10+11.1 L/mol-­s. This is the fastest known reaction in water.3 Even for acetic acid, the proton-­transfer reactions are very fast:

HAc   H 2 O

Ac

H 3 O              k2, f

10  

5.9

Ac

HAc   H 2 O              k2,r

10  

10.7

H3O

L / mol -s L /mol -s

If sodium acetate (NaAc) is added to water at a constant pH of 7.0, what is the half-­time of the conversion of Ac– to HAc? The analysis in this section has been conducted with first-­order reactions approaching equilibrium. Therefore, you need to convert the second-­order rate constants into pseudo-­first-­order rate constants. For the forward reaction: kf = k2,f [H2O] = (10+5.9 L/mol-­s)(55.56  M) = 10+7.6 s–1. For the reverse reaction: kr = k2,r[H3O+] = (10+10.7 L/mol-­s)(10–7 M) = 10+3.7 s–1. The half ln 2 ln 2 ­time is 1.7 10 8 s or 17  nanoseconds. The equilibrium k f kr 10 7.6 s 1 10 3.7 s 1 assumption for fast reactions clearly is justified at reasonable times. Another example of the approach to equilibrium is shown in Worked Example 23.6.

Worked Example 23.6 

Approach to Equilibrium

In wastewater that has been disinfected by chlorination, ammonia reacts with hypochlorous acid to form monochloramine (NH2Cl) as follows: HOCl NH3

NH 2 Cl H 2 O

with kf = 4.2×106 L/mol-­s at 25°C. Monochloramine also can hydrolyze: NH 2 Cl H 2 O

HOCl NH 3

with kr = 2.1×10 L/mol-­s at 25°C. Find the HOCl and monochloramine concentrations over time if 0.1 mg/L as Cl2 of chlorine is added to 10 mg/L as N of ammonia. –5

Solution The initial ammonia and chlorine concentrations are

1 10

2

g L = 7.1×10–4 M

g 14 g mol 1 10 L = 1.4×10–6 M, respectively. Since the ammonia is in great excess, and g 70.9 mol 4

 This reaction is so fast that the rate of the reaction is limited only by the diffusion of the reactants through the media (water). Reactions such as this are called diffusion-­controlled reactions. The theoretical upper limit for second-­order rate constants of diffusion-­controlled reactions is in the 109 – 1011 L/mol-­s range.

3

Key idea: The sum of the forward and reverse rate constants determines the apparent reaction rate constant in the approach to equilibrium

ISTUDY

580   Chapter 23  Chemical Kinetics of Aquatic Systems assume that [NH3] is constant. You can calculate a pseudo­first-­order rate constant = kf ´ = kf [NH3] = (4.2×106 L/mol-­s)(7.1×10–4 M) = 3.0×103 s–1. Plugging into eq. 23.10 with [A]0 = initial chlorine concentration = 1.4×10–6 M and [B]0 = initial monochloramine concentration = 0, you will obtain the plot shown below. Concentration (μM)

1.5 NH2Cl

1.0 0.5

HOCl 0.0 0.0

1.0

2.0

Time (ms)

The reaction is very fast, with a half-­time of about 0.23 ms.

23.5.3  Kinetics of Sequential Reactions In the environment, it is not uncommon for the products of one reaction to be reactants in a second reaction and subsequently converted into new products. The overall reaction sequence is A → B → C, or: A → B; kA,1 and B → C; kB,1. If the two reactions are elementary, then the corresponding rates are (you may wish to try writing the rate expressions before proceeding): d A dt d B dt d C dt



kA,1 A kA,1 A

kB,1 B

kB,1 B

Solving for [A] first, then [B], and finally [C], you can show that for [B]0 = [C]0 = 0: A

A 0e

B

A

0

C

A

0

A

0

kA ,1t

kA,1 kB,1 kA,1 A 1

e

kA ,1t

e

kB,1t

kB,1e

kA ,1t

B 1 kB,1 kA,1

kA,1e

kB,1t

An example is shown in Figure 23.5. Note that [A] decreases exponentially, whereas [B] shows a maximum and [C] exhibits a sigmoidal (S–­shaped) growth. You can show (Problem 6) that the maximum value of [B] and the time at which the maximum value of [B] occurs depend on both rate constants. If kB,1 is much greater than kA,1, then [B] will be very small at all times and [C] will increase without a lag. Under these conditions, we say that A → B is the

ISTUDY

23.5  |  More Complex Kinetic Forms   581 120

Concentration (mM)

100 [A]

[C]

80 [B]

[A]0 = 100 mM, [B]0 = [C]0 = 0

60

kA,1 = 0.1 s–1, kB,1 = 0.025 s–1

40 20 0 0

25

50

75

100

125

150

Time (s) FIGURE 23.5  Sequential First-­Order Reactions (A → B; kA,1 and B → C; kB,1)

120 [C]

Concentration (mM)

100 [A] 80 [A]0 = 100 mM, [B]0 = [C]0 = 0

60

kA,1 = 0.1 s–1, kB,1 = 1.0 s–1

40

Dotted line is C produced 

20

from A with k1 = 0.1 s–1

[B]

0 0

25

50

75

100

125

150

Time (s) FIGURE 23.6  Sequential Reactions with

k B,1 k A, 1

10 Compared to A → C

rate-­determining step in the sequence A → B → C. In other words, kA,1 will determine the concentrations of the species. Under such conditions, it is easy to miss B since the concentration profiles are consistent with A → C (see Figure 23.6). Another set of sequential chemical reactions is shown in Worked Example 23.7.

rate-­determining step: the slowest step in a reaction sequence and thus the step that sets the overall rate of reaction (also called the rate-­limiting step)

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582   Chapter 23  Chemical Kinetics of Aquatic Systems

Sequential Reactions

Worked Example 23.7 

Thiocyanate (SCN ) is used in a number of industrial processes and is a byproduct of coke manufacturing. It can be removed from water by oxidation with ozone. The oxidation of thiocyanate produces cyanide, which is subsequently oxidized by ozone. The pertinent reactions are: –

SCN

2 O3

CN

2OH

O3

CN

CNO

products; k A

products; k B

The species CNO– is called cyanate. At pH 7, Tuan and Jensen (1993) found that kA[OH–]2 = 5×104 L2/mmol2-­min and kB = 2 L/mmol-­min. If the dissolved ozone concentration is fixed at 0.002 mM, how do the thiocyanate, cyanide, and cyanate concentrations change over time? Assume that the initial thiocyanate concentration is 8 mM and the initial cyanide and cyanate concentrations are zero.

Concentration (mM)

Solution The equations in the text for sequential first-­order reactions are valid if pseudo-­ first-­order rate constants are calculated. Use the equations with A = thiocyanate, B = cyanide, C = cyanate, k1 = (5×104 L2/mmol2-­min)(0.002 mM)2 = 0.2 min–1 and k2 = (2 L/mmol-­min) (0.002 mM) = 4×10–3 min. The concentration profiles are shown in the plot below. 10 8

CN–

6 4 2 0

SCN– CNO– 0 25 50 75 100 Time (min)

Note that the cyanide concentration remains high because the kB process is slow relative to the kA process.

23.6  EFFECTS OF TEMPERATURE AND IONIC

STRENGTH ON KINETICS

23.6.1  Effects of Temperature There are many approaches to determining or modeling the effects of temperature on rate constants. All of the most common approaches can be summarized in the following dependency (Espenson 1981):

k

(constant )T n e

energy term RT



eq. 23.11

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23.6  |  Effects of Temperature and Ionic Strength on Kinetics   583

In eq. 23.11, T is the temperature in K and R is the ideal gas constant. The values of the constant n and the energy term will be discussed for two models. The Swedish chemist Svante August Arrhenius (1859–1927; see the Historical Note at the end of this chapter) reasoned that rate constants should change with temperature in the same fashion that equilibrium constants change with temperature (Arrhenius 1896). His based this idea on the fact that an equilibrium constant sometimes can be thought of as the rate of two rate constants (see Section 23.5.2). Relating eq. 23.11 with the van’t Hoff equation (Section 21.3.1), he found n = 0. In Arrhenius’s approach, the constant in eq. 23.11 is given the symbol A (also called the pre­-exponential factor). The energy term in eq. 23.11 is given the symbol Ea (also called the activation energy or the Arrhenius activation energy. Note that A has the same units as the rate constant. Thus, eq. 23.11 becomes: k



Ae

Ea RT



eq. 23.12

This is called the Arrhenius equation. Although the Arrhenius equation is empirical, the activation energy, Ea, is related to the energy required to overcome the energy “hill” (ΔG‡ in Figure 23.2). The linearized form of the Arrhenius equation is:

ln k

ln A

Key idea: One model for the effect of temperature on rate constants is the Arrhenius equation: k = Ae

Ea RT .

pre-­exponential factor: the parameter in the Arrhenius equation that multiplies the exponential term activation energy (Arrhenius activation energy): the energy term in the Arrhenius equation

Ea 1 R T

1 . T Another approach was adopted by Henry Eyring (1901–1981) and is called transition-­state theory. Transition-­state theory assumes that all chemical reactions proceed through a complex called the activated complex. Reactants are considered  to  be in equilibrium with the activated complex. Thus, according to transition-­state theory, the reaction A + B → C is more properly written as: A + B = AB‡ and AB‡ → C. The superscript “‡” denotes the properties of the activated complex. Thus, ΔG‡ in Figure  23.2 refers to the free energy of reaction for the activated complex. In transition-­state theory, the reaction rate is assumed to be proportional to the concentration of the activated complex, [AB]‡. If the equilibrium constant for A + B = AB‡ is K‡ so that the rate is proportional to [AB‡] = K‡[A][B], then the second-­order rate constant should be proportional to K‡. The proportionality constant is R , where Nh κ = transmission coefficient (usually taken as 1), N = Avogadro’s constant, and h = Planck’s constant. Expressing the equilibrium constant in terms of the corresponding standard Gibbs free energy, you can show that the values in eq.  23.11 are: n = 1, ‡ R RS , and the energy term = ΔH°‡. Here, ΔS°‡ and ΔH°‡ are the standard constant e Nh entropy of reaction and standard enthalpy of reaction for A + B = AB‡. In practice, the Arrhenius equation and transition-­state theory give similar fits to plots of lnk versus 1 . As an example of the use of the Arrhenius equation, we can T explore the science behind the observation that many biochemical reactions double in rate for a 10°C increase in temperature. In other words, the lnk should be linearly related to

transition-­state theory: a reaction theory based on the assumption that the reactants equilibrate with an activated complex, which subsequently decomposes to form products

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584   Chapter 23  Chemical Kinetics of Aquatic Systems

Thoughtful Pause Using the Arrhenius equation, what can you say about the activation energy if a rate constant doubles with a 10°C increase in temperature? Evaluating the Arrhenius equation at two temperatures, T1 and T2, yields the following: kT1 kT2



e

Ea R

1 1 T1 T2



Rearranging:

Ea

R

ln

kT1

kT2 1 1 T1 T2

1 1 varT1 T2 ies only slightly for a constant value of T1 – T2. For T1= 10°C and T2= 20°C, the term kT 1 1 is 1.20×10–4 K–4. In this case, 1 = ½, so: kT2 T1 T2

Over temperatures of interest with aquatic systems (0 to 30°C), the term



Ea

ln

kT1

kT2 R 1 1 T1 T2

1 ln kJ 2 8.314 10 3 mol-K 1.20 10 4 K 1 48 kJ / mol

In other words, reactions with activation energies of about 48 kJ/mol will experience about a doubling in their rate constants with a 10°C increase in temperature. Another illustration showing the effect of temperature on rate constants is shown in Worked Example 23.8.

Worked Example 23.8 

Effects of Temperature

For the 1,2-­dibromoethane case in Worked Example 23.3, the value of the preexponential factor (A) is about 1010.5 s–1 and Ea is about 105 kJ/mol. If the hydrolysis rate constant is 6×10–9 s–1 at 25°C (and pH 7), what is the half-­time at an average groundwater temperature of 10°C? Solution From the text:

kT1

kT2

e

Ea R

1 T1

1 T2

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23.6  |  Effects of Temperature and Ionic Strength on Kinetics   585

If T1 = 10°C (= 283.15 K) and T2 = 25°C (= 298.15 K), then: ln

So:

105 kJ

k10 k25

8.314 10

3

1 283.15

kJ mol-K

1 298.15

2.24 and

k10 k25

0.106.

k10 = 0.106 and k10 = (0.106)(6×10–9 s–1) = 6.4×10–10 s–1. The half-­time is k25

ln 2 6.4 10 10 s

1

= 34.6 years.

Note that the relatively large Ea value means that the rate constant (and thus the half-­time) is sensitive to temperature. Another common way of correcting rate constants for temperature in environmental engineering is given by:

kT

k20

T 20



eq. 23.13

where kT = rate constant value at temperature T (in °C), k20 = rate constant value at 20°C, and θ = constant near 1.0 that differs in value for each type of reaction. From the Arrhenius equation kT1



kT2

e

 

e

Ea R

1 1 T1 T2

Ea   RT1T2

T2 T1

e

Ea R

T2 T1 T1T2



eq. 23.14

Ea

e RT1T2 . If T2 = 20°C and T1 is expressed in °C, then, comparing eqs. 23.13 and 23.14: (Recall that T2 – T1 has the same value whether temperatures are expressed in K or °C.) Now for temperatures between 0 and 30°C, T1T2 does not change much: T1T2 = 8.59×104 K2 at T1= T2 = 20°C. Reported values of θ for biological treatment processes range from 1.00 to 1.10 (Metcalf and Eddy 2003). Values of θ between 1.00 and 1.10 correspond to Ea between about 0 and 69 kJ/mol. If θ = 1.07, then the rate constant would double when the temperature is increased from 20°C to 30°C.

23.6.2  Effects of Ionic Strength The effects of ionic strength can best be seen by revisiting the transition-­state theory introduced in Section 23.6.1. Recall that in transition-­state theory, the reaction rate is assumed to be proportional to the concentration of the activated complex. Thus, the second-­order rate constant is proportional to K‡ written as concentrations, not activities. As a result, you can use the ionic strength corrections for equilibrium constants (developed in Section 21.2.6) to describe the effect of ionic strength on rate constants.

ISTUDY

586   Chapter 23  Chemical Kinetics of Aquatic Systems The result is that the rate constant for the reaction A + B → products depends on ionic strength in the following fashion:

kI

kI

0

A B



AB ‡

where kI is the rate constant at ionic strength I, kI→0 is the rate constant as ionic strength approaches zero, and γi is the activity coefficient of species i. The activity coefficients can be calculated from one of the single-­ion activity coefficient approximations (see Table 21.3). As an example, consider the work by Weil and Morris (1949) on the reaction between hypochlorous acid and ammonia to form monochloramine, NH2Cl (see also Worked Example 23.6). At the time of their work, it was not known if the chlorine-­containing reactant was HOCl or OCl– and it was not known if the nitrogen-­containing reactant was NH3 or NH4+. Four reactions are possible:

HOCl NH 3

NH 2 Cl H 2 O

eq. 23.15a



OCl

NH 2 Cl OH

eq. 23.15b

NH 2 Cl H 3 O

eq. 23.15c

NH 2 Cl H 2 O

eq. 23.15d



NH 3

HOCl NH 4 OCl

NH 4

Weil and Morris attacked the problem in two ways. First, they varied the pH and observed the reaction rate. The four mechanisms in eq. 23.15a–d have somewhat different pH dependencies. If, for example, the mechanism in eq. 23.15a is correct, then the reaction rate should be proportional to [HOCl][NH3] = α0,Clα1,NClTNT, where the alpha values are for the acid-­dissociation reactions of HOCl and NH4+ (pKa = 7.5 and 9.3, respectively), ClT = [HOCI] + [OCl–], and NT = [NH3] + [NH4+]. The reaction rate showed a maximum at a pH equal to about the average of the pKa values. This dependency of the rate on pH allowed for the elimination of the mechanism in eq. 23.15b (which would show an increased rate with increasing pH) and the mechanism in eq.  23.15c (which would show an increased rate with decreasing pH). However, the effects of pH on the rate could not be used to differentiate between the mechanisms in eqs. 23.15a and 23.15d.

Thoughtful Pause How could you differentiate between the mechanisms in eqs. 23.15a and 23.15d? The rate constants from the reactions in eqs. 23.15a and 23.15d should show very different dependencies on ionic strength. If the mechanism is as shown in eq. 23.15a, then the rate constant should be relatively independent of I (assuming that the activated complex is uncharged and recalling that the activity coefficients for uncharged species are near 1.0). However, if the mechanism is as shown in eq.  23.15d, then the rate

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Chapter Key Ideas   587

constant should decrease significantly with increasing I. The data showed that the measured rate constant was nearly independent of ionic strength between I = 0.01 and 0.21 M. Thus, Weil and Morris concluded that the formation of monochloramine occurs between uncharged molecules (HOCl and NH3), as in eq. 23.15a.

23.7  CHAPTER SUMMARY In this chapter, you added several kinetic modeling tools to your repertoire to allow you to calculate species concentrations when the system is controlled by kinetics rather than equilibrium. Nonequilibrium conditions may exist for species that react slowly. The rate of a reaction is related to the energy “hill” that must be overcome before reactants can be converted into products. The number of species participating in the reaction, called the molecularity, is given by the reaction stoichiometry. Another useful characteristic of a chemical reaction is its reaction rate. The reaction rate is calculated by the rate of change of any species with respect to time divided by the species stoichiometry in the reaction. Thus, the rates of reactions of all species in a reaction are related through their stoichiometry. The expression describing the rate of change of a species is called the rate expression or rate law. The molecularity determined through experiments is called the reaction order. For certain reactions, called elementary reactions, the reaction order and molecularity are identical, and thus the reaction order (and the rate expression) can be determined from the reaction stoichiometry. The rate expressions for elementary reactions are a function of only the reactant concentrations, not the product concentrations. These observations about individual reactions can be applied to a system of elementary reactions. The overall rate of change of a species is the sum of the rate expressions from each reaction in the system in which the species participates. As in all systems, rates of change of conservative substances are additive. Simple rate expressions, such as zero-­, first-­, and second-­order kinetics, can be integrated easily to give equations showing species concentrations over time. Integration also is possible for simple systems of reactions, such as reversible and sequential reaction schemes. Reversible reactions are controlled by the sum of the forward and backward rate constants. In sequential reaction systems, products are degraded to form new products. The common models describing the effects of temperature and ionic strength on rate constants were inspired by models describing the effects of temperature and ionic strength on equilibrium constants. Temperature effects often are described by the Arrhenius equation and transition-­state theory. In the Arrhenius equation, the log of the rate constant is linear with the reciprocal of the temperature (in K). Ionic strength effects can be handled by single-­ion activity coefficient approximations.

CHAPTER KEY IDEAS • Thermodynamics tells you what reactions are possible but not how fast the reactions occur. • Some reactions require energy input to overcome an energy barrier and therefore are under kinetic control and not at equilibrium. • In a reaction, the rate of change of each species concentration divided by the species stoichiometry is the same for each species.

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588   Chapter 23  Chemical Kinetics of Aquatic Systems • All the reaction rates are written as a function of reactant concentrations only. • Rates are additive because mass is conserved. • To write rate expressions with a system of reactions, remember that (1) the reaction rates for individual reactions are determined from reactant concentrations only, (2) the reaction rates for individual reactions can be determined directly from the reaction stoichiometry for elementary reactions, and (3) rates are additive. • In the zero-­order rate law, the exponent for each species is zero and the species will show a linear change with time. • No elementary reaction can exhibit zero-­order kinetics. • In the first-­order rate law, the exponent for the species is 1 and the species will show an exponential change with time. • Second-­order kinetics typically come from reactions that are second order in one reactant or from reactions that are first order in two reactants. • The kinetic equations for a reversible system must give the equilibrium concentrations as t → ∞. • An equilibrium constant can be expressed as the ratio of the forward and reverse rate constants. • The sum of the forward and reverse rate constants determines the apparent reaction rate constant in the approach to equilibrium. • One model for the effect of temperature on rate constants is the Arrhenius equation: k

Ae

Ea RT

.

CHAPTER GLOSSARY activation energy (Arrhenius activation energy):  the energy term in the Arrhenius equation chemical mechanism:  the nature of the interaction between chemical species in a chemical reaction elementary reactions:  reactions for which the reaction order and molecularity are identical and thus the reaction order (and the rate expression) can be determined from the reaction stoichiometry equilibrium control (thermodynamic control):  the situation in which it takes little or no energy to start the process of converting reactants to products and therefore species concentrations are determined by thermodynamics half-­time of the reaction (t½):  the time required to reduce a species concentration to one-­half its initial value kinetic control:  the situation in which it takes energy to start the process of converting reactants to products and therefore species concentrations are determined by kinetics molecularity:  the total number of reactant molecules (or ions) participating in the reaction pre-­exponential factor:  the parameter in the Arrhenius equation that multiplies the exponential term rate constant:  the proportionality constant in the rate expression that relates the rate of change of a species concentration with a function of species concentration rate expression (rate law):  a mathematical expression for the rate of change of a species concentration rate of a chemical reaction:  the rate of change of any species in the reaction over time divided by the signed stoichiometric coefficient of the species

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Historical Note: Arrhenius, Chick, and Foote   589

reaction order:  the apparent molecularity, calculated as the sum of the exponents on the species concentrations in the rate expression transition-­state theory:  a reaction theory based on the assumption that the reactants equilibrate with an activated complex, which subsequently decomposes to form products

HISTORICAL NOTE: ARRHENIUS, CHICK, AND FOOTE What do a Swedish Nobel laureate, a British microbiologist, and an American scientist and woman’s rights advocate have in common? You met the first two people in this chapter. Svante Arrhenius developed the idea that salts dissolve into electrolytes in his doctoral dissertation of 1884. This was the main reason he won Sweden’s first Nobel Prize in 1903. You learned of him in this chapter from his work on proposing the concept of an activation energy for reactions (as quantified by the Arrhenius equation). You also met Harriette Chick in this chapter (see Section 23.4.2). In environmental engineering and science, she is known for Chick’s law (the first-­order model of disinfection). Nutritionists know Harriette Chick through her work elucidating the connection between nutrition and the disease rickets. American scientist Eunice Newton Foote (1819–1888) may be unknown to you. Who was Eunice Foote? And what in the world connects Arrhenius, Chick, and Foote? Eunice Foote is the correct answer to two questions that I had been answering incorrectly for years with the answers “Arrhenius” and “Foote.” The two questions are “Who made the first scientific contributions to understanding global climate change?” (Foote, not Arrhenius) and “Who was the first woman to make significant contributions to environmental engineering?” (Foote, not Chick.) Foote was born as Eunice Newton in 1819. (Perhaps foreshadowing her scientific career, her father was named Isaac Newton Jr.). She attended the Troy Female Semi-

Eunice Newton Foote (Carlyn Iverson/NOAA Climate.gov) nary, purportedly the first girls’ school in the United States that provided an education equivalent to that offered to boys. Foote is best known for her experimental work with gases. She measured the temperature increase when glass cylinders containing air, hydrogen, or carbon dioxide were heated with sunlight. She found that carbon dioxide absorbed more heat. Her conclusion contains the basis of our current thoughts about global climate change: An atmosphere of that gas [carbon dioxide] would give to our earth a high temperature; and if as some suppose, at one period of its history the air had mixed with it a larger proportion than at present, an increased temperature from its own actions as well as from increased weight must have necessarily resulted. (Foote 1856)

Foote’s work was presented at the Annual Meeting of the American Association for the Advancement of Science (AAAS) in Albany, New York, on August 23, 1856. . . but not by Foote herself. Her work was presented by Smithsonian Institute Secretary

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590   Chapter 23  Chemical Kinetics of Aquatic Systems Joseph Henry.4 (The AAAS allowed female members, but having a woman present her own work at the Annual Meeting evidently was a bridge too far in 1856.) Henry did acknowledge that the work was hers. To add to the slight, Foote’s paper was not published in the proceedings of the meeting (although it was eventually published in the American Journal of Science and Arts: Foote 1856.) (For more on Foote and the role of women in the sciences in her time, see McNeill 2016.) Eunice Foote’s work on heat absorption by carbon dioxide is the first published experimental work supporting what we now call global climate change. Her work predates Irish physicist John Tyndall’s (1820–1893) first paper on the subject by three years, although Tyndall often is cited as the “father” of climate change. Arrhenius’s work came later (Arrhenius 1896), and was the first mathematical model for the calculation of temperature from carbon dioxide concentrations. Foote also was very involved in the woman’s rights efforts of her day. She signed the Declaration of Sentiments at the famed Seneca Falls Convention in 1848 as a member of the editorial committee. Foote was friends with Elizabeth Cady Stanton, a leading suffragist and fellow Troy Female Seminary alumna. So, no, “Arrhenius, Chick, and Foote” is not a new indie rock band from Buffalo. Eunice Newton Foote deserves her place among the founders of global climate change science.

PROBLEMS 1. A catalyst is a substance that increases the rate of a chemical reaction without being consumed. Catalysts do not change the equilibrium position of a reaction if the reaction is allowed to proceed to equilibrium. What characteristic energy of the system is affected by a catalyst? (Hint: Refer to Figure 23.1) 2. One reaction scheme not discussed in the text is the parallel scheme, where one species reacts in parallel with several other species. Suppose A reacts with both B and C by the following elementary reactions: A B A C

products; k B products; kC

Write the rate expressions for [A], [B], and [C]. 3. It is useful to determine rate constants from experimental data (i.e., from values of C versus t). a. How would you determine a zero-­order rate constant from experimental data? b. You suspect a reaction is first order because the species concentration looks vaguely like an exponential decrease over time. How can you find the best-­fit first-­order rate constant? Can you linearize the data [that is, find a function of C so that f(C) = mt + b]? c. How can you find the best-­fit second-­order rate constant? Can you linearize the data [that is, find a new function of C so that g(C) = mt + b]?

d. It is a truism in chemical kinetics that it is difficult to determine the kinetic order of a reaction by using the methods you developed in parts A, B, and C if the kinetic data collected are only for the first one-­ half of the reaction. Create some “perfect” data for zero-­, first-­, and second-­order kinetics for the disappearance of species A, but limit the data to [A] > ½[A]0. Fit the data to zero-­, first-­, and second­-order models. Is it difficult to determine the kinetic order under these conditions? (In other words, is it difficult to decide which model fits the data best when the data set is limited?) 4. Data were collected on the degradation of pollutant as shown below. Estimate the first-­order rate constant. Time (days)

Conc. (mg/L)

0 1 2 3 4 5

1.00 0.65 0.48 0.28 0.22 0.13

5. A population growth model was discussed in dN = +rN, where N is the population Section 23.4.3: dt at time t and r is a first-­order growth rate constant. In population dynamics, the “rule of 70” states that the

 The SI unit of electrical inductance was named after Henry (1797–1878).

4

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Problems   591 doubling time of a population undergoing exponential growth is about 70 divided by the growth rate constant r if r is expressed in % per time. Justify the “rule of 70.” 6. For the reactions shown in Section 23.5.3, show that the maximum value of [B] (= [B]max) and the time at which the maximum value of [B] occurs (= tmax) are given by:

B

max

A

0

kB,1 kA ,1

kB,1 kA ,1 kB,1

and tmax

ln kB,1

kB,1 kA ,1 kA ,1

7. For a reversible reaction (see eq. 23.10), verify that the 2 ln k f kr half-­time of the reaction is: t½ = . If kf >> kr, k f kr ln(2) . then show that this expression becomes t½ ≈ kf Does this make sense? 8. Several systems of reactions discussed in the text yield an exponential decrease in the concentrations of certain species over time, as with a simple first-­order reaction. a. For a reversible system, the reactant concentration decreases exponentially over time initially. Is its initial rate of decrease identical to that which would be observed if the reverse reaction did not occur? Why? (The initial rate is

d A at t = 0.) dt

b. For a sequential reaction system, the first reactant concentration decreases exponentially over time. Is its rate of decrease at all times identical to that which would be observed if the second reaction did not occur? Why? 9. Derive the equations for the concentrations of A, B, and C in the sequential reactions discussed in Section 23.5. 10. For very reactive intermediate species in sequential reactions, we sometimes assume that the concentration of the intermediate species is small and relatively constant over time. This is called the steady­-state assumption. A conservative approach to the steady-­state assumption d B = 0 (in the sequence: A → B → C). is to assume that dt a. How does your analysis of the sequential system change if you apply the steady-­state assumption to species B? (Hint: The mass balance equation is [A]+ [B] + [C] = [A]0. Differentiate the mass balance equation with respect to time.)

b. In the ozone example (Worked Example 23.2), write the rate expressions for all the species in the system (except H+, OH⁻, H2O, and O2: assume their concentrations to be fixed). Apply the steady-­state assumption to HO2⁻, HO2, O2⁻, O3⁻, and OH. Show d O3 = that with the steady-­state assumptions: dt ⁻ –3(k [OH ]+k )[O ]. 1

5

3

11. A common analysis of dissolved oxygen in rivers gives rise to the Streeter-­Phelps equation. This equation has two rate constants: a deaeration rate constant (kd) and a reaeration rate constant (kr). Temperature corrections often are applied in the Streeter-­Phelps equation by using θ = 1.048 for kd and θ = 1.024 for kr. a. At what temperatures will kd and kr double over their values at 20°C? b. Estimate the Ea values for kd and kr from their θ values. 12. For the monochloramine example in Example 23.6, the Arrhenius parameters for the forward reaction are Ea = 1510 K. The Arrhenius R parameters for the reverse reaction are A = l.38×108 s–1 E and a = 8800 K (values from Morris and Isaac 1983). R Plot the forward and reverse rate constants over temperature from 0 to 30°C. A = 6.6×108 L/mol-­s and

13. In the example in Section 23.6.2, the lack of dependency of the rate constant on ionic strength was used to infer a reaction mechanism. What would the expected change in the rate constant have been from I → 0 to I = 0.01 M if the reaction mechanism in eq. 23.15d was the true reaction mechanism? 14. You can apply your knowledge of chemical kinetics to other areas. a. Flocculation kinetics are characterized by particle– particle collisions. If the particle number concentration is N, write an expression for the flocculation rate. What are the units of the flocculation rate constant if N has units of number per cm3? b. Epidemics sometimes are modeled by the SIR model. A susceptible population, S, interacts with an infected population, I, and eventually recover to form a recovered population R. Write the rate expressions for S, I, and R if the dynamics are described by: Transmission: S I Recovery: I R; kr

I ; kt

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592   Chapter 23  Chemical Kinetics of Aquatic Systems

CHAPTER REFERENCES Arrhenius, S. (1896). On the influence of carbonic acid in the air upon the temperature of the ground. Phil. Mag. J. Sci., Series 5, Vol. 41: 237–276. Espenson, J.H. (1981). Chemical Kinetics and Reaction Mechanisms. New York: Mc-­Graw Hill Book Co. Foote, E.N. (1856). Circumstances affecting the heat of the Sun’s rays: Art. XXXI. Am. J. Sci. Arts, 2nd Series, XXII (LXVI): 382–383. Haag, W.R. and T. Mill (1988). Some reactions of naturally occurring nucleophiles with haloalkanes in water. Environ. Toxicol. Chem. 7: 917–924. McNeill, L. (2016). This suffrage-­supporting scientist defined the greenhouse effect but didn’t get the credit, because sexism. smithsonianmag.com, December 5, smithsonianmag.com/science-­nature/ lady-­scientist-­helped-­revolutionize-­climate-­science-­didnt-­g et-­ credit-­180961291. Metcalf and Eddy (2003). Wastewater Engineering: Treatment and Reuse. 4th ed. New York: McGraw-­Hill Book Co.

Morris, J.C. and R.A. Isaac (1983). A critical review of kinetic and thermodynamic constants for the aqueous chlorine-­ammonia system. In: Water Chlorination: Environmental Impact and Health Effects, Vol. 4, Book 1, R.L. Jolley, W.A. Brungs, J.A. Cotruvo, R.B Cummings, J.S. Mattice, and V.A. Jacobs (Eds.), Ann Arbor, MI: Ann Arbor Science Publ. Staehelin, J. and J. Hoigné (1983). Decomposition of ozone in water in the presence of organic solutes acting as promoters and inhibitors of radical chain reactions. Environ. Sci. Technol. 19(12): 1206–1213. Tuan, Y.-­J. and J.N. Jensen (1993). Chemical oxidation of thiocyanate by ozone. Ozone Sci. Eng. 15(4): 343–360. Weil, I. and J.C. Morris (1949). Kinetic studies on the chloramines. 1. The rate of formation of monochloramine, N-­ chloromethylamine and N-­chlorodimethylamine. J. Amer. Chem. Soc. 71: 1664–1671.

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C H A P T E R   24

Putting It All Together: Integrated Case Studies in Aquatic Chemistry 24.1 INTRODUCTION In this capstone chapter of the text, you will have a chance to apply your knowledge of aquatic chemistry to five challenging problems. The case studies in this chapter were selected to integrate the major tools from the text for the solution of environmentally interesting problems. The first case study involves complexation and precipitation chemistry in an industrial setting. You will calculate equilibrium concentrations of chemical species in a metal plating bath. The reaction conditions used in practice necessitate temperature and ionic strength corrections. The second case study will allow you to integrate equilibrium and kinetics to understand the rate of oxidation of ferrous iron by oxygen in natural waters. Fast reactions will be considered to be at equilibrium. Rate expressions will be developed that contain both rate constants and equilibrium constants. The third case study is a detailed analysis of mercury chemistry in natural waters. In addition to dissolved phase and precipitation chemistry, you will need to consider the complexation of mercury by surfaces. The fourth case study is about how to construct a pH buffer with a fixed ionic strength. It provides a model for creating pH buffers in the laboratory. The final case study is a more realistic evaluation of ocean acidification caused by increased carbon dioxide loading to the environment. The ionic strength of the oceans will be included in the analysis. The case studies are structured in the same fashion. A background section will introduce you to the problem. Analysis goals are established. The goals are a series of questions or the desire to explain experimental and field observations. You then can work through the analysis and interpreted to address the analysis goals. The value of the case studies goes beyond the application studied. As you work through the case studies, check whether the analysis matches the analysis goals. Pick your own analysis approach and see how it compares to the choices made in the text. Be sure to question the assumptions and the impact of the assumptions on the conclusions.

593

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594   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry

24.2  INTEGRATED CASE STUDY 1:

METAL FINISHING

24.2.1  Background In metal plating, thin coats of metals are applied to metal or plastic parts to enhance the appearance and physical characteristics of the parts. In the typical plating process, dissolved metal is reduced and plated out as zero-­valent metal. It is desirable to maintain high concentrations of dissolved metal in solution to increase the efficiency of the process. High dissolved metal concentrations usually are maintained in the plating bath by using acidic conditions or adding a strongly binding ligand such as cyanide.

Thoughtful Pause Why does maintaining low pH or high ligand concentrations help to keep the metals in solution? Recall that many metals precipitate as hydroxide solids. Thus, low pH conditions or the presence of other ligands generally increase metal solubility. Copper is one of the most common metals to be plated. In a copper strike bath, very high cyanide concentrations are used to plate out a thin copper coating. Usually, copper or other metals are used to form the final finish on top of the copper strike. Typical chemical additions to the copper strike bath are (from Lowenheim 1978): CuCN s : 15 g / L NaCN s : 28 g / L Na 2 CO3 s : 15 g / L



The strike bath commonly is maintained at 50–63°C. The typical pH is 10–10.5. Sodium carbonate is added for pH control. As carbonates (carbonate ion and carbonate-­containing species) build up in the bath, it is necessary to remove them. One common process is to refrigerate the entire bath and precipitate carbonates as sodium carbonate decahydrate (Na2CO3·10H2O).

24.2.2  Analysis Goals The typical conditions of the plating bath are designed to maintain nearly all the copper as soluble cyanide complexes. Copper is added in the form of CuCN(s). While CuCN(s) is not very soluble, it should be more soluble in the presence of excess cyanide.

Thoughtful Pause Why should excess cyanide aid in the solubilization of CuCN(s)?

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24.2  |  Integrated Case Study 1: Metal Finishing   595

Excess cyanide should help form higher concentrations of copper-­cyanide complexes and thus increase the solubility of CuCN(s). In addition to enhancing the solubility of CuCN(s), the NaCN(s) dose is supposed to be sufficient to form higher copper-­cyanide complexes, Cu(CN)32– and (Cu(CN)43–. Finally, during carbonate freeze-­out, the temperature of the refrigeration unit is supposed to be sufficiently low to precipitate carbonates, but not copper. The specific goals of the analysis are to answer the following questions: • Is CuCN(s) soluble in the strike bath? • Is nearly all the copper soluble under standard operating conditions? • Is most of the soluble copper present as Cu(CN)32– and Cu(CN)43–, or is it present as other species? • Will carbonate precipitate at lower temperature (say, 0°C) without precipitating copper?

24.2.3  Analysis For this analysis, assume the system is at equilibrium and open to the atmosphere. Although the system is complicated, use the same steps you have followed throughout this text. Start with a species list. A potential list of species is given in Table 24.1. You know that carbon dioxide will be present (since the system is open to the atmosphere), but you do not know if the solid phases exist at equilibrium. Note that all the copper-­ containing species in the species list are in the +I oxidation state. This is the usual state of affairs in a copper-­cyanide bath. Table 24.1 contains some unusual species. We normally do not include sodium-­bicarbonate or sodium-­carbonate complexes such as NaHCO30 and NaCO3–. These species are included here because of the large added sodium concentration. The second step in the analysis process is to write the pertinent equilibria. A list of equilibria may be found in Table 24.2. (As usual, the equilibrium list was used to generate the species list in Table 24.1.) Sources of the thermodynamic data are listed in Table 24.3. The logK values at 25°C and I → 0 were manipulated to allow you to draw conclusions at strike bath conditions (about 57°C and I = 2 M) and refrigerated bath conditions (about 0°C and I = 2 M). Ionic strength corrections were made using the Davies approximation with C = 0.2. It should be noted that I = 2 M is outside of the range

Table 24.1 Species List for the Metal Plating Case Study Total Mass Species

Cu(+I)T CNT CT NaT

Amount Added (M)

soluble: Cu , CN(CN) , Cu(CN)2 , Cu(CN) , Cu(CN)4 insoluble: CuCN(s), Cu2O(s) soluble: HCN, CN–, Cu(CN)0, Cu(CN)2–, Cu(CN)32–, Cu(CN)43– insoluble: CuCN(s) soluble: H2CO3*, HCO3–, CO32–, NaHCO30, NaCO3– insoluble: CO2(g), see text soluble: Na+, NaHCO30, NaCO3– insoluble: see text +

0

–

2– 3

3–

0.17 0.74 see text 0.85

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596   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry of the Davies approximation (I ≤ 0.5 M). Values of A and Ba in the Davies equation (see Table 21.3) were 0.5 and 1, respectively. They were not adjusted for temperature because preliminary calculations showed little impact of temperature. Preliminary calculations employed an equation of state for the effect on temperature on the relative permittivity of water from Klein and Swift (1977). See Section 21.2.3 for the temperature dependencies of A and B. It should be noted that the relative permittivity of water is strongly dependent on I (Klein and Swift 1977), but was not corrected for I in this case study. The single-­ion activity coefficient for all uncharged species was assumed to be 1. The equilibrium constants for the Cu-­CN complexes and the Ks0 for CuCN(s) were determined by at I = 1, 3, and 5 M. Values were interpolated to I = 2 M. See the references in Table 24.3 for details. The van’t Hoff equation was used to adjust equilibrium constants at 25°C and I = 2 to 0 and 57°C and I = 2. It was assumed that Hrxn values were constant over the temperature range of 0 to 57°C. The temperature dependence of the Ks0 value for sodium carbonate solids is a little more complicated. Based on the work of Haynes (2003), it was assumed that sodium carbonate decahydrate (Na2CO3·10H2O) would control the sodium carbonate solubility at 0°C and NaHCO3(s, Nahcolite) would control the sodium carbonate solubility at 57°C. The Ks0 values in Table 24.2 were calculated using expressions for Ks0 as a function of temperature (Haynes 2003) and then adjusted for ionic strength using the Davies equation. The third step in the analysis process is writing the mass balances. The appropriate mass balances on soluble species are:

Cu

I

T

[Cu ] [Cu CN

2

] [Cu CN

3

2

] [Cu CN

4

3

]

eq. 24.1

Table 24.2 Equilibria for the Metal Plating Case Study No. Equilibrium

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Cu+ + CN– = Cu(CN)20 Cu+ + 2CN– = Cu(CN)2– Cu+ + 3CN– = Cu(CN)32– Cu+ + 4CN– = Cu(CN)23– ½Cu2O(s) + ½H2O = Cu+ + OH– HCN = CN– + H+ CuCN(s) = Cu+ + CN– H2O = H+ + OH– CO2(g) + H2O = H2CO3* H2CO3* = HCO3– + H+ HCO3– = CO32– + H+ Na+ + HCO3– = NaHCO30 Na+ + CO32– = NaCO3– NaHCO3(s, Nahcolite) = Na+ + HCO3– Na2CO3·10H2O = 2Na+ + CO32–

Hrxn (kJ/mol)

N/A –121.6 –168.0 –214.9 –6.3 43.6 N/A 79.89 8.35 36.4 58.9 36.2 5.1 N/A N/A

logK (T in °C, I in M) (25,0)

(25,2) (57,2)

(0,2)

N/A N/A N/A N/A –14.7 –9.2 –19.49 –14.0 –1.5 –6.3 –10.3 3.35 1.28 0.38 0.14

16.4 23.8 29.4 32.1 –14.0 –8.5 –18.8 –14.0 –1.5 –5.6 –8.9 2.6 –0.1 1.4 2.2

16.4 25.4 31.7 35.0 –13.9 –9.1 –18.8 –15.1 –1.6 –6.1 –9.7 2.2 –0.2 –­ 2.1

16.4 22.0 27.0 29.0 –14.1 –7.9 –18.8 –12.8 –1.4 –5.1 –8.0 3.2 0.0 1.8 –­

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24.2  |  Integrated Case Study 1: Metal Finishing   597

Table 24.3 Sources of Thermodynamic Data in Table 24.2 Species

Data (25°C, I → 0)

Cu-­CN complexes and solids





Source

Hrxn and logKa

CN-­containing

Martell and Smith (1974)

logK and Ks0

Akilan et al. (2016)

NaCO3(s,Nahcolite) and Na2CO3·10H2O(s)

H f and logKs0

Haynes (2003)

NaHCO3 and NaCO3–

Hrxn and logK

Chung (1978)

Cu2O(s,Cuprite)

H f and G f

Korzhavyi and Johansson (2010)

All others

H f and logK

Stumm and Morgan (1996)

CNT

HCN

[CN ] 2[Cu CN

3[Cu CN CT NaT

3

2

] 4[Cu CN

2 4

] 3

]



[H 2 CO3* ] [HCO3 ] [CO32 ] [NaHCO30] [NaCO3 ] [Na ] [NaHCO30] [NaCO3 ]

eq. 24.2

eq. 24.3

At this point, the values of Cu(+I )T , CNT , CT , and NaT are unknown.

Thoughtful Pause Why are the values of Cu(+I)T, CNT, CT, and NaT unknown? The values of the total masses are unknown because some portion of each may precipitate (and exchange with the atmosphere, in the case of CT). The system was solved first at 57°C and I = 2 M. A pH of 10.25 was assumed. To solve the system, a spreadsheet was set up with a cell for each species containing the equation for its concentration as a function of the concentrations of Cu+, CN–, and Na+; fixed species concentrations (H+, OH–, and CO2(g) = 10–3.38 atm); and equilibrium constants. As a starting point, it was assumed that no solids precipitated at 57°C (the expected situation in a plating bath). Using Excel’s Solver, the concentrations (actually, pC) of Cu+, CN–, and Na+ were fitted to satisfy the objective function with constraints. The objective function was to set the copper total mass to 0.17 M (eq. 24.1), with the mass balances of cyanide and sodium as constraints (eqs. 24.2 and 24.3). Results at 57°C showed that no solids precipitated. All of the added CuCN(s) dissolved. Most of the soluble copper (89%) and most of the soluble cyanide (82%) was in the form of Cu(CN)43–. This makes sense, given the high cyanide dose. As expected, Na2CO3(s) did not precipitate at 57°C and I = 2  M. The primary form of dissolved sodium was NaHCO30 (91%). The total dissolved carbonate was very large (indicating

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598   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry significant atmospheric contribution to CT at the high pH) and mostly carbonate ion (99%). Thus, carbonate removal may be necessary. Results at 0°C showed that no solids precipitated. This was unexpected, as “carbonate crystals” have been observed. All of the added CuCN(s) dissolved. Most of the soluble copper (99%) and most of the soluble cyanide (91%) again was in the form of Cu(CN)43–. The primary form of dissolved sodium was NaHCO30 (94%). Significantly, the total dissolved carbonate concentration decreased by a factor of almost 300 compared to 57°C. At 0°C, the total dissolved carbonate was mainly NaHCO30 (54%) and CO32– (35%). So why did the total dissolved carbonate concentration decrease so significantly at the lower temperature without the predicted precipitation of carbonate solids? The key appears to be the large positive Hrxn values for K1 and K2 equilibria for the H2CO3*/ HCO3–/CO32– system, making H2CO3* and HCO3– significantly less acidic at the lower temperature. (In contrast, the calculated equilibrium concentrations were not very sensitive to the KH value.)

24.2.4  Implications and Model Improvements The equilibrium calculations summarized in this case study shed light on the speciation of copper and cyanide at equilibrium in the plating bath. The speciation information was used to answer the questions posed in Section 24.2.1 and could be used to optimize the plating process further. Knowledge of the species concentrations also could be used to develop and optimize treatment processes for removing cyanide and residual copper from plating waste streams. The calculations indicate that lowering the bath temperature should result in significantly lower total dissolved carbonate under equilibrium conditions even without precipitation of carbonate solids. The model suffers from the use of the Davies equation outside its acceptable ionic strength range. In addition, Hrxn were assumed to be constant over a large temperature range.

24.3  INTEGRATED CASE STUDY 2: OXIDATION

OF Fe(+II) BY OXYGEN

24.3.1  Background The oxidation of ferrous iron, Fe(+II), by oxygen to form ferric iron, Fe(+III), is an important process in the environment. In natural systems, the oxidation of ferrous iron is a critical step in the production of very acidic waters from abandoned mines. These streams, called acid mine drainage, can be extremely acidic, with pH values less than 1. The acidic conditions are created when the sulfur in pyrite, FeS(s), is oxidized by atmospheric oxygen, with the subsequent release of ferrous iron. Ferrous iron is oxidized by oxygen (often catalyzed by bacteria) to ferric iron. The oxidation of Fe(+II) to Fe(+III) is the rate-­determining step in the dissolution of pyrite and thus in the generation of acid mine drainage (Singer and Stumm 1970). The oxidation of ferrous iron by oxygen also is important in engineered systems. Ferrous iron is fairly soluble in water, but ferric iron is much less soluble (see Figure  24.1). If ferric iron precipitates in the home, then reddish brown staining of laundry and cooking utensils may occur. Iron precipitates can clog pipes and water

ISTUDY

24.3  |  Integrated Case Study 2: Oxidation of Fe(+II) by Oxygen   599 pH 0

0

2

4

6

8

2

10

12

14

Fe(OH)2(s)

pC

4 6 Fe(OH)3(s)

8 10 12 14

FIGURE 24.1  Comparison of the Solubility of Fe(+II) and Fe(+III) [In equilibrium with Fe(OH)2(s), logKs0 = –14.4, and Fe(OH)3(s), logKs0 = –38.8, respectively; it assumes OH– is the only ligand and ignores the formation of iron polymers.]

heaters and encourage the growth of iron-­reducing bacteria. A solution to iron precipitation in potable water is to oxidize ferrous iron in the drinking water treatment plant and remove Fe(+III) precipitates by filtration. The least expensive oxidant is atmospheric oxygen. In this application, the rate of ferrous iron oxidation is important because it will drive the sizing of basins in drinking water treatment facilities where ferrous iron is to be oxidized.

24.3.2  Analysis Goals Oxidation experiments have revealed the following information about the rate of oxidation of ferrous iron by oxygen: • The rate is slow at pH values below 8. • The rate increases by a factor of 100 for a one unit increase in pH. • At fixed oxygen partial pressure and pH, the initial disappearance of Fe(+II) is first order in the Fe(+II) concentration. It is desirable to develop a kinetic scheme consistent with the observations listed above. For applications to natural waters, it is of interest to explore the effects of ligands on the oxidation of ferrous iron by oxygen.

24.3.3  Analysis The second and third observations suggest second-­order dependence on [OH–] and first-­order dependence on Fe(+II)T. A possible rate law consistent with kinetic data is: dFe



dt

II

T

kFe

II T [OH ]2 PO2

The rate constant k is known to be 8.0 1013

L2 . At PO2 atm-mol 2 - min

0.209 atm

(atmospheric conditions), the pseudo-­first-­order rate constant for ferrous iron oxidation is: k′ = k[OH–]2PO2 =

1.6 10 [H ]2

15

min–1. Thus, you would expect that a plot of log(k′)

ISTUDY

log(k′)(k′ in min–1)

600   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry 1 0 –1 –2 –3 –4 –5 –6 –7 –8 –9

Millano and Izaguirre (1989) Singer and Stumm (1970) line: log(kʹ ) = 2pH ‐ 14.8

5.0

5.5

6.0

6.5

7.0

7.5

8.0

pH FIGURE 24.2  Dependency on pH of the Pseudo-­First-­Order Rate Constant for Ferrous Iron Oxidation by Oxygen (pH 5–8)

(with k′ in min–1) against pH would have a slope of 2.0 and an intercept of log(1.6×10–15) = –14.8. Literature data collected between pH 5 and 8 are consistent with this rate constant and rate law (see Figure 24.2). The inclusion of data collected at pH < 5 reveals a different story (see Figure 24.3). The plot of log(k′) against pH appears to have a slope of about 2 between about pH 5 and 8, a slope of about 1 between about pH 3 and 5, and a slope of about zero below pH 3. In other words, the dependency of k′ on [OH–] appears to change from [OH–]2 to [OH–]1 to [OH–]0 as the pH decreases.

Thoughtful Pause

log(k′) (k′ in min–1)

What is a possible reason for the change in the dependency of k′ on [OH–] as the pH decreases?

1 0 –1 –2 –3 –4 –5 –6 –7 –8 –9

Millano and Izaguirre (1989)

Singer and Stumm (1970)

line: k′ given by eq. 24.5b

1.0

2.0

3.0

4.0

5.0

6.0

pH FIGURE 24.3  Model Predictions for Ferrous Iron Oxidation by Oxygen

7.0

8.0

9.0

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24.3  |  Integrated Case Study 2: Oxidation of Fe(+II) by Oxygen   601

One explanation for the change in the dependency of k′ on [OH–] as the pH decreases is that dissolved oxygen reacts at different rates with different ferrous iron species. The overall reaction, Fe(+II) + O2(aq) → products, should be separated into three reactions:

Fe 2



FeOH



Fe OH

O2 aq O2 aq 2

0

products; k1 products; k2

O2 aq

products; k3

The rate law becomes: dFe



II

T

k1 [Fe 2 ] k2 [FeOH ] k3 [Fe OH

dt

2

0

] PO2

eq. 24.4

Now, recall from Section  23.5.2 that acid–base reactions are very fast. Thus, you might expect that the reactions between Fe2+ and OH– to form FeOH+ and Fe(OH)20 are very fast and the hydroxy ferrous iron complexes may be in equilibrium with OH–. The appropriate mass balance and equilibrium expressions are:

Fe

II

Fe 2



T

Fe 2

[Fe 2 ] [FeOH ] [Fe OH OH 2OH

FeOH ; K1 Fe OH

2

0

; K2

2

0

]

10 4.5 10 7.4

Solving: 1 Fe 1 K1 [OH ] K 2 [OH ]2

[Fe 2 ]



K1 [OH ] Fe 1 K1 [OH ] K 2 [OH ]2

[FeOH ]



[Fe OH

2

0

]

K 2 [OH ]2 Fe 1 K1 [OH ] K 2 [OH ]2

II

T

II

T

II

T



Substituting into the rate expression (eq. 24.4), you can show that:

dFe

dt

II

T

k1 k2 K1 [OH ] k3 K 2 [OH ]2 PO2 Fe 1 K1 [OH ] K 2 [OH ]2

II T

eq. 24.5a

ISTUDY

602   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry The pseudo-­first-­order rate constant, k′, is:

k1 k2 K1 [OH ] k3 K 2 [OH ]2 PO2 1 K1 [OH ] K 2 [OH ]2

k

eq. 24.5b

The values of the individual rate constants are: k1 = 1.0×10–8 atm–1s–1, k2 = 3.2× 10 atm–1s–1, and k3 = 1.0×104 atm–1s–1 (Wehrli 1990). The calculated values of log(k′) are shown by the solid line in Figure 24.3. The revised rate law suggests that complexation of Fe(+II) with ligands other than hydroxide ion may slow down the rate of Fe(+II) oxidation (see also Millero 1985). The effects of a ligand (e.g., chloride) can be quantified with the model. In the presence of chloride, the mass balance becomes: –2



Fe

II

T

0

[Fe 2 ] [FeOH ] [Fe OH 2 ] [FeCl ]

An additional equilibrium is required: Fe2+ + Cl– = FeCl+; K3. Assuming FeCl+ reacts very slowly with dissolved oxygen compared to the hydroxy complexes:

kCl

k1 k2 K1 [OH ] k3 K 2 [OH ]2 PO2 2 1 K1 [OH ] K 2 [OH ] K3 [Cl ]

Here, kCl is the pseudo-­first-­order rate constant for ferrous iron oxidation by oxygen in the presence of chloride. With the assumption that FeCl+ reacts very slowly with dissolved oxygen, the ratio of the rate in the presence of chloride to the rate in the absence of chloride is:

kCl k

1 K1 [OH ] K 2 [OH ]2 1 K1 [OH ] K 2 [OH ]2 K3 [Cl ]



For pH values less than about 9, the term 1 + K1[OH–] + K2[OH–]2 is about equal to 1. Thus, for pH less than about 9:

kCl k

1 . 1 K3 [Cl ]

If chloride is present in great excess over ferrous iron, then [Cl–] ≈ ClT = constant. Therefore:

kCl k

1 1 K3ClT

eq. 24.6

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24.4  |  Integrated Case Study 3: Inorganic Mercury Chemistry in Natural Waters   603 1.0 0.8

k′

k′CI

0.6 0.4 0.2 0.0 0.0

1.0

2.0

3.0

4.0

5.0

6.0

ClT FIGURE 24.4  Effect of Chloride on the Rate Constant for the Oxidation of Fe(+II)

This model for the effects of chloride on the rate of oxidation is tested against the data of Millero and Izaguirre (1989) in Figure  24.4.1 The line is the prediction from eq. 24.6 using K3 = 10–0.2 from Appendix D. The model predicts the data fairly well.

24.4  INTEGRATED CASE STUDY 3:

INORGANIC MERCURY CHEMISTRY IN NATURAL WATERS

24.4.1  Introduction Mercury is an important pollutant because of its well-­known adverse human and environmental health effects. Dissolved mercury forms complexes with many common ligands in the environment, including hydroxide, chloride, bicarbonate, carbonate, and nitrate ions. Mercury can form environmentally important organic complexes as well, including methyl-­ and dimethyl-­mercury (see the Part I case study). The focus in this integrated case study is on the inorganic chemistry of mercury; organic complexes will not be considered here. Mercury also can precipitate as hydroxide and carbonate solids. Many cations, including Hg2+ and its cationic complexes, can adsorb onto negatively charged surfaces. A common surface in the environment is formed at the interface of water and hydrous ferric oxides or HFOs (see Section 22.2.2). As discussed in Section 22.4, charged surfaces often exhibit acid–base chemistry. The uncharged surface species on HFO (denoted here by the symbol ≡S-­OH) is amphoteric. It can both accept a proton from water and donate a proton to water. A detailed model of inorganic Hg(+II) chemistry must take into account complexation, precipitation, and adsorption. Such a model was developed by Tiffreau and colleagues (1995). The approach summarized in this integrated case study is based on their work.  You might be concerned about the effects of ionic strength in Figure 24.4. The k′ and kCl′ values from Millero and Izaguirre (1989) were obtained with matching ionic strengths: NaClO4 was used in the k′ experiments and NaCl was used in the kCl′ experiments. Sodium perchlorate is generally considered to be an inert salt.

1

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604   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry

24.4.2  Analysis Goals The overall analysis goal is to determine the speciation of mercury-­containing species as a function of pH in a typical natural water containing HFO. In particular, it is desired to determine the adsorption edge for mercury on HFO.

24.4.3  Analysis The analysis approach is similar to the metal plating case study in Section 24.2.3. A list of potential species is shown in Table 24.4. The system has 41 species (including H+, OH–, and H2O). The analysis will be conducted for a water containing the total masses shown in Table 24.4. The amorphous HFO concentration used represents 50 mg/L of mol Fe mol sites and 0.205 (Tiffreau et al. 1995). The low total HFO containing 0.011 g HFO mol Fe mercury concentration was selected to minimized mercury precipitation. The pertinent equilibria are summarized in Table 24.5. In the calculations, the system was assumed to be in equilibrium with the atmosphere. Initially, no corrections were made for the effects of surface charge. Similar to the first case study, a spreadsheet was set up with a cell for each species containing

Table 24.4 Species List for the Mercury Case Study Total Mass

Species

Amount Added (M)

Hg(+II)T soluble: Hg2+, HgOH+, Hg(OH)20, Hg(OH)3–, Hg(OH)42–, Hg2OH3+, 1.84×10–7 + 0 3+ 0 – 2– 0 Hg3(OH)3 , HgCl , HgCl2 , HgCl3 , HgCl4 , HgOHCl , HgOHCl2 , HgOHCl3–, Hg(OH)2Cl–, Hg(OH)3Cl2–, Hg(OH)2Cl22–, HgNO3+, Hg(NO3)20, HgCO30, Hg(CO3) 22–, HgHCO3–, Hg(OH)CO3– insoluble: HgCO3(s), Hg(OH)2(s), 2HgO·HgCO3(s) surface: ≡S-­OHg+, ≡S-­OHgOH, ≡S-­OHgCl soluble: Cl–, HgCl+, HgCl20, HgCl3–, HgCl42–, HgOHCl0, 5.64×10–4 (20 mg/L) ClT – 2– – 2– HgOHCl3 , Hg(OH)2Cl , Hg(OH)3Cl , Hg(OH)2Cl2 insoluble: None surface: ≡S-­OHgCl soluble: H2CO3*, HCO3–, CO32–, HgCO30, Hg(CO3)22–, HgHCO3–, see text CT Hg(OH)CO3– insoluble: CO2(g), HgCO3(s), Hg(OH)2(s), 2HgO·HgCO3(s) surface: None soluble: NO3–­, HgNO3+, Hg(NO3)20 7.86×10–5 (1.1 mg/L) NT insoluble: None surface: None soluble: None 1.15×10–4 (see text) ≡ST insoluble: None surface: ≡S-­OH2+, ≡S-­OH0, ≡S-­O–, ≡S-­OHg+, ≡S-­OHgOH, ≡S-­OHgCl

ISTUDY

24.4  |  Integrated Case Study 3: Inorganic Mercury Chemistry in Natural Waters   605

Table 24.5 Equilibria for the Mercury Case Study Equilibria

logK

Reference

Hg(+II) Hydroxo Complexes Hg2+ + OH– = HgOH+

10.6

Stumm and Morgan (1996)

0

21.8

Stumm and Morgan (1996)

Hg2+ + 3OH– = Hg(OH)3–

20.9

Stumm and Morgan (1996)

19.2

Tiffreau et al. (1995)

2Hg2+ + 3OH– = Hg2OH3+

10.7

Tiffreau et al. (1995)

35.6

Tiffreau et al. (1995)

Hg2+ + Cl– = HgCl+

7.2

Stumm and Morgan (1996)

Hg2+ + 2Cl– = HgCl20

14.0

Stumm and Morgan (1996)

Hg + 2OH = Hg(OH)2 2+

–

Hg2+ + 4OH– = Hg(OH)42–

3Hg + 3OH = Hg3(OH) 2+

–

3+ 3

Hg(+II) Chloro Complexes

Hg + 3Cl = HgCl3 2+

–

–

Hg2+ + 4Cl– = HgCl42–

15.1

Stumm and Morgan (1996)

15.4

Stumm and Morgan (1996)

Hg(+II) Chloro-­Hydroxo Mixed-­Ligand Complexes

Hg2+ + OH– + Cl– = HgOHCl0

18.1

Tiffreau et al. (1995)

Hg + OH + 2Cl = HgOHCl2

0

17.5

Tiffreau et al. (1995)

Hg2+ + OH– + 3Cl– = HgOHCl3­–

17.1

Tiffreau et al. (1995)

Hg2+ + 2OH– + Cl– = Hg(OH)2Cl–

19.4

Tiffreau et al. (1995)

18.9

Tiffreau et al. (1995)

17.9

Tiffreau et al. (1995)

2+

–

–

Hg2+ + 3OH– + Cl– = Hg(OH)3Cl2–

Hg + 2OH + 2Cl = Hg(OH)2Cl 2+

–

–

2– 2

Other Hg(+II) Complexes

Hg2+ + NO3– = HgNO3+

Hg2+ + 2NO3– = Hg(NO3)20 Hg + CO 2+

= HgCO3

2– 3 2– 3 – 3 –

Hg2+ + 2CO

0

= Hg(CO3)22–

Hg2+ + HCO = HgHCO3–

Hg2+ + OH + CO32– = Hg(OH)CO3–

0.39

Tiffreau et al. (1995)

0.42

Tiffreau et al. (1995)

12.1

Tiffreau et al. (1995)

15.5

Tiffreau et al. (1995)

6.0

Tiffreau et al. (1995)

19.2

Tiffreau et al. (1995)

Hg(+II) Solids

HgCO3(s) = Hg2+ + CO32–

Hg(OH)2(s) = Hg2+ + 2OH–

2HgO⋅HgCO3(s) + 6H+ = 3Hg2+ + CO2(g) + 3H2O

–16.1

Stumm and Morgan (1996)

–25.4

Stumm and Morgan (1996)

7.0

Tiffreau et al. (1995)

Surface Equilibria

≡S-­OH2+ = ≡S-­OH0 + H+

≡S-­OH0 = ≡S-­O– + H+

–7.29

Tiffreau et al. (1995)

–8.93

Tiffreau et al. (1995)

≡S-­OH0 + Hg2+ = ≡S-­OHg+ + H+

6.9

Tiffreau et al. (1995)

≡S-­OH0 + Hg2+ + H2O = ≡S-­OHgOH + 2H+

–0.9

Tiffreau et al. (1995)

9.8

Tiffreau et al. (1995)

–14

Stumm and Morgan (1996)

–1.5

Stumm and Morgan (1996)

≡S-­OH0 + Hg2+ + Cl– = ≡S-­OHgCl + H+ Other Equilibria H2O = H+ + OH–

CO2(g) + H2O = H2CO3* H2CO3* = HCO3 + H –

HCO3– = CO32– + H+

+

–6.3

Stumm and Morgan (1996)

–10.3

Stumm and Morgan (1996)

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606   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry

Adsorbed Hg(+II) as % of Hg(+II)T

the equation for its concentration as a function of the concentrations of Hg2+, Cl–, ≡S-­OH, and NO3–; fixed species concentrations (H+, OH–, and CO2(g)); and equilibrium constants. As a starting point, it was assumed that no solids would precipitate. Using Excel’s Solver, the concentrations (pC) of Hg2+, Cl–, ≡S-­OH, and NO3– were fitted to satisfy the objective function with constraints. The objective function was to minimize the square of the difference between the mercury total mass and 1.84×10–7 M, with the mass balances of chloride, total surface, and nitrate as constraints (see Table  24.3). The results indicated that no solids would precipitate in the pH of 4 to 12 under these conditions. The absorption edge for mercury on HFO is shown in Figure 24.5. Mercury does not adsorb to a very extent at low and high pH values. Between about pH 6.6 and 8.3, over 90% of the mercury is adsorbed. To explain the effects of pH on adsorption, it is necessary to look at the speciation of mercury and at the surface species. Figures 24.6 and 24.7 show the pC-­pH diagrams for the predominant mercury-­containing species and surface species, respectively. At low pH, most of the mercury is complexed with chloride (since the ligands OH– and ≡S-­O– are in low concentration). Most of the surface is protonated, since the metal H+ is at much higher concentration than mercury species. This makes sense because the surface pKa values are 7.3 and 8.9. Closer to neutral pH, the concentration of the ligand ≡S-­O – increases and outcompetes chloride and hydroxide for mercury. Note that near neutral pH, mercury-­ containing surface complexes account for a large amount of the mercury (Figures 24.5 and  24.6), but represent only a small fraction of the total surface concentration (Figure  24.7). This is because the total mercury concentration is much less that the total site concentration. At higher pH values, the carbonate concentration increases significantly, due to the high solubility of carbon dioxide at high pH. Most of the mercury is complexed with carbonate, while the surface is mainly uncomplexed. This analysis was conducted without accounting for surface charge. For the HFO in this work, A = 600 m2/g and Cs = 0.05 g/L, so s = 30 m2/L. Surface charge effects can be important. As an example, at pH 4, the surface charge density was calculated to be 0.320 C/m2 and Ψ = +0.146 V. When including surface charge, the amount of mercury adsorbed at pH 4 increases from 0.02% to 1.7%. This is due to increased formation of ≡S-­OHgOH and ≡S-­OHgCl. 100 80 60 40 20 0

4

5

6

7

8 pH

9

10

11

FIGURE 24.5  Adsorption Edge for Mercury on a Hypothetical Hydrous Ferric Oxide

12

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24.5  |  Integrated Case Study 4: Phosphate Buffers   607

3

4

5

6

pH 8

7

9

10

11

12

[Hg(OH)CO3–]

5 [HgCl20]

[≡S-OHg+]

[Hg(CO3)22–]

[≡S-OHgOH]

pC

7 9

[≡S‐OHgCl]

11 13 FIGURE 24.6  pC-­pH Diagram Showing the Major Mercury-­Containing Species

pH 3

4

5

6

7

[≡S‐OH2+]

8 [≡S‐OH]

9

10

11

12

[≡S‐OH–]

5 [≡S‐OHg+]

[≡S‐OHgOH]

pC

7 9

[≡S‐OHgCl]

11 13 FIGURE 24.7  pC-­pH Diagram Showing the Surface Species

24.4.4  Model Implications and Improvements This integrated case study demonstrates that knowledge of individual species concentrations greatly increases your understanding of the behavior of pollutants in the environment. The model could be modified to consider other ligands and more sophisticated surface complexation models.

24.5  INTEGRATED CASE STUDY 4:

PHOSPHATE BUFFERS

24.5.1  Introduction A common laboratory activity is conducting experiments at fixed pH and fixed ionic strength. To buffer the pH near 7, it is common to use a phosphate buffer, since pKa,2 for the phosphate system is 7.2. You could make a weak phosphate buffer and add a swamping electrolyte to keep the ionic strength constant. Sodium perchlorate commonly is used as a swamping electrolyte.

ISTUDY

608   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry Suppose you want to make as strong a buffer as possible. This means a system with the highest buffer intensity (β). From the discussion in Section 12.5.3, the highest buffer intensity occurs when the total phosphate concentration, PT, is as large as possible. This case study is about designing a buffering system at a fixed pH and ionic strength with the maximum buffer intensity.

24.5.2  Analysis Goal The overall analysis goal is to create the strongest possible pH 7 buffer at I = 0.1 M using NaH2PO4 and NaOH. Based on the discussion in the previous section, the goal can be restated as: Create the pH 7 buffer at I = 0.1 M with the maximum PT using NaH2PO4 and NaOH.

24.5.3  Analysis The system is fairly simple. Species are H2O, OH–, H+, H3PO4, H2PO4–, HPO42–, PO43–, and Na+. The four equilibria are the KW equilibria and three Ka equilibria for the phosphate system, with pKa values of 2.2, 7.2, and 12.4. In the analysis, {H+} is fixed at 10–7 M and I is fixed at 0.1 M. It is attractive to perform the calculations with concentrations rather than activities. The equilibria were converted into their concentration forms. For example, consider the equilibrium H2PO4– = HPO42– + H+ with K2 = 10–7.2 {HPO 4 2 }{H } = . Here: {H 2 PO 4 }

K2

HPO 4 2

[HPO 4 2 ]

H 2 PO 4

[H ] [H 2 PO 4 ] H

So:

K2

c

[HPO 4 2 ][H ] [H 2 PO 4 ]

10

7.2



H 2 PO 4 HPO 4 2

H

The Davies equation (C = 0.2) was used to calculate the single-­ion activity coefficients for the charged species. Alpha values were calculated in terms of the H+ concentration and cK values. The total sodium ion concentration, CB, is equal to the NaOH added plus PT. Excel’s Solver was used to maximize the value of PT by varying the NaOH added and PT. The constraints were that the ionic strength calculated from species concentrations was 0.1 M and the charge balance was satisfied. The maximum PT was found to be 0.0694 M, with 0.0435 M NaOH added. The buffer intensity was estimated in two ways. First, the system was approximated as a monoprotic acid system (H2PO4–/HPO42–) since the concentrations of the other phosphate-­ containing species is very low at pH 7. Equation 12.5.a becomes:

2.303

H 2 PO 4 HPO24 PT

[OH ]

H



ISTUDY

24.5  |  Integrated Case Study 4: Phosphate Buffers   609 0.120

CB(eq/L)

0.115 0.110 0.105 0.100 6.90

6.95

7.00 pH

7.05

7.10

FIGURE 24.8  CB and pH Values Near the Optimum Solution

For the optimum solution, β ≈ 0.0374 eq/L per pH unit. dCB was estimated numerically. The value of CB (= [Na+]) was varied dpH around the optimum value of 0.113  M and the pH was found that that satisfied the charge balance. The slope of a plot of CB versus pH is β. Calculated pH values are plotted in Figure 25.8. The slope of the best-­fit line through the calculated values is 0.0373  eq/L per pH unit, nearly identical to the value estimated by the H2PO4– and HPO42– concentrations. Second,

24.5.4  Model Extension The approach taken in this integrated case study represents one way to make a pH 7 phosphate buffer: Make a NaHPO4 solution and titrate with NaOH to pH 7. Another common approach to is to make the buffer with a mixture of NaHPO4 and Na2PO4 salts.

Thoughtful Pause Can you make a stronger buffer using this second approach? The intuition you developed thus far should tell you: No. Remember that NaHPO4, Na2PO4, and NaOH dissociate completely. The original approach gave the optimal solution for PT and CB (here, [Na+]) regardless of their starting materials. Starting with NaHPO4 and NaOH gave the optimal solution: a M NaHPO4 and b M NaOH, where a = 0.0694 and b = 0.0435. Starting with NaHPO4 and Na2PO4 would give the solution: c M NaHPO4 and d M Na2PO4. From a mass balance on phosphate and sodium ion, you can show: c = a – b and d = b. In other words, the second approach has the same PT and pH as the first approach, and therefore it has the same buffer intensity. The second approach gives a mixture of 0.0259 M NaHPO4 and 0.0435 M Na2PO4 for a pH 7 phosphate buffer with I = 0.1 M and maximum buffer capacity.

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610   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry

24.6  INTEGRATED CASE STUDY 5: GLOBAL

CLIMATE CHANGE

24.6.1  Introduction The final case study in this text concerns one of the most important water quality issues of the twenty-­first century: the effects of increasing carbon dioxide partial pressure on ocean chemistry. The oceans serve as a significant sink for atmospheric carbon dioxide. As the atmosphere and oceans warm, will the oceans continue to serve as a CO2 sink? You can do a “back-­of-­the-­envelope” calculation. Simply determine if the Henry’s law constant for carbon dioxide increases or decreases with temperature. In fact, all you need for this simple approach is the sign of Hrxn for CO2 dissolution: CO2(g) + H2O = H2CO3*; KH. You can calculate:

 Hrxn

H f ,H2 CO3 *

H f ,CO2 g H f ,H2 O

20.37 kJ / mol

K2 Hrxn 1 1 , a negative value K1 R T1 T2 of Hrxn means that KH deceases with increasing temperature. As a result, the oceans should hold less dissolved carbon dioxide as the water warms. From the integrated van’t Hoff equation, ln

Thoughtful Pause What do you think about this simple model?

At this point in the text, the simple approach should seem woefully inadequate to you. The solubility of CO2 and carbonate solids depends on pH. As discussed in Chapter 18, changing atmospheric CO2 concentrations will affect the pH of marine systems. Any calculations must consider the high ionic strength of the ocean. But fear not: You have the tools to conduct a more nuanced analysis.

24.6.2  Analysis Goals The first goal of the analysis is to determine whether the ability of the world’s oceans to absorb carbon dioxide is affected by temperature. To address this goal, the total carbonate concentration in the oceans will be calculated at different temperatures. The second goal of the analysis concerns dissolved oxygen. Most gases are less soluble at higher temperatures. Some aquatic organisms are sensitive to dissolved oxygen concentrations. The second goal of the analysis is to determine the effects of temperature on dissolved oxygen concentrations in seawater.

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24.6  |  Integrated Case Study 5: Global Climate Change   611

24.6.3  Analysis A simple equilibrium model for seawater chemistry was developed. Processes included the self-­ionization of water, CO2(g) and O2(g) dissolution, precipitation of CaCO3(s), and seawater composition. The values of the equilibrium constants must be calculated for seawater temperatures and ionic strength. You know that the van’t Hoff equation is of the form lnKT

= A0 + A1 , where KT is an equilibrium constant evaluated at temperature T and A0 T and A1 are constants. More general empirical equations have been developed of the A1 form lnKT A0 A2 T A3 lnT . In seawater, a common approach is to account for T salinity (S) by an empirical expression comprising powers of S:

lnKT

A0

A1 T

A2 T

A3 lnT B1 S

B1 S B3 S1.5



Expressions of the equilibrium constants used in the model are listed in Table 24.6. The units for KW, K1, K2, and the two Ks0 values are mole per kg of seawater. The units of K H ,CO2 are

mol and the units of K H ,O2 are mol . kg seawater-atm L-atm

The heart of the analysis, as always, is the alkalinity equation in seawater:

2[Ca 2 ] C B C A

[HCO3 ] 2[CO32 ] [B OH

4

] [OH ] [H ]

Note the inclusion of borate, B(OH)4–. With the usual substitutions (see Chapter 19), the alkalinity equation becomes:

2

Ks0 2 cT

CB C A

1

2

2

K H PCO2 0

KB BT K B [H ]

[OH ] [H ]

eq. 24.7

The total dissolved borate, BT, was assumed constant at 4.16×10–4 mol/kg (Uppström  1974). The salinity was assumed constant at 35‰. (See Section  2.4.3 for the definition of salinity.) The value of CB′  – CA was calculated from the standard mean chemical composition of seawater using the concentrations from Dickson and Goyet (1994):

CB C A

[Na ] 2[Mg2 ] [K ] 2[Sr 2 ] 0.01816 mol / kg

[Cl ] 2[SO 4 2 ] [Br ] [F ]

The value of CB′ – CA was assumed to be constant. The value of –log[H+]seawater until varied to satisfy eq. 24.7. The total dissolved carbonate (CT) and dissolved oxygen (DO) concentrations were then calculated. Three scenarios were explored: (1) present day, (2) year 2100 with present-­day CO2, and (3) year 2100  with elevated CO2. The present-­day and 2100 temperatures were 13.9°C and 16.4°C, respectively. These values are based on a March 2020 temperature

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612   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry Table 24.6 Expressions for  the Equilibrium Constants in  the Global Climate Change Model (see text for units) Equil.

lnKW

Expression

148.9802 13847.26T 5.977 118.67T

pK1

pK2

lnKB

Reference 1

1

1.0495 lnT

126.34 13.4191 S

Millero (1995)

23.6521 lnT S

0.01615

0.0331S 5.33 10 5 S 2 1

6320.813 530.123 S

6.103S T

90.18333 21.0894 S

0.1248S 3.687 10 4 S 2

19.568224 2.06950 S ln T

1

5143.692 772.43 S

20.051S T

8.966.90 2890.51 S

77.942S 1.726S1.5 0.0993S 2 T

ln T logKs0,Cal

logKs0,Arg

0.2474 S

0.053105 ST 2839.319T

1

0.77712 0.0028426T 178.34T

1

171.9065 0.077993T

171.945 0.077993T

60.2409 93.4517

100 T

0.023517 0.023656 lnK H,O2

Millero (1995)

1

24.4344 25.085 S

2903.293T

1

ln K H0 ,O

2

S

Ks0,C = Ks0 for calcite, Ks0,A = Ks0 for aragonite, ln K 0 H , O2 oxygen, see the reference for details)

S

Millero (1995)

1

0.10018S 0.0059415S

1.5

Millero (1995)

T 100

0.0047036

2.026S 2.555 10 4 S 2 T

0.07711S 0.0041249S1.5

71.595 log T

23.3585 ln T 100

Millero (1995)

71.595 log T

1

0.068393 0.0017276T 88.135T lnK H,CO2

Millero et al. (2006)

19.14.61336 3.336 S ln T

148.0248 137.194 S 1.62247S

Millero et al. (2006)

T 100

2

S Valderrama et al. (2016)

714.88 5.4728T (using 0.022 as the acentric factor for 1 0.4587T

of 1.2°C above the twentieth-­century average of 12.7°C and an expected rise of 1–4°C by 2100 (NOAA 2021). The present-­day CO2 level was 420 ppm. The 2100 CO2 level used was a mid-­range level of 700 ppm (IPCC 2021; Box TS5, Figure 1).

24.6.4  Analysis Results One concern about the rise in seawater temperature is that the seawater will hold less oxygen. The analysis quantifies this concern. The model predicts a 4.4% reduction in dissolved oxygen (DO). For a constant PO2 of 0.209, the calculation DO decreases from 9.09 mg/L to 8.69 mg/L. The results of the calculations are shown in Table  24.7. The effects of temperature increase on CO2 storage can be estimated by comparing the results at fixed PCO2

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24.7  |  Chapter Summary   613

Table 24.7 Results from the Seawater Model Scenario

T(°C)

Scenario 1 Scenario 2 Scenario 3

13.9 16.4 16.4

PCO2 (ppm) 420 420 700

CT (mM)

p[H+]

[Ca2+] (mM)

1.13 1.07 1.36

7.76 7.75 7.63

9.68 9.65 9.79

(420 ppm) with current (13.9°C) and future (16.4°C) temperatures (Scenarios 1 and 2, respectively). The total dissolved carbonate concentration is calculated to decrease by 7.0% at the higher temperature. This estimate is similar to the crude model of looking only at the Henry’s law constant: KH decreased by 5.0% at the higher temperature using the van’t Hoff equation. Current conditions (Scenario 1: 420  ppm, 13.9°C) were compared to anticipated future conditions (Scenario 3: 700  ppm, 16.4°C). The calculated p[H+] decreased slightly in future conditions, from 7.76 to 7.63. In both scenarios, calcite controlled the solubility of Ca2+. The [Ca2+] was calculated to increase from 9.68×10–3 M under current conditions to 9.79×10–3 M at 700 ppm and 16.4°C. As expected, the CT was much higher with the large PCO2 . The PCO2 increased 1.7 times

700 ppm 420 ppm

, but the calculated CT increased only 1.2 times: 1.13×10–3 M under

current conditions to 1.36×10–3 M at 700 ppm and 16.4°C. This is probably due to the decreased value of KH and the lower pH.

24.6.5  Limitations and Implications It should be noted that the model predicts a p[H+] of 7.76 under current conditions, smaller that the measured average p[H+] of about 8.2. The analysis here makes some simplifying assumptions about seawater temperatures and constancy of CB′ – CA and BT. With these simplifications, the DO and pH are expected to decrease and the [Ca2+] is expected to increase. The increased solubility of calcium-­containing solids does not bode well for organisms with shells. The results indicate that the oceans will be less-­important sinks for atmospheric carbon dioxide as the planet warms. This suggests a positive (here, unwanted) feedback loop where higher CO2 concentrations in the atmosphere lead to planetary warming, which leads to a higher percentage of CO2 retained in the atmosphere, and therefore more warming.

24.7  CHAPTER SUMMARY The integrated case studies in this chapter show how your knowledge of aquatic chemistry can be applied to a wide range of challenging problems. When your aquatic chemistry odyssey began, the analysis techniques in this chapter probably were far out of reach. Now, at the end of the text (but, as the Churchill quote at the beginning of Part VI suggests, at the start of the next phase of your journey), you should feel comfortable with the conceptualization of complex problems and trust that you have the tools

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614   Chapter 24  Putting It All Together: Integrated Case Studies in Aquatic Chemistry for their solution. It is hoped that your evolving intuition and mastery of the tools will allow you to face aquatic chemistry problems with confidence and, more importantly, with some small element of pleasure.

HISTORICAL NOTE: STUMM AND MORGAN If anyone can be called the founder of aquatic chemistry, it is Werner Stumm (1924– 1999). Professor Stumm and his students made aquatic chemistry into a field of study and integrated it into environmental engineering. As Jim Morgan wrote in a moving tribute: “Two traditions were joined: that of van’t Hoff, Goldschmidt, and Sillén in explaining natural water compositions; and that of Langelier, Buswell, Larson, and Black in improving corrosion control and water treatment.” (Morgan 2002). Stumm’s legacy is vast. Five PhD students and postdoctoral fellows who worked with him in the 1960s were elected to the National Academy of Engineering (Morgan 2002). Stumm received a PhD in chemistry from the University of Zürich in 1952. After a brief stint at the Institute for Water Resources and Water Pollution Control (EAWAG, see also the Historical Note to Chapter 14), Stumm did a postdoctoral fellowship at Harvard, becoming an assistant professor of sanitary chemistry at Harvard in 1956. Stumm returned to his homeland to become the director of EAWAG in 1970. James J. Morgan (1932–2020) was Stumm’s first doctoral student at Harvard, earning his PhD in 1964. He was hired as an associate professor of environmental health engineering at Caltech in 1965, where he spent his remaining academic career. Among their many achievements, Werner Stumm and Jim Morgan jointly were awarded the prestigious Stockholm Water Prize in 1999, celebrating their 40-­year partnership. I have avoided using the first person in this text, but when you live long enough, you find yourself living through a Historical Note. I was a third-­quarter freshman in the spring of 1977 looking for a major when I stumbled across Jim Morgan’s introductory environmental science course. Jim’s infectious enthusiasm opened up a new world to me. His superlative teaching and patient research mentoring provided me the best foundation in environmental science and engineering imaginable. Among Professors Stumm and Morgan’s many, many contributions to water chemistry are the three editions of the classic text Aquatic Chemistry. My doorway to water chemistry started with page 1 of the first edition of “Stumm and Morgan.” Professor Stumm spent a sabbatical in Pasadena working of the second edition. I took my first aquatic chemistry class from Professor Morgan using the proof sheets of the second edition. And so, dear reader, at the conclusion of this text, if you have learned anything (or daresay, been inspired), it is only because of the groundwork and legacy of Jim Morgan and Werner Stumm.

CHAPTER REFERENCES Akilan, C., E. Königsberger, J.S. Solis, P.M. May, J.H. Kyle, and G. Hefter (2016). Investigation of complexation and solubility equilibria in the copper(I)/cyanide system at 25°C. Hydrometallurgy 164: 202–207. Chung, Y.-­C. (1978). Thermodynamics of ion association in aqueous solutions of sodium sulfate, sodium carbonate and sodium

bicarbonate. Master’s thesis. Department of Chemistry, Eastern Illinois University. Dickson, A.G. and C. Goyet (eds.) (1994). Handbook of Methods for the Analysis of the Various Parameters of the Carbon Dioxide System in Sea Water, version 2. US Department of Energy, ORNL/CDIAC-­74.

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Chapter References   615 Haynes, H.W. (2003). Thermodynamic solution model for trona brines. AIChE J. 49(7): 1883–1894. IPCC (Intergovernmental Panel on Climate Change) (in press). Technical Summary. In Climate Change 2021: The Physical Science Basis. Contribution of Working Group I to the Sixth Assessment Report of the Intergovernmental Panel on Climate Change. Masson-­Delmotte, V., P. Zhai, A. Pirani, S.L. Connors, C. Péan, S. Berger, N. Caud, Y. Chen, L. Goldfarb, M.I. Gomis, M. Huang, K. Leitzell, E. Lonnoy, J.B.R. Matthews, T.K. Maycock, T. Waterfield, O. Yelekçi, R. Yu, and B. Zhou (eds.). Cambridge University Press. Klein, L.A. and C.T. Swift (1977). An improved model for the dielectric constant of sea water at microwave frequencies. IEEE Trans. Antennas Propag. AP-­25(1): 104–111. Korzhavyi, P.A. and B. Johansson (2010). Thermodynamic properties of copper compounds with oxygen and hydrogen from first principles. Technical Report TR-­10-­30, Svensk Kärnbränslehantering AB (Swedish Nuclear Fuel and Waste Management Co.). Lowenheim, F.A. (1978). Electroplating. New  York: McGraw‑Hill Book Co. Martell, A.E. and R.M. Smith (1974). Critical Stability Constants. New York: Plenum Press. Millero, F.J. (1985). The effect of ionic interactions on the oxidation of metals in natural waters. Geochim. Cosmochim. Acta 49(2): 547–553. Millero, F.J. and M. Izaguirre (1989). Effect of ionic strength and ionic interactions on the oxidation of iron(II). J. Solution Chem. 18(6): 585–599. Millero, F.J. (1995). Thermodynamics of the carbon dioxide system in the oceans. Geochim. Cosmochim. Acta 59(4): 661–677.

Millero, F.J., T.B. Graham, F. Huang, H. Bustos-­Serrano, and D. Pierrot (2006). Dissociation constants of carbonic acid in seawater as a function of salinity and temperature. Marine Chem. 100(1–2) 80–94. Morgan, J.J. Werner Stumm (2002). In: National Academy of Engineering Memorial Tributes (Vol. 10) Washington, DC: National Academy Press, pp. 223–227. NOAA (National Oceanic and Atmospheric Administration) (2020). National Centers for Environmental Information. Global Climate Report – March 2020. ncdc.noaa.gov/sotc/global/202003. Singer, P.C. and W. Stumm (1970). Acid mine drainage: The rate determining step. Science 167: 1121–1123. Stumm, W. and J.J. Morgan (1996). Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters. 3rd ed. New  York: John Wiley & Sons, Inc. Tiffreau, C., J. Lützenkirchen, and P. Behra (1995). Modeling the adsorption of mercury(II) on (hydr)oxides. I. amorphous iron oxide and α-­quartz. J. Colloid Interface Sci. 172: 82–93. Uppström L.R. (1974). Boron/chlorinity ratio of deep-­sea water from the Pacific Ocean. Deep-­Sea Res. 21: 161–162. Valderrama, J.O, R.A. Campusano, and L.A. Forero (2016). A new generalized Henry-­Setschenow equation for predicting the solubility of air gases (oxygen, nitrogen and argon) in seawater and saline solutions. J. Mol. Liquids 222: 1218–1227. Wehrli, B. (1990). Redox reactions of metals at mineral surfaces. In: Aquatic Chemical Kinetics: Reaction Rates of Processes in Natural Waters. W. Stumm (ed.) New  York: John Wiley & Sons, Inc.

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APPENDIX A

Background Information A.1 INTRODUCTION It is assumed that you has some background in chemistry and mathematics prior to using this text. A chemistry background equivalent to freshman chemistry is required. Good skills in the manipulation of algebraic equations are highly desirable. A few basic concepts in chemistry, mathematics, and the use of spreadsheet programs are reviewed in this appendix.

A.2  CHEMICAL PRINCIPLES A.2.1  Background Atoms share electrons to form molecules. Molecules (from the diminutive of the Latin moles mass) are the smallest division of a substance that retains the properties of the substance. Charged molecules are called ions (from the Greek ienai to go, since ions migrate to electrodes). Cations are positively charged and anions are negatively charged (from, respectively, katienai to go down and anienai to go up: to differentiate the movement of ions in an electric field). Charges are indicated by a superscript at the end of the chemical formula of the ion, as in SO42– for sulfate ion. Formally, the names of ions include the word “ion”. For example, HCO3– is called “bicarbonate ion,” not “bicarbonate.” In this text, the word “ion” frequently is omitted. Molecules containing atoms with unpaired electrons are called radicals. Atoms, molecules, ions, and radicals are referred to as chemical species. This text is concerned with chemical equilibria involving molecules and/or ions, sometimes called ionic equilibria. Radical chemistry is important in some environmental systems, but is outside the scope of this text. The concentration of chemical species will be indicated by square brackets: [NO3–] = nitrate concentration.

ion: an atom or molecule carrying a net electric charge cation: a positively charged ion anion: a negatively charged ion radicals: molecules containing atoms with unpaired electrons chemical species: the collection of atoms, molecules, ions, and radicals in a chemical system ionic equilibria: chemical equilibria involving molecules and/or ions

A.2.2  Species Types You will work with five types of chemical species defined by their phase: dissolved species, pure liquids, pure solids, gases, and surface-­bound species. Each type is indicated by a symbol following the species name in parentheses. The species type labels are summarized in Table A.1. The dissolved species type are indicated by the symbol (aq). For example, dissolved chloride ion will be written as: Cl–(aq). Since most of the

617

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618  Appendix A  Background Information Table A.1 Labels for Species Types Species Type

Dissolved species Pure liquids Solids     Pure solids     Other solids Gases Surfaces

Key idea: Dissolved species, pure liquids, pure solids, and gases are labeled (aq), (l), (s), and (g), respectively, while surfaces are labeled with the prefix ≡

Species Label

 

(aq) (l) (s) (am) (g) ≡ (prefix)

Common Label

 

no label no label (s) (am) (g) ≡ (prefix)

species you will work with in this text are dissolved, the (aq) label frequently will be omitted. It will be used explicitly when the type of species is ambiguous. Pure liquids are indicated with the symbol (l). In most of the text, the only pure liquid you will encounter is water. Thus, the (l) symbol usually will be omitted and only used to emphasize water as a pure phase. The symbols H2O(l) and H2O are used interchangeably to indicate pure condensed water. Pure solids are indicated by the symbol (s), as in CaCO3(s). The (s) symbol always will be used with pure solids. A species name lacking the symbol (s) is a dissolved species. For example, CaSO4(s) indicates the pure solid calcium sulfate, while CaSO4 [or unambiguously, CaSO4(aq) or CaSO40] represents the dissolved calcium sulfate complex. Solid phases that are not pure solids will be indicated by the symbol (am) for amorphous. Gases are indicated by the symbol (g), as in CO2(g). A species name lacking the symbol (g) is a dissolved species, not a gas. For example, O2(g) indicates gaseous oxygen, while O2 [or, unambiguously, O2(aq)] represents dissolved oxygen. Finally, surfaces will be indicated by the symbol ≡, as in ≡S and ≡FeOH. Note that the symbol ≡ appears as a prefix before the solid.

A.2.3  Chemical Reactions reactants: reacting species products: species formed through the reaction of reactants

Several common symbols and abbreviations are used throughout this text. Reactions are the interactions of some chemical species (called reactants) to form other chemical species (called products). Three symbols are used to emphasize the direction of the reaction. The symbol → is used to emphasize the progression of the reaction from left to right, as in: H2O → H+ + OH–. Note the common approach of writing reactants on the left side of the reaction and products on the right side. Similarly, the symbol ← is used to emphasize the progression of the reaction from right to left, as in: H+ + OH– ← H2O. Note in this case, the reactant (H2O) is written on the right side. The symbol = is used to indicate the state of chemical equilibrium. The “equals sign” signals that reactions have occurred both to the right and left until the species concentrations no longer change with time. For example, writing the equilibrium expression H2O = H+ + OH– indicates that the concentrations of the three species H2O, H+, and OH– do not change with time. Since the “equals sign” indicates that reactions occur both to the right and left, you may wonder which species should be called reactants and which should be called

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A.3  |  Mathematical Principles   619

products. Indeed, at equilibrium, all reacting species are both reactants and products. By convention, the species on the left of the equilibrium expression are called reactants and species on the right of the equilibrium expression are called products.

A.2.4  Oxidation State The oxidation state (or oxidation number) refers to the formal or effective charge on an atom in a molecule or ion. Oxidation states usually are indicated by signed Roman numerals (e.g., +I and –III). To determine the oxidation state, you sometimes need a starting point. For aquatic systems, the oxidation state of H usually is +I and the oxidation state of O usually is –II. [Common counterexamples are O in O2(g) and H in H2(g).] These starting points allow the calculation of the oxidation state of other atoms in a molecule or ion. The sum of the oxidation states (weighted by the number of atoms of each type in the chemical species) must equal the charge on the species. You can write: n i 1

i

OS i

z

where νi = stoichiometric coefficient of atom i in the chemical species, (OS)i = ­oxidation state of atom i, z = overall charge on the chemical species, and n = the number of different elements in the species. The steps in calculating the oxidation state are as follows (repeated in Appendix C, Section C.1.1): Step 1: Decide if the oxidation state of H is +1 (+I) and the oxidation state of O is −2 (−II). n Step 2: Calculate the oxidation state of other atoms by vi OS i z. i 1

In this equation, νi = stoichiometric coefficient of atom i in the chemical species, (OS)i = oxidation state of atom i, z = overall charge on the chemical species, and n = the number of different elements in the species. An example will make the calculation of oxidation state clearer. Consider the oxidation state of sulfur in sulfate, SO42−. Here, n = 2 (two elements: S and O), z = −2 (overall charge of −2), νS = 1 (one sulfur atom in sulfate), and νO = 4 (four oxygen atoms in sulfate). If you assume that the oxidation state of O is −II (or −2), then: (1)(OS)S + (4) (−2) = −2. Solving, the oxidation state of sulfur, OSS, is +6 or +VI.

A.3  MATHEMATICAL PRINCIPLES In this text, a fluency with the algebraic manipulations of equations is assumed. A knowledge of calculus is useful for understanding how thermodynamic properties change with other thermodynamic properties (Chapter 3), but not essential. In terms of nomenclature, log(x) means log10(x) and ln(x) means loge(x). Recall the meaning of the summation and product signs:

3

a i 1 i

= a1 + a2 + a3 and

4

a i 1 i

= a1a2a3a4. Also,

remember the properties of the log function: log(ab) = log(a) + log(b) and log(ab) = blog(a).

Key idea: In equilibrium expressions, species on the left are called reactants and species on the right are called products oxidation state: (oxidation number): the effective charge on an atom needed to account for the overall charge on the ion or molecule

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620  Appendix A  Background Information Key idea: The p function is defined by: pX = −log(X)

or X = 10−pX

Key idea: The value of pX increases as X decreases

You will use another function in this text: the p function. The p function is defined by: pX = −log(X). Thus: X = 10−pX. The p function can be applied to constants. For example, if K = 0.001 = 10−3, then pK = 3. In addition, the p function can be applied to species concentrations. For example, pNH3 = −log([NH3]). There are two important aspects of the p function to remember. First, the value of pX increases as X decreases. As an example, pX = 7 if X = 10−7, and pX increases to 9 if X decreases to 10−9. Second, when applying the p function, you sometimes ignore the usual rule that you cannot take the logarithm of a number with units. As developed in Chapter 2, concentrations in aqueous systems usually are written in units of moles per L (= mol/L). We write: [Cl−] = 10−3 mol/L and pCl = 3 without difficulty. Always make sure you know the units of a number when applying the p function. It is a good idea to become fluent with interconverting between a value and p(value). Thoughtful Pause If x = 10

, what is px? If py is –3.5, what is y?

–6.2

If x = 10–6.2, then px = –log10x = –log10(10–6.2) = –(–6.2) = 6.2. If you write x as “ten to the something,” then px is “–1×something.” If py is –3.5, then y = 10–y = 10–(–3.5) = 103.5 = 3.16×103.

A.4  SPREADSHEET SKILLS A.4.1  General Spreadsheet Skills It is assumed that you are familiar with the use of standard spreadsheet programs (i.e., Microsoft Excel). The examples in this text use the Excel format. It is expected that you can enter a formula in a cell and copy the formula to other cells. You should be comfortable with the difference between relative and absolute cell referencing. Suppose you enter a value in cell A1, and the equation =A1 in cell B1. If you copy this formula to cell B2, the formula in B2 will be: =A2. You used relative cell referencing in the formula, so Excel incremented the reference when you copied the formula down one cell. If you want to write a formula that references only column A or only row 1 or only cell A1, then you must use an absolute cell reference. This is done by the use of dollar signs before the column or row (or both) that is fixed. For example, to reference cell A1 each time, the proper formula is: =$A$1.

A.4.2  Spreadsheet Skills for Equilibrium Calculations You want to create spreadsheets that produce accurate results. You also want to create spreadsheets that are reusable and valuable to you in the future. To be reusable, it is good practice to put important system parameters as values in their own cells. Then you can reference those cells (using absolute cell referencing) rather than typing the values into each formula. If you follow this approach, then you can use your spreadsheet to solve a similar problem easily. Here is an example. You will find in Section 11.3.6 that you can calculate the concenH tration of certain acids by C . (If you are reading this section before learning K H

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A.4  |  Spreadsheet Skills   621

about acids and bases, do not worry about the meaning of the symbols for now. Here, [H+] is about equal to 10–pH.) Suppose you want to calculate the acid concentration at a number of pH values. You could set up your spreadsheet as in Figure A.1. In this example, C = 1×10–3 (ignoring units for a moment) and K = 1×10–4. The formulas in columns B and E are shown in columns C and F, respectively.

FIGURE A.1  Calculation Example with Hard-­Coded Constants

This spreadsheet will produce accurate results, but it is difficult to reuse. To solve another problem with a different K or C value, you would have to find the all instances of the values in the formulas, retype them, and recopy the formulas. In more complexed spreadsheets, this can be a tedious and assuredly error-­prone task. A much better approach is to put the values of C and K in their own cells, as in Figure A.2. In this approach, you can refer to the values when you want to use absolute cell referencing. As an example, note the use of the absolute cell referencing (dollar signs) in the formulas in column E (formulas shown in column F). Using this approach, you can simply type new values in the cells referred to (B1 and B2 in Figure A.2) and solve a similar problem. Another way to make spreadsheets more valuable in the future is to label everything. Labeling columns and important cells is akin to including comments when writing code. Labels will remind you later how you designed the spreadsheet and make it more readable. Note in Figure A.2 that the units of C are entered in cell C1.

FIGURE A.2  Calculation Example with Referenced Constants

A.4.3  Solving Nonlinear Equations with Spreadsheets As you will learn, equilibrium systems generate nonlinear equations. Therefore, it is useful to know how to solve nonlinear equations with spreadsheets. As an example, suppose you want to find x that satisfies the nonlinear equation: 5x = x2 + 6. There are

Key idea: Make your spreadsheets more useful by using absolute referencing and labels

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622  Appendix A  Background Information three ways to use a spreadsheet program to solve this equation. For each method, it is useful to rewrite the equation with zero on one side: x2 − 5x + 6 = 0. In the first method, you guess x and iterate until x2 − 5x + 6 = 0. To accomplish this, enter any number (say 1) in cell A1 and enter the formula =A1^2−5*A1+6 in cell B1. If the value of cell A1 is 1, then the value of cell B1 will be 2. Now manually change the value of cell A1 until the value of cell B1 is as close to zero as you like. This method can be implemented easily on any programmable device. Writing the equation as equal to zero will save you a lot of time with manual solutions. You can change your guess and follow the sign of the result to hone in on the answer (to your desired number of significant figures) quickly. In the second method, use a nonlinear optimization approach such as the Excel Solver function. Fill cells A1 and B1 as describe above. Now access Solver (Tools•Solver). If Solver is not shown, then you will have to install it. Set the target cell to B1 (the cell containing the formula you wish to solve), select the “Value of:” option in “Equal To:” and type 0 in the box, and set the “By Changing Cells:” to A1 (the cell containing the value you is to change). Now click “Solve.” The approximate answer (2.00...) will appear in cell A1. In the third method, we make use of the auto-­calculation feature. In Excel, enable the auto-­calculation mode (select File•Options•Formulas, and check the Enable iterative calculation checkbox). Now rearrange the original equation and solve for x. In the example: x =

x2

6

. Type the right-­hand side in cell A1: =(A1*A1+6)/5. Excel will 5 interpret this as a command to find a number that equals one-­fifth of itself squared plus six. (If you try this and get a circular reference error, make sure the “Iteration” box is checked as describe above.)

A.4.4  Pitfalls with Using Spreadsheets to Solve Nonlinear Equations

Key idea: In chemical equilibrium problems, it often is useful to vary p(concentration) rather than the concentration

None of these methods is foolproof. For example, each method breaks down when multiple roots are present. In the example above, each method returns an answer of 2, when actually both 2 and 3 are roots of the equation. Care should be taken when multiple roots are expected. In such cases, it is useful to plot the function against the independent variable and look for roots (i.e., values of the independent variable where the function equals zero). A method may fail if the function and/or cells to be varied are extremely small. This is often the case in equilibrium calculations. In such instances, it is useful to multiply the function by a large number so that its value is reasonable. For example, if the equation to be set equal to zero is x2 − 5×10−8x (with x in cell A1), you might enter =(A1^2−5e−8*A1)*1e10. In addition, it is useful to transform the variable to be changed so that its values are reasonably large. For example, you may choose to vary log(x) rather than x. In chemical equilibrium problems, it often is useful to vary p(concentration) rather than the concentration. The third method may fail if the formula entered gives an error with the variable value of zero. For example, to solve the equation x = 1/x (which has roots 1 and −1), one would enter into cell A1: =1/A1. This will give a division by zero error, since Excel begins the calculation with the value of A1 equal to zero. One possible workaround is to tweak the formula to avoid division by zero when x = 0. In the example above, you could enter the formula as: =1/(a1+1e−20).

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Useful Physical Constants and Conversions   623

CHAPTER KEY IDEAS • Dissolved species, pure liquids, pure solids, and gases are labeled (aq), (l), (s), and (g), respectively, while surfaces are labelled with the “prefix” ≡. • In equilibrium expressions, species on the left are called reactants and species on the right are called products. • The p function is defined by: pX = −log(X) or X = 10−pX. • The value of pX increases as X decreases. • Make your spreadsheets more useful by using absolute referencing and labels. • In chemical equilibrium problems, it often is useful to vary p(concentration) rather than the concentration.

CHAPTER GLOSSARY anion:  a negatively charged ion cation:  a positively charged ion chemical species:  the collection of atoms, molecules, ions, and radicals in a chemical system ion:  an atom or molecule carrying a net electric charge ionic equilibria:  chemical equilibria involving molecules and/or ions products:  species formed through the reaction of reactants radicals:  molecules containing atoms with unpaired electrons reactants:  reacting species

USEFUL PHYSICAL CONSTANTS AND CONVERSIONS Atomic Numbers and Relative Atomic Masses (g/mol) for Selected Elements Sym. Name

H He Li Be B C N O F Ne Na Mg Al Si

hydrogen helium lithium beryllium boron carbon nitrogen oxygen fluorine neon sodium magnesium aluminum silicon

At. No. Rel. At. Mass

1 2 3 4 5 6 7 8 9 10 11 12 13 14

1.008 4.0026 6.94 9.0122 10.81 12.011 14.007 15.999 18.998 20.180 22.990 24.305 26.982 28.085

Sym. Name

Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba

Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon Cesium Barium

At. No.

Rel. At. Mass

43 44 45 46 47 48 49 50 51 52 53 54 55 56

-­ 101.07 102.91 106.42 107.87 112.41 114.82 118.71 121.76 127.60 126.90 131.29 132.91 137.33

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624  Appendix A  Background Information Sym. Name

P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo

At. No. Rel. At. Mass

phosphorus sulfur chlorine argon potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton rubidium strontium yttrium zirconium niobium molybdenum

15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

Sym. Name

30.974 32.06 35.45 39.95 39.098 40.078 44.956 47.867 50.942 51.996 54.938 55.845 58.933 58.693 63.546 65.38 69.723 72.630 74.922 78.971 79.904 83.798 85.468 87.62 88.906 91.224 92.906 95.95

La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi

Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth

Constants kJ L-atm R = ideal gas constant = 8.314×10 –3  = 0.082057 K-mol K-mol N = Avogadro’s number = 6.022×1023 F = Faraday constant = 96,485 ln10

1 mol

C A-s = 96,485 mol mol

RT = 59.16 mV at 25°C F

k = Boltzmann constant = 1.380649×10−23

J K

Conversions 1 atm = 101.325 kPa T(°C) = T(K) – 273.15 Alk(mg/L as CaCO3) = 50×Alk(meq/L) = 50,000×Alk(eq/L)

At. No.

Rel. At. Mass

57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83

138.91 140.12 140.91 144.24 -­ 150.36 151.96 157.25 158.93 162.50 164.93 167.26 168.93 173.05 174.97 178.49 180.95 183.84 186.21 190.23 192.22 195.08 196.97 200.59 204.38 207.2 208.98

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APPENDIX B

Equilibrium Revisited B.1 INTRODUCTION Most of this text is devoted to the chemistry of aqueous systems at equilibrium. This appendix explores two aspects of equilibrium that go beyond the material presented in Chapter 3. The first extension of the equilibrium concept concerns systems that exchange mass or energy with their surroundings. Recall in Chapter 3, equilibrium was defined as the state where all thermodynamic functions are invariant with time in a system that does not exchange mass or energy with its surroundings. If the system does exchange mass with its surroundings, then a different approach is required. In fact, a different name is given to the state where all thermodynamic functions are invariant with time in a system that does exchange mass or energy with its surroundings. This state is called the steady state. At first glance, the steady state and the equilibrium state appear very similar. In this appendix, the differences between steady-­state solutions and equilibrium solutions will be explored. The second extension of the equilibrium concept stems from the two approaches to determining the equilibrium position of a system: minimization of free energy and the algebraic approach. The two approaches are really the same, since the concept of K comes from setting ∆Grxn = 0. However, you may question whether the two approaches will result in calculation of the same equilibrium concentrations. In this appendix, an example will be presented to show that the two approaches are equivalent.

steady state: the condition in which the species concentrations are invariant with time in a system that does exchange mass or energy with its surroundings

B.2  EQUILIBRIUM AND STEADY STATE B.2.1  Problem Statement To analyze a system that exchanges mass with its surroundings, it is necessary to have a basic understanding of chemical equilibria and kinetics. The material in this appendix is best understood after completing Chapters 3 and 23 of this text. Consider a system receiving and discharging water so that the volume of the system is constant. Constant system volume means that the flow of water into the system equals the flow of water out of the system (assuming constant water density). This is a good model for a lake (or tank) of constant volume. Imagine that the influent stream delivers water with a constant molar concentration, [A]in, for species A. Use the 625

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626  Appendix B  Equilibrium Revisited symbol [A] for the concentration of A inside the system. If the system is completely mixed, then [A] is the same everywhere inside the system. This means that the effluent concentration also will be equal to [A].

B.2.2  Mole Balance A mole balance on species A reveals: accumulation rate of moles inside the system = rate of moles into the system – rate of moles out of the system + rate of change of moles inside the system



eq. B.1

Each term in the mole balance has units of mole/time. The accumulation rate is d A given by V  , where V is the system volume (lake volume here). The rates of flow dt of moles in and out of the system are Q[A]in and Q[A], respectively, where Q = flow in = flow out. To determine the rate of change inside the system, assume that A is transformed into another species B. Two reactions are of interest: A

B; k

kf

B

A; k

kr

Assuming the reactions are elementary (Section  23.3.3), the techniques of Section 23.3.4 can be used to write the rate of change of [A] inside the system: rate of change of [A] inside the system = –kf[A] + kr[B] Substituting into the mole balance (eq. B.1): V

d A dt

Q A in Q A

V

kf A

kr B ,

or: d A dt

A in

A

kf A

kr B

V = hydraulic residence time. Similarly, for B (assuming it is not present in Q the influent stream: [B]in = 0):

where: θ =



d B dt

B

kf A

kr B

eq. B.2

We seek the steady-­state solution; that is, the solution when the species concend A d B trations do not change with time. At steady state: = 0. The steady-­state dt dt ­concentrations are denoted [A]ss and [B]ss. From eq. B.2: 0

B ss

k f A ss

kr B ss

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B.2  |  Equilibrium and Steady State   627

Solving: B SS A SS



kf kr

kf kr 1

1 1 kr



eq. B.3

B.2.3  Comparison of Steady-­State and Equilibrium Solutions What is the ratio of the species concentrations at equilibrium? At equilibrium (with no mass exchange with the surroundings): A = B; K = equilibrium constant Thus: B eq



A eq

K

eq. B.4

Recall from Section 23.5.2 that for a system at equilibrium with no exchange with its surroundings: K =



kf

kr

. Thus, eq. B.3 becomes: B SS A SS

kf kr 1

1 kr

K 1

1 kr



eq. B.5

Comparing eqs. B.4 and B.5, it is clear that the steady-­state solution for a system exchanging mass with its surroundings and the equilibrium solution for a system that does not exchange mass with its surrounding are different. Given enough contact time (that is, as θ → ∞), the two solutions become identical, as expected. In fact, the two solutions are within 1% of each other if krθ > 100.

B.2.4  Environmental Examples Consider two examples from Section 23.5.2. The first example is monochloramine synthesis (Worked Example 23.6): HOCl + NH3 → NH2Cl + H2O; kf NH2Cl + H2O → HOCl + NH3; kr In this case, kr = 2.1×10–5 s–1. The steady-­state and equilibrium solutions will be 100 within 1% of each other when krθ > 100, or when θ > 100  = or after kr 2.1 10 5 s 1 about 55  days. At hydraulic residence times of interest in engineered systems, the steady-­state and equilibrium concentrations clearly will be different. In this case, an ­equilibrium approach may be inappropriate.

Key idea: Steady-­state solutions approach equilibrium solutions as the residence time increases

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628  Appendix B  Equilibrium Revisited On the other hand, consider the example of acetic acid hydrolysis, where kr = 10+3.7 s–1 at pH 7. In this case, for the steady-­state and equilibrium solutions will be within 1% of 100 100 each other when θ > = or after about 20 milliseconds. For this chemistry, kr 10 37 s 1 the steady-­state and equilibrium concentrations are likely to be very similar and an equilibrium approach may be appropriate even if mass is exchanged with the surroundings.

B.3  ENERGY MINIMIZATION AND

ALGEBRAIC SOLUTIONS

B.3.1  Introduction Chemical engineers frequently calculate equilibrium activities by determining the set of activities that minimize the free energy of the system, subject to mass balance constraints. Does this approach give the same equilibrium activities as the algebraic approach employed by environmental engineers?

B.3.2  Example Problem To address this issue, consider the addition of a monoprotic acid, HA, to water. The pertinent equilibria are: HA = A– + H+; K H2O = OH– + H+; KW

B.3.3  Algebraic Solution The species list is: H2O, OH–, H+, HA, and A–. Assume the system is dilute and activities and concentrations are interchangeable. In the algebraic approach, you form the mathematical equation from the equilibrium: A

K



H HA

eq. B.6

The charge balance is: [A–] + [OH–] = [H+]. The solution is expected to be acidic, so [OH–] pKa. The [HA] line segments go through the system point if extended, as shown by the dashed lines. Similarly, tap the [A-­] button to sketch the [A–] line far from the system point. The – [A ] is about equal to AT at pH >> pKa and parallel to the [OH–] line at pH