Physical Chemistry Laboratory Manual: An Interdisciplinary Approach 9789389583106

Citation preview

PHYSICAL
CHEMISTRY
LABORATORY
MANUAL

TM

An
Interdisciplinary
Approach

Salient Features • Interdisciplinary and innovative methods. • For undergraduate students of science, engineering and environmental sciences. • Useful experiments, beyond the syllabus, are also given under 'Miscellaneous Experiments'. • Wherever possible, two or more methods are given. • Brief account of the principles required for each experiment is provided. Amirtha Anand is Associate Professor of Chemistry in Maitreyi College, University of Delhi. She completed her PhD degree from the University of Madras. She has 34 years of experience in teaching undergraduate students of Honours and General courses. She has published three books and many papers in both national and international journals of high repute. She has been conferred the Best Teacher Award by the Government of Delhi in 2012. Currently, she is associated with CIET, NCERT for developing the e-contents for Massive Open Online Courses (MOOCs) in Chemistry.

978-93-89583-10-6

` 395/-

PHYSICAL
CHEMISTRY
LABORATORY
MANUAL
An
Interdisciplinary
Approach Amirtha Anand Ramesh Kumari

Amirtha Anand • ramesh Kumari

Ramesh Kumari did her graduation in Chemistry (Honours) from University of Delhi (1987) , Masters in Chemistry (M.Sc.) from IIT Delhi (1989) and PhD in Physical Chemistry from IIT Delhi in 1994. She is Associate Professor in the Department of Chemistry, Maitreyi College, University of Delhi. She has over 24 years of experience in teaching undergraduate courses in University of Delhi. She has also been the Deputy Coordinator for PG Diploma in Nanotechnology under the innovative programme of the UGC and Academic Coordinator for Undergraduate Programmes of IGNOU at Maitreyi College. She has authored 2 books for undergraduate courses and has been the content writer for study modules for e-PG Pathshala.

PHYSICAL
CHEMISTRY
LABORATORY
MANUAL


This book covers the latest syllabus of CBCS pattern of Delhi and other universities for both B.Sc. Programme and Honours courses. A large number of Physical Chemistry, Environmental Chemistry, Nanoscience, Polymer Chemistry and Analytical Chemistry experiments have been covered using interdisciplinary and innovative methods. The contents include some fundamental chemical concepts, measurement of surface tension and viscosity, colorimetry, determination of order of a reaction, hetrogeneous equilibria, adsorption on solid surfaces, thermochemical measurements, conductometric and potentiometric measurements, pH metry, environmental parameter analysis, etc. Wherever possible, two or more methods are given. So, the teachers and students will have a choice to make depending on the availability of chemicals, apparatus, instruments, time, etc. This book will give them the opportunity to relate theory and practicals for a better understanding of the subject.

Distributed by:

9 789389 583106 TM

PHYSICAL CHEMISTRY LABORATORY MANUAL An Interdisciplinary Approach

PHYSICAL CHEMISTRY LABORATORY MANUAL An Interdisciplinary Approach

Amirtha Anand

Associate Professor Department of Chemistry Maitreyi College, University of Delhi Delhi

Ramesh Kumari

Associate Professor Department of Chemistry Maitreyi College, University of Delhi Delhi

©Copyright 2019 I.K. International Pvt. Ltd., New Delhi-110002. This book may not be duplicated in any way without the express written consent of the publisher, except in the form of brief excerpts or quotations for the purposes of review. The information contained herein is for the personal use of the reader and may not be incorporated in any commercial programs, other books, databases, or any kind of software without written consent of the publisher. Making copies of this book or any portion for any purpose other than your own is a violation of copyright laws. Limits of Liability/disclaimer of Warranty: The author and publisher have used their best efforts in preparing this book. The author make no representation 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 of any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. The accuracy and completeness of the information provided herein and the opinions stated herein are not guaranteed or warranted to produce any particulars results, and the advice and strategies contained herein may not be suitable for every individual. Neither Dreamtech Press 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. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade names of their respective holders. Dreamtech Press is not associated with any product or vendor mentioned in this book. ISBN: 978-93-89583-10-6 EISBN: 978-93-89698-56-5

Edition: 2019

To the memory of my father Prof. K.S. Nagarajan. — Dr. Amirtha Anand To the memory of my father Shri. T.R. Kapoor. —Dr. Ramesh Kumari

PREFACE

This book covers the syllabus of Credit Based Choice System (CBCS) pattern of Delhi and other universities for both B.Sc., Programme and Honours courses. A large number of Physical Chemistry, Environmental Chemistry and Nanoscience experiments have been covered using interdisciplinary and innovative methods. This book will be guiding undergraduate students of science, engineering and environmental sciences of Indian universities due to its approach of teaching the practical subject in the simplest possible way. The contents include some fundamental chemical concepts, molar mass determination, measurement of surface tension and viscosity, colorimetry, determination of order of a reaction, heterogeneous equilibria, adsorption on solid surfaces, thermochemical measurements, conductometric and potentiometric measurements, pH metry, soil analysis, etc. A number of innovative and useful experiments, beyond the syllabus, are also given under ‘Miscellaneous Experiments’. Students can perform these out of their interest and curiosity. These can be taken for some project work also. This book will meet the requirement of the latest syllabus of various universities. It is written in a simple to complex approach. Wherever possible, two or more methods are given. So the teachers and students will have a choice to make, depending on the availability of chemicals, apparatus, instruments, time, etc. There are many books with similar titles. But they are of general nature and are written topic-wise. In this book, the theory of each experiment is dealt with in detail. The teachers and students can easily access what they need. Since the students study the concepts of these experiments in great detail in theory papers, brief account of the principles required for each experiment is provided. This book will give them the opportunity to relate theory and practical for a better understanding of the topics easily. Now, in a semester system, the students do not have much time and they want only the essential things. This book is a concise one but without diluting the subject. Many teachers across the Delhi University with whom we interacted during the meetings and practical examinations expressed the view that such a book is necessary. Comments and suggestions for further improvement of the book would be highly appreciated. Amirtha Anand Ramesh Kumari

ACKNOWLEDGEMENTS

Dr. Amirtha Anand would like to thank her husband Mr. Anand Srinivasan for his patience, support and constant encouragement during the preparation of the manuscript. Dr. Ramesh Kumari sincerely thanks her family members for their patience and support. Amirtha Anand Ramesh Kumari

CONTENTS

Preface Acknowledgements

vii ix

1. Introduction 1.1. Chemistry Laboratory Apparatus and their Uses 1.2 Preparation of Standard Solutions 1.3 Units for Reporting Concentration Terms 1.4 Errors in Measurements 1.5 Safety Rules in Chemistry Lab 1.6 First-Aid Procedures Viva Questions

1       

2. Surface Tension Measurement 2.1 To determine the surface tension of a given liquid at room temperature using stalagmometer by “drop weight” method. 2.2 To determine the surface tension of a given liquid at room temperature using stalagmometer by “Drop Number” method. 2.3 To study the variation of surface tension with different concentrations of a detergent and to determine its Critical Micelle Concentration (CMC). Viva Questions

19 19

3. Viscosity Measurement 3.1 To determine the absolute and relative viscosities of the given liquid taking water as reference and using Ostwald viscometer. 3.2 To study the variation of viscosity with different concentrations of sugar solution. 3.3 To study the variation of coefficient of viscosity with different concentrations of Polyvinyl alcohol (PVA) in water and to determine the molar mass of PVA. Viva Questions

27 27

22 24

26

 31

33

xiiȳ˜—Ž—œȱ

4. pH Measurement 4.1 To measure the pH of given samples of milk, juice, soap, detergent and shampoo. 4.2 To prepare the following buffer solutions and to measure the pH of each of these solutions using pH meters. ŗǯȲ˜’ž–ȱŠŒŽŠŽȬŠŒŽ’ŒȱŠŒ’ȱo acidic buffer. ŘǯȲ––˜—’ž–ȱŒ‘•˜›’ŽȬŠ––˜—’ž–ȱ‘¢›˜¡’Žȱo basic buffer. 4.3 To determine the strength of the given strong acid (hydrochloric acid) by titrating it pH-metrically and potentiometrically with a strong base (sodium hydroxide). 4.4 To determine the strength of a weak acid (acetic acid) by titrating it pH metrically and potentiometrically with a strong base (sodium hydroxide). 4.5 To study the effect of addition of HCl /NaOH to solutions of acetic acid, sodium acetate and their mixtures. Viva Questions

34

5. Thermochemical Measurement 5.1 To determine the water equivalent (or heat capacity) of a calorimeter using different volumes of water. 5.2 To determine the water equivalent (or heat capacity) of the calorimeter by back calculation method (using enthalpy of solution of sulphuric acid). 5.3 To determine the water equivalent (or heat capacity) of the calorimeter by back calculation method (using enthalpy of neutralization). 5.4 To determine the heat of neutralization of hydrochloric acid and sodium hydroxide using a thermos flask or beaker as calorimeter. 5.5 To determine the molar enthalpy of ionization (ionization energy) of acetic acid calorimetrically. 5.6 To determine the integral heat of solution (enthalpy of solution) of the given salt at a specified salt-water ratio (KNO3 in 1: 200 molar ratio). 5.7 To determine the integral heat of solution (enthalpy of solution) of the given salt at a specified salt-water ratio (CuSO4 in 1: 400 molar ratio). 5.8 To determine the enthalpy or heat of hydration of copper sulphate.

50

5.9

To determine the heat of solution of benzoic acid by solubility method.

 

38



 

 52

54

54 58 60

62

63 64

˜—Ž—œȱȳxiii

5.10 To determine the basicity of the given acid (phosphoric acid) calorimetrically. Viva Questions

67

6. Heterogeneous Equilibria 6.1 To construct the phase diagram of the given mixture (Urea : Benzoic Acid) using cooling curves method. 6.2 To find the critical solution temperature or the upper consolute temperature (CST ) for the phenol-water system by plotting the mutual miscibility curve. 6.3 To study the effect of impurities (sodium chloride and succinic acid) on the critical solution temperature of phenol-water system. Viva Questions

70 70

7. Potentiometric Measurement 7.1 To study the titration curve for a polyprotic acid with alkali potentiometrically and to determine the normality of the given acid. 7.2 To find out the strength of the given potassium dichromate solution by titrating it potentiometrically with the standard Mohr’s salt solution (0.1 N). 7.3 ˜ȱŽŽ›–’—Žȱ‘Žȱ‘Ž›–˜¢—Š–’Œȱ™Š›Š–ŽŽ›œǰȱNJ ǰȱNJ ǰȱNJȱŠ—ȱ‘Žȱ equilibrium constant KC of the following reactions from emf measurements. (i) Cu2+ + Zn(s) ֖ Zn2+ + Cu(s) (Daniel cell) (ii) Zn(s) + 2AgCl(s) ֖ ZnCl2(s)+ 2Ag(s) (iii) Pb(s) + 2AgCl(s) ֖ PbCl2(s)+ 2Ag(s) (iv) Zn(s) + Pb2+ ֖ Zn2+ + Pb(s) 7.4 To determine the concentrations of iodide and chloride ions in a given mixture. Viva Questions

80 80

8. Conductometric Measurement 8.1 To determine the cell constant of a given conductivity cell at 25°C.

90 90

8.2

8.3

To study the variation of conductance of (i) a strong electrolyte, and (ii) a weak electrolyte with concentration and to verify DebyeHuckel-Onsager equation. To determine the equivalent and molar conductances, degree of dissociation and dissociation constant of a weak acid (acetic acid), and to verify the Ostwald’s dilution law for a given weak electrolyte.





 

81

84

87 

92

94

xivȳ˜—Ž—œȱ

8.4

To determine the strength of the given strong acid (hydrochloric acid) by titrating it conductometrically with the given strong base (sodium hydroxide). 8.5 To determine the strength of the given weak acid (acetic acid) by titrating it conductometrically with the given strong base (sodium hydroxide). 8.6 To estimate the amount of hydrochloric acid (a strong acid) and acetic acid (a weak acid) present in the given mixture by titrating it conductometrically against a strong base (sodium hydroxide). Viva Questions 9. Chemical Kinetics 9.1 To study the kinetics of hydrolysis of methyl acetate, an ester, in presence of an acid (hydrochloric acid) at room temperature. 9.2 To study the hydrolysis of methyl acetate in the presence of an acid (sulphuric acid) at room temperature. 9.3 To compare the strengths or relative avidity of two acids, hydrochloric acid and sulphuric acid used in hydrolysis of methyl acetate. 9.4 To study the kinetics of saponification of an ester (ethyl acetate) with sodium hydroxide. 9.5 To study the kinetics of reaction between potassium iodide and potassium persulphate (peroxodisulphate) by integrated rate law method. 9.6 To study the kinetics of potassium iodide - potassium persulphate reaction using initial rate method (or van’t Hoff differential method or ratio variation method). 9.7 To study the kinetics of the reaction between acidified propanone (acetone) and iodine colorimetrically (or spectrophotometrically). 9.8 To study the kinetics of interaction of crystal violet with NaOH colorimetrically. Viva Questions 10. Colorimetric and Spectrophotometric Measurements 10.1 To verify Lambert Beer’s law for CuSO4, KMnO4 and K2Cr2O7 solutions using colorimeter and measure their concentrations in the given solutions. 10.2 To determine the dissociation constant of the given indicator (phenolphthalein) colorimetrically. 10.3 To determine the amount of iron present in a given solution colorimetrically using the complex formation between Fe2+ and 1,10-phenanthroline.

96

99

101

103 105 105 108 112

112 115

119

121 124 127 128 129

133 136

˜—Ž—œȱȳxv

10.4 To determine the composition of ferric ions-salicylic acid complex colorimetrically using Job’s method. 10.5 To determine the concentration of chromium and manganese in a given mixture colorimetrically. 10.6 To study the absorbance spectra of KMnO4 (in water) and K2Cr2O7 (in 0.1 M H2SO4) in the range 200–500 nm and to determine the Omax values. Also, to calculate the energies of these transitions in different units (J molecule–1, kJ mol–1, cm–1, eV). 10.7 To study the pH-dependence of UV-Visible spectrum of potassium dichromate. 10.8 To record the uv spectra (180–400 nm) of the given compounds— acetone, acetaldehyde, 2-propanol and acetic acid, all in cyclohexane and in water and to calculate the energy involved in the electronic transitions in the units of : wave no., kJ mol–1, kcal mol–1, and eV mol–1. Also, to comment on the effect of structure on the uv spectra of these compounds. 10.9 To analyze the given rotational/vibration-rotation spectrum of HCl (g) and to determine the bond length of HCl. Viva Questions

138 140 142

143 144

147 149

11. Adsorption and Distribution Measurements 11.1 To verify the Freundlich and Langmuir isotherms for adsorption of acetic acid on activated charcoal. 11.2 To study the distribution of benzoic acid between water and benzene and to determine the association factor of benzoic acid in benzene. 11.3 To study the distribution of benzoic acid between (i) water and chloroform, (ii) water and cyclohexane. 11.4 To study the distribution of acetic acid between (i) water and chloroform, (ii) water and cyclohexane. 11.5 To study the equilibrium reaction KI + I2 ֖ KI3 (Potassium tri iodide) and to determine its equilibrium constant. 11.6 To study the formation of a complex between copper and ammonia and to determine the molecular formula of the complex by partition coefficient method. Viva Questions

150 150

12. Environmental Parameter Measurement 12.1 To estimate the amount of suspended particulate matter (SPM) in air samples.

165 165

153

155 156 157 160

163

xviȳ˜—Ž—œȱ

12.2 To determine the amount of dissolved oxygen in the given sample of water. 12.3 To estimate the amount of sulphate ion present in the given sample of water by turbidimetry. 12.4 To determine the salinity of the given sample of water.

166

12.5 To determine the amount of chloride ions in a given sample of water. 12.6 To determine the acidity of a given sample of water and the amount of dissolved carbon dioxide present in it. 12.7 To determine the chemical oxygen demand of a given sample of water. 12.8 To determine the biochemical oxygen demand of the given water sample. 12.9 To estimate the total alkalinity of the given water sample.

173

168 171

175 178 179 182

12.10 To collect a soil sample in small quantity that truly represents the field for which the soil test is required. 12.11 To qualitatively detect the presence of carbonate, ammonium, nitrite, nitrate, phosphate and potassium in a given soil sample. 12.12 To determine the pH value of a given soil sample.

184

12.13 To determine the amount of total soluble salts present in a given sample of the soil. 12.14 To estimate the amount of calcium and magnesium in a given sample of the soil. Viva Questions

190

13. Miscellaneous Experiments 13.1 To determine the concentration of sodium, potassium, lithium and calcium ions in the given solutions by using flame photometric method. 13.2 To index the given powder diffraction pattern of a cubic crystalline system and to identify the lattice type/kind of unit cell and the size of the unit cell. 13.3 To determine the free acidity (as H2SO4) in ammonium sulphate fertilizer. 13.4 To determine the amount of calcium in calcium ammonium nitrate (CAN) fertilizer. 13.5 To determine available chlorine from a given sample of bleaching powder.

196 196

185 188

191 195

198

201 202 204

˜—Ž—œȱȳxvii

13.6 To estimate the acidity and the alkalinity in a given sample of pesticide formulation. 13.7 To estimate the amount of free phosphoric acid (H3PO4) in a given sample of superphosphate fertilizer. 13.8 To determine the concentration of benzoic acid and sorbic acid present in soft drinks. 13.9 To determine the concentration of sulphur dioxide in soft drinks.

206 208 209 210

13.10 To detect adulterants in various foodstuffs.

212

13.11 To synthesize nail polish.

219

13.12 To synthesize nail polish remover.

220

13.13 To prepare talcum powder.

221

13.14 To prepare face cream.

221

13.15 To prepare shampoo.

222

13.16 To prepare phenol formaldehyde resin.

224

13.17 To synthesize gold nanoparticles.

225

13.18 To synthesize silver nanoparticles.

227

13.19 To synthesize titanium dioxide nanoparticles.

228

13.20 To synthesize and characterize zinc oxide nanoparticles.

229

Viva Questions 14. Programming with Basic Language 14.1 Operators

229 231 231

14.2 Basic Statements

232

14.3 Some Built-In Function

234

14.4 Basic Programs

235

14.5 Ms Excel

243

Viva Questions Appendices Bibliography

251 253 262

1

INTRODUCTION

Chemistry, the study of matter, is an experimental science. Most of the theoretical concepts are better understood by doing practical work in chemistry lab. The indispensable role of practical work in chemistry education is universally accepted. Increasing cost of chemicals, handling large number of students, handling toxic chemicals, maintenance of instruments and accessories, etc., are few of the problems faced in carrying out the practical work. The desired modifications to adapt to cost effective , safe and efficient techniques, green and innovative ideas which are student and environment friendly is the need of the hour. The first and the foremost rule in a chemistry lab is to be safe, and so one should wear safety goggles, latex gloves and a lab apron before performing any experiment. One should have a sound knowledge of apparatus to be used under different experimental conditions. In most of the chemistry labs the following basic apparatus are required to carry out the experimental work

1.1. CHEMISTRY LABORATORY APPARATUS AND THEIR USES Beaker A beaker is a common glass shaped container which is used for transferring, mixing, stirring, and heating chemicals. The spouts on the rims of beaker aid in pouring. The markings on the beaker help to measure the volume they contain, although these marks are not intended for obtaining a precise measurement of volume. Beakers come in a wide range of sizes 50 mL, 100 mL, 150 mL, 250 mL, 400 mL, etc. (Fig. 1.1).

Fig. 1.1: Beaker

2ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Erlenmeyer flask / Conical flask / Titration flask An Erlenmeyer flask, also known as conical or titration flask is a type of laboratory flask which has a flat bottom, a conical body and a cylindrical neck. It has been named after its inventor, the Ž›–Š—ȱŒ‘Ž–’œȱ–’•ȱ›•Ž—–Ž¢Ž›ǰȱ ‘˜ȱŒ›ŽŠŽȱ’ȱ in 1861. It has a narrow neck and expands toward its base that allows easy mixing and swirling of the flask without too much risk of spilling during titrations. The marks on the conical flask give rough estimation only (Fig. 1.2).

Fig. 1.2: Titration flask

Volumetric Flask Ȋȱ A volumetric flask is also known as measuring flask or standard flask to make standard solution. It is a round flask with a long neck and flat bottom. It is calibrated to contain a particular and exact volume of solution at 20°C. Volumetric flasks are available in the following capacities: 5, 10, 25, 50, 100, 200, 250, 500, 1000 and 2000 mL. There is a small mark on the neck of the flask that indicates how far to fill the bottle (adjust the bottom of the meniscus to the centre of Fig. 1.3: Volumetric flask the mark). It comes with a special cap that does not let anything in or out. The temperature of the container is relevant because liquids expand when heated; therefore avoid using liquids that will fluctuate with temperature (hot water that will cool, for example) (Fig. 1.3). Ȋȱ To use a volumetric flask, dissolve desired mass of reagent in the flask by swirling with less than final volume of the liquid and then add more liquid and swirl. Add the final drops of liquid with pipette to the correct level and cap it. Invert the flask several times to complete mixing.

Burette Burettes are one of the most accurate apparatus in the lab used for measuring exact volume of liquid. A burette is a cylindrical glass tube that is open at the top and comes to a narrow pointed opening at the bottom. Right above the bottom opening is a stopcock that can be turned to control the amount of liquid being released. There are graduations along the length of the glass tube that indicate the volume of liquid delivered through the stopcock. By adjusting the stopcock, the amount of liquid that is released can be slowed to a drop every few seconds. Burettes are set up by using a burette clamp in combination with a stand as shown in Fig. 1.4. Burettes are usually of 50 mL capacity with a least count of 0.1 mL. Burettes of 5, 10, and 25 mL are also

—›˜žŒ’˜—ȳ3

available. When reading the liquid level in a burette, eye level should be at the same height as top of the liquid. If the eye level is too high, the liquid seems to be higher than it really is and vice versa. The error, called parallax, occurs when the eye level is not at the same height as the liquid. In burette, the liquid around the edges will be higher than the liquid in the centre, sloping down. This is called meniscus (remember to measure from the bottom of the meniscus). The lower meniscus should be used for burette readings of colourless solution while upper meniscus should be used for burette readings of a coloured solution such as KMnO4. To determine how much liquid is delivered, the initial volume of liquid is subtracted from the final volume. The main use of burette is in volumetric analysis and titrations.

Fig. 1.4: Burette

Operating a Burette 1. 2. 3. 4. 5. 6. 7.

Wash burette with distilled water and rinse with the new solution Eliminate air bubbles before use Drain solution slowly Read bottom of the concave meniscus Estimate reading to one-tenth of a division Avoid parallax For colourless liquid, note the reading for lower meniscus, and for coloured solution, note the reading for upper meniscus

Pipette Pipettes are of two types: Transfer pipettes and ŽŠœž›’—ȺȦȺ ›ŠžŠŽȱ ™’™ŽŽœǯ Pipettes are extremely accurate for measurement of a volume of solution (Fig. 1.5). Transfer pipette, as shown in Fig. 1.5 is calibrated to deliver one fixed volume. It has a large bulb with a long narrow portion above with a single graduation mark (like a volumetric flask). Never blow the last drop out of a transfer pipette. Typical volumes of transfer pipettes are 1, 2, 5, 10, 20, 25, 50 Fig. 1.5: Transfer pipette and 100 mL. Transfer pipette is quite accurate. ŽŠœž›’—ȺȦȺ ›ŠžŠŽȱ ™’™ŽŽœȱ œ‘˜ —ȱ ’—ȱ ’ǯȱ ŗǯŜȱ Š›Žȱ ‘˜œŽȱ ’—ȱ  ‘’Œ‘ȱ ‘Žȱ œŽ–œȱ Š›Žȱ graduated and are used to deliver a variable volume. It is usually intended for the delivery of predetermined variable volumes of liquid. It does not find wide use in accurate work for which a burette is generally preferred.

4ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Fig. 1.6: Graduated pipette

Funnel A funnel is usually made of plastic or glass and can have either a short stem or a long stem, depending on the need of its use. A funnel is used to transfer liquid or fine-grained substances into containers with a small opening, without spillage, for preparation of standard solutions. With the addition of a filter paper, funnel is žœŽȱ ˜ȱ œŽ™Š›ŠŽȱ œ˜•’ȺȦȺŒ›¢œŠ•œȱ ›˜–ȱ œ˜•ž’˜—ȱ Ÿ’Šȱ ‘Žȱ ™›˜ŒŽœœȱ ˜ȱ filtration in organic analysis (Fig. 1.7).

Fig. 1.7: Funnel

Watch Glass A watch glass is just a concave piece of glass. This is used to keep small amount of liquid or solid which are hygroscopic in nature in inorganic analysis. It can be used for evaporation purposes and as a lid for a beaker (Fig. 1.8).

Graduated Cylinder / Measuring Cylinder

Fig. 1.8: Watchglass

It is a cylinder with several markings up and down the length of the Œ˜—Š’—Ž›ȱ  ’‘ȱ œ™ŽŒ’’Œȱ ’—Œ›Ž–Ž—œǯȱ ›ŠžŠŽȱ Œ¢•’—Ž›œȱ Œ˜–Žȱ ’—ȱ many sizes: 10 mL, 25 mL, 50 mL, 100 mL, 250 mL, etc. The smaller they are in diameter, the more specific the volume measurements will be. When reading the volume from a graduated cylinder, line the lowest point of the meniscus up with the nearest marking on it. This is used for measuring rough estimate of volume of liquid (Fig. 1.9).

Wash Bottle Wash bottles usually are made up of polyethylene, which is a flexible solvent-resistant petroleum-based plastic. It is most commonly used to store and dispense distilled water whenever needed in the experimental procedure like washing electrodes in pH metric and potentiometric titrations. Wash bottles contain an Fig. 1.9: Measuring cylinder internal dip tube allowing upright use and are sealed with a screw-top lid. When hand pressure is applied to the bottle, the liquid inside becomes pressurized and is forced out of the nozzle into a narrow stream of liquid. It is used to rinse various pieces of lab glassware such as conical flask, beaker, etc. It is used for delivering a fine stream of distilled water or any other liquid for use in the transfer and washing of precipitates (Fig. 1.10).

—›˜žŒ’˜—ȳ5

Fig. 1.10: Wash bottle

Wire Gauze A wire gauze is a sheet of thin metal that has net-like crosses or nichrome wire mesh. The purpose of wire gauze is to be placed on the tripod stand between the Bunsen burner and the beakers to support the beakers or other glassware or flasks during heating. Wire gauze when placed between glassware and heat source diffuses the heat somewhat and is therefore safer than the direct flame. The glassware has to be flat bottomed to stay on the wire gauze. There are two types of wire gauze: woven wire gauze and a wire gauze with a ceramic centre. Both have same ability to transmit the heat efficiently, but the one that has a ceramic centre will allow the heat to disperse more evenly. The ceramic at the centre of the wire gauze is enmeshed at high pressure to prevent it from peeling. There are three sizes of the wire gauze with ceramic centre such as 4 inches, 5 inches, and 6 inches (Fig. 1.11).

Water Bath

Fig. 1.11: Wire gauze

Fig 1.12: Copper water bath

Water baths are used to heat organic solution as they are flammable. The temperature of water can be maintained below 100°C in copper water bath (Fig. 1.12) and electric water bath (Fig. 1.13).

Fig 1.13: Electric water bath

6ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Type of Glass used in Laboratory Apparatus •Šœœ Š›Žȱ žœŽȱ ’—ȱ Œ‘Ž–’œ›¢ȱ •Š‹ȱ ’œȱ žœžŠ••¢ȱ –ŠŽȱ ˜ȱ ‹˜›˜œ’•’ŒŠŽȱ •Šœœǯȱ ˜›˜œ’•’ŒŠŽȱ glass is a type of glass with silica and boron trioxide as the main glass-forming constituents. Borosilicate glasses are known for having very low coefficients of thermal expansion (~3 u 10ƺŜ Kƺŗ at 20°C), making them resistant to thermal shock, more so than any other common glass. Such glass is less subject to thermal stress and is commonly used for the construction of laboratory glassware such as reagent bottles, conical flask, test tubes, volumetric flask, beaker, etc. Borosilicate glass is known for its extremely high chemical resistance and good optical clarity. While more resistant to thermal shock than other types of glass, borosilicate glass can still crack or shatter when subjected to rapid or uneven temperature variations. When broken, borosilicate glass tends to crack into large pieces rather than shattering compared to ordinary glass that can shatter into many razor-sharp pieces with a sudden change in temperature. But the glass can react with sodium hydride upon heating to produce sodium borohydride, a common laboratory reducing agent.

Cleaning of Glass Apparatus ˜˜ȱ •Š‹˜›Š˜›¢ȱ ŽŒ‘—’šžŽȱ Ž–Š—œȱ Œ•ŽŠ—ȱ •Šœœ Š›Žǰȱ ‹ŽŒŠžœŽȱ ‘Žȱ –˜œȱ ŒŠ›Žž••¢ȱ executed piece of work may give an erroneous result if dirty glassware is used. In all instances, glassware must be physically clean; it must be chemically clean; and in many cases, it must be bacteriologically clean or sterile. All glassware must be absolutely grease-free. The safest criteria of cleanliness is uniform wetting of the surface by distilled water. This is especially important in glassware used for measuring the volume of liquids. ›ŽŠœŽȱ Š—ȱ ˜‘Ž›ȱ Œ˜—Š–’—Š’—ȱ –ŠŽ›’Š•œȱ  ’••ȱ ™›ŽŸŽ—ȱ ‘Žȱ •Šœœȱ ›˜–ȱ ‹ŽŒ˜–’—ȱ uniformly wetted. This in turn will alter the volume of residue adhering to the walls of the glass container and thus affect the volume of liquid delivered. Furthermore, in pipettes and burettes, the meniscus will be distorted and the correct adjustments cannot be made. The presence of small amounts of impurities may also alter the meniscus. Ȋȱ Wash labware as quickly as possible after use. If a thorough cleaning is not possible immediately, put glassware to soak in water. If labware is not cleaned immediately, it may become impossible to remove the residue. Ȋȱ For general washing, soap, detergent or cleaning powder may be used. Ȋȱ Chromic acid is a commonly used glassware cleaning reagent. It can be prepared by dissolving 20 g of K2Cr2O7 in little distilled water to make a paste and then slowly adding approximately 300 mL of conc. sulphuric acid to produce a total volume of 1 L chromic acid. Chromic acid solution is strongly acidic and will burn the skin severely. Take care in handling it. Ȋȱ A more efficient cleaning liquid is a mixture of conc. sulphuric acid and fuming nitric acid. This may be used if the vessel is very greasy and dirty.

—›˜žŒ’˜—ȳ7

Ȋȱ ›ŽŠœŽȱ –Š¢ȱ Š•œ˜ȱ ‹Žȱ ›Ž–˜ŸŽȱ ‹¢ȱ ’••’—ȱ Š™™Š›Šžœȱ  ’‘ȱ  Š›–ȱ œ˜Š™ȱ œ˜•ž’˜—ǰȱ leaving for 15 min., rinsing with water, followed by conc. HCl and finally with distilled water. Ȋȱ Special types of precipitates may require removal with nitric acid, aqua regia (1 part conc. HNO3 and 3 parts conc. HCl) or fuming sulphuric acid. These are very corrosive substances and should be used only when required. Ȋȱ Apparatus of borosilicate glass may be dried in an oven at 100–120°C.

1.2 PREPARATION OF STANDARD SOLUTIONS A standard solution is a solution containing a precisely known concentration of an element or a substance. A known weight of solute is dissolved to make a specific volume. It is prepared using a standard substance, such as a primary standard or secondary standard. Primary standard: A primary standard is a chemical or reagent which has certain properties such as it is extremely pure, highly stable, anhydrous, less hygroscopic, has very high molecular mass, can be weighed easily, non-toxic and it must be stable during long-term storage. A primary standard substance should satisfy the following requirements: Ȋȱ A primary standard material should be extremely pure which means that it should be a chemical of high grade of purity, preferably 99.98%. Ȋȱ It should be highly stable, which means it usually does not react easily when kept in its pure form. Or in other words it should have very low reactivity. This is important because if a reagent reacts easily with atmospheric oxygen or water or is affected by carbon dioxide or changes its property over time, then it is unreliable. We can never use such unstable and unreliable chemicals as standard. Ȋȱ It should be anhydrous which means that it does not contain any water molecule in its molecular structure. For example, the chemical name of Epsom salt is magnesium sulphate. So we write the formula as MgSO4. But the chemical Epsom salt which is found in grocery or drug store is a chemical with formula MgSO4.7H2O. Therefore, if you want to prepare a primary standard of magnesium sulphate, should purchase an anhydrous MgSO4 preferably an analytical reagent grade chemical with purity greater than 99.98%. Ȋȱ Just being anhydrous is not sufficient. The chemical preferably should be less hygroscopic, that is on opening the container it should not absorb water molecules from atmosphere. Ȋȱ ȱ œ‘˜ž•ȱ ‘ŠŸŽȱ ‘’‘ȱ –˜•ŽŒž•Š›ȱ –ŠœœȺȦȺŽšž’ŸŠ•Ž—ȱ –Šœœǰȱ œ˜ȱ ‘Šȱ Ž››˜›ȱ ž›’—ȱ weighing will be minimum. Ȋȱ The substance should be readily soluble in the given solvent. Examples of primary standard substances are oxalic acid, Mohr’s salt, potassium dichromate, sodium bicarbonate, etc.

8ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Secondary standard: A secondary standard is used by standard laboratories such as companies involved in preparation of reagents, kits or laboratories responsible for producing quality control material for other labs. They use primary standard as the primary calibrator or primary reference material. Secondary standard in turn is used for the purpose of calibration of control material in smaller lab for analysis of unknown concentration of a substance. So basically, secondary standard serves the purpose of external quality control for smaller labs. This makes it essential that the secondary standard must first be standardized against the primary standard. Examples of secondary standard substance are NaOH, KMnO4, etc. Solutions of known concentration can be prepared in a number of different ways depending on the nature of the analyte and the concentration required: Ȋȱ Weighing out a solid material of known purity, dissolving it in a suitable solvent and diluting to the required volume. Ȋȱ Weighing out a liquid of known purity, dissolving it in a suitable solvent and diluting to the required volume. Ȋȱ Diluting a solution previously prepared in the laboratory. Ȋȱ Diluting a solution from a chemical supplier.

1.3 UNITS FOR REPORTING CONCENTRATION TERMS Concentration is a general measurement unit stating the amount of solute present in a known amount of solution: Amount of solute Concentration = Amount of solution Although the terms “Solute” and “Solution” are often associated with liquid samples, they can be extended to gas-phase and solid-phase samples as well. The actual units for reporting concentration depend on how the amount of solute and solution are measured. Table 1.1 shows the most common practical units used for expressing the concentration of solute. Table 1.1: Common practical units for reporting concentration Name

Units

Symbol

Molarity

No. of moles of solute per litre of solution

M

Formality

No. of formula wt. of solute per litre of solution

F

Normality

No. of gram equivalent of solute per litre of solution

N

Molality

No. of moles of solute per kg of solvent

M

Weight %

g of solute per 100 gm of solution

%w / w

Volume %

mL of solute per 100 mL of solution

%v / v

Weight to Volume %

g of solute per 100 mL of solution

%w / v

—›˜žŒ’˜—ȳ9

1. Molarity and Formality Both molarity and formality express concentration as moles of solute per litre of solution. There is, however, a difference between molarity and formality. Molarity is the concentration of a particular chemical species in solution. Formality, on the other hand, is a substance’s total concentration in solution without regard to its specific chemical form. There is no difference between a substance’s molarity and formality if it dissolves without dissociating into ions. The molar concentration of a solution of glucose, for example, is the same as its formality. For substances that ionize in solution, such as NaCl, molarity and formality are different. For example, dissolving 0.1 mol of NaCl in 1 L of water gives a solution containing 0.1 mol of Na+ and 0.1 mol of Cl–. The molarity of NaCl, therefore, is zero since there is essentially no undissociated NaCl in solution. The solution, instead, is 0.1 M in Na+ and 0.1 M in Cl–. The formality of NaCl, however, is 0.1 F because it represents the total amount of NaCl in solution. But, however, when we state that a solution is 0.1 M NaCl, we understand it to consist of Na+ and Cl– ions. The unit of formality is used only when it provides a clearer description of solution chemistry. Moles of solute Molarity = Volume of solution (L) Moles of solute = Molarity u Volume of solution (L) Weight of solute (g) ⎛ g ⎞ Molecular mass ⎜ ⎝ mol ⎟⎠ Note that it is the final volume of the solution that is important, not the starting volume of the solvent used. The final volume of the solution might be a bit larger than the volume of the solvent because of the additional volume of the solute. In practice, a solution of known molarity is prepared by weighing an appropriate amount of solute and placing it in a volumetric flask. Enough solvent is added to dissolve the solute, and further solvent is added until an accurately calibrated final volume is reached. The solution is then shaken until it is uniformly mixed as shown in Fig. 1.14. Molarity can be used as a conversion factor to relate a solution’s volume to the number of moles of solute. If we know the molarity and volume of a solution, we can calculate the number of moles of solute. If we know the number of moles of solute and the molarity of the solution, we can find the solution’s volume. Moles of solute =

Example 1: Hydrochloric acid is sold commercially as a 12.0 M solution. How many moles of HCl are in 300.0 mL of 12.0 M solution? Solution: The number of moles of solute is calculated by multiplying the molarity of the solution by its volume. Moles of HCl = (Molarity of solution) u (Volume of solution in L) Moles of HCl = 12.0 mol L–1 u 0.300 L = 3.60 mol HCl There are 3.60 moles of HCl in 300.0 mL of 12.0 M solution.

10ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Use spatula to weigh required amount from starting reagent

0.0000

3.9567

Add watch glass to balance and re-zero balance reading (a)

Transfer powder through the funnel to volumetric flask

(b)

Add solvent and complete upto mark (c)

Fig. 1.14: Preparing a stock solution of known molarity

2. Normality Normality is an older unit of concentration. It is the concentration unit used in standard methods for the examination of water and wastewater, and in some standard EPA methods, commonly used sources of analytical methods, for environmental laboratories. Normality makes use of the chemical equivalent, which is the amount of one chemical species reacting stoichiometrically with another chemical species. Note that this definition makes an equivalent, and thus normality, a function of the chemical reaction in which the species participates. Although a solution of H2SO4 has a fixed molarity, its normality depends on how it reacts. The number of equivalents, ‘n’, is based on a reaction unit, which is that part of a chemical species involved in a reaction. In a precipitation reaction, for example, the reaction unit is the charge of the cation or anion involved in the reaction. Thus, for the reaction

—›˜žŒ’˜—ȳ11

→ PbI2(s) Pb2+(aq) + 2I– (aq) ⎯⎯ 2+ n = 2 for Pb because each ion takes two electrons and n = 1 for I-because each ion donates only one electron. In an acid-base reaction, the reaction unit is the number of H+ ions donated by an acid or accepted by a base. For the reaction between sulphuric acid and ammonia → 2NH4+ (aq) + SO42–(aq) H2SO4(aq) + 2NH3(aq) ⎯⎯ We find that n = 2 for H2SO4 because each molecule donate two ions of H+ and n = 1 for NH3 because each ion accept one H+. For a complexation reaction, the reaction unit is the number of electron pairs that can be accepted by the metal or donated by the ligand. In the reaction between Ag+ and NH3 → Ag(NH3)2+(aq) Ag+ (aq) + 2NH3(aq) ⎯⎯ The value of n for Ag+ is 2 because each ion accepts a pair of electrons in covalent bonds with Ammonia NH3. For NH3 n = 1 because each molecule of ammonia donates one electron in each covalent bond it forms with Ag+. Finally, in an oxidation–reduction reaction, the reaction unit is the number of electrons released by the reducing agent or accepted by the oxidizing agent; thus, for the reaction → Sn4+(aq) + 2Fe2+(aq) 2Fe3+(aq) + Sn2+(aq) ⎯⎯ n = 1 for Fe3+ because each ion accepts one electron in the reduction step, and n = 2 for Sn2+ because each ion donates two electrons in the oxidation step. Clearly, determining the number of equivalents for a chemical species requires an understanding of how it reacts. Consequently, the following simple relationship exists between normality and molarity. N=nuM The following example illustrates the relationship among chemical reactivity, equivalent mass, and normality. Example 2: Calculate the equivalent mass and normality for a solution of 6.0 M H3PO4 given the following reactions: → PO43–(aq) + 3H2O(l) (a) H3PO4(aq) + 3OH–(aq) ⎯⎯ → HPO42–(aq) + 2NH4+(aq) (b) H3PO4(aq) + 2NH3(aq) ⎯⎯ → H2PO4–(aq) + HF(aq) (c) H3PO4(aq) + F–(aq) ⎯⎯ Solution: For phosphoric acid, the number of equivalents is the number of H+ ion donated to the base. For the reactions in (a), (b), and (c) the number of equivalents are 3, 2, and 1, respectively. Thus, the calculated equivalent masses and normalities are ȱ ǻŠǼȱ ȱ ƽȱȺȦȺ— ƽȱşŝǯşşŚȺȦȺřȱƽȱřŘǯŜŜŚȱȱ ȱƽȱ— u M = 3 u 6.0 = 18.0 N ȱ ǻ‹Ǽȱ ȱ ƽȱȺȦȺ—ȱƽȱşŝǯşşŚȺȦȺŘȱƽȱŚŞǯşşŝȱȱ ȱƽȱ— u M = 2 u 6.0 = 12.0 N ȱ ǻŒǼȱ ȱ ƽȱȺȦȺ—ȱƽȱşŝǯşşŚȺȦȺŗȱƽȱşŝǯşşŚȱȱ ȱƽȱ— u M = 1 uȱŜǯŖȱƽȱȲŜǯŖȱ MM = Molar mass of H3PO4

12ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

3. Molality Molality is used in thermodynamic calculations where a temperature independent unit of concentration is needed. Molarity, formality and normality are based on the volume of solution in which the solute is dissolved. Since density is a temperature dependent property, a solution’s volume, and thus it’s molar, formal and normal concentrations will change as a function of its temperature. By using the solvent’s mass in place of its volume, the resulting concentration becomes independent of temperature. ˜•Š•’¢ȱ’œȱŽę—ŽȱŠœȱ‘Žȱ—˜ǯȱ˜ȱ–˜•Žœȱ˜ȱœ˜•žŽȱ’—ȱŗȱ”ȱ˜ȱœ˜•ŸŽ—ȱŠœȱ™Ž›ȱŽšžŠ’˜—DZ ˜•Š•’¢ȱƽȱ˜ǯȱ˜ȱ–˜•Žœȱ˜ȱœ˜•žŽȺȦȺ–Šœœȱǻ”Ǽȱ˜ȱœ˜•ŸŽ— Example 3: What is the molality of solution made by dissolving 25 g of NaCl in 2.0 litre ˜ȱ ŠŽ›ǵȱœœž–Žȱ‘ŽȱŽ—œ’¢ȱ˜ȱ ŠŽ›ǰȱȱƽȱŗǯŖȱȺȦȺ–ȱǻƽȱ”ȺȦȺǼǯ Solution: Molar mass of NaCl = (1 uȱŘŘǯşşȱȺȦȺ–˜•ȱǼȱƸȱǻŗȱuȱřśǯŚśȱȺȦȺ–˜•ȱǼȱƽȱśŞǯŚŚȱ 25 g NaCl uȱŗȺȦȺśŞǯŚŚȱƽȱŖǯŚŘŞȱ–˜•ȱŠ• 2.0 litre water uȱŗȺȦȺŗȱƽȱŘǯŖȱ” Molality = No. of moles of solute per kg of solvent. ŖǯŚŘŞȱ–˜•ȺȦȺŘȱ”ȱƽȱŖǯŘŗŚȱ–˜•ȱ”–1. The solution has concentration of NaCl equal to 0.214 mol kg–1.

4. Weight, Volume, and Weight-to-Volume Ratios Ž’‘ȱ ™Ž›ŒŽ—ȱ ǻƖȱ  ȺȦȺ Ǽǰȱ Ÿ˜•ž–Žȱ ™Ž›ŒŽ—ȱ ǻƖȱ ŸȺȦȺŸǼȱ Š—ȱ  Ž’‘Ȭ˜ȬŸ˜•ž–Žȱ ™Ž›ŒŽ—ȱ ǻƖȱ ȺȦȺŸǼȱŽ¡™›ŽœœȱŒ˜—ŒŽ—›Š’˜—ȱŠœȱž—’œȱ˜ȱœ˜•žŽȱ™Ž›ȱŗŖŖȱž—’œȱ˜ȱœ˜•ž’˜—ǯȱȱœ˜•ž’˜—ȱ ’—ȱ ‘’Œ‘ȱŠȱœ˜•žŽȱ‘ŠœȱŠȱŒ˜—ŒŽ—›Š’˜—ȱ˜ȱŘřƖȱ ȺȦȺŸȱŒ˜—Š’—œȱŘřȱȱ˜ȱœ˜•žŽȱ™Ž›ȱŗŖŖȱ–ȱ˜ȱ solution. Parts per million (ppm) and parts per billion (ppb) are mass ratios of grams of solute to one million or one billion grams of sample, respectively. For example, steel that is 450 ppm in Mn contains 450 mg of Mn for every kilogram of steel. If we approximate ‘ŽȱŽ—œ’¢ȱ˜ȱŠ—ȱŠšžŽ˜žœȱœ˜•ž’˜—ȱŠœȱŗǯŖŖȱȺȦȺ–ǰȱ‘Ž—ȱœ˜•ž’˜—ȱŒ˜—ŒŽ—›Š’˜—œȱŒŠ—ȱ‹Žȱ expressed in parts per million or parts per billion (Refer Appendix J). For gases a part per million usually is a volume ratio. Thus, a helium concentration of 6.3 ppm means that one litre of air contains 6.3 mL of He.

Dilution of Concentrated Solutions Dilution is one of the main preparation processes which is used in daily routine analysis in laboratories. For convenience, chemicals are sometimes bought and stored as concentrated solutions that must be diluted before use. Aqueous hydrochloric acid, for example, is sold commercially as a 12.0 M solution, yet it is most commonly used in the laboratory after dilution with water to a final concentration of 6.0 M or 1.0 M. → Diluted Solution Concentrated Solution + Solvent ⎯⎯ The key fact to remember when diluting a concentrated solution is that the number of moles of solute is constant; only the volume is changed by adding more solvent.

—›˜žŒ’˜—ȳ13

Because the number of moles of solute can be calculated by multiplying molarity times volume, we can set up the following equation: Moles of solute (constant) = Molarity u Volume = Mi u Vi = Mf u Vf where, Mi is the initial molarity, Vi is the initial volume, Mf is the final molarity, and Vf is the final volume after dilution. Rearranging this equation into a more useful form shows that the molar concentration after dilution (Mf ) can be found by multiplying the initial concentration (Mi) by the ratio of initial and final volumes (ViȺ/ȺVf ) Mf = Mi u (ViȺ/ȺVf ) where (VfȺ/ȺVi) is called dilution factor. Mf = MiȺ/ȺDF where dilution factor (DF) = VfȺ/ȺVi = MiȺ/ȺMf Suppose, for example, that we dilute 50.0 mL of a solution of 2.00 M to a volume of 200.0 mL. The solution volume increases by a factor of four (from 50 mL to 200 mL); so the concentration of the solution must decrease by a factor of four (from 2.00 M to 0.500 M): Mf = 2.0 M uȱśŖȱ–ȺȦȺŘŖŖȱ–ȱƽȱŖǯśŖŖȱ DF = VfȺ/ȺViȱƽȱŘŖŖȱ–ȺȦȺśŖȱ–ȱƽȱŚ Dilution factor of value 4 means the solution concentration has been diluted by factor of 4. Example 4: What volume of 10 M acetic acid is required to prepare 1.0 L of 0.50 M acetic acid? Solution: 10 M u Vreagent = 0.50 M u 1.0 L Vreagent = 0.050 L = 50 mL A volume of 50 mL of 10 M acetic acid is required to prepare 1.0 L of 0.50 M acetic acid. ˜›ȱ ‘Žȱ ŒŠ•Œž•Š’˜—ȱ ˜ȱ –˜•Š›’¢ȱ ˜ȱ Œ˜—ŒŽ—›ŠŽȱ ŠŒ’œȺȦȺ‹ŠœŽœǰȱ –ŠœœƖȱ Š—ȱ Ž—œ’¢ȱ along with molar mass of the liquid compound is required. The given formula can be used to calculate the molarity: M = Mass% u density uȱŗŖŖŖȺȦȺ Table 1.2: Density, mass%, molarity of stock solutions of commonly used acids and bases S. No.

Acid / Base

Density (g/mL)

Mass%

Molarity

1.

Acetic acid (CH3COOH)

1.05

100

17 M

2.

Hydrochloric acid (HCl)

1.18

37

12 M

3.

Nitric acid (HNO3)

1.42

70

16 M

4.

Phosphoric acid (H3PO4)

1.70

85

15 M

5.

Sulphuric acid (H2SO4)

1.84

96

18 M

6.

Ammonium hydroxide (NH4OH)

0.90

29

15 M

Note:ȱ ••ȱ ›ŽŠŽ—œȱ Š—ȱ Œ‘Ž–’ŒŠ•œȱ žœŽȱ œ‘˜ž•ȱ ‹Žȱ ˜ȱ —Š•¢’ŒŠ•ȱ ›ŠŽȱ Š—ȱ ’—ȱ Š••ȱ experiments using instruments like pH meter, conductometer, potentiometer, colorimeter, etc., double distilled or deionized water must be used to maintain accuracy.

14ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

1.4 ERRORS IN MEASUREMENTS The function of a student or researcher is to use his sound knowledge of chemistry to get result close to the true value. Any deviation from the true value is known as error. The errors in an experiment can be classified as Systematic (determinate) errors and Random (Indeterminate) errors. Systematic errors are caused due to operational and personal error due to insufficiency of human sensory organs to make observation accurately; instrumental and reagent errors due to faulty use of balances, uncalibrated glassware and instruments, errors in methods and additive and proportional error. Random errors in an experiment are caused by unknown and unpredictable changes in the experiment. The student does not have a control over this like the environmental conditions like the room temperature and pressure, and changes taking place within the instrument. Random errors can be eliminated by recalibration of instrument, carrying out a blank experiment, and carrying out large number of observations. The accuracy of an experimental measurement is how close the observed value is to the true or correct value. The accuracy of a measurement is reduced by systematic errors. The precision of a measurement is how close a number of observations of the same quantity agree within each other or repeatability of the experiment. The precision is limited by the random errors. This is illustrated in Fig. 1.15. XX XX

X

XX XX

X

X

X

Accurate and precise

Not accurate, but precise

Not accurate, not precise

Fig. 1.15: Accuracy and precision

Most of the experimental errors can be eliminated by minimizing the systematic and random errors. Installing an electronic laboratory notebook like the Accelrys notebook, which can take the data directly from the digital equipment, will reduce the transmission mistake by the student between the reading of the measurement and the recording manually. Spreadsheet softwares can be used to eliminate the calculation mistakes and help in calculating the various parameters like slope and intercept using regression analysis. The improper chemical regent use can be reduced by maintaining chemical inventory management. This provides information about the associated safety data sheet, reagent age or shelf life and its proper use.

1.5 SAFETY RULES IN CHEMISTRY LAB The main objective of this section is to provide information about the safety rules to be followed during handling of hazardous and toxic chemicals in the laboratory, dos and

—›˜žŒ’˜—ȳ15

don’ts in the chemistry lab and the first aid procedures. The proper procedure to be employed while disposing the chemicals, complying with the safety regulations, is also very crucial. The safety concerns while dealing with chemicals in the chemistry lab are: Ȋȱ Developing awareness about the possible hazards about certain toxic chemicals. Ȋȱ Paying attention on the labels written on the bottles of the chemicals. Ȋȱ ›˜™Ž›ȱž—Ž›œŠ—’—ȱ˜ȱ‘Žȱ›’œ”ȱ™‘›ŠœŽœȺȦȺŠ—Ž›ȱœ’—œȱ–Ž—’˜—Žȱ˜—ȱ‘Žȱ‹˜•Žœǯ Ȋȱ Using common sense and alertness to avoid accidents in the laboratory. Ȋȱ Avoiding unwanted exposure to hazardous chemicals by eliminating or modifying procedures. Ȋȱ Being aware of emergency procedures to deal with accidents efficiently. Ȋȱ Being well versed with the dos and don’ts of the chemistry lab. Categories of hazards in the chemistry lab: The hazards in the chemistry lab can be generally categorized as Ȋȱ Toxic Ȋȱ Explosive Ȋȱ Flammable Ȋȱ Corrosive Ȋȱ Environmental hazards Table 1.3: Common signs and risk phrases on the chemical bottles Hazard Pictograms

Information / Understanding Irritant and Harmful: This covers a wide range of (sometimes relatively minor) hazards Precautions: Avoid contact with the skin, do not breathe, etc.

Toxic: This pictogram represents “Lethal” nature of the chemical that causes severe damage to health. Precautions: Avoid contact with human body.

Environmental Hazard: The chemicals pose some environmental hazard if not got rid of correctly. Precautions: Proper care to be taken on disposal.

(Contd.)

16ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• (Contd.) Explosive: Explosion can be triggered by light, heat, friction or mechanical shock. Precautions: Keep them away from all ignition sources such as open flames, hot surfaces, spark sources and direct sunlight.

Flammable or extremely flammable: The chemicals can burn easily at normal temperature; have flashpoint below 37.8°C (100°F). Precautions: Store in flame resistant cupboards away from ignition sources, in properly sealed containers.

Corrosive: Common acids (HCl, H2SO4, HF, etc.) and bases (NaOH, KOH, NH4OH, etc.) are corrosive. Damage body tissue, corrode metals. Precautions: Avoid contact with human body, do not inhale. 8

Oxidizing / fire promoting: Oxidizing chemicals spontaneously evolve oxygen at room temperature or with slight heating that promote combustion. Precautions: To be kept away from flammable chemicals.

Spontaneously combustible: Ignite (without an external ignition source), within few minutes after coming in contact with air. Precautions: Prevent contact with air, if exposed to air cover with sand or other non combustible material.

Health hazard: Cause serious and prolonged health effects on short or longterm exposure. Fatal if swallowed or inhaled. Precautions: Avoid contact with skin or inhalation.

Dos and don’ts in chemistry lab Ȋȱ Do not shout, play, push or run while working in chemistry lab. Ȋȱ Treat every chemical as hazardous. Ȋȱ Do not eat, drink and smoke while working in chemistry lab. Ȋȱ Don’t sniff or taste chemicals as it can be dangerous. Ȋȱ Keep chemicals away from the source of heat and light. Ȋȱ Don’t throw broken glassware and filter paper in sink. Ȋȱ Any chemical waste must be disposed of properly.

—›˜žŒ’˜—ȳ17

Ȋȱ Ȋȱ Ȋȱ Ȋȱ Ȋȱ Ȋȱ Ȋȱ Ȋȱ Ȋȱ

All trash should be disposed of in waste cans. Always pour acid into water, do not pour water into acid. Read the labels of the chemicals carefully before using. ˜—Ȃȱ žœŽȱ œŠ–Žȱ œ™Šž•ŠȺȦȺ™’™ŽŽȱ ˜ȱ ›Ž–˜ŸŽȱ Œ‘Ž–’ŒŠ•œȱ ›˜–ȱ  ˜ȱ ’Ž›Ž—ȱ containers. Wear gloves while handling corrosive chemicals. Spills of chemicals (acids, bases) must be properly cleaned.

—˜ ȱ‘Žȱ•˜ŒŠ’˜—ȱŠ—ȱ‘˜ ȱ˜ȱžœŽȱœŠŽ¢ȱŽšž’™–Ž—ȱǻ’›ŽȱŽ¡’—ž’œ‘Ž›ǰȱŽ¢ŽȺȦȺŠŒŽȱ wash station, chemical spill kits, etc.) Wear splash proof goggles while heating materials, to avoid eye injury. Don’t use inflammable materials like hair spray, deodorants during working with open flames.

1.6 FIRST-AID PROCEDURES Ȋȱ Chemicals in the eyes: Flush your eyes with water and wash immediately. Do not ŠĴŽ–™ȱ˜ȱ˜ȱ˜ȱ‘ŽȱŒ•’—’Œȱ‹Ž˜›Žȱ̞œ‘’—ȱ¢˜ž›ȱŽ¢Žœǯȱȱ’œȱ’–™˜›Š—ȱ‘Šȱ̞œ‘’—ȱ with water be continued for a prolonged time—l0 or 15 minutes. Ȋȱ Chemicals in the mouth: Any chemical taken into the mouth should be spat out and the mouth rinsed thoroughly with water. If someone swallows a chemical, note the name of the chemical and notify the clinic immediately. Ȋȱ Chemical spills on the skin: For a small area, flush the skin with water first. For a small acid or base spill on the skin, neutralize an acid with baking soda; neutralize a base with boric acid. Ȋȱ Fire-clothing or hair: A person whose clothing or hair catches on fire will often run around hysterically in an unsuccessful effort to get away from the fire. This only provides the fire with more oxygen and makes it burn faster. It is the responsibility of the closest person to bring the fire blanket to the victim as quickly as possible. Smother the fire by wrapping the victim in the blanket. For fire in the room itself, use the fire extinguisher, turn the burners off at the source and turn the gas main valve to shut off gas flow; unplug appliances, evacuate the room. Ȋȱ Bleeding from a cut: For minor cuts, apply pressure to the wound with sterile gauze, wash with soap and water, and apply a sterile bandage. If the victim is bleeding badly, raise the bleeding part, if possible, and apply pressure to the wound with a piece of sterile gauze. Ȋȱ Breathing smoke or chemical fumes: If smoke or chemical fumes are present in the laboratory, all persons—even those who do not feel ill—should leave the laboratory immediately. Since smoke rises, stay low while evacuating a smokefilled room. Thoroughly ventilate the room before going back to work. Ȋȱ Fainting: If a person faints, lie the person down on the back. Position the head lower than the legs and provide fresh air. Loosen restrictive clothing.

18ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

VIVA QUESTIONS 1. Name the following apparatus

ȱ

2. 3. 4. 5. Ŝǯȱ 7.

________ g of oxalic acid is required to prepare 200 mL of 1 N solution. Name some primary standard solutions. Which reagent is used for cleaning of glassware? What is the least count of a burette? ‘¢ȱŠ›Žȱœ˜–Žȱœ˜•ž’˜—œȱ”Ž™ȱ’—ȱŠ–‹Ž›ȺȦȺŠ›”ȱ‹˜•Žœǵ This Hazard sign stands for:

8. What is the use of beaker? 9. What is the difference between titration, volumetric and round bottom flask? 10. When do you use measuring cylinder, burette and pipette? Which is most accurate? 11. Name the type of glass used to make the glassware used in chemistry lab. 12. What is the difference between primary and secondary standard? 13. What is the difference between molarity, normality and molality? 14. When do we use ppm and ppb units? 15. What are the various errors incurred in chemistry experiments? 16. What is the difference between precision and accuracy?

2

SURFACE TENSION MEASUREMENT

EXPERIMENT: 2.1 Aim: To determine the surface tension of a given liquid at room temperature using stalagmometer by “drop weight” method. Requirements: Stalagmometer, beaker, weighing bottle, given liquid, distilled water. Theory: Surface tension of a liquid is defined as the force acting per unit length along the surface of the liquid at right angles to any imaginary line taken on the surface of the liquid in any direction. It is denoted by symbol “JȄȱŠ—ȱ’œȱ–ŽŠœž›Žȱ’—ȱ ȱž—’ȱ˜ȱ dyne per cm, i.e., dyn cm–1 or in SI unit of Nm–1. The apparatus used for the determination of surface tension is called stalagmometer (Fig. 2.1).

A

Capillary tube Capillary B Surface tension acts along the circumference

Weight of drop

Fig. 2.1: Stalagmometer

20ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

There are two methods employed for the measurement of surface tension, namely: 1. Drop weight method 2. Drop number method In this experiment, drop weight method is used. It is based on the fact that the size of a drop of a liquid falling off from the end of the capillary tube depends on the surface tension of the liquid and the size of the capillary end which provides the outer line of the attachment for the drop. Thus, when a liquid is allowed to flow through a capillary tube, a drop will increase in size up to a certain point and then fall off. The total surface tension force supporting the drop is 2 SrJ, where J is the surface tension of liquid as defined above, and r is the radius of the outer circumference of the drop at the end of the capillary tube. It is along this line that the liquid, glass and air meet and the force of surface tension operates. The drop will fall when its weight force, ‘w’, just exceeds the surface tension force acting along the circumference. Hence, w = mg = 2SrJ ...(1) where w = weight of the drop m = mass of the drop g = acceleration due to gravity 2Sr = outer circumference of contact of the drop with the capillary tip. Using equation (1), the surface tension of a liquid may be determined if the mass of a single drop, and the outer circumference of the drop at the place of hanging from the tip of the capillary are known. When the liquid completely wets the bottom surface of the capillary, drop circumference can be taken to be same as that of the tip of the capillary. If the same stalagmometer is used for two different liquids, then ‘r’ will be same in both the cases. For the two liquids 1 and 2, we have w1 = m1g = 2SrJ1 ...(2) w2 = m2 g = 2SrJ2 ...(3) where m1 and m2 are the masses of single drop of liquids 1 and 2; and J1 and J2 are the surface tensions of the liquids. Dividing equation (2) by (3), we get w1Ⱥ/Ⱥw2 = m1Ⱥ/Ⱥm2 = J1Ⱥ/ȺJ2 ...(4) If one of the two liquids is water, then at a particular temperature, the surface tension of the other liquid will be Jl = (mlȺ/Ⱥmw) u Jw ...(5) where Jl = surface tension of liquid at a particular temperature Jw = surface tension of water at that temperature ml = mass of the liquid mw = mass of water Water is taken as reference liquid since it is an universal solvent. It is colourless and has a density of 1 g mL–1. The temperature coefficient with respect to various properties such as surface tension, viscosity, etc., is very low. Surface tension decreases with increase in temperature since the kinetic energy of the molecules increases, and so the intermolecular force of attraction decreases.

ž›ŠŒŽȱŽ—œ’˜—ȱŽŠœž›Ž–Ž—ȳ21

Procedure: 1. Clean and wash the stalagmometer with distilled water and dry it. 2. Set the stalagmometer vertically with the help of a clamp and boss. 3. Take distilled water in a beaker and adjust the height of the stalagmometer so that its lower end is immersed in distilled water. 4. Suck up distilled water through a rubber tube attached to the upper end of the stalagmometer until the water level reaches above the mark ‘A’. Adjust the rate of flow of water with the help of the pinch cock so that 10 to 15 drops of water come out per minute. 5. Now suck up water once again without disturbing the adjusted pinch cock until the water level reaches above the mark ‘A’. Allow the drops of water to fall off, and when the water level reaches ‘A’, collect 20 drops of water in a clean dry pre-weighed weighing bottle. 6. Weigh the collected water along with the weighing bottle. 7. Repeat the above procedure to get three concordant readings. 8. Empty and dry the stalagmometer or rinse it thoroughly with the given liquid. 9. Repeat steps 2 to 7 with the given liquid. 10. Calculate the Jl of the given liquid using equation (5), and knowing Jw at room temperature (Appendix D). Observation: Room temperature = ........°C Surface tension of water at ........ °C = ........ dyn cm–1 ȱ ȱ ȱ Šœœȱ˜ȱŽ–™¢ȱ Ž’‘’—ȱ‹˜Ĵ•Žȱƽȱ 1 g Table 2.1: Data recording S. No.

Liquid

Number of drops

1.

Water

20 20 20

2.

Given liquid

20 20 20

Mass of weighing bottle + 20 drops, w2 (g)

Mass of 20 drops (w2 w1) (g)

Average mass of 20 drops (g)

Mass of 1 drop (g)

Calculation: Mass of 1 drop of water, mw = ........ g ȹŠœœȱ˜ȱŗȱ›˜™ȱ˜ȱ•’šž’ǰȱ–l = ........ g Surface tension of the given liquid is = Jl = (mlȺȦȺ–w) u Jw dyn cm–1 Result: The surface tension of the given liquid at .......°C = ....... dyn cm–1 Precautions: 1. Stalagmometer should be washed thoroughly with chromic acid, distilled water and ether. This is to remove any trace of grease or dust that may remain in the capillary, particularly on its tip.

22ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

2. The apparatus should be set vertically. 3. There should be no air bubbles in the apparatus when the liquids are filled. 4. Drops should be allowed to fall under their own weight and not due to kinetic force of flow, i.e., they should not be pushed away. 5. Rate of flow of the liquids must not exceed 15 drops per minute. 6. The apparatus should not be disturbed while counting the number of drops, as this may cause the drops to fall before they develop to their full size.

EXPERIMENT: 2.2 Aim: To determine the surface tension of a given liquid at room temperature using stalagmometer by “Drop Number” method. Requirements: Stalagmometer, beaker, weighing bottle, graduated pipette, given liquid, distilled water. Theory: Surface tension of a liquid is defined as the force acting per unit length along the surface of the liquid at right angles to any imaginary line taken on the surface of the liquid in any direction. It is denoted by symbol “JȄȱŠ—ȱ’œȱ–ŽŠœž›Žȱ’—ȱ ȱž—’ȱ˜ȱ dyn cm–1 or in SI unit of Nm–1. There are two methods employed for the measurement of surface tension, namely: 1. Drop weight method 2. Drop number method In this experiment drop number method is used. It is easier to count the number of drops formed by equal volumes of two liquids than to find the masses of single drops. For two different liquids, masses of equal volume are directly proportional to their densities. If n1 and n2 are the number of drops produced from the same volume, ‘v’, of two liquids, then Volume of a single drop of liquid 1 = vȺ/Ⱥn1 ...(1) ? mass of a single drop, m1 = (vȺ/Ⱥn1) u d1 ...(2) where d1 is the density of liquid 1. Since 2SrJ1 = mg, where J = surface tension of liquid, for liquid 1, 2SrJ1 = m1g = (vȺ/Ⱥn1) u d1 u g ...(3) Similarly, for liquid 2, we have 2SrJ2 = m2g = (vȺ/Ⱥn2) u d2 u g ...(4) where J2 is the surface tension and d2 is the density of liquid 2. Dividing equation (3) by (4), we get J1Ⱥ/ȺJ2 = (n2 d1)Ⱥ/Ⱥ(n1 d2) ...(5) One of the liquids is usually water as its surface tension data for different temperatures are easily available. Therefore,  JlȺ/ȺJw = (nw dl)Ⱥ/Ⱥ(nl dw) or Jl = [nw dlȺ/Ⱥnl dw] u Jw ...(6) Procedure: 1. Clean, dry and set the stalagmometer vertically with the clamp and boss. 2. Adjust the level so that the lower end of the apparatus is immersed in distilled water taken in a beaker.

ž›ŠŒŽȱŽ—œ’˜—ȱŽŠœž›Ž–Ž—ȳ23

3. Suck up the water through the rubber tube till the water level reaches somewhere above the upper mark A. 4. Adjust the rate of fall of water drops in such a way that 10–15 drops fall off per minute by adjusting the pinch cock. 5. Now, without disturbing the pinch cock and apparatus suck up water again till the level reaches above mark ‘A’ (Fig. 2.1). 6. Allow water to flow down till it reaches mark ‘A’. Immediately thereafter start counting the number of drops coming out of the capillary tip, until the water level touches mark ‘B’, i.e., count the number of drops of water as it flows from mark ‘A’ to mark ‘B’. 7. Repeat this procedure to get at least three concordant readings. 8. Dry the stalagmometer or rinse it with the given liquid. 9. Repeat the steps 2 to 7 with the given liquid. 10. Weigh accurately a clean and dry weighing bottle. 11. Clean and wash a pipette with distilled water, and pipette out 5 mL of it into the weighing bottle (step 10), and weigh this accurately. 12. Repeat the steps 10 and 11 with the given liquid. 13. Calculate the densities of distilled water and the liquid with the readings obtained in steps 10, 11 and 12. 14. Determine the surface tension of the given liquid using the equation (6) and knowing the surface tension of the used distilled water at room temperature (Appendix D). Observation and Calculation: Room Temperature = ........°C Surface tension of the used distilled water at ........°C ; (Jw) = ........ dyn cm–1. Table 2.2: Data recording S. No.

No. of drops of water from ‘A’ to ‘B’ (nw)

No. of drops of liquid from ‘A’ to ‘B’ (n1)

1. 2. 3.

Mass of empty weighing bottle, w1 = ........ g Mass of weighing bottle + 5 mL water, w2 = ........ g Mass of weighing bottle + 5 mL liquid, w3 = ........ g. Density of water, dw = (w2 – w1)Ⱥ/Ⱥ5 = ........ g mL–1 Density of liquid, dl = (w3 – w1)Ⱥ/Ⱥ5 = ........ g mL–1 Surface tension of given liquid = Jl = [nw dlȺ/Ⱥnl dw] u Jw = ........ dyn cm–1 Result: The surface tension of the given liquid at ........°C = ........ dyn cm–1. Precautions: Write as in previous experiment.

24ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 2.3 Aim: To study the variation of surface tension with different concentrations of a detergent and to determine its Critical Micelle Concentration (CMC). Requirements: Stalagmometer, standard flasks, beakers, weighing bottle, graduated pipettes, detergent solutions (0.01 M, 0.001 M, 0.002 M, 0.003 M, 0.005 M, 0.006 M, 0.007 M and 0.008 M ), distilled water. Theory: Detergents are good examples of surface active materials called surfactants. These are substances that get adsorbed on to a liquid or solid surface and reduce the surface tension of the medium. The aqueous solutions of surfactants are colloidal in nature and therefore with an increase in concentration, the molecules tend to aggregate. These aggregates are called micelles or association colloids. The concentration at which micelles begin to form is the Critical Micelle Concentration (CMC). In other words, the concentration above which these micelles become detectable is known as Critical Micelle Concentration (CMC). It is defined as the concentration of surfactants above which micelles form and all additional surfactants added to the system go to micelles. With an increase in concentration of the surfactant, the surface tension of water decreases gradually and becomes constant beyond the CMC. The CMC can be useful in determining the efficiency and economy in selecting the proper detergent for a particular purpose. Detergents have applications in the fields of domestic cleaning, personal hygiene and cosmetic, pharmaceutical and food products, as well as in industrial and commercial cleaning, and product formulations. The surface tension of water decreases with the addition of detergent as a solute. It implies that the surface tension of a solution decreases with increase in concentration. This is due to the fact that the number of molecules of detergent increases with an increase in concentration of the solution and these molecules form clusters in between the water molecules on the surface. Consequently, the inward pull decreases. Moreover, the attraction between water and detergent molecules is stronger than that between the molecules of water or the molecules of detergent (molecules of the same kind). This tends to reduce the inward pull of the water molecules at the surface by those water molecules in the bulk. Figure 2.2 shows the relationship between the surface tension and the concentration of the surfactant. Soaps and detergents work in the following manner: Three types of energy are required for good cleaning: 1. Chemical energy provided by the detergent or soap. 2. Thermal energy provided by warm or hot water. 3. Mechanical energy provided by hands or a machine. These types of energy interact and should be in proper balance. Let us see how they work together. Let us assume that we have oily, greasy dirt on clothing. Water alone will not remove this soil. One important reason is that oil and grease present in dirt repel the water molecules.

–1

Surface tension (dyn cm )

ž›ŠŒŽȱŽ—œ’˜—ȱŽŠœž›Ž–Ž—ȳ25

60 50 40 30 CMC 20 10 2

4

6

8

10

12

14

Concentration of the surfactant 3 (M × 10 )

Fig. 2.2: Plot of surface tension against concentration of the surfactant

Now we add soap or detergent. The surfactant’s water-hating (hydrophobic) end is repelled by water but attracted to the oil in the dirt. At the same time, the water-loving (hydrophilic) end is attracted to the water molecules. These opposing forces loosen the dirt and suspend it in the water. Warm or hot water helps dissolve grease and oil in dirt. Washing machine agitation or hand rubbing helps pull the dirt free. We often use the words, soap and detergent, interchangeably, but they are quite different. A detergent is a mixture of synthetic chemicals and additives, while soap is a kind of detergent which is made up of natural ingredients. In this experiment, we use sodium lauryl sulphate (SLS). Its molecular formula is (CH3–(CH2)11–SO4– Na+ ). Its molar mass is 288.37 gmol–1. Procedure: 1. Weigh 0.7209 g of SLS, dissolve it in distilled water and make up to 250 mL in a measuring flask. This is a stock solution of concentration 0.01 M. 2. From this solution, prepare a series of solutions of concentrations 0.001 M, 0.002 M, 0.003 M, 0.005 M, 0.006 M, 0.007 M and 0.008 M. 3. Determine the surface tension of each of these solutions using stalagmometer, as described in Experiment 2.2. 4. Find out the surface tension of 0.01 M also. Observation: Table 2.3: Data recording Concentration of SLS (M)

0.001

0.002

0.003

0.005

0.006

0.007

0.008

0.01

–1

Surface tension (Nm )

Plot the surface tension versus concentration of SLS as shown in Fig. 2.2. Draw two straight lines so as to intersect each other. The concentration corresponding to the point of intersection gives the Critical Micelle Concentration of the detergent, SLS.

26ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Result: The Critical Micelle Concentration of the detergent, sodium lauryl sulphate = ........ M. Interpretation: The surface tension of water decreases with the addition of detergent as a solute. It implies that the surface tension of a solution decreases with an increase in concentration. This is due to the fact that the number of molecules of detergent increases with an increase in concentration of the solution and these molecules form clusters in between the water molecules on the surface. Consequently, the inward pull decreases. Moreover, the attraction between water and detergent molecules is stronger than that between the molecules of water or the molecules of detergent (molecules of the same kind). This tends to reduce the inward pull of the water molecules at the surface by those water molecules in the bulk.

Surface tension, N/m

80 70 60 50 40 30 CMC 20 0

0.0025

0.005

0.0075

0.01

0.0125

0.015

SLS concentration, M

Fig. 2.3: Plot of surface tension versus SLS

VIVA QUESTIONS

ȱ

1. 2. 3. Śǯȱ 5. 6. 7. 8. 9. 10. 11.

Write Ramsay and Shield equation. Explain all the terms involved. Write Eotvos law. Name the apparatus used to find the surface tension of a given liquid. Ž’—Žȱœž›ŠŒŽȱŽ—œ’˜—ǯȱ‘ŠȱŠ›Žȱ’œȱȱŠ—ȱ ȱž—’œǵ Write the equation for drop weight and drop number method for determination of surface tension. What is the effect of concentration on the surface tension of a given solution? What is the effect of temperature on the surface tension of a solution? What is detergency? What is the effect of addition of a detergent on the surface tension of a given liquid? Why is water used as a reference liquid to find relative surface tension? Can any other liquid be used in place of water as a reference liquid? Explain. Why a flow rate of falling drops is maintained at 13–15 drops per minute?

3

VISCOSITY MEASUREMENT

EXPERIMENT: 3.1 Aim: To determine the absolute and relative viscosities of the given liquid taking water as reference and using Ostwald viscometer. Requirements: Ostwald viscometer, rubber tube, graduated pipette, beaker, stop watch, weighing bottle, given liquid, distilled water. Theory: Viscosity is a property which measures the resistance to flow offered by a fluid layer when another adjacent and parallel fluid layer tends to flow past the first layer. It can also be defined as the force of internal friction between two layers of a liquid moving past one another with different velocities. The coefficient of viscosity or absolute viscosity of a liquid is defined as the tangential force per unit area required to maintain a unit difference of velocity between two parallel layers of the liquid held apart at unit distance. This is denoted by the symbol K. Relative viscosity is defined as the ratio of the absolute viscosities of the liquid (Kl) and water (Kw) at the same temperature, i.e., Kr = KlȺ/ȺKw and hence it is unitless. If the distance between the layers is dx, the required difference of velocities between these layers is du, and the area of each layer is A, then at a given temperature, the force, ‘f’, required for maintaining the selected velocity difference between the two layers is related to the other factors by the following proportionality: f D A. duȺ/Ⱥdx (at constant T) ...(1) ȱƽȱ΋ȱǯȱžȺ/Ⱥdx (at constant T) ...(2) ? K = fȺ/ȺǯȱǻžȺȦȺ¡Ǽȱ ǻŠȱŒ˜—œŠ—ȱǼȱ ǯǯǯǻřǼ K o coefficient of viscosity of the fluid at given temperature. ’–Ž—œ’˜—œȱ’—ȱ ȱœ¢œŽ–DZȱ¢—Ⱥ/Ⱥcm2 uȱǻŒ–ȺȦȺœǼȺ/Ⱥcm o dyn.s.cm–2 Dimensions in SI system: newtonȺ/Ⱥmetre2 uȱǻ–Ž›ŽȺȦȺœǼȺ/Ⱥmetre o N. s. m–2  ȱž—’ȱ˜ȱK is the Poise. Centipoise is the smaller unit of K, i.e., 10–2 Poise ȳȲȲȹȹŗȱ˜’œŽȱƽȱŗȱ¢—ǯœǯŒ––2 = 0.1 N.s. m–2 = 1 g.cm–1.s–1 = 0.1 kg.m–1.s–1

28ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

The coefficient of viscosity (K) can be measured by using the following equations: Poiseulle’s equation for laminar flow in a horizontal capillary tube is: v = (SȺȦȺŞǼǯȱ™›4Ⱥ/Ⱥl. K ...(4) v o volume of liquid flowing out of the capillary in time ‘t’. t o time in which the volume ‘v’, flows out. p oȱŒ˜—œŠ—ȱŸŽ›’ŒŠ•ȱ™›Žœœž›Žȱ‘ŽŠȱ˜›ȱ–Š’—Š’—’—ȱ‘Žȱž—’˜›–ȱ•˜ ȱ›ŠŽȱŸȺȦȺǯ r o radius of the capillary tube l o length of capillary tube. K o coefficient of viscosity of the liquid. Considering water as a reference,   Krelative = KliquidȺ/ȺKreference (water) ... (5) 4 4 = (S r h Ul g .tlȺ/ȺŞ•ŸǼȺȦȺǻS r h Uw g .twȺ/Ⱥ8lv) = Ul .tlȺ ȦȺUw .tw ... (6) where Ul and Uw are the densities of liquid and water respectively. From equation (5),   Kliquid = Krelative u Kreference (water) The value of Kreference (water) at the temperature of the experiment can be taken from literature. Different types of viscometers: 1. Ostwald viscometer. 2. Suspended level type viscometer. Ostwald viscometer: It consists of a fine capillary tube XY (about 10 cm long and 0.4 mm diameter), a bulb A at its upper end and a U tube and bulb B at the lower end. D

E

Upper mark (F) A

X

Lower mark (G)

B

Y

Fig. 3.1: Ostwald viscometer

’œŒ˜œ’¢ȱŽŠœž›Ž–Ž—ȳ29

Procedure: 1. First clean and dry the viscometer and set it vertically in the burette stand. 2. Introduce a definite measured volume of water (15 mL) through E into the larger bulb B. 3. Then suck the water into the bulb A through a rubber tube attached at D to a level somewhat above the upper mark. 4. Allow the water to flow freely through the capillary and note the time ‘t’ using Šȱœ˜™ ŠŒ‘ȱ˜›ȱ‘Žȱ•˜ ȱ˜ȱ ŠŽ›ȱ›˜–ȱ‘Žȱž™™Ž›ȱ–Š›”ȱǻǼȱ˜ȱ‘Žȱ•˜ Ž›ȱ–Š›”ȱǻ Ǽǯȱ 5. Then remove water and rinse the viscometer with the given liquid. 6. Repeat the experiment by taking exactly the same volume of the liquid in bulb B as was the volume of water taken. 7. Determine the densities of water and the liquid using 10 mL of each. 8. Note down the room temperature. Observation: Room temperature = ........ °C Coefficient of viscosity of water at ........ °C, Kwater = ........ centipoise (Refer Appendix D) Table 3.1: Data recording S. No.

Liquid

Time of flow (sec)

Average time of flow (sec)

1.

Water

(i) (ii) (iii)

tw = ........

2.

Given liquid

(i) (ii) (iii)

t1 = ........

Calculation: Mass of empty weighing bottle = w1 g Mass of weighing bottle + 10 mL of water = w2 g Mass of weighing bottle + 10 mL of liquid = w3 g w − w1 g mL–1 Density of water, Uw = 2 10 w 3 − w1 Density of liquid, Ul = g mL–1 10 The relative viscosity of given liquid, Krelative = Ul .tlȺ/ȺUw .tw The absolute viscosity of given liquid, Kliquid = Krelative u Kwater Result: The relative viscosity of given liquid is found to be ......... The absolute viscosity of given liquid is found to be ......... Precautions: 1. The viscometer should not be held very tightly. 2. The time taken should be noted very accurately. 3. The experiment should be carried out at a constant temperature. 4. The viscometer should be clamped perfectly vertical.

30ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 3.2 Aim: To study the variation of viscosity with different concentrations of sugar solution. Requirements: Ostwald viscometer, rubber tube, standard flasks, graduated pipettes, beakers, stop watch, weighing bottle, sugar solutions (1%, 0.1%, 0.2%, 0.3%, 0.4%, 0.6% and 0.8%), distilled water. Theory: As in Experiment 3.1 Procedure: 1. Weigh 2.5 g of solid sugar and dissolve it in a small quantity of water taken in a 250 mL measuring flask and make the volume up to the mark. This is the stock solution of 1% concentration. 2. Prepare solutions of six other concentrations such as 0.1%, 0.2%, 0.3%, 0.4%, 0.6% and 0.8% using the stock solution. 3. Follow the procedure given in Experiment 3.1. 4. Measure the time of flow of distilled water by taking a fixed volume, say 15 or 20 mL. 5. Take 3 or 4 readings and get the average. 6. Similarly, find the time of flow of each of the sugar solutions by taking the same volume as that of water. 7. Determine the densities of water and the sugar solutions. Determine the absolute viscosities of each of the sugar solutions using the following equations: Krelative = KliquidȺ/ȺKreference (water) = Ul .tl / Uw .tw Kliquid = Krelative u Kreference (water) Observation and calculation: Room temperature = ........ °C Time of flow of water = ........ sec. Table 3.2: Data recording S. No.

Concentration (%)

Time of flow (sec)

Kliquid (centipoise)

Plot the graph between coefficient of viscosity of sugar solutions versus concentration (%) and analyze the result. Result: The plot of the graph shows that as concentration increases Kliquid ........

’œŒ˜œ’¢ȱŽŠœž›Ž–Ž—ȳ31

Interpretation: With an increase in the concentration of sugar solution, the coefficient of viscosity decreases. It is due to the fact that the number of molecules of the solute (sugar, in this case) increases with an increase in the concentration of the solution and hence the intermolecular force of attraction between the molecules of solute and solvent (sugar and water) increases. This, in turn, increases the viscosity of the solution thereby decreasing the rate of flow.

EXPERIMENT: 3.3 Aim: To study the variation of coefficient of viscosity with different concentrations of Polyvinyl alcohol (PVA) in water and to determine the molar mass of PVA. Requirements: Ostwald viscometer, graduated pipettes, beakers, standard flasks, glass rod or stirrer, polyvinyl alcohol, distilled water. Theory: Based on Poiseulle’s equation, the ratio of the viscosities of two liquids, called relative viscosity can be expressed as (refer Experiment 3.1). Kr = K1ȺȦȺK2 = (S r4 h U1 g .t1ȺȦȺŞ•ŸǼȺȦȺǻS r4 h U2 g .t2ȺȦȺŞ•ŸǼȱ ȱ ǯǯǯǻŗǼ ...(2) = U1 .t1ȺȦȺU2 .t2 where U1 and U2 are the densities and t1 and t2 are times of flow of liquids 1 and 2 respectively. It was suggested by H. Staudinger that the viscosity of a dilute solution of a polymer in a suitable solvent varies with its molar mass in a regular manner. The higher the molar mass in a homologous series of linear polymers, the larger is the increase in viscosity for a given polymer concentration. It is due to the fact that the large molecules present in a liquid medium disturb the homogeneity thereby increasing the resistance to flow of the molecules of the liquid. Thus, the viscosity increases. If Ks and K0 are the viscosities of the the solution and pure solvent respectively, the relative viscosity is given by Kr = KsȺȦȺK0= Us .tsȺȦȺU0 .t0 ...(3) For solutions of low concentrations, the densities of solutions are almost equal to the density of the pure solvent and therefore equation (3) may be modified as Kr = KsȺȦȺK0= tsȺȦȺ0 ...(4) The specific viscosity, Ksp = (Ks – K0ǼȺȦȺK0 = KsȺȦȺK0 – 1 ...(5) = Kr – 1 The term KspȺȦȺȱ ’œȱ ŒŠ••Žȱ ›ŽžŒŽȱ Ÿ’œŒ˜œ’¢ȱ ǻ›Ž•Š’ŸŽȱ ’—Œ›ŽŠœŽȱ ’—ȱ Ÿ’œŒ˜œ’¢ȱ ™Ž›ȱ ž—’ȱ concentration of polymer) and is a function of molar mass of the polymer, i.e., Kred = KspȺ/ȺC ...(6) where C is the concentration of the polymer. The values of Kred change regularly and linearly with concentration but those of Kr and Ksp change rapidly. Thus, a plot of KspȺ/ȺC (i.e., Kred) against C is almost a straight line which on extrapolation to infinite dilution gives the value of intrinsic viscosity, Kint . Thus, Lt ⎯⎯→ 0 (KspȺ/ȺC) Kint = C ...(7)

32ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

As Ksp is dimensionless, Kint has the unit of (concentration)–1. The relationship between the intrinsic viscosity and molar mass of the polymer is given by MarkHouwink equation as Kint = KMD ...(8) where K and D are constants for a polymer in a suitable solvent at a specified temperature. This equation is valid for polymers having molar masses greater than 10,000. The value of D depends on the shape of the molecule. It was theoretically predicted that D should be zero for large rigid spheres, 0.5 - 0.8 for randomly coiled (flexible) linear chains and 2 for rigid rods. The values of K and D for some systems are given in Table 3.4. In order to obtain molar mass from viscosity data, K and D must be known. These can be determined only if the molar masses of samples of the polymer are known by other methods such as osmotic pressure and light scattering measurements. Thus, the viscosity method is a secondary method but it is a very convenient method if the values of K and D are known. The molar mass obtained by viscosity method is called viscosity average molar mass. Due to some uncertainty in K and D values, this is slightly different from number average molar mass and weight average molar mass. Procedure: 1. Weigh 2.5 g of dry polyvinyl alcohol and dissolve it in 200 mL of hot distilled water taken in a beaker. This should be done by gradually adding the polymer in small lots at a time and slowly stirring the solution without the bubbles of foam being formed. 2. After the polymer is dissolved completely, cover the solution and allow it to cool. 3. Then transfer it slowly to a 250 mL measuring flask through the sides. Rinse the beaker with small quantities of distilled water at a time and transfer the rinsings in to the measuring flask and make the volume up to the mark. This is the stock solution of 1% concentration. 4. Prepare solutions of six other concentrations such as 0.1%, 0.2%, 0.3%, 0.4%, 0.6% and 0.8% using the stock solution. 5. Follow the procedure given in Experiment 3.1. Measure the time of flow of distilled water by taking a fixed volume, say 15 or 20 mL. Take 3 or 4 readings and get the average. 6. Similarly, find the time of flow of each of the polymer solutions by taking the same volume as that of water. 7. Determine the reduced viscosities of each of the polymer solutions using equations (4), (5) and (6). Observation and calculation: Room temperature: ......... Ԩ Solvent: ........; Solute: ........ K = ........ ; D = ........

’œŒ˜œ’¢ȱŽŠœž›Ž–Ž—ȳ33

Table 3.3: Data recording S. No.

Concentration (%)

Time of flow (sec)

Kr = tsȺ/Ⱥt0

Ksp

Kred

Plot a graph between Kred and C and find out Kint from the intercept on the y-axis. Kint = KMD Calculate M, the molar mass of the polymer. Result: The molar mass of the polyvinyl alcohol = ........ Note: Similar experiments can be performed for (i) polystyrene in benzene or toluene (ii) polymethyl methacrylate in acetone. Table 3.4: Values of the constants K and D for various polymer-solvent systems Solvent

Polymer

K u 105

D

S. No.

Temperature

1.

25

Water

Polyvinyl alcohol

20.0

2.

25

Acetone

Polyvinyl acetate

18.8

0.69

3.

25

Acetone

Polymethyl methacrylate

7.5

0.70

4.

25

Toluene

Polymethyl methacrylate

7.5

0.71

5.

25

Benzene

Polystyrene

10.2

0.74

0.76

6.

30

Benzene

Polyisobutylene

61.0

0.56

7.

30

Benzene

Natural rubber

18.5

0.74

VIVA QUESTIONS

ȱ ȱ

1. 2. řǯȱ Śǯȱ 5. 6. 7. 8. 9. 10.

Write Poiseuille’s equation. Explain all the terms involved. Name the apparatus used to find viscosity of a given liquid. Ž’—ŽȱŸ’œŒ˜œ’¢ǯȱ‘ŠȱŠ›Žȱ’œȱȱŠ—ȱ ȱž—’œǯ ’ŸŽȱŠ›”Ȭ ˜ž ’—”ȱŽšžŠ’˜—ǯ What is the effect of concentration on viscosity in cases of glycerol and sodium chloride solution? What is the effect of temperature on viscosity of a solution? What is the ideal time of flow of water? Why is water used as a reference liquid to determine relative viscosity? Can any other liquid be used in place of water? What are the other types of viscometers used to find viscosity of different liquids?

4

pH MEASUREMENT

EXPERIMENT: 4.1 Aim: To measure the pH of given samples of milk, juice, soap, detergent and shampoo. Requirements: pH meter, glass and calomel electrodes, beakers, standard flasks, standard buffer solutions of pH 4 and 9, samples of milk, juice, soap detergent and shampoo, distilled water. Theory: The pH of a solution describes the degree of acidity and alkalinity of the given solution. The pH measurements are important in agronomy, environmental science, nutrition, food science and medicine. Normal healthy skin of humans is slightly acidic in the pH range of 5.4–5.9. Use of detergent and shampoo with high pH can cause dehydration of skin and irritability. Therefore, the pH determination is very important with relevance to the prevention and treatment of skin diseases. Based on the fruit and brand of juice, the pH of juice varies from 2.2 to 4.19. Acidic juices can erode enamel of the teeth thereby making more and more cavities and can also effect the intestinal lining of stomach. The term pH stands for ‘potential of hydrogen’ and is defined as the negative logarithm of the activity of Hydrogen ions. pH = – log [aH+] ˜›ȱȱȱ ™ ȱƽȱ •˜ȱǻŗȺȦȺŠH+) Since in dilute solutions, activity of hydrogen ion is very closely equal to the molar concentration of hydrogen ions to a first approximation, pH of a solution may also be defined by the equation ȱ ȱ ™ ȱƽȱ •˜ȱǻŗȺȦȺŒH+) The pH value = 7 indicates that the solution is neutral, pH value < 7 means that it is acidic and pH value > 7 implies that it is alkaline in nature. A solution stronger than 1 M H+ ions can have a negative pH value. A pH value above 14 is quite rare to meet as solutions of alkalies much stronger than 14 are difficult to obtain due to solubility limitations.

™ ȱŽŠœž›Ž–Ž—ȳ35

Procedure: 1. Turn on the pH meter at least 20 minutes before starting the experiment. This is to stabilize the instrument (Fig. 4.1). Burette

Combined glass electrode Ring stand pH meter Beaker Stirrer

Fig. 4.1: pH meter

2. Calibrate the pH meter as follows: Rinse the electrodes with distilled water and dry them and place them in a beaker containing the standard buffer of pH 9. 3. Wait for 3 minutes and adjust the calibration knob in such a way that the pH meter reads 9. 4. Wash the electrodes with distilled water, dry them and place them in the standard buffer solution of 4. The reading should show a value of 4. If not, slightly adjust the calibration knob to get the pH value of 4. The instrument is now calibrated. 5. Take each of the solutions of juice, shampoo, detergent, soap and milk one by one in a clean and dry 100 mL beaker and measure their pH values after about 3 minutes. 6. Take care to wash the electrode and beaker with distilled water and dry them before taking the reading for each solution. Table 4.1: Data recording Item No.

Product

1.

Juice

2.

Shampoo

3.

Detergent

4.

Soap

5.

Milk

pH value

Result: The nature of sample 1, 2, 3, 4, 5 is ---,---,---,---,---, respectively. Precaution: The solutions of soap, detergent and shampoo must be very dilute, otherwise due to high alkalinity or acidity the electrodes might get damaged.

36ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 4.2 Aim: To prepare the following buffer solutions and to measure the pH of each of these solutions using pH meters. Solution 1: Sodium acetate-acetic acid o acidic buffer. Solution 2: Ammonium chloride-ammonium hydroxide o basic buffer. Requirements: pH meter, glass and saturated calomel electrodes or combined electrode, beakers, standard flasks, graduated pipettes, burette, conical flasks, droppers, buffer solutions of pH 4 and 9, standard solutions of acetic acid (0.1 M), sodium acetate (0.1 M), sodium hydroxide (1 M), oxalic acid (1 N), phenolphthalein indicator, distilled water. Theory: Same as Experiment 4.1. Buffers or Buffer Solutions: It is sometimes necessary to have solutions of definite pH to be used and stored for long. The use of such a solution is an important part of many industrial processes such as electroplating, manufacture of leather, photographic material, dying and in bacteriological research. Such solutions have their difficulty for exposure in atmosphere. They become acidic as they absorb carbon dioxide from the air. If the solution is stored in glass bottle, it becomes alkaline. A buffer solution is defined as the solution whose pH remains practically constant even on addition of small amounts of acid or base. These are of two types: (i) Solution of single substances: the solution of the salt of a weak acid and a weak base, e.g., ammonium acetate; (ii) Solution of mixtures: This can be classified as acidic buffer and basic buffer. Acidic buffer solution: It can be made by mixing a solution of a weak acid with a solution of one of its salts with a strong base (acetic acid and sodium acetate). Alkaline buffer solution: It can be made by mixing solutions of a weak base with a solution of one of its salts with a strong acid (ammonium hydroxide and ammonium chloride). The pH of a buffer is given by Henderson-Hasselbalch equation, or simply Henderson equation, i.e., pH = pKa + log (CsaltȺ/ȺCacid) = pKa + log (Cbase formȺ/ȺCacid form) where pKa = log 1Ⱥ/Ⱥ[Ka] and Ka is the dissociation constant of the acid. The equation above suggests that pH of a mixed solution of an acid and its salt depends on the ratio of their concentrations and not on the absolute values. This is however not strictly true. For making a buffer solution of a specified pH value, we choose an acid with a pH value near the desired pH value. A further modified pH is obtained by adjusting the ratio of the Csalt to Cacid. Suppose we want to make a buffer solution of pH 5.5. Acid chosen for this can be acetic acid with a pKa value of 4.76. pH = pKa + log (CsaltȺ/ȺCacid) ...(1) pH = 5.5 = 4.76 + log (CsaltȺ/ȺCacid)

™ ȱŽŠœž›Ž–Ž—ȳ37

log (CsaltȺ/ȺCacid) = 5.5 – 4.76 = 0.74 CsaltȺ/ȺCacid = 5.495. If equimolar concentrations are used, then the ratio of the concentrations will be equal to the ratio of volumes. Hence, equation (1) becomes pH = pKa + log (VsaltȺ/ȺVacid) ...(2) This means that the salt concentration should be 5.495 times the concentration of the acid. We may mix equal volumes of 1 M acetic acid and 5.495 M sodium acetate solution. The pH values of the buffers made from weak base are given by: pH = 14 – pKb – log (CsaltȺ/ȺCbase) ...(3) This is based on Henderson-Hasselbalch equation. If equimolar concentrations are used, then the ratio of the concentrations will be equal to the ratio of volumes. Hence, equation (3) becomes pH = 14 – pKb – log (VsaltȺ/ȺVbase) ...(4) In this experiment, an acid buffer is prepared, the pH values measured and compared with the expected values. Procedure: 1. Standardize the sodium hydroxide solution by titrating it against standard oxalic acid using phenolphthalein as indicator. 2. Then standardize the acetic acid solution by titrating it with the standardized sodium hydroxide solution using phenolphthalein as indicator. 3. Turn on the pH meter at least 20 minutes before starting the experiment. This is to stabilize the instrument. 4. Calibrate the pH meter as given in Experiment 4.1. For acidic buffer: (i) Prepare 100 mL of 0.2 M acetic acid solution and 100 mL of 0.2 M sodium acetate solution in distilled water. These are stock solutions. (ii) Take six clean and dry standard flasks (25 mL) and label them with numbers 1 to 6. (iii) Prepare various solutions of the buffer by mixing specified volumes of acetic acid and sodium acetate solutions as given in Table 4.2. Table 4.2: Preparation of acetic acid-sodium acetate buffer solutions Total volume = 25 mL S. No.

Volume (mL) of acetic acid solution

Volume (mL) of sodium acetate solution

1.

22.5

2.5

2.

20.0

5.0

3.

15.0

10.0

4.

10.0

15.0

5.

5.0

20.0

6.

2.5

22.5

Observed pH

Expected pH

38ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

(iv) Measure the pH of each of these solutions using digital pH meter. (v) Calculate the pH values using equation (4) and these are the expected ones. Note: Take buffers in small beakers (50 mL) for pH measurement so that the electrodes would be immersed properly. Result: On comparing the observed and expected pH values using equation (4), it is observed that the two ........ with each other. Note: Observed values can be compared with the literature values also. For basic buffer: (i) Prepare 0.2 M solution of ammonium chloride and 0.2 M solution of ammonium hydroxide. (ii) Make various solutions of buffer by mixing specified amounts of ammonium chloride and ammonium hydroxide solution as shown in Table 4.3. (iii) Follow the same procedure as in the case of acidic buffer. (iv) Calculate the pH values using equation (6) and these are the expected ones. Table 4.3: Preparation of ammonium chloride-ammonium hydroxide buffer solutions Total volume = 25 mL S. No.

Volume (mL) of ammonium chloride solution

Volume (mL) of ammonium hydroxide solution

1.

22.5

2.5

2.

20.0

5.0

3.

15.0

10.0

4.

10.0

15.0

5.

5.0

20.0

6.

2.5

22.5

Observed pH

Expected pH

Result:ȱ˜–™Š›Žȱ‘Žȱ˜‹œŽ›ŸŽȱŸŠ•žŽœȱ ’‘ȱ‘ŽȱŒŠ•Œž•ŠŽȱ˜—ŽœȺȦȺ•’Ž›Šž›ŽȱŸŠ•žŽœǯ

EXPERIMENT: 4.3 Aim: To determine the strength of the given strong acid (hydrochloric acid) by titrating it pH-metrically and potentiometrically with a strong base (sodium hydroxide). Requirements:ȱ ™ ȺȦȺ–ȱ –ŽŽ›ǰȱ •Šœœȱ Š—ȱ ȱ Ž•ŽŒ›˜Žœȱ ˜›ȱ Šȱ Œ˜–‹’—Žȱ Ž•ŽŒ›˜Žǰȱ beakers, standard flasks, burette, graduated pipettes, standard oxalic acid solution (0.1 N), sodium hydroxide solution (0.1 N), hydrochloric acid solution (approximately 0.1 N), phenolphthalein indicator, distilled water. Theory: The basic principle of all pH and potentiometric titrations is that one solution is taken in a titration flask (or beaker) and the other solution (titrant) is added from the burette when the concentration of some ion in the titration flask keeps on changing. If an electrode reversible with respect to this ion is set up in the titration flask, its potential will keep on changing. Such electrodes are known as indicator electrodes. For

™ ȱŽŠœž›Ž–Ž—ȳ39

measuring the change in the potential of the indicator electrode, it is to be combined with a standard (or reference electrode). The standard electrode should be such that its potential remains unchanged during titration. In most of the titrations, saturated calomel electrode (SCE) is used as a reference. This electrode is represented as Cl–°Hg2Cl2°Hg and the half-cell reaction is Hg2Cl2 + 2e– o 2Hg + 2Cl–. The appropriate Nernst equation is ESCE = E°SCE – (2.303 RTȺ/Ⱥ2F)log[Cl–]2 = E°SCE – (2.303 RTȺ/ȺF)log[Cl–]. In the case of acid-base titrations, the change in the potential of the indicator electrode is due to the change in the pH of the solution. For acid-base titrations, quinhydrone, hydrogen and glass electrodes (reversible to hydrogen ions) are commonly used. In this experiment, a glass electrode is used as an indicator electrode. According to Nernst equation, E = E ° + 0.0591 log [H+] where E ° is a constant depending on the nature of the glass and the pH of the solution taken inside the bulb. It includes the potential of Ag–AgCl electrode. It is clear from the equation that E decreases exponentially as the concentration of hydrogen ions decreases. When the glass electrode is combined with SCE to form the cell for titration, the cell emf will also change depending on the change in E . If the pH of the analyte is plotted against the volume of NaOH solution added from the burette, a titration curve as shown in Fig. 4.2 would be obtained. 12 11 10 9

pH

8 7

Equivalence point

6 5 4 3 2 Volume of NaOH solution (mL)

Fig. 4.2: Titration curve for strong acid versus strong base

40ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

In the initial stage, when no base was added, the pH is low which is due to the presence of H3O+ from the dissociation of hydrochloric acid. ZZX H3O+ + Cl– HCl + H2O YZZ As the titration proceeds, the OH– ions produced by the dissociation of NaOH start reacting slowly with H3O+. So, the pH gradually increases till the equivalence point is reached. Still the solution is acidic due to the presence of a large excess of H3O+ ions. Once the neutralization is complete, the pH rises steeply as H3O+ ions get completely neutralized. The corresponding reaction is HCl + NaOH o NaCl + H2O The solution has only salt and water and therefore it is neutral and the pH is 7. After the equivalence point, addition of NaOH continues and the pH increases indicating that the solution is basic due to the presence of excess OH– ions from NaOH. Since the emf of any electrode reversible to hydrogen ions is proportional to the pH, the cell emf will exhibit the same trend during titration and therefore the titration curve will be identical to the pH titration curve. The concentration of the titrant need not be higher than that of the solution to be titrated in potentiometric titrations unlike in conductometric titrations. The first and second derivative plots give more precise equivalence point than the ˜—Žȱ’ŸŽ—ȱ‹¢ȱ™ ȺȦȺŽ–ȱŸŽ›œžœȱŸ˜•ž–Žȱ™•˜œǯȱ¢™’ŒŠ•ȱ™•˜œȱŠ›Žȱœ‘˜ —ȱ’—ȱ’œǯȱŚǯřȱŠ—ȱŚǯŚǯ

70

Equivalence point

4000

60

3000

D pH/DV

40

2000 1000

2

DpH/DV

2

50

30 20

Equivalence point

0 –1000

10

–2000 Volume of NaOH solution (mL)

Fig. 4.3: First derivative plot for acid-base titration

Volume of NaOH solution (mL)

Fig. 4.4: Second derivative plot for acid-base titration

Advantages of pHȺ/Ⱥpotentiometric titrations: 1. No indicator is required. 2. Can be employed for coloured solutions. 3. Useful for mixture of acids and bases, including weak acids and weak bases. 4. Useful for titrations involving polybasic acids where it is difficult to use the indicators. 5. No prior knowledge of pH range is required for acid-base titration.

™ ȱŽŠœž›Ž–Ž—ȳ41

Procedure: 1. Standardize the sodium hydroxide solution by titrating it against standard oxalic acid using phenolphthalein as indicator. 2. Switch on the instrument (pHȺ/ȺmV meter) and let it stabilize for 30 minutes. Calibrate it as given in Experiment 4.1 . 3. Take 20 mL of the standard HCl solution in a 100 mL beaker. 4. Dip the indicator (glass) and reference (SCE) electrodes in the solution (the electrodes must be immersed completely) and connect the leads of the electrodes to the instrument. Instead, a combined electrode can also be used. If the electrodes are not immersed completely, add 10 or 20 mL of distilled water. 5. Add NaOH solution from the burette in 1 mL lots. 6. After each addition, stir the solution well and measure the pH of the solution and emf of the cell. 7. From the rough titration, find out the range of volume of NaOH solution in which the equivalence point lies, i.e., corresponding to a sharp change (maximum change) in pH and emf. 8. Repeat the titration by adding the titrant in 0.1 mL lots in this range and also before and after the range to a volume of about 1 mL. 9. Plot (i) pH and emf versus volume of NaOH solution respectively. The equivalence point is the point of inflexion of the curve, i.e., where the curve changes its direction (Refer Fig. 4.1). ȱ ǻ’’Ǽȱ NJ™ Ⱥ/ȺNJȱŠ—ȱNJȺ/ȺNJȱŸŽ›œžœȱŠŸȺǰȱ‘ŽȱŠŸŽ›ŠŽȱŸ˜•ž–Žȱ˜›ȱŽŠŒ‘ȱNJ™ ȱŠ—ȱNJȱ value. This is the first derivative plot (Fig. 4.3). The equivalence point is given by the peak of the curve. 10. Determine the volume of NaOH solution corresponding to the equivalence point and knowing its normality, find out the normality and hence the strength of the given HCl solution. 11. Plot a graph between the second derivative of the pH and, cell emf versus volume of alkali used, respectively (Fig. 4.4). This is the second derivative plot. The first and second derivative plots give more precise equivalence point than the one given by pHȺ/Ⱥemf versus volume plots. Observation: Table 4.4: Data recording S. No.

Volume of NaOH added (mL)

pH of solution

E.M.F. of the cell (mV)

1. 2. 3. 4. 5. (Contd.)

42ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• (Contd.) 6. 7. 8. 9. 10.

Table 4.5: Data recording S. No.

'pH

'E (mV)

'V (mL)

'pH / ΔV (mL–1)

'E / 'V (mV mL–1)

Vav (mL)

Calculation: Volume of NaOH solution required for neutralization = V1 Normality of sodium hydroxide = N1 Volume of hydrochloric acid solution taken = 40 mL V × N1 Normality of hydrochloric acid solution = 1 40 Strength of hydrochloric acid solution = Normality u Equivalent mass of Hydrochloric acid

= ........ g L–1 Result: The strength of the given hydrochloric acid solution = ........ g L–1

EXPERIMENT: 4.4 Aim: To determine the strength of a weak acid (acetic acid) by titrating it pH metrically and potentiometrically with a strong base (sodium hydroxide). Requirements: pHȺ/ȺmV meter, glass and SCE electrodes or a combined electrode, beakers, standard flasks, burette, graduated pipettes, standard oxalic acid solution (0.1 N), sodium hydroxide solution (0.1 N), acetic acid solution (approximately 0.1 N), phenolphthalein indicator, distilled water. Theory: Refer Experiment 4.3. If the pH of the analyte is plotted against the volume of NaOH solution added from the burette, a titration curve as shown in Fig. 4.5 would be obtained. In the initial stage, when no base was added, the pH is low which is due to the presence of H3O+ from the dissociation of acetic acid. ZZX CH3COOȹȮ + H3O+ CH3COOH + H2O YZZ

™ ȱŽŠœž›Ž–Ž—ȳ43

12 11

pH

10 9

Equivalence point

8 7 6

Volume of NaOH solution (mL)

Fig. 4.5: Titration curve for weak acid versus strong base

Since acetic acid is a weak acid, the initial pH is higher than the pH in the case of a strong acid, HCl. As the titration proceeds, the OH– ions produced by the dissociation of NaOH slowly start reacting with H3O+. So, the pH gradually increases till the equivalence point is reached. Still the solution is acidic due to the presence of an excess of H3O+ ions. Once the neutralization is complete, the pH rises steeply as H3O+ ions get completely neutralized. The corresponding reaction is: CH3COOH + NaOH oCH3COONa + H2O The important difference between the titrations of strong acid-strong base and weak acid-strong base is that in the former, the pH is 7 (neutral) at the equivalence point, whereas in the latter, the pH is around 9 (basic) at the equivalence point. This is due to the fact that sodium acetate is a salt which is basic in nature. According to the reaction ZZX CH3COOȹȮ + Na+ CH3COONa YZZ CH3COOȹȮ + H2O o CH3COOH + OH– acetate ion, the strong conjugate base of the weak acetic acid reacts with water to give OH– ion and this makes the solution basic in nature. After the equivalence point, NaOH is in excess and this portion will be identical to that of the titration curve of strong acidstrong base titration. Figure 4.6 compares both the curves. Since the emf of any electrode reversible to hydrogen ions is proportional to the pH, the cell emf will exhibit the same trend during titration and therefore the titration curve will be identical to the pH-titration curve. The concentration of the titrant need not be higher than that of the solution to be titrated in potentiometric titrations unlike in conductometric titrations.

44ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

12 11 10 9

pH

8

Weak acidstrong base

7 6 5

Strong acid-strong base

4 3 2 Volume of NaOH solution (mL)

Fig. 4.6: Titration curves for strong acid versus strong base and weak acid versus strong base

Procedure: 1. Standardize the sodium hydroxide solution using standard oxalic acid solution and phenolphthalein indicator. ȱ Řǯȱȱ ’Œ‘ȱ ˜—ȱ ‘Žȱ ’—œ›ž–Ž—ȱ ǻ™ ȺȦȺmV meter) and let it stabilize for 30 minutes. Calibrate it as given in Experiment 4.1. 3. Take 20 mL of the standard CH3COOH solution in a 100 mL beaker. 4. Dip the indicator (glass) and reference (SCE) electrodes in the solution (the electrodes must be immersed completely) and connect the leads of the electrodes to the instrument. Instead, a combined electrode can also be used. If the electrodes are not immersed completely, add 10 or 20 mL of distilled water. 5. Add NaOH solution from the burette in 1 mL lots. 6. After each addition, stir the solution well and determine the cell emf and pH of the solution. 7. From the rough titration, find out the range of volume of NaOH solution in which the equivalence point lies, i.e., corresponding to a sharp change (maximum change) in emf. 8. Repeat the titration by adding the titrant in 0.1 mL lots in this range and also before and after the range to a volume of about 1 mL. 9. Plot (i) pH and emf versus volume of NaOH solution respectively. The equivalence point is the point of inflexion of the curve, i.e., where the curve changes its direction. (ii) 'pHȺ/Ⱥ'V and 'EȺ/ /Ⱥ'V versus Vav, the average volume for each 'pH and 'E value. The equivalence point is given by the peak of the curve. This is called first derivative plot (Refer Fig. 4.3).

™ ȱŽŠœž›Ž–Ž—ȳ45

10. Determine the volume of NaOH solution corresponding to the equivalence point and knowing its normality, find out the normality and hence the strength of the given CH3COOH solution. 11. Plot a graph between the second derivative of the cell pH and, emf and, volume of alkali used respectively. This is the second derivative plot (Refer Fig. 4.4). 12. The first and second derivative plots give more precise equivalence point than ‘Žȱ˜—Žȱ’ŸŽ—ȱ‹¢ȱ™ ȺȦȺŽ–ȱŸŽ›œžœȱŸ˜•ž–Žȱ™•˜ǯ

Observation: Table 4.6: Data recording S. No.

Volume of NaOH added (mL)

pH of solution

E.M.F. of the cell (mV)

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Table 4.7: Data recording S. No.

'pH

'E (mV)

'V (mL)

'pH / ΔV (mL–1)

'E / 'V (mV mL–1)

Calculation: Volume of NaOH solution required for neutralization = V1 Normality of sodium hydroxide = N1 Volume of acetic acid solution taken = 40 mL V × N1 Normality of acetic acid solution = 1 40

Vav (mL)

46ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Strength of acetic acid solution = Normality u Equivalent mass of acetic acid = ........ g L–1 Result: The strength of the given acetic acid solution = ........ g L–1

EXPERIMENT: 4.5 Aim: To study the effect of addition of HClȺ/ȺNaOH to solutions of acetic acid, sodium acetate and their mixtures. Requirements: pH meter, glass and saturated calomel electrodes or combined electrode, beakers, standard flasks, graduated pipettes, burette, conical flasks, droppers, buffer solutions of pH 4 and 9, standard solutions of acetic acid (0.1 M), sodium acetate (0.1 M), hydrochloric acid (1 M), sodium hydroxide (1 M), oxalic acid (1 N), phenolphthalein indicator, distilled water. Theory: A buffer is a solution that resists a change in pH on the addition of small quantities of acid or base. The change in pH will be insignificant. Two common types of buffer solutions are: (1) a weak acid together with a salt of the same acid with a strong base. These are called acid buffers, e.g., CH3COOH + CH3COONa. (2) a weak base and its salt with a strong acid. These are called basic buffers, e.g., NH4OH + NH4Cl. In this experiment, we will deal with acid buffer. The pH of the acid buffer is given by Henderson-Hasselbalch equation pH = pKa + log (CsaltȺ/ȺCacid) It can be used for calculating the pH of solutions containing a pair of acid and its conjugate base such as HAȺ/ȺA–, HA–Ⱥ/ȺA2– or B+Ⱥ/ȺBOH. The buffer action can be explained as follows for CH3COOH - CH3COONa buffer: ZZX H+ + CH3COOȹȮ CH3COOH YZZ CH3COONa o Na+ + CH3COOȹȮ Since the salt is completely ionized, it provides the common ions CH3COO– in excess. According to Le Chatelier’s principle, it facilitates the backward reaction and thus suppresses the ionization of acetic acid. This reduces the concentration of H+ ions and raises the pH of the solution. Thus, a 0.1 M acetic acid solution has a pH of 2.87 but a solution of equal volumes 0.1 M acetic acid and 0.1 M sodium acetate has a pH of 4.74. Thus, 4.74 is the pH of the buffer. On addition of 0.01 mole of HCl, the pH changes from 4.74 to 4.66, while on the addition of 0.01 mole of NaOH, the pH changes from 4.74 to 4.83. Obviously, the changes in pH are insignificant and the buffer solution maintains a fairly constant pH. This can be explained as follows: 1. Addition of HCl: On adding a small quantity of HCl, the H+ ions produced react with acetate ions to form unionized CH3COOH. Thus, the added H+ ions are neutralized and the pH of the buffer solution should remain unchanged. However, the concentration of CH3COOH is increased and the equilibrium is

™ ȱŽŠœž›Ž–Ž—ȳ47

shifted slightly to the right to increase H+ ions. This explains the slight decrease of pH (pH = –log10 [H+]) of the buffer solution on addition of HCl. 2. Addition of NaOH: When NaOH is added to the buffer solution, the OH– ions produced combine with H+ ions of the buffer to form water molecules. As a result, the equilibrium shifts to the right to produce more and more H+ ions and the OH– ions are neutralized. Thus, the pH should remain the same. But, another equilibrium is set up as ZZX H O + CH COOȹȮ CHCOOH + OH– YZZ 2 3 according to which the concentration of CH3COOH decreases and consequently [H+] is also slightly less. So, the pH slightly increases. The buffering capacity of a buffer (the amount of acid or base that it can neutralize without a significant change in pH) is limited by the nature and concentration of its components. 1. Effect of addition of HCl on the pH of acetic acid: Since HCl as a strong acid dissociates completely, the concentration of H+ ions increases on adding HCl to acetic acid. Thus, the pH decreases. 2. Effect of addition of NaOH on the pH of acetic acid: The 0.1 M solution of acetic acid has a pH of 2.88. When 0.01 mole of NaOH is added, the pH increases to 3.85 and further addition of NaOH raises the pH further. The reason for this is that the two react to give acetate ion which is a base and this gets hydrolyzed to produce OH–. So, the basic nature increases. CHCOOH + OH– o CH3COOȹȮȱ+ H2O 3. Effect of addition of HCl on the pH of sodium acetate: Both HCl and sodium acetate react to form acetic acid and sodium chloride. CH3COONa + HCl o CH3COOH + NaCl Sodium acetate being basic in nature has a pH greater than 7. On adding HCl, the pH decreases due to the formation of acetic acid. 4. Effect of addition of NaOH on the pH of sodium acetate: The pH increases further as the basicity of the solution increases. Note: The general trends and explanation are given. However, the exact nature and extent of change of pH depends on the concentration and volume of acid and base added and also those of acetic acid, sodium acetate and their mixtures. For example, it was found out experimentally that on the addition of 1 mL of 1 M NaOH solution in drops to acetic acid-sodium acetate buffer (20 mL of 0.1 M acetic acid + 20 mL of 0.1 M sodium acetate), the pH increases gradually from 4.77 to 4.89. But when NaOH was further added, the pH steeply increases to 12.2. Procedure: 1. Standardize the sodium hydroxide and hydrochloric acid solutions by the usual procedure. 2. Switch on the pH meter at least 20 minutes before starting the experiment. This is to stabilize the instrument.

48ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

3. Calibrate the pH meter as given in Experiment 4.1. 4. Measure 20 mL of 0.1 M CH3COOH solution and transfer the solution to a clean 50 mL beaker. 5. Measure its pH. Add 6 drops of 1 M HCl solution one drop at a time. Swirl the solution after each addition and measure the pH. 6. Measure 20 mL of 0.1 M CH3COONa solution and transfer the solution to another clean 50 mL beaker and measure its pH. 7. Add 6 drops of 1 M HCl solution one drop at a time. Swirl the solution after each addition and measure the pH. 8. Repeat the same procedure with 1 M NaOH solution instead of HCl solution. 9. Mix 20 mL of 0.1 M CH3 ȱ œ˜•ž’˜—ȱ Š—ȱ ŘŖȱ –ȱ ˜ȱ ȹŖǯŗȱ ȱ  3COONa solution and measure the pH of the resulting buffer solution. 10. Add 6 drops of 1 M HCl solution one drop at a time. Swirl the solution after each addition and measure the pH. 11. Repeat the same procedure with 1 M NaOH solution instead of HCl solution. Observation: Table 4.8: Data recording S. No

Solution

1.

CH3COOH

2.

CH3COONa

3.

CH3COOH + CH3COONa

pH of solution

Solution + 1 M HCl + 1 drop

+ 1 drop

+ 1 drop

+ 1 drop

+ 1 drop

+ 1 drop

+ 1 drop

+ 1 drop

Table 4.9: Data recording S. No.

Solution

1.

CH3COOH

2.

CH3COONa

3.

CH3COOH + CH3COONa

pH of solution

Solution + 1 M NaOH + 1 drop

+ 1 drop

+ 1 drop

+ 1 drop

Result: 1. The effect of addition of HCl: (i) increasesȺ/Ⱥdecreases the pH of CH3COOH (ii) increasesȺ/Ⱥdecreases the pH of CH3COONa (iii) increasesȺ/Ⱥdecreases the pH of the buffer 2. Write the effect of addition of NaOH on the same lines. Interpretation: Analyze the result on the basis of the theory given above.

™ ȱŽŠœž›Ž–Ž—ȳ49

VIVA QUESTIONS ȱ

ŗǯȱ Ž’—Žȱ™ ȱ˜ȱŠȱœ˜•ž’˜—ǯȱ ’ŸŽȱ’œȱž—’ǯ 2. Is it possible for a solution to have zero or negative pH value? Justify your answer. 3. Calculate the pH of a solution of 2 M HCl. 4. Why is it necessary to determine the pH of juices, milk, soaps, detergents and shampoos? ȱ śǯȱ ‘ŠȱŠ›Žȱ‹žŽ›œǵȱ ’ŸŽȱ‘Ž’›ȱŒ‘Š›ŠŒŽ›’œ’Œœǯ 6. What are acidic and basic buffers? Describe with examples. 7. Write Henderson-Hasselbalch equation. Explain the terms involved in it. 8. What is the use of the above equation? 9. How do you calibrate the pH meter? 10. Why should the pH meter be calibrated? 11. Name the reference and indicator electrodes used in pH meter. 12. Name the solution in which you keep the reference and indicator electrodes after use.

5

THERMOCHEMICAL MEASUREMENT

EXPERIMENT: 5.1 Aim: To determine the water equivalent (or heat capacity) of a calorimeter using different volumes of water. Requirements: Thermos flask of 500 mL capacity, fitted with a two-holed rubber cork, two thermometers (0.1 °C), burette, glass or metal stirrer with loop, stopwatch, beakers, watch glass, distilled water. Theory: Heat capacity of a calorimeter is the amount of heat required to raise the temperature of calorimeter by 1°C. Water equivalent of a calorimeter is defined as the amount of water that would absorb the same amount of heat as the calorimeter for 1° rise in temperature. In other words, it is the number of grams of water with the same heat capacity as the calorimeter since 1 calorie of heat raises the temperature of 1 g of water by 1 degree (specific heat of water = 1). In case of glass vessel, the water equivalent value is found for such parts of the calorimeter, that are actually in contact with the reacting system. In this case, the usual method of obtaining the water equivalent by multiplying together the mass and specific heat of the material of the vessel is not practically feasible. The water equivalent of a glass vessel is found by carrying out an experiment similar to the experiment to be performed later on in the vessel and where heat change is known. Equal volumes are used so that area of the calorimeter in contact with the system does not change. Thermos flask can be used as a calorimeter. A typical graph of temperature versus time for determining the water equivalent of the calorimeter is given in Fig. 5.1. Procedure: Set I (with 100 mL of distilled water): 1. Take a thermos flask of 500 mL capacity, fitted with a two-holed rubber cork— one hole for a thermometer and the other for a stirrer. Measure 100 mL of distilled water into the flask. Wait for a few minutes so that the water attains the temperature of the flask. Note this temperature of water. 2. Measure another 100 mL of distilled water in a beaker covered with a watch glass and heat it.

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ51

Temperature (°C)

Hot water

Mixing time

th tm

Mixture

Cold water tc

0.5 1.0

2.0

3.0 3.5 4.0

6.0

5.0

7.0

8.0

Time (minutes)

Fig. 5.1: Water equivalent of the calorimeter

3. When the temperature of the water rises about 15°C above the room temperature, remove the burner and note down the temperature after every half a minute for about 3 minutes (to be done when the temperature starts falling.) 4. At 3.5th minute, quickly pour the hot water into the water present in the thermos flask. 5. This must be the time when half of the hot water has been poured into the thermos flask. 6. Stir the mixture gently and again note its temperature from 4th minute for every half a minute up to 6 minutes. 7. Plot a graph of temperature versus time and from it find out the temperature of hot water (th), cold water (tc) and the mixture (tm) at the time of mixing. 8. Draw a vertical line on the graph corresponding to the time of mixing and extend the lines of hot water, cold water and the mixture to this vertical line. The points of intersection will be the required temperatures th, tc and tm. Observation: Table 5.1: Data recording Time (in minutes)

Temperature (°C) of Cold water (tc)

Hot water (th)

Mixture (tm)

0.0

-

0.5

-

1.0

-

1.5

-

2.0

(Contd.)

52ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• (Contd.) 2.5

-

3.0 3.5

Time of mixing---------

--------------------------

4.0

-

-

4.5

-

-

5.0

-

-

--------------------------

5.5 6.0

Calculation: Mass of cold water = 100 g Temperature of cold water = tc °C Mass of hot water = 100 g Temperature of hot water = th °C Temperature of the mixture = tm °C Let the water equivalent of calorimeter and its parts be w g Heat lost by hot water = ms't = 100 u 1 u (th – tm) cal Heat gained by cold water = ms't = 100 u 1 u (tm – tc) cal Heat gained by calorimeter = w u (tm – tc) cal Heat lost = Heat gained 100 u 1 u (th – tm) = 100 u 1 u (tm – tc) + w u (tm – tc) = (w + 100) (tm – tc) w + 100 = 100(th – tmǼȺȦȺǻm – tc) w = [100(th – tmǼȺȦȺǻm – tc)] – 100 g The heat capacity of the calorimeter also has the same value but with the unit of cal degȹȮŗ. Set II: Repeat the above procedure with 150 mL each of hot and cold water and determine the water equivalent as described above. Result: The water equivalent of a calorimeter was found to be, Set I = ........ g Set II = ........ g

EXPERIMENT: 5.2 Aim: To determine the water equivalent (or heat capacity) of the calorimeter by back calculation method (using enthalpy of solution of sulphuric acid). Warning: Although it is given in the syllabus and it is possible to carry out this experiment, it is not advisable to do so. It is very dangerous as the dissolution of sulphuric acid in water is highly exothermic.

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ53

Requirements: Thermos flask with two-holed rubber cork, two 0.1°C thermometers, guarded burette, beakers, watch glass, glass or metal stirrer with loop, stopwatch, sulphuric acid, distilled water. Theory: Same as given in Experiments 5.1, 5.6 and 5.7. In this experiment, the sulphuric acid and water are taken in 1:200 molar ratio. The enthalpy of solution of sulphuric acid at this ratio was found to be – 17.9 kcal. Procedure: Determine the change in temperature, i.e., the increase in temperature since the process is exothermic, as given below. 1. Measure out exactly 3 mL of concentrated sulphuric acid (of density 1.84 g mL–1). 2. Take 200 mL of cold water in a thermos flask and note down its temperature every half a minute for 3 minutes. 3. At 3.5th minute, add very carefully (without spilling), the concentrated acid to the water in the thermos flask and stir thoroughly but gently. 4. Continue recording the temperature of the solution from the 4th minute for every half a minute till it becomes constant. 5. Plot a graph of temperature versus time and determine the temperature of water (t1°C) and the final temperature of the solution (t °C) from the graph. 6. Determine the increase in temperature. Observation and graph: As in Experiment 5.7. Calculation: Mass of water in the thermos flask = 200 g Mass of H2SO4 = m g Temperature of water = t1 °C Final temperature of solution from graph = t °C Heat evolved is, Q = (w + 200) (t – t1) cal M Heat of solution per mole of solute = Q u m where M stands for molar mass of solute. Heat of solution of H2SO4 (known) = 17.9 kcal = 17,900 cal M = 17,900 Thus, Q u m m Q = 17,900 u M m Therefore, (w + 200) (t – t1) = 17,900 u M m w = [(17,900 u ǼȺȦȺǻȱȮȱ1)] – 200 M = ........ g The heat capacity of the calorimeter will have the same value but with the unit of cal deg–1. Result: The water equivalent was found to be ........ g.

54ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 5.3 Aim: To determine the water equivalent (or heat capacity) of the calorimeter by back calculation method (using enthalpy of neutralization). Requirements: Thermos flask with two-holed rubber cork, two 0.1°C thermometers, standard flasks, burette, conical flask, beakers, watch glass, glass or metal stirrer with loop, stopwatch, 1 N HCl and 1 N NaOH solutions, standard oxalic acid solution, phenolphthalein indicator, distilled water. Theory: Same as given in Experiments 5.1 and 5.4 . Procedure: Standardize the HCl and NaOH solutions by the usual procedure. Make sure that they are of 1 N concentration. Determine the change in temperature, i.e., the increase in temperature since the process is exothermic, as given in Experiment 5.4. Observation and graph: As in Experiment 5.4. Calculation: Heat of neutralization of HCl and NaOH, 'Hn = 13.7 kcal = 13,700 cal 1 ȱ ȱ ȱNJ n = Q u 0.1 ȱ ȱ ȱȱƽȱŗřŝŖŖȺȦȺŗŖȱƽȱŗřŝŖȱŒŠ• Substituting for Q in the equation given below (refer to Experiment 5.4), calculate the value of w. ta + t b ⎞ ⎛ cal Q = qsolution = (w + 200) u 1 u ⎜ t m – 2 ⎟⎠ ⎝ Result: The water equivalent was found to be ........ g.

EXPERIMENT: 5.4 Aim: To determine the heat of neutralization of hydrochloric acid and sodium hydroxide using a thermos flask or beaker as calorimeter. Requirements: Thermos flask with two-holed rubber cork, two 0.1°C thermometers, standard flasks, burette, conical flask, beakers, watch glass, glass or metal stirrer with loop, stopwatch, 1 N HCl and 1 N NaOH solutions, standard oxalic acid solution, phenolphthalein indicator, distilled water. Theory: When aqueous solutions of acids and bases react with each other, heat is evolved and the reaction is known as neutralization. The change in heat accompanying a neutralization reaction is termed enthalpy or heat of neutralization. It is defined as the amount of heat evolved when one gram equivalent of an acid is neutralized by one gram equivalent of a base, both being taken as dilute aqueous solutions. The heat of neutralization for strong acids and strong bases is a constant value of 13.70 kcal (i.e., 57.27 kJ) in almost all cases. This constancy of heat of neutralization in the case of strong acids and strong bases is explained on the basis of ionic theory of

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ55

Arrhenius. According to this, strong acids and bases are completely ionized in aqueous solutions. For example, in the case of HCl and NaOH, the neutralization is represented as H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) o Na+(aq) + Cl–(aq) + H2O(l) After cancelling the common ions on both sides, the net reaction is H+(aq) + OH–(aq) o H2O(l) Thus, the heat of neutralization is simply the heat of formation of water from hydrogen and hydroxyl ions which is fixed at a given temperature and pressure. This ionic process is the same for all neutralization reactions involving strong acids and bases. This implies that the heat of neutralization of strong acids and bases is constant, being equal to the enthalpy of formation of water. A typical graph of temperature versus time for determining the heat of neutralization is given in Fig. 5.2. t

Temperature (°C)

Mixture

Alkali t2 t1 Acid 0.5

1.0

1.5

2.0

2.5

3.0 3.5 4.0 Time (minutes)

4.5

5.0

5.5

6.0

6.5

Fig. 5.2: Heat of neutralization

Energy is neither created nor destroyed during a chemical reaction and therefore, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), qreaction, plus the heat absorbed or lost by the solution (the “surroundings”), qsolution, must be equal to zero, i.e., qreaction + qsolution = 0 This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution: qreaction = – qsolution This is the basic concept of all calorimetric experiments and calculations. Here, the “system” is the substance or substances undergoing physical or chemical change and the “surroundings” implies the solution and the calorimeter which absorbs heat from the system or loses heat to the system.

56ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Procedure: SET I : Determination of heat capacity or water equivalent of the calorimeter: Follow the procedure described in Experiment 5.1. Observation calculation and graph: As given in Experiment 5.1. Table 5.2: Determination of water equivalent of the calorimeter Time (in minutes)

Temperature (°C) of Cold water (tc)

Hot water (th)

Mixture (tm)

0.0 0.5 1.0 1.5 2.0

-

2.5

-

3.0

-

3.5

Time of mixing---

---------------------

4.0

-

-

4.5

-

-

5.0

-

-

5.5

-

-

6.0

-

-

Calculation: Mass of cold water = 100 g Temperature of cold water = tc °C Mass of hot water = 100 g Temperature of hot water = th °C Temperature of the mixture = tm °C Let the water equivalent of calorimeter and its parts be w g Heat lost by hot water = ms't = 100 u 1 u (th – tm) cal Heat gained by cold water = ms't = 100 u 1 u (tm – tc) cal Heat gained by calorimeter = w u (tm – tc) cal Heat lost = Heat gained 100 u 1 u (th – tm) = 100 u 1 u (tm – tc) + w u (tm – tc) = (w + 100) (tm – tc) w + 100 = 100(th – tmǼȺȦȺǻm – tc) w = [100(th – tmǼȺȦȺǻm – tc)] – 100 g

--------------------------

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ57

The heat capacity of the calorimeter also has the same value but with the unit of cal deg–1. SET II : Determination of heat of neutralization: Standardize the HCl and NaOH solutions by the usual procedure. Make sure that they are of 1 N concentration. 1. Take a thermos flask of 500 mL capacity, fitted with a two-holed rubber cork one hole for a thermometer and the other for a stirrer. 2. Measure 100 mL of sodium hydroxide solution into the flask. Wait for a few minutes so that the alkali attained the temperature of the flask. Note down this temperature. 3. Measure 100 mL of hydrochloric acid solution in a beaker covered with a watch glass and note down the temperature after every half a minute for 3 minutes. Simultaneously, note down the temperature of the alkali solution also for every half a minute. 4. At 3.5th minute, quickly pour the acid solution into the alkali solution present in the thermos flask. This must be the time when half of the acid solution has been poured into the thermos flask. 5. Stir the mixture gently and again note its temperature after every half a minute up to 6 minutes. 6. After the experiment is completed, add a drop of phenolphthalein indicator to check whether the acid or the base is completely neutralized or not. 7. Plot a graph of temperature versus time and from it find out the temperature of HCl (ta), of NaOH (tb) and of the mixture (tm) at the time of mixing. 8. Draw a vertical line on the graph corresponding to the time of mixing and extend the lines of HCl , NaOH and mixture to this vertical line. The points of intersection will be the required temperatures ta, tb and tm. Observation: Table 5.3: Determination of heat of neutralization Time (in minutes)

Temperature (°C) of HCl (ta)

NaOH (tb)

Mixture (tm)

0.0

-

0.5

-

1.0

-

1.5

-

2.0

-

2.5

-

3.0

-

3.5

Time of mixing----

--------------------

4.0

-

-

4.5

-

-

--------------------------

(Contd.)

58ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• (Contd.) 5.0

-

-

5.5

-

-

6.0

-

-

Calculation: Mass of HCl solution = 100 g Temperature of acid = ta °C Mass of NaOH solution = 100 g Temperature of base = tb °C Temperature of the mixture = tm °C Let the water equivalent of calorimeter and its parts be w g Heat absorbed by the solution and calorimeter, qsolution = ms't ⎛ t + tb ⎞ = (w + 200) u 1 u t m – a cal ⎜⎝ 2 ⎟⎠ The specific heats of solutions are assumed to be as that of water, i.e., 1. This is due to the fact that the increase in density almost compensates the decrease in specific heat. The heat of neutralization reaction is qreaction = – qsolution where qsolution is the quantity of the heat absorbed, by the solution and calorimeter, when 0.1 g equivalent each of acid and base are neutralized (since 100 mL of 1 N solutions are taken). Therefore, Heat or enthalpy of neutralization reaction is 'Hn = qreactionȺ/Ⱥ0.1 = – qsolutionȺ/Ⱥ0.1 = (–) ........ cal = (–) ........ kcal. The negative sign indicates that the process is exothermic. Result: The heat or enthalpy of neutralization of hydrochloric acid and sodium hydroxide = ........ kcal.

EXPERIMENT: 5.5 Aim: To determine the molar enthalpy of ionization (ionization energy) of acetic acid calorimetrically. Requirements: Thermos flask with two-holed rubber cork, two 0.1°C thermometers, standard flasks, burette, conical flask, beakers, watch glass, glass or metal stirrer with loop, stopwatch, 1 N HCl, 1 N CH3COOH and 1 N NaOH solutions, standard oxalic acid solution, phenolphthalein indicator, distilled water. Theory: Refer Experiment 5.4. H+(aq) + OH–(aq) o H2O(l) The heat of neutralization is simply the heat of formation of water from hydrogen and hydroxyl ions which is fixed at a given temperature and pressure. This ionic process is the same for all neutralization reactions involving strong acids and bases. This implies that the heat of neutralization of strong acids and bases is constant being equal to the enthalpy of formation of water.

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ59

But it is not the case with the neutralization of weak acids and strong bases, weak bases and strong acids or weak acids and weak bases. The constant value of 13.7 kcal is not observed in such cases and found to be lower. The reason for lower heat of neutralization in such cases is, apart from the formation of water by the combination of hydrogen and hydroxyl ions, ionization or dissociation of a weak acid or a weak base also occurs. A part of the heat of formation of water (13.7 kcal) is utilized for the ionization of the weak acid or the weak base involved. Hence, the difference between the two heats of neutralization (one of a strong acid and a strong base, and the other ‹Ž ŽŽ—ȱŠȱ ŽŠ”ȱŠŒ’ȺȦȺ‹ŠœŽȱ Š—ȱœ›˜—ȱ‹ŠœŽȺȦȺŠŒ’Ǽȱ ’ŸŽœȱ‘Žȱ‘ŽŠȱ˜ȱ’˜—’£Š’˜—ȱ˜ȱ‘Žȱ weak acid or base. If both the acid and base are weak, then the heat of neutralization between them will be still lower. For example, heat of ionization of weak acid (acetic acid) will be equal to the difference between the heat of neutralization of NaOH-HCl pair and acetic acid-NaOH pair. This is shown as CH3COOH ֖ CH3COOȹȮ + H+ 'Hionization=? (i) NaOH + CH3COOH o CH3COONa + H2O 'H1 kcal (ii) NaOH + HCl o Na+ + ClȹȮ + H2O 'H2 = –13.7 kcal Subtract equation (ii) from equation (i) (iii) CH3COOH – HCl o CH3COONa – Na+ – ClȹȮ 'H3 = 'H1 – (–13.7) kcal ȱ ȱ ȱ ȱ ȱ ȱ ȱ ȱȱȱȱȱȱȱȹȹƽȱǻ'H1 + 13.7) kcal Rearranging, we get CH3COOH – H+ – ClȹȮ o CH3COOȹȮ + Na+ – Na+ – ClȹȮ CH3COOH ֖ CH3COOȹȮ + H+ 'Hionization = 'H3 kcal The sign of enthalpy must be taken into consideration. Neutralization is an exothermic process whereas ionization is an endothermic process. The heat of neutralization of HCl and NaOH may slightly vary from 13.7 kcal. For procedure, observation, graph and calculation, refer Experiment 5.4 and use 1 N acetic acid instead of 1 N hydrochloric acid. Determine: (i) the water equivalent of the calorimeter (ii) the heat of neutralization of HCl and NaOH (as described in Experiment 5.4) and, (iii) the enthalpy of neutralization of acetic acid and sodium hydroxide ȱ Ȋȱ Š•Œž•Š’˜—ȱ˜ȱŽ—‘Š•™¢ȱ˜ȱ’˜—’£Š’˜—ȱ˜ȱŠŒŽ’ŒȱŠŒ’DZ Enthalpy of neutralization of HCl and NaOH = x kcal Enthalpy of neutralization of CH3COOH and NaOH = y kcal Therefore, enthalpy of ionization (ionization energy) of CH3COOH = (y – x) kcal Result: Molar enthalpy of ionization (ionization energy) of acetic acid = ........ kcal Note: Both the neutralization reactions are exothermic and hence, proper sign must accompany the values of enthalpy. Comment on the result.

60ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 5.6 Aim: To determine the integral heat of solution (enthalpy of solution) of the given salt at a specified salt-water ratio (KNO3 in 1: 200 molar ratio). Requirements: Thermos flask with two-holed rubber cork, 0.1°C thermometer, beakers, watch glass, glass or metal stirrer with loop, weighing bottle, stopwatch, potassium nitrate, distilled water. Theory: Most of the chemical changes occur in aqueous or other solutions. Integral heat of solution is the heat change involved when a specified quantity of solute is dissolved in a specified quantity of a specified solvent at a constant temperature and pressure. The state of solution of a substance is described in terms of moles of solvent used per mole of solute dissolved at a constant temperature and pressure. Integral heats of solution of most salts in per mole quantities of solute are tabulated for 100, 200 Š—ȱŚŖŖȱ–˜•Žœȱ˜ȱ ŠŽ›ȱǻ™™Ž—’¡ȱ Ǽǯȱ‘ŽœŽȱŸŠ•žŽœȱ˜›ȱŠȱœž‹œŠ—ŒŽȱŠ›Žȱ’Ž›Ž—ȱ˜›ȱ different amount of solvents. During the dissolution of a salt, heat may be absorbed (endothermic) or evolved (exothermic). If the amount of solvent used is so large that a further addition of solvent no longer causes a change in the integral heat of solution per mole of solute, the integral heat of solution is said to be the heat of solution of substance at infinite dilution. It is a limiting value of the integral heat of solution. It is sometimes called total heat of solution. Heat change effects will be observed when a given solution is diluted to another solution, and the heat change accompanying the dilution of a solution from one concentration to another concentration is called integral heat of dilution which further equals the difference between two integral heats of solution for the substance for two specified concentration states of solutions. For an infinitely dilute solution, integral heat of further dilution will be equal to zero. An example is given below. HCl (g) + 200 H2O o HCl.200 H2O; 'H = – 17.735 kcal mol–1 ...(1) –1 ...(2) HCl (g) + 10 H2O o HCl.10 H2O; 'H = –16.608 kcal mol Subtracting equation (2) from equation (1) and rearranging, we get equation (3) as ...(3) HCl. 10H2O + 190 H2O o HCl.200 H2O ; 'H= –1.127 kcal mol–1 The integral heat of dilution from 1:10 solution to 1:200 solution is –1.127 kcal mol–1. Differential heat of solution of a substance at a specified concentration of a solution, at a constant temperature and pressure, is the heat change observed per mole of solute dissolved in such a large quantity of solution that no significant change in concentration of solution occurs. In this experiment, the integral heat of solution (enthalpy of solution) of potassium nitrate in 1: 200 molar ratio is to be determined. Heat is absorbed when KNO3 dissolves in water. A typical graph of temperature versus time for determining the integral heat of solution for an endothermic process is given below.

Temperature (°C)

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ61

Water t1

Solution

t

0

1

2

3 3.5 4

5

6

7

8

9

Time (minutes)

Fig. 5.3: Integral heat of solution

Energy is neither created nor destroyed during a chemical reaction and therefore, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), qreaction, plus the heat absorbed or lost by the solution (the “surroundings”), qsolution, must be equal to zero, i.e., qreaction + qsolution = 0 This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution: qreaction = – qsolution This is the basic concept of all calorimetric experiments and calculations. Here, the “system” is the substance or substances undergoing physical or chemical change and the “surroundings” implies the solution and the calorimeter which absorb heat from the system or lose heat to the system. Procedure: 1. Determine the water equivalent (w) of the calorimeter using 100 mL of hot water and 100 mL of cold water. 2. Weigh about 5.6 g of KNO3 solid accurately. 3. Take 200 mL of cold water in a thermos flask and note down its temperature every half a minute for 3 minutes. 4. At 3.5th minute, add carefully (without spilling) KNO3 to the water in the thermos flask and stir thoroughly but gently. 5. Continue recording the temperature of the solution from the 4th minute for every half a minute till it becomes constant. This indicates that the salt is dissolved completely. 6. Open the thermos flask and make sure that no salt is left undissolved. 7. Plot a graph of temperature versus time and determine the temperature of water (t1°C) and the final temperature of the solution (t °C) from the graph. 8. Calculate the integral heat of solution of KNO3 (1: 200 molar ratio).

62ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Observation: Mass of water in the thermos flask = 200 g Mass of KNO3 = m g Temperature of water = t1°C Final temperature of the solution from graph = t°C Water equivalent of calorimeter = w g Calculation:

Heat lost by water and the calorimeter is: qsolution = (w + 200) (t – t1) cal (This will be negative since t1 > t). Hence, the heat of the dissolution process is: qreaction = – qsolution and this will be positive. M Ⱥ ‘Ž›ŽȱȱœŠ—œȱ˜›ȱ m molar mass of the solute. The positive sign indicates that the process is endothermic. Result: The integral heat or enthalpy of solution of KNO3 (1 : 200 molar ratio) = ........ kcal mol–1 Therefore, heat of solution per mole of the solute = – qsolution u

EXPERIMENT: 5.7 Aim: To determine the integral heat of solution (enthalpy of solution) of the given salt at a specified salt-water ratio (CuSO4 in 1: 400 molar ratio). Requirements: Thermos flask with two-holed rubber cork, 0.1°C thermometer, beakers, watch glass, glass or metal stirrer with loop, weighing bottle, stopwatch, anhydrous copper sulphate, distilled water. Theory: Same as in Experiment 5.6. The dissolution of anhydrous copper sulphate in water is an exothermic process. A typical graph of temperature versus time for determining the integral heat of solution for an exothermic process is given in Fig. 5.4. Procedure: 1. Determine the water equivalent (w) of the calorimeter using 100 mL of hot water and 100 mL of cold water. 2. Weigh about 4.43 g of anhydrous copper sulphate accurately. 3. Take 200 mL of cold water in a thermos flask and note down its temperature every half a minute for 3 minutes. 4. At 3.5th minute, add carefully (without spilling) the copper sulphate to the water in the thermos flask and stir thoroughly but gently. 5. Continue recording the temperature of the solution from the 4th minute for every half a minute till it becomes constant. This indicates that the salt is dissolved completely.

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ63

Solution

Temperature (°C)

t

Water t1

0

1

2

3 3.5 4

5

6

7

8

9

Time (minutes)

Fig. 5.4: Integral heat of solution

6. Open the thermos flask and make sure that no salt is left undissolved. 7. Plot a graph of temperature versus time and determine the temperatures of water (t1°C) and the final temperature of the solution (t °C) from the graph. 8. Calculate the integral heat or enthalpy of solution of copper sulphate (1: 400 molar ratio). Observation and calculation are similar to those given in Experiment 5.6. Result: The integral heat of solution of copper sulphate (1: 400 molar ratio) = ........ kcal mol–1. Note: In this case, qsolution is positive and qreaction is negative. So, the heat or enthalpy of solution is negative indicating that the process is exothermic.

EXPERIMENT: 5.8 Aim: To determine the enthalpy or heat of hydration of copper sulphate. Requirements: Thermos flask, weighing bottle, 0.1 degree thermometer, stopwatch, two-holed rubber cork, stirrer, anhydrous copper sulphate, distilled water. Theory: Enthalpy or heat of hydration of a salt is the amount of energy released when one mole of the anhydrous salt combines with requisite number of moles of water to form the hydrate. The reaction is CuSO4(s) + 5H2O(l) o CuSO4.5H2O(s); 'H = ? ...(1) The enthalpy change for the above reaction cannot be measured directly. Utilizing the heat of solution data and applying Hess’ law, the enthalpy change of the reaction, which is the enthalpy of hydration of copper sulphate, can be calculated. CuSO4(s) + nH2O(l) o CuSO4(aq); 'H = –15.89 kcal ...(2) CuSO4.5H2O(s) + nH2O(l) o CuSO4(aq) + 5H2O; 'H = +2.80 kcal ...(3)

64ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

The measurement consists of the determination of heats of solution of anhydrous and hydrated salts in water for the same molar ratio. The difference between the two values will give the heat of hydration of the salt. For example, the heat of solution of hydrated copper sulphate is found to be +2.80 kcal (endothermic) and that of anhydrous copper sulphate is found to be –15.89 kcal (exothermic). Subtracting equation (3) from equation (2), we get equation (1). Thus, the heat of hydration of copper sulphate will be = [–15.89 – (+2.80)] = –18.69 kcal mol–1. It is an exothermic process. In general, if the heat of hydration of a salt is represented by 'H1 and the heats of solution of anhydrous and the hydrated salts are given by 'H2ȱŠ—ȱNJH3 respectively, 'H1 = 'H2 – 'H3 with appropriate signs. Procedure: Same as in Experiment 5.7. Observation and calculation: As given in Experiment 5.7. Determine the integral heats of solution of anhydrous copper sulphate ('H2) and hydrated copper sulphate ('H3) in 1 : 400 molar ratio and calculate the heat of hydration of copper sulphate ('H1). Result: The enthalpy or heat of hydration of copper sulphate = ........ kcal mol–1

EXPERIMENT: 5.9 Aim: To determine the heat of solution of benzoic acid by solubility method. Requirements: Beakers, standard flasks, glass rod or stirrer, thermometers, burette, graduated pipettes, conical flask, benzoic acid, standard oxalic acid solution (0.025 N), sodium hydroxide solution (0.025 N), phenolphthalein indicator, ice, distilled water. Theory: Solubility is the property of a solid, liquid or gaseous chemical substance called solute to dissolve in a solid, liquid or gaseous solvent to form a homogenous solution of the solute in the solvent. It may also be defined as the amount of substance that will dissolve in a given amount of solvent to make a saturated solution. It is usually expressed as amount in grams of the solute in 100 g of solvent. According to thermodynamics, the effect of temperature on solubility is given by the equation log S1Ⱥ/ȺS2 = ('HsolutionȺ/Ⱥ2.303 R) [(1Ⱥ/ȺT2 ) – (1Ⱥ/ȺT1)] Le Chatelier’s principle indicates that the solubility of a substance increases with increase in temperature if the dissolution process is endothermic. Similarly, the solubility of a substance increases with decrease in temperature if the dissolution process is exothermic. An equilibrium is maintained between the solution and the undissolved solid benzoic acid, if a saturated solution of benzoic acid is prepared at a higher temperature and is cooled with constant stirring, and if some of the substance is undissolved. Such saturated solutions can be quickly pipetted out at various temperatures and the concentration of the pipetted solutions of benzoic acid can be readily measured by titration against a standard sodium hydroxide solution using phenolphthalein as indicator. Temperatures about 10 degrees above and below room temperature are chosen besides the room temperature for measurement of solubilities. Cooling may be done by putting the solution container in an ice bath containing cold water.

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ65

Procedure: 1. Standardize the solution of sodium hydroxide with standard oxalic acid solution using phenolphthalein as indicator. 2. Heat about 50 mL of distilled water in a 250 mL beaker to a temperature which is about 15°C above room temperature. 3. Add solid benzoic acid with constant stirring to prepare a saturated solution. Make sure that some solid remains undissolved. 4. Now note down the temperature. 5. Quickly pipette out 10 mL of the solution into a 100 mL clean conical flask with a guarded pipette with a cotton plug at the lower end (this is to avoid any solid particle entering into the pipette) and titrate against standardized NaOH solution (0.025 N) using phenolphthalein as indicator. 6. Repeat the titration for at least three concordant readings. 7. Wash and dry the pipette between titrations. 8. Cool down the solution to room temperature and follow the same procedure. 9. For measuring the solubility at a temperature lower than room temperature, start with a saturated solution (about 50 mL) at room temperature and cool it in an ice bath containing cold water. When desired temperature(about 10Ԩ below room temperature) is reached, note down the temperature. 10. Quickly pipette out 10 mL of the solution into a 100 mL clean conical flask with a guarded pipette and titrate against standardized NaOH solution(0.025 N) using phenolphthalein as indicator. 11. Repeat the titration for at least three concordant readings. The pipette need not be washed and dried between titrations. 12. Cool down the solution to 0°C by adding more ice to the bath and carry out the same procedure. Note: The titrations must be carried out very quickly in a minimum time possible as otherwise, the change in temperature of the solution will affect the results. Alternatively, prepare the saturated solutions at room temperature or at a higher temperature by immersing the beaker in a thermostat maintained at that particular temperature. Observation: At ........ °C, concentration of the NaOH solution = 0.025 N Volume of saturated benzoic acid solution pipetted out = 10 mL ȱ Ȳȱ’›ŽȱŸŠ•žŽȱƽȱNaOH Vacid u Nacid = VNaOH u NNaOH 10 u Nacid = VNaOH u 0.025 Nacid = VNaOH uȱŖǯŖŘśȺȦȺŗŖ

66ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Calculate the normality of benzoic acid solution for all temperatures and record in Table 5.4. Table 5.4: Data recording S. No.

Temperature (°C)

1.

-------- (Higher)

2.

Room temperature

3.

--------- (Lower)

4.

0

Titre value (mL)

Normality of benzoic acid solution

Calculation: Solubility at ........ °C: Molecular mass of benzoic acid = 122 g mol–1 Normality of benzoic acid solution = N1 Molarity of solution, M1 = N1 since the acid is monobasic. ›Š–œȱ˜ȱŠŒ’ȱ™Ž›ȱ•’›Žȱ˜ȱ‘Žȱœ˜•ž’˜—ȱƽȱ1 u 122 M1 × 122 × 100 1000 Benzoic acid is only sparingly soluble in water and therefore, the density of its aqueous solution can be taken as equal to that of pure water. Hence, 100 mL of solution may be considered 100 g of water. Solubilities at other temperatures can be calculated in the same manner. Solubility = grams per 100 gram of water =

Table 5.5: Data recording S. No.

Temperature (°C)

Solubility (S) (g per 100 g of water)

1. 2. 3. 4.

Calculate the heat of solution, 'Hsolution, using the solubilities at any two temperatures and the following equation: log S1Ⱥ/ȺS2 = ('HsolutionȺȦȺ2.303 R )[(1Ⱥ/ȺT2) – (1Ⱥ/ȺT1)] Verify with the data at other temperatures also. Result: The heat of solution of benzoic acid = ........ kcal. Precautions: 1. Benzoic acid solution should be saturated. 2. Pipette must be rinsed with the respective liquid before each titration. 3. Pipette should be plugged with cotton at the lower end. 4. Care must be taken to ensure that benzoic acid does not separate out inside the pipette.

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ67

EXPERIMENT: 5.10 Aim: To determine the basicity of the given acid (phosphoric acid) calorimetrically. Requirements: Thermos flask with two-holed rubber cork, two 0.1°C thermometers, standard flasks, burette, beakers, watch glass, glass or metal stirrer with loop, stopwatch, 1 M solution of given acid (phosphoric acid), 1 M NaOH solution, standard oxalic acid solution, phenolphthalein indicator, distilled water. Theory: The basicity of an acid is defined as the number of replaceable hydrogen ions in a particular acid. It may be mono, di or tri basic acid. 100 mL of 1M solution of the acid, where basicity is “B” will neutralize 100 u B mL of 1 M NaOH solution. This fact is utilized to find “B” (basicity) of an acid calorimetrically. If B = 1, then 100 mL of 1 M acid will neutralize 100 mL of 1 M NaOH. Let the heat of reaction per mole of acid be Q1. If B = 2, then 66.7 mL of 1 M acid will neutralize 133.3 mL of 1 M NaOH. Let the heat of reaction per mole of acid be Q2. If B = 3, then 50 mL of 1 M acid will neutralize 150 mL of 1 M NaOH. Let the heat of reaction per mole of the acid be Q3. ( Since 200 ml of water is used to get the water equivalent, total volume of acid and alkali must be 200 mL). If it is found that Q1 | Q2 | Q3, then the acid is monobasic. If it is found that Q3 | Q2 > Q1, then the acid is dibasic (Note: There may be a small difference between Q2 and Q3 due to dilution.) If it is found that Q3 > Q2 >> Q1, then the acid is tribasic [Q1 is the heat evolved for 1 2 the rd neutralization, Q2 is the heat evolved for the rd neutralization and Q3 is the 3 3 heat evolved for the complete neutralization]

Method I: Procedure: 1. Determination of water equivalent: As mentioned in Experiment 5.1. Use 100 mL each of cold and hot water. 2. Determination of Q1 : Take 100 mL of 1 M acid solution and 100 mL of 1 M NaOH solution and determine the heat evolved as done in Experiment 5.4 (Determination of heat of neutralization) 3. Determination of Q2 : Take 66.7 mL of 1 M acid solution and 133.3 mL of 1 M NaOH solution and follow the same procedure to find out Q2. 4. Determination of Q3 : Take 50 mL of 1 M acid solution and 150 mL of 1 M NaOH solution and follow the same procedure to find out Q3. Observation: As in Experiments 5.1 and 5.4. Calculation: ⎛ t + t b ⎞ 1000 Q1 = (w + 200) × 1 × ⎜ t m – a × cal mol –1 2 ⎟⎠ 100 ⎝

68ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

where, ta, tb and tm are the temperatures of acid, NaOH and the mixture respectively and w is the water equivalent of the calorimeter. 100 mL in the denominator is the volume of acid used. Similarly, determine Q2 and Q3. Change the volume of acid in the equations for Q2 and Q3. Based on the values of Q1, Q2 and Q3, find out the basicity of the given acid. Result: The basicity of the given acid = ........ Interpretation: Write in terms of Q1, Q2 and Q3.

Method II: Theory: It is the same as in Method I but can be put forward in a different way as follows. Normally, 100 mL of a 1 M solution of the acid neutralizes 100 mL of 1 M solution of NaOH and the heat evolved will correspond to neutralization of only 1 mole of hydrogen ions (H+). The basicity of the acid (B) is a measure of available hydrogen ions in an acid. It implies that B moles of hydrogen ions are present in the acid. If we add 100 mL of 1 M NaOH solution, only one mole of H+ ions available from the acid will be neutralized and the heat evolved will correspond to only one step of neutralization. If we add another 100 mL of 1 M NaOH solution to the resulting solution, heat is again evolved of about the same order as in the first step (but somewhat less). This suggests that a second mole of hydrogen ions is available from one mole of acid. Proceeding stepwise we will ultimately come to a step in which no more heat is evolved. At each step, the heat of neutralization must be calculated. The observed change in temperature alone will not yield the correct result. No more alkali is added after this step. The basicity of acid is one less than the steps of addition of NaOH to a stage when no more evolution of heat is observed. Procedure: 1. Carry out the experiment with 100 mL of acid. 2. Add 100 mL of 1 M NaOH solution in each step. 3. Calculate the heat of neutralization at each step. 4. Pay due attention to the volume of solution at each step of neutralization. 5. Determine the basicity of the given acid from the number of steps. Result: The basicity of given acid is found to be ----

VIVA QUESTIONS 1. 2. 3. 4. 5.

What is the difference between colorimetry and calorimetry? How do you differentiate an isothermal system from an adiabatic system? Define state variable. Define water equivalent of a calorimeter. Define heat capacity of a calorimeter.

‘Ž›–˜Œ‘Ž–’ŒŠ•ȱŽŠœž›Ž–Ž—ȳ69

6. 7. 8. 9. ȱ ŗŖǯȱ 11. 12. ȱ ŗřǯȱ 14. 15. 16. 17.

How do we find the time of mixing and the temperature at the time of mixing? Define integral heat of solution. What is difference between differential and integral heats of solution? State Hess’s law. ȱ’Ž›Ž—’ŠŽȱ‹Ž ŽŽ—ȱŽ¡Ž—œ’ŸŽȱŠ—ȱ’—Ž—œ’ŸŽȱ™›˜™Ž›¢ǯȱ ’ŸŽȱŽ¡Š–™•Žȱ˜ȱŽŠŒ‘ǯ Name the vessel used to carry out thermochemical experiments. Define (i) exothermic, and (ii) endothermic reaction. ‘’•ŽȱŠ––˜—’ž–ȱŒ‘•˜›’ŽȺȦȺ™˜Šœœ’ž–ȱ—’›ŠŽȱ’œȱŠŽȱ˜ȱ ŠŽ›ǰȱ‘Žȱ›Žœž•’—ȱ solution feels cool to touch. Comment. Define (i) entropy, and (ii) enthalpy. How can we find the heat of hydration of a given salt? How will you determine the heat of neutralization and heat of ionization for a given weak acid? State first, second and third laws of thermodynamics.

6

HETEROGENEOUS EQUILIBRIA

EXPERIMENT: 6.1 Aim: To construct the phase diagram of the given mixture (Urea : Benzoic Acid) using cooling curves method. Requirements: Boiling tube fitted with thermometer and glass stirrer, large mouthed conical flask, burner, benzoic acid and urea. Theory: The solid-liquid equilibria are studied at constant atmospheric pressure because the effect of pressure on these equilibria is very small. These systems are called condensed systems. In the case of condensed system, one of the variables is fixed (as measurements are carried out at constant pressure). Therefore, the degrees of freedom is reduced by one and is written as F = C – P + 1. The equation is known as condensed or reduced phase rule equation. For a two component system, the maximum degrees of freedom are two. Therefore, phase diagram of a two-component system is drawn by choosing the rectangular axes representing temperature and composition. Eutectic system is a two-component system in which the components are completely miscible in liquid state but they do not form compound. The solid phase of the eutectic system consists of pure components. The phase diagram of different solid-liquid equilibria can be drawn by thermal analysis where the system is heated till molten state and then allowed to cool and its temperature after regular interval of time is recorded. The cooling curves of temperature versus time are drawn for different compositions (mixture of two components), and from the breaks and arrest in the cooling curves, the useful information regarding the initial and final solidification temperatures is determined. Cooling curve of pure component: For a pure component (solid A or B), cooling will be smooth till the solidification commences. After the solidification starts the system will have two phases in equilibrium and the degrees of freedom, F = C – P + 1 = 1 – 2 + 1 = 0 (System is invariant), i.e., solidification takes place at constant temperature (complete arrest in the rate of cooling). The curve will retain the smooth cooling after whole liquid has solidified.

ŽŽ›˜Ž—Ž˜žœȱšž’•’‹›’Šȳ71

Cooling curves of mixture: For a mixture (let B in A), cooling is smooth when the system is in liquid phase with degrees of freedom, F = C – P + 1 = 2 – 1 + 1 = 2 (system is bi-variant). The break point or a temperature halt occurs when the solid A starts solidifying from the liquid mixture as the rate of cooling is retarded. The curve moves along with more and more solidification of A and consequently the liquid becomes richer in B. The degrees of freedom along the curve is F = C – P + 1 = 2 – 2 + 1 = 1 (system is uni-variant). The continued cooling causes start of the solidification of component B along with A. At this point, the system has degrees of freedom F = C – P + 1 = 2 – 3 + 1 = 0, i.e., system is invariant. This results in the complete arrest of the cooling curve as the solidification takes place at constant temperature. The temperature will remain constant until entire liquid phase is solidified. After this smoothing, cooling curve is obtained (F = 1). Cooling curve for a solid solution.

Temperature (°C)

Liquid phase Phase change start temp. Phase change finish temp. Solid phase

Time (sec)

Fig. 6.1: Cooling curve for a solid solution

q1 I′′ I

Liquid

q3 q4

Solid A + liquid

q5

I′ S′

S

q2

–

q6

S′′

Solid B + liquid

Temperature (°C)

Using these cooling curves the phase diagram of eutectic system is drawn by plotting the break point temperatures (temperature halt) with respect to the percentage composition of the mixture of two components.

Two solids (1)

(2)

(3)

(4)

(5)

(6)

A

B XB % Composition

Fig. 6.2: Use of thermal analysis. Demonstration of how thermal analysis can be used to determine a phase diagram. Cooling curves (1) and (6) represent behaviour for pure A and B, respectively © Physical Chemistry (e-book); Laidler, Meiser, Sanctuary | www.mchmultimedia.com

72ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Eutectic Point: The eutectic point of the phase diagram is a point where the two corresponding curves (A in B and B in A) intersect. At this point the two solid phases A and B and the liquid phase exist in equilibrium. The degrees of freedom at eutectic point is F = C – P + 1 = 2 – 3 + 1 = 0, i.e., the system is invariant. This is the lowest temperature at which the liquid phase can exist. The temperature corresponding to this point is the eutectic temperature and the corresponding composition is the eutectic composition. The eutectic temperature is the lowest possible melting point over all the compositions of A and B. TmA

Tm B

1

T1

Temperature (°C)

Liquidus

X

All liquid

A + Liquid a

b

2

T2 c T3 TE

d

B + Liquid

3 E A + B (all solid)

A

% Composition

Solidus

B

Fig. 6.3: Solid-liquid phase diagram with an eutectic point

Procedure I: 1. Weigh 5 g of benzoic acid and transfer it to the boiling tube fitted with thermometer and glass stirrer. 2. Heat the boiling tube (heating should be done by rotating the burner around the base of the boiling tube) till the benzoic acid melts. 3. Allow the temperature to rise at least 15–20°C above the melting point of benzoic acid. 4. Keep the boiling tube in a wide mouthed conical flask for cooling. 5. Note the fall in temperature after every 30 seconds (with continuous stirring of the mixture in the boiling tube). 6. Continue this till the temperature has fallen down to 70–75°C (The readings will be for 100% benzoic acid). 7. Now add 0.5 g of urea and heat again till the mixture melts. 8. Allow the temperature to rise 15–20°C above the melting point of the mixture. 9. Note the fall in temperature for an interval of 30 seconds. These values are obtained for 90% benzoic acid.

ŽŽ›˜Ž—Ž˜žœȱšž’•’‹›’Šȳ73

10. Carry out similar steps for 80%, 70%, 60%, 50% of benzoic acid. Similar observations are to be noted for 100%, 90%, 80%, 70%, 60% of urea. 11. The break points and complete arrest of cooling is determined by drawing the temperature versus time curves of the above mentioned compositions and the corresponding phase diagram is drawn by plotting the break points against the composition of the mixtures. Result: The phase diagram obtained from the above system shows this to be ........ type. Procedure II: 1. Set up the apparatus as shown in Fig. 6.4. Thermometer Stirrer Clamp Test tube Mixture of two solids Stand

Burner

Fig. 6.4: Apparatus to study cooling curves

2. Prepare the following mixtures of A and B components by weighing requisite quantities in test tubes numbered from 1 to 10. Table 6.1: Preparation of various mixtures Component

Mass (g)

A

10

9

8

7

6

5

4

3

2

1

0

B

0

1

2

3

4

5

6

7

8

9

10

3. Check the ice point of the thermometer before and after the experiment and use the average of these readings for correcting all the other temperatures. 4. Check the melting points of the pure components on the thermometer. 5. Melt the first mixture by placing the test tube 1 in the water bath. 6. When all the crystals of the solids have disappeared and a homogeneous liquid phase is obtained, remove the tube from the bath, wipe it clean and fix it, along

74ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

with the thermometer and ring stirrer, into the boiling tube which provides an air jacket. 7. Stir the mixture slowly with the stirrer and note the temperature every half a minute. 8. Tap the thermometer near the point where mercury thread stands. 9. Note the temperature at which the first crystals of the solid are formed. 10. Allow the cooling to continue and find out the temperature or temperatures at which the liquid-solid system shows temperature halt(s), before complete solidification. 11. Lowest temperature halt will be the eutectic temperature. 12. Proceed in a similar manner with other test tubes. Plot the cooling curves for all the mixtures. Result: The phase diagram for the given two component system is of ........ type. Any of the following systems can be studied by this method. Table 6.2: Various two component systems S. No.

A

B

1.

Benzoic acid

Cinnamic acid

2.

Naphthalene

Benzoic acid

3.

Naphthalene

β-naphthol

4.

Resorcinol

Urea

5.

Acetamide

Salicylic acid

EXPERIMENT: 6.2 Aim: To find the critical solution temperature or the upper consolute temperature (CST ) for the phenol-water system by plotting the mutual miscibility curve. Requirements: Boiling tube fitted with a two-holed cork, beaker, metal stirrer with a loop, 0.5°C thermometer, wire gauze, phenol, distilled water. Theory: Phenol and water are nearly immiscible liquids or partially miscible liquids. When taken together, these separate into two distinct layers. These are conjugate layers and each layer contains a small amount of the other liquid in it, i.e., the layers are phenol containing a very slight trace of water and water containing a very slight trace of phenol. If we heat this system, it will be seen that at a definite temperature, the two layers merge into each other resulting in a homogeneous layer. This temperature is known as the miscibility temperature. The value of miscibility temperature depends on the relative proportion of phenol and water in the system. However, for this system, there exists a characteristic temperature at and above which the phenol and water are miscible completely whatever may be their relative proportions. This is known as critical solution temperature or upper consolute temperature.

ŽŽ›˜Ž—Ž˜žœȱšž’•’‹›’Šȳ75

Thus, if we plot the miscibility temperature against % composition for phenol-water system, we get a curve with a maximum, which corresponds to the CST (Fig. 6.6). Procedure: 1. Take 5 mL of liquid phenol in a clean dry boiling tube fitted with a cork carrying 0.5°C thermometer and a stirrer (Fig. 6.5).

Thermometer Utility clamp

Stirrer Test tube

250 mL beaker Water Iron ring with wire gauze

Fig. 6.5: Apparatus to find mutual miscibility temperature

2. Put 3 mL of distilled water into this. 3. Suspend this boiling tube with the contents in a beaker containing ordinary water with the help of a stand. 4. When the mixture is stirred, milkiness (or turbidity) is seen and start heating the beaker of water. 5. Constantly stir the mixture in the boiling tube while heating. 6. Note the exact temperature at which the milkiness disappears. 7. Cool the mixture and note the temperature again when the turbidity just reappears (These two readings should not preferably differ by more than 0.5°C). ȱ Şǯȱ ˜ȱ ˜—ȱ Š’—ȱ –ŽŠœž›Žȱ Š–˜ž—œȱ ˜ȱ  ŠŽ›ȱ Š—ȱ —˜’—ȱ ‘Žȱ Ž–™Ž›Šž›Žœȱ ˜ȱ milkiness disappearing and reappearing every time. 9. Change the water in the outer beaker after taking four readings. 10. Plot a graph between the miscibility temperature (T on Y-axis) and composition (X-axis) and determine the CST corresponding to the maximum point on the curve and the composition of phenol at this temperature (Fig. 6.6).

76ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• 78 76 74 72 35%, 68.5°C

Temperature (°C)

70 68 66

Water

64 62 60 58 10

20

30

40

50

60

70

80

90

Mass percent of phenol

Fig. 6.6: Miscibility temperature-composition curve for phenol-water system

Observation and calculation: Volume of phenol taken = 5 mL Mass of phenol taken = Volume u density = 5 u density = ........ g Table 6.3: Data recording S. No. Vol. of water added each time (mL) 1. 2. 3. 4. 5. 6. 7. 8.

+3 +2 +2 +2 +1 +1 +4 +4

% of Phenol

Temperature (°C) at which turbidity disappeared (T1)

reappeared (T2)

T1 + T2 (° C) 2

64.0 51.7 43.3 37.3 34.8 32.7 26.3 22.0

Result: CST of phenol-water system = ........°C. The composition at the CST = ........ % of phenol. Precautions: 1. Mixture in the boiling tube should be constantly stirred. 2. Thermometer bulb should dip completely in the liquid in the boiling tube. 3. The level of water in the beaker should be such as to completely surround the liquid in the boiling tube.

ŽŽ›˜Ž—Ž˜žœȱšž’•’‹›’Šȳ77

4. Water in the beaker must be changed before every reading or at least after taking four readings.

EXPERIMENT: 6.3 Aim: To study the effect of impurities (sodium chloride and succinic acid) on the critical solution temperature of phenol-water system. Requirements: Boiling tube fitted with a two-holed cork, beaker, metal stirrer with a •˜˜™ǰȱ Ŗǯśǚȱ ‘Ž›–˜–ŽŽ›ǰȱ  ’›Žȱ Šž£Žǰȱ ™‘Ž—˜•ǰȱ œ˜’ž–ȱ Œ‘•˜›’Žȱ œ˜•ž’˜—ȱ ǻŗƖȱ –ȺȦȺŸǼǰȱ œžŒŒ’—’ŒȱŠŒ’ȱœ˜•ž’˜—ȱǻŗƖȱ–ȺȦȺŸǼǰȱ’œ’••Žȱ ŠŽ›ǯȱ Theory: The critical solution temperature is sensitive to the presence of a trace of impurities, i.e., any foreign material. If the impurity is soluble only in one of the solvents, the upper critical solution temperature is increased. This is due to the fact that the molecules of the impurity get surrounded by the molecules of the solvent in which it is soluble and thus get solvated. This, in turn, decreases the miscibility of the two solvents. Hence, the upper CST increases. For example, the miscibility of phenol and water is reduced by addition of many common salts such as alkali and alkaline-earth halides (like NaCl). This is because of the tendency of water molecules to associate with ions thereby hydrating them. In this way, simple ions reduce the tendency of water to solvate phenol and therefore the miscibility of water and phenol decreases. Thus, the upper CST is enhanced. If the impurity is soluble in both the solvents, its molecules move to and from the two solvents. This is similar to cementing the two solvents or we can say that the molecules of the impurity form a bridge between the solvents. Thus, the miscibility of the two solvents increases thereby decreasing the upper CST. Succinic acid is a well-known example for decreasing the upper CST of phenol-water system. The reverse will be the effect of lower CST of the systems. The change in CST is directly and linearly proportional to the percentage of impurity. The impurities will have similar effect on the CST of a mixture of fixed composition of components. This forms the basis for Crismer test for detecting and estimating water in alcohols. This has applications in various industries. In pharmaceutical industry, salt is added to make the organic material form a phase to separate from the salty aqueous phase. This procedure is known as “salting out”. It also helps in choosing the best solvent for a drug and overcome the problems that arise during preparation of pharmaceutical products. Procedure: 1. Procedure: For sodium chloride as impurity: Same as in Experiment 6.2, but instead of distilled water, add measured amounts of 1% sodium chloride solution in successive steps. Observation and calculation: As given in Experiment 6.2. 2. Procedure: For succinic acid as impurity:

78ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Same as in Experiment 6.2. But instead of distilled water add measured amounts of 1% succinic acid solution in successive steps. Observation and calculation: As given in Experiment 6.2 (Figs. 6.7 & 6.8). Interpretation: Write for both the impurities based on your observation whether the CST of phenol-water system increases or decreases on the addition of each impurity. 78 (40%, 76.5°C)

76 74

NaCl solution

72

Temperature (°C)

70 68 66 64 62 60 58 10

20

30

40

50

60

70

80

90

Mass percent of phenol

Fig. 6.7: Effect of addition of sodium chloride on the CST of phenol-water system 78 76 74 72

Temperature (°C)

70 68 38.5%, 64.5°C

66 64 62 Saccinic acid solution

60 58

10

20

30 40 50 60 70 Mass percent of phenol

80

90

Fig. 6.8: Effect of addition of succinic acid on the CST of phenol-water system

ŽŽ›˜Ž—Ž˜žœȱšž’•’‹›’Šȳ79

VIVA QUESTIONS

ȱ ȱ ȱ

1. What is a cooling curve? Explain its significance. 2. Draw the phase diagrams of (a) simple eutectic system, (b) congruent melting compound with eutectic, and (c) incongruent melting compound with eutectic. řǯȱ Ž’—Žȱǻ’Ǽȱ–˜•Žȱ›ŠŒ’˜—ǰȱǻ’’Ǽȱ–ŠœœȺȦȺŸ˜•ž–Žȱ›ŠŒ’˜—ǰȱŠ—ȱǻ’’’ǼȱŸ˜•ž–Žȱ›ŠŒ’˜—ǯ Śǯȱ ‘Šȱ’œȱ–ŽŠ—ȱ‹¢ȱŒ›’’ŒŠ•ȱœ˜•ž’˜—ȱŽ–™Ž›Šž›Žǵȱ ’ŸŽȱ’œȱ˜‘Ž›ȱ—Š–Žǯ śǯȱ ’ŸŽȱŽ¡Š–™•Žœȱ˜›ȱœ¢œŽ–œȱ‘ŠŸ’—ȱž™™Ž›ȱǰȱ•˜ Ž›ȱȱŠ—ȱ‹˜‘ȱž™™Ž›ȱŠ—ȱ lower CST. 6. What is the effect of adding sodium chloride and succinic acid on the CST of phenol-water system? 7. Write the statement of phase rule. Explain the terms involved in it.

7

POTENTIOMETRIC MEASUREMENT

EXPERIMENT: 7.1 Aim: To study the titration curve for a polyprotic acid with alkali potentiometrically and to determine the normality of the given acid. Requirements:ȱ˜Ž—’˜–ŽŽ›ȱ˜›ȱ™ ȺȦȺ–ȱ–ŽŽ›ǰȱ•ŠœœȱŽ•ŽŒ›˜ŽȱŠ—ȱœŠž›ŠŽȱŒŠ•˜–Ž•ȱ electrode (SCE) or a combined electrode, beakers, standard flasks, burette, graduated pipettes, standard oxalic acid solution (0.1 N), sodium hydroxide solution (0.1 N), sulphuric acid solution (approximately 0.1 N), phenolphthalein indicator, distilled water. Theory: Titration of polyprotic acids (or bases) requires more attention than titration of monoprotic ones (refer Chapter 4). There are two reasons for that: First, polyprotic acid can have more than one inflection point on the titration curve. œȱ’—ȱ•Š‹˜›Š˜›¢ȱ™›ŠŒ’ŒŽǰȱ™˜œœ’‹•Žȱ™ ȺȦȺ–ȱŒ‘Š—Žȱž›’—ȱ’›Š’˜—ȱ’œȱ•’–’Žȱ˜ȱŠ‹˜žȱ 7–8 units at most and that means steep part of the titration curve—when split between two end points—must be short. This does not help to achieve good accuracy of titration. Second, quite often without precise analysis, the stoichiometry of the reaction is not obvious. For example, when we titrate phosphoric acid with a strong base, we will never be able to observe the third end point (titration curve is completely flat in this area). Thus, we have to select the indicator that will allow titration of either only first proton, or two protons and calculate reaction stoichiometry accordingly. Even then determination would not be easy. Procedure: 1. Take 0.1 N sulphuric acid solution in a beaker with calomel and glass electrodes, or combination electrode, and titrate it against 0.1 N NaOH solution, as described in Experiment 4.3. 2. Note the value of emf for each addition of alkali from the burette. 3. Plot emf versus volume of alkali added and find the equivalence point. Sulphuric acid with pKa1 = –3 and pKa2 = 2 shows only one steep rise. Although its second proton is much less acidic than the first one, sulphuric acid is strong enough

˜Ž—’˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ81

that both protons get titrated together. There is only one, high steep part of the titration curve. Observation: Table 7.1: Data recording S. No.

Volume of alkali added (mL)

E.M.F. of Cell (mV)

'E (mV)

'V (mL)

'E / 'V (mV mL–1)

Vav (mL)

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Calculation: N1V1 = N2V2 where N1 is normality of acid (unknown) V1 is the volume of acid taken in beaker (20 mL) N2 Normality of alkali V2 Volume of alkali from graph Result: The normality of given acid = ........ N Note: In the case of phosphoric acid, first two protons are similar in strength to protons in maleic acid, thus initially titration curve looks similar, with two separate end points. Third dissociation constant is so small, that even after adding large excess of 0.1 M titrant, over 20% of the acid is in the form of HPO42–. pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.35.

EXPERIMENT: 7.2 Aim: To find out the strength of the given potassium dichromate solution by titrating it potentiometrically with the standard Mohr’s salt solution (0.1 N). Requirements:ȱ˜Ž—’˜–ŽŽ›ȱ˜›ȱ™ ȺȦȺ–ȱ–ŽŽ›ǰȱ™•Š’—ž–ȱŽ•ŽŒ›˜ŽǰȱœŠž›ŠŽȱŒŠ•˜–Ž•ȱ electrode (SCE), beakers, standard flasks, burette, graduated pipettes, Mohr’s salt solution (0.1 N), potassium dichromate solution (approximately 0.2 N), 4 N sulphuric acid solution, distilled water.

82ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Theory: Emf measurements are useful for: (1) calculating the thermodynamic parameters such as ' ǰȱ'H, 'S, and equilibrium constant of the reaction; (2) determining the transport number of ions; (3) finding out the solubility and solubility product of sparingly soluble salts; (4) determining the pH of a solution; (5) determining the valency of ions; (6) calculating ionic activity coefficients of solutions; (7) locating the equivalence points in titrations involving acids and bases, redox systems, complexes and precipitation. The basic principle of all potentiometric titrations is that one solution is taken in a titration flask (or beaker) and the other solution is added from the burette when the concentration of some ion in the titration flask keeps on decreasing. If an electrode reversible with respect to this ion is set up in the titration flask, its potential will keep on changing. Such electrodes are known as indicator electrodes. For measuring the change in the potential of the indicator electrode, it is to be combined with a standard (or reference electrode) using a salt bridge. The standard electrode should be such that its potential remains unchanged during titration. In most of the titrations, saturated calomel electrode (SCE) is used as a reference. This electrode is represented as Cl–°Hg2Cl2°Hg and the half-cell reaction is Hg2Cl2 + 2e– o 2Hg + 2Cl–. The appropriate Nernst equation is ESCE = E°SCEȱȮȱǻŘǯřŖřȱȺȦȺŘǼ•˜ǽ•–]2 = E°SCEȱȮȱǻŘǯřŖřȱȺȦȺǼ•˜ǽ•–]. Indicator Electrodes: These are used to measure the concentrations of ionic species in solutions with which they can establish equilibrium rapidly. Mostly these are metal electrodes for the corresponding ions in solution. Thus, Zn, Cu, and Ag can be used as indicator electrodes for Zn2+, Cu2+, and Ag+ ions respectively. Inert electrodes such as Pt and Au can also be used when the oxidized and reduced forms of an electrode system are both present in solution. The electron transfer process occurs on the inert metal which connects the redox system to the external circuit. The change in emf results from the tendency of the ion to pass from one oxidation state to another more stable state. In this experiment, a Pt electrode is used as indicator electrode for Fe2+ – Fe3+ system. The reactions involved in this titration are: Cr2O72– + 14H+ + 6e– o 2Cr3+ + 7H2O 6Fe2+ o 6Fe3+ + 6e– Cr2O72– + 14H+ + 6Fe2+ o 2Cr3+ + 6Fe3+ + 7H2O (Since E0 is the standard potential which is reduction potential by convention, the electrode reaction must be written as reduction reaction, Fe3+ + e– o Fe2+). Advantages of potentiometric titrations: 1. No indicator is required. 2. Can be employed for coloured solutions. 3. Useful for mixture of acids and bases, including weak acids and weak bases. 4. It is possible to carry out the titrations involving polybasic acids where it is difficult to use the indicators. 5. No prior knowledge of pH range is required for acid-base titrations.

˜Ž—’˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ83

Procedure: Standardize the Mohr’s salt solution by the usual procedure. ȱ ŗǯȱ  ’Œ‘ȱ˜—ȱ‘Žȱ’—œ›ž–Ž—ȱǻ™˜Ž—’˜–ŽŽ›ȱ˜›ȱ™ ȺȦȺ–ȱ–ŽŽ›ǼȱŠ—ȱ•Žȱ’ȱœŠ‹’•’£Žȱ for 30 minutes. 2. Take 40 mL of the standard Mohr’s salt solution (or 20 mL of the solution and 20 mL of water) in a 100 mL beaker and add one test tube of dilute H2SO4 (4 N). 3. Dip the indicator (Pt) and reference (SCE) electrodes in the solution (the electrodes must be immersed completely) and connect the leads of the electrodes to the instrument. 4. Add K2Cr2O7 solution from the burette in 1 mL lots. 5. After each addition, stir the solution well and determine the cell emf. 6. From the rough titration, find out the range of volume of dichromate solution in which the equivalence point lies, i.e., corresponding to a sharp change (maximum change) in emf. 7. Repeat the titration by adding the titrant in 0.1 mL lots in this range and also before and after the range to a volume of about 1 mL. 8. Plot (i) emf versus volume of K2Cr2O7 solution. The equivalence point is the point of inflection of the curve, i.e., where the curve changes its direction. (ii) 'ȺȦȺ'V versus Vavǰȱ‘ŽȱŠŸŽ›ŠŽȱŸ˜•ž–Žȱ˜›ȱŽŠŒ‘ȱNJȱŸŠ•žŽǯȱ‘ŽȱŽšž’ŸŠ•Ž—ŒŽȱ™˜’—ȱ is given by the peak on the curve. 9. Determine the volume of K2Cr2O7 solution corresponding to the equivalence point and calculate its normality and hence its strength. Observation: Table 7.2: Data recording S. No.

Volume of K2Cr2O7 solution (mL)

Emf, E (mV)

ΔE (mV)

ΔV (mL)

ΔE / ΔV (mV mL–1)

Vav (mL)

Result: The strength of the given K2Cr2O7 solution = ........ gL–1 Note: 1. The cell potential depends on the concentration of H+ ions, besides the concentration of Fe2+ and Fe3+ ions. In order to make it dependent only on [Fe2+] and [Fe3+], an excess of H2SO4 is added. 2. There is no restriction on the concentration of the titrant as in the case of conductometric titrations. Both the solutions can be of the same concentration also.

84ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

dE dV

E(volt)

Vol. (mL) Vol. (mL)

Fig. 7.1: Plot of emf versus Volume of titrant

Fig. 7.2: Plot of dE / dV versus Volume of titrant (first derivative plot)

2

dE 2 dV

Vol. (mL) 2

2

Fig. 7.3: Plot of d E / dV versus Volume of titrant (double derivative plot)

EXPERIMENT: 7.3 Aim: To determine the thermodynamic parameters, ' ǰȱ'H, 'S and the equilibrium constant KC of the following reactions from emf measurements. (i) Cu2+ + Zn(s) ֖ Zn2+ + Cu(s) (Daniel cell) (ii) Zn(s) + 2AgCl(s) ֖ ZnCl2(s)+ 2Ag(s) (iii) Pb(s) + 2AgCl(s) ֖ PbCl2(s)+ 2Ag(s) (iv) Zn(s) + Pb2+ ֖ Zn2+ + Pb(s) Requirements: For reaction (i): Potentiometer or voltmeter or multimeter, beakers, thermostat or hot water bath, cold water bath, thermometers, zinc and copper electrodes, 1 M solutions of copper sulphate and zinc sulphate, saturated KCl salt bridge, ice, distilled water. Theory: The electrical energy produced by a cell can be defined as the product of the quantity of electricity passing through it and the emf of the cell. For every mole of electrons transferred in a reaction, the quantity of electricity that flows through the cell is one Faraday (1 F = 96,500 coulombs). Therefore, if n electrons are transferred in a reaction, the quantity of electricity passing through the cell is nF Faradays. If E is the emf of the cell, then

˜Ž—’˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ85

Electrical energy = nFE ...(1) This energy is utilized to perform electrical work which will be maximum when the cell operates reversibly. Hence, ...(2) Net electrical work, (–wmax) = nFE where the negative sign indicates that the work is done by the cell. The electrical work performed by the cell is at the expense of the free energy of the reaction. The decrease in free energy equals the maximum work done in a reversible process. ǯǯǯǻřǼ (–wmax) = –' ȱȱ

Ž—ŒŽǰȱ ’ȱ  Šœȱ œžŽœŽȱ ‹¢ȱ ’‹‹œȱ Š—ȱ Ž•–‘˜•£ȱ ‘Šȱ ‘Žȱ Ž•ŽŒ›’ŒŠ•ȱ Ž—Ž›¢ȱ Š—ȱ therefore, the electrical work produced in a reversible cell equals the decrease in free energy during the cell reaction. Therefore, from equations (2) and (3), –' ȱƽȱ—ȱ˜› ' ȱƽȱȮ—ȱȱ ǯǯǯǻŚǼ The equation (4) bridges thermodynamics and electrochemistry. From this, various relationships can be derived as ...(5) 'H = nF [T(wȺȦȺwT)P – E] ...(6) 'S = nF(wȺȦȺwT)P ...(7) nFE° = RT ln KC where the term (wȺȦȺwT)P is known as the temperature coefficient of the emf of the cell. Using these relationships, the thermodynamic parameters can be determined. Procedure: Preparation of salt bridge: Salt bridge is a U shaped glass tube containing agar jelly mixed with a relatively inert electrolyte like KCl, NaCl and KNO3. To make this, prepare a saturated solution of KCl in 100 mL of cold water. Keep in hot water bath and add 3–4 g of agar powder so that it dissolves without froathing. The U tube is filled using rubber tube and suction. Allow this to cool and set in U tube. 1. Set up the Daniel cell as shown in Fig. 7.4. 2. Clean the Zn and Cu electrodes by rubbing with emery or sand paper, wash well with water and dry. Fill the electrode vessels with respective electrolyte solutions and dip the electrodes. 3. Combine the vessels through a salt bridge. 4. Connect the electrodes to the appropriate poles (ends) of the potentiometer or voltmeter. 5. Measure the cell potentials at room temperature. 6. Determine the cell potentials at higher temperatures, say, 35°C, 45°C, 55°C and 65°C by placing the cell in a thermostat. 7. Measurement at lower temperatures can be made by keeping the cell in water bath containing cold water and adding ice.

86ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• Electrons

Salt bridge Sulphate ions

Zinc ions

Zinc electrode

Zinc sulphate

Copper electrode

(–) Anode oxidation

(+) Cathode reduction

Copper (II) sulphate

Fig. 7.4: Daniel cell

Observation: Table 7.3: Data recording S. No.

Temperature (°C)

1.

25

2.

35

3.

45

4.

55

5.

65

Emf (mV)

Calculation: From the value of emf at 25°C, ' ȱŠ—ȱ C can be evaluated (equations 4 and 7). Using the measured values of emf at different temperatures, plot a graph between emf (E) versus temperature (T), and determine the temperature coefficient (wȺȦȺwT)ȹ, 'H, and 'S can then be calculated (equations 5 and 6). Result: 1. ' ȱƽȱǯǯǯǯǯǯǯǯ 2. 'H = ........ 3. 'S = ........ 4. KC = ........

˜Ž—’˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ87

EXPERIMENT: 7.4 Aim: To determine the concentrations of iodide and chloride ions in a given mixture. Requirements:ȱ˜Ž—’˜–ŽŽ›ȱ˜›ȱ™ ȺȦȺ–ȱ–ŽŽ›ǰȱœŠž›ŠŽȱŒŠ•˜–Ž•ȱŽ•ŽŒ›˜ŽȱǻǼȱ˜›ȱ glass electrode, silver electrode, burette, graduated pipettes, conical flasks, beakers, KNO3 salt bridge, silver nitrate solution (0.1 M), standard sodium chloride (0.1 M), potassium chromate indicator, a mixture containing solutions of potassium iodide and potassium chloride (approximately 0.05 M each). Theory: The concentrations of chloride and of iodide ions in an unknown solution can be determined by potentiometric titration using a standard solution of silver nitrate. The potential of the solution is measured with respect to a reference electrode. In this experiment, a silver electrode immersed in the solution is used as the indicator electrode and a saturated calomel electrode will serve as a reference electrode. Or, glass electrode can be used as the reference electrode since the pH of the solution is invariant during the titration . On adding AgNO3 solution from the burette to the solution in the beaker, a reversible electrode Ag _AgCl(s),Cl– is obtained. The cell is represented as SCE ¨KNO3 salt bridge ¨(KCl + KI) Ag+ ¨Ag The emf values will increase with the addition of AgNO3 from the burette since the concentration of Ag+ will increase as the titration proceeds. This is in accordance with the Nernst equation, i.e., ° + 0.0591 log [Ag+] at 25°C. Ecell = Ecell –16 AgI ( Ksp = 1 u 10 ) precipitates first since it is less soluble than AgCl (Ksp = 1 u 10–10). Near the point when the titration of iodide ions get completed, there will be a sharp change in the cell emf and then silver chloride starts precipitating. Again there will be a sharp change in emf when the titration of chloride ions is about to be over. ΔE ›Š™‘œȱŒŠ—ȱ‹Žȱ™•˜Žȱ‹Ž ŽŽ—ȱǻ’ǼȱŽ–ȱǻǼȱŸŽ›œžœȱŸ˜•ž–Žȱ˜ȱ‘Žȱ’›Š—ǰȱŠ—ǰȱǻ’’Ǽȱ ΔV versus average volume of the titrant. From the inflection points, the concentrations of chloride and iodide ions in the mixture can be evaluated. The data obtained can also be used for the calculation of the solubility and the solubility product of AgCl and AgI. Procedure: 1. Switch on the instrument and let it stabilize for 30 minutes. 2. Standardize the silver nitrate solution using a standard solution of sodium chloride and potassium chromate indicator. 3. Pipette out 25 mL of the mixture into a 100 mL beaker and immerse a silver electrode in it. 4. Combine this half-cell with the SCE through the salt bridge (Refer Experiment 7.3). 5. Complete the circuit by connecting the electrodes to appropriate terminals of ‘Žȱ™˜Ž—’˜–ŽŽ›ȱ˜›ȱ™ ȺȦȺ–ȱ–ŽŽ›ǯ 6. Titrate the mixture by adding the standard silver nitrate solution from the burette in 1 mL lots.

88ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

7. After each addition, stir the solution well and determine the cell emf. 8. From the rough titration, find out the range of volume of AgNO3 solution in which the two equivalence points lie, i.e., corresponding to a sharp change (maximum change) in emf at two points. 9. Repeat the titration by adding the titrant in 0.1 mL lots in this range and also before and after the range to a volume of about 1 mL. 10. Plot (i) emf versus volume of AgNO3 solution. The graph will have two equivalence points—the first one corresponding to iodide ions and the second one due to chloride ions. The equivalence point is the point of inflection of the curve, i.e., where the curve changes its direction (ii) 'ȺȦȺ'V versus Vav, the average volume of AgNO3 solution for each 'E value. The equivalence points are given by two peaks in the graph. 11. Determine the exact volume of AgNO3 solution corresponding to the equivalence points and knowing its concentration, find out the concentration of iodide and chloride ions. 12. Repeat the titration to check the reproducibility. Observation: Table 7.4: Data recording S. No.

Volume of AgNO3 solution (mL)

Emf, E (mV)

ΔE (mV)

ΔV (mL)

ΔE / ΔV (mV mL–1 )

Vav (mL)

Calculation: Since the molar mass and equivalent mass are the same for silver nitrate, potassium chloride and potassium iodide, their molarity = normality. The volume (V1) of AgNO3 solution corresponding to the first equivalence point is that required only for iodide ions and the volume (V2) corresponding to the second equivalence point is the total volume required for both iodide and chloride ions. Therefore, (V2 – V1 ) is the volume of AgNO3 solution required for chloride ions. By the volumetric principle, Vmixture u Niodide = V1 u NAgNO3 Vmixture u Nchloride = (V2 – V1) u NAgNO3 From these equations, the normality, which is also the molarity, can be calculated for the ions and hence the number of moles can also be determined.

˜Ž—’˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ89

Result: The concentration of iodide in the given mixture = ........ The concentration of chloride in the given mixture = ........ Note: 1. Silver nitrate will gradually stain skin and clothing black. Protect yourself. 2. Continue the titration past the second end point by a few mL. 3. At some point during the titration, the potential may go from positive to negative and it should be noted down. 4. The ratio of Ksp of AgI and AgCl can be calculated from the measured potentials at the equivalence points. 5. Beyond the second break on the titration curve (the equivalence point for precipitation of silver chloride), the silver ion concentration is in excess and can be calculated from the Nernst equation and free silver ion concentration.

VIVA QUESTIONS 1. What is the principle involved in potentiometric titrations? 2. Explain why cannot we use voltmeter to measure the emf of a cell. 3. Why do we require reference electrode and an indicator electrode in the determination of emf? 4. Name the various reference electrodes. Write the cell representation of calomel electrode. 5. What do you understand by standard electrode potential? 6. Why cannot we measure the absolute value of electrode potential? ȱ ŝǯȱ ’ŸŽȱ Ž¡Š–™•Žœȱ ˜ȱ ŸŠ›’˜žœȱ ™ ȱ Ž•ŽŒ›˜Žœǯȱ ‘Šȱ Š›Žȱ ‘Žȱ ŠŸŠ—ŠŽœȱ ˜ȱ •Šœœȱ electrode? 8. Why quinhydrone electrode can be used for pH of approximately 8.5 only? 9. Under which titration antimony-antimony trioxide electrode is used? 10. What are the advantages of potentiometric titrations over conventional volumetric titrations? ȱ ŗŗǯȱ ’ŸŽȱ ‘Žȱ ŠŸŠ—ŠŽœȱ ˜ȱ ’›œȱ Ž›’ŸŠ’ŸŽȱ Š—ȱ œŽŒ˜—ȱ Ž›’ŸŠ’ŸŽȱ ™•˜œȱ ˜ŸŽ›ȱ Ž–ȱ versus volume plot. 12. Name the indicator electrode used for acid versus base titration. 13. What is the indicator electrode used in ferrous ion versus dichromate titration. 14. Name the indicator electrode used in silver nitrate versus sodium chloride titration. 15. Why cannot we use the same indicator electrode for all types of titrations? 16. How can you determine various thermodynamic parameters using emf measurement? 17. How can emf studies be used to determine simultaneously two ions in a given mixture?

8

CONDUCTOMETRIC MEASUREMENT

Important: Calibrate the conductivity meter using a standard solution of KCl (0.1 N) before performing any experiment.

EXPERIMENT: 8.1 Aim: To determine the cell constant of a given conductivity cell at 25°C. Requirements: Conductometer (conductivity meter), conductivity cell, standard flasks, beakers, pipettes, potassium chloride (KCl), conductivity water. Theory: A conductivity cell consists of two platinum electrodes coated with platinum black (finely divided platinum), kept parallel to each other and sealed into a glass tube. A thin coating with platinum black helps to minimise errors due to current imperfection and polarisation at the electrodes. The spacing between the two electrodes is wider for solutions of high conductance and narrower for solutions of low conductance. The electrodes are fixed firmly so that their positions and hence the spacing do not change accidentally during the measurement. The assembly is shown in Fig. 8.1 (a).

(a) (b)

Fig. 8.1: (a) Conductivity cell (b) Cell immersed in a test solution

˜—žŒ˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ91

The ratio of the length, i.e., the distance between the two electrodes, to the area of cross section of an electrode is known as cell constant. It is a constant for a given cell and may vary with different cells. It is not possible to measure accurately the area of an electrode or the distance between the two electrodes in normal conductivity cells. Hence, the cell constant is found using a standard solution of known specific conductance (also known as conductivity) from literature and determining its conductance experimentally using the given cell. By definition, Measured conductance = Cell constant =

Specific conductance Cell constant

...(1)

Specific conductance Measured conductance

...(2)

Thus, cell constant is also defined as the ratio of the specific conductance (N) of a solution to its measured conductance in the cell. It has the unit of cm–1. In SI system, its unit is m–1. The specific conductance (N) of a solution is the conductance of the solution in a cell in which the area of cross section of each electrode is 1 cm2 and the distance between the electrodes is 1 cm. Its unit is ohm–1 cm–1 or S cm–1 (S stands for Siemens). It can also be defined as the conductance of unit volume of the solution. Conductivity water For the precise measurement of conductivity, specially purified water or deionized water with a specific conductance of 0.05 u 10–6 ohm–1 cm–1 is used instead of ordinary distilled water. Such water can be prepared in lab by redistilling distilled water after adding potassium permanganate and potassium hydroxide in pyrex glass distillation unit or using ion exchange resin. It should be stored in completely sealed pyrex glass flask. All solutions should be prepared using conductivity water only for conductance measurement. Ž—Ž›Š••¢ǰȱœŠ—Š›ȱœ˜•ž’˜—œȱ˜ȱ™˜Šœœ’ž–ȱŒ‘•˜›’Žǰȱ˜›ȱ ‘’Œ‘ȱœ™ŽŒ’’ŒȱŒ˜—žŒŠ—ŒŽȱ values are known, are used to measure the conductance values. Procedure: 1. Prepare 0.001 M, 0.01 M and 0.1 M solutions of KCl by accurately weighing the required quantity of the salt and making up to 100 mL in a measuring flask with conductivity water. 2. Wash the conductivity cell thoroughly and rinse with conductivity water and then with 0.001 M KCl solution. 3. Take sufficient quantity of 0.001 M KCl solution in the conductivity cell in such a way that the electrodes are completely immersed in the solution. 4. Make sure that the solution level is well above the electrode plates and the electrodes do not touch the bottom or sides of the beaker (Fig. 8.1 (b)). 5. Connect the cell to the conductometer and measure the conductance. 6. Wash the cell carefully and repeat the above procedure with the other two solutions.

92ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Observation: Table 8.1: Data recording S. No

[KCl] (M)

1.

0.001

2.

0.01

3.

0.1

Conductance (ohm–1)

Cell constant (cm–1)

Mean = ........ cm–1

Calculation: For 0.001 M KCl solution: Specific conductance = ........ ohm–1 cm–1 (take from the Table given in the Appendix H) Measured conductance = ........ ohm–1 Specific conductance Measured conductance = ........ cm–1 Similarly, calculate for other two solutions and take the average. Result: The cell constant of the given cell = ........ cm–1 Note: 1. Handle the instrument, cell and glassware with care. 2. Avoid parallax error while taking the readings. ȱ řǯȱ ‘ŽȱŒ˜—žŒ’Ÿ’¢ȱŒŽ••ȺȦȺŽ•ŽŒ›˜Žœȱœ‘˜ž•ȱ—˜ȱ˜žŒ‘ȱ‘Žȱ Š••œȱŠ—ȱ‘Žȱ‹˜˜–ȱ˜ȱ the beaker. Using equation (2), cell constant =

EXPERIMENT: 8.2 Aim: To study the variation of conductance of (i) a strong electrolyte, and (ii) a weak electrolyte with concentration and to verify Debye-Huckel-Onsager equation. Requirements: Conductometer (conductivity meter), beakers, standard flasks, solutions of 0.1 N KCl and 0.1 N CH3COOH, conductivity water. Theory:ȱ˜‘ȱ‘Žȱœ™ŽŒ’’ŒȱŒ˜—žŒŠ—ŒŽȱǻŒ˜—žŒ’Ÿ’¢ǼȱǻΎǼȱŠ—ȱŽšž’ŸŠ•Ž—ȱŒ˜—žŒŠ—ŒŽȱ (equivalent conductivity) (/) vary with dilution. The former decreases with dilution due to the fact that the number of ions per unit volume decreases whereas the latter increases with dilution because the number of ions increases. Moreover, the increase in volume during dilution more than compensates the decrease in specific conductance. A limiting high value is obtained at infinite dilution and this is known as equivalent conductance at infinite dilution or at zero concentration. This is denoted by /0 or /Lj. The Debye-Huckel-Onsager equation for dilute solutions of strong electrolytes is /C = /0 – (a + b /0) C , where ‘a’ and ‘b’ are the characteristic constants and /C is the equivalent conductance of the electrolyte at the concentration ‘C’ in terms of normality. The equivalent conductance can be calculated using the equation /C = N uȱŗŖŖŖȺȦȺǯ

˜—žŒ˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ93

When the values of /C are plotted against C , a straight line with a negative slope will be obtained. This line can be extrapolated to meet the y-axis and the intercept would give the value of /0 of the strong electrolyte. Thus, strong electrolytes obey Debye-Huckel-Onsager equation. The weak electrolytes do not obey this equation. The equivalent conductance would gradually increase with dilution and steeply increase at very high dilutions. The line would almost be parallel to y-axis and therefore cannot be extrapolated. Hence, /0 for a weak electrolyte is not possible to be determined by this method. So, such electrolytes do not obey this equation. The equivalent conductance depends on two factors: number of ions and speed of ions. In the case of strong electrolytes, the number of ions remain the same at all concentrations since they are completely dissociated. So, only the speed of the ions play a vital role. As dilution increases, the force of attraction between the ions decreases and hence the speed of the ions increases gradually. This explains the behaviour of strong electrolytes. In the case of weak electrolytes, the speed of the ions increases gradually as in the case of strong electrolytes. So, the equivalent conductance increases gradually in the beginning with decrease in concentration. The number of ions also increases with dilution because at high concentrations the degree of dissociation is less and increases with increase in dilution. Therefore, both the factors operate at high dilutions and hence the equivalent conductance steeply rises. Procedure: 1. Determine the cell constant of the cell as described in Experiment 8.1. 2. Prepare a stock solution of 1 N KCl quantitatively with conductivity water. 3. Dilute this solution appropriately to get solutions of 0.1 N, 0.05 N, 0.025 N, 0.0125 N, 0.00625 N and 0.003175 N concentration. 4. Measure the conductance of each of these solutions on a conductivity meter (wash the cell every time with distilled water, rinse it with conductivity water and then with the solution under investigation before measuring the conductance) and calculate specific conductance values. 5. If digital conductivity meter is used, the specific conductance is directly obtained. 6. Determine the equivalent conductance of each solution and plot a graph between the values of /C and C . 7. Similarly, prepare the solutions of various concentrations of CH3COOH and measure the conductance of each solution. 8. Calculate the values of /C and plot the graph. Table 8.2: For, KCl, a strong electrolyte S. No. Concentration, C (N)

C u 103

Conductance Specific conductance, Equivalent conductance, /C N (ohm–1 cm–1) (ohm–1) (ohm–1 cm–1eq–1)

94ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• Table 8.3: For, CH3COOH, a weak electrolyte Concentration, C (N)

C u 103

Conductance (ohm–1)

Specific conductance, Equivalent conductance, /C N (ohm–1 cm–1) (ohm–1 cm–1eq–1)

Calculation: Cell constant = ........ cm–1 Measured conductance = ........ ohm–1 Specific conductance = Cell constant u Measured conductance = ........ ohm–1 cm–1 1000 Equivalent conductance = Specific conductance u C Result: Explain the pattern of the graph 1. For KCl : 2. For CH3COOH : Interpretation: Interpret the result based on the ‘Theory’. Note: 1. Instead of equivalent conductance, molar conductance can also be used. 2. Handle the instrument, cell and glassware with care. 3. Avoid parallax error while taking the readings. 4. The conductivity cell / electrodes should not touch the walls and the bottom of the beaker.

EXPERIMENT: 8.3 Aim: (i) To determine the equivalent and molar conductances, degree of dissociation and dissociation constant of a weak acid (acetic acid), and (ii) to verify the Ostwald’s dilution law for a given weak electrolyte. Requirements: Conductivity meter, conductivity cell, measuring flasks, beakers, pipettes, 0.1 N acetic acid (stock solution), conductivity water. Theory: Acetic acid is a weak electrolyte. Weak electrolytes are only partially dissociated in solution. Hence, for such electrolytes the dissociation constant (K) is given by Ostwald’s law as K=

Cα 2 (1 − α )

˜—žŒ˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ95

where C is molar concentration and D is the degree of dissociation. This can be written α2 K = . For acetic acid, molarity is the same as normality. The value of D is as (1 − α ) C given as the ratio of the equivalent conductance of the electrolyte at a particular concentration to that at infinite dilution, i.e., 

D=

Λ eq Λ0

.

However, in such cases /0 may be determined only by the application of Kohlrausch’s law of independent migration of ions. Procedure: 1. Prepare solutions of different concentrations, say 0.05 N, 0.02 N, 0.01 N, 0.005 N and 0.002 N, of the given weak electrolyte (acetic acid) by accurately diluting the given stock solution in conductivity water. 2. Using conductivity cell of known cell constant, measure the conductance for each of the above solutions. 3. Calculate the specific conductance, equivalent and molar conductances, and using /0 value (taken from literature) for the electrolyte, compute the dissociation constant, at the experimental temperature. 4. If digital conductivity meter is used, the specific conductivity is directly obtained. 1 α2 versus . A straight line passing through the origin will indicate C (1 − α ) that Ostwald’s law is verified. The slope equals the dissociation constant, K. 6. Compare the calculated and graphical K values. Observation: 5. Plot

Table 8.4: Data recording Normality, N Conductance Specific (or molarity, C ) ohm–1 conductance (N) of the solution ohm–1 cm–1

Equivalent conductance (/eq) ohm–1 cm2 eq–1

Degree of dissociation Λ eq D= Λ0

Dissociation constant Cα 2 K= 1− α

0.002 0.005 0.010 0.020 0.050 0.100 Mean dissociation constant = ........

96ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• Table 8.5: Data recording 1 (L mol–1) C

Molarity, C of the solution

α2 (1− α )

0.002 0.005 0.010 0.020 0.050 0.100

Calculation: ȱŽ••ȱŒ˜—œŠ—ȱǻ•ȺȦȺŠǼȱ˜ȱ‘ŽȱŒŽ••ȱƽȱǯǯǯǯǯǯǯǯȱŒ––1.  /0 for acetic acid = 390.7 ohm–1 cm2 eq–1 (using Kohlrausch’s law). Specific conductance, N = Cell constant u Measured conductance Equivalent conductance, /eq = N u Volume in mL containing 1 g.eq., of the electrolyte 1000 where N is the normality of the solution. =N u N Molar conductance, /m = N u Volume in mL containing 1 mole of the electrolyte 1000 where M is the molarity of the solution. M For acetic acid, the equivalent and molar conductances are equal as N = M. Result: The dissociation constant value for the given weak electrolyte, acetic acid, is (i) calculated : (ii) graphical : Interpretation: Write this based on your result. Note: 1. Handle the instrument, cell and glassware with care. 2. Avoid parallax error while taking the readings. ȱ řǯȱ ‘ŽȱŒ˜—žŒ’Ÿ’¢ȱŒŽ••ȺȦȺŽ•ŽŒ›˜Žœȱœ‘˜ž•ȱ—˜ȱ˜žŒ‘ȱ‘Žȱ Š••œȱŠ—ȱ‘Žȱ‹˜˜–ȱ˜ȱ the beaker. =N u

EXPERIMENT: 8.4 Aim: To determine the strength of the given strong acid (hydrochloric acid) by titrating it conductometrically with the given strong base (sodium hydroxide). Requirements: Conductometer, conductivity cell, beakers, burette, standard flasks, graduated pipettes, conical flask, stirrer or glass rod, sodium hydroxide solution (0.1 N), hydrochloric acid solution (approximately 0.01 N), standard oxalic acid (0.1 N), phenolphthalein indicator, distilled water.

˜—žŒ˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ97

Theory: The advantages of conductometric titrations over volumetric titrations are: 1. There is no need for any indicator. 2. Applicable for weak acids and bases. 3. Applicable for very dilute solutions. 4. Determination of equivalence point and not end point. Conductivity measurements can be employed for other useful purposes: 1. For calculating /0 for weak electrolytes. 2. For comparing relative strengths of acids (H+ ion concentration in equinormal solutions). 3. For determining solubility and solubility product of sparingly soluble salts. 4. For calculating the hydrolysis constant of salts. 5. For studying kinetics of reactions involving ionic solutions as reactants or products. 6. For determining the ionic product of water. 7. For determining equivalence points of different types of titrations such as acidbase, redox, precipitation, etc. The basic principle of all conductometric titrations is that when one solution from the burette is added to another solution in the titration flask or beaker, some ions of the latter are replaced by those of the former and hence conductance changes. According to Kohlrausch, the electrical conductance of a solution depends on the number and mobility of ions. In the titration of a strong acid like HCl with a strong base, like NaOH, before the NaOH solution is added, the acid solution has a high conductance particularly due to the highly mobile H+ ions. As alkali solution is added, the fast moving H+ ions are replaced by combination with the hydroxyl ions forming feebly ionized H2O molecules and their place is taken by the slow moving Na+ ions as given by the equation [H + Cl–] + [Na+ + OH–] o Na+ + Cl– + H2O (feebly ionized). Consequently, the conductance of the solution decreases and continues falling off with every addition of NaOH solution until the equivalence point is reached, i.e., till all the H+ ions are replaced by Na+ ions. Any subsequent addition of alkali means increase of more Na+ ions and the fast moving hydroxyl ions, and thus the conductance begins to increase. If we plot the conductance of the solution against the volume of alkali added, the points will lie on two lines which are almost straight. The point of intersection of interpolated lines will be the equivalence point of the titration and the conductance will be minimum at this point. A typical plot is shown in Fig. 8.2. If a digital conductivity meter is used, conductivity (specific conductance) can be measured directly and can be plotted in place of conductance because it is nothing but the conductance of unit volume of the solution. It is advisable to use the titrant of concentration about 5–10 times that of the solution to be titrated because the decrease in the conductance due to dilution is negligible in such a case.

–1

Conductance (ohm )

98ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Equivalence point

Volume of alkali added (mL)

Fig. 8.2: Titration curve for strong acid versus strong base

Procedure: 1. Standardize the sodium hydroxide solution with the standard oxalic acid solution using phenolphthalein as the indicator. 2. Switch on the conductivity meter and let it stabilize for 20 minutes. Connect the conductivity cell to the instrument. 3. Take 50 mL of the given HCl solution in a 100 mL beaker and dip the conductivity cell in the solution. 4. Add NaOH solution from the burette in 0.5 mL lots. 5. Stir the solution thoroughly and determine the conductance after each addition. 6. The conductance will go on decreasing till a certain point and will start increasing. 7. Continue the titration and take five or six more readings. 8. Plot the conductance of the solution versus volume of NaOH solution added. 9. Locate the equivalence point and determine the volume of NaOH solution required for complete neutralization of the acid. 10. Determine the strength of the given HCl solution. Observation: Table 8.6: Data recording S. No.

Volume of NaOH solution (mL)

1.

0.5

2.

1.0

3.

1.5

4.

2.0

-

-

-

-

-

-

Conductance (ohm–1)

˜—žŒ˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ99

Calculation: Volume of NaOH solution required for neutralization = V1 mL Normality of NaOH solution = N1 Volume of HCl solution = 50 mL V × N1 Normality of HCl solution = 1 = N2 50 Strength of HCl solution = Normality (N2) u Equivalent mass of HCl Result: The strength of given HCl solution = ........ g L–1 Comment on the shape of the curve and justify your result. Note: Handle the instrument, cell and glassware with care. Avoid parallax error while taking the readings. The conductivity cell / electrodes should not touch the walls and the bottom of the beaker.

EXPERIMENT: 8.5

–1

Conductance (ohm )

Aim: To determine the strength of the given weak acid (acetic acid) by titrating it conductometrically with the given strong base ( sodium hydroxide). Requirements: Conductivity meter (with cell), burette, beakers, standard flasks, conical flask, graduated pipettes, beakers, stirrer or glass rod, standard oxalic acid solution (0.1 N), acetic acid solution (approximately 0.01 N), sodium hydroxide solution (0.1 N), phenolphthalein indicator, distilled water. Theory: Advantages of conductometric titrations over volumetric titrations and the usefulness of conductance measurement (Refer Experiment 8.4). The basic principle of all conductometric titrations is that when one solution from the burette is added to another solution in the titration flask or beaker, some ions of the latter are replaced by those of the former and hence conductance changes. According to Kohlrausch, the electrical conductance of a solution depends on the number and mobility of ions. In the titration of a weak acid like CH3COOH with a strong base, like NaOH, if conductance is plotted with the volume of alkali added, we will get a curve as shown in Fig. 8.3. Conductivity can also be plotted in place of conductance since it is nothing but the conductance of unit volume of the solution.

Equivalence point

Volume of alkali added (mL)

Fig. 8.3: Titration curve for a weak acid versus strong base

100ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Since the acid is weak, the hydrogen ions available will be low and so the initial conductance will be low. As alkali solution is added, highly ionized salt, sodium acetate, is formed and the conductance of the solution increases. CH3COOH + [Na+ + OH–] o CH3COO– + Na+ + H2O (feebly ionized). (weakly ionized) A slight fall in conductance in the initial stage of the titration is due to the removal of a few H+ ions present and the suppression of further ionization of acetic acid by common-ion effect (this depends on the concentrations of the acid and the base). Then the conductance increases, as NaOH neutralizes the un-dissociated acid, until the equivalence point is reached. Any subsequent addition of alkali means increase of more Na+ ions and the fast moving hydroxyl ions and thus the conductance begins to increase rapidly. If we plot the conductance (or conductivity) of the solution against the volume of alkali added, the point of intersection of interpolated lines will be the equivalence point of the titration. It is advisable to use the titrant of concentration about 5–10 times that of the solution to be titrated because the decrease in the conductance due to dilution is negligible in such a case. Procedure: 1. Standardize the sodium hydroxide solution with the standard oxalic acid solution using phenolphthalein as the indicator. 2. Switch on the conductivity meter and let it stabilize for 20 minutes. 3. Connect the conductivity cell to the instrument. 4. Take 40 mL of the given CH3COOH solution in a 100 mL beaker and dip the conductivity cell in the solution. 5. Add NaOH solution from the burette in 0.2 mL lots. 6. Stir the solution thoroughly and determine the conductance after each addition. 7. The conductance may show a slight dip in the beginning and then will gradually increase till a certain point after which it will increase sharply. 8. Continue the titration and take five or six more readings. 9. Plot conductance of the solution versus volume of NaOH solution added. 10. Locate the equivalence point and determine the volume of NaOH solution required for complete neutralization of the acid. 11. Determine the strength of given CH3COOH solution. Observation: Table 8.7: Data recording S. No.

Volume of NaOH solution added (mL)

1.

0.2

2.

0.4

3.

0.6

Conductance (ohm–1)

˜—žŒ˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ101 4.

0.8

-

-

-

-

-

-

Calculation: Volume of NaOH solution required for neutralization = V1 mL Normality of NaOH solution = N1 Volume of CH3COOH solution = 40 mL V × N1 Normality of CH3COOH solution = 1 40 Strength of CH3COOH solution = Normality u Equivalent mass of CH3COOH = ........ gL–1 Result: The strength of given CH3COOH solution = ........ gL–1 Comment on the shape of the curve and justify your result. Note: 1. Handle the instrument, cell and glassware with care. 2. Avoid parallax error while taking the readings. ȱ řǯȱ ‘ŽȱŒ˜—žŒ’Ÿ’¢ȱŒŽ••ȺȦȺŽ•ŽŒ›˜Žœȱœ‘˜ž•ȱ—˜ȱ˜žŒ‘ȱ‘Žȱ Š••œȱŠ—ȱ‘Žȱ‹˜˜–ȱ˜ȱ the beaker.

EXPERIMENT: 8.6 Aim: To estimate the amount of hydrochloric acid (a strong acid) and acetic acid (a weak acid) present in the given mixture by titrating it conductometrically against a strong base (sodium hydroxide). Requirements: Conductivity meter (with cell), burette, standard flasks, conical flask, graduated pipettes, beakers, stirrer or glass rod, standard oxalic acid solution (0.1 N), hydrochloric acid solution (approximately 0.01 N), acetic acid solution (approximately 0.01 N), sodium hydroxide solution (0.1 N), phenolphthalein indicator, distilled water. Theory: Upon adding a strong base to the mixture of acids, the conductance first falls until the strong acid is neutralized completely, then rises as weak acid is converted into its salt and finally rises more steeply as excess alkali is added. Such a titration curve consists of three lines which intersect at two particular points, known as the equivalence points. In the titration of a mixture of hydrochloric acid and acetic acid against sodium hydroxide, on plotting the conductance (or conductivity) values against the volume of titrant (sodium hydroxide) added, we get three straight lines, and the first point of intersection gives the equivalence point corresponding to hydrochloric acid and the second one to acetic acid (Fig. 8.4).

H Na O

–1

Conductance (ohm )

102ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

H

C

CH

l

V1

CO

OH

3

V2

Volume of alkali added (mL)

Fig. 8.4: Titration curve for a mixture of strong and weak acids versus a strong base

Procedure: 1. Standardize the NaOH solution against standard oxalic acid (0.1 N) using phenolphthalein as the indicator. 2. Switch on the conductivity meter and let it stabilize for 20 minutes. 3. Connect the conductivity cell to the instrument. 4. Take the given mixture (40 mL of about 0.01 N acetic acid + 20 mL of about 0.01 N hydrochloric acid) in a 100 mL beaker and dip the conductivity cell in the solution. 5. Titrate against 0.1 N NaOH solution by adding 0.5 mL each time. Stir the solution in the beaker thoroughly during titration. Determine the conductance after each addition. 6. The conductance of the solution decreases till the first equivalence point corresponding to the complete neutralization of HCl is observed. 7. After this, on continuing the addition of NaOH, there will be a small rise in conductance values till the equivalence point corresponding to the complete neutralization of CH3COOH is reached. 8. After that, the conductivity value increases steeply due to the conductance of excess hydroxide ions of NaOH. 9. Plot the graph with the volume of NaOH solution consumed versus conductance values. 10. Locate the two breaks 1 and 2 respectively.

˜—žŒ˜–Ž›’ŒȱŽŠœž›Ž–Ž—ȳ103

11. Volume at break 1 (V1) gives the volume of NaOH solution corresponding to the neutralization of hydrochloric acid, while the difference between the volumes at two breaks (V1 and V2) will give the equivalent volume of NaOH solution required to neutralize acetic acid. Observation: Table 8.8: Data recording S. No.

Volume of NaOH solution (mL)

1.

0.5

2.

1.0

3.

1.5

4.

2.0

Conductance (ohm–1)

-

Calculation: Normality of NaOH solution = N1 Volume of NaOH solution required for neutralization of HCl = V1 mL Volume of the mixture of acids = 60 mL V × N1 Normality of HCl solution = 1 = N2 60 Strength of HCl solution = Normality (N2) u Equivalent mass = ........ gL–1 Volume of NaOH solution required for neutralization of HCl and CH3COOH = V2 mL Volume of NaOH solution required for neutralization of CH3COOH = (V2 – V1) mL (V − V1 ) × N1 Normality of CH3COOH solution = 2 = N3 60 Strength of CH3COOH solution = Normality (N3) u Equivalent mass = ........ gL–1 Result: Amount of HCl present in the given mixture = ........ gL–1 Amount of CH3COOH present in the given mixture = ........ gL–1 Comment on the nature of the graph. Note: Handle the instrument, cell and glassware with care. Avoid parallax error while Š”’—ȱ‘Žȱ›ŽŠ’—œǯȱ‘ŽȱŒ˜—žŒ’Ÿ’¢ȱŒŽ••ȺȦȺŽ•ŽŒ›˜Žœȱœ‘˜ž•ȱ—˜ȱ˜žŒ‘ȱ‘Žȱ Š••œȱŠ—ȱ the bottom of the beaker.

VIVA QUESTIONS ȱ

1. What is conductivity? What are its units? Řǯȱ Ž’—ŽȱŒŽ••ȱŒ˜—œŠ—ǯȱ ’ŸŽȱ’œȱž—’œǯ

104ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

ȱ

3. 4. 5. Ŝǯȱ 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

How do you calibrate a conductivity meter? Name the electrode used in the determination of conductance of a solution. Why is the electrode used in conductometric titration platinized? ‘Šȱ ˜ȱ ¢˜žȱ ž—Ž›œŠ—ȱ ‹¢ȱ –˜•Š›ȱ Š—ȱ Žšž’ŸŠ•Ž—ȱ Œ˜—žŒŠ—ŒŽǵȱ ’ŸŽȱ ‘Ž’›ȱ mathematical expression. What are their units? What do you mean by degree of dissociation and dissociation constant of a weak acid? State Kohlrausch’s law. Write down the mathematical expression of Ostwald’s law. What is the principle involved in conductometric titrations? While plotting the titration curve, what do you have on y-axis? What is the nature of curve for the titration of a strong acid versus a strong base? Why does the conductance of a solution increase after the equivalence point of a titration? What is the nature of curve in the case of a weak acid versus a strong base titration? Can we plot equivalent or molar conductance instead of conductance? Justify your answer. What are the advantages of conductometric titrations over conventional volumetric titrations? State Debye-Huckel-Onsager equation. What is the effect of concentration on conductance of strong electrolyte and of weak electrolyte? What are the factors that affect the conductance of a solution? Why should the concentration of the titrant taken in the burette be about 5-10 times higher than the solution to be titrated (kept in beaker)?

9

CHEMICAL KINETICS

EXPERIMENT: 9.1 Aim: To study the kinetics of hydrolysis of methyl acetate, an ester, in presence of an acid (hydrochloric acid) at room temperature. Requirements: Conical flasks with corks, standard flasks, beakers, burette, graduated pipettes, a thermostat or a water bath, 0.5 N hydrochloric acid solution, 0.05 N sodium hydroxide solution, pure methyl acetate, stopwatch, ice cold distilled water (ice must be made from distilled water), phenolphthalein indicator. Theory: The hydrolysis of methyl acetate in presence of an acid is given below. H+ CH3COOCH3 + H2O ⎯⎯→ CH3COOH + CH3OH This is a pseudo first order reaction or a pseudo unimolecular reaction as the concentration of water is high and hence it is assumed to remain constant during the course of reaction. The H+ ions, which catalyze the reaction, contributed by the acid (here HCl), also remains constant. Hence, the rate of the reaction is determined by the concentration of methyl acetate only. Therefore, the rate equation can be written as: ȱ ¡ȺȦȺȱƽȱ”ȱǽ 3COOCH3]1 By integrating this equation, the following equations for a first order reaction are obtained. k = 2.303 log a t a–x 2.303 a log k a–x where ‘k’ is the rate constant or velocity constant of the reaction or specific reaction rate; ‘a’ is the initial concentration of ester and ‘x’ is the concentration of ester hydrolyzed up to time ‘t’. The half life of a reaction is the time required for the reactant concentration ǻŠǼȱ˜ȱŽŒ›ŽŠœŽȱ˜ȱ˜—Žȱ‘Š•ȱǻŠȺȦȺŘǼȱ˜ȱ’œȱ’—’’Š•ȱŒ˜—ŒŽ—›Š’˜—ǯ

or

t=

k= tŗȺȦȺŘ =

2.303 a log t1/ 2 (a/ 2) 0.693 k

106ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

As more and more acetic acid is produced during the course of reaction, the rate of the reaction is studied by titrating known volumes of the reaction mixture with a standard alkali at various intervals of time from the start of the reaction. The volume of alkali consumed will increase with the progress of the reaction. From the titre values, the amount of acetic acid formed and hence the amount of methyl acetate hydrolyzed can be evaluated. V − V0 would give a straight line passing through the origin A plot of ‘t’ versus log ∞ V∞ − Vt indicating that it is a first order reaction. The constancy of the values of ‘k’ at different intervals of time will confirm this. Procedure: 1. Take 50 mL of 0.5 N hydrochloric acid in a clean 100 mL reagent bottle and cork it. 2. Take about 10 mL of pure methyl acetate in a boiling tube and loosely cork it. ȱ řǯȱ •ŠŒŽȱ‹˜‘ȱ‘Žȱ•Šœ”ȱŠ—ȱ‘ŽȱŽœȱž‹Žȱ’—ȱŠȱ‘Ž›–˜œŠȺȦȺ ŠŽ›ȱ‹Š‘ȱ–Š’—Š’—ŽȱŠȱ or near room temperature. Note down the temperature. 4. Keep about 200 mL of ice cold distilled water ready (ice must be made from distilled water). 5. After 30 minutes, when both the acid and the ester attain the thermal equilibrium (constant temperature), pipette out 2 mL of methyl acetate from the boiling tube into the reagent bottle containing 50 mL of hydrochloric acid solution. 6. Start the stopwatch when half of the ester in the pipette has been discharged (in the first order reaction, any stage of the reaction can be taken as the initial stage and so the stopwatch can be started at any point of time and not necessarily in the beginning). 7. Shake the mixture and at once, pipette out 2 mL of the aliquot (reaction mixture) into a conical flask containing 25 mL of the ice cold water. 8. This freezes or arrests the hydrolysis reaction (the lowering of temperature of reaction mixture slows down the reaction very much but does not stop it completely). 9. Now, titrate it quickly against 0.05 N sodium hydroxide solution using phenolphthalein as the indicator. The end point is the first appearance of the light pink colour which does not vanish for 10 seconds. The colour will fade away with time. 10. Note down the volume of sodium hydroxide solution required for the titration. 11. This volume V0 of alkali corresponds to the concentration of hydrochloric acid in the reaction mixture before the hydrolysis started. 12. Similarly, pipette out 2 mL of the aliquot at successive intervals of 10, 20, 30, 40, 50, 60 and 70 minutes into ice cold water in a conical flask and titrate against standard sodium hydroxide solution as before.

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ107

13. Add the bulk of expected alkali volume (known from the previous titration) quickly in each succeeding titration and then complete the titration carefully with shaking to the first appearance of pink colour which does not fade away in 10 seconds. 14. Note down the volume of alkali required for each titration. The titre value corresponding to the titration after ‘t’ minutes, from the start of the reaction, is denoted as Vt . 15. Infinite time reading, i.e., the volume of sodium hydroxide solution required, when the hydrolysis is complete, is taken as follows: (i) Pipette out 10 mL of the reaction mixture in a clean and dry small conical flask and loosely cork it. (ii) Place it in a water bath maintained at about 60°C for about an hour so that the reaction is over. (iii) Now, cool it to room temperature. Pipette out 2 mL of this mixture and titrate as before against the standard alkali. (iv) Let the titre value be VLj. This corresponds to the total acid concentration after the hydrolysis is complete. Observation: Table 9.1: Data recording Temperature of the reaction = ........°C Time in minutes (t)

Titre value (mL) (Vt)

(V∞ – Vt) (mL)

log

(V∞ – V0 ) (V∞ – Vt )

k=

2.303 (V – V ) log ∞ 0 (min–1) t (V∞ – Vt )

0 10 20 30 40 50 60 70 f

Calculation: Amount of HCl in the reaction mixture at t = 0 D V0 (when no acetic acid is formed yet) Amount of HCl + amount of CH3COOH formed when hydrolysis is complete DVLj Therefore, amount of CH3COOH formed when hydrolysis is complete D (VLjȱ– V0) Hence, amount of ester hydrolyzed when hydrolysis is complete D (VLjȱ– V0) Since hydrolysis is complete, amount of ester hydrolyzed = Initial concentration of ester, a

108ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

So, (VLjȱ– V0) D a ...(1) At time ‘t’, amount of HCl + amount of CH3COOH formed D Vt Amount of HCl (constant as it is a catalyst) D V0 Hence, amount of CH3COOH formed at time ‘t’, ½ ...(2) ¾ i.e., x D (Vt – V0) (or amount of ester hydrolyzed) ¿ Therefore, amount of ester unhydrolyzed at ‘t’, i.e., (a – x) D [(VLjȱ– V0) – (Vt – V0)] D (VLjȱ– Vt) ...(3) Substituting the titre values as in equations (1) and (3) in the formula for the first order reaction, we get (V – V0 ) 2.303 k= log ∞ t (V∞ – Vt ) Calculate the values of k using the above formula for different values of ‘t’. (V – V0 ) Plot a graph of ‘t’ versus log ∞ (V∞ – Vt ) Determine the order of the reaction from calculated and graphical methods. Calculate the half-life period of the reaction using the formula, tŗȺȦȺŘȱƽȱŖǯŜşřȺȦȺ”ǯ Result: At ........ °C: 1. Order of the reaction = ........ 2. Rate constant of the reaction: (i) calculated = ........ (ii) graphical = ........ 3. Half-life period, tŗȺȦȺŘ : (i) calculated = ........ (ii) graphical = ........ Interpretation: Write based on your result. Take the hint from the following ‘Note’. Note: 1. The values of ‘k’ calculated for different degrees of completion of reaction should be reasonably constant. They should not differ in order. The attached number may vary. 2. The graph should be a straight line passing through the origin. 3. The calculated and the graphical values of ‘k’ must reasonably agree with each other.

EXPERIMENT: 9.2 Aim: To study the hydrolysis of methyl acetate in the presence of an acid (sulphuric acid) at room temperature. Requirements: Conical flasks with corks, standard flasks beakers, burette, pipette, a thermostat or water bath, 0.5 N sulphuric acid solution, 0.05 N sodium hydroxide solution, pure methyl acetate, stopwatch, ice cold distilled water (ice must be made from distilled water), phenolphthalein indicator.

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ109

Theory: The hydrolysis of methyl acetate in presence of an acid may be represented as H+ CH3COOCH3 + HOH ⎯⎯→ CH3COOH + CH3OH This is an example of a pseudo unimolecular and pseudo first order reaction. The concentration of water is high and so can be assumed to remain practically constant during the course of reaction. Concentration of H+ ions (which catalyze the reaction) also remains constant. Hence, the rate of the reaction is determined by the concentration of methyl acetate only. Therefore, the rate equation can be written as: ȱ ȱ ȱ ȱ ¡ȺȦȺȱƽȱ”ȱǽ 3COOCH3]1 By integrating this equation, the following equations for a first order reaction are obtained. 2.303 a 2.303 a log or t= k= log t a–x k a–x where ‘k’ is the rate constant, velocity constant or specific reaction rate of the reaction, ‘a’ is the initial concentration of ester and ‘x’ is the concentration of ester hydrolyzed up to time ‘t’. As more and more acetic acid is produced during the course of reaction, the progress of the reaction is studied by titrating known volumes of the reaction mixture with a standard alkali at various intervals of time from the start of the reaction. The volume of alkali consumed will increase with the progress of the reaction. From the titre values, the amount of acetic acid formed and hence the amount of methyl acetate hydrolyzed can be evaluated. V − V0 would give a straight line passing through the origin A plot of ‘t’ versus log ∞ V∞ − Vt indicating that it is a first order reaction. The constancy of the values of ‘k’ at different intervals of time will confirm this. Procedure: 1. Take 50 mL of 0.5 N H2SO4 in a clean 100 mL reagent bottle and loosely cork it. 2. Take about 10 mL of pure methyl acetate in a boiling tube, loosely cork it. 3. Place both of them in a thermostat maintained at or near room temperature. Note down the temperature. 4. Keep about 200 mL of ice cold distilled water ready (ice must be made from distilled water). 5. After 30 minutes, when both the acid and the ester attain the thermal equilibrium (constant temperature), pipette out 2 mL of methyl acetate from the boiling tube into the reagent bottle containing 50 mL of sulphuric acid solution. 6. Start the stopwatch when half of the ester in the pipette has been discharged (in the first order reaction, any stage of the reaction can be taken as the initial stage and so the stopwatch can be started at any point of time and not necessarily in the beginning). 7. Shake the mixture and at once pipette out 2 mL of the aliquot (reaction mixture) into a conical flask containing 25 mL of the ice cold water(ice made from distilled water).

110ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

8. This freezes or arrests the hydrolysis reaction (the lowering of temperature of reaction mixture slows down the reaction very much but does not stop it completely). 9. Now titrate it quickly against 0.05 N sodium hydroxide solution using phenolphthalein as indicator. The end point is the first appearance of the light pink colour which does not vanish for 10 seconds. The colour will fade away with time. 10. Note down the volume of sodium hydroxide required for the titration. 11. This volume V0 of the alkali corresponds to the concentration of sulphuric acid in the reaction mixture before the hydrolysis started. 12. Similarly, pipette out 2 mL of the aliquot at successive intervals of 10, 20, 30, 40, 50, 60 and 70 minutes into ice cold water in a conical flask and titrate against standard alkali as before. 13. Add the bulk of expected alkali volume (known from previous titration) quickly in each succeeding titration and then complete the titration carefully with shaking to the first appearance of pink colour which does not fade away in 10 seconds. 14. Note down the volume of alkali required for each titration. The titre value corresponding to the titration after t minutes, from the start of the reaction, is denoted as Vt . 15. Take the infinite time reading, i.e., the volume required when the hydrolysis is complete, as follows: (i) Pipette out 10 mL of the reaction mixture in a clean and dry small conical flask and loosely cork it. (ii) Place it in a water bath maintained at about 60°C for about an hour so that the reaction is over. Now cool it to room temperature. (iii) Pipette out 2 mL of this mixture and titrate as before against the standard alkali. (iv) Let the titre value be VLj. This corresponds to the total acid concentration after the hydrolysis is complete. Observation: Table 9.2: Temperature of the reaction: ........ °C Time in minutes (t) 0 10 20 30 40

Titre value (Vt ) (mL)

(V∞ – Vt ) (mL)

log

(V∞ – V0 ) (V∞ – Vt )

k=

2.303 (V – V ) log ∞ 0 (min–1) t (V∞ – Vt )

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ111 50 60 70 f

Calculation: Amount of H2SO4 in the reaction mixture at t = 0 D V0 (when no acetic acid is formed yet) Amount of H2SO4 + amount of CH3COOH formed when hydrolysis is complete DVLj Therefore, amount of CH3COOH formed when hydrolysis is complete D (VLjȱ– V0). Hence, amount of ester hydrolyzed when hydrolysis is complete D (VLjȱ– V0). Since hydrolysis is complete, amount of ester hydrolysed = Initial concentration of ester, a So, (VLjȱ– V0) D a ...(1) At time ‘t’, amount of H2SO4 + amount of CH3COOH formed D Vt Amount of H2SO4 (constant as it is a catalyst) D V0 Hence, amount of CH3COOH formed at time ‘t’, i.e., x D (Vt – V0) ...(2) (or amount of ester hydrolyzed ) Therefore, amount of ester unhydrolyzed at ‘t’, i.e., (a – x) D [(VLjȱ– V0) – (Vt – V0) ] D (VLjȱ– Vt) ...(3) Substituting the titre values as in equations (1) and (3) in the formula for the first order reaction, we get (V – V0 ) 2.303 k= log ∞ t (V∞ – Vt ) Calculate the values of k using the above formula for different values of ‘t’. (V – V0 ) Plot a graph of ‘t’ versus log ∞ (V∞ – Vt ) Determine the order of the reaction from calculated and graphical methods. Calculate the half-life period of the reaction using the formula, tŗȺȦȺŘȱƽȱŖǯŜşřȺȦȺ”ǯ Result: At ........°C: 1. Order of the reaction = ........ 2. Rate constant of the reaction: (i) calculated = ........ (ii) graphical = ........ 3. Half-life period, tŗȺȦȺŘ: (i) calculated = ........ (ii) graphical = ........ Interpretation: Write based on your result. Take the hint from the following ‘Note’. Note: 1. The values of ‘k’ calculated for different degrees of completion of reaction should be reasonably constant. They should not differ in order.

112ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

2. The attached number may vary. 3. The graph should be a straight line passing through the origin. 4. The calculated and the graphical values of ‘k’ must reasonably agree with each other.

EXPERIMENT: 9.3 Aim: To compare the strengths or relative avidity of two acids, hydrochloric acid and sulphuric acid used in hydrolysis of methyl acetate. Requirements: Conical flasks, standard flasks reagent bottles with lids, beakers, burette, pipettes of various capacities, a water bath or a thermostat, stopwatch, pure methyl acetate, 0.05 N sodium hydroxide solution, 0.5 N solutions of hydrochloric acid and sulphuric acid, ice cold distilled water (ice must be made from distilled water), phenolphthalein indicator. Theory: The hydrolysis reaction of methyl acetate in the presence of an acid is represented as CH3COOCH3 + H2O ⎯⎯→ CH3COOH + CH3OH Strong acids such as hydrochloric acid and sulphuric acid act as catalysts in this reaction. The rate of reaction, i.e., the hydrolysis of methyl acetate in the presence of a strong acid is found to be directly proportional to the concentration of the acid. The H+ ions from the acid play the catalyst role and thus contribute directly to the rate of reaction. Since different acids with equal strengths contribute different number of H+ ions, the rate of reaction for each of the acids will vary. Hence, the specific reaction rate of acid catalyzed hydrolysis of methyl acetate (an ester), can be used to compare the available concentration of hydrogen ions in equinormal solutions of two acids. Thus, if equal volumes of the two acids HCl and H2SO4 of same normality are used for the hydrolysis of methyl acetate and if kHCl and kH2SO4 are the respective rate constants of the reaction, then the ratio of the strengths, i.e., their relative avidity (S) is given by S = SHClȺȦȺH2SO4 = kHClȺȦȺ” H2SO4 where SHCl and SH2SO4 are the strength of HCl and H2SO4 respectively. Procedure: Perform the experiment for kinetics of acid hydrolysis of ester in presence of HCl and H2SO4, at the same temperature, as given in Experiments 9.1 and 9.2. Observation and Calculation: As given in Experiments 9.1 and 9.2. Result: The relative strength or relative avidity of two acids is found to be .........

EXPERIMENT: 9.4 Aim: To study the kinetics of saponification of an ester (ethyl acetate) with sodium hydroxide. Requirements: Standard flasks, conical flasks, reagent flasks with lids, burette, ›ŠžŠŽȱ ™’™ŽŽœǰȱ ‘Ž›–˜œŠȱ ˜›ȱ ™•Šœ’Œȱ ›˜ž‘ǰȱ œ˜™ ŠŒ‘ǰȱ ȺȦȺŚŖȱ Ž‘¢•ȱ ŠŒŽŠŽȱ œ˜•ž’˜—ǰȱ ȺȦȺŚŖȱ ŒŠ›‹˜—ŠŽȬ›ŽŽȱ œ˜’ž–ȱ ‘¢›˜¡’Žȱ œ˜•ž’˜—ǰȱ ȺȦȺŚŖȱ ‘¢›˜Œ‘•˜›’Œȱ ŠŒ’ȱ solution, phenolphthalein indicator, distilled water.

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ113

Theory: The saponification of ethyl acetate by sodium hydroxide, also known as alkaline hydrolysis of ethyl acetate, is represented as CH3COOC2H5 + NaOH o CH3COONa + C2H5OH The rate equation for this reaction is dx = k (CH3COOC2H5)1 (NaOH)1 ...(1) dt The integrated rate law equations for a second order reaction, when the concentrations of reactants are equal, are: k=

1⎛ x ⎞ ta ⎜⎝ a – x ⎟⎠

...(2)

k=

1⎛ 1 1⎞ – ⎟ ⎜ t ⎝ a – x a⎠

...(3)

⎛ 1 ⎞ 1 ⎜⎝ a – x ⎟⎠ = + kt a tŗȺȦȺŘ =

1 ka

...(4) ...(5)

where ‘k’ is the rate constant of the reaction, ‘a’ is the initial concentration of reactants, ‘x’ is the concentration of reactant reacted up to a time ‘t’ and (a – x) is the concentration of the reactants left unreacted at time ‘t’. As the reaction proceeds, sodium hydroxide is consumed and therefore, the progress of the reaction is studied by determining the amount of alkali left unreacted in the reaction mixture at any instant of time. Saponification is a fast reaction and hence the ›ŽŠŒŠ—œȱŠ›Žȱ—˜›–Š••¢ȱŠ”Ž—ȱ’—ȱ•˜ ȱŒ˜—ŒŽ—›Š’˜—œȱœžŒ‘ȱŠœȱ˜ȱȺȦȺŚŖȱ˜›ȱŽŸŽ—ȱ•Žœœȱ ‘Ž—ȱ room temperature exceeds 25°C. The reaction is followed by the back-titration method, i.e., by adding standard hydrochloric acid solution to the aliquot and titrating the excess of acid with the standard sodium hydroxide solution. Procedure: ȱ ŗǯȱ Š”ŽȱśŖȱ–ȱ˜ȱȺȦȺŚŖȱŽ‘¢•ȱŠŒŽŠŽȱœ˜•ž’˜—ȱŠ—ȱśŖȱ–ȱ˜ȱȺȦȺŚŖȱœ˜’ž–ȱ‘¢›˜¡’Žȱ solution in two separate clean and dry reagent bottles with lids. 2. Place them in a plastic trough containing water or a thermostat maintained at room temperature. Note down the temperature. 3. When both the solutions attain thermal equilibrium, (after about 30 minutes), transfer the sodium hydroxide solution to ethyl acetate solution. 4. Start the stopwatch when half the volume of alkali has been transferred into the ester. This is taken as the zero time of the reaction. 5. At intervals of 5, 10, 15, 25, 40, 60 and 80 minutes, pipette out 5 mL of the aliquot ŽŠŒ‘ȱ ’–Žȱ ’—˜ȱ Šȱ Œ˜—’ŒŠ•ȱ •Šœ”ȱ Œ˜—Š’—’—ȱ ŘŖȱ –ȱ ˜ȱ ȺȦȺŚŖȱ ‘¢›˜Œ‘•˜›’Œȱ ŠŒ’ȱ solution to arrest the reaction. 6. Quickly titrate the excess of acid, left unreacted from the reaction mixture after —Žž›Š•’£Š’˜—ȱ‹¢ȱ‘Žȱœ˜’ž–ȱ‘¢›˜¡’Žȱœ˜•ž’˜—ǰȱ ’‘ȱœŠ—Š›ȱǻȺȦȺŚŖǼȱŠ•”Š•’ǯ

114ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

7. As the reaction proceeds the alkali remaining behind in the reaction mixture ŽŒ›ŽŠœŽœȱŠ—ȱ‘ŽȱŸ˜•ž–Žȱ˜ȱȺȦȺŚŖȱŠ•”Š•’ȱ›Žšž’›Žȱ˜›ȱ‘Žȱ‹ŠŒ”Ȭ’›Š’˜—ȱ˜ȱŽ¡ŒŽœœȱ of acid increases. Observation: Table 9.3: Data recording Temperature of the reaction = ........ °C Time (min)

(a – x) (M)

x (M)

1⎛ x ⎞ ⎜ ⎟ ta ⎝ a – x ⎠

1 a– x

k=

(M–1)

L mol–1min–1

5 10 15 25 40 60 80

Calculation: ⎛ 50 M ⎞ M × = Initial concentration of reactants = ⎜ ⎝ 100 40 ⎟⎠ 80 Volume of aliquot taken = 5 mL M NaOH required for back-titration of excess acid at time ‘t’ = Vt mL Volume of 40 M HCl added to it = 20 mL Volume of 40 M M 20 mL of HCl { 20 mL of NaOH 40 40 Excess acid { Vt mL of

M NaOH 40

Acid used for unreacted NaOH in 5 mL of aliquot { (20 – Vt) mL of Concentration of unreacted NaOH in the reaction mixture = Thus, (a – x) at time ‘t’ = a=

M NaOH 40

20 – Vt 1 . M 5 40

20 – Vt M 200 1 M since equal volumes of solutions of ester and sodium 80 40 hydroxide are mixed and the concentration becomes half.

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ115

⎛ 1 20 – Vt ⎞ Therefore, x = ⎜ – M 200 ⎟⎠ ⎝ 80 Calculate values of ‘k’ by substituting values of ‘a’, ‘x’ and (a – x) in the integrated rate law equation. 1 against ‘t’ and obtain the value of ‘k’ from the slope of the Plot a graph of (a – x) straight line. [Alternatively, the concentrations can be converted in terms of volume and then, the kinetic parameters can be determined]. Important: The concentration of sodium hydroxide solution must be checked and verified M , then appropriate values should be by standardization. If it is not exactly 40 used to calculate ‘a’. Calculate the halt-life period of the reaction using the formula, 1 tŗȺȦȺŘ = ka Result: At ........ °C: 1. Order of the reaction = ........ 2. Rate constant of the reaction (i) calculated = ........ (ii) graphical = ........ 3. Half-life period of the reaction (i) calculated = ........ (ii) graphical = ........ Interpretation: Write based on your result. Note: 1. If the values of ‘k’ obtained are reasonably constant for different stages of the reaction, the order of the reaction can be taken as two. 1 2. If a plot of versus ‘t’ gives a straight line, it also confirms that the reaction (a – x) is of second order.

EXPERIMENT: 9.5 Aim: To study the kinetics of reaction between potassium iodide and potassium persulphate (peroxodisulphate) by integrated rate law method. Requirements: Conical flasks, reagent bottles with lids, standard flasks, graduated pipettes, burette, thermostat or plastic trough, stopwatch, 0.05 M potassium persulphate (K2S2O8) solution, 0.1 M potassium iodide (KI) solution, 0.02 M sodium thiosulphate (Na2S2O3) solution, 2 M acetic acid, freshly prepared starch solution, crushed ice, distilled water. (Molecular and equivalent masses are: K2S2O8 : 270 and 135, KI : 166 and 166, Na2S2O3.5H2O : 248 and 248).

116ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Theory: The acid catalyzed reaction between potassium persulphate and potassium iodide may be written as K2S2O8 + 2KI o 2K2SO4 + I2 or S2O82– + 2I– o 2SO42– + I2 The rate of the reaction may be represented as d[I2ǾȺȦȺȱƽȱ”ǽ2O82–][I–]2 If it is a single-step reaction as suggested above, the order with respect to persulphate must be one, that with respect to iodide must be two and the total order of the reaction must be three. But it is experimentally found out that the total order is two, one with respect to each reactant. This can be explained using the most probable two-step mechanism as follows: S2O82– + I– o (S2O8I)3– (S2O8I)3– + I– o 2SO42– + I2 The first step is slow and hence, rate-determining. Therefore, the rate of the reaction is d[I2ǾȺȦȺȱƽȱ”ǽ2O82–]1 [I–]1 This reaction, therefore, is of the second order as the concentrations of both reactants appear in the rate equation of the reaction to first power. In this experiment, the initial concentrations of both the substances are taken in their stoichiometric ratios. Hence, the integrated rate law equations used for second order kinetics are 1 x k= t a(a – x) tŗȺȦȺŘ =

1 ka

where ‘k’ is the rate constant of the reaction, ‘a’ is the initial concentration of persulphate and ‘x’ is the concentration of it that has reacted till the time ‘t’. The unit of ‘k’ is (conc–1) (time–1) which is expressed as L mol–1 min–1 or M–1 min–1. During the reaction, iodine is liberated and the progress of the reaction can be followed by titrating the liberated iodine in the specified volume of reaction mixture against standardized thiosulphate solution at different intervals of time. The titre values are proportional to iodine formed and therefore to the amount of persulphate which has reacted from the specified volume of reaction mixture. They give the values of ‘x’ at different intervals of time. Procedure: 1. Prepare 0.05 M solution of K2S2O8 and 0.1 M solution of KI by standard procedure. 2. Prepare 2 M solution of CH3COOH by diluting 11.6 mL of glacial acetic acid to 100 mL with distilled water. 3. Take 25 mL of 0.05 M K2S2O8 solution and 25 mL of 0.1 M KI solution + 5 mL of 2 M CH3COOH solution in two separate reagent bottles, loosely cork them and place them in a thermostat or a plastic trough containing water, maintained at room temperature. Note down the temperature.

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ117

4. When the two reactants have attained the thermal equilibrium (after about 30 minutes), pour the iodide-acetic acid solution quickly into the persulphate solution. 5. Start a stopwatch when half of the solution is transferred. 6. Cork the reagent bottle and shake it gently but thoroughly from time to time. 7. At intervals of 5 minutes, pipette out 5 mL of the aliquot into a conical flask containing 25 mL ice cold distilled water (or distilled water to which ice pieces have been added) in order to freeze the reaction. 8. Quickly titrate the liberated iodine against 0.02 M sodium thiosulphate solution using starch as the indicator. 9. Normally, 1 mL of starch solution is added when the solution becomes pale yellow in colour a little before the end point is reached. But if the colour of the aliquot is light, starch can be added in the beginning itself. The end point is the ’›œȱ’œŠ™™ŽŠ›Š—ŒŽȱ˜ȱ‘ŽȱŸ’˜•ŽȺȦȺ‹•žŽȱŒ˜•˜ž›ǯȱ‘ŽȱŒ˜•˜ž›ȱ›ŽŠ™™ŽŠ›œȱ•ŠŽ›ȱ‹ŽŒŠžœŽȱ the reaction is not stopped completely on cooling with ice cold water. 10. At a convenient time in the early part of the experiment, pipette out 10 mL of the reaction mixture into a clean conical flask, add 2 g of solid potassium iodide and place the stoppered flask in a thermostat or a hot water bath maintained at 60°C for about 30 minutes, i.e., till the reaction is complete. 11. Cool this mixture and maintain it at room temperature for at least 10 minutes. 12. Pipette out 5 mL of the reaction mixture and titrate it with 0.02 M sodium thiosulphate solution using starch as the indicator. 13. This titre value corresponds to the initial concentration of potassium persulphate, which after the completion of the reaction would have yielded its equivalent amount of iodine. 14. Calculate ‘a’, ‘x’ and (a – x), and determine the value of ‘k’. Plot ‘t’ versus ǻŗȺȦȺǻŠ – x)) and determine ‘k’, graphically. Observation: Temperature of the reaction = ........ Initial concentration of potassium persulphate = ........ Table 9.4: Data recording Time (min)

5 10 15 20 25 30 35

Titre value, x (mL)

a–x (M)

1 x t a(a – x)

1 a– x

k=

(M–1)

L mol–1min–1

118ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Calculation: Initial concentration of potassium persulphate and the potassium iodide may be calculated as follows: Since 25 mL of 0.05 M potassium persulphate , 25 mL of 0.1 M potassium iodide and 5 mL of acetic acid solutions are mixed, the concentration of potassium persulphate in the mixture becomes 25 0.05 u = 0.0227 M or 0.0454 N 55 25 = 0.0454 M or 0.0454 N Initial concentration of potassium iodide = 0.1 u 55 Check these initial concentrations with the values experimentally determined. In this experiment the initial concentrations of both the substances are in their stoichiometric ratios. Hence, the following equations can be used for second order kinetics. 1 x k= ...(1) t a(a – x) t= kt = tŗȺȦȺŘ =

1 1 – k(a – x) ka

...(2)

1 1 – a–x a

...(3)

1 ka

...(4)

Result: At ........°C: 1. Order of the reaction = ........ 2. Rate constant of the reaction = ........ 3. Half-life period = ........ Interpretation: Write based on your result. Note: 1. If reaction between potassium persulphate and potassium iodide follows second order kinetics, then the value of ‘k’ calculated using above relationship should be almost a constant at different stages of the reaction. 1 2. The graph between ‘t’ and should be a straight line with slope = k and a–x 1 intercept = . a 3. The experimentally determined order is consistent with the two-step mechanism given.

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ119

EXPERIMENT: 9.6 Aim: To study the kinetics of potassium iodide - potassium persulphate reaction using initial rate method (or van’t Hoff differential method or ratio variation method). Requirements: Conical flasks, reagent bottles with lids, standard flasks, pipettes, burette, beakers, thermostat or plastic trough, stopwatch, 0.1 N potassium persulphate solution (equivalent mass = 135), 0.1 N potassium iodide solution (equivalent mass = 166), 0.01 N sodium thiosulphate solution (equivalent mass = 248 for pentahydrate), 1 N acetic acid, freshly prepared starch solution, crushed ice, distilled water. Theory : This reaction involves oxidation of iodide ions to iodine by persulphate ions. The persulphate ions oxidize iodide ions to iodine in preference to oxidizing thiosulphate ions since the latter are present in low concentration. The overall reaction can be expressed as 2I– + S2O82– o 2SO4 2– + I2 The reaction can be monitored by titrating the iodine produced, at different intervals of time, with sodium thiosulphate solution using starch as the indicator. The initial slope of a plot between the volume of thiosulphate solution (titre value) and the time gives the initial rate of the reaction (volume of thiosulphate is directly proportional to the volume of iodine which, in turn, is directly proportional to the concentration of iodide). The rate law is given as dx = k [I–]m [S2O82–]n ...(1) dt dx where is the rate of the reaction, and, ‘m’ and ‘n’ are the orders of the reaction with dt respect to iodide and persulphate respectively. If two reactions are carried out in which the concentration of potassium persulphate is kept constant and the concentration of potassium iodide is varied by fixed ratio, the order with respect to iodide is determined. Similarly, if two other reactions are performed by keeping the concentration of potassium iodide constant and varying that of potassium persulphate, the order with respect to potassium persulphate is determined. For example, if [S2O82–] = 0.025 M in two reactions and, [I–] = 0.05 M in the first reaction and [I–] = 0.025 M in the second reaction, then (rate)1 = k u (0.05)m u (0.025)n ...(2) and (rate)2 = k u (0.025)m u (0.025)n ...(3) m

(rate)1 ⎛ 0.05 ⎞ = = 2m (rate)2 ⎜⎝ 0.025 ⎟⎠

...(4)

⎡ (rate)1 ⎤ log ⎢ ...(5) ⎥ = m log 2 ⎣ (rate 2 ) ⎦ The order with respect to potassium iodide, ‘m’, can be calculated using equation (5). Similarly, the order with respect to potassium persulphate also can be found out. The total order of the reaction is (m + n).

120ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

For potassium iodide, the molar mass and equivalent mass are identical and so, molarity = normality of the solution. But for potassium persulphate, molar mass is  ’ŒŽȱ’œȱŽšž’ŸŠ•Ž—ȱ–ŠœœȱŠ—ȱ‘Ž›Ž˜›Žǰȱ–˜•Š›’¢ȱƽȱǻ—˜›–Š•’¢ȺȦȺŘǼȱ˜ȱ‘Žȱœ˜•ž’˜—ǯ Procedure: 1. Standardize the sodium thiosulphate solution by titrating it with standard potassium dichromate in presence of acidified potassium iodide solution using starch as the indicator. 2. Take three clean and dry reagent bottles with stoppers and number them. 3. Measure exactly 5 mL of 1 N acetic acid, 50 mL of 0.1 N potassium iodide solution and 20 mL of distilled water in bottle no. 1 (Table 9.5), shake well and keep it in a thermostat or a water trough. Note down the room temperature. 4. Place the potassium persulphate solution (0.1 N) also in the thermostat or a water trough. After about 30 minutes (when the solutions attained the thermal equilibrium), pipette out 25 mL of potassium persulphate solution into reagent bottle no.1 containing the potassium iodide solution. Start the stopwatch when half of the persulphate solution is transferred to the bottle. 5. Shake the mixture thoroughly and immediately pipette out 5 mL of the aliquot into a conical flask containing about 20 mL of ice cold water to quench the reaction. Quickly titrate the liberated iodine with the standardized thiosulphate solution using starch as the indicator. Continue this at every 5 minutes for at least 30 minutes. 6. Repeat the procedure for other two reaction sets in bottles no. 2 and 3 (Table 9.5). Observation: Room temperature = ........ °C Table 9.5: Data recording Bottle no.

Volume of 1 N acetic acid (mL)

Volume of 0.1 N KI solution (mL)

Concentration Volume of 0.1 N Concentration of KI solution K2S2O8 solution of K2S2O8 (M) (mL) solution (M)

1.

5

50

0.05

25

0.0125

20

2.

5

25

0.025

12.5

0.00625

57.5

3.

5

25

0.025

25

0.0125

45

Volume of water (mL)

Table 9.6: Data recording S. No.

Time (min)

Burette reading Initial

1. 2. 3. 4. 5. 6.

5 10 15 20 25 30

Final

Volume of thiosulphate solution (mL)

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ121

Calculation: Plot the graphs of volume of thiosulphate solution versus time for all the three reaction sets. Determine the slope at the initial stage of the reaction (at 5th or 10th minute) for each of these three reactions and this will give the respective rates. Therefore, ⎛ dx ⎞ m n ⎜⎝ dt ⎟⎠ = (rate)1 = k u (0.05) u (0.0125) 1

...(6)

⎛ dx ⎞ m n ⎜⎝ dt ⎟⎠ = (rate)2 = k u (0.025) u (0.00625) 2

...(7)

⎛ dx ⎞ m n ...(8) ⎜⎝ dt ⎟⎠ = (rate)3 = k u (0.025) u (0.0125) 3 Dividing the equation (6) by equation (8), the order with respect to potassium iodide, ‘m’, can be calculated. Similarly, dividing the equation (7) by equation (8), the order with respect to potassium persulphate, ‘n’, can be calculated. Result: The order with respect to (i) potassium iodide = ........ (ii) potassium persulphate = ........ The total order = ........

EXPERIMENT: 9.7 Aim: To study the kinetics of the reaction between acidified propanone (acetone) and iodine colorimetrically (or spectrophotometrically). Requirements: Colorimeter, cuvettes (cells), graduated pipettes, beakers, standard flasks, stopwatch, 0.5 M hydrochloric acid solution, 0.5 M propanone solution, 0.01 M iodine solution, de-ionised water or distilled water. Theory: Solid iodine (I2) crystals have a dark shiny purple colour. An aqueous iodine solution is brown or yellow, depending on the concentration. When an aqueous iodine solution reacts with propanone in the presence of an acid, the brown or yellow colour slowly fades away as the iodine is consumed. The products of the reaction are monoiodoacetone and hydrogen iodide. The hydrogen ion is a catalyst for this reaction. The reaction is represented as CH 3 COCH 3 (aq) + I 2 (aq) + H + → CH 3 COCH 2 I(aq) + H + (aq ) + I – (aq) colourless

brown/yellow

colourless

colourless

The rate law for this reaction is of the form rate = k [(CH3)2CO]D [I2]E [H+]J The rate of this reaction will be followed by observing the decrease in the intensity of colour (solution’s colour fades over time). As the concentration of iodine decreases, ‘Žȱ’—Ž—œ’¢ȱ˜ȱ‹›˜ —ȺȦȺ¢Ž••˜ ȱŒ˜•˜ž›ȱ˜ȱ‘Žȱ›ŽŠŒ’—ȱœ˜•ž’˜—ȱŠ•œ˜ȱŽŒ›ŽŠœŽœǯȱ‘ŽȱŒ‘Š—Žȱ in the intensity of colour allows the use of colorimetry to study the kinetics of the reaction. For dilute solutions, absorbance is proportional to the concentration and the

122ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

change in absorbance gives a measure of the decrease in the concentration of substance. The rate of disappearance of iodine is given in terms of decrease in absorbance which is equivalent to the actual decrease in its concentration. This method is simpler than the time-consuming conversion to concentration units using a calibration curve. The initial rate method can be used to determine the value of the rate constant, k, and the orders of the reaction with respect to I2 (the value of D), H+ (the value of E), and (CH3)2CO (the value of J). Four runs will be performed using different concentrations of reactants. For each run, an absorbance versus time plot will be drawn. The slope of each plot can be used to find the initial rate of the reaction for that particular run. Order of reaction with respect to each reactant is determined by separately doubling the concentration of each of CH3COCH3(aq), H+(aq) and I2(aq). A first order reaction with respect to CH3COCH3(aq) is confirmed when, while keeping the concentrations of I2(aq) and H+(aq) constant, the initial rate of decrease in concentration of I2(aq) (or the initial rate of decrease in absorbance) doubles as the concentration of CH3COCH3(aq) is doubled. Same is the case with other reactants. Procedure: 1. Switch on the colorimeter at least 30 minutes before the start of the experiment. 2. Set the blue filter (O = 450 – 460 nm) since the aqueous solution used here is yellow in colour. 3. Otherwise find the value of the wavelength at which the solution has maximum absorption (Omax) and set the corresponding filter. 4. Note down the room temperature. 5. Prepare four different reaction mixtures in numbered 10 mL standard flasks, as given in Table 9.7. Always add iodine at last. Shake and mix the solutions well. Table 9.7: Data recording S. No.

Volume of CH3COCH3 solution (mL)

Volume of I2 solution (mL)

Volume of HCl solution (mL)

Volume of distilled water (mL)

1.

1.0

1.0

5.0

3.0

2.

1.0

2.0

5.0

2.0

3.

2.0

2.0

5.0

1.0

4.

2.0

2.0

2.5

3.5

6. Take the reaction mixture one by one in the colorimetric cell and measure the absorbance every 2 minutes till the absorbance value becomes 10% of the initial value. 7. Record the data for all four sets as given in Table 9.8. 8. Plot a graph between absorbance and time for each reaction mixture.

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ123

Observation: Table 9.8: Absorbance values with time Time (min) Absorbance

Calculation: Using the plotted graph between absorbance and time for each reaction mixture, calculate the rate of the reaction, at the initial stage (between 10% and 20% of the reaction), from the slope of the graph. For reaction mixtures 1 and 2, applying the rate equation given under ‘Theory’, we get rate2ȺȦȺ›ŠŽ1 = [I2]2EȺȦȺǽ2]1E Since the concentrations of CH3COCH3 and HCl are the same for these two reactions, they cancel each other. The concentration of I2 in reaction 2 is double that of I2 in reaction 1. Therefore, rate2ȺȦȺ›ŠŽ1 = 2EȺȦȺŗE = 2E log (rate2ȺȦȺ›ŠŽ1) = E log 2 log (rate 2 / rate1 ) E= log 2 So, the order, E, with respect to I2 can be calculated. Similarly, from the ratios of the initial rates of reactions 2 and 3, the order with respect to acetone (D) and using the rates of the reactions 3 and 4, the order with respect to HCl (J) can be determined. The total order of the reaction is the sum of the individual orders. Calculate the concentrations of the reactants from the respective volumes. Substituting the individual orders and the values of concentration in the rate equation, the rate constant, k, can be evaluated. Result: At ........ °C: 1. The order with respect to propanone (acetone) = ........ 2. The order with respect to iodine = ........ 3. The order with respect to acid = ........ 4. Total order of the reaction = ........ 5. The rate constant of the reaction = ........ Interpretation: Write based on your result. Note: 1. Deduce the experimental rate equation for the reaction. 2. Based on the individual orders, propose a possible mechanism of the reaction. 3. The order of the reaction with respect to acetone and acid is supposed to be one and that with respect to iodine is supposed to be zero. Verify with your results. 4. The reaction can also be studied by following the progress of the reaction volumetrically.

124ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 9.8 Aim: To study the kinetics of interaction of crystal violet with NaOH colorimetrically. Requirements: Spectrophotometer or colorimeter, standard flasks, cuvettes, test tubes, stopwatch, crystal violet solid, 0.1 M and 0.05 M solutions of NaOH, oxalic acid, phenolphthalein indicator, distilled water. Theory: Chemical kinetics is the study of reaction rates. In this experiment, the kinetics of the reaction between crystal violet and NaOH will be studied. A spectrophotometer or colorimeter can be used to monitor the crystal violet concentration as a function of time. The structures of the reactant and the product, and the reaction stoichiometry are shown in Fig. 9.1.

..

(CH3)2N C

+

H H C C C H H C C

C C H H C

..

(CH3)2N C

C C C H H

H H C C

C N(CH3)2

+

OH –

C C C H H (Crystal violet)

..

H H C C

(CH3)2N C

C OH

C C H H

C

..

(CH3)2N C

H H C C C C C H H

H H C C

..

C N(CH3)2

C C C H H

Fig. 9.1: Reaction between crystal violet and NaOH

All the reactants and products are colourless except for crystal violet which has an intense violet colour. Thus, during the course of the reaction, the colour of the reaction mixture becomes less and less intense, ultimately becoming colourless when all the crystal violet has been consumed. The crystal violet colour is due to the extensive system of alternating single and double bonds which extends over all the three benzene rings and central carbon atom. This alternation of double and single bonding is termed conjugation and molecules which have extensive conjugation are usually highly coloured. In the product, the conjugation is lost and the three rings are no longer in conjugation with each other, and hence the product is colourless. The reaction is, therefore, followed by measuring the absorbance of the reaction mixture at successive intervals of time. The absorbance is directly proportional to the intensity of colour. The concentration of crystal violet and

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ125

the intensity of colour decrease as the crystal violet - NaOH reaction proceeds and absorbance will also decrease. The rate of the crystal violet - NaOH reaction is given by the following generalized rate law. Rate = k [OH–]x [CV]y ...(1) where ‘k’ is the rate constant of the reaction, CV is an abbreviation for crystal violet (C25H30N3+), ‘x’ is the reaction order with respect to OH–, and ‘y’ is the reaction order with respect to CV. The values of ‘x’ and ‘y’ will be determined experimentally. In this experiment, the initial concentration of OH–, is made much greater than the initial concentration of CV. Thus, the [OH–] change, during the time that CV is consumed, is negligible. For this reason [OH–]x can be treated as a constant, and equation (1) can be written as Rate = k’ [CV]y ...(2) where k’ = k[OH–]x; k’ is termed pseudo rate constant. The integrated form of pseudo rate law (2) depends on reaction order with respect to CV. The integrated rate laws for y = 1 (1st order) and 2 (2nd order) are given by equations (4) and (5). ln [CV]t = –k’t + ln[CV]0 ...(3) log [CV]tȱƽȱȮ”ȂȺȦȺŘǯřŖřȱƸȱ•˜ȱǽǾ0 ...(4) ȱ ŗȺȦȺǽǾtȱƽȱ”ȂȱȮȱŗȺȦȺǽǾ0 ...(5) where [CV]0 is concentration of crystal violet in the reaction mixture at time zero, before any reaction has occurred, and [CV]t is concentration at any time ‘t’ during the course of reaction. If a plot of log [CV]t versus time is linear, y = 1 and reaction is of first ˜›Ž›ǯȱȱ•’—ŽŠ›ȱ™•˜ȱ˜ȱŗȺȦȺǽǾt versus time indicates a second order reaction in CV. It is established that the reaction is first order with respect to crystal violet and first order with respect to NaOH. Crystal violet solution obeys Lambert Beer’s law : A = Hlc where A is the absorbance of the solution, H is the molar absorptivity of the substance, ‘l’ is the path length, i.e., the length of the cell through which light passes and ‘c’ is the concentration of the solution. The path length is normally 1 cm. For crystal violet, H = 5.0 u 104 M–1cm–1. Using this and the measured absorbance at different time intervals, [CV]tȹǰȱ ‘Žȱ concentration of crystal violet at various stages of the reaction can be calculated. Procedure: 1. Prepare a stock solution of 1.5 u 10–3 M crystal violet by weighing exactly 0.0612 g of the solid and dissolve it in a little distilled water. Make the solution up to 100 mL with distilled water. Take 1 mL of this solution and dilute it to 100 mL with distilled water. This is 1.5 u 10–5 M solution of crystal violet. Use this for the experiment. 2. Standardize the prepared NaOH solutions (0.1 M and 0.05 M) using a standard solution of oxalic acid and phenolphthalein indicator.

126ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

3. Note down the room temperature. 4. Set the colorimeter at 600 nm wavelength. 5. Take 1 mL of 0.1 M NaOH solution and place it in a 10 mL standard flask and make up to the mark with crystal violet solution, and mix thoroughly. 6. Fill the cuvette with this reaction mixture and immediately start measuring the absorbance every two minutes for 30 minutes (the intensity of colour decreases with time and the solution becomes colourless indicating that the reaction is complete). 7. Then remove the cuvette and discard the solution. 8. Repeat the above procedure with 0.05 M NaOH and crystal violet solutions. 9. Plot (i) log [CV]tȱŸŽ›œžœȱ’–ŽǰȱŠ—ȱǻ’’ǼȱŗȺȦȺǽǾt versus time, for both the sets and determine the order with respect to crystal violet. 10. From the appropriate graph, determine the rate constants for both the sets. 11. From the ratio of the two ‘k’ values, find out the order with respect to NaOH. 12. Estimate the true rate constant from each of the ‘k’ values. 13. Calculate the half-life time of the reaction. Observation: Room temperature = ........ °C Table 9.9: Set I S. No.

[NaOH] = 0.1 M; Data recording Time (min)

Table 9.10: Set II S. No.

Absorbance

[CV]

1 / [CV]

ln[CV]

log[CV]

1 / [CV]

ln[CV]

log[CV]

[NaOH] = 0.05 M; Data recording

Time (min)

Absorbance

[CV]

‘Ž–’ŒŠ•ȱ ’—Ž’Œœȳ127

Calculation: A linear plot of log [CV]t versus time indicates that the reaction is first order with respect to CV. Show that the plot for second order equation is not linear. Rate constant for Set I = k’I Rate constant for Set II = k’II k’IȺȦȺ”ȂII = k[OH–]xIȺȦȺ”ǽ –]xII = [OH–]xIȺȦȺǽ –]xII Substitute the values of k’s and the concentrations of NaOH in the reaction mixture for Sets I and II, and evaluate the value of ‘x’, the order with respect to NaOH. – k’I = k[OHȹ ]xI Substitute the values of k’ and the concentration of NaOH for Set I and calculate the value of ‘k’, the true rate constant of the reaction. Verify using the data for Set II. Determine the half-life time of the reaction using the first order equation tŗȺȦȺŘȱƽȱŖǯŜşřȺȦȺ”ǯ Result: At ........ °C: 1. The reaction is ........ order with respect to crystal violet and ........ order with respect to NaOH. 2. The overall order of the reaction is ......... 3. The value of the true rate constant = ........ 4. The half-life time, tŗȺȦȺŘ = ........ Interpretation: Interpret the result based on the theory already discussed.

VIVA QUESTIONS 1. 2. ȱ řǯȱ 4. 5. 6. 7. 8. 9. 10.

Define chemical kinetics. Why do different reactions have different rates? ’ŸŽȱŽ¡Š–™•Žȱ˜ȱœ•˜ ǰȱŠœȱŠ—ȱ’—Ž›–Ž’ŠŽȱ¢™Žœȱ˜ȱ›ŽŠŒ’˜—ǯ What is the difference between order and molecularity of a reaction? What are the different methods for finding the order of a reaction? Describe various methods to determine the rate of reaction. Mention the methods to stop the reaction at a given time. Why do you use ice cold water in acid hydrolysis of a given ester? How do you find the value of Vλ in various reactions? Write the differential and integrated rate expressions for the following reactions: (a) Acid hydrolysis of ester (b) Saponification of ester (c) Reaction between potassium iodide and potassium persulphate (d) Iodination of acetone ȱ ŗŗǯȱ ’ŸŽȱŽ¡Š–™•Žœȱ˜ȱ£Ž›˜ȱ˜›Ž›ǰȱ’›œȱ˜›Ž›ǰȱœŽŒ˜—ȱ˜›Ž›ǰȱ‘’›ȱ˜›Ž›ȱŠ—ȱ›ŠŒ’˜—Š•ȱ order reactions. 12. Write the units for rate constant for zero, first , second and third order reactions. 13. Why is the reaction with order and molecularity greater than three not known? 14. Why should we measure the rate only at the initial stage, and not at any given time, in the determination of the order of the reaction?

10

COLORIMETRIC AND SPECTROPHOTOMETRIC MEASUREMENTS

PRINCIPLE OF PHOTOMETRY Photometric methods provide a simple means for determining minute quantities of constituents. The variation in the intensity of colours of the system with variation in Œ˜—ŒŽ—›Š’˜—ȱ ˜ȱ œ˜–Žȱ Œ˜–™˜—Ž—ȱ ’œȱ ‘Žȱ ‹Šœ’œȱ ˜ȱ ‘’œȱ Š—Š•¢œ’œǯȱ Ž—Ž›Š••¢ǰȱ Œ˜•˜ž›ȱ ’œȱ developed due to the formation of a coloured compound when an appropriate reagent is added. Colorimetric method involves determining the concentration of a substance by measuring its relative absorption of light with respect to a known concentration of the same substance. Therefore, this is also regarded as absorption photometry in the visible region of light. The instruments normally used for this purpose are filter photometer and spectrophotometer. Filter photometer, which is also known as photoelectric colorimeter, is employed with light contained within a narrow range of wavelength (O) provided by passing white light through filters. Filters are materials in the form of plates of gelatin, glass, etc., and these transmit light of only a limited spectral region. In spectrophotometer, light of definite wavelength (O) extending to the ultraviolet region of the spectrum constitutes the source of light. Here, the photoelectric cells are used to measure the light transmitted from the solution. These analyses are governed by Lambert’s law and Beer’s law. The Lambert’s law states that when monochromatic light passes through a transparent medium, the rate of decrease in intensity of the incident light with the thickness of the medium, is proportional to the intensity of light. Beer’s law states that the intensity of a beam of monochromatic light decreases exponentially as the concentration of absorbing substance increases arithmetically. Combining both the laws, log (I0Ⱥ ȦȺt) = Hlc where Io and It are the intensities of incident and transmitted lights, ‘c’ is the concentration in mol L–1, ‘l’ is the thickness of the medium or length of cell in cm and ‘H’ is a constant called the molar absorptivity. log (I0Ⱥ ȦȺt) is known as absorbance (A) of the medium.

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ129

A = Hlc or A D c since for a particular substance ‘H’ is a constant and ‘l’ is a constant if the same cuvette is used for solutions of different concentrations. This is called the Beer-Lambert law or Lambert Beer’s law and is the fundamental equation used in colorimetry. Photoelectric colorimeters or spectrophotometers are used for the measurement of concentrations of coloured solutions. These instruments have either a set of filters to be chosen visually or in-built filters. A number of companies manufacture these instruments and they provide the instruction manuals for use of their respective instruments. In the following experiments, spectrophotometer can also be used in place of colorimeter.

EXPERIMENT: 10.1 Aim: To verify Lambert Beer’s law for CuSO4, KMnO4 and K2Cr2O7 solutions using colorimeter and measure their concentrations in the given solutions. Requirements: Colorimeter, cuvettes (cells), graduated pipettes, standard flasks, 0.004 M CuSO4 solution, 0.001 M KMnO4 solution in water and in 1 M sulphuric acid, 0.001 M K2Cr2O7 solution in water and in 1 M sulphuric acid, 1 M sulphuric acid solution, distilled water. Theory: The absorption of photons of light is described by the Lambert Beer’s law. It is a relationship that indicates a decrease in intensity as a beam passes through a medium that can absorb it. Consider a parallel beam of monochromatic light of initial intensity, I0, passing through a homogeneous absorbing medium (Fig. 10.1). It

Io l

Cuvette

Fig. 10.1: Representation showing the passage of incident light of intensity, Io, through a medium in a cuvette of light path, ‘l’. The transmitted light has an intensity, It.

According to this law, when a beam of monochromatic light passes through a solution, the decrease in the intensity of light transmitted is directly proportional to the concentration of the solution and the path length (l) of the solution. Mathematically, it is expressed as log (IoȺ ȦȺt) = Hcl ...(1) where Io and It are the intensities of incident and transmitted lights, ‘c’ is the concentration in mol L–1, ‘l’ is the thickness of the medium or length of cell (path length) in cm and ‘H’ is a constant called the molar absorptivity of the substance. The ratio, log (IoȺ ȦȺt) is known as absorbance (A) of the medium. The ratio (ItȺ ȦȺo) is called transmittance, T.

130ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Hence, A = log (IoȺȦȺt) = Hcl ...(2) and ADc ...(3) •œ˜ǰȱȱ ȱƽȱ•˜ȱǻŗȺȦȺǼȱ ȱ ǯǯǯǻŚǼ The log scale should be noted. An absorbance of 1 will have a transmittance of 0.1 and a % transmission of 10 but an absorbance of 2 will have the transmittance of 0.01 and the % transmission of 1. This Lambert Beer’s law is the fundamental equation used in colorimetry. Any coloured solution can be used to test the validity of this law, for example, copperammonia complex. The wavelength of maximum absorbance of such coloured soutions can be selected from Table 10.1. Table 10.1: Colours at various wavelengths S. No.

Colour of the solution

Complimentary colour (colour of the filter)

Wavelength (nm) of the transmitted colour

1.

Yellowish-green

Violet

400–420

2.

Orange

Indigo

420–440

3.

Yellow

Blue

440–500

4.

Purple

Green

500–570

5.

Blue

Yellow

570–585

6.

Greenish-blue

Orange

585–620

7.

Bluish-green

Red

620–780

A plot of absorbance versus concentration is a straight line passing through the origin (equation (2)). This is known as the calibration curve. From the slope, the molar absorptivity of the absorbing species in the solution can be determined. This plot is used to find out the concentration of the unknown solution using its absorbance. A typical calibration curve is shown for your reference in Fig. 10.2.

Absorbance

Standard 1 Standard 2 Standard 3 Standard 4 Concentration of unknown

Concentration (mol/L)

Fig. 10.2: Typical calibration curve for Lambert Beer’s law

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ131

The law may fail under the following circumstances: 1. The light used is not monochromatic in nature. 2. Stray lights reach the sensor of the colorimeter causing irregular effects. 3. There may be an equilibrium between different coloured substances involving the solute. The other coloured substances may be the products of hydrolysis, ’œœ˜Œ’Š’˜—ȺȦȺ’˜—’£Š’˜—ȱ˜›ȱŠœœ˜Œ’Š’˜—ȱ˜ȱ‘Žȱœ˜•žŽǯ However, even if the calibration plot is non-linear, it can be used for the estimation of the concentration of the given solution, by the interpolation method. Procedure: 1. Prepare a series of standard solutions of copper sulphate by pipetting out 1, 2, 3, 4, 5 and 6 mL of stock (0.004 M) solution in separate numbered 10 mL standard flasks and making the solution up to the mark with distilled water. 2. Shake the solutions well for proper mixing. 3. Switch on the colorimeter 30 minutes before starting the experiment so that the instrument gets stabilized, i.e., acquires stable temperature. ȱ Śǯȱ ’••ȱ‘ŽȱŒŽ••ȺȦȺŒžŸŽŽȱ ’‘ȱ ŠŽ›ȱŠ—ȱœŽȱ‘ŽȱŒŠ•’‹›Š’˜—ȱ ’‘ȱŠ‹œ˜›‹Š—ŒŽȱŠȱ£Ž›˜ȱ or transmittance at 100%. 5. Now fill the cell, after rinsing, with any one of the standard solutions of copper sulphate and find the absorbance at different wavelengths, i.e., with each of the filters. 6. At each wavelength, adjust the colorimeter with water for 0% absorbance or 100% transmittance. 7. Determine the wavelength, Omax at which the absorbance is maximum. Use this wavelength for further work. 8. Measure the absorbance of each of the standard solutions of copper sulphate at the Omax value. It is advisable to start with the dilutest solution and go in the order of increasing concentration as this will minimize the error in the results. 9. Plot a graph of absorbance versus concentration in ppm (parts per million). This is called the calibration curve. The conversion of molarity to ppm is explained in Appendix J. 10. Measure the absorbance of the given solution and read out the corresponding concentration from the calibration curve. Observation: Table 10.2: Absorbance and concentration for CuSO4 solution S. No.

Volume of copper sulphate solution mL)

Volume of water (mL)

1.

0

10

2.

1

9

3.

2

8

Concentration of solution (ppm)

Absorbance

(Contd.)

132ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• (Contd.) 4.

3

7

5.

4

6

6.

5

5

7.

6

4

11. For KMnO4 and K2Cr2O7 solutions, this law can be checked in water as well as sulphuric acid as solvents. 12. Prepare a standard stock solution of 0.001 M KMnO4 in distilled water. Let this be solution A. From this stock solution prepare the standard solutions of varying concentrations as given in Table 10.3. Table 10.3: Absorbance and concentration for aqueous KMnO4 solution S. No.

Volume of solution A (mL)

Volume of water (mL)

1.

0

10

2.

1

9

3.

2

8

4.

3

7

5.

4

6

6.

5

5

7.

6

4

8.

7

3

9.

8

2

10.

10

0

Concentration of solution (ppm)

Absorbance

13. Determine the value of Omax with one of these solutions as described in the earlier part of the procedure. 14. Measure the absorbance of each of these solutions at this wavelength. 15. Draw a plot of absorbance versus concentration. This is the calibration curve. Now take the absorbance of the given unknown solution and find its concentration from the calibration curve. 16. Repeat the same procedure with 0.001 M KMnO4 solution in 1 M sulphuric acid (solution B). 17. Follow the same procedure for 0.001 M K2Cr2O7 solution in water (A) and in 1 M sulphuric acid (B). Result: Lambert Beer’s law is obeyed by the following solution(s): The law is not obeyed by the solution(s): Concentration of solutions of (i) CuSO4 = ........ (ii) KMnO4 = ........ (iii) K2Cr2O7 = ........

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ133

Interpretation: Examine the calibration curves. While potassium permanganate obeys the law both in water and in acid medium, an aqueous solution of potassium dichromate does not obey the law. It is because dichromate gets hydrolyzed in aqueous medium to yield chromate ions. This hydrolysis is prevented in acidic medium by adding an excess of sulphuric acid. Cr2O72– + H2O ֖ 2CrO42– + 2H+

EXPERIMENT: 10.2 Aim: To determine the dissociation constant of the given indicator (phenolphthalein) colorimetrically. Requirements: Colorimeter, cuvettes, pH meter, beakers, test tubes, 0.2 M sodium hydroxide solution (free from carbonate), 0.2 M boric acid, sodium carbonate, freshly prepared solution of phenolphthalein. Theory: Acid base indicators have different colours in the acidic and the alkaline solutions. The change in colour is due to an exchange of H+ ions between the two forms. If the acid form of the indicator is represented by HIn, its dissociation can be expressed as HIn ֖H+ + In– The dissociation constant of the acid form of indicator is given by [H + ][In – ] ...(1) [HIn] If the concentration of the indicator in a solution is C moles per litre and its degree of dissociation is D, then concentrations of various species are [HIn] = C(1 – D) [H+] = CD [In–] = CD The degree of dissociation can be varied not only by changing the concentration of the indicator but also by changing the concentration of the hydrogen ion, i.e., the pH of the solution. The pH can be adjusted using appropriate buffer solutions. The important factor in the use of indicators is that the colour changes are perceptible from a 10% dissociation to a 90% dissociation of indicator. This entire range is controlled by a change in pH of solution in the range of pKIn ± 1 (pKIn = –log KIn). Therefore, in this experiment, the buffers having the pH range of pKIn ± 1 must be used. When hydrogen ion concentration is controlled, we can write that KIn =

[H + ] × α C[for conc. of In – ] [H + ]α = (1 – α ) C[for the conc. of HIn] (1 – α ) Inverting equation (2), we get 1 1 1– α × = + K In α [H ] KIn =

...(2)

...(3)

134ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Taking logarithm of both sides of equation (3), it becomes α – log KIn = – log[H+ ] – log 1– α α i.e., pKIn = pH – log 1– α

...(4) ...(5)

α = pH – pKIn ...(6) 1– α The second term on the right is a constant. As the pH of the solution increases, D also increases. The determination of KIn involves adjusting the [H+] and measuring the absorbance of either In– or HIn, one or both of which may be coloured. If we have a filter which can absorb light of colour corresponding to only one form of indicator, colorimetric method can be performed. Procedure: 1. Switch on the colorimeter 30 minutes before starting the experiment so that the instrument gets stabilized, i.e., acquires stable temperature. 2. Prepare boric acid (H3BO3) - sodium hydroxide buffer solutions with different amounts of sodium hydroxide and boric acid in seven different 25 mL standard flasks (Table 10.5). 3. Prepare a saturated solution of sodium carbonate in a beaker. 4. Using the pH meter, measure the pH of these eight different solutions and record the values. 5. Label seven different test tubes and take 10 mL of the first buffer solution in the test tube No. 1 and second solution in test tube No. 2 and so on. 6. Let the eighth tube contain the saturated sodium carbonate solution. 7. Add two drops of indicator solution to each of the eight test tubes. 8. The saturated sodium carbonate solution has high enough pH to cause almost complete dissociation of phenolphthalein. 9. Maximum colour will be developed in test tube No. 8. 10. Using this solution, determine the wavelength for maximum absorbance (Omax). 11. At this Omax, measure the absorbance for all the eight solutions. Observation: log

Table 10.4: Absorbance of indicator in saturated Na2CO3 solution at different wavelengths S. No. 1. 2. 3. 4. 5. 6.

O (nm)

Absorbance

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ135 7. 8. -

Table 10.5: Data recording S. No.

0.2 M Boric acid (mL)

0.2 M NaOH (mL)

1.

10.0

0.0

2.

9.0

1.0

3.

8.0

2.0

4.

7.5

2.5

5.

7.0

3.0

6.

6.5

3.5

7.

6.0

4.0

8.

pH

Ai

Di = Ai / A8

log

α 1– α

pKIn= pH – log

α 1– α

Saturated Na2CO3 solution Mean pKIn = ........

Calculation: If C is the total concentration of phenolphthalein in each tube, the concentration of ionized phenolphthalein will be DC. In tube No. 8, D = 1. According to Lambert Beer’s law, the absorbance is directly proportional to the concentration of the absorbing species, i.e., AvC Therefore, DiC v Ai for each tube where i = 1, 2, 3, 4, 5, 6 and 7. C v A8 for tube No. 8 since D= 1. Hence, for any tube Ai   Di = A8 From these values of D and pH values, pKIn can be calculated as α pKIn = pH – log 1– α pKIn = – log KIn KIn = antilog (–pKIn) Result: Dissociation constant of given indicator = ........ α Note: A graph can also be plotted with pH on y-axis and log on the x-axis. The 1– α intercept will give the value of pKIn. This should be in reasonable agreement with the one calculated above.

136ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 10.3 Aim: To determine the amount of iron present in a given solution colorimetrically using the complex formation between Fe2+ and 1,10-phenanthroline. Requirements: Colorimeter, cuvettes, standard flasks, graduated pipettes, beakers, ferrous ammonium sulphate (Mohr’s salt), 1,10-phenanthroline monohydrate, sodium acetate, hydroxyl ammonium chloride, sulphuric acid, acetic acid, distilled water. (a) Stock iron solution: Weigh 0.7g of pure dry ferrous ammonium sulphate (molar mass = 392.13 g mol–1) and transfer into a 100 mL standard flask. Add about 10-20 mL of distilled water. Dissolve the solid completely before diluting to volume. Add 2.5 mL of dilute sulphuric acid into the flask carefully and then dilute the solution with distilled water up to the mark. This is the stock solution of 1000 ppm Fe2+. ȱ ǻ‹Ǽȱ ŗǰŗŖȬ™‘Ž—Š—‘›˜•’—Žȱ œ˜•ž’˜—DZȱ ŖǯŘśƖȱ ǻ–ȺȦȺǼȱ œ˜•ž’˜—ȱ ˜ȱ ‘Žȱ –˜—˜‘¢›ŠŽȱ ’—ȱ distilled water. (c) Acetic acid-sodium acetate buffer solution: Mix 100 mL each of 0.2 M acetic acid and 0.2 M sodium acetate solutions. (d) Hydroxylamine hydrochloride solution (10%): Dissolve 10 g solid in 100 mL of distilled water. Theory: Iron(II) reacts with 1,10-phenanthroline or o-Phen to form an orange-red complex [(C12H8N2)3Fe]2+.

N N

N Fe 2+ N

N N

Fig. 10.3: Fe2+— 1,10–phenanthroline complex

This property provides an excellent and sensitive method for the determination of iron in aqueous solution. The molar absorptivity (H) of this complex is 11,000 L mol–1cm–1 at the Omax = 508 nm (wavelength of maximum absorption). The intensity of the colour is independent of pH in the range 2–9 and is stable for a long period. Acetic acidsodium acetate buffer is used to maintain the pH in the range 4.5-5. If Fe(III) is present in the solution, then it can be reduced with hydroxylamine hydrochloride. A = Hlc where A is the absorbance of a substance at a specified wavelength O (nm), ‘l’ is the length of the light path through the sample (cm), ‘H’ is the molar absorptivity of absorbing species.

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ137

Procedure: 1. Switch on the colorimeter 30 minutes before starting the experiment so that the instrument gets stabilized, i.e., acquires stable temperature. 2. Pipette out 1, 2, 3, 4, 6, 8 and 10 mL of standard stock solution of Fe(II) into numbered (1-7) standard flasks of 50 mL capacity. 3. Add to each flask, 5 mL of acetic acid-sodium acetate buffer solution and 5 mL of phenanthroline solution. 4. Make the solution up to the mark with distilled water, mix well and allow to stand for 20 minutes so that the colour develops completely. 5. Find out the value of Omax and measure the absorbance of each solution, at this wavelength, in a colorimeter. 6. Draw the calibration curve with absorbance versus concentration in ppm. 7. Pipette out 5 mL of the solution to be analyzed into the eighth flask. 8. Add 5 mL of acetic acid-sodium acetate buffer solution and 5 mL of phenanthroline solution. 9. Make the solution up to the mark with distilled water, mix well and allow to stand for 20 minutes. Measure the absorbance of this solution at Omax. 10. Find concentration of unknown solution from its corresponding absorbance. Table 10.6: Preparation of solutions of complex Test tube no.

1

2

3

4

5

6

7

Ferrous ion solution (mL)

1

2

3

4

6

8

10

Acetic acid solution (mL)

5

5

5

5

5

5

5

Phenthroline solution (mL)

5

5

5

5

5

5

5

Water (mL)

39

38

37

36

34

32

30

Observation: Table 10.7: Determination of Omax Wavelength, O (nm)

Omax = ....... nm

Absorbance

138ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• Table 10.8: Measurement of absorbance of various mixtures Solution of test tube number

Absorbance

1. 2. 3. 4. 5. 6. 7.

Result: The amount of iron present in the given solution = ........ ppm. Note: If Fe(III) is present in the solution, then it can be reduced with hydroxylamine hydrochloride. Add 5 mL of 10% hydroxylamine hydrochloride solution, adjust the pH of the slightly acid solution to 3–6 with the buffer solution, add 5 mL of 1,10phenanthroline solution and dilute to 50 mL, mix well and measure the absorbance after 10 minutes.

EXPERIMENT: 10.4 Aim: To determine the composition of ferric ions-salicylic acid complex colorimetrically using Job’s method. Requirements: Colorimeter, cuvettes, standard flasks, graduated pipettes, beakers, 0.002 M HCl solution, 0.001 M ferric chloride hexahydrate (molar mass = 270) solution in 0.002 M HCl (this is a solution of 0.001 M Fe3+ ions), 0.001 M salicylic acid (molar mass = 138) in 0.002 M HCl, distilled water. Theory: The complex between ferric ions (Fe3+ ) and salicylic acid is quite stable in acid solution in pH range 2.6–2.8. Hydrochloric acid (0.002 M) is normally used to maintain this range . The phenolic ‘OH’ group and the –COOH group of salicylic acid remain unionized at this pH range. Since the complex is coloured, its composition can be determined by measuring its absorbance using the colorimeter. The two bonding species combine in integral numbers during the complex formation. xFe3+ + yC6H4 (OH) COOH ֖Complex According to Job’s method, when equimolar solutions of the reactants are mixed in different proportions of the complex, maximum amount at equilibrium is formed when the two reactants are present in the same mole ratio as required for the complex formation. If the reactants are colourless and the complex is coloured, the absorbance of the complex can be measured at various concentrations. But, if two solutions are coloured, then a filter that can absorb any one and not the other can be chosen. In other words, the wavelength at which only one of the solutions absorbs is selected. In this case, the complex formed is deep violet in colour and hence, appropriate wavelength is selected to determine the volume ratio of the reactants for which absorbance of the mixture is maximum.

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ139

Procedure: 1. Switch on the colorimeter 30 minutes before starting the experiment so that the instrument gets stabilized, i.e., acquires stable temperature. 2. Prepare solutions of different compositions using ferric ion and salicylic acid solutions in 10 mL standard flasks as given in Table 10.6. Table 10.9: Preparation of solutions of the complex Ferric ion solution (mL)

9

8

7

6

5

4

3

2

1

Salicylic acid solution (mL)

1

2

3

4

5

6

7

8

9

3. Take any one of the prepared solutions and determine the wavelength of maximum absorption (Omax). 4. Then measure the absorbance of each of these solutions at Omax. 5. Plot absorbance versus volume of ferric ions in the mixture. 6. Locate the maximum point on the curve. Observation: Table 10.10: Determination of Omax. Wavelength, O (nm)

S. No.

Absorbance

Omax = ........ nm

Table 10.11: Measurement of absorbance of various mixtures S. No.

Volume of ferric ion solution (mL)

Volume of salicylic acid solution (mL)

1.

9

1

2.

8

2

3.

7

3

4.

6

4

5.

5

5

6.

4

6

7.

3

7

8.

2

8

9.

1

9

Absorbance

140ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Result: The composition of ferric ions-salicylic acid complex is found to be ......... Note: 1. Instead of ferric chloride, ferric nitrate or ferric alum can also be used. In the case of ferric alum, 0.0005 M solution will yield 0.001 M solution of ferric ions. 2. By the same method, the composition of complex between ferric ions and sulphosalicylic acid (1:1) and between nickel ions and 1,10-phenanthroline (1:3) can be determined. 3. The data can also be used to find out the equilibrium constant, dissociation constant or stability constant of the complex.

EXPERIMENT: 10.5 Aim: To determine the concentration of chromium and manganese in a given mixture colorimetrically. Requirements: Colorimeter, cuvettes, standard flasks, graduated pipettes, burettes, potassium dichromate, potassium permanganate, distilled water. Theory: The individual concentration of chromium (species 1) and manganese (species 2) present in a mixture can be determined only if both of them obey Lambert Beer’s law over their concentration in the mixture. The absorbances of two solutes are additive at a given wavelength, provided there is no reaction between the two solutes, i.e., ...(1) At wavelength O1, AO = A1 + A2 1 ...(2) and at wavelength O2, AO = A1 + A2 2 where AO1 and AO2 are the measured absorbances of the mixture at two wavelengths O1 and O2 and A1 and A2 are the absorbances of the individual species 1 and 2 at respective wavelengths. Both the coloured species would absorb at all wavelengths in the visible range but to different extent. The Omax of the two species should not overlap. Therefore, the wavelengths are selected in such a way that species 1 absorbs strongly at one wavelength and the species 2 absorbs strongly at the other wavelength. The absorbance spectra of potassium dichromate and potassium permanganate solutions show that the wavelengths of maximum absorption (Omax) for dichromate and permanganate are 440 nm and 525 nm respectively. According to Lambert Beer’s law, absorbance is given by A = Hcl where ‘H’ is molar absorptivity at a particular wavelength, ‘c’ is the concentration expressed in mol L–1 and ‘l’ is the thickness of the medium in cm. Therefore, ...(3) at wavelength O1, AO1 = H11 c1l + H21 c2l ...(4) and at wavelength O2, AO2 = H12 c1l + H22 c2l where H11 and H12 are molar absorptivities of species 1 (chromium) at O1 and O2 respectively. Similarly, H21 and H22 are molar absorptivities of species 2 (manganese) at O1 and O2 respectively.

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ141

Normally, the length of the cell, i.e., the thickness of the medium is 1 cm. Solving these simultaneous equations (3) and (4), we get ...(5) c1 = (AO1 H22 – AO2 H21ǼȺȦȺǻH11 H22 – H12 H21) ...(6) c2 = (AO1 H12 – AO2 H11ǼȺȦȺǻH12 H21 – H11 H22) The values of the molar absorptivities H11, H12, H21 and H22 can be deduced from the plots of the absorbance against concentration of pure solutions of substances 1 and 2 (from the slope, ‘H’ (l = 1 cm), of the straight line passing through the origin). By measuring the absorbance of the mixture at wavelengths O1 and O2, the concentrations of the two components can be calculated. Procedure: 1. Switch on the colorimeter 30 minutes before starting the experiment so that the instrument gets stabilized, i.e., acquires stable temperature. 2. Prepare 0.006 M, 0.004 M, 0.002 M, 0.001 M and 0.0005 M solutions of potassium dichromate and potassium permanganate in 1 M sulphuric acid using appropriate stock solutions. 3. Using the solution of maximum concentration (0.006 M ), select the wavelength for maximum absorption (Omax). The values must be around 440 nm for dichromate and 525 nm for permanganate solutions. 4. Measure the absorbance of all the solutions of potassium dichromate at 440 nm and that of the solutions of potassium permanganate at 525 nm using 1 cm cell. 5. Plot the absorbance against concentration for each of the solutions at both the wavelengths and calculate the molar absorptivities H1 and H2 at both 440 nm and 525 nm. 6. Now, record the absorbance of the given mixture, containing unknown concentrations of potassium dichromate and potassium permanganate, at both the wavelengths. 7. Substitute the appropriate values of absorptivities and absorbances in equations (5) and (6) and determine the concentrations of chromium and manganese in the given mixture. 8. Prepare a series of mixtures of the solutions, using one of the solutions of dichromate and permanganate, as given in Table 10.12 and measure the absorbance of each of the mixtures at 440 nm or at 525 nm. 9. Calculate absorbance of the mixtures using the molar absorptivities determined above and the respective concentration (use equation 3 or 4). 10. Compare the observed and calculated absorbance values to prove that the additivity principle is obeyed.

142ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Observation: Table 10.12: Data recording S. No.

Volume of K2Cr2O7 solution (mL)

Volume of KMnO4 solution (mL)

1.

25.0

0.0

2.

22.5

2.5

3.

20.0

5.0

4.

17.5

7.5

5.

15.0

10.0

6.

12.5

12.5

7.

10.0

15.0

8.

7.5

17.5

9.

5.0

20.0

10.

0.0

25.0

Absorbance observed

Absorbance calculated

Result: 1. The values of observed and calculated absorbance at wavelength ........ nm show that the additivity principle is ......... 2. The concentration of chromium and manganese in the given mixture are found to be ........ and ........ respectively. Note: If the mixture is of complex form like an alloy steel or ferro alloy, then, in order to prevent the interference of other metals, add 0.7 M of phosphoric acid along with sulphuric acid.

EXPERIMENT: 10.6 Aim: To study the absorbance spectra of KMnO4 (in water) and K2Cr2O7 (in 0.1 M H2SO4) in the range 200–500 nm and to determine the Omax values. Also, to calculate the energies of these transitions in different units (J molecule–1, kJ mol–1, cm–1, eV). Requirements: Spectrophotometer, standard flasks, beakers, cuvettes, 0.001 M KMnO4 solution (in water) and 0.001 M K2Cr2O7 solution (in 0.1 M H2S04). Theory: A plot of absorbance versus wavelength is known as an absorbance spectrum. It is recorded to determine the optimal wavelength, i.e., the wavelength at which the absorbance is maximum, for analyzing a particular compound. This wavelength is denoted by the symbol Omax. The energy involved in the electronic transition can be calculated in various units using the value of Omax and appropriate inter-conversions. Energy in: 1. Wave number unit (cm–1ǼȱDZȱȱƽȱŗȺȦȺOmax (cm). 2. J molecule–1DZȱȱƽȱ‘ŒȺȦȺOmax ; ‘h’ in J.s, ‘c’ in ms–1, Omax in m. 3. J mol–1: E = hcNAȺȦȺOmax ; NA = Avogadro number. 4. kJ mol–1: E in J mol–1 u 10–3.

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ143

5. kcal mol–1: 4.184 kJ = 1 kcal; Therefore, E = E in kJ mol–1ȺȦȺŚǯŗŞŚǯ 6. eV : 1.6021 u 10–19 J = 1 eV ; Therefore, E = E in J molecule–1ȺȦȺŗǯŜŖŘŗȱu 10–19 7. eV mol–1: 98.484 kJ mol–1 = 1 eV mol–1; E = E in kJ mol–1ȺȦȺşŞǯŚŞŚǯ Procedure: 1. Switch on the spectrophotometer 30 minutes before starting the experiment so that the instrument gets stabilized, i.e., acquires stable temperature. 2. Take 0.001 M KMnO4 solution (in water) in a cuvette and record the absorbance spectrum in the wavelength range 200–500 nm using the spectrophotometer. If the instrument is of manual type, measure the absorbance at different wavelengths. 3. Plot the absorbance versus wavelength and get the absorption spectrum. 4. Find out the wavelength corresponding to the maximum absorbance and this is Omax. 5. Determine the energy of transition in different units using the equations given under ‘Theory’. 6. Repeat the same procedure with 0.001 M K2Cr2O7 solution (in 0.1 M H2SO4). Calculation: Using the measured Omax value, calculate the energy in various units for both KMnO4 and K2Cr2O7, by applying the equations 1–7. Result: The energies of the transitions in different units (J molecule–1, kJ mol–1, cm–1, eV) are: 1. For KMnO4 : 2. For K2Cr2O7: Interpretation: Analyze the spectra, compare the energies and give appropriate interpretation.

EXPERIMENT: 10.7 Aim: To study the pH-dependence of UV-Visible spectrum of potassium dichromate. Requirements: Spectrophotometer, standard flasks, graduated pipettes, potassium dichromate, potassium hydroxide solution (0.1 N), hydrochloric acid solution (0.1 N), distilled water. Theory: The effect of pH on chemical equilibria is quite interesting and well known. This is often used to influence the equilibrium processes. The potassium ’Œ‘›˜–ŠŽȺȦȺŒ‘›˜–ŠŽȱŽšž’•’‹›’ž–ȱ’œȱ›Ž–Š›”Š‹•¢ȱ™ ȱœŽ—œ’’ŸŽǯȱ‘ŽȱŽšž’•’‹›’ž–ȱ’—ȱŠŒ’ȱ solution forms mainly dichromate ions and the one in basic solution mainly forms chromate ions according to the equation Cr2O72– + H2O ֖ 2 CrO42– + 2H+ Both the intensity and the Omax will be different for dichromate and chromate ions in acidic and basic media. It is observed that the Omax values shift to the higher range in the basic medium compared to those in the acidic medium. Similarly, there is an increase in absorbance also for both the ions. This is of practical importance because potassium dichromate in acid solution is used as a standard to check the accuracy and the reliability of spectrophotometers.

144ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Procedure: 1. Switch on the spectrophotometer 30 minutes before starting the experiment so that the instrument gets stabilized, i.e., acquires stable temperature. 2. Prepare a stock solution of potassium dichromate containing 5 mg of solid in 100 mL of distilled water. 3. Prepare a solution by taking 8 mL of the stock solution in a 10 mL standard flask and making it up to the mark with 0.1 N hydrochloric acid solution. 4. Record the spectrum of this solution with distilled water as reference in the spectrophotometer in the wavelength range 200–500 nm. Follow the standard procedure. 5. Prepare a solution by taking 8 mL of the stock solution in a 10 mL standard flask and making it up to the mark with 0.1 N potassium hydroxide solution. 6. Record the spectrum as before. Observation: Table 10.13: Absorbance maxima of potassium dichromate Potassium dichromate in

Omax (1) (nm)

Omax (2) (nm)

Acid solution Basic solution

Result: Omax (1) and Omax(2) values in (i) acid solution = ........ (ii) basic solution = ........ Discussion: Take the spectra and analyze. Interpret the shift in wavelengths and intensities of the peaks (from your result), based on the structures of the anions and the intensities of colours.

EXPERIMENT: 10.8 Aim: To record the uv spectra (180–400 nm) of the given compounds—acetone, acetaldehyde, 2-propanol and acetic acid, all in cyclohexane and in water and to calculate the energy involved in the electronic transitions in the units of : wave no., kJ mol–1, kcal mol–1, and eV mol–1. Also, to comment on the effect of structure on the uv spectra of these compounds. Requirements: Spectrophotometer, cuvettes, standard flasks, beakers, graduated pipettes, acetone, acetaldehyde, 2-propanol and acetic acid, cyclohexane, distilled water. Theory: The measurement of absorption spectra of an organic molecule in solution is carried out on an uv and vacuum uv range of absorption spectra. Absorbance at 280 nm corresponds to n o S* transition and other at 180 nm corresponds to S o S* transition in solution phase. The n o S* transition corresponds to promotion of one of the non-bonding electrons on oxygen to high energy anti-bonding or S* orbital, ‘žœȱ Ž•˜ŒŠ•’£’—ȱ ’ȱ ˜ŸŽ›ȱ ǁȹȱ ƽȱ ȱ ›˜ž™ǰȱ ‘˜ž‘ȱ ‘Žȱ ›Š—œ’’˜—ȱ ’œȱ •Žœœȱ ’—Ž—œŽȱ Š—ȱ

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ145

symmetrically disallowed. The transition occurs at a longer wavelength compared to S o S* transition. The transition is symmetry - forbidden because the n-orbital lies in nodal plane (xy) of S* orbital plane (xz), whereas S and S* orbitals are in same plane (xz) and have considerable overlap; consequently S o S* transitions are allowed while n o S* transitions are forbidden. In acetone, the lowest unoccupied orbital of the carbonyl group is the anti-bonding or S* orbital of CO group. The electrons in the highest occupied orbitals are the lone pair of electrons on oxygen. This is accepted since non-bonding electrons are generally held more loosely than any other electron used in bonding. The excitation energy is for promotion of an electron from non-bonding py orbital to anti-bonding S* orbital. The intensity of an absorption band can be obtained experimentally by application of Beer’s law: Integrated Intensity = ∫ ε ν cl , where ε ν is the molar absorptivity at the wave number ν. ν1 and ν2 are the lower and higher wave number limits of absorption band under consideration. Effects of different solvents: The position and intensity of an absorption band may shift when the spectrum is recorded in different solvents. It is directly related to the degree of interaction of solvent and solute. Nonpolar solvents such as cyclohexane normally do not interact with solute either in the ground or in the excited state (because of the presence of weaker dipole - induced dipole forces) and the absorption spectrum of a compound in these solvents is similar to the one in pure gaseous state. However, polar solvents such as water and alcohol may stabilize or destabilize the molecular orbitals of a molecule either in the ground state or in the excited state and the spectrum may vary significantly from the one recorded in nonpolar solvent. Polar solutes, such as the compounds given above, can interact with polar solvents through stronger dipole-dipole forces. In the presence of hydrolytic solvents, the lone pair of electrons on carbonyl oxygen acts as an electron donor to hydrogen of solvent to form the H-Bond. The formation of H-bond lowers the energy of non-bonding orbital by an amount approximately equal to H-bond. In n o S* transition, which is responsible for longer wavelength, one of the electrons is removed from the n-orbital and promoted to an empty anti-bonding S* orbital and one ‘n’ electron is remaining on oxygen, which is not adequate to sustain H-bond. Hence, the H-bond must be broken in the process of formation and polar solvent should not affect the energy of excited state. The difference in the position of wavelength maxima, when solvent is changed from hexane to ethanol or water should be a measure of strength of H-bond in latter solvent. Procedure: 1. Switch on the spectrophotometer 30 minutes before starting the experiment so that the instrument gets stabilized, i.e., acquires stable temperature. 2. Make approximately 0.05 M acetone solution in water. 3. Fill water in one cuvette and set the absorbance to 0 % in the spectrophotometer. 4. Take the given solution of acetone in the cuvette. 5. Record the absorption spectrum in the wavelength range 180–400 nm.

146ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

6. If the instrument is of manual type, measure the absorbance and transmittance in the wavelength range 180–400 nm. 7. Plot a graph of absorbance versus wavelength. This is the uv spectrum for acetone. 8. The energy of transition, at the Omax, can be determined in various units using appropriate formulae given under Experiment 10.6. 9. Record the spectrum of the approximately 0.05 M solution of acetone in cyclohexane. Carry out the same procedure to record the spectra of acetaldehyde, 2-propanol and acetic acid. Observation: Table 10.14: Data recording Wavelength (nm)

Absorbance

Transmittance

Calculation: Using the measured Omax value, calculate the energy in various units by applying the equations given under Experiment 10.6, for acetone, acetaldehyde, 2-propanol and acetic acid, all in cyclohexane and in water. Result: ’ŸŽ—ȱ’—ȱ‘Žȱ˜••˜ ’—ȱŠ‹•ŽȱŗŖǯŗśǯ Table 10.15: The energies of the transitions in different units S. No.

Compound

In cyclohexane J molecule–1

1.

Acetone

2.

Acetaldehyde

3.

2-propanol

4.

Acetic acid

kJ mol

–1

cm

In water –1

eV

J molecule–1

kJ mol–1

cm–1

eV

When the solvent is changed from cyclohexane to water while recording the spectrum of acetone, absorptions due to n o S* transitions are shifted to shorter wavelength (from ........ to ........) while those due to S o S* transitions are shifted to longer wavelength (from ........ to ........). Interpretation: Analyze the spectra and comment on the effect of structure on the uv spectra of these compounds. Compare the energies and give appropriate interpretation. Note: 1. S o S* transition: In the case of S o S* transitions, the excited state is more polar than the ground state, and, dipole-dipole interactions and hydrogen bonding with solvent molecules lower the energy of excited state more than that of the

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ147

ground state. Therefore, a polar solvent decreases the energy of S o S* transition and such transitions are shifted to longer wavelength as solvation on dielectric Ž—Ÿ’›˜—–Ž—ȱ˜ȱŠ‹œ˜›‹’—ȱ–˜•ŽŒž•Žȱ’œȱ’—Œ›ŽŠœŽǯȱ‘Žȱœ‘’ȱ’œȱ‹Š‘˜Œ‘›˜–’ŒȺȦȺ›Žȱ shift. 2. n o S* transition: These transitions are affected by solvents such as water, which are capable of forming hydrogen bond. These associations involve non-bonding electrons of heteroatom, i.e., oxygen atom in acetone. Thus, involvement of non-bonding electrons in hydrogen bonding lowers the energy of n-orbital. Hence, the energy required for n o S* increases and such transitions Š›Žȱœ‘’Žȱ˜ȱ•˜ Ž›ȱ ŠŸŽ•Ž—‘ǯȱ‘Žȱœ‘’ȱ’œȱ‘¢™œ˜Œ‘›˜–’ŒȺȦȺ‹•žŽȱœ‘’ǯ

EXPERIMENT: 10.9 Aim:ȱ˜ȱŠ—Š•¢£Žȱ‘Žȱ’ŸŽ—ȱ›˜Š’˜—Š•ȺȦȺŸ’‹›Š’˜—Ȭ›˜Š’˜—ȱœ™ŽŒ›ž–ȱ˜ȱ •ȱǻǼȱŠ—ȱ˜ȱ determine the bond length of HCl. Theory: Spectrum is the record of intensity of light transmitted or scattered by a molecule as a function of wavelength (OǼǰȱ  ŠŸŽȱ —ž–‹Ž›ȱ ǻЌǼȱ ˜›ȱ ›ŽšžŽ—Œ¢ȱ ǻX). The rotational and vibration-rotation spectrum can be used to determine the bond length of a diatomic molecule. Application of the Schrodinger equation and the quantum mechanical calculations give the selection rule to predict the feasibility of transitions. According to this, for a diatomic molecule, only those rotational transitions are allowed in which the rotational quantum number, J, changes by unity. All other transitions are forbidden. This can be expressed as 'J = ±1. Therefore, we have to consider only the transitions J = 0 o 1, J = 1 o 2, J = 2 o 3 and so on. The energy of transition is derived as ȱ ȱ ЌJ oJ+1 = 2B(J + 1) cm–1 (J = 0, 1, 2, 3, . . . . .) ...(1) where B is the rotational constant of the molecule and is given by ...(2) ȱ ȱ ȱƽȱ‘ȺȦȺŞS2Ic Here, I is the moment of inertia of the molecule and ‘c’ is the velocity of light. For various transitions, the energies are 2B, 4B, 6B, ........ cm–1 and the spectrum will consist of lines at 2B, 4B, 6B, ........ cm–1. The separation between any two adjacent lines is 2B cm–1, i.e., all the lines are equally spaced. The distance between any two adjacent lines can be measured and ‘I’ can be found out. By definition, I = Pr2 ...(3) P = the reduced mass of the molecule and r = the bond length of the molecule. From equation (3) r = I/μ ...(4) Substituting the values of ‘I’ and ‘P’ in equation (4), the bond length ‘r’ can be evaluated. If vibration-rotation is considered, it can be utilized to determine all the parameters that can be determined from pure vibrational and pure rotational spectra. The selection rules are 'v = ±1, ±2, etc. 'J = ±1

148ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Normally, 'J = 0 is not allowed and so it is clear that a vibrational change must always be accompanied by a simultaneous rotational change. But 'J = 0 is allowed under rare circumstances, i.e., when the ground state or the excited state or both have orbital angular momentum along the internuclear axis. Such lines with 'J = 0 constitute Q branch of the spectrum. If both the states do not have the orbital angular momentum, then, 'J = ±1 only. The spectrum will consist of a band centre or band origin or Q branch corresponding to the vibrational transition and equally spaced lines (separation of 2B) on each side of Q branch (if Q branch is not possible, the gap of 4B is observed between the first lines of P and R branches). The lines on the low wave number side of the band origin correspond to 'J = –1 and are called P branch of the spectrum. The lines on the high wave number side of the band origin correspond to 'J = +1 and are termed R branch of the spectrum. By measuring the spacing on both P branch and R branch, average value of B can be determined and the bond length can be calculated following the method described above. Procedure: Measure the distance between any two adjacent lines in the spectrum, using a scale. To get more accurate result, measure four or five similar distances and take the mean value. For example, measure the distances between lines 1 & 2, 3 & 5, 7 & 8, etc. Find out the average. This will be in cm. Convert it into wave number unit of cm–1. It can be done by measuring the distance in cm between any two wave number values and equating the two (refer the sample spectrum).

Q-Gap

R

P

Q

3000

2900

2800

7.5 cm

Fig. 10.4: Vibration-rotation spectrum of HCl (g)

2700 cm

–1

˜•˜›’–Ž›’ŒȱŠ—ȱ™ŽŒ›˜™‘˜˜–Ž›’ŒȱŽŠœž›Ž–Ž—œȳ149

Calculation: Substitute the values of B in cm–1, h = 6.626 u 10–34 Js and c = 3 u 1010 cm.s–1 in equation (2), and calculate the value of I in the unit of kg.m2. Then, substituting I and P in kg in equation (4), determine the value of ‘r’ in m. Result: The bond length of HCl molecule = ........ m = ........ pm. Note: You can get spectra of different molecules such as CO, NO, etc., from standard Ž¡‹˜˜”œǰȱ’—Ž›—Žȱœ’Žœȱ˜›ȱ›˜–ȱ›Ž™žŽȱ’—œ’žŽœȺȦȺ’—œ’ž’˜—œȱŠ—ȱ›¢ȱ˜ȱŠ—Š•¢£Žȱ‘Ž–ȱ to have the practical knowledge of spectroscopy.

VIVA QUESTIONS 1. What is colorimetry? 2. What is the difference between visual and photoelectric colorimetry? 3. How is light radiation characterized in terms of wavelength, wave number and energy? ȱ Śǯȱ ’ŸŽȱ‘Žȱ›Ž•Š’˜—ȱ‹Ž ŽŽ—ȱ‘Žȱž—’œȯ—œ›˜–ǰȱ–Ž›ŽǰȱŒŽ—’–Ž›ŽȱŠ—ȱ—Š—˜–Ž›Žǯ 5. State Lambert Beer’s law. Mention the conditions under which Lambert Beer’s law is applicable. 6. What are the causes of deviations of above law? 7. What is a cuvette? Name the type of glass used to make it. 8. What are filters? Why are they used? ȱ şǯȱ ‘Šȱ’œȱ‘Žȱ›Ž•Š’˜—ȱ‹Ž ŽŽ—ȱ›Š—œ–’Š—ŒŽȱŠ—ȱŠ‹œ˜›‹Š—ŒŽȺȦȺ˜™’ŒŠ•ȱŽ—œ’¢ǵ 10. How do you plot a calibration curve? 11. What is the necessity to find Omax? 12. How is optical density related to the concentration of the solution? 13. What are the advantages of photoelectric or colorimetric determination? 14. What are the important criteria for a satisfactory colorimetric analysis? 15. How can you determine the concentration of unknown coloured and colourless solutions colorimetrically? 16. While Lambert Beer’s law is applicable to aqueous and acidified potassium permanganate solution, it is obeyed only by acidified potassium dichromate solution and not by aqueous solution. Justify. 17. Name the various colour developing solutions for iron, nickel, phosphate and copper metals. 18. What is a blank solution? What is its role in colorimetry? 19. What is the physical significance of molar absorptivity? 20. Why do we use ammonia in the colorimetric determination of copper? 21. Why are strong acids such as HCl and HNO3 added to ferric alum before adding KSCN solution? 22. Why is sulphuric acid not recommended in colorimetry? 23. Red colour appears when potassium thiocyanate is added to ferric alum. What is it due to? 24. How can you find the composition of a complex colorimetrically?

11

ADSORPTION AND DISTRIBUTION MEASUREMENTS

EXPERIMENT: 11.1 Aim: To verify the Freundlich and Langmuir isotherms for adsorption of acetic acid on activated charcoal. Requirements: Standard flasks, beakers, burettes, graduated pipettes, reagent bottles, funnels, conical flasks, 0.5 M acetic acid solution, 0.1 M sodium hydroxide solution, oxalic acid, phenolphthalein indicator, activated charcoal, pieces of glazed paper, distilled water. Theory: The phenomenon of attracting and retaining the molecules of a substance on the surface of a liquid or a solid resulting in a higher concentration of the molecules on the surface is called adsorption. The substance which gets adsorbed at the surface is known as adsorbate. The substance on which adsorption occurs is known as adsorbent. Adsorption is a surface phenomenon while absorption is a bulk phenomenon. In absorption, the molecules of one substance are passed from the surface into the bulk of the solid or liquid. For example, ink is absorbed by the blotting paper whereas gases are adsorbed by charcoal. When there is a doubt as to whether adsorption or absorption is taking place, the term ‘sorption’ is used. The amount of adsorbate adsorbed at the surface depends on the temperature and pressure. At constant temperature, there exists a specific relationship between the amount of substance adsorbed and its equilibrium concentration. Such a relationship is called adsorption isotherm. Various scientists have given different adsorption isotherms. Freundlich has given an empirical relation known as Freundlich adsorption isotherm : ȱ ȱ ȱ ¡ȺȦȺ–ȱƽȱ ȱȹǻŗȺȦȺ—Ǽȱȱ˜›ȱȹȹ¡ȺȦȺ–ȱƽȱ ȱȱǻŗȺȦȺ—Ǽ ...(1) where x is the amount of adsorbate (gas or liquid substance), in moles or grams, adsorbed by ‘m’ grams of adsorbent at the equilibrium concentration (C) or pressure ǻǼDzȱ ȱŠ—ȱŗȺȦȺ—ȱŠ›ŽȱŒ‘Š›ŠŒŽ›’œ’ŒȱŒ˜—œŠ—œȱ ‘˜œŽȱŸŠ•žŽœȱŽ™Ž—ȱ˜—ȱ‘ŽȱŽ–™Ž›Šž›Žȱ Š—ȱ—Šž›Žȱ˜ȱŠœ˜›‹ŠŽȬŠœ˜›‹Ž—ȱœ¢œŽ–ǯȱ‘ŽȱŸŠ•žŽȱ˜ȱŗȺȦȺ—ȱ•’Žœȱ‹Ž ŽŽ—ȱŖȱŠ—ȱŗǯȱ The logarithmic form of equation (1) is ȱ ȱ ȱ •˜ȱǻ¡ȺȦȺ–Ǽȱƽȱ•˜ȱ ȱƸȱŗȺȦȺ—ȱ•˜ȹȱȺȺ˜›ȺȺȱ•˜ȱǻ¡ȺȦȺ–Ǽȱƽȱ•˜ȱ ȱƸȱŗȺȦȺ—ȱ•˜ȱȱ ǯǯǯǻŘǼ

œ˜›™’˜—ȱŠ—ȱ’œ›’‹ž’˜—ȱŽŠœž›Ž–Ž—œȳ151

ȱ Žȱ™•˜ȱŠȱ›Š™‘ȱ‹Ž ŽŽ—ȱ•˜ȱǻ¡ȺȦȺ–ǼȱŸŽ›œžœȱ•˜ȱȱ˜›ȱ•˜ȱǰȱŠȱœ›Š’‘ȱ•’—Žȱ ’••ȱ‹Žȱ ˜‹Š’—Žǯȱ›˜–ȱ‘Žȱœ•˜™ŽȱŠ—ȱ’—Ž›ŒŽ™ǰȱ‘ŽȱŸŠ•žŽȱ˜ȱǻŗȺȦȺ—ǼȱŠ—ȱ ȱŒŠ—ȱ‹ŽȱŒŠ•Œž•ŠŽǯ Langmuir derived another isotherm: ȱ ȱ ¡ȺȦȺ–ȱƽȱǻŠȱȺȦȺǻŗƸ‹ǼǼȱȱ ǯǯǯǻřǼ where ‘a’ and ‘b’ are Langmuir constants. Equation (3) can be modified as ȱ ȱ ȺȦȺǻ¡ȺȦȺ–ǼȱƽȱǻŗȺȦȺŠǼȱƸȱǻ‹ȺȦȺŠǼȱȱȱ ǯǯǯǻŚǼ ȱ™•˜ȱ˜ȱȺȦȺǻ¡ȺȦȺ–ǼȱŸŽ›œžœȱȱ ˜ž•ȱ‹ŽȱŠȱœ›Š’‘ȱ•’—Žȱ ’‘ȱœ•˜™ŽȱŽšžŠ•ȱ˜ȱǻ‹ȺȦȺŠǼȱŠ—ȱ ’—Ž›ŒŽ™ȱ ŽšžŠ•ȱ ˜ȱ ǻŗȺȦȺŠǼǯȱ ‘žœǰȱ ‘Žȱ ŸŠ•žŽœȱ ˜ȱ Œ˜—œŠ—œȱ ȁŠȂȱ Š—ȱ ȁ‹Ȃȱ ŒŠ—ȱ ‹Žȱ ˜‹Š’—Žȱ experimentally. In the case of a gas adsorbate, the concentration, C can be replaced by the pressure, P. Typical plots of these two isotherms for the adsorption of a gas are shown below:

1 Slope = n

b Slope = a

log x/m

b a

P x/m Intercept = log K

Intercept =

1 a

P

log P

Fig. 11.1: Freundlich adsorption isotherm

Fig. 11.2: Langmuir adsorption isotherm

For liquid-solid interface, the pressure is replaced by the equilibrium concentration of the adsorbate, Ce. Procedure: 1. Standardize the solutions of acetic acid and sodium hydroxide by usual procedure. 2. Take six dried and cleaned 100 mL reagent bottles and number them. 3. Set up two burettes, one for distilled water and other for 0.5 M acetic acid solution. 4. Now prepare the following mixtures in various reagent bottles and mix them well. Table 11.1: Composition of various mixtures 1

2

3

4

5

6

Acetic acid solution (mL)

Bottle No.

50

40

30

20

10

0

Distilled water (mL)

0

10

20

30

40

50

5. To each flask add 2 g of powdered activated charcoal. 6. Place them in a thermostat at room temperature for about 30 minutes. Shake the contents from time to time.

152ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

7. Calculate the initial concentration of the acid for the five flasks (from the original concentration) and correct them, if necessary, using the titre value for the sixth (blank) reagent bottle. 8. Set up six funnels with dry filter papers of equal size over six conical flasks and filter the solutions in separate conical flasks i.e., one in each flask. 9. Reject about 5 mL of initial filtrate in each case and collect the remaining filtrates in separate conical flasks. 10. Titrate 10 mL each of the solutions against standardized sodium hydroxide solution. 11. Tabulate the data as shown below. Observation: Room temperature = ........ °C Mass of charcoal = 2 g Concentration of acetic acid = a M Table 11.2: Data recording Bottle Volume of NaOH Initial Concentration of Amount of (x / m) log No. required for concentration acetic acid after acid adsorbed, (mol g–1) (x / m) of acetic acid, adsorption, Ce x = (Ci – Ce) / 20 10 mL of the solution (mL) Ci (mol L–1) (mol L–1) (moles)

log Ce

1. 2. 3. 4. 5. 6.

Calculation: 1. First calculate the initial concentration (Ci) of acetic acid in mol L–1 for each flask. For example, for bottle no. 1, CiȱƽȱŠȱǰȱŠ—ȱ˜›ȱ‹˜•Žȱ—˜ǯȱŘǰȱǻŚŖȱ¡ȱŠǼȺȦȺśŖȱȱŠ—ȱœ˜ȱ˜—ǯ 2. From the titration data, obtain the value of CŽȹ, the concentration of acetic acid after adsorption, i.e., at equilibrium (using V1N1 = V2N2 and normality = molarity for both acetic acid and sodium hydroxide). 3. Calculate the amount adsorbed as follows: Total volume of the solution = 50 mL Change in concentration due to adsorption = (Ci – Ce) mol L–1 Therefore, the amount, i.e., the number of moles of acid adsorbed is x = (Ci – CeǼȱśŖȺȦȺŗŖŖŖȱƽȱǻi – CeǼȺȦȺŘŖȱ–˜•ȱ–1. 4. From this calculate the amount of acid adsorbed per gram of adsorbent, i.e., ǻ¡ȺȦȺ–Ǽȱ’—ȱ–˜•ȱ–1.

œ˜›™’˜—ȱŠ—ȱ’œ›’‹ž’˜—ȱŽŠœž›Ž–Ž—œȳ153

ȱ

śǯȱ •˜ȱŠȱ›Š™‘ȱ‹Ž ŽŽ—ȱ•˜ȱǻ¡ȺȦȺ–ǼȱŸŽ›œžœȱ•˜ȱe. If the plot is straight line, this will prove the validity of the Freundlich isotherm over the given conditions of temperature and concentration range. ȱ Ŝǯȱ ›˜–ȱ ‘Žȱ œ•˜™Žȱ Š—ȱ ’—Ž›ŒŽ™ȱ ŒŠ•Œž•ŠŽȱ ‘Žȱ ›Žž—•’Œ‘ȱ Œ˜—œŠ—œȱ ŗȺȦȺ—ȱ Š—ȱ ȱ respectively. 7. Similarly, test Langmuir’s adsorption isotherm by plotting Ce ȺȦȺǻ¡ȺȦȺ–ǼȱŸŽ›œžœȱe. 8. From the slope and intercept calculate Langmuir constants ‘a’ and ‘b’. Result: The adsorption of acetic acid on activated charcoal ȱ ŗǯȱ ‹Ž¢œȺȦȺ˜Žœȱ—˜ȱ˜‹Ž¢ȱ‘Žȱ›Žž—•’Œ‘ȱ’œ˜‘Ž›–ȱ ȱ Řǯȱ ‹Ž¢œȺȦȺ˜Žœȱ—˜ȱ˜‹Ž¢ȱ‘ŽȱŠ—–ž’›ȱ’œ˜‘Ž›– ȱ˜‹Ž¢œǰȱ‘ŽȱŸŠ•žŽœȱ˜ȱ›Žž—•’Œ‘ȱŒ˜—œŠ—œȱŗȺȦȺ—ȱŠ—ȱ ȱŠ›ŽDZȱǯǯǯǯǯǯǯǯ and Langmuir constants ‘a’ and ‘b’ are: ........ Note:ȱ ‘Žȱ Šœ˜›™’˜—ȱ ˜ȱ ŸŠ›’˜žœȱ Šœ˜›‹ŠŽȺȦȺŠœ˜›‹Ž—ȱ œ¢œŽ–œȱ œžŒ‘ȱ Šœȱ ˜¡Š•’Œȱ ŠŒ’ȺȦȺȱ Œ‘Š›Œ˜Š•ǰȱ ’˜’—ŽȺȦȺŒ‘Š›Œ˜Š•ǰȱ ™’Œ›’Œȱ ŠŒ’ȺȦȺŒ‘Š›Œ˜Š•ȱ ŒŠ—ȱ ‹Žȱ œž’Žȱ žœ’—ȱ ’›’–Ž›’Œȱ ˜›ȱ colorimetric methods.

EXPERIMENT: 11.2 Aim: To study the distribution of benzoic acid between water and benzene and to determine the association factor of benzoic acid in benzene. Requirements: Cleaned and dried stoppered reagent bottles (250 mL), standard flasks, ‹ž›ŽŽœǰȱ›ŠžŠŽȱ™’™ŽŽœǰȱŒ˜—’ŒŠ•ȱ•Šœ”œǰȱ‘Ž›–˜œŠȺȦȺ™•Šœ’Œȱ›˜ž‘ǰȱ‹Ž—£˜’ŒȱŠŒ’ǰȱ benzene, sodium hydroxide solutions (0.02 N and 0.1 N), phenolphthalein indicator, distilled water. Theory: According to Nernst’s distribution law, when a solute is added to a mixture of two immiscible or partially miscible solvents at a given temperature and the solute exists in the same molecular state in both the solvents, then it distributes between the two solvents in such a way that the ratio of its concentrations in the two solvents is a constant. This ratio is independent of the amount of the solute added and is known as distribution coefficient or partition coefficient, KD. This law does not hold good if the solute is associated or dissociated in one of the solvents. When benzoic acid is distributed between water and benzene, it remains almost as simple molecules in water (negligible ionization may be ignored) but is mostly associated in benzene. Benzoic acid forms dimers, (C6H5COOH)2, in benzene. Hence, Nernst’s distribution law is not obeyed. The law with a modification is obeyed. Let us consider a solute A which exists as simple molecules in solvent 1 and mainly as associated molecules An, in solvent 2. Let the concentration in solvent 1 and solvent 2 be C1 and C2 respectively. In solvent 2, there is an equilibrium between the simple and associated molecules according to the equation nA ֖ An The equilibrium constant is Kc = CA /C nA ...(1) n

154ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

where CA is the concentration of simple molecules and CAn that of associated molecules, both in solvent 2. Thus, ...(2) CA = (CAn ȺȦȺ c)ŗȺȦȺ— ŗȺȦȺ— As Kc is a constant, equation (2) can be written as CA = a constant u (CAn )ŗȺȦȺ— or

CA = a constant u

n

CA

n

...(3)

Since the solute is mostly in the associated form in solvent 2, CAn may be taken as the total concentration C2 and therefore, CA = a constant u

n

C2

...(4)

The distribution law is applicable only for the same molecular state of the solute, simple molecules A in this case. So, C1ȺȦȺA = a constant ...(5) Substituting for CA from equation (4) into equation (5), we get C1 / n C2 = a constant = KD

...(6)

This modified law can be used to verify the law and to find out the association factor of benzoic acid in benzene. Procedure: 1. Take 1 g of benzoic acid in a well stoppered reagent bottle. 2. Add 100 mL of distilled water and 50 mL of benzene. ȱ řǯȱ ‘Š”Žȱ‘Žȱ–’¡ž›Žȱ‘˜›˜ž‘•¢ȱǻŠ’Š˜›ȺȦȺœ‘Š”Ž›ȱŒŠ—ȱŠ•œ˜ȱ‹ŽȱžœŽǼȱŠ—ȱ™•ŠŒŽȱ’ȱ’—ȱ a thermostat or a plastic trough containing water for about 30 minutes. 4. Note down the temperature of the experiment. 5. Shake the mixture every three to four minutes taking care to see that the drops adhering to the sides of the bottle settle down. 6. When the equilibrium is attained (the two layers get separated completely), pipette out 5 mL of benzene layer into a conical flask containing 10 mL of distilled water and titrate against 0.1 N sodium hydroxide solution using phenolphthalein as an indicator. 7. Try to get at least three concordant titre values. 8. Now, pipette out 20 mL of aqueous layer and titrate against 0.02 N sodium hydroxide solution using phenolphthalein as an indicator. ȱ şǯȱ ŽȱŠȱ•ŽŠœȱ‘›ŽŽȱŒ˜—Œ˜›Š—ȱ’›ŽȱŸŠ•žŽœǯ 10. Repeat the same procedure taking 1.5 g and 2 g of benzoic acid respectively. Observation and calculation: Room Temperature = ........ °C

œ˜›™’˜—ȱŠ—ȱ’œ›’‹ž’˜—ȱŽŠœž›Ž–Ž—œȳ155

Table 11.3: Data recording Bottle No.

Volume of NaOH for 5 mL of benzene layer (mL)

C1 (M)

1.

1. 2. 3.

1. 2. 3.

2.

1. 2. 3.

1. 2. 3.

3.

1. 2. 3.

1. 2. 3.

Volume of NaOH for 20 mL of aqueous layer (mL)

C2 (M)

Table 11.4: Data recording Bottle No.

C1 (M)

C2 (M)

C1 / C2

KD = C1 / C2

1. 2. 3. Mean = ........

Plot a graph of C1 against C2 . A straight line passing through the origin is obtained. From the slope of the line, find out the value of KD and compare with the calculated one. Result: The value of n = ........ Hence, the association factor of benzoic acid in benzene = ........

EXPERIMENT: 11.3 Aim: To study the distribution of benzoic acid between (i) water and chloroform, (ii) water and cyclohexane. Requirements: Cleaned and dried stoppered reagent bottles (250 mL), standard flasks, ‹ž›ŽŽœǰȱ›ŠžŠŽȱ™’™ŽŽœǰȱŒ˜—’ŒŠ•ȱ•Šœ”œǰȱ‘Ž›–˜œŠȺȦȺ™•Šœ’Œȱ›˜ž‘ǰȱ‹Ž—£˜’ŒȱŠŒ’ǰȱ chloroform, cyclohexane, sodium hydroxide solutions (0.02 N and 0.1 N), phenolphthalein indicator, distilled water. Theory: Same as in Experiment 11.2. Procedure: Same as in Experiment 11.2, but instead of benzene, use chloroform for Part (i). Repeat the same procedure with water and cyclohexane for Part (ii). Plot a graph between C1 and n C2 for different ‘n’ values. Compare the graphical and calculated results. Observation and calculation: Same as in Experiment 11.2.

156ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Result: The value of n = ........

Ž—ŒŽǰȱ‘ŽȱŠœœ˜Œ’Š’˜—ȱŠŒ˜›ȱ˜ȱ‹Ž—£˜’ŒȱŠŒ’ȱ’—ȱ ǻ’Ǽȱ ŠŽ›ȺȦȺŒ‘•˜›˜˜›–ȱƽȱǯǯǯǯǯǯǯǯ ȱ ȱ ǻ’’Ǽȱ ŠŽ›ȺȦȺŒ¢Œ•˜‘Ž¡Š—Žȱƽȱǯǯǯǯǯǯǯǯ Compare the results for different organic solvents. Note: Butanol may also be used as one of the solvents.

EXPERIMENT: 11.4 Aim: To study the distribution of acetic acid between (i) water and chloroform, (ii) water and cyclohexane. Requirements: Cleaned and dried stoppered reagent bottles (100 mL), standard flasks, ‹ž›ŽŽœǰȱ›ŠžŠŽȱ™’™ŽŽœǰȱŒ˜—’ŒŠ•ȱ•Šœ”œǰȱ‘Ž›–˜œŠȺȦȺ™•Šœ’Œȱ›˜ž‘ǰȱŘȱȱŠŒŽ’ŒȱŠŒ’ȱ (58 mL of glacial acetic acid in 500 mL of the solution), chloroform, cyclohexane, sodium hydroxide solutions (0.5 N and 0.1 N), phenolphthalein indicator, distilled water. Theory: Same as in Experiment 11.2. Procedure: 1. Prepare the mixtures as given in Table 11.5. Table 11.5: Composition of mixtures S. No.

Volume of chloroform (mL)

Volume of 2 M acetic acid (mL)

Volume of distilled water (mL)

1.

20

10

40

2.

20

20

30

3.

20

30

20

4.

20

50

00

2. Shake each of these mixtures thoroughly and place them in a thermostat or a plastic trough containing water for about 20 minutes. 3. Note down the temperature of the experiment. 4. Shake the mixture every two minutes taking care to see that the drops adhering to the sides of the bottle settle down. 5. When the equilibrium is attained (the two layers get separated completely), pipette out 5 mL of aqueous layer from bottle 1 and titrate with 0.5 N sodium hydroxide solution using phenolphthalein as the indicator. 6. Similarly, pipette out 5 mL of chloroform layer, add about 10 mL of distilled water (this is to extract the acetic acid completely into aqueous layer in order to obtain the end point correctly) and titrate with 0.1 N sodium hydroxide solution using phenolphthalein as the indicator. 7. Repeat the procedure for other three bottles and tabulate the observation and calculation as given in Tables 11.3 and 11.4 in Experiment 11.2. 8. Plot a graph between C1 and

C2 for different values of ‘n’. Compare the graphical and calculated results for different values of n. n

œ˜›™’˜—ȱŠ—ȱ’œ›’‹ž’˜—ȱŽŠœž›Ž–Ž—œȳ157

9. Carryout the experiment with cyclohexane instead of chloroform by following the same procedure. Result: The value of ‘n’ = ........

Ž—ŒŽǰȱ‘ŽȱŠœœ˜Œ’Š’˜—ȱŠŒ˜›ȱ˜ȱŠŒŽ’ŒȱŠŒ’ȱ’—ȱ ǻ’Ǽȱ ŠŽ›ȺȦȺŒ‘•˜›˜˜›–ȱƽȱǯǯǯǯǯǯǯǯ ȱ ȱ ǻ’’Ǽȱ ŠŽ›ȺȦȺŒ¢Œ•˜‘Ž¡Š—Žȱƽȱǯǯǯǯǯǯǯǯ Compare the results for different organic solvents.

EXPERIMENT: 11.5 Aim: To study the equilibrium reaction KI + I2 ֖ KI3 (Potassium tri iodide) and to determine its equilibrium constant. Requirements: Cleaned and dried stoppered reagent bottles (250 mL), standard flasks, ‹ž›ŽŽœǰȱ ›ŠžŠŽȱ ™’™ŽŽœǰȱ Œ˜—’ŒŠ•ȱ •Šœ”œǰȱ ‘Ž›–˜œŠȺȦȺ™•Šœ’Œȱ ›˜ž‘ǰȱ ’˜’—Žǰȱ chloroform, potassium iodide solution (0.1 M), sodium thiosulphate solution (0.01 N), starch, distilled water. Theory: The partition study experiment can be used for the evaluation of equilibrium constant, K for this reversible reaction KI + I2 o KI3 – The reaction can ionically be expressed as I– + I2 ֖ I 3 . This reaction takes place in an aqueous solution and according to the law of mass action, the equilibrium constant K is given by K = [KI3ǾȺȦȺǽ Ǿȱǽ2] ...(1) where [KI3], [KI] and [I2] represent the equilibrium concentrations in any single aqueous solution. The equilibrium condition may be investigated by studying the distribution of iodine between some organic solvent and aqueous solution of potassium iodide. The first study enables us to evaluate the distribution coefficient KD from the relationship CorgȺȦȺaq = KD ...(2) where C’s are the concentrations of free iodine in the two solvents which are determined by titrations. Since the distribution law applies only to a species common to both layers, the Œ˜—ŒŽ—›Š’˜—ȱ ˜ȱ ›ŽŽȱ ȹ’˜’—Žȱ ’—ȱ ‘Žȱ ŠšžŽ˜žœȱ ™˜Šœœ’ž–ȱ ’˜’Žȱ •Š¢Ž›ȱ aq can be determined from the concentration of iodine in the organic layer Corg , where even other species are present, using Caq = CorgȺȦȺ D ...(3) The titration of the aqueous layer with sodium thiosulphate solution, however, gives the total of the concentrations of iodine - free I2 as well as that combined with potassium iodide to give KI3. Subtraction of calculated concentration of free iodine from this total result gives the concentration of KI3. The concentration of KI3 in the aqueous layer gives the concentration of combined KI.

158ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Knowing the equilibrium values of [KI3], [KI] and [I2] for any aqueous layer, the equilibrium constant may be calculated from the above equation (1). Table 11.6: Equivalent and molar masses Gram equivalent mass (g)

Gram molecular mass (g)

I2

127

254

KI

166

166

KI3

210

420

Procedure: A. Determination of distribution coefficient KD: 1. Prepare a saturated solution of iodine in chloroform (CHCl3) (about 5%) at room temperature and filter it. 2. Take three 250 mL reagent bottles and label them with numbers 1, 2 and 3. 3. Place 25, 30 and 40 mL respectively of this filtered solution in these bottles and add 15, 10 and 0 mL of chloroform so that the total volume in each bottle is 40 mL (Table 11.7). 4. Now add 150 mL of distilled water in each bottle, put stoppers for all the three Š—ȱœ‘Š”Žȱ‘Ž–ȱ‘˜›˜ž‘•¢ȱǻŠ’Š˜›ȺȦȺœ‘Š”Ž›ȱŒŠ—ȱŠ•œ˜ȱ‹ŽȱžœŽǼȱŠ—ȱ”ŽŽ™ȱ‘Ž–ȱ in a thermostat maintained at room temperature. 5. Shake the bottles from time to time for about an hour. You will then notice complete separation of two liquid layers in each bottle. The upper layers will be aqueous solutions and the lower ones are organic solutions. 6. Pipette out 10 mL of the aqueous layer from bottle no. 1 into a conical flask. 7. Titrate this with 0.01 N sodium thiosulphate solution using starch as an indicator. 8. The solubility of iodine in aqueous layer is very low and therefore the titre value will be low. Take three concordant readings. 9. Repeat the same procedure with the upper aqueous layers from the bottle no. 2 and bottle no. 3 separately. 10. Now, pipette out 2 mL of chloroform layer from the bottle no. 1 into a conical flask containing 25 mL of distilled water. 11. Titrate with 0.01 N thiosulphate solution using starch as an indicator. 12. Repeat the titration procedure with the lower organic layers from bottle no. 2 and bottle no. 3 separately. 13. Calculate the value of distribution coefficient of I2 between chloroform and water for each bottle as KD = CCHCl3ȺȦȺ H2O 14. Take the mean value of KD. B. Determination of equilibrium constant K: 1. Take three reagent bottles and label them 4, 5 and 6. 2. Perform a similar experiment using 0.1 M potassium iodide solution in place of water.

œ˜›™’˜—ȱŠ—ȱ’œ›’‹ž’˜—ȱŽŠœž›Ž–Ž—œȳ159

3. Determine the concentrations of I2 in potassium iodide layer and chloroform layer from each of these bottles. Table 11.7: Preparation of solutions with water Bottle No.

Volume of I2 in CHCl3 (mL)

Volume of CHCl3 (mL)

Volume of water (mL)

1

25

15

150

2

30

10

150

3

40

0

150

Table 11.8: Preparation of solutions with potassium iodide Bottle No.

Volume of I2 in CHCl3 (mL)

Volume of CHCl3 (mL)

Volume of 0.1 M KI solution (mL)

4

25

15

150

5

30

10

150

6

40

0

150

Observation: Room temperature = ........ °C Table 11.9: Distribution of I2 between CHCl3 and water Volume of aqueous layer = 10 mL Volume of organic layer = 2 mL Bottle No.

Volume of thiosulphate for aqueous layer (mL)

CH2O (M)

Volume of thiosulphate for organic layer (mL)

CCHCl3 (M)

KD = CCHCl3 / CH2O

1 2 3 Mean KD = ........

Table 11.10: Distribution of I2 between CHCl3 and aqueous KI solution Volume of aqueous layer = 10 mL Volume of organic layer = 2 mL Bottle No.

Initial concentration of KI solution (M)

4

0.1

5

0.1

6

0.1

Volume of thiosulphate for aqueous layer (mL)

Calculation: 1. Determination of KD : Normality of thiosulphate solution = 0.01 N Molar mass Equivalent mass of I2 = 2

Volume of thiosulphate for organic layer (mL)

160ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Normality 2 Bottle No. 1: Molarity =

Concentration of I2 in aqueous layer (CH2O) =

Vthio × 0.01 = ........ M 10 × 2

Concentration of I2 in organic layer (CCHCl3) =

Vthio × 0.01 = ........ M 2×2

Therefore, KD = CCHCl3ȺȦȺH2O Calculate KD for bottles 2 and 3 and take the mean value. 2. Determination of K : Bottle No. 4: Concentration of free I2 in aqueous layer, i.e., [I2] free = C1 KD = CCHCl3ȺȦȺ1 Vthio × 0.01 = ........ M 2×2 C1 = CCHCl3ȺȦȺ D Concentration of I2 in aqueous KI layer = C2 = [KI3] + [I2] free CCHCl3 =

Vthio × 0.01 = ........ M 10 × 2 Therefore, concentration of KI3 in aqueous KI layer at equilibrium, i.e., [KI3] = C2 – C1 = ........ M Initial concentration of KI = 0.1 M Therefore, concentration of unreacted KI, i.e., free KI = C3 = 0.1 – [KI3] = ........ M Substituting the appropriate concentrations calculated as above in equation (1), K = (C2 – C1ǼȺȦȺǻ3 u C1) Similarly, calculate the values of K for bottles 5 and 6 and take the mean value. =

EXPERIMENT: 11.6 Aim: To study the formation of a complex between copper and ammonia and to determine the molecular formula of the complex by partition coefficient method. Requirements: Standard flasks, burette, graduated pipettes, conical flasks, stoppered reagent bottles (250 mL), thermostat or a plastic trough, 1 M aqueous solution of ammonia, 0.05 M hydrochloric solution, 0.1 M aqueous solution of copper sulphate, chloroform, methyl red indicator, distilled water. Warning: Open the liquor ammonia bottle very carefully while cooling in ice. Do not open without cooling in ice. Even dilute solutions of ammonia must not be pipetted out directly as this will cause burns in mouth and lungs. Sucker or guarded pipette should be used.

œ˜›™’˜—ȱŠ—ȱ’œ›’‹ž’˜—ȱŽŠœž›Ž–Ž—œȳ161

Theory: Ammonia distributes between water and chloroform when an aqueous ammonia solution is shaken with chloroform. At equilibrium, the ratio of the concentration of ammonia in water to that in chloroform is a constant known as distribution constant, KD . This distribution law, known as Nernst distribution law, is a consequence of thermodynamic requirements of equilibrium, phase rule and Henry’s law. Nernst distribution law can be expressed as KD = CH2OȺ ȦȺCHCl3 where CH2O is the concentration of ammonia in water and CCHCl3 is the concentration of ammonia in chloroform. KD is also known as partition coefficient. When copper sulphate is added to a mixture of ammonia and chloroform, ammonia forms a coordination complex with copper(II) ions which is a d9 species. Ammonia has a lone pair of electrons and so behaves like a ligand. Cu2+ + xNH3 ֖ [Cu(NH3)x]2+ This experiment is done in two parts: Determination of partition coefficient of ammonia between water and chloroform and determination of formula of the complex. In the second part, the concentration of ammonia in chloroform layer can be found out by titrating it against a standard acid. Knowing this and the partition coefficient (from first part), the concentration of free ammonia in aqueous layer can be determined. We know that the total concentration (initial) of ammonia is the sum of the concentrations of ammonia in chloroform layer, that of free ammonia in the aqueous layer and that of ammonia combined with copper ions in aqueous layer. Using all these quantities, the number of moles of ammonia needed to form the complex can be found out. Procedure: 1. Standardize the 1 M ammonia solution (stock) with 0.05 M HCl solution. This can be done by diluting the 1 M solution ten times and using methyl red indicator. 2. Prepare 0.5 M ammonia solution from the standardized stock solution. A. For the determination of partition coefficient of ammonia between water and chloroform: 1. Prepare the mixtures in well stoppered reagent bottles as given below. (i) 50 mL of chloroform and 50 mL of 1 M ammonia solution in bottle 1. (ii) 50 mL of chloroform and 50 mL of 0.5 M ammonia solution in bottle 2. ȱ Řǯȱ ‘Š”Žȱ ‘˜›˜ž‘•¢ȱ ǻŒŠ—ȱ ”ŽŽ™ȱ ’—ȱ Š—ȱ Š’Š˜›ȺȦȺœ‘Š”Ž›ȱ ˜›ȱ ŗŖȱ –’—žŽœǼȱ Š—ȱ ™•ŠŒŽȱ these bottles in a thermostat or a plastic trough containing water for about 20 minutes to attain the thermal quilibrium. 3. Note down the temperature. 4. From the first bottle, carefully withdraw 10 mL of chloroform layer into a conical flask containing 30 mL of distilled water. This is done to extract ammonia into aqueous layer. 5. Titrate the solution with 0.05 M HCl solution using methyl red indicator. 6. The colour change from yellow to red indicates the end point. 7. Add a slight excess of indicator, preferably towards the end point, and shake thoroughly to observe the colour change in aqueous layer.

162ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

8. Try to get at least three concordant readings. 9. Similarly, withdraw 10 mL of aqueous layer and determine the concentration of ammonia in it by titrating with the standard acid using the same indicator. ȱ ŗŖǯȱ ŽȱŠȱ•ŽŠœȱ‘›ŽŽȱŒ˜—Œ˜›Š—ȱ›ŽŠ’—œǯ 11. Repeat the same procedure with the second bottle. B. Determination of the molecular formula of the complex: 1. Pipette out 25 mL of 1 M ammonia solution, 25 mL of 0.1 M copper sulphate solution and 75 mL of chloroform into a well stoppered reagent bottle. ȱ Řǯȱ ‘Š”Žȱ  Ž••ȱ ǻŒŠ—ȱ ”ŽŽ™ȱ ’—ȱ Š—ȱ Š’Š˜›ȺȦȺœ‘Š”Ž›ȱ ˜›ȱ ŗŖȱ –’—žŽœǼȱ Š—ȱ ™•ŠŒŽȱ ‘ŽœŽȱ bottles in a thermostat or a plastic trough containing water for about 20 minutes to attain the thermal equilibrium. 3. Note down the temperature. 4. Carefully withdraw 20 mL of chloroform layer. 5. Add 40 mL distilled water and titrate with 0.05 M HCl solution using methyl red indicator, taking the precautions as given in the first part. 6. Try to get at least three concordant readings. 7. Carry out a duplicate experiment under the same conditions. Observation: Room temperature: ........°C Table 11.11: Data recording Bottle No.

Volume of acid for 10 mL of aqueous layer (mL)

CH2O (M)

1

1. 2. 3.

1. 2. 3.

2

1. 2. 3.

1. 2. 3.

Volume of acid (mL) for 10 mL of chloroform layer

CCHCl3 (M)

KD = CH2O / CCHCl3

Mean = ........

Table 11.12: Data recording Bottle No.

Volume of acid (mL) for 20 mL of chloroform layer

1

1. 2. 3.

2

1. 2. 3.

Calculation: Distribution constant, KD = ........

CCHCl3 (M)

œ˜›™’˜—ȱŠ—ȱ’œ›’‹ž’˜—ȱŽŠœž›Ž–Ž—œȳ163

Concentration of ammonia in chloroform layer (CCHCl3) = X mol L–1 (from second part) 75 Total moles of ammonia in chloroform = ×X 1000 Total moles of ammonia taken initially =

25 ×1 1000

Therefore, total moles of ammonia in aqueous layer =

25 75X − 1000 1000

Concentration of free ammonia in aqueous layer = KD u X mol L–1 50 × KD × X 1000 Moles of ammonia combined with Cu2+ ions in aqueous layer Moles of free ammonia in aqueous layer (50 mL) =

75 X ⎞ ⎛ 50 ⎛ 25 ⎞ – –⎜ K D X⎟ = y = ⎜ ⎟ ⎝ 1000 1000 ⎠ ⎝ 1000 ⎠ Total moles of Cu2+ ions in aqueous layer =

25 × 0.1 1000

⎛ 25 ⎞ 1 Ratio of Cu2+ to NH3 in the complex is Cu2+ : NH3 = ⎜ × 0.1⎟ ÷ y = ⎝ 1000 ⎠ X Therefore, x can be calculated to the nearest whole number. This value of x is substituted in the formula [Cu(NH3)x]2+. (Note: Copper ions are assumed to be completely complexed with ammonia at such high concentrations). Result: The molecular formula of the complex between copper and ammonia =....... Note: This experiment can be extended to determine the formation and dissociation constants of the complex using different quantities of ammonia.

VIVA QUESTIONS 1. Define ‘adsorbate’ and ‘adsorbent’. 2. Differentiate between adsorption, absorption and sorption. 3. What is effect of concentration, temperature and surface area on adsorption phenomena? 4. What is an adsorption isotherm? 5. Write the expression for Freundulich adsorption isotherm. Explain all the terms involved. 6. Write the expression for Langmuir adsorption isotherm. Explain all the terms involved. 7. What is meant by BET adsorption isotherm? Explain.

164ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

8. Write the conditions under which Langmuir and BET adsorption isotherms are applicable. 9. State Nernst distribution law. What are its limitations? 10. Define partition coefficient. What are its units? 11. How do you find out whether a solute is associated or dissociated in a particular solvent? 12. How is the Nernst distribution law modified, if the system involves the dissociation or association of a solute? ȱ ŗřǯȱ Ž’—ŽȱŽšž’•’‹›’ž–ȱŒ˜—œŠ—ǯȱ ’ŸŽȱ’œȱž—’œǯ 14. Explain the term ‘equilibrium constant’ with an example. 15. What are the applications of Nernst distribution law?

12

ENVIRONMENTAL PARAMETER MEASUREMENT

EXPERIMENT: 12.1 Aim: To estimate the amount of suspended particulate matter (SPM) in air samples. Theory: Atmospheric particulate matter – also known as particulate matter (PM) or particulates – are minute solid particles or liquid droplets suspended in the earth’s atmosphere. Sources of particulate matter can be natural or man-made. They have huge impact on climate and precipitation which adversely affect human health and vegetation. Subtypes of atmospheric particulate matter include suspended particulate matter (SPM), thoracic and respirable particles (diameter 4–10 Pm), inhalable coarse particles, which have a diameter between 2.5 and 10 Pm (PM10), fine particles with a diameter of 2.5 Pm or less (PM2.5), ultrafine particles and soot. ‘ŽȱȱŠ—ȱ ȱŽœ’—ŠŽȱŠ’›‹˜›—Žȱ™Š›’Œž•ŠŽœȱŠȱ ›˜ž™ȱŗȱŒŠ›Œ’—˜Ž—ȱǻŒŠ—ŒŽ›ȱ causing). Particulates are the deadly form of air pollution due to their ability to penetrate deep into the lungs and bloodstreams unfiltered leading to lung cancer, heart attacks, permanent DNA mutations and premature death.

Filter Air inlet

Electric motor

Pump

Fig. 12.1: High volume sampler

Particulates are generally estimated using a high volume sampler (Fig. 12.1). The sampler consists of an electric motor and a pump. A gas meter is attached to this unit to indicate the volume of air pumped through the sampler. The outlet of the pump is

166ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

connected to a filter which is made up of glass fibre. About 2 u106 litres of air can be handled at a time by this sampler. When air passes through the filter, particles of diameter in the range 10–100 Pm are collected on the filter and fine particles of diameter lesser than 10 Pm pass through the filter. The difference in the mass of the filter before and after the passage of air gives the amount of particulate matter in the air. Procedure: 1. Sampling site should be chosen at about a height of 10 m from the ground level and away from any disturbance or obstacle. Select the region to be studied, such as residential, industrial, disposal, highways, etc. 2. Maintain the filter at 25°C and at the relative humidity of 50% for 20 hours. 3. Weigh it accurately. 4. Place the filter in the sampler and switch on the sampler (do not switch it on without the filter paper). Let it stabilize for 15 minutes. 5. Then pass 2 u 106 litres of air through it at the rate of 1500 L min–1. 6. After the passage of air (it will take about 20–22 hours) remove the filter. 7. Again leave it for 20 hours at 25°C and at 50% relative humidity. 8. Weigh it again. 9. Calculate the difference in the mass of the filter. 10. Report the amount of particulates as mg m–3 or Pg m–3 of air. Calculation: Initial mass of the filter = w1 mg Final mass of the filter = w2 mg Mass of total suspended particulates (SPM) = (w2 – w1) mg Volume of air = 2 u 106 L = 2 u 106 u 103 cm3 = 2 u 103 m3 Amount of total suspended particulates (SPM) = ((w2 – w1ǼȺȦȺŘȱu 103) mg m–3 = w3 mg m–3 Since, 1 mg = 103 Pg, w3 mg = w3 u 103 Pg Hence, the amount of SPM can also be expressed as w3 u 103 Pg m–3 Result: Amount of total suspended particulates (SPM) = ........ mg m–3 = ........ Pg m–3 Note: 1. Repeat the experiment two or three times and take the mean value, as the single value does not give the correct picture. 2. Regular monitoring must be done twice a week. 3. The measurement should not be done on a rainy day in open atmosphere.

EXPERIMENT: 12.2 Aim: To determine the amount of dissolved oxygen in the given sample of water. Requirements: Standard flasks, beakers, graduated pipettes, conical flasks, burette, airtight stoppered reagent bottles, sodium thiosulphate solution (0.01 N), potassium

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ167

dichromate solution (0.001 N), manganese sulphate (MnSO4.4H2O), potassium fluoride (KF.2H2O), sodium hydroxide, sodium iodide, sodium azide (NaN3), concentrated sulphuric acid, freshly prepared starch solution, water sample, distilled water. Reagents: 1. Manganese sulphate solution: Weigh about 25 g of MnSO4.4H2O and dissolve in 50 mL of distilled water and filter to get a clear solution. 2. Potassium fluoride solution: Dissolve 20 g of KF.2H2O in 50 mL of distilled water. 3. Alkaline iodide-azide solution: Dissolve 25 g of NaOH and 7 g of NaI in distilled water and dilute to 500 mL. Dissolve 0.5 g of NaN3 in 25 mL of distilled water and add this solution with stirring to the alkaline iodide solution. Theory: The dissolved oxygen (DO) is the amount of atmospheric oxygen dissolved in water and is expressed in parts per milion (ppm), i.e., mg L–1. Its solubility in water depends on the temperature and pressure of the atmosphere. The amount of oxygen present in an aqueous solution is 8.32 ppm if the solution is saturated with air at 1 atm pressure and 298 K, and it may range from 7 to 15 ppm with a slight variation in Œ˜—’’˜—œǯȱ‘Žȱœ˜•ž‹’•’¢ȱ˜ȱ˜¡¢Ž—ȱ’œȱ–˜›Žȱ’—ȱ›Žœ‘ȱ ŠŽ›ȱ‘Š—ȱ’—ȱœŽŠ ŠŽ›ȺȦȺ‹›ŠŒ”’œ‘ȱ water which contains a large amount of dissolved salts. The solubility of oxygen decreases in polluted water too. As the DO is essential for the survival of aquatic organisms, fish, etc., the water with less than 4 ppm DO is unsuitable for aquatic life. Again, in order to enable bacteria to oxidize the organic matter present in water, DO is required. Absence or low concentration of DO indicates that the water is polluted. Hence, the determination of DO is necessary. The Winkler method used in this experiment is based on the fact that the DO oxidizes Mn2+ to a higher oxidation state under basic conditions and the oxidized manganese then liberates iodine from potassium iodide or sodium iodide in acid medium. The iodine liberated is titrated with sodium thiosulphate solution. The amount of DO can be determined from the amount of iodine liberated. The appropriate chemical equations are: 1 Mn2+ + 2OH– + O2 o MnO2 + H2O 2 MnO2 + 2I– + 4H+ o Mn2+ + I2 + 2H2O I2 + 2S2O32– o 2I– + S4O62– Water collected from the field must be “fixed” immediately to get the reliable result. This is because the sample when brought from the field to the laboratory may undergo a change in DO value due to a change in temperature and the occurrence of biological reactions with time. The water sample is “ fixed” by adding the solutions of KF, MnSO4, alkaline iodide-azide and concentrated sulphuric acid. Once the sample is “fixed”, estimation can be done at a convenient time in the laboratory. Procedure: 1. Standardize the thiosulphate solution using the standard dichromate solution as usual.

168ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

2. Exactly measure 250 mL of sample water into a stoppered reagent bottle. 3. Quickly add to this, 1 mL of KF solution, 2 mL of MnSO4 solution and 2 mL of alkaline iodide-azide solution. 4. Stopper the bottle tightly and shake the contents thoroughly. 5. Then add 2 mL of concentrated sulphuric acid and shake gently to dissolve the precipitate formed. 6. From this solution, exactly measure out 100 mL of the clear solution into a conical flask and titrate the liberated iodine with the standardized thiosulphate solution using starch as an indicator. 7. Repeat the experiment to get concordant values. Calculation: From the chemical equations given under “Theory”, we get 1000 mL of 1 N thiosulphate solution { 8 g oxygen ? Dissolved oxygen (ppm or mg L–1) =

V2 × N × 1000 × 8 V1

V1 = volume of water sample in mL V2 = titre value in mL N = normality of thiosulphate solution Result: The amount of dissolved oxygen in the given sample of water = ........ ppm (mg L–1) Note: 1. If the titre value is very low (low DO), a large volume of the water sample may be taken. 2. Fe3+, if present in water, will liberate iodine from iodide thereby interfering in the estimation. This is masked by adding fluoride ions in the form of KF. 3. Nitrite ion, if present in water, can also liberate iodine and this is prevented by the addition of NaN3 which destroys the nitrite ion: NaN3 + H+ o HN3 + Na+ HN3 + NO2– + H+ o N2 + N2O + H2O

EXPERIMENT: 12.3 Aim: To estimate the amount of sulphate ion present in the given sample of water by turbidimetry. Requirements:ȱ™ŽŒ›˜™‘˜˜–ŽŽ›ȺȦȺŒ˜•˜›’–ŽŽ›ǰȱŒžŸŽŽœǰȱœŠ—Š›ȱ•Šœ”œǰȱ›ŠžŠŽȱ pipettes, sodium sulphate, barium chloride solution (10% m/V), concentrated hydrochloric acid, sodium chloride, distilled water. Reagents: 1. Standard sulphate solution: Dissolve 0.1480 g of sodium sulphate in distilled water and make up to 100 mL. This is a standard solution of 1000 ppm (mgL–1) sulphate.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ169

2. NaCl-HCl conditioning reagent: Dissolve 24 g of NaCl in 100 mL of distilled water containing 2 mL of concentrated HCl. This is used to stabilize the suspension and minimize interferences. Theory: Both turbidimetry and colorimetry involve the measurement of intensity of the light transmitted through a medium. Turbidimetry is a method in which white light is passed through a finely divided suspension of the substance to be estimated. Using the turbidimeter, the light transmitted by the suspensions of solutions of known concentration is determined and a calibration curve is drawn. The light transmitted by the suspension of unknown sample is measured and then the concentration of the sample is determined from the calibration curve. The law followed in turbidimetry is very similar to Lambert Beer’s law followed in colorimetry. It is expressed as I S = log = kbc Io I = transmittance,T; where Io = Incident intensity; I = transmitted intensity; Io S = turbidence due to scattering and this is a measure of scattering power; k = turbidity constant; b = path length; c = concentration of the suspended material. The determination of transmittance of suspensions can also be carried out in spectrophotometer or colorimeter and the concentration of the sample is determined from the calibration curve. Also, the scattering power of the suspension can be used for this purpose. Since there will be no absorption but only scattering of light in the case of a suspension, the absorbance values in spectrophotometer or colorimeter should be taken as the scattering power of the suspension. Selecting the wavelength is important because the sample must not absorb radiation at that wavelength. Moreover, it depends on the colour of the suspension medium. If it is white, it is customary to use O = 420 nm, and if it is yellowish, then O = 550 nm. The amount of the light scattered is proportional to the concentration of the insoluble particle in the suspension. Sulphate in water comes from dissolved minerals, namely, sodium sulphate (salt cake), magnesium sulphate (epsom salt) and calcium sulphate (gypsum). Sulphate can also come from fertilizer or sewage treatment. If the sulphate concentration in a water source is less than 10 mgL–1, it is an indication that the water source is fresh and unpolluted. Higher levels of sulphate in any water source can be indicative of some form of pollution. Typical pollution sources are mine drainage and effluent return flows, which can contain sulphate concentration of as high as 500 mg L–1. When new water users (new residents or travellers) are initially exposed to higher levels of sulphate in drinking water, it can have adverse health effects, causing diarrhoea. Elevated sulphate levels in water can impart a bitter, astringent taste and a “rotten egg” smell to water.

170ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

High sulphate concentration in water can accelerate corrosion of metals, especially iron. Children, travellers and people who drink large volumes of such water can be at risk and suffer from adverse health effects, such as “Traveller’s diarrhoea.” In this experiment, the sulphate ions present in the water sample are converted to a suspension of barium sulphate using barium chloride reagent. Then, by turbidimetry, the concentration of barium sulphate is estimated. SO42–(aq) + Ba2+(aq)o BaSO4(s) Procedure: 1. Pipette out 2, 4, 6, 8, 10, 12 mL of standard sulphate solution into six numbered 50 mL standard flasks. 2. Add 10 mL of NaCl-HCl conditioning reagent to each flask and shake well. 3. Then add 10 mL of barium chloride solution to the first flask and quickly make up to the volume with distilled water. 4. Shake the flasks thoroughly for 1 minute. 5. Quickly transfer the solution to the cell of the spectrophotometer or colorimeter. 6. Read the scattering intensity of the suspension at 420 nm. 7. Similarly, add 10 mL of barium chloride solution to other flasks one by one and measure the scattering intensity in each case. 8. Draw the calibration graph between the scattering intensity and sulphate concentration. 9. Into another 50 mL standard flask, pipette out 25 mL of the given water sample and prepare the suspension in the same manner. 10. Measure its scattering intensity and read the corresponding concentration of sulphate from the calibration graph. 11. Carry out a duplicate experiment with another 25 mL of the water sample. Result: The concentration of the sulphate ions in the given water sample = ........ mgL–1(ppm). Note: 1. Many other methods such as titrimetry, gravimetry and colorimetry are available for the determination of sulphate in water samples. However, it has been observed that the turbidimetric method is sensitive, accurate and permits the determination of microgram amounts of sulphate present in water samples. 2. Turbidimeter is cost-effective and portable. It can be easily carried to the site of investigation. So, it is widely used in the analysis of water. 3. If turbidimeter is not available in laboratory, spectrophotometer or colorimeter can be used but with the same technique and principle of turbidimetry as this will avoid the large number of expensive chemicals used to produce colour required for colorimetric experiments.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ171

EXPERIMENT 12.4 Aim: To determine the salinity of the given sample of water. Theory: Salinity is a measure of the total dissolved salts in seawater. Although this sounds simple, measuring the salinity is a problem when we realize that some of the salts do not simply dissociate into ions but chemically react with water to form complex ions. Some of the ions include dissolved gases such as CO2 (which is converted to carbonic acid (H2CO3), bicarbonate (HCO3–), and carbonate (CO32–)). Salinity is an important characteristic of natural waters and it is the key factor in determining what species of plants and animals will be found in a certain area. Duxbury (1971) defined salinity as: “The total amount in grams of solid material dissolved in 1 kilogram of seawater when all the carbonate has been converted to oxide, all of the iodine and bromine have been replaced by chlorine, and all the organic matter has been completely oxidized.” Here, we may define salinity in a simple way as the total amount in grams of solid material dissolved in 1 kg of seawater. Normally, salinity is expressed in parts per thousand (ppt) and is written as ‰. For example, if we have 1000 g of seawater that contains 30 g of dissolved salt, then the salinity is 30 parts per 1000 parts of water (30‰). The salinity is generally measured as the molarity of chloride ions because the composition of the major ions in seawater is constant. This corresponds to about 0.0 M - 0.5 M of chloride ions. For our purpose, we require a water sample of known salinity and our salt water in the laboratory contains only sodium chloride and water, unlike seawater which has dozens of salts and minerals. However, the methodology adopted for seawater would work for other waters too. There are many methods available for salinity determination for water such as evaporation, measurement of conductivity, measurement of water density by hydrometer, refraction of light through water and titration of the chloride ion (CI–). It was found that the titration method is very precise and reproducible. The results obtained by this method are very close to the true values of salinity. This is the reason that it is the standard technique, for decades, in oceanography for determining salinity. This was earlier known as Boyle’s method and popularly called Mohr’s method. (I) Boyle’s method: First, we take a water sample of known salinity (say, 35 ppt) and titrate it with silver nitrate solution using potassium chromate as indicator. By following the same procedure with the unknown sample and using the salinity of known sample, that of unknown sample can be determined. When we add a solution containing silver nitrate to a solution containing sodium chloride, the chloride ions in the solution will react with the silver ions to produce a solid precipitate, silver chloride, leaving the sodium and nitrate ions in solution. Potassium chromate (K2CrO4) is used as an indicator to tell us exactly when all of the chloride in the water sample has reacted to form AgCl. The key is that the silver ions

172ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

prefer to combine with the chloride ions rather than the chromate ions. Thus, the silver chromate precipitate will not appear until all the chloride ions have reacted. The potassium chromate indicator remains colourless as long as Cl– is present, but the instant the last of the Cl– is bound up as silver chloride, the solution turns orange. The orange colour occurs when the extra chromate ion (CrO 24 − ) combines with the silver ion to form a complex ion called silver chromate. (II) Using the same procedure and data, salinity can also be found out with a slight variation in calculation. As in normal titrimetric methods, the normality of silver nitrate solution is determined by titrating it with the water sample of known salinity. This is nothing but the standard sodium chloride solution, the normality of which is known. Then, using the normality of silver nitrate solution, that of unknown water sample and hence, the amount of sodium chloride in it can be calculated. Requirements: Standard flasks, burette, graduated pipettes, conical flasks, silver nitrate solution (0.1 M), sodium chloride solution (5 ppt) (known water sample), potassium chromate, given water sample, distilled water. Reagent: Potassium chromate indicator: Dissolve 5 g of the solid in 100 mL of distilled water. Procedure: 1. Pipette out 10 mL of the water sample of known salinity into a clean conical flask. 2. Add 1 mL of potassium chromate indicator and titrate with the silver nitrate solution taken in the burette. 3. First a white precipitate of silver chloride is formed. 4. Continue the titration with constant swirling till the first permanent red-orange or reddish brown colour appears (this is silver chromate, and it only appears when all the chloride has reacted). The colour may be faint depending on the concentrations of the solutions. 5. Repeat the titration to get three concordant readings. 6. Repeat the above procedure with the given water sample (for which the salinity is to be determined). Calculation: Method I: Volume of AgNO3 solution consumed by known water sample = x mL Salinity of known water sample = 5 ppt Volume of AgNO3 solution consumed by given water sample = y mL Volume of AgNO 3 consumed by known water sample Salinity of known water sample =

Volume of AgNO 3 consumed by given water sample Salinity of given water sample

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ173

Therefore,

y x = 5 Salinity of given water sample

Rearranging this equation, we get y ×5 x Method II: Mass of NaCl dissolved to make a 5 ppt solution = 5 g in 1000 mL of distilled water 5 = 0.0855 Normality of this solution = 58.5 From the titration of known water sample with AgNO3 solution, using the titrimetric principle VNaCl u NNaCl = VAgNO3 u NAgNO3 Calculate the normality of AgNO3 solution. Apply the same principle to the titration of given water sample with AgNO3 solution and determine the normality of the given water sample (in terms of NaCl). Find out the strength and hence, the salinity of given water sample. Result: The salinity of given water sample = ........ ppt. Note: 1. The measurement of salinity is a special case of measuring total dissolved solids (TDS) in solutions with very high salt content, such as brackish water or seawater. Salinity is a dimensionless number that expresses the mass of dissolved salts in a given mass of solution. Mass of dissolved salts Salinity = Mass of solution 2. Surface water has low TDS, high turbidity and more bacteria, whereas groundwater has high TDS, low turbidity and less bacteria. Salinity of given water sample =

EXPERIMENT: 12.5 Aim: To determine the amount of chloride ions in a given sample of water. Requirements: Burette, graduated pipettes, conical flasks, standard flasks, beakers, silver nitrate solution (0.005 M), standard sodium chloride solution (0.005 M), potassium chromate (K2CrO4) indicator, water sample, distilled water. Reagents: 1. Standard sodium chloride solution (0.005 M): Dry NaCl at 140°C for an hour and cool it in a desiccator. Dissolve 0.0731 g of it in distilled water and make this solution up to 250 mL. 2. Silver nitrate solution (0.005 M): Dissolve 0.2124 g of AgNO3 in distilled water and make this solution up to 250 mL. Store this in a brown bottle. Standardize this solution using standard NaCl solution and 1 mL of K2CrO4 indicator. Make indicator correction in the titre value.

174ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

3. Potassium chromate indicator: Dissolve 2.5 g of solid in 50 mL of distilled water. Theory: Chloride ion is essential for the balance of electrolytes in our bodies. There is a continuous intake and excretion of chlorine from all animals and therefore, it is one of the abundant anions found in wastewater. It acts as a good marker ion for pollution sources. Chloride gives water a salty taste and is detectable at a level of 250 ppm if its cation is sodium. If calcium and magnesium are present as counter ions, this salty taste is detectable only at chloride concentrations of 1000 ppm or higher than that. High chloride levels can pose a threat to crops and freshwater aquatic plants as these do not have a mechanism to excrete the excess amount of salt. Such high levels of chloride in water increase the rate of corrosion of metallic pipes. Mohr’s method is used to determine the concentration of chloride ions in water. In this method, the chloride ion is titrated with silver nitrate solution using potassium chromate as the indicator. Ag+ (aq) + Cl–(aq) ֖ AgCl (s) (Ksp = 1.77 u 10–10); Ksp is known as the solubility product. A white precipitate of AgCl is formed till the end point is reached. Based on the solubility product of silver chloride, the silver ion concentration at the end point is 1.33 u 10–5 M. The chromate ion concentration required to initiate formation of silver chromate under this condition is calculated from the solubility product of silver chromate 6.6 u10–3 M. But at this concentration, chromate ion imparts very intense yellow colour as a result of which the red silver chromate cannot be easily detected. So, a low concentration of chromate ions is normally used. Hence, the concentration of silver ions should be increased. When excess of silver ions are added in the form of silver nitrate at the end point and all the chloride ions in solution have been precipitated, the silver ion concentration increases to a level at which the solubility product of silver chromate is exceeded. Then, silver chromate is precipitated as a reddish-brown product. 2Ag+ + CrO42– ֖ Ag2CrO4 (Ksp = 1.12 u 10–12) The appearance of reddish-brown colour indicates the end point. A small excess of Ag+ is required to produce a visible quantity of Ag2CrO4. Therefore, an indicator correction has to be made in the titre value by performance of a blank titration, described in the procedure. The optimum pH of the titration is in the range 7–8. At low pH, CrO42– is converted to Cr2O72– and at high pH, Ag+ is precipitated as AgOH. Hence, it is ideal to carry out the titration in a neutral medium. For all the titrations, a definite volume of the indicator should be used. Procedure: 1. Standardize the silver nitrate solution using standard sodium chloride solution and 1 mL of K2CrO4 indicator. 2. Pipette out 100 mL of water sample into a clean conical flask. 3. Add 1 mL of K2CrO4 indicator.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ175

4. Titrate this solution with the standardized silver nitrate solution taken in the burette. 5. The end point is the appearance of reddish brown colour. Carry out a duplicate titration in an identical manner. 6. Carry out a blank titration using 100 mL of deionized and chloride-free water and 1 mL of indicator. Subtract this titre value from that obtained for the water sample for indicator correction. Calculation: For standardization, VNaCl u NNaCl = VAgNO3 u NAgNO3 NAgNO3 = VNaCl u NNaClȺȦȺAgNO3 = N Volume of water sample = V1 mL Normality of chloride in water sample = NCl– Volume of AgNO3 solution = V2 mL (corrected titre value) Normality of AgNO3 solution = NAg+ V1 u NCl– = V2 u NAg+ NCl– = V2 uȱȺȦȺ1 Amount of chloride ion in the water sample = 35.45 u NCl– gL–1 = 35.45 u NCl– u 1000 mgL–1 (ppm) Result: The amount of chloride ions in the given sample of water = ........ mgL–1 (ppm) Note: 1. If the pH of water sample is less than 7, neutralize with 1 M NaOH, and if it is greater than 7, neutralize with 1 M H2SO4. 2. Intense colour of water will interfere in this estimation. Water can be decolourized by adding 3 mL of Al(OH)3 suspension to each 100 mL of water. Stir and keep it aside for 30 minutes. Then filter and wash the floc. Collect the filtrate (including the washings) quantitatively for titration. 3. Sulphite, sulphide and cyanide, if present in water, will also interfere in this estimation. These can be eliminated using appropriate standard procedure. 4. If the chloride content of the water sample is very low, 0.001 M AgNO3 can be used for titration.

EXPERIMENT: 12.6 Aim: To determine the acidity of a given sample of water and the amount of dissolved carbon dioxide present in it. Requirements: Standard flasks, burette, graduated pipettes, conical flasks, beakers, sodium hydroxide solution (0.02 N), standard oxalic acid solution (0.02 N), sodium thiosulphate (2.5% m/V) solution, methyl orange indicator, phenolphthalein indicator, water sample, distilled water. All the solutions must be prepared in previously boiled and cooled distilled water. Theory: Acidity is a measure of the capacity of water to neutralize bases. It is the sum of all titrable acids present in the water sample. Strong mineral acids and weak acids

176ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

such as carbonic acid, and acetic acid present in the water sample contribute to acidity of the water. Usually, dissolved carbon dioxide (CO2) is the major acidic component present in the unpolluted surface waters. Environmental Significance Ȋȱ Acidity interferes in the treatment of water. Carbon dioxide is of important consideration in determining whether removal by aeration or simple —Žž›Š•’£Š’˜—ȱ  ’‘ȱ •’–ŽȺȦȺ•’–Žȱ œ˜Šȱ Šœ‘ȱ ˜›ȱ Š ȱ  ’••ȱ ‹Žȱ Œ‘˜œŽ—ȱ Šœȱ ‘Žȱ  ŠŽ›ȱ treatment method. The size of the equipment, chemical requirements, storage spaces and cost of the treatment depend on the carbon dioxide present. Ȋȱ Aquatic life is affected by high water acidity. The organisms present are prone to death with low pH of water. Ȋȱ High acidity water is not used for construction purposes especially in reinforced concrete construction due to the corrosive nature of high acidity water. Ȋȱ Water containing mineral acidity is not fit for drinking purposes. Ȋȱ Industrial wastewaters containing high mineral acidity must be neutralized before they are subjected to biological treatment or direct discharge to water sources. Ȋȱ When water has a low pH, it is often referred to as “soft water.” Soft water is more acidic, therefore, it can be corrosive and harmful to any metal it comes into contact with. When the water corrodes the metal, this corrosion can then seep into the water. Ȋȱ Another problem with soft water is that it can cause stains in any clothing that is washed in it. Natural water normally contains both dissolved carbon dioxide and mineral acids. The amount of mineral acids in water is determined by titration with sodium hydroxide to a pH of 4.5 using methyl orange indicator. This acidity is known as methyl orange acidity. The total acidity due to both dissolved carbon dioxide and the mineral acid can be determined by titration with sodium hydroxide using phenolphthalein indicator to a pH of 8.3. This total acidity is known as phenolphthalein acidity. The presence of chlorine in water interferes and makes the methyl orange end point difficult to detect. This interference is prevented by the addition of a few drops of sodium thiosulphate solution. The acidity is expressed in terms of CaCO3 (mg L–1). If the water sample contains heavy metal salts, the sample is boiled to accelerate the hydrolysis of the metal salts and allowing the titration to be completed quickly. Procedure: Standardize the sodium hydroxide solution with the standard oxalic acid solution using phenolphthalein indicator. I. Methyl orange acidity (due to mineral acids): Pipette out 100 mL of the water sample into a conical flask. Add a few drops of sodium thiosulphate solution. Then add two drops of methyl orange indicator and titrate with the standardized sodium hydroxide solution. The end point is the appearance of faint orange

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ177

colour which is characteristic of pH 4.5. Repeat the titration to obtain at least three concordant values. Calculation: Volume of NaOH solution consumed = V1 mL (titre value) Normality of NaOH solution = N1 M 1000 mL of 1 N NaOH solution { g of CaCO3 = 50 g of CaCO3 2 V × N1 V1 mL of N1 NaOH solution { 1 u 50 g of CaCO3 1000 × 1 Volume of water sample taken = 100 mL V × N1 u 50 g of CaCO3 100 mL of water sample { 1 1000 × 1 V1 × N1 V × N1 1000 = 1 g of CaCO3 × 50 × 1000 × 1 100 2 V × N1 Therefore, the methyl orange acidity of water sample = 1 g of CaCO3 per litre 2 V × N1 = 1 u 1000 mg L–1 2 = X mg L–1 II. Phenolphthalein acidity (total acidity due to mineral acids and dissolved CO2): Pipette out 100 mL of the water sample into a conical flask. Add a few drops of sodium thiosulphate solution. Then add two drops of phenolphthalein indicator and titrate with the standardized sodium hydroxide solution. The end point is the appearance of pale pink colour. Repeat the titration to obtain at least three concordant values. Calculation: Following exactly the above calculation, phenolphthalein acidity in terms of CaCO3 V × N1 = 1 u 1000 mg L–1 = Y mg L–1 2 where V1 = Volume (mL) of NaOH solution consumed in the titration N1 = Normality of NaOH solution 1000 mL of water sample {

⎛ Y – X⎞ III. The amount of dissolved CO2 = ⎜ × 44 ⎝ 100 ⎟⎠ where molar mass of CaCO3 is 100 and that of CO2 is 44. Result: 1. Methyl orange acidity (due to mineral acids) = ........ mg L–1 2. Phenolphthalein acidity (total acidity due to mineral acids and dissolved CO2) = ........ mg L–1 3. The amount of dissolved CO2 = ........ mg L–1

178ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Note: Stirring the analyte during the titration must be avoided as it would lead to loss of dissolved carbon dioxide. Therefore, after a preliminary titration, a second titration may be carried out by adding a little less than the required amount of titrant to the water sample and then completing the titration quickly.

EXPERIMENT: 12.7 Aim: To determine the chemical oxygen demand of a given sample of water. Requirements: Burette, whole and graduated pipettes, standard flasks, beakers, conical flasks, heating mantle, potassium dichromate solution (0.01 N), Mohr’s salt solution (0.01 N), concentrated sulphuric acid, mercuric sulphate, silver sulphate, 1,10-phenanthroline monohydrate, ferrous sulphate heptahydrate, water sample, distilled water. Reagent: Phenanthroline-ferrous sulphate (ferroin) indicator: Dissolve 1.48 g of 1,10phenanthroline monohydrate and 0.7 g FeSO4.7H2O in 100 mL of distilled water. Theory: The chemical oxygen demand (COD) is a measure of the amount of oxygen in milligrams required to oxidize all the organic pollutants present in water to carbon dioxide and water by a strong oxidizing agent such as acidified potassium dichromate. This can be represented by the general equation: a + 8c CnHaOb + c Cr2O72– + 8c H+ o n CO2 + H2O + 2c Cr3+ 2 2 a b where c = n+ – 3 6 3 The determination of COD of water is necessary since the ratio of COD to BOD (Biochemical or Biological Oxygen Demand) helps in assessing the biodegradability of wastewater. The estimation of BOD is described in Experiment 12.8. In this experiment, a known amount of potassium dichromate solution is added, in presence of an acid, to the water sample. Some dichromate would be utilized for the oxidation of organic pollutants and the unreacted dichromate is titrated with the standard Mohr’s salt solution using ferroin indicator. Correction is to be carried out for the presence of chloride : Chloride ion, if present in water, gets itself oxidized to chlorine by potassium dichromate. Various experiments have shown that for each 1 mg L–1 of chloride ion, 0.23 mg of oxygen (COD) must be subtracted from the total COD measured. Therefore, the chloride content of the sample must be determined with a separate aliquot of the water sample (refer Experiment 12.5) to make this correction in COD. The interference by chloride ions can be avoided by the addition of mercuric sulphate to the sample. The mercuric ion combines with the chloride ion to form an inactive, weakly ionized HgCl2. There is no correction required in the calculated COD value, if the chloride ions are thus “masked” using mercuric sulphate.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ179

For the oxidation of aliphatic compounds, aromatic hydrocarbons and pyridine, silver sulphate is used as a catalyst. Procedure: 1. Standardize the Mohr’s salt solution by the usual procedure. 2. Exactly measure out 50 mL of the given water sample into a 500 mL conical flask. 3. Pipette out 25 mL of potassium dichromate solution (0.01 N) into this. 4. Add carefully, with gentle mixing, 50 mL of concentrated sulphuric acid. 5. Then add about 1 g of silver sulphate and 1 g of mercuric sulphate. ȱ Ŝǯȱ ›˜™ȱŒŠ›Žž••¢ȱŘȱ˜›ȱřȱ‹˜’•’—ȱŒ‘’™œȺȦȺ™ž–’ŒŽȱœ˜—ŽȺȦȺ™˜›ŒŽ•Š’—ȱ‹’œȱ’—˜ȱ‘Žȱ•Šœ”ȱ and then attach a reflux condenser. 7. Reflux the mixture for about two hours on a heating mantle. 8. Cool the flask and rinse the inside of the condenser with a small amount of distilled water so that the washings run into the flask. 9. Then add 5 drops of ferroin indicator and titrate the unreacted dichromate with standardized Mohr’s salt solution (0.01 N) in the burette. 10. The end point is the sharp change of colour from blue to reddish brown. 11. Simultaneously conduct a duplicate experiment with the same quantity of the water sample . 12. Perform a blank using 50 mL of distilled water (instead of the water sample), following the same procedure including refluxing, titrating, etc. Calculation: (V – V3 ) × N Normality of potassium dichromate used = 2 V1 Equivalent mass of oxygen = 8 Therefore, COD in terms of oxygen =

(V2 – V3 ) × N × 8 × 1000 mg L–1 V1

V1 = volume of water sample taken in mL V2 = blank titre value in mL V3 = titre value in mL with the water sample N = normality of the ferrous ammonium sulphate solution. Result: The chemical oxygen demand of the given sample of water = ....... mg L–1

EXPERIMENT: 12.8 Aim: To determine the biochemical oxygen demand of the given water sample. Theory: Biochemical oxygen demand (BOD), also called biological oxygen demand is defined as a measurement of the amount of dissolved oxygen needed or demanded by

180ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

aerobic biological organisms to break down organic material present in a given water sample at a certain temperature over a given time period. The general equation is ⎛ ⎛a 3 ⎞ a b 3 ⎞ CnHaObNc + ⎜ n + – – c⎟ O 2 o n CO2 + ⎜ – c⎟ H2O + c NH3 ⎝ ⎝2 2 ⎠ 4 2 4 ⎠ –1 BOD values may range from 30 to 100 mg L depending on the quality of water. Clean water samples have low BOD values while polluted water samples have high BOD values. The determination of BOD of water is essential because the ratio of COD to BOD helps in assessing the biodegradability of wastewater. An aliquot of the water sample is maintained in an incubator at 20°C for five days in a closed BOD bottle, without allowing air to enter. During this period, the bacterial decomposition gets completed and the water sample is said to be stabilized. Measuring the dissolved oxygen in the water sample before and after incubation would indicate the amount of oxygen used for bacterial and chemical action and for stabilizing the water. This represents the BOD. It is expressed in mg L–1. A water sample with a low BOD can be used as such for the BOD determination. But a water sample with a high BOD must be diluted accurately and pretreated before the determination of BOD. The pretreatment is essential because certain constituents present in a water sample can inhibit biochemical oxidation and interfere with the BOD analysis. Interferences include caustic alkalinity or acidity; the presence of residual chlorine; or the presence of toxic elements, including trace elements such as copper, lead, chromium, mercury and arsenic; or compounds such as cyanide. Procedure for pretreating samples for some common interferences is described in this experiment. Because bacterial growth requires nutrients such as nitrogen, phosphorus and trace metals, these are added to the dilution water, which is buffered to ensure that the pH of the incubated sample remains in a range (6.5–7.5) suitable for bacterial growth. Complete stabilization of a sample may require a period of incubation too long for practical purposes. Hence, five day period has been accepted as the standard incubation period. If the dilution water is of poor quality, the BOD of the dilution water will affect the results. Requirements: Incubator or a constant temperature water bath, BOD bottles (100 mL capacity), standard flasks, beakers, graduated pipettes, conical flasks, burette, airtight stoppered reagent bottles, potassium dihydrogen phosphate (KH2PO4), di-potassium hydrogen phosphate (K2HPO4), di-sodium hydrogen phosphate (Na2HPO4.7H2O), ammonium chloride, ferric chloride, calcium chloride, magnesium sulphate, manganese sulphate (MnSO4.4H2O), potassium fluoride (KF.2H2O), sodium hydroxide, sodium iodide, sodium azide (NaN3), sodium thiosulphate solution (0.01 N), potassium dichromate solution (0.001 N), concentrated sulphuric acid, freshly prepared starch indicator, distilled water. Reagents: 1. Manganese sulphate solution: Weigh about 25 g of MnSO4.4H2O and dissolve in 50 mL of distilled water and filter to get a clear solution.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ181

2. Potassium fluoride solution: Dissolve 20 g of KF.2H2O in 50 mL of distilled water. 3. Alkaline iodide-azide solution: Dissolve 25 g of NaOH and 7 g of NaI in distilled water and dilute to 500 mL. Dissolve 0.5 g of NaN3 in 25 mL of distilled water and add this solution with stirring to the alkaline iodide solution. 4. Phosphate buffer solution: Dissolve 0.85 g of KH2PO4, 2.18 g of K2HPO4, 3.34 g of Na2HPO4.7H2O and 0.17 g of NH4Cl in 100 mL of distilled water. 5. Ferric chloride solution : Dissolve 0.025 g of FeCl3-6H2O in distilled water and dilute the solution to 100 mL. 6. Calcium chloride solution: Dissolve 2.75 g of CaCl2 in 100 mL of distilled water. 7. Magnesium sulphate solution: Dissolve 2.25 g of MgSO4.7H2O in 100 mL of distilled water. 8. Water for dilution : Take 1 litre of distilled water in a 2-litre bottle. Shake this for about ten minutes so that the distilled water gets saturated with atmospheric oxygen. To this, add 1 mL of phosphate buffer solution, 1 mL of magnesium sulphate solution, 1 mL of calcium chloride solution and 1 mL of ferric chloride solution. Procedure: Procedure A-BOD determination without dilution: 1. If the water sample is fairly clean, it will have a low BOD value and it can be used as such directly. Fill four BOD bottles (100 mL capacity) up to the brim and tightly stopper these. 2. Make sure that there are no air bubbles inside the bottles. 3. Immediately determine the dissolved oxygen (DO) in two of the bottles (refer Experiment 12.2). 4. Seed the other two bottles with 1 mL of domestic wastewater and keep these bottles in an incubator or a constant temperature water bath at 20°C. After five days, determine the dissolved oxygen in these two bottles. BOD (mg L–1) = DO (before incubation) – DO (after incubation) Procedure B-BOD determination after dilution: 1. This method is used when the BOD value is quite high. Pipette out 50 mL of water sample into each of the two BOD bottles. 2. Fill the bottles up to the brim with the dilution water prepared and stopper the bottles tightly. 3. Immediately determine the DO of the diluted water using one of the bottles. 4. Incubate the other bottle for five days at 20°C and then determine the DO of the incubated water. Calculation: BOD (mg L–1) = DO (before incubation) – DO (after incubation) (without dilution)

182ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

BOD (mg L–1) = [DO (before incubation) – DO (after incubation] u f (with dilution) where ‘f ‘ is the dilution factor = Volume of the bottle yVolume of water sample Result: The biochemical oxygen demand of the given water sample = ........ mg L–1. Note: 1. BOD bottles of fixed volumes, provided with cup seal and vinyl tubing fixed at the mouth of the bottle, are commercially available. 2. If the BOD of the water sample is likely to be greater than the DO level, the sample must be accurately diluted before estimation. 3. Sometimes, the water sample may be very low in bacteria content and therefore, the water may not get stabilized in five days. In such a case, the water has to be seeded with a small, known volume of domestic wastewater. When this is done, a seed correction which is determined by measuring the BOD value of the wastewater used for seeding should be made. The waste water may be diluted and then its BOD value determined. 4. The ratio of COD to BOD indicates the ease with which a wastewater sample can be biologically degraded. ȱ ǻŠǼȱ ȱ ‘Žȱ ›Š’˜ȱ ȺȦȺȱ ’œȱ •Žœœȱ ‘Š—ȱ ŗǯŝǰȱ ‘Žȱ  ŠœŽ ŠŽ›ȱ ŒŠ—ȱ ‹Žȱ Œ˜–™•ŽŽ•¢ȱ biodegraded easily. (b) If the ratio is between 1.7 and 10, the wastewater can be biodegraded but not completely. (c) If the value is closer to 1.7, the sample can be slowly biodegraded. (d) If the ratio is more than 10, the wastewater cannot be biodegraded. Any attempt to biodegrade such wastewaters would be of no use. ‘žœǰȱ‘Žȱ›Š’˜ȱȺȦȺȱ‘Ž•™œȱ’—ȱŠœœŽœœ’—ȱ‘Žȱ‹’˜Ž›ŠŠ‹’•’¢ȱ˜ȱ ŠœŽ ŠŽ›ǯ

EXPERIMENT: 12.9 Aim: To estimate the total alkalinity of the given water sample. Requirements: Standard flasks, beakers, burette, graduated pipettes, conical flasks, hydrochloric acid solution (0.01 N), standard sodium carbonate solution (0.01 N), phenolphthalein indicator, methyl orange indicator, water sample, distilled water. Theory: In pure water, addition of even small amounts of a strong acid results in a drastic change in pH. For example, if one drop of 0.1 M of HCl is added to 250 mL of pure water, the pH will change from 7.0 to 4.7. Most natural waters are protected from such a drastic change by the presence of ionic compounds that react quickly with acid in order to buffer the system. This capacity of a solution to neutralize the acid is known as acid neutralizing capacity or total alkalinity. It is expressed in terms of CaCO3 (mgL–1). The measurement of alkalinity is useful for water and wastewater treatments. Major part of alkalinity in natural waters is due to hydroxides, carbonates and bicarbonates although trace amounts of borates, silicates and phosphates and free hydroxyl ions may be present.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ183

The alkalinity of the water can be determined by titrating it with standard hydrochloric acid solution. If phenolphthalein is used as an indicator, the end point (pH = 8.3) corresponds to complete neutralization of OH– ions and half neutralization of CO32– ions. This partial alkalinity is called phenolphthalein alkalinity. The reactions are NaOH + HCl o NaCl + H2O Na2CO3 + HCl o NaHCO3 + NaCl During methyl orange end point (pH = 4.5), HCO3– originally present and that obtained from CO32– are neutralized according to the equation NaHCO3 + HCl o NaCl + CO2 + H2O The total alkalinity corresponds to the amount of acid required to react with the hydroxide, carbonate and bicarbonate present in water. Procedure: 1. Standardize the hydrochloric acid solution using standard sodium carbonate solution and phenolphthalein indicator. Phenolphthalein alkalinity: 2. Pipette out 100 mL of the water sample into a clean conical flask. 3. Add 2 drops of phenolphthalein indicator and if pink colour develops, titrate with the standardized hydrochloric acid taken in the burette. If pink colour does not appear on adding phenolphthalein (it implies that hydroxides are absent in the water sample), add methyl orange and titrate as given under “Total alkalinity”. 4. The end point is the disappearance of pale pink colour. Note down the volume of the acid consumed. Total alkalinity: 5. To the same solution, add 2 drops of the methyl orange indicator. 6. Check the burette reading and continue the titration with the same hydrochloric acid in the burette. 7. The end point is the colour change from yellow to red-orange. Calculation: V × N × 50 × 1000 Phenolphthalein alkalinity (mgL–1) in terms of CaCO3 = 2 V1 (Equivalent mass of CaCO3 = 50) where V1 = volume of water sample (mL) V2 = volume of hydrochloric acid used (mL) (titre value) N = normality of hydrochloric acid V × N × 50 × 1000 Total alkalinity (mgL–1) in terms of CaCO3 = 3 V1 where V1 = volume of water sample (mL)

184ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

V3 = total volume of hydrochloric acid used (mL) (including the first titration) N = normality of hydrochloric acid If all the three, hydroxides, carbonates and bicarbonates, are present in the water sample, only phenolphthalein and total alkalinities could be determined and not individual alkalinity. But, if one of them is absent, then individual alkalinities with respect to the other two can be determined. Result: Phenolphthalein alkalinity in terms of CaCO3 = ....... mgL–1 Total alkalinity in terms of CaCO3 = ........ mgL–1 Note: 1. The total alkalinity can also be determined potentiometrically. 2. If pH of water sample is found to be less than 4.5, there is no need to determine alkalinity since the sample does not have any acid neutralizing capacity. 3. The total alkalinity is also referred to as m value and phenolphthalein alkalinity is called p value. A plot of p, m values and pH is normally used to determine the amount of calcium in drinking water (Tilliman’s curve). 4. The values of p and m are used to determine various types of alkalinities such as hydroxides (OH–), carbonates (CO32–) and bicarbonates (HCO3–). (a) If p = m, water sample contains only OH– alkalinity ȱ ǻ‹Ǽȱ ȱ™ȱƽȱ–ȺȦȺŘȱ˜›ȱ–ȱƽȱؙȱ ŠŽ›ȱœŠ–™•ŽȱŒ˜—Š’—œȱ˜—•¢ȱ32– alkalinity (c) If p = 0; m > 0, water sample contains only HCO3– alkalinity ȱ ǻǼȱ ȱ™ȱǁȱ–ȺȦȺŘȱ˜›ȱؙȱǁȱ–ǰȱ ŠŽ›ȱœŠ–™•ŽȱŒ˜—Š’—œȱ – and CO32– alkalinities ȱ ǻŽǼȱ ȱ™ȱǀȱ–ȺȦȺŘȱ˜›ȱؙȱǀȱ–ǰȱ ŠŽ›ȱœŠ–™•ŽȱŒ˜—Š’—œȱ32– and HCO3– alkalinities 5. A water sample with high alkalinity is resistant to changes in pH. 6. The difference between the pH and alkalinity is that pH is an intensity factor (which measures the concentration of acid or alkali immediately available for reaction) while alkalinity is a capacity factor (it measures the ability of water to sustain reaction with added acid or base).

EXPERIMENT: 12.10 Aim: To collect a soil sample in small quantity that truly represents the field for which the soil test is required. Requirements: Screw type, tube type and post-hole type augers, spade, a trough or bucket to collect the dugout soil, 8 to 10 inches diameter sieve having 6 meshes per inch (big meshes), sieve with 2 mm mesh (very small mesh), air dryer with humidity controller, simple grinder, pestle with mortar, pieces of clean cloth (approximately 0.5 m u 0.5 m), white paper, flat metal spatula and polythene bags for storing the final samples. Procedure: For taking out soil samples from the field: 1. Mark the overall boundary of the field. Leaving 80 cm to 100 cm from the boundary, mark the inner area for taking out the soil samples.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ185

2. The field should have uniform texture, colour, same crop grown over the area and slope. If any of these parameters vary, then mark the areas accordingly and soil samples are to be taken separately from each area and referenced. 3. Avoid taking samples near trees, recently fertilized areas, composite piles or stored areas and the areas which are not to be considered for testing. 4. Select an area of the field (refer point 2) and mark points on the area for taking out the soil. Samples are to be taken randomly (in a zig-zag manner) from each of the marked points taking care of the points mentioned in point 3. ȱ śǯȱ ž’Ž•’—Žȱ˜›ȱžœ’—ȱ‘Žȱ˜˜•œDZȱǻŠǼȱ˜›ȱ›¢ȱŠ—ȱ‘Š›ȱœ˜’•ǰȱžœŽȱœŒ›Ž ȱ¢™ŽȱŠžŽ›ȱ (b) For moist and soft soil, use tube auger or spade (c) For wet areas, use post hole auger. 6. Make a V-shaped cut up to a depth of 15 to 20 cm. Take out a slice of 2 to 3 cm thick. 7. Samples taken out from each point must have equal volume and mass to achieve a truly representative sample. 8. Collect the samples so taken in a trough or a bucket. Processing the sample: 1. Sieving: Pass the sample through the sieve by gently rubbing over the mesh of the sieve to remove stones and gravels using the sieve with bigger mesh (6 meshes per inch). 2. Drying: Keep aside a portion of the sieved fresh moist sample for carrying out some analyses. Dry the remaining portion in shade at about 25°C and at the relative humidity of 25–60%. Some analyses require the sample to be oven dried at 100°C. ȱ řǯȱ ›’—’—ȱ Š—ȱ œ’ŽŸ’—DZ Using a roller or pestle with mortar, break the large lumps of soil into fine soil and pass it through a 2 mm sieve before the soil gets dried. 4. Mixing: The sample has to be thoroughly mixed. To achieve this, place the soil after grinding, sieving and drying on a clean square cloth. Hold any two opposite corners. Bring one corner over the sample keeping the other corner fixed so that the sample gets mixed. Repeat this with the other corner keeping the first corner fixed. This process is to be repeated for the other two corners. The entire process is to be repeated 8 to 10 times so that the sample gets thoroughly mixed. 5. Store the soil sample in a polythene bag or in an airtight container with screw cap. Label properly with the date of sampling and the name of location from where the soil is collected.

EXPERIMENT: 12.11 Aim: To qualitatively detect the presence of carbonate, ammonium, nitrite, nitrate, phosphate and potassium in a given soil sample.

186ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Requirements: Standard flasks, conical flasks or reagent bottles with lids, beakers, test tubes, Whatman filter papers No. 42 and 1, sodium acetate tri hydrate, glacial acetic acid, mercuric chloride, potassium iodide, sodium hydroxide, D-naphthylamine, sulphanilic acid, ferrous sulphate, ammonium nitrate, ammonium molybdate, stannous chloride, tin metal pieces, cobalt nitrate hexa hydrate (Co(NO3)2.6H2O), sodium nitrite, concentrated and dilute acids - HCl, HNO3, H2SO4, freshly prepared starch indicator, soil sample, distilled water. Reagents: 1. Extracting (Morgan’s) solution: Dissolve 25 g of sodium acetate tri hydrate in about 100 mL of distilled water. Add 8 mL of glacial acetic acid and make up to 250 mL with distilled water. 2. Nessler’s reagent: Dissolve 6 g of HgCl2 and 4 g of KI together in ammonia-free distilled water and make up to 25 mL. To this add 25 mL of sodium hydroxide (6 M) solution. Allow it to stand overnight and then decant. 3. DȬ—Š™‘‘¢•Š–’—Žȱ œ˜•ž’˜—DZȱ ’œœ˜•ŸŽȱ Ŗǯśȱ ȱ ˜ȱ œ˜•’ȱ ’—ȱ ŗŖŖȱ –ȱ ˜ȱ ŘƖȱ ǻŸȺȦȺŸǼȱ hydrochloric acid. 4. Sulphanilic acid solution: Dissolve 1 g of solid in 125 mL of 5 N acetic acid. 5. Ammonium molybdate solution: Dissolve 26 g of ammonium nitrate in 25 mL of distilled water and add 40 mL of concentrated nitric acid. Dissolve separately 2.35 g of ammonium molybdate in 15 mL of distilled water. Add this slowly to the nitrate solution with constant stirring. Make up the volume to 100 mL with distilled water. Keep the solution at 60°C for 6 hours in a water bath. Allow the solution to stand overnight. Then, filter through a Whatman No. 42 filter paper, if required. 6. Stannous chloride solution: Prepare a stock solution by dissolving 10 g of crystalline SnCl2 in 25 mL of concentrated HCl. Warm, if necessary. Keep this solution in an amber coloured bottle. Just before use, dilute 1 mL of stock solution to 100 mL with distilled water. To prevent oxidation and to keep the stock solution for long, add a piece of tin metal . 7. Sodium cobaltinitrite solution: Prepare 25 mL of a saturated solution of cobalt nitrate hexa hydrate and 25 mL of a saturated solution of sodium nitrite. Mix both the solutions with constant stirring. Then add 2.5 mL of glacial acetic acid and dilute to 100 mL with distilled water. Allow to stand overnight and filter, if required. Theory: Carbonate is present in soil as calcium and magnesium carbonate. Calcium carbonate is present mainly in the free state. Nitrogen is normally present as ammonium, nitrate or nitrite ion and phosphorus is present as orthophosphate. Potassium exists as its silicate. For the qualitative analysis, the nutrients in the soil are extracted in to the solution. Morgan’s extracting solution, which is a mixture of sodium acetate tri hydrate and acetic acid solutions, is used for this purpose. The soil extract is then tested with the respective reagents for the presence of various ions. The micronutrients such as Zn, Cu, Fe and Mn are present in very small quantities in soil and so, the qualitative detection is not successful for these elements.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ187

Procedure: 1. Weigh accurately about 5 g of soil sample and transfer it to a glass bottle or a conical flask carrying a lid. 2. Add 50 mL of the extracting solution and shake thoroughly for about an hour. ˜žȱ–Š¢ȱžœŽȱŠ’Š˜›ȺȦȺœ‘Š”Ž›ǰȱ’ȱŠŸŠ’•Š‹•Žǯȱ 3. Filter through Whatman No. 1 filter paper. 4. Analyze the soil extract (i.e., the filtrate) as given in Table 12.1. Table 12.1: Qualitative analysis of the soil sample S. No. Experiment 1. To a few drops of the extract, add a few drops of dilute hydrochloric acid (1:1) 2. To a small portion of the extract, add a few drops of Nessler’s reagent. 3.

4.

5.

6.

a. b.

7.

8.

Treat a few mL of the extract with a few drops of dilute acetic acid. Then add a few drops each of sulphanilic acid and D-naphthylamine solutions. Add a small quantity of dilute sulphuric acid to a few drops of the extract. Then add a few crystals of KI followed by 2 or 3 drops of fresh starch indicator Brown-ring test for nitrate (if NO2 is absent): To a part of the extract, add a few drops of dilute sulphuric acid and freshly prepared ferrous sulphate solution. Then add slowly and carefully concentrated sulphuric acid in drops along the sides of the test tube. If nitrite is present, eliminate it as below: Acidify a portion of the extract with dilute sulphuric acid and heat with solid ammonium chloride or urea until no more gas is evolved. This treatment decomposes nitrite nitrogen which volatilizes off. Test a small portion of this solution with dilute sulphuric acid, KI and starch to ensure the complete removal of nitrite. Then divide the nitritefree solution into two parts. Perform brown ring test with one part of the solution. To another part, add a pinch of zinc dust and heat. Then add a few drops each of dilute sulphuric acid, KI and starch. Now add a pinch of zinc dust and heat in order to reduce nitrate to nitrite. To a little of the extract add a small quantity of concentrated nitric acid, ammonium molybdate and dilute stannous chloride solutions. To a small quality of the extract add a small amount of sodium cobaltinitrite solution.

Observation Effervescence takes place Reddish-brown colouration or precipitate Red or pink colour Blue colour

Inference Presence of carbonate (as CaCO3 and MgCO3) Presence of nitrogen as ammonium ion Presence of nitrite nitrogen Presence of nitrite nitrogen

Presence of nitrate A brown-ring at the junction nitrogen of two layers

Formation of a Presence of nitrate brown-ring nitrogen No blue colour Indicates that nitrite has been removed completely. Presence of nitrate Blue colour nitrogen Blue colour Presence of phosphorus (phosphate) Yellow precipitate

Result:ȱ‘Žȱ’ŸŽ—ȱœ˜’•ȱœŠ–™•ŽȱŒ˜—Š’—œȱǯǯǯǯǯǯǯǯȱ’˜—ȺȦȺ’˜—œǯ

Presence of potassium

188ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 12.12 Aim: To determine the pH value of a given soil sample. Requirements: The pH meter, glass electrode and saturated calomel electrode or combined electrode, standard flasks, beakers, potassium hydrogen phthalate (KHC8H4O4), sodium tetraborate (borax), soil sample, distilled water. 1. For buffer solution of pH 4.0: Dissolve 1.021 g of potassium hydrogen phthalate (KHC8H4O4) in distilled water and make up to 100 mL with distilled water. 2. For buffer solution of pH 9.2: Dissolve 1.91 g of sodium tetraborate (borax) in distilled water and make up to 100 mL with distilled water. The pH of the soil indicates its alkaline or acidic or neutral nature. It is very important for the growth of various crops. Strong alkaline or acidic nature of the soil will kill the crop. The pH value must be known to take suitable measures for the reclamation of the soil. The pH of the soil is given by the equation: pH = –log10aH+, where aH+ is the activity of hydrogen ions in gram ions L–1. It can also be expressed in terms of the concentration of H+ ions as pH = –log10 [H+]. These ions in the soil are formed by the dissociation of soluble acids and soil particles. Electrometry method is used for the determination of soil pH. This method uses voltaic cell principle with the test solution acting as electrolyte. It consists of two Ž•ŽŒ›˜Žœǰȱ˜—ŽȱŠœȱ›ŽŽ›Ž—ŒŽȱŽ•ŽŒ›˜ŽȱŠ—ȱ‘Žȱ˜‘Ž›ȱŠœȱ–ŽŠœž›’—ȺȦȺ’—’ŒŠ˜›ȱŽ•ŽŒ›˜Žǯȱ The reference electrode is maintained at a constant voltage with respect to solution and is not affected by pH changes. The voltage of the other electrode, i.e., the measuring Ž•ŽŒ›˜ŽȱŒ‘Š—Žœȱ ’‘ȱ‘ŽȱŒ˜—žŒ’Ÿ’¢ȱŠ—ȱ™ ȱ˜ȱ‘Žȱœ˜•ž’˜—ǯȱ •ŠœœȱŽ•ŽŒ›˜Žȱ™ ȱ meter with saturated calomel as reference electrode is usually used for determining the soil pH. Digital pH meter can also be used. It has a combined electrode. Normally, the voltage indicator is calibrated directly in pH units. The soil pH is determined at the moisture saturation percentage due to the following reasons: 1. Different soils can be conveniently wetted to equipotential moisture status. 2. The moisture films of the soil solutions form with required thickness to make good contact with the glass electrode. 3. At saturation condition, the moisture content is the highest possible in any field under study. 4. Small changes in dilution of the soil solution will have very small effect on the pH readings. Procedure: 1. Set the temperature knob of the instrument at 25°C , as the pH meter reading 0 to 14 is valid only at 25°C. This is because the dissociation of a solution increases with temperature. 2. Calibrate the pH meter with the buffer solutions of pH 4.0 and 9.2 prepared as given above.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ189

3. Take the soil sample in a 50 mL beaker to approximately half its height and add distilled water slowly in small quantities. 4. Do not stir the soil and continue to add water until it just wets the entire soil quantity taken in the beaker. 5. Add a few more drops till the surface shines with traces of water and then slowly stir the soil with a glass rod (only after the soil is completely wet and the surface shines with traces of water. Otherwise, puddle mass of soil will be formed, obstructing and making the water to flow slowly). 6. Make a thin paste of the soil by slow stirring. If needed, add a few more drops of water to make it a thin paste. In case paste becomes too watery, add a small quantity of soil to make a thin paste. When the glass rod is withdrawn, the paste should flow to close the hole. If not add little more water. 7. The soil paste has now attained the moisture saturation percentage and is known to have equipotential moisture content for various soils. The surface of such a soil paste will shine with water and is said to be at the flow point or liquid limit. 8. Insert slowly and carefully the electrodes in the soil paste and note down the pH meter reading after the reading becomes steady and constant. Move the electrodes a little bit so that the electrodes do not have water film around them. 9. Repeat the measurement 4 to 5 times to get a consistent reading. 10. Before taking the measurement each time, ensure that the electrodes are washed  ’‘ȱ’œ’••Žȱ ŠŽ›ȱŠ—ȱ ’™Žȱ›¢ȱ ’‘ȱŠȱ’•Ž›ȺȦȺ’œœžŽȱ™Š™Ž›ǯȱ’›ȱ‘Žȱ™ŠœŽȱ˜›ȱ soil suspension well before proceeding with every measurement. 11. Frequently check the calibration of the instrument with buffers of known pH. Result: The pH value of the given soil sample = ........ Note : 1. The soil solution must be stirred whenever reading is taken. 2. The electrodes should be kept in distilled water when not in use. 3. Terminology commonly used to describe the acid-base status of soils is as follows: pH less than 4 strongly acidic 4 to 5 moderately acidic 5 to 6 slightly acidic 6 to 8 neutral 8 to 9 slightly alkaline 9 to 10 moderately alkaline greater strongly alkaline than 10

190ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 12.13 Aim: To determine the amount of total soluble salts present in a given sample of the soil. Requirements:ȱ˜—žŒ’Ÿ’¢ȱ–ŽŽ›ǰȱŒŽ••ǰȱ‹ŽŠ”Ž›œǰȱœŠ—Š›ȱ•Šœ”œǰȱŒ˜—’ŒŠ•ȱ•Šœ”ǰȱœ’››Ž›ȺȦȺȱ glass rod, standard potassium chloride solution (0.05 M) (this solution has a specific conductivity of 6.669 millimhos cm–1 at 25°C), soil sample, distilled water. Theory: Soil contains both organic and inorganic soluble salts which will have to be determined to understand the nature of the soil and to carry out further necessary treatment to make the soil cultivable. The strength of the soluble or ionic salts in a soil is determined from the water extract of the soil. The ionic strength of the water extract of a soil is directly proportional to the salt content of the soil. Due to the electrical nature of the ions, the electrical conductivity of the water extracts and hence that of the soil, increases with the amount of soluble salts present in the soil. Thus, the concentration of soluble salts in a soil sample can be determined by measuring the electrical conductivity of its water extracts. The determination of conductance of saturated paste (or water extract) of the soil provides a more representative measurement of total soluble salts in the soil solution because it more closely approximates the water content of the soil under field conditions. However, this is more time-consuming and more susceptible to error due to variability between analysts in the preparation of the saturated paste. In most soil testing laboratories where large numbers of samples are to be analyzed, measurement of conductance is done with a fixed soil : water ratio ranging from 1:1 to 1:5. This method is fast, easily done and reproducible across a wide range of salts and soils. The ratio may be in terms of volume or mass. This method is more useful to compare the salt contents and salinity of different types of soils. The conductance is measured in millimhos (1 mho = 1000 millimhos). Specific conductance or conductivity, N is the conductance of 1 cm3 of an electrolyte solution at 25°C. This is measured with a cell of known cell constant. The conductivity of the solution is normally expressed in millimhos cm–1, which is equivalent to the SI unit of decisiemens per metre (dS m–1). The unit dS m–1 is the official international unit. Cell constant (K) is usually determined with a standard KCl solution. Nsoil = K u Csoil where Nsoil = specific conductance or conductivity of water extract of the soil Csoil = conductance of water extract of soil measured K = cell constant The cell constant K is determined with the standard KCl solution using K = NKClȺ ȦȺKCl where NKCl = specific conductance or conductivity( known) of standard KCl solution CKCl = measured conductance of standard standard KCl solution Procedure: 1. Switch on the conductivity meter 20 minutes before starting the experiment.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ191

2. 3. 4. 5.

Adjust the temperature to 25°C. Calibrate it using the standard KCl solution (0.05 M). Then measure the conductance of this solution using the given cell. Take 30 g of the soil sample in a 100 mL conical flask and add 60 mL of distilled water. 6. Shake the flask well for about an hour (can use mechanical shaker) and allow it to stand for 10 minutes. 7. Take out the supernatant liquid and measure its conductance using the same cell. 8. The temperature knob should be set at 25°C. Calculation: Conductance of 0.05 M KCl solution, CKCl = ........ millimhos Cell constant, K = NKClȺ ȦȺKCl ƽȱŜǯŜşşȺȦȺKCl cm–1 (NKCl = 6.699 millimhos cm–1) Measured conductance of the soil extract = Csoil = ........ millimhos Specific conductance or conductivity of the soil extract, Nsoil = K u Csoil The concentration of salts in the given soil sample = 12.5 u Nsoil = ........ ppm (mg L–1) Based on experiments, it was found that the concentration of salts in the soil sample is directly proportional to its specific conductivity. The proportionality constant varies for different salts. However, for most of the salts the proportionality constant is found to be around 12.5 (Soil Chemical Analysis, M L Jackson, Prentice-Hall of India Pvt. Ltd., New Delhi, 1973). Result: The amount of total soluble salts present in a given sample of the soil = ........ ppm (mg L–1) Note: 1. The concentration of salts in the soil can also be estimated by independent methods using elaborate procedures and compared with this method. 2. Soils are classified based on their electrical conductivity (EC) of the saturated extract and exchangeable sodium percentage (ESP). Exchangeable sodium percentage is the fraction, expressed as a percentage of exchangeable sodium ions compared to total exchangeable ions. (i) Normal soils have EC values less than 4 dS m–1 and ESP values less than 15%. (ii) Saline soils have EC values greater than 4 dS m–1 with an ESP value less than 15%. (iii) Sodic soils have EC values less than 4 dS m–1 but ESP greater than 15%. (iv) Saline-sodic soils have both high EC (>4 dS m–1) and high ESP (>15%).

EXPERIMENT: 12.14 Aim: To estimate the amount of calcium and magnesium in a given sample of the soil.

192ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Requirements: Pestle and mortar, standard flasks, burette, graduated pipettes, conical flasks with stoppers, ammonium hydroxide, sodium acetate (CH3COONa.3H2O) glacial acetic acid, sodium salt of ethylenediamine tetra acetic acid (Na2EDTA.2H2O), zinc sulphate (ZnSO4.7H2O), Eriochrome black T, dilute H2SO4, methanol, ammonium chloride, sodium hydroxide, murexide, sodium chloride, distilled water, Whatman No. 42 filter paper. Reagents: 1. Extracting reagent: Dissolve 25 g of sodium acetate tri hydrate in about 100 mL of distilled water. Add 8 mL of glacial acetic acid and make up to 250 mL with distilled water. This is called Morgan’s extracting solution. 2. EDTA solution (0.01 M): Dissolve 1.85 g of solid Na2EDTA.2H2O in distilled water and make this up to 500 mL. 3. Standard Zn2+ solution (0.01 M): Weigh 0.7185 g of ZnSO4.7H2O solid and dissolve it in a minimum quantity of dilute H2SO4 and make the solution up to 100 mL with distilled water. 4. Eriochrome black T indicator: Dissolve 0.5 g of solid in 100 mL of methanol. 5. Ammoniacal buffer solution (pH 10): Dissolve 18 g of ammonium chloride in 75 mL of about 30% ammonium hydroxide and dilute to 500 mL with distilled water. 6. Sodium hydroxide solution(1 M): Dissolve 4 g of solid in water and make up to 100 mL with distilled water. This has a pH of 12. 7. Murexide indicator: Mix 0.1 g of murexide with 9 g of sodium chloride and grind this into a fine mixture using a clean mortar and pestle. Theory: Calcium and magnesium are present in the dissolved state in the saturated soil extract and are very important constituents for the growth of the crop. Although ŠȺȦȺȱ›Š’˜ȱŸŠ›’Žœȱ˜›ȱ’Ž›Ž—ȱœ˜’•œȱŠ—ȱŒ›˜™œǰȱ’ȱ’œȱ‘Žȱ–˜œȱ’–™˜›Š—ȱŠŒ˜›ȱ˜›ȱŒ›˜™ȱ growth and yield. The amounts of soluble Ca and Mg are obtained by extracting the soil with a solution of sodium acetate and measuring their concentrations in the extract by titration with EDTA (ethylenediamine tetra acetic acid). These are normally expressed as mass per 100g of the soil. Alternatively, Ca and Mg in the extracts can also be measured by atomic absorption spectrophotometry. Eriochrome black T is not a suitable indicator for the titration of calcium ions since the complex formed between calcium and Eriochrome black T is weak and this leads to premature dissociation of the complex which results in a premature end point. Murexide is found to be a suitable indicator for calcium titration with EDTA. It forms a red complex with calcium at pH = 12 and higher. At this pH, magnesium precipitates as Mg(OH)2. So, for simultaneous determination of Mg and Ca two titrations are to be carried out: one at pH = 10 and using Eriochrome Black T (the sum of moles of both metals is obtained) and second, at pH = 12 and using Murexide (only calcium is titrated).

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ193

Procedure I: (for estimation of calcium only) 1. Take 25 g (accurately weighed) of soil sample in a conical flask and add 250 mL of extracting reagent. Place the stopper on the flask and shake the flask thoroughly for about 15 to 20 minutes. 2. Filter the solution using a Whatman No. 42 filter paper. 3. Collect the filtrate in another conical flask for analysis. This filtrate is the soil extract. 4. Standardization of EDTA solution: (i) Pipette out 10 mL of standard Zn2+ solution into a clean conical flask, add 3 mL of ammoniacal buffer and 2 drops of Eriochrome black T indicator. (ii) Titrate this solution with the EDTA solution. The end point is the change of colour from wine red to steel blue. (iii) Repeat the titration to get at least three concordant values. (iv) Calculate the molarity of EDTA solution using the equation Vzn2+ u M zn2+ = VEDTA u MEDTA 5. Estimation of calcium in soil: (i) Pipette out 50 mL of the soil extract into a clean conical flask. (ii) Add 5 mL of 1 M sodium hydroxide solution and 50 mL of distilled water. (iii) Then add 1 g of murexide indicator and titrate the solution quickly with the EDTA taken in the burette. (iv) The end point is the colour change from pink to purple. (v) The titration should be completed quickly to avoid the possible formation of calcium carbonate precipitate. (vi) Repeat the titration to obtain at least three concordant values. Calculation: (for the estimation of calcium only) Volume of soil extract = V1 mL Volume of EDTA solution = V2 mL (titre value) Molarity of EDTA solution = M V2 × M =X Therefore, molarity of calcium in soil extract = V1 So, the amount of calcium in gL–1 =

V2 × M × 40 = A V1

where 40 is the atomic mass of calcium. A × 50 = B g 1000 i.e., 25 g of soil (originally taken) contains B g of calcium B × 100 g of calcium Hence, 100 g of soil will contain 25 The amount of calcium in 50 mL of extract (titrated) =

194ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Procedure II: (for estimation of calcium and magnesium in soil): 1. Pipette out 50 mL of the soil extract into a clean conical flask. 2. Add 3 mL of ammoniacal buffer and 2 drops of Eriochrome black T indicator. 3. The solution becomes wine red. Titrate the solution with EDTA taken in the burette. 4. The end point is the disappearance of the last shade of red and formation of steel blue colour. 5. Repeat the titration to obtain at least three concordant values. Calculation: (for the estimation of calcium and magnesium in soil) Volume of soil extract = V1 mL Volume of EDTA solution = V2 mL (titre value) Molarity of EDTA solution = M V ×M =Y Therefore, molarity of calcium and magnesium in soil extract = 2 V1 i.e., the number of moles of calcium and magnesium per litre = Y The number of mole of calcium per litre of the soil extract = X (from part I) So, the number of mole of magnesium in the soil extract = (Y – X) The amount of magnesium in the soil extract = (Y – X) u 24.3 gL–1 where 24.3 is the atomic mass of magnesium. (Y – X) × 24.3 The amount of magnesium in 50 mL of extract (titrated) = × 50 1000 =Cg i.e., 25 g of soil (originally taken) contains C g of magnesium Hence, 100 g of soil will contain

C × 100 g of magnesium 25

Result: ‘ŽȱŠ–˜ž—ȱ˜ȱŒŠ•Œ’ž–ȱ’—ȱ‘Žȱœ˜’•ȱƽȱȱƽȱǯǯǯǯǯǯǯǯȱȺȦȺŗŖŖȱ ‘ŽȱŠ–˜ž—ȱ˜ȱ–Š—Žœ’ž–ȱ’—ȱ‘Žȱœ˜’•ȱƽȱȱƽȱǯǯǯǯǯǯǯǯȱȺȦȺŗŖŖȱ The Ca : Mg ratio = ........ Note: ȱ ŗǯȱ ŠȺȦȺȱ ›Š’˜ȱ ŸŠ›’Žœȱ ˜›ȱ ’Ž›Ž—ȱ œ˜’•œȱ Š—ȱ Œ›˜™œȱ ‹žȱ ’ȱ ’œȱ ‘Žȱ –˜œȱ ’–™˜›Š—ȱ factor for crop growth and yield. 2. If there is interference by the ions of Fe, Zn, Al and Ca, they are inhibited or masked by the addition of masking agents, such as potassium cyanide and hydroxylamine hydrochloride. 3. During calcium estimation, instead of murexide, hydroxynaphthol blue indicator (end point-colour change from blue to violet) or calconcarboxylic indicator (end point-colour change from wine red to pale blue) can be used.

—Ÿ’›˜—–Ž—Š•ȱŠ›Š–ŽŽ›ȱŽŠœž›Ž–Ž—ȳ195

VIVA QUESTIONS 1. 2. 3. 4. 5.

ȱ

ȱ ȱ

ȱ

What is SPM? What are different types of particulates? What is meant by ‘dissolved oxygen’ and how is it expressed? Why do we have to fix the water sample? How do we do it? What is the basis of Winkler method? What is the role of sodium azide in the determination of dissolved oxygen in water? 6. What is turbidimetry? How is it compared with colorimetry? What are its advantages over other methods? 7. Indicate the equation used in turbidimetric technique. Şǯȱ Ž’—ŽȱȁœŠ•’—’¢ȱ˜ȱ ŠŽ›Ȃǯȱ ’ŸŽȱ’œȱž—’ǯ 9. Describe a method to find out the ‘salinity of water’. 10. What are the adverse effects of particulates on humans? 11. What is the principle behind Mohr’s method? 12. What is EDTA? 13. Why is Eriochrome black T not a suitable indicator for calcium ions? 14. Which indicator is normally used for calcium titration with EDTA? 15. What is a masking agent? 16. What do you mean by phenolphthalein and methyl orange acidities? 17. Why do you add a few drops of sodium thiosulphate solution during the estimation of acidity of water? 18. How is the acidity of water expressed? 19. What are COD and BOD? 20. What is ferroin indicator? 21. How is BOD related to pollution of water? 22. What is meant by incubation period? 23. What is the use of determination of COD and BOD? 24. What is ‘total alkalinity’ of water? 25. Differentiate between total alkalinity and phenolphthalein alkalinity. How are these expressed? 26. What is an extracting solution? Name one. 27. What important ions are usually present in the soil? ŘŞǯȱ ’ŸŽȱ˜—ŽȱŽœȱ˜ȱŽŽŒȱ‘ŽȱŸŠ›’˜žœȱ’˜—œȱ™›ŽœŽ—ȱ’—ȱœ˜’•ȱœŠ–™•Žœǯ 29. Define pH of the soil. řŖǯȱ ’ŸŽȱ‘Žȱ™›’—Œ’™•Žȱ˜ȱŽ•ŽŒ›˜–Ž›¢ȱ–Ž‘˜ǯ 31. What is meant by moisture saturation percentage of a soil? 32. Why is it necessary to determine the total soluble salts in a soil? řřǯȱ ’ŸŽȱ‘Žȱ›Š—Žȱ˜ȱŽ•ŽŒ›’ŒŠ•ȱŒ˜—žŒ’Ÿ’¢ȱŸŠ•žŽœȱ˜›ȱŸŠ›’˜žœȱ¢™Žœȱ˜ȱœ˜’•œǯ

13

MISCELLANEOUS EXPERIMENTS

EXPERIMENT: 13.1 Aim: To determine the concentration of sodium, potassium, lithium and calcium ions in the given solutions by using flame photometric method. Requirements: Flame photometer, standard flasks, graduated pipettes, beakers, sodium chloride, potassium chloride, lithium chloride, calcium chloride, deionized water or double distilled water. Theory: Flame emission spectroscopy was formerly known as flame photometry. Flame photometry is an atomic emission for the routine detection of metal salts, principally Na, K , Li and Ca. Quantitative analysis of these species is performed by measuring the flame emission of solutions containing the metal salts. Solutions are aspirated into flame. The hot flame evaporates the solvent leaving behind the solid salt. The salt vaporizes to gaseous state when the molecules get dissociated into free neutral atoms. Some of these atoms get excited by the thermal energy of the flame. As the excited atoms return to the ground state, light is emitted at the characteristic wavelength for each metal. Filters are used to select the emission wavelength monitored for the analyte species. The intensity of emitted radiation is directly proportional to the amount of elements present in the analyte (given sample). Since the intensity of radiation depends on the temperature of the flame but increase in temperature leads to ionization instead of excitation, in the case of alkali metals, therefore, temperature control is essential. This is done by suitably adjusting the fuel-oxidant composition. In this experiment, a series of standard solutions of a salt is prepared and the intensity of emitted radiation is measured for each solution. A calibration curve is drawn with intensity versus concentration. By measuring the intensity of emitted radiation of the given sample, the concentration of the particular ion can be determined. Instrumentation: (a) Flames: For flame spectroscopy an essential requirement is that the flame used shall produce temperature in excess of 2000K. In most cases, this requirement can only be met by burning the fuel gas in an oxidant gas which is usually air. (b) The Nebuliser–burner system: The purpose of the nebuliser-burner system is to convert the test solution to gaseous atoms; the function of the nebuliser is to

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ197

produce a mist of the test solution. The solution to be nebulised is drawn up a capillary tube by the action of a jet of air blowing across the top of capillary. A gas flow at a high pressure is necessary to produce a fire mist. (c) Monochromator: The purpose of the monochromator is to select a given emission line and to isolate it from other lines, and, occasionally from molecular band emissions. In a simple flame emission photometer an interference filter can be used. In a more sophisticated flame emission spectrophotometer, which requires better isolation of the emitted frequency, a prism or a grating monochromator is employed. (d) Detectors: For the simple flame emission photometer, a barrier-layer cell is a sufficiently good detector because an intense wide band of energy reaches the detector. In the photovoltaic cell or barrier-layer cell, light striking the surface of a semiconductor such as selenium mounted upon a base plate (usually iron) leads to the generation of an electric current, the magnitude of which is governed by the intensity of light beam. The output from the detector is fed into a suitable readout system. Interferences: Various factors may affect the flame emission of a given element and lead to interference with the determination of the concentration of a given element. These factors may be classified as: (a) Spectral interferences (b) Chemical interferences Procedure: Use always deionized or double distilled water for all preparations and in all stages of the experiment. 1. Prepare a standard solution of sodium chloride by weighing 0.6359 g of NaCl in 250 mL of water. This is 1000 ppm solution of Na+ ions. 2. Prepare solutions of various concentrations, namely, 10, 20, 30 , 40, 50 and 60 ppm by diluting the required amount of 1000 ppm solution to 50 mL with water in a standard flask. 3. Switch on the flame photometer and make sure that the flame reaches the steady state. Adjust the air and fuel control knobs in order to have a blue flame. Insert the Na filter. Dip the capillary in water and set the intensity reading to zero. Set the reading to 100, using calibration knob, for the 60 ppm (maximum concentration) solution of NaCl. 4. Dip the capillary again in water, till the reading comes back to zero. 5. Then dip the capillary into the 50 ppm solution of NaCl and note down the intensity. 6. Repeat the steps 4 and 5 for the remaining solutions, every time using deionized or double-distilled water to bring the reading back to zero. 7. Plot a calibration curve between intensity and concentration. It must be a straight line passing through the origin. 8. Carry out the steps 4 and 5 for the solution of unknown concentration and note down the intensity.

198ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

9. Read the concentration of the sodium ions in the given (unknown) sodium chloride solution, corresponding to its intensity, from the calibration curve. 10. Repeat the whole procedure with appropriate filters to determine the concentrations of K+, Li+ and Ca2+ ions in the given solutions of potassium chloride, lithium chloride and calcium chloride. Observation: Table 13.1: Intensities at various concentrations of NaCl solution S. No.

Concentration of Na+ ions (ppm)

1.

10

2.

20

3.

30

4.

40

5.

50

6.

60

7.

Unknown solution

Intensity

Result: The concentration of Na+ ions in the given solution = ........ ppm. Similarly, report the results for other ions. Interpretation:ȱ ’ŸŽȱ’—ȱŽ›–œȱ˜ȱŒŠ•’‹›Š’˜—ȱŒž›ŸŽǰȱ¢™Žœȱ˜ȱ’•Ž›œǰȱ ŠŸŽ•Ž—‘ȱ›Š—Žȱ˜ȱ emission of radiation, magnitude of intensities and accuracy of results.

EXPERIMENT: 13.2 Aim: To index the given powder diffraction pattern of a cubic crystalline system and ˜ȱ’Ž—’¢ȱ‘Žȱ•Š’ŒŽȱ¢™ŽȺȦȺ”’—ȱ˜ȱž—’ȱŒŽ••ȱŠ—ȱ‘Žȱœ’£Žȱ˜ȱ‘Žȱž—’ȱŒŽ••ǯ Theory: X-ray diffraction is one of the important tools used in solid state chemistry and material science in the determination of crystal structure. The diffraction pattern of a crystal is analysed using Bragg’s equation and the expression for interplanar distance in terms of Miller indices. Indexing refers to the naming of the planes of a crystal or giving Miller indices to various planes. In order to know whether a cubic crystal is primitive, face-centred or body-centred, the intensity of diffracted X-ray is plotted against 2T. Such a plot will show a series of reflection maxima. The planes which produce these maxima are indexed as follows: According to Bragg’s equation, nO = 2d sinT ...(1) sinT = nOȺȦȺ؍ȱȱ ǯǯǯǻŘǼ 2 2 2 2 sin T = n O ȺȦȺŚ ...(3) For a cube of length ‘a’, the interplanar distance is given by d2 = a2ȺȦȺ‘2 + k2 + l2 ...(4) 2 2 2 2 2 2 2 From equations (3) and (4), sin T = (n O ȺȦȺŚŠ )(h + k + l ) ...(5)

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ199

Normally, n = 1, but if higher planes are included, automatically we will get secondorder, third-order, etc., reflections. For example, if n = 1, sin2T = O2ȺȦȺŚŠ2 for 100 plane If n = 2, sin2T = 4O2ȺȦȺŚŠ2 for 100 plane But for the same n = 1, sin2T = 4O2ȺȦȺŚŠ2 for 200 plane Therefore, these planes get automatically included. Hence, for n = 1 and for any ‘h’, ‘k’, ‘l’ values, equation (5) can be written as sin2T = (O2ȺȦȺŚŠ2)(h2 + k2 + l2) ...(6) As mentioned above, the 2T values corresponding to maxima can be determined from the diffraction pattern (plot) and from T values and using equation (6), the values of Miller indices ‘h’, ‘k’, ‘l’ can be determined. Sometimes, the diffraction pattern as given in the following diagram may be provided (refer the experimental procedure taught in theory).

Beam entry

Beam exit

Fig. 13.1: A typical diffraction pattern

If x is the distance of each line from the central spot and ‘r’ is the radius of the film, the circumference 2Sr of the film corresponds to a scattering angle of 360°. Therefore, ‘x’ will correspond to x × 360 = 2T (experimental). 2πr ⎛ x ⎞ So, T= ⎜ × 360⎟ / 2 ⎝ 2πr ⎠

200ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Thus, from distances of various lines from the central spot, the values of T and hence sin2T can be determined. Then using equation (6), and the reference Table 13.3, the values of Miller indices ‘h’, ‘k’, ‘l’ can be calculated. Following is the procedure for a typical example. Procedure: A powder diffraction photograph of KCl gave the following lines from the central spot: 13.0, 18.2, 22.8, 26.0, 29.6, 32.0, 37.4, 39.4, 41.6, 44.0, 46.2 all in mm. The wavelength of X-rays used and the radius of the film were 70.8 pm and 5.74 cm respectively. How to index the lines, identify the kind of unit cell and determine its size? Table 13.2: Data recording X (mm)

13.0

18.2

22.8

26.0

29.6

32.0

37.4

39.4

41.6

44.0

46.2

Θ

6.5

9.1

11.4

13.0

14.8

16.0

18.7

19.7

20.8

22.0

23.1

2

sin T

0.0128 0.0250 0.0391 0.0506 0.0652 0.0760 0.1028 0.1136 0.1261 0.1403 0.1539

The common factor or least no., is 0.0128. Divide sin2T by 0.0128 (h2 + k2 + l2) (hkl)

1

2

3

4

5

6

8

9

10

11

12

100

110

111

200

210

211

220

221 or 300

310

311

222

Thus, the lines have been indexed. The absence of 7 for (h2 + k2 + l2) indicates that the cell is primitive or simple cubic. The values of 7 and 15 are not possible since the sum of three squares can not be equal to 7 or 15. The common factor equals O2ȺȦȺŚŠ2 as shown in equation (6). O2ȺȦȺŚŠ2 = 0.0128. O = 70.8 pm Size of the unit cell, a = 312.9 pm Table 13.3: Calculated and observed data hkl

100

110

111

200

210

211

-

220

221 or 300

310

311

222

320

h 2 + k2 + l2

1

2

3

4

5

6

7

8

9

10

11

12

13

Simple cubic or primitive

√

√

√

√

√

√

-

√

√

√

√

√

√

Body-centred cubic (bcc)

x

√

x

√

X

√

-

√

x

√

x

√

x

Face centred cubic (fcc)

x

X

√

√

X

x

-

√

x

x

√

√

x

Result: The powder diffraction show the given system as ........ with ........ lattice type with ........ size of unit cell.

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ201

EXPERIMENT: 13.3 Aim: To determine the free acidity (as H2SO4) in ammonium sulphate fertilizer. Requirements: Standard flasks, burette, graduated pipettes, conical flasks, beakers, pestle and mortar, sieve (1 or 0.4 mm), ammonium sulphate fertilizer (grade IS: 8261967), sodium hydroxide solution (0.02 N), standard oxalic acid solution (0.02 N), phenolphthalein indicator, methyl red indicator, methyl red-methylene blue mixed indicator, distilled water. Reagents: 1. Methyl red indicator: Dissolve 0.15 g of water soluble methyl red in 500 mL. water. 2. Methyl red—methylene blue mixed indicator solution—prepared by mixing equal volumes of 0.2 per cent solution in rectified spirit of methyl red and 0.1 per cent solution in rectified spirit of methylene blue. Theory: Ammonium sulphate is primarily used as a fertilizer for alkaline soils. In the soil the ammonium ion is released and a small amount of acid is formed. Thus, it lowers the pH balance of the soil while contributing essential nitrogen for plant growth. The determination of free acidity in a fertilizer is important because it changes the acidity of the soil which is essential for plant growth. Different plants and crops require different acidity of soil. When soil acidity changes, the solubility of a number of metal ions also changes. Plant growth is really affected by the varying concentration of these metal ions in solution. Under acidic conditions, many soil minerals dissolve and increase the concentration of metal ions to toxic levels thereby inhibiting the growth of the plants. The nutrients, phosphorus and molybdenum are less available in acidic soils and ŒŠ•Œ’ž–ȱŠ—ȺȦȺ˜›ȱ–Š—Žœ’ž–ȱ–Š¢ȱŠ•œ˜ȱ‹ŽȱŽ’Œ’Ž—ǯ Under alkaline conditions, the solubility of minerals decrease to such an extent that nutrient deficiencies occur. Plant growth is therefore prevented by the deficiencies in zinc, copper, iron, manganese and boron. Phosphorus is also less available in alkaline soils and high levels of calcium may prevent the uptake of potassium and magnesium. If the availability of free acid in a fertilizer is estimated, the application of it to the soil and management of the soil acidity and alkalinity can be achieved. This is done to adjust the acidity in such a way that there are no toxic metals in solution and the availability of nutrients is at its maximum. Procedure: ›Ž™Š›Žȱ‘ŽȱœŠ–™•Žȱ˜ȱŠ––˜—’ž–ȱœž•™‘ŠŽȱŠœȱ˜••˜ œDZȱ ›’—ȱŠ‹˜žȱŗȱȱ˜ȱ ammonium sulphate fertilizer as rapidly as possible to avoid loss or gain of moisture and pass through a sieve (1 mm, if it is moist, and 0.4 mm if it is dry). Mix thoroughly and store in an air tight bottle. 1. Dissolve about 50 mg of prepared sample, accurately weighed, in about 50 mL of cold natural water which should not contain even a trace of alkaline solution (would neutralize free acid). 2. Filter and make up the volume to about 100 mL.

202ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

3. Titrate with standard sodium hydroxide solution, using methyl red as indicator. The end point is the change in colour from red to yellow. 4. If satisfactory end point with methyl red is not obtained, methyl red-- methylene blue mixed indicator may be used. The end point is the change in colour from violet-blue to green. 5. Use preferably a micro burette for this titration. Calculation: Mass of sample taken = W g Volume of NaOH solution used in titration = V mL Normality of NaOH solution = N 1000 mL of 1 N NaOH solution { 49 g of H2SO4 V mL of N NaOH solution { (V uȱǼȱŚşȺȦȺŗŖŖŖȱȱ˜ȱ 2SO4 = w g of H2SO4 w Free acidity as H2SO4 by mass % = × 100 W Result: The free acidity (as H2SO4) in ammonium sulphate = ....... %

EXPERIMENT: 13.4 Aim: To determine the amount of calcium in calcium ammonium nitrate (CAN) fertilizer. Requirements: Pestle and mortar, standard flasks, burette, graduated pipettes, conical flasks, calcium ammonium nitrate fertilizer (IS specifications of CAN IS: 2409-1963), magnesium sulphate (MgSO4.7H2O), zinc sulphate (ZnSO4.7H2O), ammonium acetate, ammonium hydroxide, glacial acetic acid, Na2EDTA.2H2O, Eriochrome black T, dilute H2SO4, methanol, ammonium chloride, N-amyl alcohol, hydrochloric acid (approximately 4 N), distilled water. Reagents: EDTA solution (0.01 M): Dissolve 1.85 g of solid Na2EDTA.2H2O in distilled water and make this up to 500 mL. Standard Zn2+ solution (0.01 M): Weigh 0.7185 g of ZnSO4.7H2O solid and dissolve it in a minimum quantity of dilute H2SO4 and make up the solution to 100 mL with distilled water. Standard Mg2+ solution (0.01 M): Weigh 0.25 g of MgSO4.7H2O solid and dissolve in a small quantity of distilled water. Make this solution up to 100 mL with distilled water. Eriochrome black T indicator: Dissolve 0.5 g of solid in 100 mL of methanol. Ammoniacal buffer solution (pH 10): Dissolve 18 g of ammonium chloride in 75 mL of about 30% ammonium hydroxide and dilute to 500 mL with distilled water. Theory: Calcium ammonium nitrate fertilizer, which contains a little magnesium,helps plants to grow wide leaves of rich green colour. It makes the harvest richer by 5–20%. Even if it is used for a long term, it does not cause soil acidification. The biological activity of the soil is not affected. Plants are enriched not only with nitrogen but also

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ203

with magnesium and calcium. Moreover, calcium ammonium nitrate provides the necessary nutrients to plants and thus compensates the lack of light and the damage of acid soil. Calcium is a multifunctional nutrient in the physiology of crop plants and its deficiency will lead to premature shedding of blossoms and buds, abnormally dark green foliage, weakened stems and death of growing points. It is important to estimate the amount of calcium in the fertilizer because the rapid plant growth, the strength of stems that hold flowers and fruits, and the quality of the fruits produced, depend strongly on the calcium availability. In this experiment, amyl alcohol is used to extract calcium ions of the fertilizer and the extracted calcium is estimated by usual EDTA titration. For calcium ions, Eriochrome black T indicator is not suitable because the complex formed between calcium and the indicator is weak and this results in a premature end point. Therefore, calcium ions can be estimated in the presence of a known volume of magnesium ions using ammonium chloride-ammonium hydroxide buffer and Eriochrome black T indicator. Magnesium forms a complex with the indicator which is more stable than that of calcium. During the titration, magnesium forms a weaker complex with EDTA and calcium ions replace magnesium ions from the complex forming a more stable Ca-EDTA complex. At the end point, when all calcium ions are consumed, one drop excess of EDTA will displace magnesium ions from the Mg-indicator complex and the colour of the solution will change from red to blue. Procedure: Step I: ›’—ȱšž’Œ”•¢ȱŠ‹˜žȱŗŖȱȱ˜ȱ‘Žȱ–ŠŽ›’Š•ǰȱŠŒŒž›ŠŽ•¢ȱ Ž’‘Žǰȱ ’‘ȱŠ‹˜žȱśŖȱ–ȱ of amyl alcohol in a pestle and mortar and transfer the contents to a conical flask. Wash the pestle and mortar with a few mL of amyl alcohol and transfer the washings to the conical flask. Shake the contents of the flask for about half an hour (a mechanical shaker may also be used) and then filter. Transfer the filtrate to a separating funnel and extract calcium ions completely with water in five or six steps using a small quantity of water in each step. Add a few drops of dilute hydrochloric acid during the extraction in order to avoid formation of an emulsion of amyl alcohol with water. Concentrate the water extract at low temperature to about half its volume. Step II: Standardization of EDTA solution: Pipette out 10 mL of standard Zn2+ solution into a clean conical flask, add 3 mL of ammoniacal buffer and 2 drops of Eriochrome black T indicator. Titrate this solution with the EDTA solution. The end point is change of colour from wine red to steel blue. Repeat the titration to get at least three concordant values. Calculate the molarity of EDTA solution using the equation. VZn2+ u MZn2+ = VEDTA u MEDTA Step III: Quantitatively transfer the concentrated extract to a conical flask and estimate calcium as described below. If the volume is very large, pipette out a definite quantity of the solution and proceed. Pipette out 10 mL of standard magnesium sulphate solution into the extract. Add 3 mL of ammoniacal buffer and 2 drops of Eriochrome black T indicator. The solution becomes wine red. Titrate the solution with EDTA taken in the burette.

204ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

The end point is the disappearance of the last shade of red and formation of steel blue colour. Repeat the titration to obtain at least three concordant values. Calculation: Mass of the fertilizer taken = W g Volume of fertilizer extract = V1 mL Volume of EDTA solution = V2 mL (titre value) Molarity of EDTA solution = M V ×M ? molarity of calcium in the extract = 2 V1 V2 × M So, the amount of calcium = u 40 = A gL–1 V1 where 40 is the atomic mass of calcium. A × 50 = B g The amount of calcium in V1 mL of extract (titrated) = 1000 i.e., W g of the fertilizer (originally taken) contains B g of calcium B × 100 Hence, 100 g of the fertilizer will contain g of calcium 25 Result: The amount of calcium in calcium ammonium nitrate fertilizer = ........ % Note: Some CAN fertilizer formulations contain a small quantity of magnesium also. In such cases, follow the procedure for EDTA titration as described in Experiment 12.14.

EXPERIMENT: 13.5 Aim: To determine available chlorine from a given sample of bleaching powder. Requirements: Standard flasks, beakers, burette, graduated pipettes, conical flasks, watch glass, soil sample, potassium iodide, hydrochloric acid or glacial acetic acid, sodium thiosulphate, starch powder, distilled water. Reagents: 1. Sodium thiosulphate. Dissolve 12.41 g of sodium thiosulphate in distilled water and make up the volume to 1 litre. To maintain the pH between 9 and 10, add 0.1 g of Na2CO3 to keep the bacterial activity minimum. 2. 10% Potassium iodide solution. ȱ řǯȱ •ŠŒ’Š•ȱŠŒŽ’ŒȱŠŒ’ȱ˜›ȱŘȱȱ •ȱœ˜ž’˜—ǯ 4. Starch solution: Take 0.1 g of starch and make a consistent paste by adding small quantity of distilled water. Pour the paste into 100 mL boiling water and boil it for 2 to 3 minutes. Cool the solution and keep it in an airtight bottle. Theory: Action of dilute acid on bleaching powder liberates chlorine and the chlorine so liberated is known as available chlorine. This liberated chlorine acts as a bleaching agent. Chlorine is present in bleaching powder as hypochlorite as well as basic chloride, CaCl2.

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ205

The available chlorine in a sample of bleaching powder is determined iodometrically, a type of indirect method of titration that involves liberated iodine solution. The bleaching powder solution is treated with excess KI solution in acidic medium. Cl2 + 2I– o I2 + 2Cl– i.e., Cl2 {I2. The liberated iodine is titrated with standard sodium thiosulphate solution using starch as indicator. OCl– + 2I– + 2H+ ֖ Cl– + I2 + H2O 2– I2 + 2S2O2– 3 o S4O6 + 2I Procedure: ȱ ŗǯȱ ›’—ȱŠ‹˜žȱŗȱȱ˜ȱœŠ–™•Žǰȱ‹•ŽŠŒ‘’—ȱ™˜ Ž›ǰȱŠŒŒž›ŠŽ•¢ȱ Ž’‘Žǰȱ’—ȱŠȱ–˜›Š›ȱ and pestle with distilled water till a smooth paste is formed. Add 15 to 25 mL of distilled water and decant the fine part into 100 mL standard flask. Again grind the residue left behind and repeat the procedure of decanting. No grit material is left, wash the paste and mortar in the same flask. Add more water till the volume is 100 mL. Place the stopper on the flask and shake the flask well. 2. Using a burette, take out 10 mL of this solution and put it in a conical flask. Add 5 mL of KI solution and 10 mL of HCl or 3 mL of glacial acetic acid solution. Cover the flask with a watch glass and keep it in dark for 5 minutes so that liberation of iodine is complete. 3. Titrate this liberated iodine with sodium thiosulphate solution (in a burette), using starch as indicator, till the blue colour disappears. Note down the initial and final readings of the burette. 4. Repeat the procedure 3 to 4 times to get concordant readings. Calculation: Initial burette reading = ........ mL Final burette reading = ........ mL ? volume of Na2S2O3 = ........ mL Nthio u Vthio = Niodine u Viodine ? Niodine = (Nthio uVthioǼȺȦȺiodine However, we know that Niodine = Nchlorine So, the strength of available chlorine in bleaching powder = (Nchlorine u 35.5) = A gL–1 The % of available chlorine is the amount of chlorine in 100 g of bleaching powder. Strength of bleaching powder solution = 10 gL–1 Hence, 10 g of bleaching powder contains A g of chlorine. ? 100 g of bleaching powder will contain 10 A g Result: The % of available chlorine in the given sample of bleaching powder = 10 A.

206ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 13.6 Aim: To estimate the acidity and the alkalinity in a given sample of pesticide formulation. Requirements: Methyl red indicator solution-aqueous-1% by volume, sodium hydroxide solution (0.05 N), hydrochloric acid (0.05 N). Theory: The acidity and the alkalinity of pesticides are important in several aspects because they are related to pH. The important practical effect of pH is its relationship to the stability of pesticides. Many commonly used pesticides are susceptible to alkaline hydrolysis in spray solutions and so, a significant amount of the pesticide decomposes to an inactive form if the spray solution is alkaline. Many areas get water supplies which has a pH range 7.5–9.0. Decomposition of pesticides in such water results in poor insect control. The decomposition may be slowed or prevented by acidifying spray solutions to a pH of 6.0 or below with an acidifying product. Acidified pesticide sprays often give improved initial pest control and longer residual control. The pH has an effect on solubility of pesticides. Nutrients must be present in watersoluble form for plant uptake. In order to make nutrients more soluble in spray solution and more readily available for plant uptake, the spray solution must be acidified to a safe level. This effect is important in fast growing crops (where quick response is desired ) or where deficiency correction is urgent. The effect of pH is also felt on leaf surfaces, foliar sprays and in suppressing mite population. The determination of acidity and alkalinity is also important to select appropriate adjuvants. Agricultural spray adjuvants are substances used to improve the performance of pesticides such as herbicides, insecticides, fungicides and other agents that control or eliminate unwanted pests. Procedure: (a) Standardize the sodium hydroxide solution using standard oxalic acid solution and phenolphthalein as indicator. (b) Similarly, standardize hydrochloric acid with standard sodium carbonate solution using methyl orange as indicator. A. Procedure for Acidity Determination 1. Weigh accurately 10 g of the pesticide into a clean and dry conical flask, add 25 mL of acetone and mix thoroughly. 2. Warm the flask gently for effective dissolution of the active ingredient present. 3. Add 75 mL of distilled water and let it stand for an hour. 4. Filter the supernatant aqueous extract and take 50 mL of the filtrate. 5. Titrate immediately with the standard sodium hydroxide solution using methyl red indicator. The end point is the colour change from red to yellow. 6. Carry out a blank determination on an aliquot of 50 mL made from 25 mL of acetone and 75 mL of water. If red colour appears on adding methyl red, then continue the titration.

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ207

Calculation (for acidity determination): Volume of NaOH solution required (titre value) = V1 mL Normality of NaOH solution = N1 Volume of pesticide extract = 50 mL V × N1 Normality of acid in the extract (N2) = 1 50 Converting into mass in terms of H2SO4 , mass of H2SO4 in w g of pesticide = N 2 × 49 ×

100 1000

? acidity of the pesticide formulation (in terms of H2SO4) = N 2 × 49 ×

100 100 × % 1000 w

Subtract the blank titre value from V1 mL, if necessary. B. Procedure for Alkalinity Determination 1. Weigh accurately 10 g of the pesticide into a clean and dry conical flask, add 25 mL of acetone and mix thoroughly. 2. Warm the flask gently for effective dissolution of the active ingredient present. 3. Add 75 mL of distilled water and let it stand for an hour. 4. Filter the supernatant aqueous extract and take 50 mL of the filtrate. 5. Titrate immediately with the standard hydrochloric acid solution using methyl red indicator. The end point is the colour change from yellow to red. 6. Carry out a blank determination on an aliquot of 50 mL made from 25 mL of acetone and 75 mL of water. If yellow colour appears on adding methyl red, then continue the titration. Calculation (for alkalinity determination): Volume of HCl solution required (titre value) = V1 mL Normality of HCl solution = N1 Volume of pesticide extract = 50 mL V1 × N1 Normality of alkali in the extract (N2) = 50 Converting into mass in terms of NaOH , mass of NaOH in w g of pesticide = N 2 × 40 ×

100 1000

? alkalinity of the pesticide formulation (in terms of NaOH) = N 2 × 40 ×

100 100 × % 1000 w

Subtract the blank titre value from V1 mL, if necessary. Result: 1. The acidity of the pesticide formulation (in terms of H2SO4) = ........ % 2. The alkalinity of the pesticide formulation (in terms of NaOH) = ........ % Note: These estimations can also be carried out by pH-metric or potentiometric titrations.

208ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

EXPERIMENT: 13.7 Aim: To estimate the amount of free phosphoric acid (H3PO4) in a given sample of superphosphate fertilizer. Requirements: Pestle and mortar, 500 P sieve, Whatman No. 1 filter paper, standard flasks, burette, graduated pipettes, conical flask, beakers, superphosphate fertilizer, acetone, standard oxalic acid solution (0.1 N), sodium hydroxide solution (0.1 N), distilled water. Phenolphthalein–1-naphtholphthalein mixed indicator: Mix two parts of 0.1% solution of phenolphthalein in ethanol with one part of 0.1% solution of l-naphtholphthalein in ethanol. Theory: Phosphorous is an essential plant nutrient and plays an active role in photosynthesis, energy transfer, cell division and enlargement. It is important in root and seed formation. It improves the quality of fruits and vegetables. It is taken up usually as dihydrogen phosphate ion (H 2 PO 4− ). This ion is derived from phosphoric acid (H3PO4). The choice of fertilizers for rectifying the phosphorous deficiencies in soil is based on the availability of phosphoric acid in a fertilizer. Various types of phosphates are used as fertilizers—Monoammonium dihydrogen phosphate and diammonium hydrogen phosphate, superphosphate (made when rock phosphate is treated with sulphuric acid to make calcium phosphate and calcium sulphate (gypsum)), triple superphosphate (made by treating rock phosphate with phosphoric acid). In this experiment, the phosphoric acid is extracted with acetone and titrated with standard sodium hydroxide solution using phenolphthalein–1-naphtholphthalein mixed indicator. Only two protons of phosphoric acid are replaceable and this mixed indicator is suitable for the titration of phosphoric acid to the diprotic stage (pH = 8.9). The third dissociation constant is very low and the pH at the third end point is 12.35. No suitable indicator is available for the third stage. The equivalent mass of phosphoric acid is 49, i.e., molar mass y 2. A mixture of phenolphthalein (2 parts of a 0.1 per cent solution in ethanol) and l-naphtholphthalein (1 part of a 0.1 per cent solution in ethanol) passes from pale rose to violet at pH = 8.9. The mixed indicator is suitable for the equivalence point at pH = 8.9. Procedure: ȱ ŗǯȱ ›’—ȱŠ‹˜žȱŗŖȱȱ˜ȱ‘Žȱœž™Ž›™‘˜œ™‘ŠŽȱŽ›’•’£Ž›ȱŠ—ȱ™Šœœȱ’ȱ‘›˜ž‘ȱŠȱśŖŖȱP sieve. 2. Weigh accurately about 5 g of the prepared fertilizer sample and transfer it to a tightly stoppered reagent bottle. 3. Add 100 mL of acetone. Shake for about an hour (mechanical shaker may be used). 4. Filter quickly through Whatman No. 1 filter paper into a conical flask and wash four or five times with acetone, using about 10 mL each time. 5. Evaporate the acetone to the maximum possible.

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ209

6. Add about 50 mL of distilled water and two drops of the mixed indicator. 7. Titrate with standard NaOH solution, until the colour changes from pale rose to green to violet. 8. Repeat the titration to get at least three concordant readings. Calculation: Mass of superphosphate taken = W g Volume of NaOH solution = V mL Normality of NaOH solution = N 1 mL of 1 N NaOH solution { 0.049 g of H3PO4 V mL of N NaOH solution { V u N u 0.049 g of H3PO4 V × N × 0.049 × 100 Free phosphoric acid (per cent by mass) = W Result: The amount of free phosphoric acid in the given sample of superphosphate = ........%.

EXPERIMENT: 13.8 Aim: To determine the concentration of benzoic acid and sorbic acid present in soft drinks. Requirements: Spectrophotometer, cuvettes, beakers, graduated pipettes, measuring cylinder, ether, anhydrous sodium sulphate, 2 M hydrochloric acid solution, 0.01 M sodium hydroxide solution, distilled water. Theory: The main preservatives allowed and used in soft drinks are sulphur dioxide, sorbic acid and its salts, and benzoic acid and its salts. Benzoic acid and its sodium salt are among the most widely used antibacterial agents in foods. Many berries (e.g., raspberries) contain appreciable amounts (0.05%) of benzoic acid as part of their natural composition. Benzene in soft drinks is of potential concern due to the carcinogenic nature of the benzene molecule. This contamination is a public health concern and has caused significant outcry among environmental and health advocates. Benzene is formed from decarboxylation of the preservative benzoic acid in the presence of ascorbic acid (vitamin C) and metal ions (iron and copper) that act as catalysts, especially under heat and light. Sorbic acid and its sodium salt are effective antimicrobial preservatives but high levels can affect the taste of a product. Sorbic acid (CH3-CH=CH-CH=CH-COOH) is present in some fruits and is a selective growth inhibitor for moulds, yeasts and bacteria. Benzoates and sorbates are often used in combination, especially where the soft drink is highly acidic. Benzoic and sorbic acids can be determined by the measurement of ultraviolet absorbance after extracting them with ether. In dilute sodium hydroxide solution, benzoic acid has a maximum absorption at 200 nm and sorbic acid has a maximum absorption at 254 nm. The molar absorptivity (H) values are 130 M–1 cm–1 and 2.48 u 104 M–1 cm–1 for benzoic acid and sorbic acid respectively.

210ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Procedure: 1. Remove the gas from the soft drink by warming it on a hot plate but do not boil. 2. Take 25 mL of degassed soft drink in a 100 mL separating funnel and add two or three drops of HCl solution. 3. Extract the mixture with three portions of ether, using volumes of 50, 30 and 20 mL. Collect the extracts and wash with 20 mL of distilled water. 4. Dry the extract by swirling with 2 g of anhydrous sodium sulphate, adding a little quantity at a time, till the turbidity disappears. 5. Evaporate the extract to dryness on a steam bath and dissolve the residue in 100 mL of 0.01 M NaOH solution. 6. Record the uv spectrum of the solution from 190 to 300 nm. Calculation: Calculate the concentrations of benzoic acid and sorbic acid using the Lambert Beer’s law as: A = Hcl where A is the absorbance, ‘H’ is the molar absorptivity, ‘c’ is the concentration (M) and ‘l’ is the cuvette length in cm. Result: The concentration of benzoic acid = ........ The concentration of sorbic acid = ........

EXPERIMENT: 13.9 Aim: To determine the concentration of sulphur dioxide in soft drinks. Requirements: Standard flasks, burette, graduated guarded pipettes, conical flasks, beakers, sodium hydroxide solution (1 N), sulphuric acid solution (4 N), 0.01 M iodine solution (dissolve 2.5 g of KI in water and add 1.27 g of I2 and stir thoroughly. Dilute to 500 mL with distilled water), standard sodium thiosulphate solution (0.01 M), sodium bicarbonate, freshly prepared starch solution, distilled water. Theory: Sulphur dioxide and sulphites have been widely used as antimicrobials for many centuries and are very effective preservatives. Sulphur dioxide is used as a food preservative and is the most effective inhibitor of the deterioration of dried fruits and fruit juices. Sulphites are also used as antioxidants, antibrowning agents and colour stabilizers. Sulphites are mainly used in soft drinks to control the growth of undesirable microorganisms such as yeasts and as antioxidants to prevent browning reactions from occurring. Sulphur dioxide and sulphites are known to cause allergic reactions in certain sensitive consumers especially those with asthmatic problems. These reactions are more common when sulphur dioxide gas is used. In this experiment, Ripper method is used in which sulphur dioxide is determined by the redox titration with iodine solution using starch as the indicator. SO2 + I2 + H2O o SO3 + 2H+ + 2I–

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ211

For free SO2, the soft drink is acidified and titrated directly, and for total SO2, first sodium hydroxide is added to the soft drink to break down the complexes containing the bound SO2, then it is acidified and titrated. Sodium bicarbonate is added prior to commencing the titration in order to create an inert blanket of carbon dioxide gas to prevent interference caused by oxygen in air. Coloured drinks may require decolorizing with activated charcoal before performing the titration so that the colour change at the endpoint is observed clearly. While determining total sulphur dioxide, the sample is pretreated with sodium hydroxide solution to adjust the pH. This causes chemically bound forms of sulphur dioxide to be released in solution as free sulphur dioxide. It is essential to titrate to an end point where the blue colour persists at least for 30 seconds to make sure that all forms of sulphur dioxide in solution have reacted. Procedure: Standardize the iodine solution using standard sodium thiosulphate solution and starch indicator. I. Determination of free SO2: 1. Pipette out 50 mL of soft drink into a 250 mL conical flask. 2. Add about 1 mL of freshly prepared starch solution. 3. Add 5 mL of 4 N sulphuric acid ( if you do not get the proper end point, try with 25%, i.e., 9 N H2SO4). Mix well. 4. Rinse and fill the burette with 0.01 M iodine solution. 5. Add about 1 g (approximately one spatula) of solid sodium bicarbonate to the flask and commence the titration immediately. 6. Titrate rapidly until the solution turns into blue colour which persists for 30 seconds. 7. Calculate the free SO2 concentration in mg L–1 using the following equation: Volume (mL) of iodine solution × Molarity of iodine × 64 × 1000 Volume (mL) of soft drink sample –1 where 64 g mol is the molar mass of SO2. II. Determination of total SO2: 1. Pipette out 20 mL of soft drink into a 250 mL conical flask. 2. Add 25 mL of 1 N sodium hydroxide solution. 3. Stopper the flask, mix well and allow to stand for 10 minutes. 4. Add about 1 mL of freshly prepared starch solution. 5. Add 10 mL of 4 N sulphuric acid (if you do not get the proper end point, try with 25% H2SO4). Mix well. 6. Rinse and fill the burette with 0.01 M iodine solution. 7. Add about 1 g solid sodium bicarbonate to the flask and commence the titration immediately. 8. Titrate rapidly until the solution turns into blue colour which persists for 30 seconds. 9. Calculate the total SO2 concentration in mg L–1 using the equation given above. SO2 (mg L–1) =

212ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Result: The concentration of sulphur dioxide = ........ mg per litre of soft drink. Note: 1. In case of coloured soft drinks such as coca cola, add a small quantity of activated charcoal in order to decolorize the soft drink so that the colour change at the end point can be clearly observed. 2. Hydrochloric acid can also be used in place of sulphuric acid.

EXPERIMENT: 13.10 Aim: To detect adulterants in various foodstuffs. Food adulteration is the intentional addition of non-permitted foreign matter to food or the removal of useful ingredients from food, to increase profits. Adulteration may also be incidental due to lack of knowledge and lack of hygiene. According to the Prevention of Food Adulteration Act of India 1954, it also includes the wilful addition of those substances, which adversely affect the quality of food and their incidental contamination during the period of growth, processing, transportation and distribution. The International Codex Alimentarius Commission has suggested many internationally adopted food standards, including the provisions in respect of food hygiene, food additives, pesticide residues, contaminants, labelling, analysis and sampling. In India too, we have two national standards—the Prevention of Food Adulteration and the Fruit Products Order, both being amended from time to time. In spite of many rules and Acts, the adulteration of food is steadily increasing. This has an impact on public health especially in the poor and under-nourished population. For example, the non-permitted dyes are used for colouring, which are not absolutely safe. Most of them are carcinogenic in nature. They may cause pathological lesions in the vital organs like liver, spleen and kidney. Similarly, lead chromate or metanil yellow may cause anaemia, paralysis, abortion, mental retardation and brain damage in children. Some examples of adulterants and their harmful effects are given here. Ž—ž’—Žȱ ŽŠȱ ’œȱ –’¡Žȱ  ’‘ȱ ‘Žȱ žœŽȱ ŽŠȱ  ‘’Œ‘ȱ ’œȱ Œ˜••ŽŒŽȱ ›˜–ȱ ŽŠȱ œ‘˜™œȱ Š—ȱ ’œȱ sprinkled with water soluble coal tar dyes. Tea is also sometimes coloured to get a dark coloured decoction for giving an impression that the tea is of good quality. Coal tar is a mixture of many chemicals, derived from petroleum. Coal tar is recognized as a human carcinogen and the main concern with individual coal tar colours is their potential to cause cancer. These colours may be contaminated with low levels of heavy metals and some are combined with aluminium compounds. Aluminium compounds and many heavy metals are toxic to the brain. Substituting pepper with papaya seeds is an act of adulteration. Apart from financial loss, a consumer may not get the value of medicinal effect from this valuable spice. In addition, the papaya seeds create or add to the digestive problems. In order to make them look fresh and attractive, the peas are artificially coloured with coal tar dyes. Sometimes, they are coloured with non-permitted dyes like malachite green and yellow aniline, which are carcinogenic.

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ213

Pure silver leaves in a very fine thinness are extensively used to give good appearance to Indian sweets. They are also used in paan (betel leaves). Silver leaves have some medicinal values also. However, because of silver leaves being costly, they are being replaced by aluminium leaves of very fine thinness. Colophony is a resin obtained as a residue after the distillation of turpentine oil and this is the major adulterant in asafoetida. The effects of exposure to colophony are classified into bronchial asthma and contact dermatitis. Milk sold in the cities is being transported from a longer distance. If the milk is not transported at a low temperature, there is every likelihood of its getting spoilt because of bacterial action. Neutralizers are added in the milk to avoid spoilage of milk. Common neutralizers used are sodium hydroxide (caustic soda), sodium carbonate and sodium bicarbonate. Sodium hydroxide causes vomiting and in severe cases can cause burns on the lips and tongue, and harms the mucosa of the oesophagus. Sodium carbonate and sodium bicarbonate disrupt hormone signalling that regulate development and reproduction, and can create gastrointestinal problems such as vomiting and diarrhoea. Urea is another adulterant added to the milk to increase its viscosity thereby giving a feeling of rich milk. The potential adverse effects of urea include indigestion, diarrhoea, acidity, malfunctioning of kidneys, damage to the intestinal tract and digestive system, ulcers and impaired vision. Formalin is added to prolong the shelf life of milk. Side effects of drinking milk adulterated with formalin include mood and balance alteration, liver and kidney damage, and abdominal pain, among others. Bura sugar is obtained by grinding plantation white sugar and finds extensive use in sweetmeat preparations and also for direct consumption. Its white colour makes it easily adulterable with washing soda. It is harmful to the body and can cause diarrhea, nausea and vomiting. ‘ŽŽȱŠ—ȱ‹žŽ›ȱ‘ŠŸŽȱ‹ŽŽ—ȱžœŽȱ’—ȱ‘Žȱ—’Š—ȱ‘˜–Žœȱ›˜–ȱ’–Žœȱ’––Ž–˜›’Š•ǯȱ‘Žȱ introduction of vanaspati into the Indian market has opened ways for adulteration of ghee and butter with vanaspati. It is a hydrogenated oil, also called trans-fat. It causes heart diseases, increases the risk of diabetes and its consumption is linked to breast cancer. Argemone oil, which is added to edible oils, contains the alkaloids sanguanarine and dihydro sanguanarine which are toxic and cause epidemic dropsy. ‘Žȱ ‘Ž——Š’ȱ ˜›™˜›Š’˜—ȱ ž—Ž›ȱ Š–’•ȱ Šžȱ ˜ŸŽ›—–Ž—ȱ ‘Šœȱ ›Ž•ŽŠœŽȱ Šȱ •’œȱ ˜ȱ adulterants in various foodstuffs and their harmful effects in the form of a Table 13.4 which is shown below. Reference is given under ‘Bibliography’. Table 13.4: Adulteration in foodstuff and its harmful effects Food Article

Adulterant

Bengal gram dhal & Kesari dhal Thoor dhal

Harmful Effects Lathyrism cancer

Tea

Used tea leaves processed and coloured Liver disorder

Coffee powder

Tamarind seed, date seed powder

Diarrhoea (Contd.)

214ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• (Contd.) Chicory powder

Stomach disorder, giddiness and joint pain

Milk

Unhygenic water & starch

Stomach disorder

Khoa

Starch & less fat content

Less - nutritive value

Wheat and other food grains (Bajra)

Ergot (a fungus containing poisonous substance)

Poisonous

Sugar

Chalk powder

Stomach - Disorder

Black pepper

Papaya seeds and light berries

Stomach, liver problems

Mustard powder

Argemone seeds

Epidemic dropsy & glucoma

Edible oils

Argemone oil

Loss of eyesight, heart diseases, tumour

Mineral oil

Damage to liver, carcinogenic effects

Karanja oil

Heart problems, liver damage

Castor oil

Stomach problem

Asafoetida

Foreign resins galbanum, colophony resin Dysentery

Turmeric powder

Yellow aniline dyes

Carcinogenic

Non-permitted colourants like metanil yellow

Highly carcinogenic

Chilli powder Sweets, Juices, Jam

Tapioca starch

Stomach disorder

Brick powder, sawdust

Stomach problems

Artificial colours

Cancer

Non-permitted coal tar dye (Metanil yellow)

Metanil yellow is toxic and carcinogenic

Jaggery

Washing soda, chalk powder

Vomiting, diarrhoea

Pulses (Green peas and dhal)

Coal tar dye

Stomach pain, ulcer

Supari

Colour and saccharin

Cancer

Honey

Molasses sugar (sugar plus water)

Stomach disorder

Carbonator water beverages

Aluminium leaves

Stomach disorder

Cloves

Cloves from which volatile oil has been Cheating, waste of money extracted

In order to prevent a large number of diseases that are disastrous to our health, it is desirable and advisable to detect various adulterants present in the foodstuffs. Requirements: Standard flasks, beakers, graduated and guarded pipettes, glass rods, droppers, separating funnel, porcelain dish, watch glass, test tubes, silica crucible, magnifying glass, magnet, forceps, spatula, filter paper, litmus paper, dilute and concentrated hydrochloric acid, sulphuric acid and nitric acid, sodium hydroxide, sodium bicarbonate, sodium carbonate, rectified spirit, ethyl alcohol, distilled water. The specific reagents are given along with their preparation at appropriate places (Table 13.5). Procedure: The tests for the detection of various adulterants are given in Table 13.5

Adulterant

Tea seed oil

(a) Vanaspati

3. Vegetable oil

4. Ghee and butter

(c) Karanjia (pungam) oil

(b) Castor oil

2. Edible oils (a) Mineral oil (groundnut and coconut oil)

1. Mustard Oil Argemone oil or castor oil

Food Article

Observation and Inference

To a little amount of the ghee or butter in a test tube, add equal amount of concentrated hydrochloric acid and a little sugar, and shake vigorously. Keep it standing for 5 minutes.

(a) Mix 0.8 mL of acetic anhydride, 1.5 mL of chloroform and 0.2 mL of concentrated sulphuric acid in a test tube. Cool it in an ice bath for 2 minutes. Then add 6 to 7 drops of the given oil. If the liquid appears cloudy, add a few more drops of acetic anhydride to make clear. Keep the test tube in the ice bath for 5 minutes. (b) Add 10 mL of dry ether to the above solution and shake.

Take about 2 mL of the sample in a test tube and add 2 mL of alcoholic potash (prepared by dissolving 4.3 g of KOH in 50 mL of alcohol) to it. Warm the sample on a low flame for about 10 minutes and add 1 mL of water to it. Dissolve the sample in petroleum ether and acidify the solution with dilute hydrochloric acid. To this, add a few drops of 1% ammonium molybdate in 2 M sulphuric acid and then shake vigorously. To 2 drops of the oil sample, add 2 mL of antimony trichloride in chloroform solution (20% W / V).

Change in colour from blue to red shows the presence of vanaspati.

(Contd.)

Brown colour appears which changes to red within a minute and then fades away. Confirms the presence of tea seed oil.

Development of green colour indicates the presence of tea seed oil.

Immediate appearance of canary yellow to orange yellow indicates the presence of karanjia oil.

Turbidity shows the presence of castor oil.

The appearance of turbidity shows the presence of mineral oil.

Take a small quantity (2–3 mL) of mustard oil in a (a) Needle-shaped brown crystals show the test tube and add 1 mL of dilute hydrochloric acid. presence of argemone oil. Mix well and add a few drops of ferric chloride (b) Turbidity shows the presence of castor oil. solution (prepared by dissolving 3 g of solid in 50 mL of distilled water) slowly through the side of the test tube.

Test

Table 13.5: Detection of adulterants in commonly used foodstuffs

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ215

Test

10. Asafoetida

(a)

9. Turmeric powder

(a) (b)

(b)

Brick powder

8. Chilli powder

(a) Add water and shake. (b) Make a paste of chilli powder and concentrated hydrochloric acid, and introduce into the flame. Aniline yellow Shake a little of the turmeric powder (suspended in water) with alcohol. dyes Non-permitted To a little of the sample add a few drops of colourants like concentrated hydrochloric acid. metanil yellow Foreign resins Powder a gram of asafoetida and shake it with a few drops of alcohol. Filter the extract and to 5 mL Galbanum of it, add a few drops of 10% ferric chloride solution colophony (prepared by dissolving 3 g of the solid in 50 mL of resin. distilled water). To a few mL of petroleum ether extract of asafoetida, add 5 mL of 5% aqueous copper acetate solution. Allow the layers to separate. Burn a small amount of powdered asafoetida on a nickel spatula or on a spoon.

(a) Tonic dyes (e.g., To the sample, add 25 mL of concentrated hydrochloric metanil yellow) acid and keep in a water bath for 15 minutes. Spread a small amount of dhal in a filter paper and (b) Kesari dhal visually examine it using a magnifying glass.

To about 5 mL of melted ghee, add 5 mL of 0.1% phloroglucinol in alcohol and 5 mL of concentrated hydrochloric acid. Shake the contents well. (a) Chicory powder Take a small amount of coffee powder in a test tube and add water. Sprinkle a small quantity of coffee powder on a (b) Tamarind filter paper and add a few drops of 1% sodium powder carbonate solution. Sawdust and some Sprinkle some tea powder on a wet filter paper. colourants

Adulterant

(b) Rancid fat

7. Dhal

6. Tea

5. Coffee powder

Food Article

(Contd.) Observation and Inference

If it burns with a bright flame like camphor, it shows that the sample is pure. (Contd.)

A bluish green colour in the petroleum ether layer shows the presence of colophony resin.

Olive green colour shows the presence of foreign resins.

(a) Colour separation indicates non-permitted colour. (b) Pinkish-red colour indicates used up tea leaves or outer coats of dhal. Pale red colour shows non-permitted colour like metanil yellow. Kesari dhal can be detected by its physical appearance. It is wedge-shaped having a slant on one side and appear square as compared to other dhals. Brick powder settles down showing contamination. Brick-red flame colour. Presence of calcium salts in brick powder. The solution turns yellow immediately, if aniline yellow dye is present. Red with no further change in colour shows the presence of metanil yellow.

Red colouration shows the presence of tamarind powder.

Red colour separation if the chicory content is high.

Deep pink or red colour indicates the presence of rancid fat.

216ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Mineral acid

(a) Invert sugar

13. Vinegar

14. Honey

Dissolve about 1 g of sample in 5 mL of distilled water and filter. To the filtrate, add 0.5 mL of concentrated hydrochloric acid and evaporate to dryness. Add a few drops of ammonium hydroxide and heat the residue to 250°C. Add a few drops of sodium hydroxide solution and introduce a filter paper dipped in Nessler’s reagent. To a small amount of the food item, add hot water and to the separated colouring matter, add a few drops of concentrated hydrochloric acid.

16. Sweets and Metanil yellow juices (colouring agent)

Dissolve 1 g of honey in 10 mL of cold water. Transfer this solution into a small separating funnel and shake the solution with 10 mL of ether. Decant the ether layer and evaporate in a porcelain dish to half the original volume and add 1 mL of 1% solution of resorcinol in concentrated hydrochloric acid. Dissolve about 1 g of sample in 5 mL of distilled water and filter. To the filtrate, add 0.5 mL of concentrated hydrochloric acid and evaporate to dryness. Add a few drops of ammonium hydroxide and heat the residue to 250°C. Add a few drops of sodium hydroxide solution and introduce a filter paper dipped in Nessler's reagent.

15. Fruit juices, Saccharin jam and wine

(b) Saccharin

Dip a red litmus paper in a small amount of an aqueous solution of bura sugar.

Washing soda

12. Bura sugar

Colour change from red to blue shows the presence of washing soda.

Colour separation indicates adulteration.

(Contd.)

Pinkish red colour shows the presence of metanil yellow.

If the paper becomes reddish brown, presence of saccharin is indicated.

If the paper becomes reddish brown, presence of saccharin is indicated.

Pink colour indicates the presence of invert sugar in honey sample.

Evaporate 2 mL of vinegar sample in a crucible with Formation of brown precipitate shows the presence excess of ammonium hydroxide and heat. Dissolve of ammonium ion indicating the presence of the residue in water and add Nessler's reagent. mineral acid in vinegar.

Keep the sample immersed in water for about half an hour and stir.

11. Pulses Colour dye stuffs (green peas and dhal)

(Contd.)

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ217

Starch

(a) Urea

20. Milk, khoa and bura sugar

21. Milk

Add a little water to a small quantity of the food item and boil for a few minutes. Cool and add iodine solution (prepared by dissolving 1.27 g of iodine and 2.4 g of potassium iodide in 100 mL of distilled water).

Add a few drops of dilute hydrochloric acid.

22. Silver leaves Aluminium or foil (in leaves sweets and paan / betel leaves)

Appearance of violet or blue ring at the junction of two layers shows the presence of formalin.

The appearance of a red colour, or a deep rose red colour indicates the presence of a neutralizer while a brownish colouration shows absence of any neutralizer.

The appearance of a distinct yellow colour indicates the presence of added urea. (if it is light yellow, it may be due to the presence of natural urea in milk).

Blue colouration shows the presence of starch.

Effervescence shows the presence of washing soda.

Pure silver leaves will be easily crushed and crumble to the powder form while aluminium leaves will only break into smaller shreds. (b) Take the suspected silver leaves and make it in the Pure silver leaves burn away completely while form of a ball and burn it with the help of a flame. aluminium leaves are reduced to grey ash.

(a) Take some portion of the leaves and crush it between two fingers.

Take 5 mL of milk in a test tube and add 5 mL of p-dimethyl amino benzaldehyde reagent (1.6 per cent in ethyl alcohol containing 10% hydrochloric acid) to it. (b) Neutralizer Take 5 mL of milk in a test tube and add 5 mL of (sodium bicar- rectified spirit to it. Then add 4 drops of rosalic acid bonate, sodium solution (prepared by dissolving 0.5 g of solid in carbonate and 50 mL ethyl alcohol). sodium hydroxide) (c) Formalin To about 10 mL of milk in a test tube, add about 1 mL of concentrated sulphuric acid through the sides of the test tube, without shaking it.

Washing soda

19. Jaggery

Add a small amount of pepper to 5 mL of rectified spirit.

(b) Rotten pepper

Pepper is black or brownish black in colour. It has wrinkled surface and has a characteristic smell and pungent taste. The papaya seeds are greenish brown in colour and have smooth surface and oval shape. They are less pungent in taste. Rotten pepper will float.

Spread a small amount of pepper in a filter paper and visually examine it using a magnifying glass.

(a) Papaya seeds

Observation and Inference

18. Black pepper

Test Add 5 mL of 20% sodium chloride solution to 1 g of The ergot affected grains will float on top. The wheat or bajra. healthy grains will settle at the bottom.

Adulterant

17. Wheat and Ergot (a fungus) bajra

Food Article

(Contd.)

218ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ219

EXPERIMENT: 13.11 Aim: To synthesize nail polish. Requirements: Beakers, stirrer, airtight container. Theory: Modern day nail polish is sold in liquid form in small bottles and is applied with a tiny brush. Within a few minutes after application, the substance hardens and forms a shiny coating, on the fingernail, that is both water- and chip-resistant. Ž—Ž›Š••¢ǰȱŠȱŒ˜Š’—ȱ˜ȱ—Š’•ȱ™˜•’œ‘ȱ–Š¢ȱ•ŠœȱœŽŸŽ›Š•ȱŠ¢œȱ‹Ž˜›Žȱ’ȱ‹Ž’—œȱ˜ȱŒ‘’™ȱŠ—ȱŠ••ȱ off. A nail polish should have the following characteristics: 1. Proper viscosity, wetting and flow properties 2. Uniform colour, good gloss and good adhesive properties 3. Sufficient flexibility so that it does not crack or become brittle 4. Sufficient hard surface which is resistant to impact and scratch 5. Reasonable drying time (1-2 minutes) without developing bloom 6. The ability to maintain the above-mentioned properties for a reasonable time Nail polish is mainly prepared by dissolving film forming polymer (nitrocellulose) in a volatile organic solvent (butyl or ethyl acetate). Dibutyl phthalate and camphor are the typical plasticizers used to give the non-brittle film. Dyes and pigments (D & C Red No. 7, calcium lake, D & C, Red No. 34, aluminium lake, bismuth oxychloride, iron oxides) are then added to give the glittery and shimmer look. The thickening agent strearalkonium hectorite is added to maintain the required viscosity. Benzophenone-1 acts as a suitable ultraviolet stabilizer to resist the changes in colour when the film comes in contact with sunlight. Formaldehyde resin is a suitable film modifier. Table 13.6: Composition of nail polish Ingredients Nitrocellulose

% (W / W) 14.90

Butyl acetate

34.04

Toluene

30.00

Formaldehyde resin

7.10

Dibutyl phthalate

4.90

Camphor

2.40

Stearalkonium hectorite

1.20

Benzophenone-1

0.20

Bismuth oxychloride D&C red no. 7 (Dye to give colour)

4.80 0.05–0.1

Procedure: Mix nitrocellulose, butyl acteate and toluene in a beaker with constant stirring till you get a clear solution. Add formaldehyde resin, dibutyl phthalate, camphor, stearalkonium hectorite and benzophenone-1 one by one. Continue the mixing until all the ingredients dissolve completely and a clear lacquer is formed. It may take several hours. Pass the lacquer through filter press or centrifuge it. Then add the pigmented chips or concentrated tinters and mix thoroughly. Nail polish, also

220ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

known as nail lacquer is formed. The product is transferred to a small glass bottle with brush and closed with lid for use. Result: The yield of the prepared product = ........ g.

EXPERIMENT: 13.12 Aim: To synthesize nail polish remover. Requirements: Beakers, stirrer, hot plate, airtight container. Theory: Nail polish remover is a mixture of solvents to remove nail polish from nails. Nail polish remover works by dissolving the chemicals present in nail polish and moisturizing the dried nail. An ideal nail polish remover should not be irritating to surrounding skin, should have pleasant odour and leave the nails with smooth touch. Table 13.7: Composition of the nail polish remover S. No.

Ingredients

% (W / W)

1.

Ethyl Acetate

90

2.

Water

10

Procedure: Mix the two ingredients, ethyl acetate and water in 9:1 ratio in a glass bottle and shake. A simple non-smearing nail polish remover has been prepared. To improve the quality of nail polish remover, try other creamy options as given below: Table 13.8: Composition of oil based nail polish remover S. No.

Ingredients

% (W / W)

1.

Acetone

83.0

2.

Diethylene glycol mono ethyl ether

14.5

3.

Castor oil

2.5

Procedure: Mix the above ingredients and shake till you get a uniform mixture. Table 13.9: Composition of cream based nail polish remover S. No.

Ingredients

% (W / W)

1.

Bees wax

3

2.

Microcrystalline wax

1

3.

Acetylated monoglycerides

10

4.

Diethylene glycol monoethyl ether

52

5.

Stearic acid

15

6.

Ethyl acetate

15

7.

Triethanolamine

4

Procedure: Mix all the ingredients from S. No. 1-5 in table 13.9 with slow heating on a hot plate. Once a melt is formed, add triethanolamine and ethyl acetate one by one to the mixture with continuous stirring. Allow this to cool and store in a glass bottle with tight cover. Result: The yield of the prepared product = ........g.

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ221

EXPERIMENT: 13.13 Aim: To prepare talcum powder. Requirements: Beakers, blender, airtight container. Theory: Talc is the main component of talcum powder. It is composed of hydrated magnesium silicate and has the chemical formula, Mg3Si4O10(OH)2. This powder has a clear or dusty appearance and it can be translucent or opaque. It renders the skin a smooth and glossy appearance. It also has other ingredients that makes its composition perfect for application over skin. It is used in many industries like pharmaceutical, ™Š™Ž›ȱ –Š”’—ǰȱ ™•Šœ’Œǰȱ ›ž‹‹Ž›ǰȱ ˜˜ǰȱ Ž•ŽŒ›’Œȱ ŒŠ‹•Žœǰȱ ŽŒǯȱ ›ŽŽ—’œ‘ȱ ˜›ȱ ›Ž¢’œ‘ȱ Š•Œȱ ’œȱ known as soapstone or steatite. This variant finds applications in construction of sinks, electrical switch boards, stoves, etc. Talc has a glazing sheen and due to this property, it is widely used in the ceramic industries. It is also used as a food additive by pharmaceutical companies. Zinc stearate and zinc oxide can act as water repellants and have astringent, antibacterial and antiviral properties. Magnesium carbonate acts as drying agent and pH adjuster. Titanium oxide acts as a UV filter. Table 13.10: Composition of talcum powder S. No. 1.

Ingredients

%(W / W)

Talc

65

2.

Zinc oxide

23

3.

Zinc stearate

5

4.

Magnesium carbonate

3

5.

Titanium oxide

3

6.

Perfume

1

Procedure: Mix ingredients no. 1–5 thoroughly in a blender. Add a few drops of desired fragrant essential oil as a perfume. Mix it well and store in a sealable container. Result: The yield of the prepared product = ........ g.

EXPERIMENT: 13.14 Aim: To prepare face cream. Requirements: Beakers (150 mL and 250 mL), hot plate, thermometer, stirrer. Theory: The function of a skin cream is to protect the skin from the harsh environment and restore the moisture content of dry skin. This will help to carry out the normal functioning of skin, allowing the elimination of waste products through the pores and maintaining the normal body temperature. The cream should not be heavy, insoluble and sticky to the skin as otherwise it will have detrimental effect to human health. The ingredients of the creams vary according to the requirement, i.e., for improving texture and softness of skin or providing some specific medicinal property. A basic and satisfactory cream may be prepared from the following procedure.

222ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• Table 13.11: Composition of face cream S. No.

Ingredients

% (W / W)

1.

Stearic acid

10

2.

Mineral oil

10

3.

Triethanolamine

2

4.

Lanolin

7

5.

Perfume

0.5 mL

6.

Distilled water

(Make to 100) 70.5

Procedure: 1. Place stearic acid, lanolin and mineral oil in a 150 mL beaker (No. 1). 2. Heat the beaker in a water bath until all the ingredients have melted. 3. Keep this mixture warm. 4. Heat the mixture of triethanolamine and distilled water in a 250 mL beaker (No. 2) to a temperature of 80–90°C on a hot plate. 5. Now mix the contents of mixture from beaker No. 1 to this beaker No. 2 with constant stirring till smooth and uniform paste is formed. Allow the contents to cool. 6. Add the perfume of your choice to get the pleasant odour of required level. 7. The properties of the cream may vary depending on variation in the ratio of ingredients. Result: The yield of prepared cream = ........ g.

EXPERIMENT: 13.15 Aim: To prepare shampoo. Requirements: Beakers, stirrer. Theory: A shampoo is a cleaning formulation of chemicals called surfactants in a suitable form, solid, liquid or powder, that have the ability to surround oily materials on surfaces and allow them to be rinsed away by water from the hair scalp. While there are numerous types of shampoos, the majority are low viscosity solution formulas delivered from a plastic bottle. Often they are marketed towards different hair types or conditions. The basic ingredients in a shampoo formulation are as follows: Water: As primary ingredient in all shampoos, it makes up about 70 to 80% of the entire formula. It helps to dilute the detergents, makes the formula easier to spread and reduces irritation. It also keeps the formula inexpensive. Surfactants: These are the next most abundant ingredient in a shampoo. These surfactants are the primary cleansing ingredients and make up about 10%–15% of the formula. They can be anionic, non-ionic, cationic and ampholytic. They are derived from natural fatty acids or petroleum derivatives. Common primary detergents include ammonium lauryl sulphate, sodium lauryl sulphate and amine oxides.

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ223

Table 13.12: Composition of the shampoo S. No.

Ingredients

% (W / W)

1.

Distilled water (make to 100)

Make to 100

2.

Detergent (Sodium lauryl sulphate)

3.

Stearic acid

2.6

4.

Sodium hydroxide

1.0

5.

Lauryl alcohol

6.

Dimethicone

1.0

7.

Cocamidopropyl betaine

5.0

8.

Carbomer

1.0

10.0

10.0

9.

DMDM Hydantoin

Q.S

10.

Lanolin

0.5

11.

Lysine hydrochloride

0.5

12.

Perfume

Q.S

QS is quantity specified

Foam boosters: Other types of surfactants are added to shampoos to improve the foaming characteristics of the formulation. These compounds, usually betaines or alkanolamides, help to increase the amount of foam and the size of the bubbles. Like primary surfactants, they are also derived from fatty acids and have both water soluble and oil soluble characteristics. Typical materials include Lauramide or Cocamidopropyl betaine. Thickeners: To some extent, the secondary detergents make shampoo formulations thicker. Simply adding salt can also increase shampoo thickness. However, other materials such as methylcellulose, which is a cellulosic polymer or carbomer, are also used to increase the viscosity. Conditioning agents: Some materials are added to shampoos to offset the harsh effect of surfactants. Typical conditioning agents include polymers, silicones and quaternary agents. These ingredients are left on the hair surface after rinsing and they modify the characteristics such as feel, softness, combability and static charge. Shampoos that specifically feature conditioning as a benefit are called 2-in-1 shampoos because they clean and condition hair in the same step. Examples of conditioning agents include žŠ›ȱ ¢›˜¡¢™›˜™¢•›’–˜—’ž–ȱŒ‘•˜›’Žȱ ‘’Œ‘ȱ’œȱŠȱ™˜•¢–Ž›ǰȱ’–Ž‘’Œ˜—Žȱ ‘’Œ‘ȱ’œȱŠȱ silicone and Quaternium 80, a quaternary agent. Preservatives: Any formula that contains water has the potential to be contaminated by bacteria and other microbes. For this reason, preservatives are added to prevent such growth. Two of the most common preservatives used in shampoos are DMDM Hydantoin (1,3-Bis (hydroxymethyl)-5,5-dimethylimidazolidine-2,4-dione) and methyl-paraben. Other ingredients: A variety of other compounds are included in shampoos, if desired. Dyes for changing the colour, fragrances for changing the odour, pH adjustment ingredients, chelating agents (EDTA), opacifying ingredients (propylene glycol) and

224ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

more (vitamins, proteins, and herbal extracts ) which are not normally expected to have any impact on the final product performance. Dandruff shampoos will include a drug active ingredient like zinc pyrithione. Procedure: 1. Take 10 g of sodium lauryl sulphate , 2.6 g of stearic acid and add 50 mL of water in a 250 mL beaker. 2. Heat on a hot plate at 80°C with constant stirring until a homogeneous mixture is obtained. 3. In a boiling tube take 1 g of sodium hydroxide and 15 mL of distilled water. 4. Dissolve this and pour this into the solution in beaker with constant stirring for 10 minutes. 5. Now add lauryl alcohol and lanolin with constant stirring. 6. Add rest of the ingredients one by one in the beaker with constant stirring to get a smooth shampoo of desired properties and purpose. Result: The yield of the prepared shampoo formulation =........ g.

EXPERIMENT: 13.16 Aim: To prepare phenol formaldehyde resin. Requirements: Beakers, glass rod, funnel, filter paper, watch glass, glacial acetic acid, phenol, formaldehyde (40%) solution, sulphuric acid (concentrated ), distilled water. Theory: Phenol formaldehyde is prepared as a condensation product of reaction between phenol and formaldehyde in acidic or alkaline medium. The first product of the resin is ortho hydroxyl methyl phenol and para hydroxy methyl phenol. Polymerization then takes place to form long chains of fully cross-linked phenol formaldehyde resin. Phenol formaldehyde has good electrical insulating properties and finds use in the manufacture of electrical switches, plugs and insulating foams. They are hard, resist abrasion and scratching, and have good adhesive properties. They find use in the production of plywood and laminating products and cation exchange resins. Procedure: 1. Place about 5 mL of glacial acetic acid and 2.5 mL of 40% formaldehyde solution in a 500 mL beaker. 2. Add about 2 g of phenol with constant stirring using a glass rod till a saturated solution is formed. 3. To this, add a few drops of concentrated sulphuric acid with constant stirring carefully. 4. A voluminous white solid product is obtained in the beaker. 5. Wash the white solid with distilled water filter and dry it in the folds of filter paper. 6. Weigh it and find the yield of product.

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ225 OH H

OH H

H + CH2

H

CH2

O

H

H Phenol

OH

OH

H Formaldehyde

H

H

H

H H n Prepolymer phenolic resin (novolac) H

H

H

CH2 H

+ H2O + CH2 O

H

OH

H CH2

+ CH2

CH2 H

OH

OH

H

H

H

H

OH

H

H

H

OH

H

OH

+ H2 O H

H

CH2

CH2

H CH2

CH2 OH

O

CH2

CH2 H

H

OH

Cross-linked phenol formaldehyde resin

Fig. 13.2: Reaction between phenol and formaldehyde

Result: The yield of phenol formaldehyde resin = ........ g Precaution: Stay slightly away from the beaker when you add sulphuric acid, as the reaction is vigorous and exothermic. Stirring should be done throughout the experiment.

EXPERIMENT: 13.17 Aim: To synthesize gold nanoparticles. Requirements: Erlenmeyer flask (50 mL) or beaker (50 mL), hot plate, magnetic stirrer, droppers, test tubes, spectrophotometer or colorimeter, double distilled water. 1. Stock solution of gold ions (10 mM): Dissolve 1.0 g of hydrogen tetrachloro aurate (HAuCl4.3H2O) in 250 mL distilled water to make a 10 mM stock solution of gold (III) ions that can be kept for years if stored in a brown bottle. The solid is hygroscopic and so purchase HAuCl4.3H2O (Aldrich 244597 or 520918) in 1 g quantities and use the entire bottle. 2. Dilute 25 mL of the stock solution to 250 mL to make 1 mM solution required for this experiment. 3. 1% tri sodium citrate (Na3C6H5O7.2H2O) solution: Dissolve 0.5 g of the solid in 50 mL of double distilled water. 4. Sodium chloride solution: Dissolve at least 0.5 g of NaCl in 10 mL of double distilled water or use a saturated solution.

226ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Theory: Research in the field of nanoparticle synthesis and determining their properties has lead to different technological applications. The control of size, shape and composition is a key factor in determining the properties of materials at the nano level. In bulk at the macroscale, the element of gold is yellow, but at the nanoscale, the element of gold turns red to purple in colour. The formation of gold nanoparticles can therefore be observed by a change in colour since small nanoparticles of gold are red. The layer of absorbed citrate anions on the surface of the nanoparticles keeps the nanoparticles separated and the presence of this colloidal suspension can be detected by the reflection of a laser beam from the particles. Switching to a smaller anion allows the particles to approach more closely and another colour change is observed. Procedure: 1. Rinse all glassware with double distilled water before starting the experiment. 2. Add 20 mL of 1 mM gold solution to a 50 mL beaker or Erlenmeyer flask on a hot plate. 3. Add a magnetic stirrer and bring the solution to a rolling boil. To the rapidlystirred boiling solution, quickly add 2 mL of a 1% solution of trisodium citrate dihydrate. 4. The gold sol gradually forms as the citrate reduces the gold (III) to gold (0). 5. Remove from heat when the solution has turned deep red or 10 minutes has elapsed. 6. Record the visible spectrum of the solution to confirm the formation of gold nanoparticles. 7. If necessary, add additional water to the cuvette to get the absorbance on scale. 8. Put a small amount of the gold nanoparticle solution in two test tubes. Use one tube as a colour reference and add 5–10 drops of NaCl solution to the other tube and find Omax. Result: The colour of the prepared solution = ........ Omax of the solution formed = ........ nm. Interpretation: If Omax is around 530 nm, this means gold nanoparticles are formed. Precautions: 1. Wear plain glasses to protect your eyes. 2. Chemical gloves are recommended throughout the preparation. 3. Never look directly into a laser or shine a laser at another person. Note: Before the addition of the reducing agent, the gold in solution is in the Au+3 form. When the reducing agent is added, gold atoms are formed in the solution and their concentration rises rapidly until the solution exceeds saturation. Particles are then formed in a process called nucleation. The remaining dissolved gold atoms bind to the nucleation sites and growth occurs. The red colour for gold indicates the formation of nanoparticles. By changing the concentration, reaction time and temperature, nanoparticles of different sizes can be obtained. The typical spectrum of gold shows Omax at about 530 nm.

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ227 Nucleation

Supersaturation

Growth Au

Au

0

+3

Time

Addition of reducer

Fig. 13.3: Formation of gold nanoparticles

EXPERIMENT: 13.18 Aim: To synthesize silver nanoparticles. Requirements: Round bottom flask, magnetic stirrer cum heater, spectrophotometer or colorimeter, 0.002 M aqueous silver nitrate solution, 0.02 M aqueous tri sodium citrate solution, double distilled water. Theory: Colloidal nanoparticles are synthesized by reduction of metal salt or acid with tri sodium citrate. The particle size, size distribution and optical properties of nanoparticles are dependent on the reaction parameters and can be controlled to achieve desired results. Mechanism of reaction could be expressed as follows: 4Ag+ + C6H5O7Na3 + 2H2O o 4Ag° + C6H5O7H3 + 3Na+ + H+ + O2n Normally, the silver nanoparticles absorb at the Omax of about 400 nm. With an increase in the concentration of silver nanoparticles, the value of Omax increases. Typical spectra are shown in Fig. 13.4. Procedure: 1. Take 50 mL of 0.002 M aqueous silver nitrate solution in a round bottom flask. 2. Keep this on a heater with a magnetic stirrer for about 15 minutes at 80°C. 3. Mix 50 mL of 0.02 M aqueous tri sodium citrate and stir for an hour at 80°C. 4. Clear solution of silver turns golden yellow colour indicating the formation of silver nanoparticles. 5. Take the absorption spectrum of the prepared solution and find Omax. 6. The Scanning Electron Microscopy (SEM) analysis further confirms the formation of nanoparticles. Result: The colour of the prepared solution = ........ The Omax of the solution = ........ nm.

228ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• 1mM

3mM

5mM

10mM

1.6 1 mM 3 mM

1.4

5 mM

Absorbance (a.u.)

1.2

10 mM

1.0 0.8 0.6 0.4 0.2 0.0 300

350

400

450 500 Wavelength (nm)

550

600

Fig. 13.4: Spectra of silver nanoparticles of various concentrations

EXPERIMENT: 13.19 Aim: To synthesize titanium dioxide nanoparticles. Requirements: 50 mL Erlenmeyer flask or 50 mL beaker, stirring hot plate, magnetic stirrer, droppers, test tubes, X-ray diffractometer, titanium chloride, ammonium hydroxide, 2-propanal, double distilled water. Theory: Synthesis of Rutile titanium dioxide can be carried out by sol gel method by hydrolysis of titanium tetrachloride in deionized water in the presence of ammonium hydroxide as hydrolysis catalyst. The size of TiO2 nanoparticles obtained by this method is of diameter 26.4 nm. The temperature has great influence on the particle size and crystalline phase of TiO2. Procedure: 1. Take 20 mL of 0.01 M titanium chloride solution in a round bottom flask. 2. To this add 60 mL of 0.1 M ammonium hydroxide solution dropwise with stirring at room temperature. 3. White colour solution indicates the formation of titanium dioxide (TiO2) nanoparticles. 4. Centrifuge and wash the precipitate with double distilled water. 5. Dry in isopropanol and take the X-ray diffraction pattern to find the size of particles. Observation:ȱ‘Žȱœ’£ŽȱŒŠ—ȱ‹ŽȱŒŠ•Œž•ŠŽȱ‹¢ȱꗍ’—ȱȱžœ’—ȱŒ‘Ž››Ž›ȱ˜›–ž•Š Result: The size of particles is found to be ........ nm

’œŒŽ••Š—Ž˜žœȱ¡™Ž›’–Ž—œȳ229

EXPERIMENT: 13.20 Aim: To synthesize and characterize zinc oxide nanoparticles. Requirements: 50 mL Erlenmeyer flask or a 50 mL beaker, stirring hotplate, magnetic stirrer, droppers, test tubes, double distilled water, zinc sulphate, alcohol, ammonium hydroxide. Theory: Zinc oxide has unique physical and chemical properties like high chemical stability, high electrochemical coupling coefficient and photostability. This can be prepared by direct precipitation method using zinc nitrate, zinc chloride or zinc sulphate and potassium hydroxide or ammonium hydroxide as precursors. The size of nanoparticles generated lie in the range of 20–40 nm. Procedure: 1. Prepare a mother solution by mixing 25 mL aqueous ethanolic solution of 0.2 M zinc sulphate and 50 mL double distilled water. 2. Prepare 25 mL of 25% ammonium hydroxide solution in deionized water. 3. Slowly add ammonium hydroxide solution into zinc sulphate solution at room temperature under vigorous stirring on a magnetic stirrer cum heater at a temperature 50–60°C. 4. Centrifuge the white powder obtained at 5000 rpm for 20 minutes. 5. Then wash it thrice with distilled water and finally with absolute alcohol. 6. Now dry it at 60°C in air atmosphere in an oven for 8 hours. Record XRD spectra and SEM images to find the size of nanoparticles. Result: The size of nanoparticles is found to be ........ nm.

VIVA QUESTIONS 1. Name the technique used to determine the concentration of sodium and potassium in a given sample. 2. Why is nebulizer burner system used in flame photometry? 3. Write the Bragg’s equation. 4. What are the different patterns of lines that differentiate between primitive , BCC and FCC cubic crystal? 5. What is the use of finding free acidity of a given fertilizer? 6. Name the indicator used in the complexometric determination of calcium. What is the role of ammonia buffer? 7. How can you determine free phosphoric acid in a given sample of fertilizer? 8. Name the preservatives used in cold drinks. 9. What are the various forms of phosphorous fertilizers? 10. Name the common adulterants present in food items. 11. Name the various constituents present in nail polish. 12. What are the ideal characteristics of nail polish remover?

230ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

13. What is the main component of talcum powder? What is the role of zinc oxide and magnesium carbonate? 14. What are the ideal characteristics of face cream? 15. What is the role of sodium lauryl sulphate and carbomer in shampoo? 16. What is the use of phenol formaldehyde resin? 17. How do you ascertain that gold nanoparticles are formed? 18. How do you synthesize silver nanoparticles? 19. How do you characterize nanoparticles of titanium oxide and zinc oxide?

14

PROGRAMMING WITH BASIC LANGUAGE

In this chapter an attempt is made to give some fundamental concepts of BASIC language and write chemistry oriented problem solving program. All the relevant information has been summarized. For details (Reference No. 7). BASIC stands for Beginners All Purpose Symbolic Instruction Code. It is a simple and powerful language for chemists to write programs. The program in BASIC high level language is written in the form of certain instructions called statements in the form of lines. This is an interpreted language and execution of program takes place line by line. In designing the program, the step-by-step procedure for solving the problem is known as algorithm. The process of finding and correcting the program error-getting the bugs out is called debugging. A syntax error is the violation of BASIC’s rule of statement structure. A logical error is one that leads to incorrect results when the program is executed. The program development cycle consists of analyzing (giving input and output data and choosing variable names), designing (giving the formula and logical process necessary to solve the problem), coding (checking for syntax and logic errors) and testing (running and debugging) the program.

14.1 OPERATORS The QBASIC language, like any other language, has its own character set, operators, constants and variables. The calculations in QBASIC are carried out with the help of following arithmetic operators (Table 14.1). Table 14.1: QBASIC arithmetic operators Name

Arithmetic Operator

Purpose

Plus

+

Addition

Minus

-

Subtraction

Asterisk

*

Multiplication

Slash

/

Division

Caret

^

Exponentiation

232ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ• Table 14.2: Relational and logical operators Operator =

Meaning Equal to Not equal to


=

Greater than or equal to

AND

Conjunction

OR

Disjunction

NOT

Logical negation

14.2 BASIC STATEMENTS Table 14.3: A summary of various BASIC statements is presented below (For Details Ref. no. 7) ClS

Clears the screen and positions the cursor in the upper left hand corner

Data C1, C2…

Stores the constants to be accessed by the program’s READ statements in the data

DEF FN name (var 1, var2,…) = expression

Creates a user defined function

DIM array1(size1), array 2(size2),…

Allocate storage space for arrays with a particular size

END

Terminates execution of program

FOR num var = initial value to final value [STEP increment]

Controls the execution of a FOR / NEXT loop

GOSUB line num or label

Transfers control to the subroutine beginning at the specified line number

GOTO line num

Transfers the control to the specified line number

IF condition THEN statement

Execute the specified statement if the condition is true, otherwise follows the given statement.

IF condition THEN statement1 ELSE statement2

If the given condition is true, statement 1 is executed; otherwise statement 2 is executed.

INPUT var1, var2

Input data from the keyboard and assigns that data to the specified list of variables

LET variable = expression

Evaluate the expression and assigns its value to variable

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ233 LOCATE (r,c)

Places the cursor on the specified row and column print position

NEXT numvar

Indicates the bottom of a FOR…NEXT loop

ON num exp GOSUB linenum1, linenum2..

Transfer the control to the subroutine beginning at the specified line numbers

ON num exp GOTO linenum1, linenum2..

Transfer the control to the one the specified line numbers

PRINT EXP1; EXP2; EXP3;….

Print the values of EXP1, EXP2, EXP3 in one zone of the screen (without space)

PRINT EXP1,EXP2,EXP3, ….

Print the values of EXP1, EXP2,EXP3 in different zones on the screen

RANDOMIZE numexp

Reseeds the random number generator

REM Comment or ’ comment

Write the comment or explanatory note wherever required in the program

RESTORE

Resets the data pointer to the beginning of the data block

RETURN

Transfers control from subroutine to the statement following the GOSUB or ON…GOSUB that is called from main program

SCREEN num

num 0 Sets the screen to default mode, i.e., TEXT mode num 1 Sets the screen to graphic mode, i.e., medium resolution num 2 Sets the screen graphics mode, i.e., high resolution graphics

STOP

Terminates execution which may be resumed by CONT statement

SWAP var1, var2

Interchange the contents of the indicated variables

WEND

Indicates the bottom of a while ….wend loop

WHILE condition

causes the WHILE…WEND to be reentered if the condition is true.

GRAPHIC Statement CIRCLE (x, y) radius, color [start, end]

Draws a circle if start and end is not provided; An arc if start and end are non-negative and a sector if start and end are negative

DRAW string variable or constant

Draw non-geometric shape depending on direction of pixel

LINE (x1, y1) –(x2, y2) , color

Draws the line between the two given points

LINE (x1, y1) – (x2, y2), color, B[F]

Draws the BOX with the given corner in the specified color; if F appears then filled box is formed

PAINT (x, y). Interior, boundary

Fills the region containing (x, y) and enclosed by the boundary

VIEW (x1, y1)-(x2, y2)

Select the given view port from x1, y1 to x2, y2 pixel coordinates

WINDOW (x1,y1)-(x2, y2) Replace the pixel numbering from screen coordinates to world coordinates

234ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

14.3 SOME BUILT-IN FUNCTION Table 14.4: QBASIC built-in function S. No.

Function

Output

1.

SIN(X)

Returns mathematical Sine of X in radians

2.

COS(X)

Returns mathematical Cosine of X in radians

3.

TAN(X)

Returns mathematical Tangent of X in radians

4.

ASIN(X)

Returns mathematical Sin-1 of X

5.

LOG(X)

Returns natural logarithm of X

6.

EXP(X)

Returns the exponential of X

7.

INT(X)

Returns the largest integer value

8.

FIX(X)

Truncate X to integer

9.

RND(X)

Generate random number between 0-1

10.

SGN(X)

Converts value of X +1 or –1 depending on positive and negative value of X

11.

SQR(X)

Calculate square root of X

12.

CINT (X)

Rounds the number with fractional part to the next whole number

13.

MOD

Returns the remainder

The steps involved in running the QBASIC program involves installing QBASIC language. The following dialog box appears. It is menu driven.

Select the file option. Open new file. Write the program with the help of various characters, operators and statements. ŠŸŽȱ‘Žȱ™›˜›Š–ǯȱŽ‹žȱ‘Žȱ™›˜›Š–ǯȱž—ȱ‘Žȱ™›˜›Š–ȱǯȱ ’ŸŽȱ›Žšž’›Žȱ’—™žœȱŠ—ȱ get the output in a desired format.

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ235

14.4 BASIC PROGRAMS For Illustration, some simple problems are written in BASIC language ȱ ȱŗ REM progm to convert Fahrenheit to Degree Celsius LET F=100 LET C = (5/9)*(F-32) PRINT “The temperature in degree Fahrenheit”; F PRINT “equivalent to degrees celsius as”; C END ȱ ȱŘ REM PROGM TO FIND FACTORIAL OF A NUMBER FACT=1 FOR I=1 TO 10 STEP 1 FACT=FACT* I PRINT FACT NEXT I END The FOR….NEXT loop will be executed to print the value of factorial 10! = ȱ ȱř REM PROGM TO FIND THE PRESSURE OF GAS USING IDEAL GAS EQUATION READ N,R,T,V P=(N*R*T )/V PRINT “PRESSURE OF GAS USING IDEAL GAS EQUN”;P DATA 1,0.082,298.15,1 END ȱ ȱŚ REM PROGM TO FIND THE PRESSURE OF GAS USING VARIOUS GAS EQUATION READ N,R,T,V,a,b P=(N*R*T)/(V-N*b)-((N^2)*a)/(V^2) PRINT “PRESSURE OF GAS USING VAN DER WAAL GAS EQUN”;P DATA 1,.082,298.15,1,137,0.0131 END ȱ ȱś REM PROG TO READ AND PRINT A MATRIX INPUT “The num of row and column”; M,N

236ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

DIM A(10,10) FOR I=1 TO M FOR J=1 TO N READ A(I,J) PRINT A(I,J); NEXT J PRINT NEXT I DATA 1,2,3,4 END RUN The num of row and column? 2, 2 1 2 3 4 ȱ ȱŜ REM PROG TO ADD TWO GIVEN MATRICES INPUT M,N REM NESTED LOOP TO READ AND PRINT MATRIX A DIM A(10,10), B(10,10), C(10,10) FOR I=1 TO M FOR J=1 TO N READ A(I,J) PRINT A(I,J); NEXT J PRINT NEXT I DATA 1,2,3,4 REM NESTED LOOP TO READ AND PRINT MATRIX B FOR I=1 TO M FOR J=1 TO N READ B (I,J) PRINT B (I,J); NEXT J PRINT NEXT I

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ237

DATA 5,6,7,8 REM NESTED LOOP TO ADD MATRIX A AND B FOR I=1 TO M FOR J=1 TO N C(I,J)=A(I,J)+B(I,J) NEXT J PRINT NEXT I REM PROG TO PRINT ADDED MATRIX C FOR I=1 TO M FOR J=1 TO N PRINT C(I,J); NEXT J PRINT NEXT I END ȱ ȱŝ REM PROG TO SUBTRACT TWO GIVEN MATRICES INPUT M,N REM NESTED LOOP TO READ AND PRINT MATRIX A DIM A(10,10), B(10,10), C(10,10) FOR I=1 TO M FOR J=1 TO N READ A(I,J) PRINT A(I,J); NEXT J PRINT NEXT I DATA 1,2,3,4 REM NESTED LOOP TO READ AND PRINT MATRIX B FOR I=1 TO M FOR J=1 TO N READ B (I,J) PRINT B (I,J); NEXT J PRINT NEXT I

238ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

DATA 5,6,7,8 REM NESTED LOOP TO SUBTRACT MATRIX A FROM B FOR I=1 TO M FOR J=1 TO N C(I,J)=B(I,J) - A(I,J) NEXT J PRINT NEXT I REM PROG TO PRINT SUBTRACTED MATRIX C FOR I=1 TO M FOR J=1 TO N PRINT C(I,J); NEXT J PRINT NEXT I END ȱ ȱŞ REM PROG TO MULTIPLY TWO GIVEN MATRICES INPUT M,N REM NESTED LOOP TO READ AND PRINT MATRIX A DIM A(10,10), B(10,10), C(10,10) FOR I=1 TO M FOR J=1 TO N READ A(I,J) PRINT A(I,J); NEXT J PRINT NEXT I DATA 1,2,3,4 REM NESTED LOOP TO READ AND PRINT MATRIX B FOR I=1 TO M FOR J=1 TO N READ B (I,J) PRINT B (I,J); NEXT J PRINT NEXT I

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ239

DATA 5,6,7,8 REM NESTED LOOP TO MULTIPLY MATRIX A AND B FOR I=1 TO M FOR J=1 TO N C(I,J)=0 FOR K=1 TO M C(I,J)=A(I,K)*B(K,J) + C(I,J) NEXT K NEXT J NEXT I REM NESTED LOOP TO PRINT PRODUCT OF TWO MATRICES FOR I=1 TO M FOR J=1 TO N PRINT C(I,J); NEXT J PRINT NEXT I END ȱ ȱş REM PROGM TO FIND TRANSPOSE OF A MATRIX INPUT M,N REM NESTED LOOP TO READ AND PRINT ANY MATRIX FOR I=1 TO M FOR J=1 TO N READ A(I,J) : PRINT A(I,J); NEXT J PRINT NEXT I DATA 2,4,6,8 REM NESTED LOOP TO PRINT TRANSPOSE FOR I=1 TO N FOR J=1 TO M B(I,J) = A(J,I) NEXT J PRINT NEXT I END

240ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

ȱ ȱŗŖ REM TO FIND ROOTS OF QUADRATIC EQUATION INPUT “ENTER VALUE OF A,B,C= ”; A, B, C LET D = (B^2)- (4*A*C) PRINT “DISCRIMINANT”; D IF D < 0 THEN PRINT “IMAGINARY ROOTS” ELSE 20 REAL = (-B/(2*A)) IMAGIN= SQR(-D)/(2*A) PRINT “ROOT1”; REAL; “+I”; IMAGIN PRINT “ROOT2”;REAL; “-I”; IMAGIN: GOTO 100 20 IF D = 0 THEN PRINT “REAL AND EQUAL ROOTS” ELSE 30 ROOT1 = (-B/(2*A)) ROOT2=ROOT1 PRINT “ROOT1 AND ROOT2”; ROOT1;ROOT2; GOTO 100 30 IF D > 0 THEN PRINT “REAL AND UNEQUAL ROOTS” ROOT1 = (-B/(2*A)+ SQR(D)/(2*A) ROOT2= (-B/(2*A)- SQR(-D)/(2*A) PRINT “ROOT1”; ROOT1 PRINT “ROOT2”; ROOT2 100 END The value of binomial coefficient is predicted using following formula n Cr = n! /( r !(n-r)!) ȱ ȱŗŗ REM TO FIND BINOMIAL COEFFICIENT USING GOSUB STATEMENT INPUT “THE VALUE OF N AND R”; N, R A = N GOSUB 50 NFAC= P A = N – R GOSUB 50 NRFAC = P A = R GOSUB 50 RFAC = P NCR = NFAC/NRFAC/RFAC PRINT “ncr” ; NCR END

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ241

50 REM SUBROUTINE PROGM STARTS P = 1 FOR I = 1 TO A P = P * I NEXT I RETURN ȱ ȱŗŘ REM PROGM TO DRAW COOLING CURVE CLS: SCREEN 1 LOCATE (1,15): PRINT “COOLING CURVE FOR THE 30% BENZOIC ACID” LOCATE (2,15) : PRINT “AND 30% NAPHTHALENE” WINDOW (0,0) - (15,150) LINE (0,0) – (0,190) LINE (0,15) – (100,15) LOCATE (6,1): PRINT “T” LOCATE (7,1): PRINT “E” LOCATE (8,1): PRINT “M” LOCATE (9,1): PRINT “P” LOCATE (10,6): PRINT “BREAK” LOCATE (15,7): PRINT “HALT” LOCATE (24,20): PRINT “TIME” I = 0.5 TIME = I DIM T(25): WHILE I = 0.0001 X3=(X1+X2)/2 D = FNF(X2)+FNF (X3) I=I+1 PRINT I, X3, D IF I< = 20 THEN 100 PRINT I STOP 100 IF D < 0 THEN X1=X3 ELSE X2=X3 WEND PRINT I; X3 ȱ ȱŗŚ REM PROGM TO FIND VALUE OF A FUNCTION USING TRAPEZOIDAL RULE INPUT “LOWER LIMIT, UPPER LIMIT, NO. OF STRIPS” A, B, N DEF FNP (X) = 3*(X^3) – 4* (X^2) +5*X +10 H = (B-A)/N SUM = (FNP (A) + FNP(B))/2 FOR = J TO (N-1) X = A+(J*H) SUM=SUM +FNP(X) NEXT J I= SUM*H PRINT “THE VALUE OF INTEGRAL IS ” ; I END ȱ ȱŗś REM PROGM TO FIND VALUE OF A FUNCTION USING SIMPSON RULE INPUT “LOWER LIMIT,UPPER LIMIT, INITIAL VALUE”; A, B, N DEF FNP(X) = (X+4) ^4 H= (B-A) / N S1= FNP(A)+FNP(B): S2=0: S4=0 FOR J = 1 TO (N-2) STEP 2 S4 = S4 + f (A+J*H) S2 = S2 + f (A+(J+1)*H) NEXT J I=(H/3)*(S1+(2*S2)+(4*S4)) PRINT “THE VALUE OF INTEGRAL”; I END

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ243

ȱ ȱŗŜ REM PROGM TO FIND VALUE OF PI BY MONTE CARLO METHOD 5 INPUT “NUMBER OF POINTS”; IF N < = 0 THEN 100 I2=0 FOR I = 1 TO N X = RND(0): Y= RND(1) IF SQR ( ( X-0.5)^2) * (Y-0.5)^2) > 0.5 THEN 10 I2 =I2+1 10 NEXT I PI = 4*(I2/N) PRINT “PI =” ; PI GOTO 5 100 END

14.5 MS EXCEL Experiment 14.1 To plot the given data using MS Excel Theory: Microsoft Excel is a spreadsheet program used to store and retrieve numerical data in a grid format of columns and rows. Excel is ideal for entering, calculating, manipulating and analyzing scientific data using various arithmetic and library function. Excel document is called a workbook. A workbook always has at least one worksheet. Worksheets are the grid where you can store and calculate data. You can have many worksheets stored inside a workbook, each with a unique worksheet name. Worksheets are laid out in columns (vertical) and rows (horizontal). The intersection of any given row and column is a cell. Cells are really where you enter any information. A cell will accept a large amount of text, or you can enter a date, number, or formula. Each cell can be formatted individually with distinct border, background colour, and font colour/size/type. Each column has a capital letter on the top to show what column it is. Each row has a number to the immediate left of the first column, to show what row it is. Each cell is identified by a unique address consisting of the column letter followed by the row number. For example, the address of the cell in the first column, first row is A1. The address of the cell in the second column, third row is B3. If you click a cell, the cell address appears just above column A. Excel includes a formula library for calculating things like Net Present Value (NPV), standard deviation, interest payments over time, and other common financial and mathematic formulae. Excel offers a wide array of charts to visualize data. They range from simple line graphs to bubble and radar charts. Excel has two main tools for charting: standard charts and pivot charts.

244ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Steps for Excel: To draw graph in MS-Excel (Absorbance versus Concentration)Lambert –Beer Law 1. Enter the data of Concentration and Absorbance ȱ ȱ ˜ȱ˜ȱ—œŽ›oScattero¢™Žȱ˜ȱ ›Š™‘ǯȱ—ȱŽ–™¢ȱ›Š™‘ȱ‹˜¡ȱ ˜ž•ȱŠ™™ŽŠ›ǯ

2. Right click on the graph box and select data option.

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ245

In the left side click Add option from legend entries in which you could write what graph you are drawing (e.g., Absorbance).

3. In Series X-values, select the select data blue button on right side and select the data from the Excel sheet for concentration.

246ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ247

4. Similarly, select the Y series values which are absorbance values and then select ok. You will get this graph.

248ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

5. You can also find the regression coefficient and the corresponding equation of the line observed from the data. Select all the data points on the graph by clicking on any single data point on the graph.

6. Right click on the points selected and select Add trendline.

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ249

7. On the basis of line you are drawing (linear, exponential, etc.), linear in this case, select the option display equation and R-squared value.

8. You will observe a trendline drawn in black colour and the R2 value (regression coefficient) and the linear equation. You can also change the colour of lines as per your choice.

250ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

For plotting of various graphs of practical data in chemistry, MS excel can be very conveniently used to plot the data in desired format for comparison. The various steps involved are: ’—Žȱǻ˜›ȱǼȱ ›Š™‘œDZȱ’—Žȱǻ˜›ȱǼȱ›Š™‘œȱŠ›Žȱ˜›ȱŠŠȱŒ˜••ŽŒŽȱ’—ȱŠ—ȱŽ¡™Ž›’–Ž—ȱ’—ȱ which the independent variable (the one that goes on the X-axis) is quantitative (numerical). Step 1: Enter the data in the cells of an Excel spreadsheet. Step 2: Use the mouse to highlight the block of cells containing your data, then click ‘Žȱ‘Š›ȱ’£Š›ȱ‹ž˜—ȱǻ’••žœ›ŠŽȱ’—ȱŽ™ȱŗȱ’—ȱ‘Žȱ™ŠœœŠŽȱ˜—ȱŠ›ȱ ›Š™‘œǼǯ Step 3: Choose the XY (Scatter) chart. Usually, the best subtype is the one in which individual data points are connected with a line. Step 4: Excel will then show you what it thinks is the correct chart, like this: If it looks right to you, click Next. Otherwise, go Back. Step 5: The Chart Wizard next gives you a series of options for changing the look of your charts. Add axis labels and a title and change the default legend. Step 6: Save the file and place the output in the desired file. One can try the following data for plotting: 1. Plot Maxwell Distribution Curve Dn = 4* 3.14*(M/(2*3.14*R*T)^(3/2) * EXP((-M*C*C)/ (2*R*T))*(C*C) C varies from 0 -1500, temperature varies from 300 to 400 step 20 2. Plot PV isotherms using ideal gas equation P= (n*R*T)/V , V varies from 0.004 to 0.4 litres step 0.004 ,? temperature varies from 350 to 500 step 50 For van der Waal gas equation, a = 4.17, b = 0.0371 P= (R*T)/(V-b)- (a/(V^2)) same conditions as above 3. Plot wave function and amplitude of wave function with n = 1 for particle in one dimension box. 4. Plot potentiometric titration curve using data for strong acid versus strong base. Plot first derivative and second derivative plots. 5. Plot conductometric titration curve using data for strong acid versus strong base and weak acid versus strong base. 6. CST plot for temperature versus composition of phenol-water system. 7. Plotting data of cooling curve and then drawing phase diagram. 8. Plot for finding rate constants for first and second order rate laws. 9. Radial distribution plots, probability and probability density for various (1s, 2s, 2p, 3d) orbitals. 10. Absorbance versus concentration plots for KMnO4 and K2Cr2O7 solutions.

›˜›Š––’—ȱ ’‘ȱŠœ’ŒȱŠ—žŠŽȱȳ251

VIVA QUESTIONS 1. 2. 3. 4. 5. 6. 7. 8.

Write full form of BASIC. What is debugging? What are the various types of errors in BASIC language? What is the function of DIM and RESTORE statement? Write the syntax for INPUT and READ…DATA statements. What is the difference between screen 0 , 1 and 2? Write the statement to draw a filled box. Write the output of the following statements: PRINT SQR (4) PRINT CINT (5.9) PRINT INT (5.9) PRINT FIX (5.9) 9. Write a program to draw concentric circles with centre at (100,100) with radii varying with 10 pixel. 10. Name the software used to plot the data.

APPENDICES

APPENDIX A Some basic units Base quantity Name Length

Base unit Symbol l, h, r

Name

Symbol

metre

m kg

Mass

m

kilogram

Time

t

second

s

Electric current

I, i

ampere

A

Temperature

T

kelvin

K

Amount of substance

n

mole

mol

Luminous

Iv

candela

cd

APPENDIX B Some useful relationships ln x = 2.3026 logx

1 dyne = 10–5 N

1 eV =1.60218 u10–19 J

1 atmosphere = 1.1032 u105 Pa

1 calorie = 4.184 joules

R = 8.314 J mol–1 K–1 = 0.082 dm3 atm mol–1K–1 = 1.987 cal deg–1mol–1

1A0 = 10–10 m = 10–1nm 0.1240 MeV = 1.987 u10–23J = 1.198 u102 kJmol–1

k = 1.38 u10–23 JK–1 NA = 6.023 u10–23

1 litre-atmosphere=101.3 J

h = 6.63 u 10–34 Js

kT = 207.226 hc

c = 2.99 u 108 ms–1

1 erg = 10–7 J

F = 96,485 C mol–1

254ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

APPENDIX C Solutions of common acids and bases Reagent

Molarity

Volume (mL) required to make 1 L of 0.1 M solution

Conc. HCl

11.6

8.6

Conc HNO3

15.4

6.5

Conc.H2SO4

17.8

5.6

Glacial CH3COOH

17.4

5.8

Ammonia

14.8

6.8

H3PO4

14.6

6.8

APPENDIX D Density, surface tension and viscosity of water against air at various temperatures Temperature (°C)

Density (g mL–1)

Surface tension (dyn cm–1)*

Viscosity(centipoise, cP)**

0

0.99987

75.6

1.6735

5

0.99999

74.9

1.5182

10

0.99973

74.22

1.3039

15

0.99913

73.49

1.1374

20

0.99823

72.75

1.0019

22

0.9978

72.44

0.9540

24

0.9973

72.13

0.9140

25

0.99707

71.97

0.8903

26

0.9968

71.82

0.8730

28

0.9962

71.50

0.8350

30

0.99567

71.18

0.7973

32

0.9950

70.86

0.7679

35

0.9941

70.38

0.7200

40

0.99224

69.56

0.6526

50

0.98807

67.91

0.5465

60

0.98324

66.18

0.4660

70

0.97781

64.40

0.4036

80

0.97183

62.6

0.3540

* In SI unit, it is millinewton per metre (mN m–1). 1dyn cm–1 = 1 mN m–1 ** In SI unit, it is expressed as milliPascal.second (mPa.s). 1 cP = 1 mPa·s

™™Ž—’ŒŽœȱȳ255

APPENDIX E Values of surface tension and viscosity of some common liquids at 20°C Surface Tension (dyn cm–1)

Viscosity (millipoise, mP)*

Acetic acid

27.8

12.22

Acetone

23.7

3.31

Benzene

28.9

6.50

n-Butyl alcohol

24.6

29.48

Carbon tetrachloride

27.0

9.70

Chloroform

27.1

5.60

Ethyl acetate

23.9

4.55

Ethyl alcohol

22.8

12.00

Ethylene glycol

47.7

19.90

Glycerol

63.4

14900

Methyl alcohol

22.6

5.93

n-Propyl alcohol

23.8

22.56

m-Toluene

28.9

6.20

o-Toluene

30.1

8.10

p-Toluene

28.4

6.48

n-Hexane

18.4

3.26

Liquid

* 10 mP = 1 cP

APPENDIX F Heats of neutralization Reaction HCI + NaOH oNaCl + H2O

ΔH (kcal (g.eq)–1) –13.84

ΔH (kJ (g.eq)–1) –57.85

HNO3 + NaOH o NaNO3 + H2O

–13.76

–57.51

CH3COOH + NaOH o CH3COONa + H2O

–13.42

–56.09

HCl + NH4OH o NH4Cl + H2O

–12.76

–53.33

CH3COOH + NH4OH o CH3COONH4 + H2O

–12.04

–50.32

1 1 1 HCI + MgO(solid) → MgCl2 + H2O 2 2 2

–17.45

–74.19

256ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

APPENDIX G Heats of solution of some salts Substance Ammonium chloride Potassium nitrate Barium chloride (anhy) Barium chloride (hyd) Copper sulphate(anhy) Copper sulphate (hyd) Sodium hydroxide

Moles of water / mole of salt 200 200 400 400 400 400 200

ΔH (kcal mol–1) +3.88 +8.52 –2.70 +4.93 –15.80 +2.75 –9.94

ΔH (kJ mol–1) +16.22 +35.61 –11.26 +20.60 –66.04 +11.49 +41.55

APPENDIX H Values of specific conductance of KCl solutions (ohm–1 cm–1) Molarity 0.001 0.010 0.020 0.100 1.000

18°C 0.0001271 0.0012250 0.0024000 0.0111900 0.0982200

20°C 0.0001326 0.0012780 0.0025000 0.0116700 0.1021000

25°C 0.0001469 0.0014150 0.0027000 0.0128860 0.1117300

APPENDIX I Specific conductance of KCl solution (ohm–1 cm–1 or siemen cm–1) Temperature(°C)

0.1 N KCl

0.02 N KCl

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

0.01167 0.01191 0.01215 0.01239 0.01264 0.01288 0.01313 0.01337 0.01362 0.01387 0.01412 0.01437 0.01462 0.01488 0.01513

0.002501 0.002553 0.002606 0.002659 0.002712 0.002765 0.002819 0.002873 0.002927 0.002981 0.003036 0.003091 0.003146 0.003201 0.003256

35

0.01559

0.003312

™™Ž—’ŒŽœȱȳ257

APPENDIX J For very dilute solutions, it is convenient to express the concentration in terms of parts per million (ppm) or parts per billion (ppb). Mass of the solute Mass of the solute × 106 ; ppb = × 109 ppm = Mass of the solution Mass of the solution If the solvent is water, the density of the solution is essentially 1 g mL–1. ‘Ž›Ž˜›Žǰȱ™™–ȱƽȱ–ȱ˜ȱ‘Žȱœ˜•žŽȺȦȺŗŖ6ȱ–ȱ˜ȱ ŠŽ›ȱƽȱ–ȱ˜ȱ‘Žȱœ˜•žŽȺȦȺŗȱȱ˜ȱ‘Žȱ solution 1 ppm corresponds to 1 PȺȦȺ–ȱ˜›ȱŗȱ–ȺȦȺȱ˜›ȱŗŖ–4 wt % ŗȱ™™‹ȱŒ˜››Žœ™˜—œȱ˜ȱŗȱ—ȺȦȺ–ȱ˜›ȱŗȱPȺȦȺȱ˜›ȱŗŖ–7 wt % The molarity of the solution can be converted into ppm of the solution as follows: Mass of solute per litre of water = Molar mass u Molarity Calculate the mass of solute per litre of water from the molarity of the solution. For aqueous solutions, mass of solute per litre of water = mass of solute per litre of solution (since the density of the solution is essentially that of water, i.e.,1 g mL-1). Let Mass of the solute × 106 the mass per litre of solution = m g. Hence, by definition, ppm = Mass of the solution =

m × 106 100

How small is the ppm? To get an idea of how small 1 part in a million (1 ppm) is, compare the following equivalents: Ȋȱ 2.5 cm in 25 km Ȋȱ One minute in 2 years Ȋȱ One paise in Rs. 10000 Ȋȱ One drop in 50 L water Ȋȱ One mg in 1 L Ȋȱ One mg in 1 kg Ȋȱ One μg in 1g Ȋȱ One μg in 1 mL Ȋȱ One Hz in 1 MHz

258ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

APPENDIX K Š‹•ŽDZȱ ŗȲ’›’ŒȱŠŒ’Ȭ™‘˜œ™‘ŠŽȱ‹žŽ›ȱǻŗŞǚǼ Range 2.2–8.0 A. Citric acid solution - 0.1 M B. Disodium hydrogen phosphate solution - 0.2 M S. No.

A mL

B mL

pH

A mL

B mL

pH

1.

19.60

0.40

2.2

9.28

10.72

5.2

2.

17.82

2.18

2.6

8.40

11.60

5.6

3.

15.89

4.11

3.0

7.37

12.63

6.0

4.

14.30

5.70

3.4

6.15

13.85

6.4

5.

12.90

7.10

3.8

4.55

15.45

6.8

6.

11.72

8.28

4.2

2.61

17.39

7.2

7.

10.65

9.35

4.6

1.27

18.73

7.6

8.

9.70

10.30

5.0

0.55

19.45

8.0

Š‹•ŽDZȱ ŘȲŒŽ’ŒȱŠŒ’Ȭœ˜’ž–ȱŠŒŽŠŽȱ‹žŽ›ȱǻŗŞǚǼ Range 3.42–5.89 A: Acetic acid = 0.2 M B: Sodium acetate solution = 0.2 M S. No.

A mL

B mL

pH

1.

9.0

1.0

3.72

2.

8.0

2.0

4.05

3.

7.0

3.0

4.27

4.

6.0

4.0

4.45

5.

5.0

5.0

4.63

6.

4.0

6.0

4.80

7.

3.0

7.0

4.99

8.

2.0

8.0

5.23

9.

1.0

9.0

5.57

10.

0.5

9.5

5.89

™™Ž—’ŒŽœȱȳ259

Š‹•ŽDZȱ řȲ‘˜œ™‘ŠŽȱ‹žŽ›ȱǻŗŞǚǼ Range 5.29–8.04 ȱ ǯȱ ’œ˜’ž–ȱ‘¢›˜Ž—ȱ™‘˜œ™‘ŠŽȱœ˜•ž’˜—ȯȺȦȺŗś B. Potassium dihydrogen phosphate solution—0.2 M S. No.

A mL

B mL

pH

A mL

B mL

pH

1.

0.25

9.75

5.29

5.0

5.0

6.81

2.

0.50

9.50

5.59

6.0

4.0

6.98

3.

1.00

9.00

5.91

7.0

3.0

7.17

4.

2.00

8.00

6.24

8.0

2.0

7.38

5.

3.00

7.00

6.47

9.0

1.0

7.73

6.

4.00

6.00

6.64

9.5

0.5

8.04

Š‹•ŽDZȱ ŚȲ––˜—’ž–ȱŒ‘•˜›’ŽȮ––˜—’ž–ȱ‘¢›˜¡’Žȱ‹žŽ›ȱǻŗŞǚǼ Range 8.31–10.22 A: Ammonium chloride = 0.2 M B: Ammonium hydroxide = 0.2 M S. No.

A mL

B mL

pH

1.

1.0

9.0

8.31

2.

2.0

8.0

8.66

3.

3.0

7.0

8.90

4.

4.0

6.0

9.09

5.

5.0

5.0

9.26

6.

6.0

4.0

9.44

7.

7.0

3.0

9.63

8.

8.0

2.0

9.86

9.

9.0

1.0

10.22

260ȳ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ȱŠ‹˜›Š˜›¢ȱŠ—žŠ•

Š‹•ŽDZȱ śȲ˜›Š¡Ȭ˜’ž–ȱ ¢›˜¡’ŽȱžŽ›ȱǻŘŖǚǼ Range 9.28–10.1 A: Solution of borax (19.05 g of sodium tetraborate decahydrate (Na2B4O7.10H2O) in 1000 mL of distilled water) = 0.05 M B: Sodium hydroxide = 0.2 M 6.2 mL of A + x mL of B, diluted to a total of 25 mL with distilled water. S. No.

x mL

pH

1.

0.0

9.28

2.

0.9

9.35

3.

1.4

9.4

4.

2.2

9.5

5.

2.9

9.6

6.

3.6

9.7

7.

4.3

9.8

8.

4.8

9.9

9.

5.4

10.0

10.

5.8

10.1

Š‹•ŽDZȱ ŜȲ •¢Œ’—ŽȮ˜’ž–ȱ‘¢›˜¡’Žȱ‹žŽ›ȱǻŘŖǚǼ Range 8.6–10.6 A: 0.2 M glycine solution B: 0.2 M sodium hydroxide solution 6.25 mL of A + x mL of B, diluted to a total of 25 mL with distilled water. S. No.

x mL

pH

1.

0.50

8.6

2.

0.75

8.8

3.

1.10

9.0

4.

1.50

9.2

5.

2.10

9.4

6.

2.80

9.6

7.

3.40

9.8

8.

4.80

10.4

9.

5.70

10.6

™™Ž—’ŒŽœȱȳ261

Š‹•ŽDZȱ ŝȲ˜›’ŒȱŠŒ’Ȯ˜’ž–ȱ‘¢›˜¡’Žȱ‹žŽ›ȱǻŘŖǚǼ Range 7.94–9.46 A: Boric acid = 0.2 M B. Sodium hydroxide solution = 0.2 M S. No.

A mL

B mL

pH

1.

9.0

1.0

7.94

2.

8.5

1.5

8.22

3.

8.0

2.0

8.55

4.

7.5

2.5

8.74

5.

7.0

3.0

8.99

6.

6.5

3.5

9.12

7.

6.0

4.0

9.46

BIBLIOGRAPHY

ȱ

ȱ ȱ

ȱ

ȱ

ȱ

ȱ

ŗǯȱ ǯǯȱ ‘˜œ•Šǰȱǯǯȱ Š›ǰȱǭȱŠ›œ‘ȱ ž•Š’ǰȱŽ—’˜›ȱ›ŠŒ’ŒŠ•ȱ‘¢œ’ŒŠ•ȱ‘Ž–’œ›¢ǰȱ R. Chand & Co., 2017. 2. B. Viswanathan, & P.S. Raghavan, Practical Physical Chemistry, Viva Books Private Limited, New Delhi, 2014. řǯȱ ǯǯȱ ›Žž•’ǰȱ ȃ Š—‹˜˜”ȱ ˜ȱ —Š•¢’ŒŠ•ȱ ‘Ž–’œ›¢ǰȄȱ ǯȱ Ž’Žœǰȱ ǯǰȱ Œ ›Š Ȭ Hill, NY, 1963. Śǯȱ ǯǯȱ Š›•Š—ǰȱ ǯȱ ’‹•Ž›ǰȱ ǭȱ ǯǯȱ ‘˜Ž–Š”Ž›ǰȱ ¡™Ž›’–Ž—œȱ ’—ȱ ‘¢œ’ŒŠ•ȱ ‘Ž–’œ›¢ǰȱ’‘‘ȱŽ’’˜—ǰȱŒ ›Š Ȭ ’••ǰȱŽ ȱ˜›”ǰȱŘŖŖřǯ 5. D.A. Skoog, D.M. West, F.J. Holler, & S.R. Crouch, Fundamental of Analytical Chemistry, Cengage Learning , 2014. Ŝǯȱ ǯ ȱ ŽŽ›¢ǰȱ ǯȱ ŠœœŽǰȱ ǯȱ Ž—‘Š–ǰȱ ǭȱ ǯǯȱ Ž——Ž¢ǰȱ ˜Ž•Ȃœȱ Ž¡‹˜˜”ȱ ˜ȱ Quantitative Chemical Analysis, Seventh edition, ELBS with Longman, 2000. 7. http://www.chennaicorporation.gov.in/departments/health/adulteration.htm 8. M.L. Jackson, Soil Chemical Analysis, Prentice-Hall of India Private Limited, 1973. 9. P.C. Jain, & M. Jain, Engineering Chemistry, Dhanpat Rai and Sons, Delhi, 2015. 10. P. Yates, Chemical Calculation. Second edition, CRC Press, 2007. 11. R, Cremlyn, Pesticides: Preparation and Mode of Action, John Wiley, 1980. ŗŘǯȱ ǯȱ ˜™Š•Š—ǰȱ–’›‘Šȱ—Š—ǰȱǭȱǯȱ’•›Žȱžž–Š›ǰȱȱŠ‹˜›Š˜›¢ȱŠ—žŠ•ȱ˜›ȱ Environmental Chemistry, I. K. International Publishing House Pvt Ltd, New Delhi, 2009. ŗřǯȱ ǯȱ ˜™Š•Š—ǰȱ ǯȱ Ž—”Š™™Š¢¢Šǰȱ ǭȱ ǯȱ ŠŠ›Š“Š—ǰȱ —’—ŽŽ›’—ȱ ‘Ž–’œ›¢ǰȱ ’”Šœȱ Publications, 2004. 14. Ramesh Kumari, Computer and their Application to Chemistry, Narosa Publishing House, 2002, 2nd revised and Enlarged Edition, 2005. ŗśǯȱ  ȱ‘•ž Š•’Šǰȱž—’Šȱ‘’—›ŠǰȱǭȱŠ›œ‘ȱ ž•Š’ǰȱ˜••ŽŽȱ›ŠŒ’ŒŠ•ȱ‘Ž–’œ›¢ǰȱ University Press (India) Limited, 2005.

PHYSICAL
CHEMISTRY
LABORATORY
MANUAL

TM

An
Interdisciplinary
Approach

Salient Features • Interdisciplinary and innovative methods. • For undergraduate students of science, engineering and environmental sciences. • Useful experiments, beyond the syllabus, are also given under 'Miscellaneous Experiments'. • Wherever possible, two or more methods are given. • Brief account of the principles required for each experiment is provided. Amirtha Anand is Associate Professor of Chemistry in Maitreyi College, University of Delhi. She completed her PhD degree from the University of Madras. She has 34 years of experience in teaching undergraduate students of Honours and General courses. She has published three books and many papers in both national and international journals of high repute. She has been conferred the Best Teacher Award by the Government of Delhi in 2012. Currently, she is associated with CIET, NCERT for developing the e-contents for Massive Open Online Courses (MOOCs) in Chemistry.

978-93-89583-10-6

PHYSICAL
CHEMISTRY
LABORATORY
MANUAL
An
Interdisciplinary
Approach Amirtha Anand Ramesh Kumari

Amirtha Anand • ramesh Kumari

Ramesh Kumari did her graduation in Chemistry (Honours) from University of Delhi (1987) , Masters in Chemistry (M.Sc.) from IIT Delhi (1989) and PhD in Physical Chemistry from IIT Delhi in 1994. She is Associate Professor in the Department of Chemistry, Maitreyi College, University of Delhi. She has over 24 years of experience in teaching undergraduate courses in University of Delhi. She has also been the Deputy Coordinator for PG Diploma in Nanotechnology under the innovative programme of the UGC and Academic Coordinator for Undergraduate Programmes of IGNOU at Maitreyi College. She has authored 2 books for undergraduate courses and has been the content writer for study modules for e-PG Pathshala.

PHYSICAL
CHEMISTRY
LABORATORY
MANUAL


This book covers the latest syllabus of CBCS pattern of Delhi and other universities for both B.Sc. Programme and Honours courses. A large number of Physical Chemistry, Environmental Chemistry, Nanoscience, Polymer Chemistry and Analytical Chemistry experiments have been covered using interdisciplinary and innovative methods. The contents include some fundamental chemical concepts, measurement of surface tension and viscosity, colorimetry, determination of order of a reaction, hetrogeneous equilibria, adsorption on solid surfaces, thermochemical measurements, conductometric and potentiometric measurements, pH metry, environmental parameter analysis, etc. Wherever possible, two or more methods are given. So, the teachers and students will have a choice to make depending on the availability of chemicals, apparatus, instruments, time, etc. This book will give them the opportunity to relate theory and practicals for a better understanding of the subject.

Distributed by:

9 789389 583106 TM