Inorganic Chemistry (for the specialty «Biotechnology»): educational-methodical handbook 9786010422223

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

O. I. Ponomarenko I. V. Matveyeva Sh. N. Nazarkulova

INORGANIC CHEMISTRY (FOR THE SPECIALTY «BIOTECHNOLOGY») Educational-methodical handbook

Almaty «Qazaq university» 2017

UDC 546 (075) Р 82 It is recommended for publication by the Academic Council of the Faculty of Chemistry and Chemical Technology and RISO of al-Farabi Kazakh National University (protocol #2 dated to 29.12.2016)

Reviewer: PhD Yarovaya Yelena

Р 82

Ponomarenko O.I. Inorganic Chemistry (for the specialty «Biotechnology»): educational-methodical handbook / O.I. Ponomarenko, I.V. Matveyeva, Sh.N. Nazarkulova. – Almaty: Qazaq university, 2017. – 162 p. ISBN 978-601-04-2222-3 The educational-methodical handbook contains a brief theoretical material, exercises, tasks for implementation of the ISW and laboratory work on the course «Inorganic Chemistry». The educational-methodical handbook can be recommended for preparation for classes and independent work of students enrolled in the specialty «Biotechnology» and will also be useful to teachers, students, undergraduates, PhD students of higher educational institutions. Published in authorial release. Учебно-методическое пособие содержит краткий теоретический материал, упражнения, задания для выполнения СРС и лабораторных работ по курсу «Неорганическая химия». Учебно-методическое пособие может быть рекомендовано для подготовки к занятиям и самостоятельной работы студентов, обучающихся по специальности «Биотехнология», а также будет полезной преподавателям, студентам, магистрантам, PhD докторантам высших учебных заведений. Издается в авторской редакции.

UDC 546 (075) ISBN 978-601-04-2222-3

 Ponomarenko O.I., Matveyeva I.V., Nazarkulova Sh.N., 2017  Аl-Farabi KazNU, 2017

INTRODUCTION The course «Inorganic chemistry» is one of the fundamental disciplines of the chemical cycle and the necessary basis for successful study of both chemical and biological disciplines. Chemistry has a central place among natural sciences. Being a fundamental science, closely associated with any branch of human activity, chemistry is an essential part of human culture. The educational-methodical handbook is intended for students of specialty «5B070100 – Biotechnology» of higher education. It should be noted that the guide discusses the most important sections of «Inorganic Chemistry». The leading ideas of educationalmethodical handbook are:  material unity of substances in nature, their genetic relationship;  causal relationships between composition, structure, properties and application of substances;  understanding of chemical compound as a link in unbroken chain of transformations of substances, participation of substances in circulation of chemical elements in chemical evolution;  interconnectedness of science and practice: practice is the driving force of development of science, and successes of practice are results of science;  humanistic nature of chemical science, focus on solving of global problems of modernity. The educational-methodical handbook is designed for a broad mastering of the main laws of relationship between structure and chemical properties of matter, chemical reactions, structure of chemical compounds and their biological activity. It makes it possible to learn to predict the conversion of inorganic compounds based on the laws of chemistry and typical properties and reactions of these compounds. 3

In each theme included in this handbook the theoretical foundations necessary for further implementation of independent work are briefly discussed. Moreover, for the independent control of studying material questions and exercises for self-training, tasks to perform independent work and methodology of laboratory work are included. Handbook begins with an introductory section, which are the basic rules of work in chemical laboratories, the students become familiar with the most common laboratory equipment and chemical glassware. The annex includes the necessary minimum of reference materials. This educational-methodical handbook is designed to help in independent work of students during study of the most difficult sections of discipline «Inorganic chemistry».

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STANDARD INSTRUCTION FOR SAFETY WORK IN A CHEMICAL LABORATORY 1. While working in a chemical laboratory exercise caution, keep order and cleanliness in the workplace, follow the safety instructions. Confusion and haste in violation of safety regulations can lead to accidents. 2. Before starting work: а) wear clothes that provide maximum protection and cover most of the skin; you must wear closed-toe shoes. Clothes should be made of natural materials, such as cotton, that do not catch fire as easily as synthetic materials; b) firmly understand the order and security rules of the experiment; c) verify the availability and reliability utensils, appliances and other items needed for the job; d) free up the workplace from unnecessary for the work objects and materials. 3. Students are not permitted in the chemistry lab without the presence of the instructor. No student may work in the lab without the instructor’s supervision. 4. Only the authorized scheduled experiment can be performed in the lab. No unauthorized experiment is allowed. You are not allowed to alter the procedure of the lab experiment. Carefully follow all instructions. 5. Absolutely no noise or disruptive behavior in the lab. No fooling around. Do not run or walk quickly through the lab. Before you back up, look behind you to make sure no student is behind you that you might bump into. 6. Do not waste chemicals; do not take more than what is required. Chemicals used in the laboratory are costly. Wasted chemicals are not only costly but also bad for our environment. 7. Carefully check the label on the bottle before using its content. Make sure it is the correct chemical and correct concentration. 5

8. All chemicals in the laboratory are to be considered dangerous. Avoid handling chemicals with fingers. Always use a tweezer, spatula or spoon. Do not taste, or smell any chemicals. 9. Never return unused chemicals to their original containers. If you do so, you will be contaminating the chemical. Dispose of the leftover chemical in the proper «waste container». Check with your instructor if you are unsure of what to do with your leftovers. 10. Do not touch, eat, or smell any chemical unless instructed to do so. When instructed to smell a chemical, you need to fan the air above the chemical toward your nose. Do not sniff the chemical directly by bringing it close to your nose. If you do so, the odor may badly irritate your nasal passage. 11. Do not touch your face, eyes, or mouth while in laboratory. If you must do so, first wash your hands. 12. All work associated with the release of vapors or gases carry out in a fume hood with properly functioning ventilation. 13. Never look into a container that is being heated. 14. Heated metals, glassware and ceramics stay hot for a long time. Allow plenty of time for a hot metal to cool before touching it. Since you cannot tell from the appearance of the metal, glass, or ceramic that it is still hot, you should test it by cautiously bringing the back of your hand close to the metal to feel if heat is radiating from it. 15. Handle hot objects like a beaker, evaporating dish, and crucible with the proper pair of tongs. Use the beaker tongs, evaporating dish tongs, and crucible tongs, respectively. 16. Do not heat a closed container. Pressure build up may cause the container to explode. 17. During the heating the liquid, do not leave it unattended, even for a short time. 18. Handle corrosive chemicals with extreme care. When diluting a concentrated acid, you must always add the acid slowly to the water while stirring to avoid splattering and releasing the heat all at once. In other words, ADD ACID. Never do the reverse, the result could be quite hazardous. 19. Never pipette by mouth. 20. In all operations to concentrated acids and alkalis be sure to use rubber gloves and protective goggles. 21. Dissolving of alkali produce in a porcelain dish by adding to the water small portions of material with continuous stirring. Pieces of alkali take only with tweezers or forceps. 6

22. Clean up all small spills immediately. If a spill is large and is expected to poses a hazard to others in the laboratory, stop the activity and tell for instructor. 23. Work with organic solvents perform in a fume hood. 24. Always work in a well-ventilated area. 25. Keep flammable liquids away from heat sources and open flames in the fume hood. If a flammable liquid is used in the laboratory, do not use an open flame at all. 26. Before using a fume hood: а) keep the sash at 5-10 cm or less from the working surface while working in the hood to ensure maximum flow rate and to protect yourself from potential chemical splashes or explosions. The sash should be closed when you are not working in the hood. b) Make sure that the sash is open to the proper operating level, which is usually indicated by arrows on the frame. c) Make sure that the air gauge indicates that the air flow is within the required range. 27. Keep your work area neat and clean. During the lab period, store your books and bags under your bench. Keep the area around your chair clear. Do not store belongings on top of the lab bench. Large bookbags belong under the lab bench, smaller purses can be stored in a drawer, each lab bench has pullout writing spaces for notebooks. 28. Keep shared areas of the laboratory clean. This includes areas such as the balance room and the supply bench where the stock bottles are kept. It is especially important to keep the balance clean and free of chemical spills. 29. Eating, drinking, and chewing gum are not allowed in the lab. No food or drink is allowed in the lab to avoid possible contamination. Chewing gum may absorb chemicals from the laboratory. 30. Wash your hands as often as possible, especially before leaving the lab. Be careful not to touch your eyes or other body areas without thoroughly washing your hands first. 31. Be alert. Notify the instructor immediately if you notice an unexpected chemical reaction during your experiment or any unsafe condition. Report all accidents (chemical spill, broken glass, etc.) or injuries (burn, cut, chemical splash etc.) no matter how minor to the instructor immediately. 7

LABORATORY GLASSWARE Laboratory glassware (Fig. 1) refers to a variety of equipment, traditionally made of glass, used for scientific experiments and other work in science, especially in chemistry laboratory. There are several types of glass, each used for different purposes. Borosilicate glass, which is commonly used in reagent bottles, can withstand thermal stress. Quartz glass, which is common in cuvettes, can withstand high temperatures and is transparent in certain parts of the electromagnetic spectrum. Darkened brown or amber (actinic) glass, which is common in dark storage bottles, can block ultraviolet and infrared radiation. Heavy-wall glass, which is common in glass pressure reactors, can withstand pressurized applications. Test tubes (Fig. 1, 1-2) have clear glass to allow monitoring and observation during an experiment. Test tubes are used by chemists in the lab to heat, hold and mix small quantities of liquid or solid chemicals during experiments. Test tubes are available in a multitude of lengths and widths, typically from 10 to 20 mm wide and 50 to 200 mm long. A test tube has either a flat bottom, a round bottom, or a conical bottom. Some test tubes are made to accept a ground glass stopper or a screw cap. A test tube with a stopper is often used for temporary storage of chemical samples. During the work the volume of reagent in the test tube should not exceed half the tube volume. Upon heating place the test tube in the test tube rack. Holding the test tube clamp at the back end, place the bottom of the test tube into the Bunsen burner flame. The test tube should be pointed away from all people at all times. It should be held at a 45° angle, and should be continuously moved, to prevent any one portion from being heated too strongly. Beakers (Fig. 1, 3) are glass cylinders used for mixing and heating chemicals. They are usually made of a type of glass that can withstand high temperatures. The diameter of a beaker is the same from top to bottom. 8

9 Figure 1. Chemical glassware

While many beakers do have markings on the side indicating volume, these markings (or graduations) are only approximate values; they should not be used for measuring exact volumes. Most also have a small spout (or «beak») to aid pouring. Heating should be carried out only through the grid of asbestos or in a water bath. Glass beakers should not be used for evaporation of solutions. Weighing bottles is bakers with grinded caps used for precise weighing of solids. Erlenmeyer flasks (Fig. 1, 4) are glass conical containers used for mixing, heating, and storing chemicals. They are usually made of a type of glass that can withstand high temperatures. The flask is shaped like a cone with the opening at the top of the flask being the narrow end of the cone. While many Erlenmeyer flasks have markings on the side indicating volume, these graduations are only approximate values; they should not be used for measuring exact volumes. The Erlenmeyer flask are useful in titration experiments, where the shape of the flask enables the contents to be swirled with a minimum of overflow. Erlenmeyer flasks are also useful in the preparation of supersaturated solutions; the conical shape slows the rate at which the contents cool, and helps to minimize premature crystallization of the solute particles. Round-bottom flask is a flask with a spherical body and one or more necks with ground glass joints (Fig. 1, 6-8). Graduated cylinders (Fig. 1, 27) they are useful for measuring liquid volumes to within about 1%. They are for general purpose use, but not for quantitative analysis. If greater accuracy is needed, use a pipette or volumetric flask. 2. Burettes is used to deliver solution in precisely-measured, variable volumes (Fig. 1, 30). Burettes are used primarily for titration, to deliver one reactant until the precise end point of the reaction is reached. 3. Volumetric flasks is used to make up a solution of fixed volume very accurately (Fig. 1, 5). To make up a solution in volumetric flask, first dissolve the solid material completely, in less water (or solvent) than required to fill the flask to the mark. After the solid is completely dissolved, very carefully fill the flask to the mark. Move your eye to the level of the mark on the neck of the flask and line it up so that the circle around the neck looks like a line, not an ellipse. 10

Then add distilled water a drop at a time until the bottom of the meniscus lines up exactly with the mark on the neck of the flask. Take care that no drops of liquid are in the neck of the flask above the mark. Finally, cover the flask by its cover and remember to mix your solution thoroughly, by inverting the flask and shaking. A pipette (Fig. 1, 28) is used to measure small amounts of solution very accurately. Two types of pipettes commonly used are transfer pipettes and measuring pipettes. a. Transfer pipettes include volumetric pipettes and are designed to deliver accurately a fixed volume of liquid such as standard solutions and non-viscous samples, and consist of a cylindrical bulb joined at both ends to narrowed glass tubing. These pipettes are allowed to drain by gravity. b. The second type of pipette is the graduated or measuring pipette. Two kinds are available: The Mohr pipette and the serological pipette. Mohr pipette is calibrated between two marks on the stem, and the other has graduation marks down to the tip. Some types of serological pipettes are marked of blow-out; these types need to blowout the remaining drops of liquid in the tip. Mohr pipette deliver between their calibration marks. Measuring pipettes are principally used for the transfer of reagents and are not generally considered accurate enough to pipette samples and standards. Volumetric pipettes are more accurate than measuring pipettes. Aqueous solutions form a curved surface when placed in a container. This surface is called a meniscus. To accurately read the level of liquid in a piece of glassware, it is important that your eye be on the same level as the surface of the liquid in order to avoid parallax errors. A Büchner (Fig. 1, 30), or sidearm flask is essentially an Erlenmeyer flask with an additional small tube extending from the side of the neck. The bottom is conical in shape, with a short neck from which the small tube extrudes. The entire flask is made of thick glass. The small side-arm tube is composed of a hose barb, which is a serrated section that catches a thick-walled flexible hose. Because of this design, a Büchner flask is well-equipped for creating vacuums with the help of a Büchner funnel. With the funnel on top of the glass neck and a tube sucking the pressure out of the side, vacuums for sluicing liquids can be created very easily within the flask. 11

Laboratory funnels (Fig. 1, 14-17) are funnels that have been made for use in the chemical laboratory. There are many different kinds of funnels that have been adapted for these specialized applications. Filter funnels, thistle funnels (shaped like thistle flowers), and dropping funnels have stopcocks which allow the fluids to be added to a flask slowly. For solids, a powder funnel with a wide and short stem is more appropriate as it does not clog easily. When used with filter paper, filter funnels, Buchner and Hirsch funnels can be used to remove fine particles from a liquid in a process called filtration. For more demanding applications, the filter paper in the latter two may be replaced with a sintered glass frit. A condenser is an apparatus or item of equipment used to condense (change the physical state of a substance from its gaseous to its liquid state). Liebig Condensers (Fig. 1, 18) are the most common condensers and the cheapest. They consist of an inner tube surrounded by an outer tube. The coolant flows through the outer tube and the substance gets condensed in the inner tube. It is only effective with substances that have boiling points of around 100 ºC or higher. It can be used in distillations as well as refluxes Graham condenser (also Grahams or Inland Revenue condenser) (Fig. 1, 19) has a coolant-jacketed spiral coil running the length of the condenser serving as the vapor/condensate path. This is not to be confused with the «coil condenser». The coiled condenser tubes inside will provide more surface area for cooling and for this reason it is most favorable to use but the drawback of this condenser is that as the vapors get condensed, it tends to move them up in the tube to evaporate which will also lead to the flooding of solution mixture. Friedrichs condenser (sometimes incorrectly referred to as «Friedrich's» condenser) (Fig. 1, 20), a spiraled finger condenser, was invented by Fritz Walter Paul Friedrichs. It consists of a large, spiraled internal cold finger-type capillary tube disposed within a wide cylindrical housing. Coolant flows through the internal cold finger; accordingly, vapors rising up through the housing must pass along the spiraled path. Desiccators (Fig. 2) are sealable enclosures containing desiccants used for preserving moisture-sensitive items such as cobalt chloride paper for another use. 12

Figure 2. Desiccators

A common use for desiccators is to protect chemicals which are hygroscopic or which react with water from humidity. The desiccator is used to store dried samples in a dry atmosphere. It should not be used to dry an object, but to maintain an already dried object indefinitely in a dry condition.

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LESSON #

1

GENETIC RELATIONSHIP BETWEEN THE MAIN CLASSES OF INORGANIC COMPOUNDS

OBJECTIVE: To summarize the main classes of inorganic compounds: oxides, acids and salts. Consider their classification, chemical properties and show the relations between classes of inorganic compounds.

HIGHLIGHTS: 1. Oxides: basic, acidic, amphoteric. 2. Chemical properties of the oxides. 3. Bases: soluble (alkali), insoluble and amphoteric. 4. Chemical properties of bases. 5. Formulas and names of acids and salts. 6. Classification of acids: by the basicity and by the content of oxygen. 7. Chemical properties of acids. 8. Salts: medium, acidic, basic, double, mixed, complex. 9. Chemical properties of salts. 10. Acid-base interactions – neutralization. 11. Classification of inorganic hydroxides. 12. Compliance of hydroxides to oxides. 13. Genetic series of inorganic compounds. THEORETICAL PART. Scheme of classification of substances: 14

Metals Simple Nonmetals Basic Oxides Acidic

Substances

Oxygenated Acids Anoxic Alkalies

Complex Bases

Insoluble Amphoteric* Neutral*

Salts

Acidic* Basic*

There is an important relationship between classes, called genetic («geneziz» in Greek means «origin»). This relationship is that from one class of substances can be obtained substances of other classes. There are two main ways of genetic relationships between substances-metals and non-metals. For example, calcium sulfate (CaSO4) can be obtained either from calcium metal, or the other way – from sulfur non-metal:

On the other hand, the salts can again decompose into metal and non-metal:

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At the same time, there are other ways of interconversion of compounds of different classes. Thus, the genetic relationships between different classes of compounds are very diverse. The acid-base interactions are neutralization reactions. Medium salt Acidic salt Basic salt

Acid + Base = Water + Salt

I. The medium salt can be prepared by the interaction of any acid and any base during the neutralization reaction. II. The acid salt can be obtained only at partial neutralization of a polybasic acid. III. The basic salt may be prepared by partial neutralization of many acidic base. Examples: I. NaOH + HCl = NaCl + H 2O Base Acid Medium salt Mg(OH)2 + H 2SO4 = MgSO4 + H 2O

Base

Acid

Medium sult

The name of salts: NaCl – sodium chloride MgSO4 – magnesium sulfate II. H3PO 4 + 1KOH = KH 2 PO4 + 1H 2O Polybasic Monoacidic Acidic salt acid base H3PO 4 + 2KOH = K 2 HPO4 + 2H 2O Polybasic Monoacidic Acidic salt acid base H3PO4 + 3KOH = K 3PO4 + 3H 2O Polybasic Monoacidic Medium salt acid base 16

The name of salts: KH2PO4 – potassium dihydrophosphate K2HPO4 – potassium hydrogenphosphate K3PO4 – potassium phosphate III. Fe(OH)3 + 1HCl = Fe(OH) 2Cl + 1H 2O Polyacidic Monobasic Basic salt base acid Fe  OH 3 + 2HCl = Fe  OH  Cl2 + 2H 2O Polyacidic Monobasic Basic salt base acid Fe(OH)3 + 3HCl = FeCl3 + 3H 2O Polyacidic Monobasic Medium salt base acid The name of salts: Fe(OH)2Cl – dihydroxoiron (III) chloride Fe(OH)Cl2 – hydroxoiron (III) chloride FeCl3 – iron (III) chloride CLASSIFICATION OF INORGANIC HYDROXIDES For each oxide corresponds its own hydroxide. For the acid oxide correspond acids hydroxide – oxoacid; for the basic oxide corresponds basic hydroxides – base, and for amphoteric oxide corresponds – amphoteric hydroxides. The oxidation state of an element in the oxide, and its corresponding hydroxide are the same. For example: Na2+1O ---- Na+1OH Al2+3O3 ----- Al+3(OH)3 P+52O5 ---- H3P+5O4 Mg+2O ---- Mg+2(OH)2 Zn+2O ----- Zn+2(OH)2 S+6O3 ---- H2S+6O4 Inorganic hydroxides are divided into three classes: Basic hydroxides are bases, which react with acid and acidic oxides, but do not react with bases and basic oxides. 17

For example, Mg(OH)2 – base Mg(OH)2 + H2SO4 = MgSO4 + 2H2O Mg(OH)2 + SO3 = MgSO4 + H2O Mg(OH)2 + NaOH = the reaction is impossible Mg(OH)2 + Na2O = the reaction is impossible  Acid hydroxides are oxoacids (oxygenates acids), which react with base and basic oxides, but do not react with acids and acidic oxides. For example, H2SO4 – oxyacids H2SO4 + 2NaOH = Na2SO4 + 2H2O H2SO4 + Na2O = Na2SO4 + H2O H2SO4 + HNO3 = the reaction is impossible H2SO4 + N2O5 = the reaction is impossible  Amphoteric hydroxides react with strong acids and acidic oxides of strong acids and with strong bases and their oxides. This is amphoteric phenomenon. For example, Zn(OH)2 = H2ZnO2 is amphoteric hydroxide Zn(OH)2 + H2SO4 = ZnSO4 + 2H2O Zn(OH)2 + SO3 = ZnSO4 + H2O Zn(OH)2 + 2NaOH = Na2ZnO2 + 2H2O Zn(OH)2 + Na2O = Na2ZnO2 + H2O Acid-basic nature of the oxides and hydroxides oxides and hydroxides Acidic Basic Amphoteric

The interaction with oxide or hydroxide With acid and acidic oxide With the base and a basic oxide + + + +

QUESTIONS AND EXERCISES FOR SELF-CHECKING: 1. Determine the class (or classes) of compounds listed below: Na2O, CO2, CuO, SO3, Fe2O3. Which of them will react with concentrated sulfuric acid? Write the reactions. 2. If alkali (NaOH, KOH) stored in not tightly closed containers, gradually changes due to the reaction with carbon dioxide CO2. Write the reaction equation.

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3. The structural formulas of five substances are listed below. Which classes of compounds are these substances? Write reaction between substances (a) and (d); between substances (a) and (b); between substances (a) and (e). a) H–O–Na b) H–O–Cl c) O=Ti=O d) O=S=O e) H–O Se=O H–O 4. Write the equation of receiving oxides by heating of the following substances: а) Ca(OH)2, b) H3PO4, c) Ca3(PO4)2. 5. Draw the structural formula of aluminum sulfate Al2(SO4)3. 6. As a result, the reaction between the solutions of two salts of silver chloride precipitate formed, and potassium nitrate remained in solution. Between which salts were there a reaction? 7. From the two following reactions, only one can proceed. Which one? What is produced? Ni + MgCl2 = Mg + NiCl2 = 8. From the two following reactions, only one can proceed. Which one? Write the equation of reaction. FeCl3 + NaOH = NaCl + KOH = 9. How do you carry out the following conversion: Ca → Ca(OH)2 →CaCO3 →CaO → CaSO4 →CaCl2 → Ca? 10. How do you carry out the following conversion: Phosphorus → phosphorus (V) chloride → hydrogen chloride → zinc chloride → silver chloride? 11. Specify the composition of the final product, call them and arrange the coefficients in the following equations: H2S + Pb(NO3)2 = Cu + HNO3 (conc.) = CuSO4 + Zn = Na2CO3 + HCl = TASKS FOR PREPARATION OF ISW «The genetic relationship between the main classes of inorganic compounds» 1. Specify the methods for preparing all substances specified in your version. 2. For specified substances list their chemical properties, write equations of reactions and indicate the name of the reaction products. 3. Derive the chemical formula of oxide corresponding to acid and base.

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#

Oxide

Acid

Base

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41

Na2O CuO ZnO Li2O CO2 N2O5 K2O MgO SO3 P2O5 Al2O3 Cr2O3 Cl2O7 Fe2O3 BaO Rb2O CaO SO2 Br2O7 Fe2O3 FeO Cr2O3 B2O3 P2O3 SrO CO2 N2O5 K2O MgO P2O5 Al2O3 Cr2O3 Cl2O7 Rb2O CaO SO2 Na2O CuO ZnO Li2O CO2

HCl HNO3 HClO4 HClO4 H2SO4 HBr HI HClO2 HClO HClO3 H2Cr2O7 H2SnO3 H2PbO3 H2CO3 HCl H2S H2SO4 HClO4 HNO3 H2CO3 HBr HI HNO3 H2SiO3 HNO2 HCl HNO3 HClO4 HClO4 H2CO3 HCl H2S H2SO4 HNO3 H2SO3 HI H2SO4 HClO4 HNO3 H2CO3 HBr

Mg(OH)2 KOH Ni(OH)2 Fe(OH)2 Fe(OH)3 Mn(OH)2 Ba(OH)2 LiOH NaOH Bi(OH)3 KOH NaOH KOH Ca(OH)2 Ba(OH)2 KOH NaOH Cd(OH)2 RbOH CsOH NaOH Mg(OH)2 Mn(OH)2 KOH Ba(OH)2 LiOH NaOH Bi(OH)3 KOH Cu(OH)2 Ba(OH)2 Co(OH)2 Mg(OH)2 KOH Ni(OH)2 Fe(OH)2 Ca(OH)2 Ba(OH)2 KOH NaOH Cd(OH)2

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Amphoteric hydroxide Zn(OH)2 Cr(OH)3 Al(OH)3 Be(OH)2 Cr(OH)3 Pb(OH)2 Sb(OH)3 Sn(OH)2 Zn(OH)2 Cr(OH)3 Pb(OH)2 Zn(OH)2 Cr(OH)3 Zn(OH)2 As(OH)3 Sn(OH)2 Sb(OH)3 Al(OH)3 Pb(OH)2 Al(OH)3 Cr(OH)3 Sb(OH)3 As(OH)3 Zn(OH)2 Pb(OH)2 Cr(OH)3 Al(OH)3 Be(OH)2 Pb(OH)2 Zn(OH)2 Cr(OH)3 Al(OH)3 Be(OH)2 Cr(OH)3 Sb(OH)3 As(OH)3 Cr(OH)3 Pb(OH)2 Sb(OH)3 Sn(OH)2 Zn(OH)2

Salt KMnO4 K2SO3 KNO3 FeSO4 CrCl3 MnSO4 FeCl3 Cu(NO3)2 KClO BiCl3 Pb(NO3)2 ZnSO4 K3PO4 ZnCl2 Na2CO3 NaCl K2S KClO4 Pb(NO3)2 Cu(NO3)2 Fe(NO3)3 K2SO3 KNO3 KNO2 Mn(NO3)2 FeSO4 CrCl3 MnSO4 FeCl3 Cu(NO3)2 KClO K3PO4 ZnCl2 Na2CO3 NaCl K2S KMnO4 K2SO3 KNO3 FeSO4

LABORATORY WORK Classes of inorganic compounds. Preparation and study of properties of hydroxides and salts Experiment #1. The preparation and study of properties of hydroxides. 1.1. In three tubes each pour separately 2ml solution of the copper (II), iron (III), nickel (II) salt and then pour into each tube an equal volume of alkali. Mark character (crystalline or amorphous) and color of precipitate. 1.2. With precipitation of hydroxides obtained in the experiment 1.1., heat with the liquid and heat to boil. Heating must be done on a small flame with frequent shaking. Mark the color change. Save the precipitates. Experiment #2. Preparation of medium salts. To the precipitate obtained in experiment 1.2., while shaking slowly, add nitric acid till precipitate dissolves completely. Write reaction equations. In the test tube add 2 ml barium chloride solution and add an equal amount of sulfuric acid. Explain precipitation and mark its color. Write equation of the reaction. Name initial and resulting products. In the test tube with 2 ml barium chloride solution, add 2 ml potassium bichromate. Explain precipitation and mark its color. Write the equation of the reaction. In the test tube with 2 ml silver nitrate solution, add 2 ml of hydrochloric acid. Explain the precipitation and mark its color. Write the equation of the reaction. Experiment #3. Preparation of the acid salts. 3.1. Pour ½ volume of lime water in test tube, and pass through it carbon dioxide produced in the Kipp apparatus. Observe a white precipitate and then flowing carbon dioxide until precipitates completely dissolve. Explain the observed. Write equations of the reactions. 3.2. Obtained in the experiment 3.1. solution, divide into 3 test tubes: In the first test tube heat to boil. In the second test tube add lime water. In the third one add sodium carbonate solution. Explain appearance of precipitates in all three test tubes. Write equations of the reactions. Experiment #4. Preparation of basic salts. Pour in two test tubes 2 mL of copper (II) sulfate solution. In the first test tube, pour excess of potassium hydroxide and observe amorphous precipitate. In the second test tube, pour potassium hydroxide by dropwise with stirring until a pale green precipitate is formed. Test tubes with precipitates cautiously heated to boil. Explain why in the first test tube precipitate turns black and in the second – the color does not change. Write equations of the reactions. Name the initial substances and reaction products.

Experiment #5. Preparation and study of properties of amphoteric hydroxides

5.1. Pour in a test tube, 2 ml aluminum chloride or aluminum sulphate solution and while stirring, pour potassium hydroxide solution until a precipitate is formed.

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The resulting precipitate with solution is poured into two test tubes. In one test tube, add dropwise a solution of potassium hydroxide and in the other – hydrochloric acid to completely dissolve the precipitate. Write equation of the the reaction. Name the initial substances and reaction products. 5.2. Pour in a test tube 2 ml of zinc sulfate solution and while stirring, pour potassium hydroxide solution until a precipitate is formed. The resulting precipitate with the solution is poured into two test tubes. To the first test tube, add dropwise of solution of potassium hydroxide, and the other – hydrochloric acid to completely dissolve the precipitate. Write equation of the reaction. Name the initial substances and reaction products. 5.3. Pour in a test tube 2 ml solution of chromium (III) chloride, and while stirring, pour solution of potassium hydroxide until a precipitate is formed. The resulting precipitate with solution is poured into two test tubes. To the first test tube add dropwise of solution of potassium hydroxide, and to the second – hydrochloric acid to completely dissolve the precipitate. Write equations of the reactions. Name the initial substances and reaction products.

22

LESSON #

2

THE BASIC CONCEPTS AND LAWS OF CHEMISTRY

OBJECTIVE: Consideration of the application of the law of equivalents, and some of the basic gas laws for laboratory practice to determining the equivalent and molecular mass of substances.

HIGHLIGHTS: 1. Atomic-molecular doctrine. 2. Atom, molecule, element, simple substance. 3. The law of conservation of mass. 4. The law of definite proportions. The relative nature of the law. 5. The law of multiple proportions. 6. The law of Avogadro. The consequences of Avogadro's law. 7. The relative atomic mass, relative molecular mass, molar mass, amount of substance. 8. The molar volume of gases. 9. The relative and absolute densities. 10. Equivalent and the equivalent weight of the element. 11. Equivalent and the equivalent weight of the substance. 12. The law of equivalents. THEORETICAL PART. The basic laws of chemistry include: the law of conservation of mass of substance and energy, the law of definite proportions, law of multiple proportions, Avogadro's law, the law of equivalents. 23

Applications of some gas laws to the determination of the molar mass of the substance The relative molecular mass is one of the main characteristics of a substance. Relative molecular mass of a substance (M) is a quantity equal to the ratio of the average mass of the molecule of the natural isotopic composition of the substance to 1/12 of the mass of an atom of carbon (C-12). As the atomic mass, relative molecular masses are expressed in atomic mass units (amu), they are numerically equal to the sum of the relative atomic masses of all atoms that make up molecules of a substance. To characterize the quantity of a substance, we use a special unit called mole. Mole is amount of substance that contains as many structural units (atoms, molecules, ions) as there are atoms in 0.012 kg of carbon C-12. Molar mass (M) is the value equal to relative mass of substance to the amount of the substance. The molar mass has the dimension kg/mol or g/mol. In general, the molar mass of a substance, is expressed in g/mol which is numerically equal to the relative atomic or relative molecular mass of the substance. Molar mass can be calculated for a substance in a molecular or atomic state. To determine the molar mass of gases, Mendeleyev-Clapeyron equation is used. (2.1) where: m is mass of gas, kg; V is volume of gas, L; P is pressure of gas, Pa; Т is temperature of gas, K; R is universal gas constant, 8.314 J/(K×mole); and М is the molar mass of gas, g/mole. Equation (2.1) is obtained from the equation of state for gases: (2.2) where: Р, V and Т are pressure, volume and temperature measured in the experiment; Ро, Vo and То are pressure, volume and temperature for normal conditions. 24

In the expression (2.2)

= R, that is the universal gas constant.

Hence PV = RT for the 1 mole of gas and PV = nRT for the n mole of gas. After such reasoning implies the relation equation (2.1), from which we can conclude that: (2.3) From Avogadro's law follows another important consequence: the weight ratio of equal volumes of the two gases are constant for these gases. This constant is called the Relative Density of the gas and is denoted by D. Since the molar volumes of gases are the same (a consequence of Avogadro's law), the ratio of the molar mass of gas vapors are also equal to this constant: D = М1/М2 (2.4) where: M1 and M2 are the molar masses of the two gaseous substances. D value is determined experimentally as the ratio of the mass of the same volume of test gas M1 and reference gas with a known molecular weight M2. According to the values of D and M2, molar mass of the gas under investigation can be found. M1 = D. M2.

(2.5)

The law of equivalents In 1793 the German chemist I.V. Richter formulated the law of equivalents that all substances react in equivalent ratios. Equivalent is a real or notional chemical particle which can add or highlight in a reaction one atom or a hydrogen ion. In essence, the equivalent is expressed in units of carbon material mass. It is the one shown by a valence or basic (acidity). To determine the equivalent of element is enough to know the atomic mass and valence: E=Аr/B (2.6) where: E is equivalent, Аr is element´s atomic mass; B is valence. 25

Unit amount of equivalent is mole. Molar (mole) equivalent mass (Em) (or equivalent mass of element) element is called mass, which adds or replaces 1.008 mass parts of hydrogen and 8.000 mass parts of oxygen (or 12 mass parts of carbon 12С). Mass of 1 mole equivalent is the molar equivalent of mass. Equivalent is used for a number of practical calculations (molar solutions, electrochemical processes). The mathematical expression of equivalents law has the form: mA/mB = Em(А)/Em(В)

(2.7)

where: mA, mB are А and В reactants mass; Em(А), Em(В) are equivalent mass of these substances. For the case where one or both substance(s) is/are in a gaseous state, it can be written as, mA/Vв= Em(А)/VE(B); VA/VB=VE(A)/VE (B),

(2.8)

where: VA, VВ are А and В reactants volume; VE(А), VE(В) are equivalent volume of these substances (l/mol). It is important to learn to calculate the equivalents of complex substances. In general, the equivalent of complex substance is determined by reaction, where substance is involved. Therefore, it is not constant and can have different meanings. For example, the oxide molar mass is defined as: – Oxide´s equivalent mass is equal to the sum of oxygen and element equivalent mass constituent of the oxide. – Acid´s equivalent mass is equal to the sum of equivalent mass of hydrogen and acid residue. To calculate the equivalent mass of acid, it is necessary to divide its molar mass on basicity of acids which for this reaction is determined by the number of hydrogen atoms replaced with metal. For example, orthophosphoric acid Н3РО4 depending on the flow of reaction conditions can be considered as a one basic, two basic or three basic. Since the formation of dihydrophosphate, hydrophosphate and phosphate of phosphoric acid equivalent will correspond to 1, 1/2 and 1/3 mole. 26

– Base´s equivalent mass is equal to the sum of the metal equivalent mass and a hydroxyl group. To calculate the base´s equivalent mass, it is necessary to divide the molar mass at the acidity of base which is determined by the number of hydroxyl groups included in the reaction. Em(NaOH) = 40 g/mole; Em(Ca(OH)2) = 74/2 = 37 g/mole. – Salt´s equivalent mass is equal to the sum of the metal mass and acid residue. Salt Em (NaNO3) = 85 g/mole, Em (Cr2 (SO4)3) = = 65.4 g/mole. From these examples, it follows that the equivalent weight of a compound is generally not constant and depends on the chemical reaction that takes part in the present compound. If in the reaction involved gaseous substances, the concept of an equivalent amount is used, i.e. volume, which takes in the given conditions one equivalent of a gaseous substance. So, under normal conditions, an equivalent amount of hydrogen is 11.2 liters, equivalent amount of oxygen – 5.6 liters. QUESTIONS AND EXERCISES FOR SELF-CHECKING: 1. Give the formulation of the mass and energy conservation law. 2. Formulate the law of definite proportions. 3. Give formulation of the multiple proportions law. 4. Formulate the simple volume relations law. 5. Give the formulation of Avogadro´s law. 6. Formulate the equivalent law. 7. What is a system? 8. What is an isolated system? 9. What is an equivalent? 10. Give the definition of a molar equivalent mass. 11. Give the mathematical expression of equivalents law. 12. Give the formula to calculate the molar equivalent mass of elements, acids, bases and salts. 13. Give the definition of Avogadro’s number and the mole of substance. 14. During decomposition, 2.17 g of mercury oxide produced 0.17 g of xygen. Find the mass of mercury. 15. How many grams of СаО and СО2 can be obtained from 100 grams of СаСО3? 16. Determine the mass of 2 moles of argon? 17. How many structural units are contained in 0.5 bromine moles? 18. What is the relative density of hydrogen by chlorine, air and helium? 19. Determine the molecular mass of a gas, the density of which by hydrogen is 14? 20. What is the volume of 6 g of hydrogen at standard conditions?

27

21. Determine the mass of 224 liters of acetylene at standard conditions. 22. The mass of 250 ml gas is equal to 0.903 g. Calculate the molecular mass of the gas? 23. What is the equivalent mass of carbon in carbon dioxide? 24. What is the equivalent mass of sodium sulfate? 25. In what compound is the equivalent mass of nitrogen equal to 2.8 g / mole? 26. What is the equivalent of sulfur in a compound containing 50% sulfur and 50% oxygen? 27. Determine the equivalent of the elements in compounds and complex substances listed below: Сl2О7, Н2МnО4, Аl(NО3)3, Na2B4О7, Са3(РО4)2, Na2HPО4, Bi(OH)2. LABORATORY WORK Determination of relative molecular mass of gaseous substances Pick a dry 250-300 ml flask with stopper and mark with a pencil on the glass the position of the lower edge of the stopper. Then weigh the dry stoppered flask with air on a chemical balance. Now, lower into the flask a glass tube attached to the Kipp gas generator producing carbon dioxide. Open the tap of the Kipp gas generator and allow carbon dioxide to come out of it slowly for 3-5 minutes. Pass carbon dioxide into the flask again for 3-4 minutes, after which weigh it once more. To verify the completeness of the displacement of air, it is necessary to pass carbon dioxide gas for another 2-3 minutes and again weigh the flask. If the latter two weighings produce identical results or results that do not differ by more than 0.02 g, the filling of the flask may be considered completed. Otherwise repeat the filling procedure, until a constant, weight of the flask is achieved. With this done, record the temperature and the pressure by referring to a room thermometer and barometer. To measure the volume of the flask, fill it to the marking with water at room temperature, then pour the water into a measuring cylinder to determine its volume, and record the results of the experiment. Record experimental data and calculation: Weight of flask with air and stopper – m1 (g) Weight of flask with carbon dioxide and stopper – m2 (g) 1st weighing 2nd weighing Volume of flask – V (ml) Temperature – t оС; Absolute temperature – Т, К. Pressure – Рo, 760 mmHg. Atmospheric pressure at the implementation experience – Р, mmHg. 1. Calculate the normal volume of air in the flask by the equation:

PV P 0V 0  T T0

28

(2.9)

2. Calculate the mass of air in the flask – m3, g (taking into account that the air density is 1.293 g/l) 3. Determine the mass of the empty flask: m4 = m1 - m3 4. Determine the mass of carbon dioxide in the flask: m5 = m2 - m4 5. Calculate the carbon dioxide density in the air. Dair(СО2) =

m5 m3

(2.10)

6. Calculate the mass of hydrogen in the volume of the flask m6, taking into account the hydrogen density – 0.09 g/l. 7. Calculate the carbon dioxide density by hydrogen: (CO2) =

m5 m3

(2.11)

8. Calculate the relative molecular mass of carbon dioxide by the three ways: М = 2.016 M = 29 Dair m5 RT М= pV

(2.12)

9. Calculate the relative error by equation: (2.13) Determining the Equivalent of a Metal 1. Weigh on technical-chemical balance 0.1 g of magnesium and wrap it in filter paper. 2. Retract the vapor tube of the Wurtz flask vertically mounted cylinder filled with water. 3. Carefully pour the 10 ml 2N HCl into a Wurtz flask without wetting the walls of its neck. 4. Place the weighed amount of the magnesium into the neck of the Wurtz flask. Close the flask opening tightly with its stopper. Turn the Wurtz flask to its vertical working position so that the metal drops into the 2N HCl. The reaction of magnesium with an acid: Mg + 2HCl = MgCl2 + H2 5. Measure the volume of the water displaced from the apparatus by the evolved hydrogen.

29

Calculations: P0V0 PV  , we find T0 T

1. Using the gas state equation

PVT 0 . P0T

V0=

(2.14)

Since hydrogen has been assembled over the water, the hydrogen pressure is joined to the vapor pressure of water (partial pressure) – h, which must be subtracted from the total pressure – P. Pressure of water vapor – h at this temperature is found in the table. Table 1 The saturated water pressure vapor in equilibrium with water Temperature, °C 15 16 17 18 19 20

Vapor pressure, mmHg 12.788 13.634 14.530 15.477 16.477 17.535

Temperature, °C 21 22 23 24 25 30

Vapor Pressure, mmHg 18.650 19.827 21.088 22.377 23.756 31.824

Substituting into the equation (2.14) find: V0 =

V ( P  h)T0 P0T

(2.15)

2. Find the released hydrogen mass:

mH 2 

V0 2.016 22400

(2.16)

3. On the basis of the equivalent law, define the equivalent of magnesium:

mMg mH

=, EMg =

mMg EH mH

(2.17)

4. Calculate the percentage of errors by the formula: ´

30

(2.18)

3

Lesson #

ATOMIC STRUCTURE. PERIODIC SYSTEM AND ITS RELATIONSHIP WITH ATOMIC STRUCTURE

OBJECTIVE: Consideration of atomic structure, theory of quantum numbers, application of the principles that determine the sequence of filling of atomic orbitals with electrons, periodic system structure and periodic nature of changing of properties of elements.

HIGHLIGHTS: 1. Atoms and their structures. 2. Components of an atom. Concept of isotopes. 3. Wave-particle duality. 4. Electronic shell of an atom. 5. Principles defining the sequence of filling of atomic orbitals by electrons. 6. Quantum numbers and their characteristics. 7. Electronic formula of atoms. 8. The excited state of atoms. The valence electrons. 9. Structure of the periodic table of elements. 10. The periodic nature of the changing of certain properties of atoms, elements and simple substances. 11. Characteristics of element, a simple substance and its compound according to position in the periodic system. THEORETICAL PART Components of an atom (nucleus and electrons). Atom is the smallest constituent unit of ordinary matter that has the properties of a chemical element. In the center of the atom is po31

sitively charged nucleus around which the moving negative electrons form the electron shells. In chemical processes atomic nuclei of elements сomprising the reactants is not changed. Changes in this undergo only electronic shells. Because nucleus size is 10,000 times smaller than atom, the geometric size of an atom is not determined by the nucleus diameter (~ 10-12 cm), but it is determined by the electron shell diameter (-10-8 cm). The electron (symbol e-), as a particle has a rest mass of 9.1 × 10-31 kg, carries a negative electric charge equal to 1.6 ×1019 Kl. The absolute value of this charge is the smallest; it is called the elementary charge and equal to 1. The nuclei of atoms have a complex structure and are composed of even smaller particles – nucleons. There are two types of nucleons: the proton (p+ symbol), which has a positive electric charge 1.6·1019 Kl (elementary charge) and has a rest mass 1.6726·10-27 kg; neutron (designated n0), which has no electric charge (electrically neutral) and has a rest mass equal to 1.6749·10-27 kg. In the atomic mass units (a.m.u.) Аr (р+) = Аг (n°) = 1 a.m.u. In fact, the atomic mass unit accepted the one nucleon mass (p+, n°). The electron mass is 1837 times less than nucleon mass, so it can be argued that the atom mass is concentrated in its nucleus, and thus the mass number (Ar) of atom of chemical element is equal to the number (N) contained therein nucleons. Ar = N(p+) + N(n°). The atoms with the same number of protons but a different number of neutrons are called isotopes. Isotopes of atoms of the same element have the same electronic structure (the same number of p+), and similar chemical properties. This is the reason why they can not be divided by chemical methods. Protons (р+) Atom

General title is «nucleons» Neutrons (n0)

Nucleus

Electronic shell – electrons (е) 32

The structure of the electron shell of an atom. Electronic formula of an atom. Electron shell of an atom is aggregate of electrons, each of which has a corpuscular-wave character. The structure of the electron shell of an atom is determined by the different energy, E content of individual electrons in an atom. The electrons of multielectron atom which have a close reserve of energy, E are grouped into energy levels (n = 1,2,3,4,5,6,7). Each energy level contains electrons that are considered as the electronic layer. The capacity of the energy level (max filling by electrons) corresponds to the formula 2n2. The first energy level of maximum filling of electrons is equal to 2, i.e, (2 = 2·12); the capacity of the second energetic level is equal to 8 е (2·22 = 8); third – 18 е; fourth – 32 е and etc. The energy of the electrons in the same energy level (except the first) is not the same, therefore each energy level (except for n = 1) is split into the energy sublevels. In the first sublevel there is one energy sublevel s. The second energy sublevel splits into two sublevels 2s and 2p; third energy sublevel – on three sublevels 3s, 3p, 3d; fourth – on four sublevels – 4s, 4p, 4d, 4f. Accordingly, the fifth energy level must contain 5 sublevels: 5s, 5p, 5d, 5f, 5g, but for all known elements, g-sublevel is not filled with electrons. Each s-sublevel contains one s-atomic orbital (s-AO), each p-sublevel contains three p-AO; in d-sublevel – five d-AO; f-sublevel – seven f-AO. s-AO

p-AO

d-AO

f-AO

The maximum filling of AO with electrons is equal to two (see below the «Pauli exclusion principle»). 17С1-

ls22s22p63s23p5

Large arabic numerals in the e-formula represent energy levels, which means in the chlorine atom the electrons are distributed on 3 energy levels. 33

The letters s, p, d, f denotes energy sublevels and small arabic numbers correspond to the number of electrons in each energy sublevel. Ground and excited state of an atom An important atom characteristic is energy, which can take only a certain (discrete) value. It corresponds to the atom stable states and only changes abruptly by a quantum transition. If the atom is in its ground state, its electrons occupy the lowest energy levels, i.e. each subsequent energy electron occupies the most advantageous position. When the external influence on the atom, associated with the transfer of power to it, for example, by heating, the electrons are transferred to the higher energy levels. This atom is said to be excited. Vacant at the lower place energy level is filled (as a prime location) by an electron from a higher energy level. In the transition electron gives energy certain amount, which corresponds to the energy difference between the levels. As a result of electronic transitions characteristics radiation arises; the spectral lines absorbed or emitted light is possible to make a quantitative conclusion on the atom energy levels. Filling the atomic orbitals in one energy sublevel is in accordance with Hund`s rule (German physicist F. Hund formulate in 1927): Atomic orbitals belonging to the same sub-layer in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin. The quantum numbers. Each electron in an atom is characterized only by its own set of four quantum numbers: main n, the orbital l, magnetic ml and spin m s. Pauli Exclusion Principle, 1925: No two electrons in the atom can have the same values of all four quantum numbers, i.e. electrons with opposite spins can occupy only one orbital. The energy level (1, 2, 3 ...) is characterized by the principal quantum number n. For all known chemical elements the values of n 34

range from 1 to 7, respectively, a period number, in which an element is located. Electrons that are in orbitals with the same value of n is spoken of as same energy level electrons. The greater the n value, the greater the electron energy and its distance from the nucleus (more than the electron cloud size). Energy sublevel (s, p, d, f) is characterized by the orbital quantum number l. It determines the difference in energy state with n = const, except n = 1. In addition, the orbital quantum number of the orbital shape defines geometric s-, p-, d-, f- sublevels. The s-AO is spherical symmetry, p-AO – dumb-bell, d-AO – four petal etc. The difference between the electrons that occupy atomic orbital’s one sublevel (n, 1 = const), except the s-sublevel, is characterized by the magnetic quantum number ml. This number is called magnetic because it characterizes the electrons behavior in an external magnetic field meaning, ml establishes mutual spatial arrangement of atomic orbitals. For the s-sublevel (n = const, 1 = 0) only one value ml = 0, which implies that s-sublevel of any (from the first to the seventh) energy level contains only one 1-AO. For p-sublevel (n> 1, l = 1) ml can take three values +1, 0, -1, hence the p-sublevel contains three p-AO. For d-sublevel (n> 2, l = 2) ml has five values: +2, +1, 0, -1, -2, and for f-sublayer (n> 3, l = 3) ml of values has seven: +3, +2, +1, 0, -1, 2, -3, so any f-sublevel has seven AO. To distinguish the two electrons on the same AO, a fourth spin quantum number ms is used. Two electrons that are on the same AO (n, l, ml = const) differ in spin quantum number, ie, two electrons in one AO have antiparallel spins. Spin is an essential feature of an electron in an atom. The electron concept spin in an atom introduced in quantum mechanics by Uhlenbeck and Goudsmit (1925), postulated on the experimental spectroscopic data analysis the possibility of considering the electron as a rotating top. Quantum numbers Quantum numbers 1 Principal Quantum Number, n

Possible values 2 Theoretical 1, 2, 3……. For known elements 1, 2, 3……7

35

Characteristics 3 Electron energy and electron cloud size. With increasing n, atomic radius increases

1 Orbital Quantum Number, L

2 0, 1, 2, 3….. (n-1) s p d f sublevel

Magnetic Quantum Number, ml

Whole numbers with - L … 0 … +L

Spin Quantum Number, ms

two values +½ and -½

3 Electron cloud shape and difference in the electrons energy state at the same energy level, except for n=1 Atomic orbitals spatial orientation (AO). s-АО – one orientation; p-АО – three; d-АО – five; f-АО – seven orientations. The direction of electron relative rotation to its imaginary axis

Periodic table structure of chemical elements Periodic table of chemical elements was developed by D.I Mendeleyev on the basis of the periodic law (1869). The modern formulation of this law is elements properties are in periodic dependence from their atomic nuclei charge. Nuclear charge Z is atomic (serial) number of the element in the periodic system. Elements arranged in ascending Z (H, He, Li, Be,…etc) form the seven periods. In the 1st is two elements, in the 2nd and the 3rd are eight elements each, in the 4th and the 5th are eighteen elemets each and in the 6th is thirty-two elements. In the 7th period is twenty-three known elements. Of the various ways to represent the periodic table of chemical elements that are used in science, the basic and most common forms are short-period and long-period form. They complement each other and are generally identical. For the long-period system form of each period corresponds one horizontal line and for the short periodic system form from the fourth period is divided into two rows. The vertical columns in the periodic table are called groups. In the short form groups contain elements of both small and large periods called the main subgroups and groups containing elements of only large periods (from the fourth) are called secondary subgroups. The group numbers are represented by roman numerals I to VIII. 36

The long-period form the main group noted by Roman numeral and the letter A, and subgroups noted by Roman numeral and letter B. This notation is convenient to use due to their brevity. Elements of some groups got their own group names: elements of IA group (Li – Fr) – alkali metals; elements of IIА group (Са – Ra) – alkaline earth elements; elements of VIA group (О – Ро) – chalcogens; elements of VIIА group (F – At) – halogens; elements of VIIIA group (Не – Rn) – noble gases; elements of IIIB group (La – Lu) – lanthanides; elements of IIIB group (Ас – Lr) – actinides. The elements are divided into groups VIIIB triad Fe-Co-Ni; Ru-Rh-Pd; Os-Ir-Pt. The first triad is called the «iron family» and two others are «Platinum Family» Between the element position in periodic table and atom structure, there is the following relationship: 1. Atoms of all elements of the same period have the same number of energy levels. Period number indicates the number of energy levels in an atom, and the value of the principal quantum number of highest energy level. Since all the atoms of known elements in the limiting case fill seven energy levels (completely go partly) in the periodic table, there are seven periods. 2. Atoms of elements of the same group have the same type of electronic configuration and similar chemical properties. The division of the groups into main (group A) and subgroups (group B) depend on the which energy level (external or internal) is completed in an atom of the corresponding element. All the s and p elements are the elements of the A-groups because in their atoms are filled external energy levels and all d and f elements are the elements of the B-group because in the atoms of these elements are filled inner energy levels. In each period, starting from the fourth, sections of s and p elements are separated by section of d-elements (in the sixth and seventh periods even by section f-elements) and so d-elements are often referred to as transition elements. The electrons that participate in the formation of chemical bond between atoms is called valence electrons. This is the weakest bonded electrons with nucleus. The maximum number of valence elec37

trons usually coincides with group number of element in the periodic table. The exceptions are the elements of Group IB and VIIIB. QUESTIONS AND EXERCISES FOR SELF-CHECKING: 1. How many protons and neutrons contain atomic nuclei of lead with mass numbers 207 and 208. 2. Write electronic formulae of atoms of Pb, Cl, Fe, Na, N. Underline in the electronic formulae thus AO which is filled by the last electron. To which e-family (s-, p-, d-, f) is each element? 3. Record electronic and electron-graphic formula of the chlorine atom in its normal and excited states. 4. Which four quantum numbers characterize the state of an electron in an atom? What value do they have for the outer electrons of an atom of calcium? 5. Which energy sublevel is filled first in atoms: 4s or 3d and 4p or 3d? Why? 6. How are the electrons energy levels and sublevels of atoms of chlorine and manganese located? 7. What distinguishes small periods from large periods? 8. How do element properties change with increasing atomic mass of (a) period (b) in the main subgroups? 9. What are the formulas of the highest oxides of the elements with atomic numbers 24, 25, 32, 74 and 82? Write the formulas of the acids corresponding to these oxides. 10. Describe the properties of bromine and its compounds based on its position in the periodic system. TASKS FOR PREPARATION OF ISW: «The chemical elements periodic system. Molecules structure» 1. For substance «a», specify the atom structure. 2. What is the difference between absolute, relative and atom molar mass? What is the values of these quantities for atoms of element «a»? 3. For element «b» write electronic and electron-graphic formula in its normal and excited state. 4. For last electron of atom of element «c» record the values of all four quantum numbers. 5. Describe the properties of the element «b» and its compounds by position in the periodic system. 6. Which of the elements of «d» refer to metals and non-metals? Justify your answer with the atom electronic structure of these elements as well as the electronegativity values. 7. According to the electronic formula fragment, «e» determine the highest oxidation state of the element, name the element and write the formula of its oxide.

38

# 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40.

«а» Pb Ca K Rb Zn Mg Fe Ar Na Al Ba Br Ga V Si Se N Ti Sb Cr O Zr S Hg Ru Cu Co Cd Mo W I Ag Ni Sc Sn Te Mn S Al Ne

«b» Se Si V Ga Br Ba Al Na Ar Fe Mg Zn Rb K Ca Pb Ne Al S Sc Sn Te Mn Mo W I Ag Ni Hg Ru Cu Co Cd N Ti Sb Cr O Zr S

«c» Mo W I Ag Ni Sc Sn Te Mn S Al Ne Pb Ca K Rb Zn Mg Fe Ar Na Al Ba Br Ga V Si Se N Ti Sb Cr O Zr S Hg N Li C P

39

«d» Na and N Li and O Mg and S Ca and O Ba and Cl Ni and Cl Cs and F Rb and Br Na and Cl K and O K and N Al and F K and I Fe and Cl Co and I K and Br Na and F K and F Zn and Cl Zn and N Al and Cl Pb and Br Al and Br Pb and Cl Sn and Cl Ni and Br Sn and O Ni and O Mg and N Li and O Li and S Al and S Zn and S Mg and Cl Na and F Na and Cl Na and S K and I Ba and N Mg and S

«e» 3s2p6d54s2 2s2p63s2p1 1s22s2p1 5s2p6d16s2 3s2p6d14s2 3s2p6d34s2 4s2p6d35s2 4s2p6d25s2 4s2p65s2 4s2p6d55s2 1s22s2p2 2s2p63s2p2 4s2p6d105s2p1 5s2p6d106s2p2 1s22s2p3 2s2p63s2p3 3s2p6d104s2p3 4s2p6d105s2p3 2s2p63s2p4 3s2p6d104s2p4 4s2p6d105s2p4 2s2p63s2p5 3s2p6d104s2p5 2s2p6d53s2p2 3s2p6d44s2 3s2p6d64s2 3s2p6d104s2p5 3s2p6d104s2p3 3s2p6d24s2 3s2p6d104s2p4 3s2p6d14s2 2s2p63s2 4s2p6d105s2p5 4s2p6d105s2p3 4s2p6d105s2p1 5s2p6d106s2p3 4s2p6d105s2 2s2p63s2p1 2s2p63s2 2s2p63s2p64s1

LABORATORY WORK The periodic law and the elements periodic system D. I. Mendeleyev Experiment #1. Chemical properties of simple substances of third period elements. Indicate which of the substances on the laboratory table is simple. What simple substances correspond to elements of the third period? Determine their reactivity towards air and water. Describe the appearance of each of these substances. Which of them react with air? Is there any regularity in the interaction of these substances with water? Write reaction equations. Experiment #2. Properties of oxides of the third period elements. In separate test tubes put a little amount of sodium oxide, magnesium oxide, aluminium oxide, silicon oxide and phosphorus oxide and add to each of them 1-2 ml of distilled water. Determine the pH of the solutions. Which of the investigated oxides dissolve in the distilled water? Write reaction equations. Experiment #3. Properties of hydroxides of the third period. In seven test tubes pour sequentially in the form of solutions or suspensions of hydroxide of the third period elements. Determine the pH of solutions. For each of suspensions of poorly soluble hydroxides divide into two parts. To one part add excess of a solution of HCl and to the other excess of NaOH. What properties show investigated hydroxides? Write reaction equations. Experiment #4. Elements properties comparison and their compounds in the periodic system groups. 1. In two test tubes put a bit of magnesium metal and calcium, add 1-2 ml of distilled water. Heat the tubes. Write observations. How do the metallic properties of elements in groups change from top to bottom? In the activity series, why is metal Li ranked first? 2. In two test tubes put a bit of magnesium and calcium metal, add 1-2 ml of hydrochloric acid solution. Explain the observed phenomena and to write reaction equations. 3. In three test tubes put a bit of powdered sodium chloride, sodium bromide and sodium iodide and add to every test tube 2-3 drops of concentrated sulphuric acid. Write observations. Taking into consideration that HCl, HBr, HI are colorless gases, conclude which hydrogen halides are easily oxidized by sulfuric acid and which are more durable? How do the properties of non-metallic elements change in groups with increasing of atomic mass? Write reaction equations. 4. In three test tubes pour solutions, chlorides or nitrates of magnesium, calcium and barium and add 2-3 ml of sodium hydroxide solution. What happens? How do the metallic properties of elements change in groups from top to bottom? Write reaction equations. Experiment #5. Iron and its properties. 1. In three test tubes pour 1-2 ml of diluted hydrochloric, sulfuric and nitric acids. Put in each tube a little iron filing. What happens? Write reaction equations.

40

2. In two test tubes pour 1-2 ml of the solution of iron (II) sulfate and iron (III) sulfate and add the same volume of a solution of sodium sulfide. What happens? Write reaction equations. 3. In a new test tube pour 1-2 ml of potassium permanganate solution and add sulfuric acid solution (to create an acidic environment) and add a few crystals of iron (II) sulphate. Heat the resulting solution. What happens? Write reaction equation. 4. In a test tube pour 1-2 ml of iron (III) chloride and add 1-2 ml of potassium iodide. What happens? Write reaction equation.

41

LESSON #

4

CHEMICAL BOND, MOLECULAR STRUCTURE, CHARACTERISTIC OF INTERACTING ATOMS

OBJECTIVE: Consider the theories of chemical bonding and its use for predicting the composition, structure and properties of different compounds. Summarize the information about the main types of chemical bonds and describe the properties of covalent and ionic bonds. Give the concept of hydrogen and metallic bonding.

HIGHLIGHTS: 1. Types of chemical bonds. 2. Valence bond theory. Overlapping of atomic orbitals. 3. Ionic bond and its properties. 4. Mechanisms of formation of covalent bonds. 5. Properties of covalent bonds (polarity, orientation, saturation). 6. Geometric structure of molecules. 7. Hybridization of atomic orbitals. 8. Metallic bond. 9. Hydrogen bond. Intermolecular and intramolecular hydrogen bonds. THEORETICAL PART. Сhemical bond (CB) formation is the desire of the atom to have a stable electron shell (8е , or for the elements of the first period 2е ) and reach in a state characterized by minimal energy. 42

Main parameters of CB are its length, strength and bond angles characterizing the structure of substances which are formed from individual atoms. The forces in chemical bonds mainly have electrical nature. Formation of CB is accompanied by a rearrangement of the electron shells of bonding atoms. Depending on the nature of the electron density distribution between the interacting atoms distinguish four main types of CB: covalent, ionic, metallic and hydrogen. Types of chemical bonds CB type Covalent nonpolar bond

EN* difference 0

Covalent polar bond

< 1.7

Ionic bond > 1.7

Metallic bond

Hydrogen bond

-

-

* EN – electronegativity

Method forming CB The total electron pair is located symmetrically with respect to the nuclei of both atoms The total electron pair is shifted in the direction of the atom having the highest EN

Examples Occurs between atoms of non-metals with the same EN value Н2, О2, Сl2 and etc. Occurs between two atoms of non-metals with different EN values NH3, H2O, HCl, H2S, HBr and etc. The total electron pair Occurs between atoms is fully transferred to of metals and nona more metals electronegative atom NaCl, BaCl2, CsF and which turns into a etc. negative ion, and turns the other atom into a positive ion Formed in metals Close to covalent in due to electrostatic nature but differs from attraction between the it in that electrons positively charged generalization carried metal ions and free out not between two electrons in the metal atoms, but many atoms. lattice. Typical for metals Occurs between the Bond stitches molecule molecules which substances in include the hydrogen associates, increasing atom and a highly EN the molecular weight of element the substance and changing the properties of substances.

43

The sequence of connections of atoms in the molecule shows a structural formula of the substance. One dash in the structural formula means a chemical bond and only atoms of elements with opposite charges can adjoin with each other. Barium hydroxide Ва(ОН)2 Ba

+2

O-2

H+1

-2

+1

O

H

Sodium hydrocarbonate NaHCO3 H+1 Na

+1

O-2 -2

O

C+4 =

O-2

Hybridization of atomic orbitals Each atomic orbital is characterized by its shape, size and energy. Sometimes the largest area of overlapping of AO is achieved through hybridization. The essence of the concept of hybridization of AO is that during the formation of certain molecules and some complex ions the change of form and energy of initial valence AO and forming on their basis same number of identical shape and energy of hybrid AO may occur. The geometric structure of molecules Molecule 1 BeCl2, HgCl2, ZnBr2 and etc. BCl3, BF3, BI3 and etc.

CH4, SiH4, CHal4 and etc.

Angle between the CB. Type of AO hybridization 2 sp – hybridization. < 180О sp2 – hybridization. < Halogen – В – – Halogen = 120О sp3 – hybridization < H – C- H = 109O28I.

44

Molecules geometry 3 Cl―E―Cl Linear, non-polar Halogen В Halogen Halogen Angled (flat), nonpolar H H H E H Tetrahedral, nonpolar

1 NH3, PH3, PCl3 and etc.

2 sp3 – hybridization < H –N – H = 107.3О

Н2О etc.

sp3 – hybridization < О – Н – О = =104.5О

3 Tetrahedral, polar. One vertex of the tetrahedron is not busy Two vertices of the tetrahedron are not busy. Polar

QUESTIONS AND EXERCISES FOR SELF-CHECKING: 1. How many electrons are involved in the formation of a chemical bond in a molecule of hydrogen chloride? 2. What is the angle of bond in the molecule BeI2? 3. What types of bonds do not exist in the ammonium chloride? 4. How change the bond polarity in in the H2O-H2S-H2Se-H2Te series? 5. Specify a geometric shape of the molecule CO2. 6. In which of the following cases of overlapping of electron clouds takes place formation of -bond? a) S – S; b) py – py; c) Px – S; d) S – Px ; e) S  d 2 2 . x y

7. The valence orbitals of the boron atom in the ВН3 molecule are hybridized as: a) sp2; b) sp; c) sp3d; d) sp3d2; e) sp3. 8. For which of the following molecules bond order is equal to three? a) O2; b) H2; c) HCl; d) N2; e) H2O. 9. Using the value of electronegativity, determine whether it is possible to consider as ionic the aluminum compound with: a) fluorine; b) chlorine; c) bromine. 10. What are the similarities and differences: a) between the metallic and ionic bonds; b) between a metal and a covalent bond? TASKS FOR PREPARATION OF ISW «Chemical bond and molecular structure» Make the structural formula of the substance «a», specify the type of chemical bond between each pair of chemically bonded atoms.

45

What is the hybridization of the electron clouds takes place in the central atom of the molecule «b»? What is the geometric structure of this molecule? Specify the type of bond in the molecule «c» and provide the bond formation scheme in these molecule. # 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40.

«a» K3PO4 Na2CO3 K2SO4 NaNO3 NaNO2 K2SO3 KHSO3 KNO3 H2SO4 K2SiO3 KHSiO3 K3AsO4 NaBiO3 HClO4 Na2SO3 KClO4 HClO3 NaClO4 Ca(OH)2 HClO3 HClO2 HClO KMnO4 K2SO4 KClO4 H3PO4 NaClO KClO3 Na3PO4 Na2SO4 Mg(OH)2 NaClO3 MgSO4 Na3PO4 NaMnO4 LiClO4 NaClO2 KBrO4 KNO2 BaSO4

«b» SiF4 BeH2 BF3 HgCl2 ZnCl2 NH3 BH3 BCl3 CCl4 CBr4 SiH4 BBr3 BF3 AlF3 CH4 SiCl4 PCl5 SCl6 BeCl2 MgH2 AlH3 GaCl3 BeF2 AlCl3 MgCl2 GaBr3 CaCl2 BeBr2 AlBr3 MgBr2 GaI3 BeI2 CaBr2 AlF3 MgI2 CaI2 SiBr4 MgF2 GeH4 GeCl4

46

«c» SO2 H2 CH4 CCl4 N2 SiH4 H2S O2 H2O BeH2 PH3 Br2 AsH3 BeCl2 KCl HgCl2 HCl BF3 NH4+ NaCl HBr KBr Cl2 BCl3 HF CsF KCl CsBr CBr4 KF H2Se NaF Na2O SbH3 K2O ZnCl2 Li2O CO2 F2 CO

5

Lesson #

FUNDAMENTALS OF CHEMICAL KINETICS. RATE OF CHEMICAL REACTIONS. FACTORS AFFECTING TO THE RATE OF CHEMICAL REACTIONS

OBJECTIVE: Consider the fundamentals of chemical kinetics, study the effect of various factors (the concentration of substance, temperature, nature of the reactants, catalyst) on the rate of chemical reactions.

HIGHLIGHTS: 1. Reaction rate concept. Mass action law. 2. Dependence of chemical reaction rate on temperature. 3. Van't Hoff Rule. 4. Activation energy concept. 5. Catalyst influence on the process speed. THEORETICAL PART. Chemical kinetics is the study of rates and mechanism of chemical processes. Chemical kinetics is the branch of chemistry studying the laws of chemical reactions with time and dependence of these laws from the external conditions as well as the mechanisms of chemical transformation. Chemical reaction rate determines the change in the number of particles (molecules, ions etc.) of any substance involved in a chemical reaction per unit time, per unit volume (for homogeneous reactions) or per unit area of surface interface (for heterogeneous reactions). 47

Vhom.r-on = ± n/Vt

(5.1)

Vhet.r-on = ± n/St

(5.2)

Factors affecting reaction rate: 1. Nature of the reactants. 2. Concentration of the reactants. 3. Temperature. 4. Catalysts. Effects of nature of the reactants. The reaction rate constant is dependent primarily on the nature of the reactants. Potassium reacts with water more quickly than sodium and lithium reacts with water slower than sodium. The following can help to understand «nature of the reactants»: 1. For the substances with molecular structure this is the type of chemical bonds in the molecules of reactants and the strength of the bonds. In order for molecules to react, the bonds should be broken. 2. For the substances with non-molecular structure (ionic or atomic crystal) this is the crystal lattice structure and its strength. 3. For the substances with single atom (e.g., metals, noble gases) this is structure of the electron shell of an atom and the binding strength of the outershell electrons. 4. For molecules of complex shape this is the likelihood of a favorable reaction, mutual arrangement of the reagents in the collision. Effect of concentration of the reactants. For elementary reactions the law of mass action (performed by Norwegian chemists C. Guldberg and P. Waage between 1864-1879) is valid. At constant temperature the rate of chemical reaction is proportional to the concentrations of the reacting substances. The rate equation includes only concentration of the initial substances in a solution or in gaseous state. Changing the concentration of substances in the solid state hardly occurs. And therefore the kinetic equation of rate does not include the concentration of solid substances and liquids in excess. 48

Elementary reaction written in general form: mA + nB = aC the rate of the reaction would be: V= k Cm (A) × Cn (B)

(5.3)

where: С(А) and С(В) are molar concentration of reactants, mol/l; m and n are stoichiometric coefficients of reactants; k is the reaction rate constant or rate coefficient of the reaction. Coefficient is equal to the rate if the concentration of substance A is equal to the concentration of substance B and equal to 1 mol/l. The reaction rate constant k may depend on conditions such as temperature, ionic strength, surface area of an adsorbent, nature of the reactants and presence of a catalyst and not depend on the concentration of the reactants. Molecularity of the chemical reaction is a number of reactant particles (atoms or molecules) participating in the elementary (simple) reaction. According to the number of reacting particles reactions can be monomolecular, bimolecular and trimolecular. These reactions are distinguished by the kinetic regularities and mechanism. The order of a chemical reaction is defined as the sum of the powers of the concentration of the reactants in the rate equation of that particular chemical reaction. Effect of temperature. The effect of temperature on the rate of chemical reaction is described by the Van't Hoff rule: the rate of the reaction increases 2-4 fold for a temperature increase of 10 °C. (5.4) where: VT2 and VT1 are rates of chemical reaction at temperature Т2 and Т1 respectively; is temperature coefficient of reaction rate. =

= 2 - 4.

=

(5.5)

Only active particles which have sufficient energy for the reaction can enter into chemical reactions. Therefore, when the tempera49

ture increases the number of active particles is greatly increased, which increases the reaction rate. The minimum amount of energy required to initiate a chemical reaction is called the Activation Energy. Arrhenius equation: K = A × e- Ea/RT

(5.6)

where: Еа is activation energy; А is constant for given reaction; е is base of natural logarithm = 2.7183; R is universal gas constant = 8.31 J/mol.K Effect of Catalysts. A catalyst is a substance that speeds up a chemical reaction, but is not consumed by the reaction, hence a catalyst can be recovered chemically unchanged at the end of the reaction it has been used to speed up or catalyze. Inhibitor is substance which slows down reaction rate. In homogeneous catalysis, catalyst and reactants are in same phase (gas or liquid). In heterogeneous catalysis, catalyst and reactants are separated by a boundary surface and process takes place on a surface of phase interface. The catalyst decreases the activation energy of the reaction, thereby increasing reaction rate, but does not change the heat of reaction. QUESTIONS AND EXERCISES FOR SELF-CHECKING: 1. Write an expression of reaction rate occurring between: a) nitrogen and oxygen; b) nitrogen and hydrogen; c) hydrogen and oxygen; g) carbon dioxide and hot coal. 2. To increase the reaction rate NO + O2 = NO2 for 100 times, how many times should the pressure be increased? 3. When the temperature rises to 50 °C, the reaction rate increases by 1200 times. Calculate the temperature coefficient. 4. For the hypothetical reaction given below, calculate the rate of change of the reaction if the concentration of each reagent is increase 2 times. 2Аaq + Вaq = C 5. Write an expression of rate for the following reaction:

50

3М + 2А2 + В = D 6. The catalyst accelerates the chemical reaction due to: 1) decreasing of the activation energy; 2) increasing of the activation energy; 3) increasing of the chaoticity of molecular motion; 4) increasing of collisions of molecules; 5) increasing of hydration energy. 7. Which factors taken together most completely affect the rate of chemical reaction? 1) the nature of the reactants, pressure and concentration; 2) the catalyst, grinding and concentration; 3) the nature of the reactants, concentration, temperature and catalyst; 4) pressure, grinding and heating; 5) cooling, catalyst and pressure. TASKS FOR PREPARATION OF ISW «Fundamentals of chemical kinetics. Rate of chemical reactions» 1. For the hypothetical reaction «а» calculate rate of change of the reaction, if the concentration of each reagent is increased by «b» times. 2. How many times does the reaction rate change if the temperature changes from «c» to «d» 0С and temperature coefficient is equal to «е»? #

«а»

«b» «c» 3 4 3 80 2 60

«d» 5 120 80

«е» Chemical Equilibrium 6 7 2 2SOg + O2g = 2SO3 g 2 2HI g = H2 g +I2 g

3 4 2 2 3 3 3 4

70 50 100 110 25 18 100 90

90 60 90 80 55 38 80 80

2 3 3 2 2 2 3 2

3 4 5 6 7 8 9 10

2 2Аaq + Вaq = D С2H4 g + 3O2 g = = 2CO2 g + 2H2 g Вs + 2А aq = С Хs + Аg = В 2Аaq + Вg = С Аaq + 2Кaq = С 2Аg = В + С Аg + В = D Кg + 2Мg = Р 3Аs + 2Вg = С

11

CO2 g + C s = 2COg

4

80

90

3

12

Аaq + Вaq = D

4

70

60

3

1 2

1

51

CO g + Cl2 g = COCl2 g N2 g + O2 g = 2NOg 2H2 g + O2 g = 2 H2O 2NO g + Cl2 g = 2NOCl g NH4Cl s = NH3 g + HCl g CaCO3 s = CaO s + CO2 g C s + CO2 g =2CO g FeO s + CO g = = Fe s + CO2 g NH4+ aq + НОН aq = NH4OH aq + Н+aq S2- aq + НОН l = НS- aq + ОН- aq

1 13 14 15 16 17 18 19 20 21

2 3Аg + В2 g = С 2Аaq = В + С 2CO2 g = 2CO g + O2 g 2HBr g = H2 g + Br2 g Аg + Ds = В 2Еg + Сs = D Мaq + 2Рaq = R SO2g + 2H2 g = Sg + 2H2Og 4Мs + 3Аg = Р

4 3 3 4 4 4 3 3 3

22 23

А2 g + 2В g =С 2Аg + В2 g + С = D

24

3

4 60 110 120 100 95 85 75 65 68

5 80 80 80 80 75 75 65 45 98

6 2 2 2 2 3 3 3 3 2

3 4

68 58

88 88

2 3

Аaq + Вg = D

4

48

68

2

25 26 27

N2 g + O2 g = 2NOg 2H2 g + O2 g = 2 H2O 2NO g + Cl2 g = 2NOCl g

4 2 3

25 30 35

45 60 75

3 2 2

28 29

CO g + Cl2 g = COCl2 g 2HI g = H2 g +I2 g

4 4

40 45

30 65

3 3

30 31 32

2Aaq + В2 g = D N2 g + 3H2 g = 2NH3 g 2SOg + O2g = 2SO3g

3 2 4

50 55 30

80 105 50

2 3 2

33 34 35

2 3 2

20 25 40

30 65 70

2 2 3

36 37

NH4Cls = NH3 g + HCl g Cs + CO2 g = 2CO g FeO s+ CO g = Fes + CO2 g H2 g +Cl2 g = 2HCl g 3Аs = 2В + С

3 4

70 30

50 70

2 3

38 39 40 41 42

А s + 2Е aq = D 2А aq + Еs = С + D 2Аg + Вg = С Мs + 2Тaq = R 2HBr g = H2 g + Br2 g

2 3 4 3 3

60 50 35 40 45

40 70 55 70 55

3 2 2 3 3

52

7 2СОg + О2 g = 2СО2 g N2 g + 3H2 g= 2NH3 g 2NО g + О2 g = 2NО2 g 2Cl2 g + 7О2 g = 2Cl2O7 g 4Ps + 5О2 g =2P2O5 s Br2g + О2g = 2Br2O7g S s + О2 g = SO2 g Na s + О2 g = Na2O s ВаСО3 s = = Ва 2+aq + СО32- aq Na2S s = 2Na+ aq + S2- aq H2SO4 aq = = 2Н+ aq + SO42- aq 4HCl g + O2 g = 2H2Oaq + 2Cl2g 2NO2 g = NOg + O2 g 2СОg + О2 g = 2СО2 g H3PO4 aq = = 3Н+ aq + РО43-aq 2HIg = H2 g + I2 g 3Fe2O3 s + H2 g = 2Fe3O4 s + H2O g H2 g + Cl2 g = 2HCl g CO2 g + C s = 2CO g Fe3O4 s + СОg = = 3FeOs + CO2 g 2HBr g = H2 g + Br2 g 2Fe s + 3Cl2 g = 2FeCl3 s SO2 g + 2H2 g = S s + 2H2O g 2CO2 g = 2CO g + O2 g С2H4 g + 3O2 g = 2CO2 g + 2H2 g 4Fe s + 3O2 g = 2Fe2O3 s 2SOg + O2g = 2SO3 g COg + Cl2g = COCl2g 2NО g + О2 g = 2NО2 g N2 g + O2 g = 2NOg

LABORATORY WORK Kinetics of Chemical Reactions Experiment #1. Dependence of the rate of the chemical reactions on the nature of reactants. In three test tubes pour 3 ml of 0.5 M solution of Na2S2O3 and in the other three: 1st – 3 ml of 0.5 M solution of Н2SO4; 2nd – 3 ml of 1 M solution of HCl; rd 3 – 3 ml of 0.5 M solution of H3PO4 By turns pairwise merge the made solutions, note the time on the stopwatch from the time of adding an acid till appearance of opalescence in the solution is observed. Sodium thiosulfate reacts with acids according to the following reactions: Na2S2O3 + Н2SO4 → Na2SO4 + H2SO3 + S + Н2О Na2S2O3 + 2НСl = 2 NaCl + SO2 + S + Н2О 3 Na2S2O3 + 2 H3PO4 = 2Na3PO4 +3SO2 + 3S +3H2O The results of observations record in the table: Nature of reactants

The time interval from the beginning of the countdown to the opalescence appearance

Sulfuric acid Hydrochloric acid Phosphoric acid

Reaction rate v= 1 t

Based on the table data, plot the reaction rate as function of nature of reactants. Make the appropriate conclusion. Experiment #2. Dependence of the rate of the chemical reactions on concentration. In three test tubes pour 1 ml of 0.25 M H2SO4 solution and in other three: 1-st - 1 ml 1N Na2S2O3 solution + 2 ml Н2О; 2-nd - 2 ml 1N Na2S2O3 solution + 1 ml Н2О; 3-rd - 3 ml of 1N Na2S2O3 solution Take turns in pairs to merge made solutions and note the time with a stopwatch from the time of addition of acid until opalescence is observed in the solution. Thiosulfate sodium reacts with sulfuric acid according to the equation: Na2S2O3 + Н2SO4 → Na2SO4 + H2SO3 + S

53

Observation results recorded in the table: Volume, ml Na2S2O3 Н2О а b

Н2SO4 d

Concentration Na2S2O3 a с= аbd

The time interval Reaction rate from the countdown v = 1 beginning to opalest cence appearance

Based on the table data, plot the reaction rate as function of concentration. Make the appropriate conclusion. Experiment #3. Dependence of the rate of the chemical reactions on temperature. The reaction between sodium thiosulfate and sulfuric acid at three temperatures will be examined. This requires repeating one of three experiments from the previous section in a simplified thermostat (thermally insulated beakers) at three temperatures and noting the reaction time in each case. Write the observed data in the table. #

Temperature, °С

The time interval from beginning of counting till appearance of opacity, t, s

Reaction rate v = 1

t

Based on the table data, plot the reaction rate as function of temperature. Make the appropriate conclusion. Experiment #4. Dependence of the rate of the chemical reactions on catalyst. The decomposition of hydrogen peroxide in aqueous solution proceeds very slowly. In three test tubes pour 2 ml of hydrogen peroxide. Note that appreciable decomposition is not observed. At the same time add to first test tube a few crystals of MnO2, to second SiO2 and in the third Fe2O3. Observe the decomposition of hydrogen peroxide. Hold smoldering torch to the orifice of the tubes smoldering kindling. Does it equally quickly decompose the peroxide? What is the catalyst for this reaction? Write the decomposition reaction of hydrogen peroxide. Make the appropriate conclusion. Experiment #5. Dependence of the rate of the chemical reactions on inhibitors. In two test tubes pour 1 ml of 1 M sulfuric acid solution and to two others: 1-st – 2 ml 1 N solution Na2S2O3 + 1 ml Н2О; 2-nd – 2 ml 1 N solution Na2S2O3 + 1 ml Н2О and 1.5 ml 0.02 М copper (II) sulfate solution. Take turns in pairs to merge made solutions and mix well, note the time on the stopwatch from the time of addition of acid until appearing opalescence in the solution is observed. Calculate and compare the reaction rate with each other, find the value of V1/V2. Draw the appropriate conclusions.

54

Lesson #

6

CHEMICAL EQUILIBRIUM. FACTORS AFFECTING CHEMICAL EQUILIBRIUM

OBJECTIVE: Study reversible and irreversible reactions. Give the concept of «chemical equilibrium». Study the effect of various factors (concentration, pressure, temperature) to the chemical equilibrium.

HIGHLIGHTS: 1. Reversibility of chemical reactions. 2. Chemical equilibrium. 3. The equilibrium constant. 4. Le Chatelier principle. 5. The effect of various factors (concentration, pressure, temperature) at the chemical equilibrium. THEORETICAL PART. Chemical reactions are divided into reversible and irreversible. Chemically irreversible reactions under given conditions go almost to the end, until complete consumption of one of the reacting substances: Zn + 2HCl → ZnCl2 + H2 No attempt to receive zinc from zinc chloride while passing hydrogen does not lead to positive results. The reaction is considered to be reversible, if the system simultaneously interacts reagents which leads to the formation of reaction products (direct reaction) and the reaction product is decomposed in the starting materials (reverse reaction). 55

Chemically reversible reactions under given conditions occur simultaneously in both the forward and reverse direction: N2 + 3H2 ↔ 2NH3 The number of irreversible reactions are less than reversible. After some time, the rate of formation of NH3 will be equal to the rate of its decomposition and there will be chemical equilibrium. In chemical equilibrium the rate of direct reaction is equal to the rate of reverse reaction: Ѵ1 = Ѵ2 (6.1)

Figure 3. The change of forward and reverse reaction rate over time

Chemical equilibrium is dynamic, that is, even when the equilibrium is established the reaction does not stop. Signs of true chemical equilibrium are: 1. The state of the system remains constant over time in the absence of external influences. 2. The state of the system changes under the influence of even small external factors. 3. The state of the system does not depend from which side it is suitable to equilibrium. On the basis of equality of rates, forward and reverse reactions at equilibrium can be written: (6.2) 56

The equilibrium constant for a particular reaction at a given temperature is equal to the ratio of product of concentrations of the reaction products to the product of the initial concentrations of the substances taken in degrees proportional to the stoichiometric coefficients. In the case of reversible heterogeneous reactions in the expression of the Кр includes only the concentrations of gaseous and dissolved substances. So, for the reaction СаСО3 = СаО + СО2 Kр = [СО2].

(6.3)

Under constant external conditions, the equilibrium is maintainned indefinitely. If the external conditions are changed, the equilibrium may change. Le Chatelier principle: If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. Factors affecting the equilibrium shift: 1. Concentration of substances. 2. Temperature. 3. Pressure. A catalyst does not have effect on the equilibrium shift because it accelerates both direct and reverse reactions. Effect of Concentration on Equilibrium. Adding or removing reagents into reaction affect equilibrium. Adding reactants or removing products increase the yield of product. On the contrary, adding products or removing reactants increase yield of reactants. For example, in the reaction: N2 + 3H2 ↔ 2NH3 Increasing the concentration of initial substances (N2 and H2) shifts equilibrium to the right and increasing concentration of products (NH3) shifts equilibrium to the left. Effect of Temperature on Equilibrium. There are endothermic and exothermic reactions. Endothermic reactions take place with absorption of heat, i.e., to undergo such 57

reactions heating is necessary. The exothermic reaction is accompanied by releasing of heat. If a direct reaction is exothermic, the reverse is endothermic and vice versa. With increasing temperature, the equilibrium shifts toward the endothermic reaction. With decreasing temperature, on the contrary, towards the exothermic reaction. For example, the folowing reaction is exothermic, i.e. comes with the release of heat, so the temperature increases the equilibrium shift towards the reaction products in the forward direction. And the temperature is lowered towards the starting materials in the opposite direction. N2 + 3H2 ↔ 2NH3 + Q Effect of Pressure on Equilibrium. With increasing pressure, equilibrium shifts toward volume reduction, ie, towards fewer gas molecules. With reducing the pressure, on the contrary, in the direction of increasing volume, ie, toward greater numbers of gaseous molecules. N2 + 3H2 ↔ 2NH3 QUESTIONS AND EXERCISES FOR SELF-CHECKING: 1. Factors affecting the chemical equilibrium. Give examples. 2. Irreversible and reversible reactions. Effect of temperature on the equilibrium. Give examples. 3. Le Chatelier principle. Give examples. 4. Effect of the pressure and concentration of substances to chemical equilibrium. Give examples. 5. Calculate the equilibrium concentration of iodine and hydrogen, if it is known that their initial concentrations were 0.02 mol/l and equilibrium concentration of HI was 0.03 mol/l. Calculate the equilibrium constant. 6. From the equality of rates of forward and reverse reactions, write expressions for the equilibrium constants of following homogeneous reactions: a) 2H2 + O2 ↔ 2H2O; b) 2SO2 + O2 ↔ 2SO3; c) 2NO2 ↔ 2NO + O2 7. In what direction will shift the equilibrium: 2CO + O2 →2CO2 + 566 kJ a) if the temperature decreases? b) if the pressure increases?

58

8. In which direction will shift the equilibrium: H2 + Cl2 → 2HCl a) if the pressure increases? b) if the concentration of HCl decreases? 9. What factors contribute to a shift of the equilibrium to the left? 2SO2 + O2



2SO3 + Q

1) increasing concentration of SO2, increasing temperature and pressure; 2) increasing concentration of SO3, decreasing temperature and increasing pressure; 3) increasing concentration of О2, decreasing temperature and pressure; 4) increasing concentration of SO2, increasing temperature and decreasing pressure; 5) increasing concentration of О2, increasing temperature and decreasing pressure. 10. Сorrect formula of the chemical equilibrium constant for the following reaction is: 3Fe(s) + 4Н2О  Fe3O4(s) + 4Н2 4  Н2  1) К р  Н 2 О4  Fe   Н 2О  2) К р   (s)  4  Fe3O4(s)   Н 2  4  Fe3O4(s)   H 2  3) К р   4  Fe(s)   H 2O   Fe  4) К р   (s)   Fe3O4  4

5) К р  Fe3O4  H 2 4 12. Under what conditions will the reaction equilibrium for the reaction SO2(g) + 2H2(g) = S(s) + 2H2O(g) shift to the right? 1) increasing the sulfur concentration; 2) decreasing the hydrogen concentration; 3) increase in pressure; 4) decrease in pressure; 5) decreasing the concentration of the sulfur (IV) oxide.

59

TASKS FOR PREPARATION OF ISW «Chemical equilibrium» 1. For equilibrium system, write the mathematical expression of the chemical equilibrium constants. Specify what constitutes the numerator and denominator of this expression. 2. Use the principle of Le Chatelier, suggest ways to shift this chemical equilibrium in the forward direction and give them the justification. # 1 1 2 3 4 5 6 7 8 9 10 11

«а» 2 2А l + В l = D С2H4 g + 3O2 g = =2CO2 g + 2H2 g В s + 2А l = С Хs+Аg =В 2А l + В g = С Аl + 2Кl = С 2Аg = В + С А2 + В = D К g + 2М g = Р 3А s + 2В g = С CO2 g + C s = 2CO g

«b» «c» «d» «е» 3 4 5 6 3 80 120 2 2 60 80 2 3 4 2 2 3 3 3 4 4

70 50 100 110 25 18 100 90 80

90 60 90 80 55 38 80 80 90

2 3 3 2 2 2 3 2 3

А aq + В aq = D 3А g + В2 g = С 2А aq = В + С 2CO2 g = 2CO g + O2 g 2HBr g = H2 g + Br2 g Аg + Ds = В 2Е g + С s = D М aq + 2Р aq = G SO2 g + 2H2 g = = S s + 2H2O g 4М s + 3А g = Р А2 g + 2В g =С 2А g + В2 g + С = D Аl+Вg=D

4 4 3 3 4 4 4 3 3

70 60 110 120 100 95 85 75 65

60 80 80 80 80 75 75 65 45

3 2 2 2 2 3 3 3 3

3 3 4 4

68 68 58 48

98 88 88 68

2 2 3 2

25 N2 g + O2 g = 2NOg 26 2H2 g + O2 g = 2 H2O 27 2NO g + Cl2 g = 2NOCl 28 CО g + Cl2 g = COCl2 29 2HI g = H2 g +I2 g

4 2 3 4 4

25 30 35 40 45

45 60 75 30 65

3 2 2 3 3

12 13 14 15 16 17 18 19 20 21 22 23 24

60

Chemical equilibrium 7 2SO2g + O2 g = 2SO3 g 2HI g = H2 g +I2 g COg + Cl2g = COCl2g N2 g + O2 g = 2NOg 2H2 g + O2 g = 2 H2O 2NOg + Cl2 g = 2NOCl g NH4Cls = NH3 g + HCl g CaCO3 s = CaOs + CO2 g Cs + CO2 g = 2CO g FeO s + CO g = Fe s + CO2 g NH4+ aq + НОН l = = NH4OH aq + Н+ aq S2-aq + НОНl = НS- aq + ОН- aq 2СО g + О2 g = 2СО2 g N2 g + 3H2 g = 2NH3 g 2NО g + О2 g = 2NО2 g 2Cl2 g + 7О2 g = 2Cl2O7 g 4Ps + 5О2 g =2P2O5 s Br2 g + О2 g = 2Br2O7g S s + О2 g = SO2 g Na s + О2 g = Na2O s ВаСО3 s = Ва 2+aq + СО32-aq Na2S s = 2Na+aq + S2-aq H2SO4 l = 2Н+aq + SO42- aq 4HCl g + O2 g = = 2H2O l + 2Cl2 g 2NO2 g = NOg + O2 g 2СОg + О2 g = 2СО2 g H3PO4 aq = 3Н+ aq + РО43-aq 2HIg = H2 g + I2 g 3Fe2O3 s + H2 g = = 2Fe3O4 s + H2O g

1 30 31 32

2 2Al + В2 g = D N2 g + 3H2 g = 2NH3 g 2SOg + O2g = 2SO3 g

3 3 2 4

4 5 50 80 55 105 30 50

6 2 3 2

33

2

20

30

2

3 2

25 40

65 70

2 3

2Fe s + 3Cl2 g = 2FeCl3 s SO2 g + 2H2 g = S s + 2H2O g

36 37

NH4Cl s = = NH3 g + HCl g Cs + CO2 g =2CО g FeO s + CO g = = Fe s + CO2 g H2 g + Cl2 g = 2HCl g 3Аs = 2В + С

7 H2 g + Cl2 g = 2HCl g CO2 g + C s = 2CO g Fe3O4 s + СОg = = 3FeOs + CO2 g 2HBr g = H2 g + Br2 g

3 4

70 30

50 70

2 3

38 39 40 41 42

А S + 2Е aq = D 2А aq + Е s = С + D 2А g + В g = С Мs + 2S aq = G 2HBr g = H2 g + Br2 g

2 3 4 3 3

60 50 35 40 45

40 70 55 70 55

3 2 2 3 3

2CO2 g = 2CO g + O2 g С2H4 g + 3O2(g) = = 2CO2 g + 2H2 g 4Fe s + 3O2 g = 2Fe2O3 s 2SOg + O2g = 2SO3 g COg + Cl2g = COCl2g 2NО g + О2 g = 2NО2 g N2 g + O2 g = 2NOg

34 35

LABORATORY WORK «Factors affecting the chemical equilibrium shift» Experiment #1. Effect of concentration of the reactants on equilibrium. In four test tubes pour 1 ml of 0.0025 M iron (III) chloride and 1 mL 0.0025 M solution of potassium thiocyanate. Gently shake the test tubes to mix solutions. All tubes are put in a rack and one test tube with the solution as a control for comparison. In solution, there is a reversible reaction: FeCl3+3KCNS  Fe(CNS)3+3KCl Iron trirodanid gives red coloring for the solution. In the first test tube add 1 mL 0.3 M FeCl3, in second 1 mL 0.6 M KCNS and third one 2-3 micro spatula of KCl. In each case note the change in intensity of color, compare these solutions with those in the control solution test tube. Write the results in the table.

1 2 3

Test tubes #

Added solution 1 mL 0.3М FeCl3 1 mL 0.6М KCNS 2-3 micro spatula of KCl

Change the color intensity (gain, weakening)

Equilibrium displacement direction (right, left)

Write a mathematical expression of the chemical equilibrium constant for the process. Which substances are in solution in the state of chemical equilibrium?

61

Which substance concentration should be changed to shift the chemical equilibrium to the right and left? Experiment #2. Effect of temperature of the reactants on equilibrium. Communicating vessels are filled with a mixture of gases: brown nitrogen dioxide NO 2 and almost colorless nitrogen tetroxide N2O4. Both gases are in equilibrium: 2 NO2 ↔ N2O4 + 62.76 kJ One of these vessels is placed in cold water and the other in hot water. Mark the change of color in each vessel. Explain the observed phenomena, and make conclusion based on the Le Chatelier’s principle. Questions and tasks to prepare for the Control #1. 1. Explain the basic provisions of the atomic-molecular theory. Give examples. 2. Find the relationship between the law of Avogadro and consequences of it. Give examples. 3. Using acid-base interactions (neutralization reactions) provide examples of formation of medium, acidic and basic salts. 4. Analyze the methods for determining the molecular mass of gaseous substances. Give examples. 5. Describe the classification of inorganic hydroxides. Give examples. 6. Analyze the factors affecting the chemical equilibrium and determine the direction of reactions. Give examples. 7. Analyze the factors affecting to the rate of chemical reactions. Give examples. 8. Specify the differences between the reactions in the heterogeneous and homogeneous systems. Give an expression for the reaction rate in these systems. 9. Compare exothermic and endothermic reactions. Assess the affect of temperature on chemical equilibrium. Give examples. 10. Compare the reversible and irreversible reactions. Explain the principle of Le Chatelier. Give examples. 11. Using the rule of van't Hoff, describe the effect of temperature on the rate of chemical reaction. Give examples. 12. Using the law of mass action, describe the effect of the concentration of substances on the rate of chemical reaction. Give examples. 13. Describe the structure of an atom and its characteristics. Give examples. 14. Analyze the structure of periodic table of chemical elements. Give examples. 15. Formulate the principles determining the sequence of filling electrons in atomic orbitals. Give examples. 16. Give the definition of a covalent bond and describe its types. Give examples. 17. Give the methods of forming of covalent bond. Give examples.

62

18. Explain the formation of an ionic bond, describe its properties and give examples. 19. Compare absolute and relative density. Provide ways to determine them. 20. Formulate the general patterns of change in the properties of the elements in the periodic system. Give examples. 21. At 220 ⁰C and 97 kPa the mass of 320 mL gas is equal to 3.56 g. Calculate the molecular mass of this gas. 22. What volume will take 4.8 g of nitrogen? 23. How many atoms consist in the mercury vapor molecules, if the density of mercury vapor by the air is equal to 6.92? 24. What is the relative density of the carbon dioxide (CO2) by the air? 25. Calculate the equivalent and a molar mass equivalent of the following substances: Na2O, H3PO4, Al(OH)3, FeCl2. 26. How many electrons are located at the level from the n = 3 in the Co atom? 27. How many electrons are in chlorine 3p sublevel in the Cl- ion? 28. Which set of elementary particle are responsible to the 92U238 atom? 29. Which sublevel will fill after 5p? 30. What elements begin and complete any energy level? 31. How can you determine the period in which the element is located by the electronic formula? 32. How will the reaction rate change CO + Cl2 = COCl while reducing the pressure to 4 times, and raising the temperature from 25 to 850 °C? The temperature coefficient is equal to 2.

63

7

Lesson #

SOLUTIONS. WAY OF EXPRESSING THE CONCENTRATION OF SOLUTIONS

OBJECTIVE: Consider ways of expressing of the solution composition, mass and mole fractions, molar, molal and equivalent (normal) concentrations and titre. Describe how to change concentration from one to another.

HIGHLIGHTS: 1. Chemical theory of solutions. 2. Concept of the solution and its components. 3. Liquid, solid and gaseous solutions. 4. Ways of expressing concentrations of solutions. 5. Saturated, unsaturated and supersaturated solutions. Concentrated and diluted solutions. THEORETICAL PART: Solution is homogeneous (single phase) systems with variable composition consisting two or more substances (components). According to state of aggregation solutions can be gaseous, liquid and solid. In a solution the dissolved substance is called the solute and the substance in which the solute is dissolved is called the solvent. In a solution consisting of the mixture of two liquids, the liquid present in the larger amount is generally regarded as the solvent. Depending on the particle size solutions divided to:

64

Solutions

True (size of the particles of the solute less than 1 nm) Colloidal (size of the particles of a solute lye from 1 to 100 nm)

In the true solutions (often simply referred as solutions) the dissolved substance is in atomic or molecular form, the particles of the solute cannot be seen even under the microscope and move freely in the environment of the solvent. The true solution is thermodynamically stable system and they are indefinitely stable over the time. Enthalpy and entropy factors are the driving forces of the formation of solutions. During dissolving crystals the entropy is increases (ΔS> 0), and conversely, when gases dissolve in liquids entropy always decreases (ΔS 0), and the interaction of the formed ion with the solvent molecules (ΔH < 0). It should be noted that regardless of the sign of the enthalpy upon dissolution always G  H – T  S  0 , because transition of a substance in the solution is accompanied by a significant increase in entropy as a result of striving the system for disordering. Solutions are also classified by the concentration of solutes. They can be unsaturated, saturated and supersaturated.

The concentration of the saturated solution is determined by the solubility of a substance at a given temperature. Solutions with a lower concentration are called unsaturated. 65

In practice, the most important solution prepared on the basis of liquid solvent. The most widely used inorganic solvent is water. Solutions with other solvent are called non-aqueous. The main provisions of the modern physical and chemical theory of solutions. 1. Solute and solvent chemically interact. 2. As a result of interaction formed unstable compounds called solvates, and the process is called solvation. In the particular case when water is the solvent, these compounds are called hydrates, and the process is called hydration. 3. Solutions themselves, and the majority of hydrates (solvates) which are formed by dissolving have a variable composition. By this they are different from chemical compounds. 4. Hydration phenomenon sometimes can be observed without using special devices. For example, anhydrous copper (II) sulfate (CuSO4) is white substance. When it dissolved in water formed a blue solution. Coloring of the solution is determined hydrated copper ions. Hydrated particles sometimes can be so strong that the allocation of the solute from the solution to the solid phase the molecules of water included in the crystal. Thus, upon evaporation of an aqueous solution of copper sulfate into the solid phase salt CuSO4∙5H2O is formed. 5. Crystalline substances composed of water molecules are called hydrates, and the water contained in them called as crystallizational. The quantitative composition of a solution determines by its concentration. Way of expressing Mathematical expression of concentration 1 2 1. Percentage m(solute) %   100% concentration or m(solution) mass fraction indicated by ω or ω% n(solute) mole 2. The molar CM  , concentration or V (solution) L molarity If V (solution)  1 L is denoted by CM  n(solute)

66

Calculation formula 3 m(solute) %   100% m(solute)  m(solvent)

n

m  M

CM 

m(solute) M (solute)  V (solution)

1 CM or M Examples. 1M H2SO4 is one molar solution of sulfuric acid 3. Molar concentration of equivalent of substances or normal is denoted by CN. Example. 1N NaOH – one normal solution of sodium hydroxide 4. Molar (mole) fraction is denoted by x or N

2 The molarity of a solution is used to represent the amount of moles of solute per liter of the solution n e (solute) mole-equiv , V (solution) L V (solution)  1 L CN  ne (solute) CN 

Normal concentration shows the number of moles of equivalents of solute contained in one liter of solution

3

ne 

m  Me

CN 

m(solute) M e (solute)  V (solution)

n(solute) m n x n(solute)  n(solvent) M n(solvent ) x(solvent)  n(solute)  n(solvent) x(solute) 

x(solvent)  x(solvent)  1

5. The molal concentration or molality is denoted

Cm 

n(solute) mole , m(solution) kg

If m(solvent)=1 kg , Cm  n(solute) Molal concentration shows how many moles of solute per one kilogram of solvent. 6. Titer of solution is m(solute) g T , denoted T V (solution) cm3 The titer of the solution indicates the number of grams of solute contained in 1 cm3 of solution.

Сm or m

n

m x M

Cm 

T

m(solute) M e (solute)  m(solution)

m(solute) V (solution)

Typical computation of the preparation of solutions. 1. Transfer percentage concentration into the molar concentration:   dsolution  10 CM  M where СМ – molar concentration of the solution, mol/l; W – percentage concentration, %; d s-on – solution density, g/сm3; М – solute molar mass, g/mol. 67

2. The diagonal circuit («rule of the cross»). It is used to calculate the concentration of the mixed solution to obtain the desired concentration. The scheme can be applied upon dilution of solution with known concentration with water to obtain the desired concentration. For this in the lower left corner put zero. а b

с

с–b а–с

а and b are great and a small initial concentration; с is given concentration of the solution; Ratio (с – b)/(а – с) shows in which the mass (for solutions with percent concentration) or volume (for solutions with molar concentration) ratio should be mixed initial solutions to obtain a solution with predetermined concentration. QUESTIONS AND EXERCISES FOR SELF-CHECKING: 1) Give a description of concepts solvent and solute. 2) How to transfer the percentage concentration into molar concentration and vice versa? 3) How to use a hydrometer? 4) List the steps of preparing of the solution a given concentration. 5) Describe the titration procedure. 6) To the 3 L of 10% nitric acid solution with density 1.054 g/mL was added 5 L of 2% solution of the same acid with density 1.009 g/mL. Calculate percent, molar and normal concentration of resulting solution. 7) How to prepare 150 g 2% copper sulfate solution? 8) How to prepare 750 mL of a 0.1 M (molar concentration) of a sodium sulfate solution? 9) How to prepare 500 mL of 1.5 N (normal concentration) of potassium carbonate solution? 10) How to prepare 450 g 18% cobalt sulfate solution? 11) How to prepare 150 mL of 0.01 M sodium chloride solution? 12) How to prepare 300 mL of 3 N ammonium carbonate solution? TASKS FOR PREPARATION OF ISW «Ways of expressing concentrations of solutions» 1. For aqueous solution of a substance «a», density «b» and concentration «c», calculate the deficient concentration:  mass fraction,  molar concentration,

68

 molar equivalent concentration  molal concentration  titer. 2. Calculations presente in the form of tasks. #

а

Data b

1 2 3

AgNO3 CaCl2 Na2SO4

W = 20% W = 5% W = 7%

1.19 1.007 1.016

4

(NH4)2SO4

W = 27%

1.115

5

NaNО3

W = 13%

1.067

6

NaNО3

T= 0.1067 g/ml

1.067

7

Na2CО3

w = 0.15%

1.160

8

Na2CО3

См= 1.66 М

1.160

9 10 11

CaCl2 Na2SО4 (NH4)2SO4

CM= 5.03 M CM = 0.14 М СN = 3.38 N

1.396 1.016 1.115

12

Na2CО3

Т = 0.18 g/ml

1.160

13 14

Na2CО3 H2SО4

СN = 3.33 N w = 96%

1.160 1.835

15

KOH

w = 24%

1.218

16

NaOH

w = 50%

1.525

17

H3PО4

w = 35%

1.216

18

H2SО4

СN = 12.5 N

1.664

19

HNO3

СN = 8 N

1.246

20

Fe2(SО4)3

См = 0.8М

1.000

21

HCl

w = 4%

1.018

22

Fe2(SО4)3

СN = 1.0 N

1.000

23

CuSО4

См = 0.1М

1.107

24

CuSО4

СN = 1.39 N

1.107

с, g/сm3

LABORATORY WORK Preparation of solutions with specified concentration Experiment #1. Preparing solutions of a percent concentration from solids. Certain volume of salt solution should be prepared (volume and mass fraction indicates by lecturer).

69

Weigh the amount of a given salt. Calculate how much water is required to dissolve a given sample. Measure out by graduated cylinder this volume of water and pour it into a glass with the sample. Mixture stirred until complete dissolution of the salt. Pour the prepared solution into the dry cylinder and determine the density of the solution using a hydrometer. Compare the density of the resulting solution with tabular data and calculate the absolute error. From the founded density, using the table determine the percentage content of the salt in solution. Compare the received value with a given concentration of the solution. Calculate normal and molar concentration of the obtained solution. Experiment #2. Preparing solutions of a percent concentration from concentrated solution. Prepare a solution of sodium hydroxide with certain density from a concentrated solution and water (volume and density of the solution indicated by lecturer). Determine the density of the concentrated solution by hydrometer. Find the percentage content of sodium hydroxide in the table. Calculate test portion , the volume of the concentrated solution and the volume of water necessary for the preparation of a given solution. By cylinder measure out calculated volumes of the concentrated solution and water, merge them, thoroughly mix and after cooling determine density of the obtained solution by hydrometer. Calculate the absolute measurement error. Calculate molarity and normality concentrations of resulting solution. Experiment #3. Preparing solution of molar concentration from solid substance. Prepare 200 mL of solution with given concentration from dry salt and water, (concentration indicates by lecturer). Calculate the salt quantity required for this. Weigh into a pre-weighed beaker a calculated amount of salt. Dissolve a sample in the small amount of water and transfer solution into a 200 mL volumetric flask. Wash the glass several times with water and pour the solution into the flask. Pour the solution with water until the mark. Stopper and mix thoroughly. Determine the density of the resulting solution by hydrometer. Find the mass fraction of the resulting solution and calculate normality concentration. Compare the density of the resulting solution with tabular data and calculate the absolute error. Experiment #4. Preparing solution of molar concentration from concentrated solution. Prepare 200 mL nitric acid (concentration of solution indicated by lecturer) from the given solution. Determine the density of this solution by hydrometer. Find its percent concentration. Calculate test portion and volume of the solution. Pour into a volumetric flask up to half distilled water and pour in it nitric acid (measured volume) through the funnel. Rinse acid from funnel, shake the solution, top up to the mark, and close the cap and mix.

70

Check the concentration of obtained solution by titration with 0.2 M NaOH solution (indicator is phenolphthalein). Titration should be regarded as complete as soon as the titrated solution is painted in pink color, no fading with a smooth circular rotating flask. Determine the resulting solution normality. Reference data mass share, % 0.5 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 12.0

Solution density, g/сm3 Na2CO3 1.0034 1.0086 1.0190 1.0294 1.0398 1.0502 1.0606 1.0711 1.0816 1.0922 1.1029 1.1244

NaCl 1.0018 1.0053 1.0125 1.0196 1.0268 1.0340 1.0413 1.0486 1.0559 1.0633 1.0707 1.0857

Na3PO4

Na2SO4

К2Сr2O7

1.0042 1.0100 1.0216 1.0335 1.0456 1.0579 1.0705 1.0832 1.0961 -

1.0027 1.0071 1.0161 1.0252 1.0343 1.0436 1.0526 1.0619 1.0713 1.0808 1.0905 1.1101

1.0052 1.0122 1.0193 1.0264 1.0336 1.0408 1.0481 1.0554 1.0628 1.0703 -

71

BaCl2 1.0026 1.0070 1.0159 1.0249 1.0341 1.0434 1.0528 1.0624 1.0721 1.0820 1.0921 1.1128

Lesson #

8

THEORY OF REDOX PROCESSES

OBJECTIVE: Review basics of redox processes, study types of redox reactions, characterize properties of the main oxidizing and reducing agents, to inculcate skills of compiling of redox reactions based on an electronic balance.

HIGHLIGHTS: 1. Characteristic features of redox reactions. 2. The most important oxidizing and reducing agents. 3. Types of redox reactions. 4. Method of compiling of redox reactions based on an electronic balance. THEORETICAL PART. From the viewpoint of electron theory, redox reactions are such reactions, during which transfer of electrons from one atom, molecule or ion to another occurs. Since the electrons in redox reactions pass only from reductant to oxidant, and the molecule of initial materials and products of reaction are electrically neutral, the number of electrons released by reductant is always equal to the number of electrons taken by the oxidant. This position is called the principle of electronic balance. It is the basis of finding the coefficients in the equations of redox reactions. In catalytic oxidation of ammonia NH3 by oxygen O2, nitrogen monoxide NO and water H2O are generated. Let us write this process using formulas: 72

NH3 + O2  NO + H2O. Above the symbols of elements, changing their oxidation states in the process of reaction, let us write their meanings: N-3H3 + O20N+2O-2 + H2O-2 From the changing of oxidation states, it can be seen that the nitrogen in the ammonia molecule is oxidized, and the oxygen molecule is reduced, i.e. reductant is ammonia, and oxygen is oxidant. From this scheme, it also follows, that the nitrogen atom, changing oxidation state from -3 to +2, gives to oxygen five electrons. Since the oxidation state of hydrogen does not change, then the ammonia molecule will give totally 5 electrons. The oxygen atom takes two electrons (oxidation state varies from 0 to -2), therefore, the oxygen molecule will accept 4 electrons. Let us write these processes in the form of scheme: N-3 - 5ē  N+2 O20 + 4ē  2O-2

4 oxidation – reducing agent 5 reduction – oxidant

According to the principle of electronic balance, the amount of molecules of oxidant and reductant should be so that the number of received and donated electrons are equal. For this common multiple should be found, and then it is divided by the number of given or acquired electrons; obtained coefficients are set before molecules of the reductant and the oxidant, respectively. The figure shows that four NH3 molecules give 20 electrons, which are taken by five O2 molecules. The coefficients of the electronic balance are primary. They shall not be subject to any changes: 4NH3 + 5O2 NO + H2O. The rest is equalized in accordance with their values: 4NH3 + 5O2 4NO + 6H2O. 73

There are three types of redox reactions: intermolecular, intramolecular and disproportionation reactions. 1. Intermolecular are reactions in which the molecules, atoms or ions of the elements, which are part of one substance (oxidizing agent), interact with molecules, ions, atoms which are part of another substance (reducing agent), for example: Mn+4O2 + 4HCl-1 = Mn2+Cl2 + Cl20 + 2H2O. 2. In intramolecular reactions oxidation state of elements of the same substance changes so that some of them are oxidized, and others are reduced. Such reactions include, for example, the decomposition of potassium chlorate and mercury (II) oxide: 2KCl+5O3-2 = 2KCl-1 + 3O20; 2Hg+2O-2 = 2Hg0 + O20. 3. In disproportionation reaction atoms of the same substance interact so with each other that some of them donate electrons (oxidize), while others attach them (reduce). For example, dissolution of chlorine in water: Cl20 + H2O = HCl+1O + HCl-1 or (Cl0Cl0 + H2O = HCl+1O + HCl-1). The most important oxidizing and reducing agents Among strong oxidants, which are widely used in practice, are halogens (Fe2, Cl2, Br2, I2), manganese oxide Mn+4O2, potassium permanganate KMn+7O4, potassium manganate K2Mn+6O4, chromium oxide Cr+6O3, potassium chromate K2Cr+6O4, potassium dichromate K2Cr2+6O7, nitric acid HN+5O3 and its salts, oxygen О2, ozon О3, hydrogen peroxide Н2О2, concentrated sulfuric acid Н2S+6О4, copper (II) oxide Сu+2О, silver oxide Ag2+1O, lead oxide Рb+4О2, hypochlorites (for example, NaCl-1O) and etc. Alkali and alkaline earth metals are strong reducing agents. Other reducing agents include: hydrogen, carbon, carbon monoxide С+2О, hydrogen sulfide Н2S-2, sulfur oxide S+4О2, sulfurous acid Н2S+4О3 and its salts, hydrogen halides (except HF), tin (II) chloride Sn+2Cl2, iron (II) sulfate Fe+2SO4. 74

QUESTIONS AND EXERCISES FOR SELF-CHECKING: 1. Specify a process, which is called oxidation, reduction? Give examples. 2. Specify redox reaction. 1) H2S + HI ; 2) MnSO4 + HCl ; 3) H2S + K2Cr2O7 ; 4) Fe(OH)3 + HCI ; 5) KClO4 + KMnO4. 3. Which of the chemical reactions is impossible? 1) 5Cl2 + Br2 + 6H2O = 10HCl + 2HBrO3; 2) 5J2 + Br2 + 6H2O = 10HJ + 2НBrO3; 3) Cl2 + I2 +6H2O = 10HCl + 2НIO3; 4) Cl2 + 2NaOH = NaCl + NaClO +H2O; 5) 3Br2 + 6NaOH = 5NaBr + NaBrO3 + 3H2O. 4. How many electrons are involved in the oxidation reaction: 2KNO3 + 3C + +S  K2S + 3CO2 + N2? 5. What is the value of the oxidation state of chlorine of potassium chlorate (KClO3)? 6. Which redox properties does sulfur in the oxidation state, equal to -2, show? 7. Which redox properties does manganese in the oxidation state, equal to +7, show? 8. Select the substance showing both oxidizing and reducing properties: 1) SnO2; 2) As2O3; 3) As2O5; 4) H2S; 5) Ag2O. 9. Which redox properties does sulfur in the oxidation state, equal to +6, show? 10. Specify the oxidizing agent in a redox reaction: 2NaNO2 + 2KMnO4+3H2SO4 = 5NaNO3 +2MnSO4+K2SO4+3H2O. 11. Identify a compound in which the nitrogen has the oxidation state +5. 1) NO; 2) HNO3; 3) NF3; 4) NO2; 5) N2O. TASKS FOR PREPARATION OF ISW: «Redox processes» 1. Calculate the oxidation state of the elements in the molecule, atom and ion «a». Which substance or ion of composition «a» plays the role of only oxidant, only reducing agent, or both in redox reactions? Explain.

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2. Using an electronic balance method, balance the redox reaction «b». # 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

«a» 2 KMnO4; H2SO3; HCl H2S; HNO3; Cl2 Br2; H2SO4; NH3 Cl2; H3PO4; HBr KI; SO2; PbO2 NaBr; HNO2; K2Cr2O7 HCl; KNO2; HClO4 H2SO4; Mg; Na2S HBr; Na2SO3; KMnO4 H2SO3; NaCl; K2SO4 NH3; MnO2; H3PO4 H3PO4; HNO2; Li2S KMnO4; S; NH3 H2S; K2CrO4; KCrO2 Cl2; K2Cr2O7; HI NH3; KNO2; CO2 H2SO4; Al; HBr Br2; BiCl3; SnCl2 H3PO4; Zn; KBr NH3; HNO2; KBiO3 SO2; K2S; K2SO4 HCl; KNO2; HClO4 HBr; K2Cr2O7; H2 Br2; H2O2; HI KI; HClO2; Mg(NO3)2 H2SO3; LiCl; NaMnO4 H2SO4; HClO; LiI Cl2; LiMnO4; MgCl2 KI; HNO2; H3PO4

31

KMnO4; HClO3; NaBr HBr; KNO2; K2SO4

32

Br2; HBrO4; CsCl

«b» 3 KMnO4+HCl  MnCl2+KCl+H2O+Cl2 K2Cr2O7+H2S+H2SO4  Cr2(SO4)3+S+K2SO4+H2O Cl2+KOH  KClO3+KCl+H2O I2+Cl2+H2O  HIO3+HCl KMnO4+H2O2  MnO2+KOH+O2+H2O Cu+HNO3  Cu(NO3)2+NO+H2O KMnO7 + HCl + H2O2 = KCl + MnCl2 +O2 + H2O Cl2+KOH  KClO3+KCl+H2O Mg + HNO3  Mg(NO3)2 +NH4NO3 Cu+HNO3  Cu(NO3)2+NO+H2O Cr2O3 + KNO3 + K2CO3  K2 CrO4+ KNO2 + CO2 H2S + H2SO3  S + H2O Сl2 + NaOH  NaCl + NaClO + H2O K2Cr2O7 + HCl  CrCl3 + Cl2 + H2O + KCl P +HIO3 +H2O  H3 PO3 + HI K2Cr2O7+H2S+H2SO4  Cr2(SO4)3+S+K2SO4+H2O Cr2O3 + KNO3 + K2CO3  K2 CrO4+ KNO2 + CO2 H2S + H2SO3  S + H2O KMnO4+NaNO2+H2SO4  MnSO4+K2SO4+ NaNO3 Сl2 + NaOH  NaCl + NaClO + H2O Cl2 + Br2 + H2O = HCl + HBrO3 KNO3 + C + S  K2S + CO2 + N2 Pb+HNO3 = Pb(NO3)2+NO2+H2O NaNO2 + NaI + H2SO4 = NO + I2 + Na2SO4 Cl2 + KOH = KOCl + KCl + H2O P + KOH + H2O = PH3 + KH2PO4 I2 + KOH = KIO3 + KI + H2O KMnO4+ KOH = K2MnO4 + O2 + H2O KMnO7 + HNO3 + H2O2 = = KNO3 + Mn(NO3)2 +O2 + H2O H2S + H2SO3  S + H2O MnSO4+PbO2 + HNO3   HMnO4+ Pb(NO3)2 + PbSO4 +H2O K2Cr2O7+H2O2 + H2SO4   Cr2(SO4)3+O2+H2O + K2SO4

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1 33 34 35 36 37 38 39 40 41

2 KI; K2Cr2O7; CrCl3 H2SO3; HI; KMnO4 SO2; HIO4; KBr NH3; PbO2; HBrO2 Br2; HClO4; HI H3PO4; NaNO3; HBrO3 PbO2; HNO2; H2CO3 HCl; HIO3; LiMnO4 PbO2; HClO4; HIO2

3 HI + H2SO4 = I2 + H2S + H2O KMnO4+H2O2  MnO2+KOH+O2+H2O KNO3 + C + S  K2S + CO2 + N2 KMnO4+K2SO3+H2SO4  MnSO4+K2SO4+H2O K2Cr2O7+H2S+HCl  CrCl3+S+KCl+H2O KMnO4+K2SO3+H2SO4  MnSO4+K2SO4+H2O I2+H2O2  HIO3+H2О Zn+HNO3  N2O + Zn(NO3)2 +H2O KMnO4+NaNO2+H2SO4   MnSO4+K2SO4+ NaNO3

LABORATORY WORK The study of redox processes Experiment #1. Reduction properties of metals. Interaction of zinc with lead acetate. Put 1-2 pieces of metallic zinc; add 2-3 ml of lead acetate (CH3COO)2Pb. What is happening? Write the reaction equation. Interaction of metals with dilute acids. In four tubes pour 2-3 ml of 2N hydrochloric or sulfuric acid, drop metal pieces of magnesium, zinc, iron, and copper therein. Test obtaing gas by a burning torch. Write reaction equations and specify the oxidant and reductant. Experiment #2. Reduction properties of complex substances containing atoms with the lowest negative oxidation state. Pour 2-3 ml of ferric (III) chloride in tube and add 1-2 drops of a solution of potassium iodide. The contents of tube dilute with water until slightly yellow color, add 1-2 drops of freshly prepared starch solution. Write the reaction equation, make electronic equations, and indicate oxidizing and reducing agents. Experiment #3. Oxidation properties of complex substances containing atoms with the highest oxidation state. The action of dilute nitric acid on metals. In two tubes pour per 2-3 ml of nitric acid solution and put the metal pieces of magnesium and zinc, and heat (not strongly!). Record the observed phenomena. Write the equation of reaction. The action of concentrated nitric acid on metals. In tube place a piece of copper and tide 1-2 ml of concentrated nitric acid. What is happening? Write the reaction equation; make electronic equations, indicate oxidizing and reducing agents. Oxidizing properties of potassium dichromate. Pour 1-2 ml of potassium dichromate solution and 1-2 drops of sulfuric acid in tube. Add a few crystals of potassium sulfate in a solution. Record the observed phenomena. Write the equation of reaction; make electronic equations, indicate oxidizing and reducing agents.

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Experiment #4. Oxidation properties of potassium permanganate. а) In three tubes pour 2-3 ml of the diluted solution of KMnO4. In the first test tube add a few drops of sulfuric acid solution, in the second - distilled water, in the third - solution of sodium hydroxide. Into each tube fill a few crystals of sodium sulphate until changing of color. Write the corresponding reaction equations, considering that MnO4- ion has a violet color, MnO42- is green, Mn2+ is colorless, and MnO2 is slightly soluble precipitate of black and mud color. Make a conclusion about the impact of media on the reduction of potassium permanganate. b) In tube pour 2-3 ml of a solution of potassium permanganate, add a few drops of sulfuric acid, followed by the dropwise addition of a solution of hydrogen peroxide until bleaching. Record the observed phenomena. Write the equation of reaction; make electronic equations, indicate oxidizing and reducing agents.

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Lesson #

9

ELECTROLYTIC DISSOCIATION OF VARIOUS CLASSES OF INORGANIC COMPOUNDS. IONIC EQUATIONS OBJECTIVE: To study the properties of acids, bases and salts from the standpoint of the theory of electrolytic dissociation; dissociation mechanism of substances with different types of chemical bonds. To extend the conception of substances and chemical reactions on the basis of knowledge of the electrical conductivity

HIGHLIGHTS: 1. The basic concepts of the theory of electrolytic dissociation. 2. The mechanism of electrolytic dissociation. 3. Acids and bases in terms of electrolytic dissociation; their classification according to various criteria. 4. Positive and negative, simple or complex, hydrated and nonhydrated ions. 5. Ion exchange reactions, various types of ion equations molecular reaction equations, shortened ionic equation of reactions, neutralization reaction. THEORETICAL PART. Background of the theory For electrolytes increasing of boiling points, decreasing of freezing points, relative lowering of saturated vapor pressure, osmotic pressure are significantly larger than the theoretical values.

, where: i is empirical coefficient (i> 1), introduced by van't Hoff. 79

(9.1)

The main positions of the Arrhenius theory: 1. Salts, acids and bases dissociate into ions by dissolving in water and other polar solvents. 2. Ions exist in solution regardless of an electric current passes through the solution or not. As a result, the number of independently moving solute particles is greater than in the absence of dissociation, and colligative properties of electrolyte increases in the i times. 3. The dissociation process is described by the law of mass action (proceeds reversibly). When the concentration decreases, dissociation is practically complete. z+

z-

K   A   ↔ ν+ K + ν – A

(9.2) 4. Van't Hoff factor (i) is related to the degree of electrolytic dissociation. i is mean cumulative number of particles (ions and molecules) obtained by the dissociation of a single molecule. i = ν + α + ν- α + (1-α) = 1 + (ν++ν–-1)α = 1 + (ν – 1)α i = 1+(ν-1)α

(9.3)

With increasing of dilution van't Hoff factor approximates to a simple integer (2, 3, 4, depending on the number of ions formed by dissolving of the molecules of the substance): NaCl = Na+ + Cl+

i→2

SO 24 

K2SO4 = 2K + AlCl3 = Al3+ + 3ClK4[Fe(CN)6] = 4K+ + [Fe(CN)6]2-

i→3 i→4 i→5

The theory of electrolytic dissociation. Aqueous solutions of salts, acids and bases conduct electricity. Substances conducting electrical current are called electrolytes. The theory of electrolytic dissociation was developed by Svante Arrhenius. When dissolved in water or other solvents consisting of a polar molecules, electrolytes are subjected to electrolytic dissociation, i.e. 80

more or less disintegrate into positively and negatively charged ions: cations (e.g., H+, Na+, Ca2+, etc.) and anions (SO4 2-, Cl- , OH-). During passing of ions into the solution, energy of electrolytic interaction of ions in the lattice is opposed to the energy of interaction of ions with dipole molecules of the solvent, which pulls ions into the solution. The ions are surrounded by solvent molecules, which form solvation shell around ions. If the energy of the interaction of ions with the solvent becomes comparable with the energy of the ions, oscillating around the equilibrium state in the crystal lattice, dissolution with dissociation occurs. Interaction with dipole molecules of solvent with elements of lattice can lead to the formation of the electrolyte even during dissolution of substances having a molecular lattice, intermediate type lattice, or in gaseous state. The process of electrolytic dissociation is represented, using chemical equations. For example, the dissociation of HCl is expressed by the following equation: HCl = H + + Cl –. Similarly, a very dilute solution of barium chloride dissociates according to the equation: BaCl2 = Ba2+ + 2Cl –. The essential is the question of the mechanism of electrolytic dissociation. The substances with ionic bonds dissociate easiest. As it is known, these substances consist of ions. During their dissolution, water dipoles are oriented around the positive and negative ions. Between ions and water dipoles forces of mutual attraction appear. As a result, bond between the ions is weakened, there is a transition of the ions from the crystal to the solution. This produces hydrated ions, i.e. ions chemically bound to the molecules of water. Electrolytes, the molecules of which are formed by the type of polar covalent bonds (polar molecules) dissociate similarly. Around each of polar molecules of substance are also oriented dipoles of 81

water, which by their negative poles are attracted to a positive pole of molecule, and positive poles are attracted to the negative pole. As a result of this interaction, bonding electron cloud (electron pair) is fully shifted to the atom with greater electronegativity; polar molecule is converted into the ionic form and then easily hydrated ions are formed.

Dissociation of polar molecules can be full or partial. Disadvantages of the Arrhenius theory. In the theory of electrolytes the question of the distribution of ions in the solution is very important. The distribution of ions is determined by the ratio of energy of the electrolytic interaction and energy of chaotic (thermal) motion of the ions. It appears that these energies are comparable in magnitude, so the actual distribution of the ions in the electrolyte is intermediate between the ordered and disordered. Arrhenius suggested that the interaction of the ions in solution does not affect the distribution and movement of which are chaotic, as in mixtures of ideal gases. Arrhenius argued that the properties of individual ions does not depend on the concentration and some properties of the solution as a whole (such as electrical conductivity) are proportional to number of ions. According to this concept, the ion mobility should not depend on the concentration of the solution, and the strong electrolyte conductivity should increase with increasing concentration of the solution. This contradicts the experimental data. According to the Arrhenius theory, the degree of dissociation of strong electrolytes α is equal to 0.75 – 0.95. Calculated from this KD depend on the concentration, i.e. are not constants. In fact, the degree of dissociation of strong electrolytes α → 1, where they are ireversibly dissociate. It was established experimentally that there is no undissociated molecules in solutions of strong electrolytes. Thus, Arrhe82

nius theory did not include the division into strong and weak electrolytes. Dissociation of acids, bases and salts. 1. all acids at dissociation form hydrogen cations and anions of acidic residues; polybasic acid dissociate stepwise: H2SO4 = H+ + HSO4 – (the 1st step) HSO4 - = H + + SO4 2- (the 2nd step) __________________________________ H2SO4 = 2 H + + SO4 2- (summarized equation) 2. all bases at dissociation form metal cation and hydroxideanions; polybasic bases dissociate stepwise: Ca(ОН)2↔Са(ОН)+ + OH- (the 1st step) Ca(OH)+ ↔Ca2++OH- (the 2nd step) _________________________________________ Ca(ОН)2 = Ca2++ 2OH- (summarized equation) 3. all salts form cations of metals and anions of acid residue during dissociation. Medium salt do not dissociate stepwise. Acidic and basic salts dissociate stepwise: (NH4)2SO4 ↔ 2NH+4 + SO42Na3PO4 ↔ 3Na+ + PO43KHSO4 ↔ K+ + HSO4HSO-4 ↔Н+ + SO42In the development of the doctrine of the solutions, the outstanding role belongs to the works of D.I.Mendeleyev. As a result of careful study of the properties of aqueous solutions of sulfuric acid and ethanol and other systems Mendeleyev found that between the molecules of the solution components occurs chemical reaction and that it plays significant role for the properties of solutions. Electrolytic dissociation is a reversible chemical reaction or dynamic equilibrium, to which the same laws as to any other chemical equilibrium are applicable. For example, to describe the process of dissociation equilibrium constant K can be used; in this case, it is called a dissociation constant KD: 83

(9.4) Thus, the dissociation constant Kd is a particular case of the equilibrium constant. There is a lot of H+ and Cl- ions in the solution (their concentrations are in the numerator); and the amount of undissociated molecules [HCl] is extremely small. Therefore, the values of Kd for strong electrolytes are very large and can not be measured directly. These values are obtained indirectly from thermodynamic data. In contrast, Kd of hydrogen fluoride is much less than one. Consequently, there is a lot of undissociated HF molecules and few H+ and F- ions in the solution. Both solutions (HCl and HF) are solutions of electrolytes, but HCl is a strong electrolyte, and HF is a weak. The example of these two substances shows that a good (or bad) dissociation in the solution, as well as a good (or bad) solubility is difficult to predict just by formula of substance. Besides the dissociation constant, a measure of power of electrolyte may be the degree of dissociation α. This is the ratio of the dissociated molecular (n1) to the total number of molecules (n0), initially appeared in the solution: (9.5) If we are not talking about the molecular, but about ionic compounds, the term «molecule» should be replaced by the term «formula unit» (for example, NaCl). For example, in experiment it was established that when acetic acid CH3COOH dissolves in water, only one of every 100 molecules dissociates into ions H+ and CH3COO-. CH3COOH

H+ + CH3COO-

This means that the degree of dissociation of acetic acid in the solution is about 1/100 or 0.01. It is possible to express α in percentage (1%). It follows from the above that the degree of electrolytic dissociation (dissociation depth) depends on: 84

type of crystal lattice of the solute; dielectric permittivity of solvent; concentration of electrolyte; temperature (dissociation degree in endothermic process increases with increasing of temperature). Based on the degree of dissociation is convenient to divide all the electrolytes as follows:  strong electrolytes (α is close to 1 or 100%)  weak electrolytes (α is about 0.01 or 1% or even less). Strong electrolytes are electrolytes, which completely dissociate into ions when dissolved in water. Among them are: Acids: HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4, HMnO4, H2SeO4 Bases: NaOH, KOH, LiOH, RbOH, CsOH, Ba(OH)2, Ca(OH)2, Sr(OH)2 Salts: watersoluble Weak electrolytes are electrolytes which partially dissociate into ions when dissolved in water. Among them are water, mineral acids (HNO2, HCN, H2S, H2SiO3, H2CO3 etc.), almost all organic acids (СН3СООН, HCOOR, H2C2O4 etc.), ammonium hydroxide (NH4OH), and all metal bases other than bases of alkali and alkaline earth metals. Such division is conditional, but it is very convenient to write ionic equations of chemical reactions: only strong electrolytes can be written in ionic form and the weak electrolytes cannot be. Since water is involved in the dissociation process, it is also possible to include it in the chemical equations. For example, the process of dissociation of acetic acid in water may be written as follows:    

CH3COOH + H2O

H3O+ + CH3COOhydronium ion

Such equations underline that the ions in the aqueous solution are not isolated but are connected with solvent molecules. For H+ ion interaction is very characteristic: because of its binding with the lone electron pair of the oxygen atoms H2O molecule form a sufficiently strong complex H2O-H+ or, as it is conventionally called, hydronium ion H3O+. 85

Since the principle of Le Chatelier is applicable to the dissociation process, it is clear from the equation that if dilute solution of acetic acid, i.e., add water to it, it should intensify the process associated with the expenditure of water, i.e., direct reaction (ions obtaining). In other words, dilution of solutions of weak electrolytes increases the degree of dissociation. If the electrolyte is strong (α = 1), the concentration of each ion is equal to the initial concentration of the electrolyte. For example, 1 liter of 0.1 M solution of NaCl contains 0.1 moles of Na + ions and 0.1 moles of Cl- ions. Let us suppose there is a solution of hydrogen cyanide HCN (hydrocyanic acid) with a concentration C mol/l. In solution hydrogen cyanide weakly dissociates into ions: HCN ↔ H+ + CN– In expression of the degree of dissociation α value n0 represents the number of molecules (or moles) of HCN, which originally apperared in the solution. In other words, n0 = C mol/l. n1 is the number of molecules (or moles) of HCN, dissociated into ions. Consequently, n1 = [H+] and n1 = [CN-]. In this case: (9.6) Hence,

[H+] = Cα, [CN–] = Cα

(9.7)

Thus, to obtain the concentration of any of the ions in solution, it is enough the total concentration of material C multiply by α (its degree of dissociation). If we now write the expression for the dissociation constant of HCN, then KD can be expressed through Сα: (9.8) The concentration of undissociated molecules [HCN] should be less than the initial concentration in equilibrium on the value, equal 86

to Сα, as a result the expression (С – Сα) is in the denominator of the fraction. If you take out the concentration C in brackets and reduce, we get the following expression: (9.9) Or, in another record: (9.10) This formula is a mathematical formulation of the law of dilution of Ostwald. Ostwald's law is valid for binary electrolytes, i.e., substances whose molecules in solution are divided into two ions (cation and anion). In the case of strong electrolytes when α is close to 1, the denominator tends to zero, and the dissociation constant KD tends to infinity. However, in the case of weak electrolytes value (1-α) differs very little from unity. Indeed, if α = 0,01, then (1-0,01) = 0.99, i.e. almost 1. Thus, for weak electrolytes law of dilution can be written in a simpler form: (9.11) According to the theory of electrolytic dissociation all reactions in aqueous solutions of electrolytes are reactions between the ions. They are called ionic reactions and equations of these reactions are ionic equations. These equations are easier than reactions recorded in the molecular form, and have more general nature. When preparing ionic equations of the reactions, you should be guided by the fact that little-dissociated substances, slightly soluble (precipitates) and gases are recorded in molecular form. Sign ↓, standing near formula of substance indicates that the substance goes away from the reaction as a precipitate; sign ↑ indicates that the substance is removed from the reaction as a gas. Strong electrolytes, as completely dissociated, are recorded in the form of ions. Rules for writing of equations of reactions in ionic form Write down the formulas of substances, entered into reaction, put the sign «equal» and write formulas of obtaining substances. Arranges the coefficients. 87

Using the table of solubility, write the soluble substances in ionic form. Insoluble substances, such as metals, metal and nonmetal oxides, water, gases, you should write in molecular form. The complete ionic equation is obtained. Reduce same ions before the «equal» sign and after it in the equation. The net ionic equation is obtained. The conditions under which the ion-exchange reactions take place till the end 1. If during the reaction weak dissociating substance, e.g. water is released. The molecular equation for the reaction of alkali with an acid: KOH + HCl = KCl + H2O Invariability of oxidation states of elements in all substances before and after reaction indicates that exchange reactions are not redox. Complete ionic equation of reaction: K+ + OH– + H+ + Cl– = K+ + Cl– + H2O. Net ionic equation of reaction: H+ + OH– = H2O. Molecular reaction equation of basic oxide with acid: CaO + 2HNO3 = Ca(NO3)2 + H2O. Complete ionic equation of reaction: CaO + 2H+ + 2NO-3 = Ca2+ + 2NO-3 +H2O. Net ionic equation of reaction: CaO + 2H+ = Ca2+ + H2O. 88

Molecular reaction equation of insoluble base with acid: 3Mg(OH)2 + 2H3PO4 = Mg3(PO4)2↓ + 6H2O. Complete ionic equation of reaction: 3Mg(OH)2 + 6H+ + 2PO43- = Mg3(PO4)2 ↓ + 6 H2O. In this case, the complete ionic equation is identical with net ionic equation. Molecular reaction equation of amphoteric oxide with acid: Al2O3 + 6HCl = 2AlCl3 + 3H2O. Complete ionic equation of reaction: Al2O3 + 6H+ + 6Cl– = 2Al3+ + 6Cl– + 3H2O. Net ionic equation of reaction: Al2O3 + 6H+ = 2Al3+ + 3H2O. 2. If as a result of reaction a water-insoluble substance is produced. Molecular reaction equation of soluble salt with alkali: CuCl2 + 2KOH = 2KCl + Cu(OH)2↓. Complete ionic equation of reaction: Cu2+ + 2Cl– + 2K+ + 2OH– = 2K+ + 2Cl– + Cu(OH)2↓. Net ionic equation of reaction: Cu2+ + 2OH– = Cu(OH)2↓. Molecular reaction equation of two soluble salts: 89

Al2(SO4)3 + 3BaCl2 = 3BaSO4↓ + 2AlCl3. Complete ionic equation of reaction: 2Al3+ + 3SO42- +3Ba2+ + 6Cl- = 3BaSO4↓ + 2Al3+ + 6Cl-. Net ionic equation of reaction: 3SO42- + 3Ba2+ = 3BaSO4↓, SO42- + Ba2+ = BaSO4↓. Molecular reaction equation of insoluble base with acid Fe(OH)3 + H3PO4 = FePO4↓ + 3H2O. Complete ionic equation of reaction: Fe(OH)3 + 3H+ + PO43- = FePO4↓ + 3H2O. In this case, complete ionic equation coincides with net. This reaction proceeds till the end, as indicated by at once two facts: formation of an insoluble in water matter and releasing of water. 3. If the gaseous substance is released as a result of reaction. Molecular reaction equation of soluble salts (sulfide) with acid: K2S + 2HCl = 2KCl + H2S↑. Complete ionic equation of reaction: 2K+ + S2– + 2H+ + 2Cl– = 2K+ + 2Cl– + H2S↑. Net ionic equation of reaction: S2– + 2H+ = H2S↑. Molecular reaction equation of soluble salts (carbonate) with acid: Na2CO3 + 2HNO3 = 2NaNO3 + H2O + CO2↑ 90

Complete ionic equation of reaction: 2Na+ + CO32- + 2H+ + 2NO3- = 2Na+ + 2NO3- + H2O + CO2↑ Net ionic equation of reaction: CO32- + 2H+ = H2O + CO2 ↑ This reaction proceeds till the end, as indicated by at once two facts: formation of water and gas – carbon (IV) oxide. Molecular reaction equation of insoluble salts (carbonate) with acid: 3СaCO3 + 2H3РO4 = Са3(PO4)2↓ + 3H2O + 3CO2↑ Complete ionic equation of reaction: 3СaCO3 + 6Н+ + 2РO43- = Са3(PO4)2↓ + 3H2O + 3CO2↑ In this case, complete ionic equation coincides with net. This reaction proceeds till the end, as indicated by at once three facts: formation of an insoluble in water matter, releasing of water and gas. QUESTIONS AND EXERCISES FOR SELF-CHECKING: 1. Which of the following reactions goes till the end? 1) hydrochloric acid + silver nitrate 2) sulphuric acid + sodium chloride 3) sodium sulfate +barium hydroxide 4) sodium chloride + potassium hydroxide 2. Write the dissociation reactions of calcium hydroxide, iron (III) sulphate, sulfuric acid, magnesium chloride. 3. For the following net ionic equations write per 3 molecular and complete ionic equations: H+ + OН- = H2O 2+ Ba + SO42- = BaSO4 2H+ + SO32- = SO2 + H2O Zn2+ + S2- = ZnS 2+ Mg + 2OH- = Mg(OH)2 4. Can the following pairs of substances be at the same time in the solution? 1) NaOH and P2O5

91

2) 3) 4) 5) 6)

Ba(OH)2 and CO2 KOH and NaOH NaHSO4 and BaCl2 HCl and Al(NO3)3 KOH and NaOH

TASKS FOR PREPARATION OF ISW: «Solutions of electrolytes» 1. Which class of inorganic compounds does «a» substances correspond to? Write for these substances equation of electrolytic dissociation and mathematical expression of constant of the electrolytic dissociation KD. For which substance KD has a meaning? 2. For the net ion-molecular exchange reaction «b» choose three molecular equation. Confirm your answer by writing of these equations in ionic form. 3. Which medium and pH value have aqueous solutions of salts «c»? Write a hydrolysis equation in molecular and ionic forms. # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

«а» HNO3; NH4OH H2CO3; Ba(OH)2 HCl; HCN H2SO4; NaOH H3PO4; KCN HNO3; KHCO3 H2SiO3; Sr(OH)2 CH3COOH; NaCl HF; K2S HCN; RbOH H2S; Na3PO4 HBr; KHS HNO2; KNO2 HClO; KClO HI; H2S K2SO3; H2CO3 KNO3; NH4OH KCl; K3PO4 FeSO4; HF Pb(NO3)2; HCNS KMnO4; HClO2 Mg(OH)2; KHSO4 KOH; MgCl2 H3AsO3; FeCl3 H2SiO3; Al2(SO4)3

«b» Cu2+ + 2OH- = Cu(OH)2 CO32- + 2H+ = H2CO3 Ca2+ + CO32- = CaCO3 Ni2+ + S2- = NiS Pb2+ + SO42- = PbSO4 Ba2+ + SO42- = BaSO4 Ag+ + Cl- = AgCl H+ + OH- = H2O NH4+ + OH- =NH4OH Zn2+ + 2OH- = Zn(OH)2 Mg2+ + 2OH- = Mg(OH)2 Fe3+ + 3OH- = Fe(OH)3 SO32- + 2H+ = H2O + SO2 2H+ + SiO32- = H2SiO3 CO32- + 2H+ = H2O + CO2 Zn2+ + CO32- = ZnCO3 Pb2+ + 2OH- = Pb(OH)2 3Ba2+ + 2PO43- = Ba3(PO4)2 Cr3+ + 3OH- = Cr(OH)3 Ag+ + I- = AgI H+ + OH- = H2O Fe2+ + 2OH- = Fe(OH)2 Ba2+ + CO32- = BaCO3 Ca2+ + SO32- = CaSO3 Mn2+ + 2OH- = Mn(OH)2

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«c» Na2S; NH4Cl K2CO3; NH4NO3 MgCl2; NaNO2 FeCl3; KNO2 Zn(NO)2; KCN ZnCl2: NaCN K2SO3; NH4Cl K3PO4, NH4NO3 FeCl2, NaCN MgCl2; NaCNS Na3PO4, NaNO2 KCNS; Fe(NO3)2 CuCl2, CH3COOK CuCl2, CH3COOK Na2CO3 , NH4Cl FeCl2, NH4Cl K2SO3 , KCN KCN , CrCl3 К2CO3 , NH4NO3 MgCl2, NH4NO3 MgSO4 , LiCN ZnCl2, КCN KNO2 , ZnCl2 FeCl3 , NH4Cl Na2S , NH4NO3

LABORATORY WORK Studying of the properties of electrolyte solutions Experiment #1. The electrical conductivity of solutions of acids, bases, salts. а) For the experiments a device consisting of a cup, two electrodes, and an electric bulb is used. A sign of good electrical conductivity is a bulb ignition. Determine the conductivity of the following substances: distilled water, sodium chloride (dry), sodium chloride solution, sugar (dry), sugar solution, hydrochloric acid solution, sodium hydroxide solution, cupric chloride solution, ammonium hydroxide, concentrated acetic acid, acetic acid solution, etc. Write observations in the following table. test substance

intensity of luminescence of light bulb

strength of electrolyte

dissociation equation

Write corresponding conclusions and equations of dissociation of compoundselectrolytes. b) In one glass pour acetic acid solution, in another pour ammonia solution. Determine the conductivity of these solutions. Pour both solutions into one glass. How does the intensity of the light bulb luminescence? Explain the observed phenomenon. Write the equation of chemical reaction. Experiment #2. The dependence of electrical conductivity (degree of dissociation) on nature of electrolyte. Pour into cups the following solutions H2SO4, H3PO4, Na2SO4, NaOH with the same concentration, check their electrical conductivity, which can be seen by (relatively) brightness of luminescence of bulb. Describe the results of observations and discuss results of experiment. Experiment #3. The dependence of electrical conductivity (degree of dissociation) on electrolyte concentration. Prepare three glasses with weak electrolyte solutions (CH3COOН, NH4OН etc.) and three glasses with strong electrolyte solutions (H2SO4, NaOH, NaCl, etc.) with different concentrations. Check their electrical conductivity. Describe the results of observations and discuss the results of the experiment. Experiment #4. The dependence of electrical conductivity (degree of dissociation) on temperature of electrolyte solution Prepare two glasses with weak electrolyte solutions (CH3COOН, NH4OН etc.) and two glasses with strong electrolyte solutions (H2SO4, NaOH, NaCl, etc.) with same concentrations. Per one glass with a solution of each of the electrolyte heat up to 60-70 °C. Check their electrical conductivity. Describe the results of observations and discuss the results of the experiment.

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Experiment #5. Ionic interactions in electrolyte solutions. а) The formation of poorly soluble substances. In three tubes add per 2-3 ml of the following solutions: first – iron (III) chloride, in the second – silver nitrate, in the third – barium chloride. Add to them by the same amount: in the first – sodium hydroxide, in the second – hydrochloric acid solution, in the third – sulfuric acid solution. Describe the observed phenomena. Write the reactions in molecular and ionic forms. b) The formation of weak acids and bases. In two test tubes add per 2-3 ml of the following solutions: in the first - sodium acetate solution, in the second - ammonium chloride. Add in the first tube few drops of sulfuric acid (1:1), mixed with a glass rod and a slightly warm solution. Determine what acid was formed by smell. Write the equation of reaction in the molecular and ionic forms. In the second tube add 4N alkali solution, and heat it. Which gas is released? Write the equation of reaction in the molecular and ionic forms. c) Neutralization reaction. Pour into two test tubes per 2-3 ml of sodium hydroxide solution and add phenolphthalein. Under the influence of which ions did solution become colored? In one tube add a solution of sulfuric acid, in the second acetic acid solution. What explains the disappearance of the painting? In which case did the discoloration of solution occur faster? Write the reactions in the molecular and ionic forms. d) Formation of volatile products of reaction Put in two test tubes per 2-3 ml of sodium carbonate solution. Check presence of carbonate ions in solution, adding a few drops of calcium chloride in one tube. What substance precipitated? Write equation of reaction in molecular and ionic forms. In the second tube add a few drops of sulfuric acid (1:1) and observe releasing of gas. Heat the tube slightly, wait until the end of releasing of gas and add a few drops of calcium chloride solution. Why is there no precipitation? Write the equation of reaction in the molecular and ionic forms. e) Experimental problem. Using solutions of reagents, which are available in laboratory, realize the reactions expressed by the following ion-molecular equations: Pb2+ + 2I- = PbI2 Fe3+ + 3OH- = Fe(OH)3 SO32- + 2H+ = H2O + SO2 Write the equation of reaction in the molecular form.

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LESSON # PH.

10

HYDROLYSIS OF SALTS

OBJECTIVE: to study ionic product of water, pH value, the most important characteristics of solutions, the transition interval of indicators; to understand the basic concepts (hydrolysis, salts classification according to acid and base strength them formed, types of hydrolysis).

HIGHLIGHTS: 1. The ionic product of water. 2. The hydrogen and hydroxyl indicators. 3. The indicators, their changing of color depending on pH. 4. Hydrolysis of the salt formed by a strong base and a weak acid. 5. Hydrolysis of the salt formed by a weak base and a strong acid. 6. Reversible hydrolysis. 7. Equations of hydrolysis reactions in the molecular, total and abbreviated form, to predict and explain the change in solution medium, the formation of acidic and basic salts. THEORETICAL PART. The ionic product of water. pH value Water is very weak electrolyte, because it dissociates into ions in a very small degree: Н2О → Н+ + ОН-. Ions OH- and H + are presented in the aqueous solution of any inorganic compound. The constant of water dissociation is as follows: 95

  KD  [ H ][OH ]

[ H 2 O]

or

KD·(Н2О)=[H+][OH-]

(10.1)

KD of water at 22°С is equal to 1.8·10-16 [Н2O] = 1000/18 = 55.56 mol/l, hence [Н+]·[ОН-] = 1.8·10-16·55.56 = 10-14 mol/l. Therefore, in water, as well as in any aqueous solution, the product of concentration of ions H+ and OH- is constant (at constant temperature), which is called the ion product of water and denoted as K W. KW = [Н+]·[ОН-] = 10-14 (10.2) For pure water the hydrogen ion concentration is equal to the concentration of hydroxide ion, as from one mole of water per one mole of H+ and one mole of OH- ions are formed. Consequently, the concentration of these ions at 22 ° C is equal to: [Н+] = [ОН-] =

= 10-7 mol/l.

(10.3)

If acid add to clean water, then [Н+] is equal to > 10-7 mol/l, and [ОН ] < 10-7 mol/l. In opposite, if the alkali add to the water, then [Н+] is equal to < 10-7 mol/l, and [ОН-] > 10-7 mol/l. However, regardless of changes of [H +] and [OH-], their product at 22 °C will always be equal to 10-14. To avoid the inconveniences associated with the usage of numbers with negative exponents, the hydrogen ion concentration is usually expressed by pH value and the concentration of hydroxide ions – through the pOH value. -

рН + рОН = 14

(10.4)

Thus, knowing the pH value, you can calculate pOH and vice versa: рН = 14 – рОН or рОН = 14 – рН (10.5) Hydrogen index pH characterizes the concentration (activity) of hydrogen ions in aqueous solutions. It is numerically equal to the 96

negative decimal logarithm of the concentration (activity) of hydrogen ions, [H+], expressed in moles per liter: pH = - lg[H+] and pH = - lg c(H+). In neutral solutions pH=7, in acidic 7. The relationship between the hydrogen ion concentration and pH value of hydroxide ions is shown in the following table: СН+ 1·100 1·10-1 1·10-2 1·10-3 1·10-4 1·10-5 1·10-6 1·10-7 1·10-8 1·10-9 1·11-10 1·10-11 1·10-12 1·10-13 1·10-14

рН 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

СОН1·10-14 1·10-13 1·10-12 1·10-11 1·11-10 1·10-9 1·10-8 1·10-7 1·10-6 1·10-5 1·10-4 1·10-3 1·10-2 1·10-1 1·100

Example battery acid sulfuric acid lemon juice, vinegar orange juice, red wine tomato juice coffee, banana milk, urine fresh water sea water baking soda emulsion of magnesia, solution ammonia soap water bleach (NaClO solution) alkali (NaOH solution)

The acidity or alkalinity medium is determined by acid-base indicator that changes its color depending on the pH of the medium. Litmus Phenolphtalein Methyl orange Universal indicator, unlike acid-base indicator changes its color over a wide pH range, namely from 1 to 14. Hydrolysis of salts. Hydrolysis of salts is a result of the polarization interaction of salt ions with polar water molecules, leading to the formation of littledissociated compound and is accompanied by a change in pH of the aqueous solution environment. Hydrolysis shifts the equilibrium of dissociation of water due to the binding of one of its ions (H+ or OH-) in a weak electrolyte salt. 97

When binding of H+ ions OH- ions accumulated in the solution, the reaction medium is alkaline, and when binding ions OH- and H+ ions accumulate, environment is acidic. A salt formed by a strong base and a strong acid does not undergo hydrolysis. A salt formed with a strong base and a weak acid (hydrolysis proceeds by the anion). This occurs during the hydrolysis of KNO2 salt. Ions of the salt NO2- and K+ react with H+ and OH- from the water. In this case nitrite ions (NO2-) are associated with the hydrogen ions (H+) in the molecule of the weak electrolyte – nitrous acid (HNO2), and OH- ions are accumulated in solution, giving it an alkaline reaction, since the ions K+ ions cannot bind OH- (KOH is strong electrolyte), pH> 7. KNO2 + H2O ↔ KОН + СН3ООН K+ + NO2− + НОН ↔ K+ + ОН− + HNO2 NO2− + НОН ↔ ОН− + HNO2

molecular equation complete ionic equation net ionic equation

The hydrolysis of salt K2S occurs stepwise. The salt formed with a strong base and a weak diacid. In this case the salt anion S 2- binds ions H+ of water; OH- ions are accumulated in solution. The equation in ionic and molecular form for the first stage has the following form the 1st step S2− + НОН ↔ HS− + ОН− К2S + Н2О ↔ NaHS + NaOH

net ionic equation molecular equation

The second step of hydrolysis takes place practically under normal conditions, since, accumulating, OH- ions reporting solution strongly alkaline reaction, which leads to the neutralization reaction equilibrium shift to the left. the 2nd step HS− + НОН ↔ H2S + ОН− КHS + Н2О ↔ КOH + H2S

net ionic equation molecular equation 98

A salt formed by a weak base and a strong acid (hydrolysis proceeds by cation). This occurs during the hydrolysis of NH4NO3 salt (NH4OH is a weak base, HNO3 is a strong acid). Let us discard NO3- ion, as it with the cation of water gives a strong electrolyte, while the equation of the hydrolysis takes the following form: NH4+ + НОН ↔ NH4OH + Н+ NH4NO3 + Н2О ↔ NH4OH + НNO3

net ionic equation molecular equation

The shortened equation shows that the OH- ions of water are bound in a weak electrolyte, H+ ions accumulate in solution and medium becomes acidic pH