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DANYLO HALYTSKY LVIV NATIONAL MEDICAL UNIVERSITY DEPARTMENT OF GENERAL, BIOINORGANIC, PHYSICAL AND COLLOIDAL CHEMISTRY MEDICAL CHEMISTRY STUDY GUIDE for the 1st year students of medical faculty (Module 2. The Equilibrium in Biological Systems Occuring On the Interfaces) L’VIV – 2012 Методичні вказівки з медичної хімії для студентів медичного факультету (Модуль 2. Рівноваги в біологічних системах на межі поділу фаз) Методичні вказівки уклали: доценти Кленіна О.В., Роман О.М., Огурцов В.В., асистент Маршалок О.І. За загальною редакцією: доцента Кленіної О.В. Методичні вказівки обговорені і схвалені до друку цикловою методичною комісією з фізико-хімічних дисциплін (протокол № 3 від 4 вересня 2011 р.). Рецензенти: проф. Й.Д. Комариця – професор кафедри фармацевтичної, органічної та біоорганічної хімії ЛНМУ імені Данила Галицького. проф. О.Я. Скляров – завідувач кафедри біологічної хімії ЛНМУ імені Данила Галицького. 2 General information on the educational process organization of “Medical Chemistry” studying within the credit-module system The educational process of medicinal chemistry studying is organized according to the requirements of credit-module system within the Bologna process. Medical Chemistry as an educational discipline is structured into 2 modules 9 practical classes in each: Module 1. Acid-Base Equilibrium and the Processes of Coordination Compounds Formation in Biological Liquids Thematic modules: 1. The chemistry of bioelements. Coordination compounds formation in biological liquids 2. Acid-base equilibrium in biological liquids Module 2. The equilibrium in Biological Systems Occurring on the Interfaces Thematic modules: 3. Thermodynamical and kinetical regularities of processes and electrokinetical phenomena in biological liquids 4. Physical-chemistry of surface phenomena. Lyophilic and lyiphobic disperse systems Laboratory records are to be kept in a bound notebook. Include in the notebook the aim of the experiments, a complete description of the work performed, all reference materials consulted, and ideas that you have related to the work, and the conclusions. Forms of the discipline assessment The maximum number of points assigned to students in each module (credit) 200, including the practice and laboratory educational activity - 120 points, and the final module control - 80 points. The assessment of trained knowledge is made on a three-point scale. For checking the student’s educational achievements are stipulated the following types and forms of the trained knowledge control: 1) the current control; 2) the practical skills gained and the laboratory experiments carrying out assessment; 3) the final module control assessment. The maximal assessment of current progress in a semester makes 60 % from a final assessment of knowledge on discipline, and the maximal assessment of examination makes 40 % from a final assessment of knowledge on discipline. 1. The current control is a regular check of educational trained achievements, spent by the teacher on current employment according to syllabus of the discipline. It is performed at each practice class according to specific objectives. 3 Theoretical students’ self-preparation control is performed in writing by answering 18 multiple choice questions in the form one-of-five, the correct answer to each is estimated at 1 point, and two numerical problems, the correct solving being estimated at 2 points. The minimum number of points that a student must gain for the crediting the theoretical part is 9 points. 2. The practical skills gained and the laboratory experiments carrying out assessment is performed after the laboratory work fulfilling by assessing the quality and fullness of its performance, the ability to interpret the obtained results. For the practical part of the lesson the student can get: 4 points if laboratory work is completely fulfilled and the student correctly explains the experiments interpret the results and make conclusions; 2 points if the laboratory work is done with some errors, the student can not fully explain and summarize the obtained results; 0 points if the laboratory work is not performed or the student can not explain and summarize the obtained results. The final score for the class is determined by the sum of the points for the current theoretical control and the laboratory experiments carrying out points as follows: Total points Grade evaluation > 21 17 – 21 11 – 16 < 11 points for the current theoretical control or 0 points for the laboratory experiments carrying out 5 4 3 2 Converting into rating points 13 11 8 0 The maximal number of points a student can get for the module is calculated by multiplying the number of points that correspond to the grade "5" to the number of topics in the module with the addition of points for individual independent work (3 points) and is equal to 120 points (13×9+3=120). The minimal number of points a student can get for the module is calculated by multiplying the number of points that correspond to the grade "3" to the number of topics in the module (8×9=72). 3. The final module control is carried out on completion of the module practical classes. The students fulfilling all types of works included in the curriculum with the points number on less than 72 are allowed. The final control is carried out in the standardized form and includes the theoretical and practical skills assessment. It should be performed in writing as 66 multiple choice questions (1 point for each correct answer) and 6 numerical problems (2 points for each in the case of being solved correctly). The maximum points for the final module control are 80. The final module 4 control is supposed to be credited if the student scored at least 50 points. The discipline assessment The discipline assessment is possible in the case of all modules credited only. The total assessment of discipline is shaped as an average of points number of the 2 modules each evaluated by summation of points for current control and experimental skills and final module control. The points on medical chemistry conversion into the ECTS scale evaluation and 4-grade evaluation The points on discipline may be conversed into the ECTS scale evaluation as follows: ECTS scale А В С D E FХ F Statistical index The top 10 % students Next 25 % students Next 30 % students Next 25 % students The last 10 % students Repeated making up The repeated course is required The points on discipline may be conversed into 4-grade evaluation as follows: The number of points on discipline The 4-grade evaluation 170 and over 140 – 169 122 – 139 less than 122 «5» «4» «3» «2» 5 THEMATIC SCHEDULE of practice and laboratory studies on medicinal chemistry Module 2. The equilibrium in Biological Systems Occurring on the Interfaces The topics Energetics of Chemical Reactions and Processes. Calculations According Thermochemical Equations and Experimental Determination of Heat Effects of Chemical Processes. Bioenergetics Kinetics of Chemical Reactions. Chemical Equilibrium. Solubility 2. Product Constant Measuring the Electromotive Forces of Galvanic Cells and Electrode 3. Potentials The Reduction-Oxidation Potentials Measuring. Potentiometry 4. Determining of pH for Solutions and Biological Liquids. Potentiometry Titration The Surface Tension and Surface-Active Substances. Adsorption on the 5. Movable Interfaces Adsorption on the Immovable Interfaces. The Adsorptive Ability of 6. Activated Charcoal Studying. Ions-Exchange Adsorption and Chromatographic Methods of Analysis 7. Lyophobic Sols Preparation and Their Properties Studying 8. The Stability of Colloidal Systems. Coagulation and Colloidal Protection High molecular compounds. The determination of the swelling degree of 9. gels and the influence of different factors on it. The determination of isoelectric point of proteins ● The final control of the acquirement the Module 2 Totally: 1. 6 Number of hours 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 25 Safety Rules The chemistry laboratory is not a dangerous place to work as long as all necessary precautions are taken seriously. In the following paragraphs, those important precautions are described. Everyone who works and performs experiments in a laboratory must follow these safety rules at all times. Students who do not obey the safety rules will not be allowed to enter and do any type of work in the laboratory and they will be counted as absent. It is the student’s responsibility to read carefully all the safety rules before the first meeting of the lab. Eye Protection: Because the eyes are particularly susceptible to permanent damage by corrosive chemicals as well as flying objects, safety goggles must be worn at all times in the laboratory. Prescription glasses are not recommended since they do not provide a proper side protection. No sunglasses are allowed in the laboratory. Contact lenses have potential hazard because the chemical vapours dissolve in the liquids covering the eye and concentrate behind the lenses. If you have to wear contact lenses consult with your instructor. If possible try to wear a prescription glasses under your safety goggles. In case of any accident that a chemical splashes near your eyes, immediately wash your eyes with lots of water and inform your instructor. Especially, when heating a test tube do not point its mouth to anyone. 4 Always assume that you are the only safe worker in the lab. Work defensively. Never assume that everyone else as safe as you are. Be alert for other’s mistakes. Cuts and Burns: Remember you will be working in a chemistry laboratory and many of the equipment you will be using are made of glass and it is breakable. When inserting glass tubing or thermometers into stoppers, lubricate both the tubing and the hole in the stopper with water. Handle tubing with a piece of towel and push it with a twisting motion. Be very careful when using mercury thermometer. It can be broken easily and may result with a mercury contamination. Mercury vapor is an extremely toxic chemical. When you heat a piece of glass it gets hot very quickly and unfortunately hot glass look just like a cold one. Handle them with a tong. Do not use any cracked or broken glass equipment. It may ruin an experiment and worse, it may cause serious injury. Place it in a waste glass container. Do not throw them into the wastepaper container or regular waste container. Poisonous Chemicals: All of the chemicals have some degree of health hazard. Never taste any chemicals in the laboratory unless specifically directed to do so. Avoid breathing toxic vapors. When working with volatile chemicals and strong acids and bases use ventilating hoods. If you are asked to taste the odor of a substance does it by wafting a bit of the vapor toward your nose. Do not stick your nose in and inhale vapor directly from the test tube. Always wash your hands before leaving the laboratory. Eating and drinking any type of food are prohibited in the laboratory at all times. Smoking is not allowed. Anyone who refuses to do so will be forced to leave the laboratory. Clothing and Footwear: Everyone must wear a lab coat during the lab and no shorts and sandals are allowed. Students who come to lab without proper clotting and shoes will be asked to go back for change. If they do not come on time it will be counted as an absence. Long hair should be securely tied back to avoid the risk of setting it on fire. If large amounts of chemicals are spilled on your body, immediately remove the contaminated clothing and use the safety shower if available. Make sure to inform your instructor about the problem. Do not leave your coats and back packs on the bench. No headphones and Walkman are allowed in the lab because they interfere with your ability to hear what is going on in the Lab. 5 Fire: In case of fire or an accident, inform your instructor at once. Note the location of fire extinguishers and, if available, safety showers and safety blankets as soon as you enter the laboratory so that you may use them if needed. Never perform an unauthorized experiment in the laboratory. Never assume that it is not necessary to inform your instructor for small accidents. Notify him/her no matter how slight it is. 7 Topic 1 Energetics of Chemical Reactions and Processes. Calculations According Thermochemical Equations and Experimental Determination Heat Effects of Chemical Processes. Bioenergetics 1. Objectives Most of the world energy is currently obtained from the combustion of fossil fuels, which are mainly hydrocarbons. We are all familiar with the idea of energy and we have a qualitative idea of what we mean by energy from everyday life. We will be particularly concerned with the energy, usually in the form of heat that can be obtained from chemical reactions. The quantitative study of the heat changes associated with chemical reactions is called thermochemistry. Thermochemistry is part of a subject of much wider scope called thermodynamics. Thermodynamics is the science of the transformations of energy. Bioenergetics is based on the basic principles of thermodynamics and describes the energy transformations in living organisms. 2. Learning Targets: − to learn the main terms and basic laws of thermochemistry; − to make thermochemical calculations for the foods fuel capacity evaluation; − to get skills of theoretical calculation and experimental determination of chemical reactions and processes heat effects; − to be able to apply the knowledge of the thermodynamics laws for the chemical processes direction prediction; − to explain the characteristics of living systems and the basic processes of energy transformations in them. 3. Self Study Section 3.1. Syllabus Content The special fields of chemical thermodynamics. Basic terms of chemical thermodynamics: thermodynamical system (isolated, closed, open, homogeneous, heterogeneous), the state variables (extensive and intensive), thermodynamical processes (reversible, irreversible). Living organisms as open thermodynamical systems. Irreversibility of life processes. The first law of thermodynamics. Enthalpy. Thermochemical equations. Standard enthalpies of formation and combustion. Hess's law. Calorimetry techniques. Biochemical processes energetic characteristics. Thermochemical calculations for the foods fuel capacity (caloricity) evaluation and making rational and therapeutic diets. 8 Spontaneous and non-spontaneous processes. The second law of thermodynamics. Entropy. Thermodynamic potentials: Gibbs’ free energy, Helmholtz’ free energy. Termodynamical equilibrium conditions. The criteria for the spontaneous processes direction. The basic principles of thermodynamics applying to living organisms. ATP as an energy source for biochemical reactions. Macroergic compounds. 3.2. Overview Definition of the first law of thermodynamics is: Energy can neither be created nor destroyed but only changed from one form to another or The energy of a system that is isolated from its surroundings is constant. If an amount of heat Q flows into a system from the surroundings, then the internal energy of the system will increase and the system can do an amount of work W on the surroundings: Q = ∆U + W. Enthalpy: The heat, QP, that flows into the system at constant pressure is equal to the enthalpy change, ∆H: Qp = – ∆Hр. Enthalpy is defined by the expression: Н = U + pV. The enthalpy of formation of the most stable form of an element in its standard state is zero. From the ∆Hf values for the reactants and products of a reaction, we can calculate enthalpy ∆H° for reaction. For a chemical reaction the enthalpy change is given by the equation: ∆Hf = Σ∆Hf(products) – Σ∆Hf(reactants), where Σ∆Hf(products) is the sum of the enthalpies of the products, and Σ∆Hf(reactants) is the sum of the enthalpies of the reactants. When the total enthalpy of the products, Σ∆Hf(products), is greater than the total enthalpy of the reactants, Σ∆Hf(reactants), the enthalpy change, ∆H, is positive There is a flow of heat Qp= –∆H from the surroundings to the reaction system. In other words, the reaction is endothermic. When the total enthalpy of the products is less than that of the reactans, the enthalpy change, ∆H, is negative. Thus, Qp= –∆H is also negative, and heat flows from the reaction system to the surrounding. In other words, the reaction is exothermic. Hess’s law (the law of constant heat summation): the energy change for any chemical or physical process is independent of the pathway or number of steps required to complete the process provided that the final and initial reaction conditions are the same. 9 ∆H1 = ∆H2 + ∆H3 = ∆H4 + ∆H5 + ∆H6 ∆Hformation = –∆Hcombustion Entropy (S) is a quantity that is a measure of the disorder of the particles (atoms and molecules) that make up the system and the dispersal of energy associated with these particles. The disorder in a system depends only on the conditions that determine the state of the system, such as composition, temperature, and pressure. The change in entropy therefore depends only on the initial and final states of the system. Entropy, like enthalpy is a state function. Definition of the second law of thermodynamics is: In any spontaneous process the total entropy of a system and its surrounding increases. For any spontaneous process we may write: Suniverse = Ssystem + Ssurroundings > 0 In other words, for any spontaneous process the total entropy change must be positive. Gibbs free energy:For any change at constant temperature and pressure we have: ∆G = ∆H – T·∆S For a spontaneous process the change in the Gibbs free energy, ∆G must be negative. In other words, the Gibbs free energy decreases during a spontaneous process. The standard free energy of formation, ∆Gf, is the free energy change for the formation of 1 mol of compound from its elements in their standard states. The standard free energies of formation, of the elements in their standard states are taken to be zero, for the free energies of formation of the substances you need to refer to a table of ∆Gf values. There are four possible combinations of ∆H and ∆S: ∆H – ∆S + Spontaneous Reaction ∆G – at all T Yes – at low T Yes – – + high T No + – – at all T No + at low T No + + – at high T Yes For any reaction nAA + nBB → nXX + nYY: The enthalpy change for the reaction is the difference between sum of the 10 standard enthalpies of formation of the products and sum of the enthalpies of formation of the reactants: ∆H° = Σn·∆H°(products) – Σn·∆H°(reactants) = = [nX·∆H°(X) + nY·∆H°(Y)] – [nA·∆H°(A) + nB·∆H°(B)] The standard entropy change for a reaction is easily calculated from the standard molar entropies, using the expression: ∆S° = Σn·S° (products) – Σn·S°(reactants) = = [nX·S°(X) + nY·S°(Y)] – [nA·S°(A) + nB·S°(B)] The standard free energy change for any reaction, ∆G may be found from the standard free energies of formation, ∆Gf of the reactants and products in just the same way as a standard enthalpy change is calculated: ∆G = ∆H – T·∆S = Σn·∆Gf (products) – Σn·∆Gf (reactants) = = [nX·∆G°(X) + nY·∆G°(Y)] – [nA·∆G°(A) + nB·∆G°(B)] 3.3. References 1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p. 2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995 p. 3. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. – Saunders College Publishing. – 1990. – 806 p. 4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001. – 1067 p. 5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGraw-Hill. – 1998. – 561 p. 6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The Central Science. – Prentice Hall. – 2000. – 1017 p. 7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular Science. – Mosby. – 1994. – 977 p. 8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. – 1997. – 1031 p. 9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991. – 625 p. 3.4. Self Assessment Exercises а) Review Questions 1. Define the basic terms of thermodynamics: a system, a phase, a component, a state variable, a state function. What is the principle of systems classification into isolated, open and closed ones? 2. Define the system type for the Earth and the cell of a living organism? Give the characteristics of the system internal energy. 3. State the first law of thermodynamics and give its mathematical expression. 4. Define the term of enthalpy, standard enthalpy, enthalpies of formation and combustion of substances? 11 5. The Hess’ law, enthalpy diagrams examples. 6. Write a mathematical expression and give several formulations of the second law of thermodynamics. Is it possible to apply this law to biological systems? 7. Define entropy, Gibbs’ and Helmholtz’ free energies. 8. What is the role of ATP in energy transformations in biological systems? Write down the thermochemical equation for the ATP hydrolysis reaction. 9. Chemical equilibrium and equilibrium constant. Write down the equilibrium constant expressions for the following reactions: а) 2SO2 + O2 ⇄ 2SO3; b) 3Н2 + N2 ⇄ 2NH3. b) Types of Numerical Problems and Their Solving Strategies Numerical problem 1. What amount of heat will flow from the reaction system where 36 g of aluminium will burn in the excess of oxygen? The heat effect of the reaction is –1676 kJ. Steps to solution: 1. 4 Al + 3 O2 → 2 Al2O3 The thermochemical equation of aluminium oxide formation is: 2Al(s) + 3/2 O2(g) → Al2O3(s) + Q, Q = –∆Hf (1 mole Al2O3) [kJ/mol] 2. The amount` of heat could be calculated from such proportion: during the burning of 2·27 g of aluminium 1676 kJ of heat flows 36 g of aluminium – Q 36 g ⋅ 1676kJ = 1117 kJ Q= 2 ⋅ 27 g Numerical problem 2. The heat effect of 1 mol blue vitriol CuSO4·5H2O dissolving is –11.5 kJ and the heat effect of the same amount of copper sulfate dissolving is 66.1 kJ. Calculate the enthalpy of hydration for copper sulfate. Steps to solution: Aqueous copper sulfate solution could be prepared in 2 ways: by dissolving of CuSO4 in water as well as by formation of CuSO4·5H2O crystalline hydrate and its further dissolving: According to the Hess’s law the enthalpy of hydration for copper sulfate is: ∆H1 = ∆H2 + ∆H3 where ∆H1 = –66.1 kJ – the enthalpy of CuSO4 dissolving, ∆H2 – the enthalpy of CuSO4 hydration, ∆H3 = 11.5 kJ – the enthalpy of CuSO4·5H2O dissolving. 12 Calculate the enthalpy of hydration for copper sulfate: ∆H2 = ∆H1 – ∆H3= –66.1 – 11.5 = –77.6 kJ Numerical problem 3. For the rusting of iron: 4Fe(s) + 3O2(g) → 2Fe2O3(s) calculate: a) the standard enthalpy change, b) the standard entropy change, c) the free Gibb’s energy change. Steps to solution: 1. The enthalpy change for this reaction is: ∆Hf0 = Σn·∆Hf0(products) – Σn·∆Hf0(reactants) = = 2·∆Hf0(Fe2O3, s) – [4·∆Hf0(Fe, s) + 3·∆Hf0(O2, g)] ∆Hf0(Fe2O3, s) = –822.2 kJ/mol, ∆Hf0(O2, g) = 0 kJ/mol, ∆Hf0(Fe, s) = 0 kJ/mol. 2. Using the values, calculate the standard enthalpy change for the reaction: ∆Hf0 = 2·(–822.2 kJ/mol) – 4·0 – 3·0 = –1644.4 kJ 3. For this reaction the standard entropy change is: ∆S0 = Σn·∆S0(products) – Σn·∆S0(reactants) = = 2·∆S0(Fe2O3) – [4·∆S0(Fe) + 3·∆S0(O2)] The standard entropy of formation for substances at 25 °C are: ∆S0(Fe, s) = 27.3 J·K–1·mol–1, ·∆S0(O2, g) = 205.0 J·K–1·mol–1, ∆S0(Fe2O3, s) = 87.4 J·K–1·mol–1 4. Using the values, calculate the standard entropy change for the reaction: ∆S0 = 2·(87.4 J·K–1·mol–1) – [4·27.3 J·K–1·mol–1 + 3·205.0 J·K–1·mol–1] = = –549.4 J·K–1·mol–1. 5. The standard free energy change for the reaction we can calculate from the equation: ∆G° = ∆H° – T·∆S° For this reaction ∆Hf° = –1644.4 kJ and ∆S° = –549.4 J·K–1·mol–1 = –0.549 kJ·K–1·mol–1. The standard free energy change for the reaction is: ∆G° = –1644.4 – 298·(–0.549) = –1480.68 kJ. 6. Alternatively, we can calculate the standard free energy change for this reaction using the standard Gibbs free energies of formation for substances at 25 °C: ∆G°f(Fe2O3, s) = –740.3 kJ/mol, ∆G°f(O2, g) = 0 kJ/mol, ∆G°f(Fe, s) = 0 kJ/mol. Using the values, calculate the standard free energy change for the reaction: ∆GO = Σn·∆G°f(products) – Σn·∆G°f(reactants) = = 2·∆Gf°(Fe2O3, s) – [4·∆Gf°(Fe, s) + 3·∆Gf°(O2, g)] = = 2·(–740.3)] – [4·0 + 3·0] = –1480.6 kJ Numerical problem 4. Calculate the standard entropy change for the reaction CO + 2Н2 ⇄ СН3ОН. Steps to Solution: 1. Write down the thermochemical equation for the reaction, pointing out the 13 states of matters of substances. Values of the standard entropies of formation for substances are given in Appendix А. СО(g) + 2 Н2(g) ⇄ СН3ОН(l) J ∆S° 197,5 130,5 126,8 mol ⋅ K 2. The equation for the enthalpy change for this reaction is: ∆Sor = Σ ∆Sfo(products) – Σ ∆Sfo(reactants) = = 1·∆Sfo(CH3OH(l)) – [1·∆Sfo(CO2(g)) + 2 ·∆Sfo(H2(g))] 3. Calculate the enthalpy change for the reaction using the values of the standard entropies of formation for substances: ∆Sor = 126.8 – (197.5 + 2⋅130.5) = –331.7 (J/К). o Answer: ∆S r ≈ – 0.33 kJ/(mol·К), ∆Sor < 0 – reaction is impossible at standard conditions. Numerical problem 5. Calculate the caloricity of 60 g of an egg which contains 12 % by mass of fats, 3.8 % of carbohydrates, and 68.5 % of proteins. Steps to Solution: 1. The caloricity of the egg can be calculates from such equation: ∆Qcomb = ∆Qcomb(fats) + ∆Qcomb(carbohydrates) + ∆Qcomb(proteins) where ∆Qcomb(fats) = Qcomb(fats)·m(fats), ∆Qcomb(carbohydrates) = Qcomb(carbohydrates)·m(carbohydrates), ∆Qcomb(proteins) = Qcomb(proteins)·m(proteins). 2. Calculate masses of fats, carbohydrates and proteins in 60 g of egg: m(fats) = m(egg ) ⋅ C p ( fats ) 60 g ⋅12% = = 7.2 g 100% 100% m(carbohydrates) ⋅ C p (carbohydrates) m(carbohydrates) = 100% m(proteins) = m( proteins ) ⋅ C p ( proteins ) 100% = = 60 g ⋅ 3.8% = 2.28 g 100% 60 g ⋅ 68.5% = 41.1 g 100% 3. Calculate the caloricity of the egg using values of oxidation heats of products in physiological conditions Qcomb(fats) = 37.8 kJ/g, Qcomb(carbohydrates) = 19.8 kJ/g, and Qcomb(proteins) = 16.8 kJ/g: ∆Qcomb(fats) = Qcomb(fats)·m(fats) = 37.8 kJ/g · 7.2 g = 272.16 kJ ∆Qcomb(carboh.) = Qcomb(carboh.)·m(carboh.) = 19.8 kJ/g · 2.28 g = 45.144 kJ ∆Qcomb(proteins)=Qcomb(proteins)·m(proteins) = 16.8 kJ/g·41.1 g = 690.48 kJ ∆Qcomb = 272.16 + 45.144 + 690.48 = 1007.78 kJ. c) Problems to Solve 1. Calculate the enthalpy change for the reaction of glucose complete oxidation in a living organism using the values of standard enthalpies of formation of 14 substances. Answer: ∆Hр = –2812.7 kJ 2. The heat of 2360.8 kJ is flowing out when phosphine PH3 burns. Calculate the phosphine standard enthalpy of formation. Answer: ∆Hf(PH3) = –5.2 kJ/mol 3. How much heat does the organism loss if it losses 650 g of water through skin? Answer: 1589 kJ 4. Calculate the entropy change for the reactions: а) CaO(т) + СO2(г) = СаСО(к); b) 2С(гр.) + СО2(г) = 2СО(г). Answer: а) –164.7 J/К; b) 170.3 J/К 4. Laboratory Activities and Experiments Section 4.1. Practical Skills and Suggested Learning Activities − to solve tasks and exercises on the thermodynamical functions calculations (enthalpy, entropy, Gibbs’ free energy) under standard conditions; − to find out the thermodynamical possibility of chemical and biochemical reactions; − to calculate the equilibrium constant and its relationship with Gibbs’ energy studying; − to calculate the temperature at which a chemical reaction can take place; − to determine experimentally the enthalpy of neutralization of a strong base with a strong acid. 4.2. Experimental Guidelines 4.2.1. Experimental Determination of the Enthalpy of Neutralization of a Strong Base With a Strong Acid Weigh a flask and fill it with 100 cm3 of the solution of KOH with the concentration of 0.5 M. Measure the temperature of the solution reading to the nearest 0.1 °C and record the initial temperature. Measure by the graduated cylinder 100 ml of 0.5 M HCl solution. Add the solution of HCl to the solution of the base, mix it and measure the highest temperature. Calculate the heat effect of the reaction: ∆Q = (msCs + mgCg) ⋅ ∆t, where ms and mg – masses of the solution and the flack, Cs – specific heat capacity of the solution, Cs = 4.18 J/(g⋅K), Cg – specific heat capacity of glass, Cg = 0.753 J/(g⋅K). Calculate the enthalpy of neutralization of the acid in the account of 1 mole of the hydrogen ions: M ∆Ηneutr= – ∆Q (J), m where M – the molar mass of the acid; m – the mass of the acid in grams. 15 Calculate the error of the experiment (∆H = -57.2 kJ is the theoretical value of the enthalpy of neutralization). 5. Conclusions and Interpretations. Lesson Summary Topic 2 Kinetics of Chemical Reactions. Chemical Equilibrium. Solubility Product Constant 1. Objectives Chemical kinetics is the field of physical chemistry, which studies the rates and mechanisms of chemical and biochemical reactions. Chemical kinetics, catalysis and equilibrium laws are of great theoretical and practical importance, since they allows to select the optimal conditions for the reactions progress. In general, reaction kinetics is the study of rate of chemical change and the way in which this rate is influenced by conditions of concentration of reactants, products and other chemical species which may be present, and the factors such as solvent, pressure and temperature. Reaction kinetics permits formulation of models for the intermediate steps through which reactants are converted into other chemical compounds and is a powerful tool in elucidating the mechanism by which chemical reactions proceed. It provides a rational approach to stabilization of drug products and prediction of shelf-life and optimum storage conditions. Study of the reactions rates is the basis for the drugs pharmacokinetics studying, clinical diagnosis, biochemistry. Assimilation characteristics and mechanism of enzymes action as biocatalysts is important for the metabolism processes understanding, diagnosis and treatment of certain diseases. Since heterogeneous equilibrium make a significant contribution to the overall homeostasis, it is important to study heterogeneous processes on the interfaces features. 2. Learning Targets: − identify all the terms in a kinetical equation. Be familiar with the terminology of order of a reaction; − giving the rate-law expression for a specific reaction at a certain temperature, calculate the initial rate of reaction at the same temperature for any initial concentrations of reactants. Use changes in concentrations to predict changes in initial rates; − perform various calculations with the integrated rate equations for zero-, first-, and second-order reactions relating rate constants, half-lives, initial concentrations, and concentrations of reactants remaining at some later time. − study the reactants concentrations influence on the reaction rate; − analyze the chemical equilibrium shifting; 16 − study the conditions of precipitates formation. 3. Self Study Section 3.1. Syllabus Content Chemical kinetics as the basis for the rates and mechanism of biochemical reactions studying. The reaction rate. Concentration affection the reaction rate. The law of mass action for the reaction rate. Rate constant. The reaction order. Kinetical equations for zero-, first- and second-order reactions. Half-life. The reaction mechanism concept and the reaction molecularity. The temperature influence the reaction rate. Van't Hoff’s rule. Activation energy. Сollision theory. Arrhenius equation. The concept of the transition state theory. The kinetics of complex reactions: parallel, successive, conjugated, chain. The concept of antioxidants. Free radical reactions in living organisms. Photochemical reactions, photosynthesis. Catalysis and catalysts. Features of catalysts. Homogeneous, heterogeneous and microheterogeneous catalysis. Acid-base catalysis. Autocatalysis. The mechanism of catalytical action. Promoters and catalytic poisons. The kinetics of enzymatic reactions. Enzymes as biological catalysts. Enzymes features: selectivity, efficiency, temperature and reaction medium affections. The concept of the enzymes action mechanism. Chemical equilibrium. Equilibrium constant and its expression. Chemical equilibrium shifting. Le Chatelier principle. Precipitation and dissolving reactions. Solubility product constant. Precipitates formation conditions. The heterogeneous equilibrium role in general homeostasis of the organism. 3.2. Overview Chemical kinetics studies the rate and the mechanism of chemical and biochemical processes. The rate of a chemical reaction is defined as the change in the concentration of a reactant (or product) in a given time interval. An expression which relates the rate of a reaction to the concentrations of the reactants, is called a rate law. In general, for a reaction aA + bB + cC + • • • → Products the rate law often has the form: v = k[A]x[B]y[C]z ... where x is called the order of the reaction with respect to A, y is the order with respect to B, z is the order with respect to C, and the sum, x+y+z+ ... , is called the overall order. For a homogeneous reaction: А(g) + В(g) → АВ(g) the expression of the reaction rate is: V = k CACB or V = k [A][B], 17 where V is the rate of the reaction, CA and CB or [A] and [B] are concentrations of reactants А і В respectively (mol/l), k is rate constant. For homogeneous reaction 2А + В → 2С kinetic equation is: V = k [A]2·[B]. In the case of heterogeneous reactions concentration of solid phase is not included to the rate equation, for example: 2Al(s) + 3Cl2 → 2AlCl3; V = k [Cl2]3. Therefore, for heterogeneous reactions the order of the reaction and molecularity are not identical. At given temperature k is a constant characteristic of the reaction. Its value is independent of the concentrations of the reactants, although it does depend on the temperature and the nature of reactants. The rate constant k is a measure of the intrinsic rate of the reaction: it is the rate when the concentrations of all the reactants are 1 mol/l. Fast reactions have large k values, while slow reactions have law k values. The rate constants expressions for zero-, first-, and second-order reactions (k0, k1, k2 – relatively) are given below: 1 k0 = (С0 – Сτ), k1 = 2,303 lg C0 , k2 = 1 ⋅ C0 − Cτ . τ τ Cτ τ C0 Cτ The Van’t Hoff’s and Arrhenius equations show the temperature affecting the reaction rate and the rate constant: V t= V 0 γ ∆T 10 E − a , k = A ⋅ e RT . Activation energy Еа may be calculated according to Arrhenius equations equation: 2,303RT1T2 k 2 . Ea = lg T2 − T1 k1 The fermentative reactions rates may be calculated using Michaelis-Menten equation: [S ] , ν V = max K M + [S ] where [S] – the substrate concentration; КМ –Michaelis-Menten constant. Few chemical reactions proceed in only one direction. Most are reversible, at least to some extend. At the start of the reversible process, the reaction proceeds towards the formation of products. As soon as some products molecules are formed, the reverse process begins to take place and reactants molecules are formed from product molecules. Chemical equilibrium is achieved when the rates of the forward and the reverse reactions are equal and the concentrations of the reactants and products remain constant. Experiments have shown that for any reaction at equilibrium the expression 18 involving the concentrations of the products and the reactants at equilibrium has a characteristic value. We can write a general equation far a reaction as follows: aA + bB + cC + ... ⇄ pP + qQ + rR + ... In this equation A, B, C, and so on, are the reactants; and P, Q, R, and so on, are the products. The letters a, b, c, . . ., p, q, r, . .. represent the number of moles of each substance involved in the balanced equation for the reaction. For this general reaction at a particular temperature the equilibrium constant is: K =( [ P]p [Q]q [ R]r ... [ A]a [ B]b [C ]c ... )eq where [A], [B], [C], ..., [P], [Q], [R], . . . are the concentrations of the reactants and the products at equilibrium. To obtain the expression for the equilibrium constant for any reaction, we raise the equilibrium concentration of each product to the power given by the number of moles of that product in the balanced equation for the reaction, and we multiply these. We then multiply the concentrations of each reactant, similarly raised to the power given by the number of moles of the reactant in the balanced equation. Finally, we divide the resulting expression for the products by that for the reactants. There is a general rule that helps us to predict the direction in which an equilibrium reaction will move when a change in concentration, pressure, volume, or temperature occurs. The rule, known as Le Chatelier's principle, states that if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset. The word "stress" here means a change in concentration, pressure, volume, or temperature that removes a system from the equilibrium state. When we use the solubility rules to predict whether or not the precipitate will form when two solutions are mixed we should strictly speaking say that the precipitate may form, rather then it will form, because if the solution were very diluted a precipitate may not form. What concentrations of ions will, in fact, give a precipitate? We can answer this question by considering the equilibrium between a solid salt and its ions in saturated solution of the salt in a more quantitative manner. In general, for an ionic compound with the formula AxBy, the equilibrium in a saturated solution can be written as: AxBy (s) ⇄ xAm+(aq) + yBn-(aq). The solubility product constant is then: Ksp = ([Am+]x × [Bn-]y)eq. In other words, the solubility product constant is equal to the product of the concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equation for the equilibrium. If we’ll express the concentration of Am+ ions in the saturated solution as xS (where S is solubility), and the concentration of Bn- ions in the saturated solution as yS, then the equation for solubility product constant may be written as: Ksp = (xS)x ⋅ (yS)y = xx ⋅ yy ⋅ Sx+y. 19 So the solubility of a slightly soluble salt (in moles/l) may be found out as: S = x+ y K sp xxy y If [A ] ⋅ [B ] > Ksp, then the solution is over-saturated. If [Am+]x ⋅ [Bn-]y < Ksp, then the solution is non-saturated. If [Am+]x ⋅ [Bn-]y = Ksp, then the solution is saturated. So, in case when [Am+]x ⋅ [Bn-]y ≥ Ksp, the precipitate will form. 3.3. References 1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p. 2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995 p. 3. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. – Saunders College Publishing. – 1990. – 806 p. 4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001. – 1067 p. 5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGraw-Hill. – 1998. – 561 p. 6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The Central Science. – Prentice Hall. – 2000. – 1017 p. 7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular Science. – Mosby. – 1994. – 977 p. 8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. – 1997. – 1031 p. 9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991. – 625 p. 3.4. Self Assessment Exercises а) Review Questions 1. On what factors does the rate of a reaction depend? Describe the effect of reactants concentrations on the reaction rate. 2. State and explain the law of mass action. 3. Explain what is meant by the kinetical equation. Write down the kinetical equations for the following reactions: a) sulfur dioxide oxidation; b) the Haber synthesis of ammonia; c) nitrogen (V) oxide decomposition. 4. What is meant by the order and the molecularity of a reaction? Give a few examples of chemical reactions with the same and different values of the reaction order and molecularity. 5. When the concentration of reactants is 1 mol/l, what is the special term by which the reaction rate is known? What are the units in which the rate constants of the first order and the second order reactions are expressed? 6. Define the following terms: effective collision, proper orientation of the m+ x n- y 20 colliding species, activation energy, activated complex. Plot and explain the potential energy profiles in the reaction progress for exothermic and endothermic reactions. 7. What do we mean by the mechanism of a reaction and its elementary step? 8. Explain the mechanism and kinetics of parallel, successive, conjugated, and chain reactions. Which kinds of reactions occur in living organism? 9. Define the following terms: catalysts, promoters and catalytic poisons. Give the examples. How does a catalyst increase the rate of a reaction? 10. Explain the term auto catalysis and its significance for biochemical processes. 11. Distinguish between homogeneous catalysis and heterogeneous catalysis and their mechanisms. Illustrate your answer with h examples. 12. Enzymes, their classification. Enzymes and chemical catalysts divergences. 13. Write down and explain Michaelis-Menten equation. 14. Define equilibrium. What is the rule for writing the equilibrium constant expressions for the reactions? Write the expressions of the equilibrium constants for homogeneous and heterogeneous reactions. 15. Explain Le Chatelier’s principle. List factors that can shift the position of an equilibrium. Give the examples. 16. What is the solubility measure for feebly soluble compounds. Write down the Ksp expressions for the following salts: CaF2, Ag2S, Mg3(PO4)2. b) Types of Numerical Problems and Their Solving Strategies Numerical Problem 1. Calculate the rate of chemical reaction: Br2 + HCOOH → 2Br– + 2H+ + CO2, if after 3 min. concentration of Br2 decreased from 0.1 mol/dm3 to 0.04 mol/l. Steps to Solution: As the rate of a chemical reaction is defined as the change in the concentration of a reactant (or product) in a given time interval, so: –4 U = – C2 − C1 = − 0.04 − 0.1 = 3.3∙10 mol/(l∙sec). τ 2 − τ1 3 ⋅ 60 Numerical Problem 2. For the homogeneous chemical reaction: N2 (g) + 3 H2 (g) → 2 NH3 (g) a) How will change the rate if the concentration of the reactants increases twice? Steps to Solution: The rate of the reaction is: 3 U 1 = k·[N2]1·[H2]1 If the concentrations of reactants increase twice, [N2]2 = 2[N2]1 and [H2]2 = 2[H2]1, the rate of the reaction will be: 3 3 U 2 = k·[N2]2·[H2]2 = k·2[N2]1·(2[H2]1) The ratio between rates is: 21 υ2 υ1 = 2[ N 2 ]1 ⋅ ( 2[H 2 ]1 ) 3 [ N 2 ]1[H 2 ]13 = 2 ⋅ 23 = 2 4 = 16 . 1 So, the rate of the reaction will increase 16 times. b) How will change the rate if the concentration of the reactants decreases 3 times? Steps to Solution: The rate of the reaction is: 3 U 1 = k·[N2]1·[H2]1 If the concentrations of reactants decrease 3 times, [N2]2 = 1/3·[N2]1 and [H2]2 = 1/3·[H2]1, the rate of the reaction will be: 3 3 U 2 = k·[N2]2·[H2]2 = k·1/3·[N2]1·(1/3·[H2]1) The ratio between rates is: υ2 υ1 = 1 / 3[ N 2 ]1 ⋅ (1 / 3[H 2 ]1 )3 = 1 / 3 ⋅ (1 / 3)3 1 1 1 = ⋅ = . 1 3 27 81 [ N 2 ]1[H 2 ]13 So, the rate of the reaction will decrease 81 times. c) How will change the rate if the pressure of system decreases in 3 times? Steps to Solution: The rate of the reaction is: 3 U 1 = k·P1(N2) ·P1 (H2) If the concentrations of reactants decrease 3 times, P2(N2) = 1/3·P1(N2) and P2(H2) = 1/3·P1(H2), the rate of the reaction will be: 3 3 U 2 = k·P2(N2)· P2 (H2) = k·1/3·P1(N2)·(1/3·P2(H2)) The ratio between rates is: υ2 υ1 = 1 / 3P1 ( N 2 ) ⋅ [1 / 3P1 (H 2 )]3 P1 ( N 2 )[ P1 (H 2 )]3 = 1 / 3 ⋅ (1 / 3)3 1 1 1 = ⋅ = . 1 3 27 81 So, the rate of the reaction will decrease 81 times. d) How did the pressure of system change if the rate of the reaction increased in 16 times? Steps to Solution: The rates of the reaction are: 3 3 U 1 = k·P1(N2)·P1 (H2) and U 2 = k·P2(N2)· P2 (H2) The ratio between rates is: υ2 υ1 = P1 ( N 2 ) ⋅ [ P2 (H 2 )]3 P1 ( N 2 )[ P1 (H 2 )]3 4 4 P = 2 . P1 P2 P = 16, 2 = 4 16 = 2 , P2 = 2P1. P1 P1 22 So, the pressure of system increased twice. e) How will change the rate if the volume of system decreases twice? Steps to Solution: The rate of the reaction is: v1 = k·[N2]1·[H2]13 The decreasing of volume of gas system in 2 times is proportional to increasing of concentrations of reactants in 2 times: [N2]2 = 2[N2]1 and [H2]2 = 2[H2]1. Therefore, the rate of the reaction will be: 3 3 U 2 = k·[N2]2·[H2]2 = k·2[N2]1·(2·[H2]1) . The ratio between rates is: υ 2 2[ N 2 ]1 ⋅ (2[H 2 ]1 ) 3 2 ⋅ 23 = = = 2 4 = 16 . 3 υ1 1 [ N 2 ]1[H 2 ]1 So, the rate of the reaction will increase 16 times. Numerical Problem 3. How many times the rate of a chemical reaction will change at the increasing of temperature from 20 to 40 OC, if the temperature coefficient γ=3? Steps to Solution: The rates of almost all chemical reactions increase with increasing temperature by a factor γ = 2 – 4 for every 10 K rise in temperature: υ2 υ1 =γ ∆T 10 =γ 40− 20 10 = 32 = 9. So, the rate of the reaction will increase 9 times. Numerical Problem 4. What is activation energy of the reaction if its rate at 100 0С is 10 times greater than at 80 0С? Steps to Solution: According to the Arrhenius equation: 2,303RT1 ⋅ T2 k 2 Ea = lg . T2 − T1 k1 k2 k2 If = 10, then log = 1, therefore: k1 k1 Ea = 2,303 ⋅ 8,31 ⋅ 353 ⋅ 373 = 125.8 (kJ/mol). 373 − 353 Numerical Problem 5. How many grams of radioactive isotope Bi will remain in 4 hours if its initial mass was 200 mg and the half-time of decomposition is 2 hours? 23 Steps to Solution: Decomposition is the reaction of 1st order, therefore the half-time of decomposition is: τ1/ 2 = 0 . 693 k1 From this equation: 0 . 693 k1 = τ 1/ 2 = 0 . 693 = 0 . 3465 2 The rate constant for the reaction of the 1st order is: k1 = 2 . 303 τ lg C0 Cτ = 2 . 303 τ lg m0 m Calculate the mass of isotope after 4 hours of decomposition: m 2 . 303 → lg m 0 = 0 . 3465 ⋅ 4 = 0 . 6018 lg 0 = 0 . 3465 4 m m 2 . 303 m0 m0 200 mg 0 . 6018 = 10 = 3 . 9976 , m = = = 50 . 03 mg m 3.9976 3 . 9976 Answer: after 4 hours 50 mg of radioactive isotope Bi will remain. Numerical Problem 6. Calculate the solubility of Fe(OH)2 in water (in mg/l) if Ksp(Mg(OH)2) = 1.0·10-15 at 25 OC. Steps to Solution: The dissociation equation for Fe(OH)2 is: Fe(OH)2 ⇄ Fe2+ + 2 OH– The solubility of Fe(OH)2 is: S = 1+ 2 K sp ( Fe ( OH ) 2 ) 1 ⋅2 1 2 = 3 1 ⋅ 10 − 15 = 4 3 0 . 25 ⋅ 10 − 15 = 0 . 63 ⋅ 10 −5 mol/l The molar mass of Fe(OH)2 is: M(Fe(OH)2) = 56 + 2·(16+1) = 90 g/mol Calculate the solubility of Fe(OH)2 in mg/l: S = 0.63·10–5 mol/l·90 g/mol = 56.7·10–5 g/l = 56.7·10–2 mg/l. Numerical Problem 7. The solubility of an electrolyte AB is 0.00714 g/l at the temperature of 25 OC. Calculate the Ksp quantity if the molar mass of the electrolyte is 100 g/mol (CaCO3). Steps to Solution: Calculate the solubility of the electrolyte in mol/l: S ( g/l ) 0 . 00714 S ( mol/l ) = = = 0 . 0000714 M ( g/mol ) 100 The dissociation equation for the electrolyte AB is: AB ⇄ A– + B+ The solubility product constant for the electrolyte AB is: Ksp = [A–]1 ⋅ [B+]1 24 mol/l In the saturated solution ions concentrations is: [A–] = [B+] = S. Calculate the solubility product constant for the electrolyte AB: Ksp = 0.0000714 · 0.0000714 = 5·10–9. Numerical Problem 8. The solubility of an electrolyte A2B is 8.59 mg/l at the temperature of 25 OC. Calculate the Ksp quantity if the molar mass of the electrolyte is 58 g/mol (Mg(OH)2). Steps to Solution: Calculate the solubility of the electrolyte in mol/l: S ( mol / l ) = S (g / l) 8 . 59 ⋅ 10 = M ( g / mol ) 58 −3 = 0 . 0001481 mol/l The dissociation equation for the electrolyte A2B is: A2B ⇄ 2A– + B2+ The solubility product constant for the electrolyte A2B is: Ksp = [A–]2 × [B2+]1 In the saturated solution ions concentrations is: [A–] = 2S and [B+] = S. Calculate the solubility product constant for the electrolyte A2B: Ksp = (2·0.0001481)2 · 0.0001481 = 1.3·10–11. c) Problems to Solve 1. How will the reaction 2NO + O2 → 2NO2 rate change if the volume in the system would be 3 times increased? Answer: the rate will increase by a factor of 27 2. The rate of a reaction increases by a factor of 1024 with 50 оС rise in temperature. Calculate the temperature coefficient γ for this reaction. Answer: γ = 3.98 3. Calculate the half-life for the radionuclide Radon-220, if the rate constant for the reaction is 1.26∙10–2 sec–1. Answer: τ1/2 = 55 sec 4. The equilibrium constant for the reaction Н2 + І2 ⇄ 2НІ is 50. Calculate the equilibrium concentrations of hydrogen and iodine if their initial concentrations were equal to 1 mol/l. Answer: [H2]eq = [I2]eq = 0.22 mol/l 5. Calculate the solubility of silver chloride AgCl in water in g/ml. Ksp(AgCl) = 1.8∙10–10. Answer: 1.9∙10–3 g/ml 6. Calculate the solubility product constant for Mg(OH)2, if its solubility equals to 1.4∙10–4 mol/l at 180 оС. Answer: Ksp = 1.1∙10–11 4. Laboratory Activities and Experiments Section 4.1. Practical Skills and Suggested Learning Activities − summarize the factors that affect reaction rates; 25 − for a given reaction, express reaction rate in terms of changes in concentrations of reactants and products per unit time; − describe simple one-step reactions in terms of collision theory and transition state theory; − describe activation energy and illustrate it graphically for exothermic and endothermic reactions; − understand, in terms of the distribution of energies of the reactant molecules, how the reaction rate depends on temperature; − be able to use the Arrhenius equation to find the rate constant from collision frequencies, A, and activation energies, Ea, and to relate rate constants at two different temperatures. Know how to represent and interpret this equation graphically; − understand, in terms of potential energy diagrams, how a reaction rate is altered by the presence of a catalyst. Give examples of homogeneous and heterogeneous catalysts. 4.2. Experimental Guidelines 4.2.1. The Reactants Concentrations Affecting the Reaction Rates Studying The reaction of sulfur formation (turbidity of solution appearance) should be studied: Na2S2O3 + H2SO4 → Na2SO4 + H2S2O3, H2S2O3 → H2O + SO2 + S↓. Using the following table, set up a series of 10 test-tubes with the quantities of 1 М Na2S2O3, 1 М H2SO4 solutions and water as indicated. Table 1 Test-tube No. Reagents Na2S2O3 H2O H2SO4 Na2S2O3 mol/l 1 2 1 4 3 2 3 5 solution concentration, 4 0.1 5 3 2 5 0.2 6 7 4 1 5 0.3 8 9 5 0 5 0.4 10 5 0.5 The time of turbidity occurrence τ, sec. The reaction rate, v = 1/τ When you would be ready to time the reactions, mix the solutions of test-tubes pairs (1 and 2, 3 and 4 and so on). Begin the timing with the stopwatch as soon as solutions have been mixed. Shake test-tubes to ensure good mixing. Stop the stopwatch when the first sign of the turbidity appears for each mixed pair. Record the measurements of time (in sec). Calculate the reaction rate as V = 1/τ. Graph the reaction rate V versus the reagent Na2S2O3 concentration. Make the conclusions about the reactants concentrations affecting the reaction rate. 26 5. Conclusions and Interpretations. Lesson Summary Topic 3 Measuring the Electromotive Forces of Galvanic Cells and Electrode Potentials 1. Objectives The mechanisms of the electrode potential, diffusion, membrane, and redox potentials origin and their magnitude affecting with different factors studying allows to realize the trends of most biochemical reactions. Commonly, a biological cell contains 25 times more K+ inside than is on the outside. The Na+-K+ pump is orientated so that it pumps Na+ out of the cell and K+ into the cell. ATP located on the inside of the pump drives the system. Decades of observations concerning membranes potentials followed, and bioelectrochemisity developed as an integral facet of the biomedical sciences. Bio potentials measurements are the basis of electrocardiography, electroencephalography and other diagnostic methods. The electromotive forces measurements allows to determine concentration of physiologically active ions (Н3О+, К+, Na+, Са2+, Сl–, NO3–etc.) in biological liquids and body tissues. 2. Learning Targets: − to learn the skills of constructing galvanic cells using different half-cells; − to measure the electromotive forces of galvanic cells produced by each cell using pH-meter; − to learn the electrode potential determining technique. 3. Self Study Section 3.1. Syllabus Content The electrochemical phenomena significance for biochemical processes. Electrodes potentials and their origin mechanisms. Nernst equation. The standard electrode potential. Half-cells potentials measurement. Indicator electrodes and reference electrodes. Silver-silver chloride electrode. Ion-selective electrodes. Glass electrode. Galvanic (electrochemical or voltaic) cells. Diffusion potential. Membrane potential. The biological role of diffusion and membrane potentials. Redox reactions significance for biochemical processes. Redox potential as a measure of the half-cell tendency to act as oxidizing or reducing agent. Peters’ equation. A standard redox potential. The spontaneity and the direction of redox reaction proceeding prediction by their redox potentials values. Equivalent factors of reduction and oxidizing agents. Redox potentials role for the biological oxidation mechanism. 27 3.2. Overview The galvanic (electrochemical) cell is a device in which the chemical energy of redox reaction is transforming into electrical one. The most common galvanic cell is constructed of two connected half-cells (electrodes). The compensatory method and multimeter of pH-meter are used for the galvanic cells electromotive forces experimental determining. All electrodes may be divided into 4 basic types: Electrodes of the 1st kind, reversible to cation. Metal plate immersed into a solution of its salt may be an example of the 1st kind electrodes. The schematical representation of the 1st kind electrodes: Men+Me, 2+ 2+ for example, Zn Zn, Cu Cu. Their potential is calculated according to the Nernst equation: ϕ =ϕ0 + or at standard conditions 2.303RT log a n+ Me nF ϕ =ϕ0 + 0.059 log a n + Me n , where ϕ0 – standard electrode potential; n – the number of electrons taking part in the electrode reaction. Electrodes of the 1st kind are often used as indicator electrodes. Since such electrodes respond rapidly to the concentration of the analyte ion in a solution, it is possible to calculate the activity of the ions in a solution. For example, the concentration of H+ ions may be determined using the hydrogen electrode (Pt) ½ H2 | H+, whose potential may be defined as: φ = 0.059 a(H+) or φ = – 0.059 pH. nd Electrodes of the 2 kind are constructed of a metal plate covered with a layer of its insoluble compound (salt, oxide, hydroxide) being in contact with a solution containing an anion of the salt. The schematical representation of the 2nd kind electrodes: MeMeA, An–, The 2nd kind electrodes potential may be calculated according to the Nernst equation: ϕ = ϕ0 − 2.303RT log a n− . A nF The examples of the 2nd kind electrodes are: − silver-silver chloride electrode Ag, AgCl | KCl; − calomel electrode Hg, Hg2Cl2 | KCl. Silver-silver chloride electrode is a silver wire covered with silver chloride and immersed into the solution of potassium chloride. Under its proceeding the following reactions occur: Ag → Ag+ + e− and Ag+ + Cl– → AgCl or Ag + Cl– → AgCl + e−. The Nernst equation for the silver-silver chloride half-cell may be given as: 2.3 R ⋅ T ϕ Ag,AgCl|KCl = ϕ o Ag,AgCl|KCl + lg a( Аg + ) or n⋅F 28 ϕ Ag, AgCl|KCl = ϕ 0 + Ag, AgCl| KCl K sp (AgCl) 2.303 R ⋅ T log n⋅ F a(Cl− ) The saturated silver-silver chloride electrode is often used as a reference electrode (φ = 0,021 V) instead of the standard hydrogen electrode, which is difficult to operate. Oxidation-reduction (redox) electrodes consist of an inert metal (platinum, gold, iridium, graphite etc.) immersed into a solution containing oxidized and reduced forms of the same substance, for example: PtFe3+, Fe2+. During the operating of redox half-cell the reactions proceed without the involving of the inert electrode material. It serves only as a conductor of electrons, oxidation or reduction products remain in the solution. The value of the redox system potential is defined according to the NernstPeters equation: ϕ red/ox = ϕ o red/ox + 2.3 RT a (Ox) , lg nF a (Re d ) 0.059 o or at standard conditions ϕ lg red/ox = ϕ red/ox + n a (Ox ) a(Re d ) . where a(Ox) and a(Red) are activities of the oxidized and reduced forms, φored/ox is standard electrode potential of a redox system, which is equal to the potential of an electrode if a(ox)=a(red). The potential of the redox system depends on the ratio a(Ox) : an increase of a(Re d ) this ratio increases the potential (intensifies oxidizing action) and a decrease of this ratio reduces the potential (strengthens reducing action). The φored/ox value serves as a measure of the oxidizing or reducing ability of a system: the greater the value φored/ox, the better its oxidizing action. Ion-Selective electrodes are electrochemical sensors their potentials magnitudes being affected the certain kind ions activity in a solution. Glass electrodes are manufactured in huge numbers for both laboratory and field measurements. They contain a built-in Ag-AgCl reference electrode in contact with the HCl solution enclosed by the membrane. The glass membrane of a pH glass electrode consists of a silicate framework containing lithium (or sodium) ions. When a glass surface is immersed in an aqueous solution then a thin solvated layer (gel layer) is formed on the glass surface in which the glass structure is softer. This applies to both the outside and inside of the glass membrane: Reference Analyte solution electrode (external) Membrane Inner solution Reference electrode (inner) Ion selective electrode Н+glass + Ме+solution ⇄ Н+solution + Ме+glass As the proton concentration in the inner buffer of the electrode is constant, a stationary condition is established on the inner surface of the glass membrane. In 29 contrast, if the proton concentration in the measuring solution changes then ion exchange will occur in the outer solvated layer and cause an alteration in the potential at the glass membrane. Only when this ion exchange has achieved a stable condition will the potential of the glass electrode also be constant. This means that the response time of a glass electrode always depends on the thickness of the solvated layer Ion selective electrodes consist of a frame 1, auxiliary electrode 2, immersed into the inner solution 3, and the membrane 4. Figure 1. Ion selective electrode Scheme 2. Glass electrode The potential of glass electrode depends on the activity of hydrogen ions: ϕ glass = ϕ 0 + RT lna H3O + (solution) F 3.3. References 1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p. 2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995 p. 3. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. – Saunders College Publishing. – 1990. – 806 p. 4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001. – 1067 p. 5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGraw-Hill. – 1998. – 561 p. 6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The Central Science. – Prentice Hall. – 2000. – 1017 p. 7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular Science. – Mosby. – 1994. – 977 p. 8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. – 1997. – 1031 p. 9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991. – 625 p. 3.4. Self Assessment Exercises а) Review Questions 1. Explain the mechanism of the electrode potential appearance. 30 2. Write Nernst equation and mention the factors which determine its magnitude. 3. Explain the construction of electrochemical (galvanic) cell and the electromotive force of a cell. Give Danielle-Jacobi cell as an example. 4. Give a few examples of the 1st and 2nd kind electrodes. Write Nernst equation for them. 5. What is meant by the standard half-cell potential? 6. What is hydrogen electrode? How is it utilized to measure the potentials of halfcells? 7. What is electrochemical series of metals? The relative strengths of the reducing and oxidizing agents. 8. Ion-selective electrodes: construction, classification, application. 9. What is meant by the redox electrode? Write Peters equation and explain it. 10. The biological significance of diffusion, membrane and redox potentials. b) Types of Numerical Problems and Their Solving Strategies Numerical problem 1. Calculate the potential of cadmium electrode in 0.01 M CdSO4 solution at standard conditions, if its standard potential is –0.40 V. Steps to Solution: According to the Nernst equation the electrode potential at standard conditions is: ϕ Cd 2 + |Cd o =ϕ0 Cd 2 + |Cd + 0.059 0.059 lg a 2+ = ϕ 0 2+ + lg([Cd 2+ ] ⋅ α ⋅ n) Cd Cd |Cd n n СdSO4 ⇄ Cd2+ + SO42–, Cd2+ + 2e ⇄ Cdo Two electrons are transferred from Cd2+ ion to cadmium metal in the balanced equation for this reaction, so n equals 2 for this electrode. One Cd2+ ion is formed at CdSO4 dissociation, so n equals 1. The standard potential of cadmium electrode is –0.40 V. Therefore, the electrode potential in 0.01 M CdSO4 solution is: ϕ Cd 2+ |Cdo = ϕ o Cd 2+ |Cdo + 0.059 log C M (Cd 2+ ) = −0.40 + 0.03⋅ log 0.01 = −0.40 + 0.03⋅ (−2) = −0.46 V 2 Numerical problem 2. Calculate the electrode potential of the Cu|CuSO4 electrode at standard conditions, if concentration of CuSO4 solution is 4 mol/L and percent of CuSO4 dissociation α=0.8. Steps to Solution According to the Nernst equation the electrode potential at standard conditions is: ϕ Cu 2+ | Cu o = ϕ o Cu 2+ | Cu o + 0.059 0.059 log a 2+ = ϕ o Cu 2+ | Cu o + log([ Cu 2 + ] ⋅ α ⋅ n) Cu n n СuSO4 → Cu2+ + SO42–, Cu2+ + 2e → Cuo Two electrons are transferred from Cu2+ ion to copper metal in the balanced equation for this reaction, so n equals 2 for this electrode. One Cu2+ ion is formed at CuSO4 dissociation, so n equals 1. The standard potential of copper electrode is 31 +0.34 V. Therefore, the electrode potential is: ϕCu 2+ |Cu o = 0.34 + 0.059 log(CM ⋅α ) = 0.34 + 0.03⋅ lg(4 ⋅ 0.8) = 2 = 0.34 + 0.03 ⋅ lg 3.2 = 0.34 + 0.015 = 0.355 V Numerical problem 3. Calculate the electromotive force of the galvanic cell consisting of two silver electrodes at standard conditions if the concentration of Ag+ ions in the electrodes solutions are 10–2 and 10–5 mol/l. Steps to Solution: The electromotive force of the galvanic cell is: EMF = φcathode – φanode According to the Nernst equation the electrode potential at standard conditions is: ϕ =ϕo + 0.059 0.059 log a + = ϕ o + log[ Ag + ] Ag n n The standard electrode potential of silver electrode is φo = 0.799 V. Calculate the electrode potential in the 1st solution: ϕ1 = 0.799 + 0.059 log10 − 2 = 0.799 + 0.059·(-2) = 0.799 - 0.118 = 0.681 V 1 Calculate the electrode potential in the 2nd solution: ϕ 2 = 0.799 + 0.059 log10 − 5 = 0.799 + 0.059·(-5) = 0.799 - 0.295 = 0.504 V 1 φ1 > φ2, therefore, the 1st electrode in this galvanic cell is cathode and the 2nd is anode. Calculate the electromotive force of the galvanic cell: EMF = 0.681 – 0.504 = 0.177 V. Numerical problem 4. Calculate the potential of the hydrogen electrode immersed in solution with pH=5. Steps to Solution: According to the Nernst equation the electrode potential at some moment of time is: + φ = φo + R ⋅ T ln [H ] = φo + 0.059 lg[H + ] . n⋅F + [H 2 ] n –5 If pH = 5, then [H ] = 10 mol/l. The standard electrode potential of hydrogen electrode φO = 0 V. Therefore, the electrode potential is: φ = 0 + 0.059·lg10–5 = 0.059·(–5) = - 0.295 V. Numerical problem 5. Calculate the electromotive force of the hydrogen galvanic cell at standard conditions if the pH values of the electrodes solutions are 4 and 10 respectively. 32 Steps to Solution: The electromotive force of the galvanic cell is: EMF = φcathode – φanode According to the Nernst equation the electrode potential at some moment of time is: + φ = φo + R ⋅ T ln [H ] = φo + 0.059 lg[ H + ] . n⋅ F n [H 2 ] The standard electrode potential of hydrogen electrode φO = 0 V. The electrode potential in the 1st solution is (if pH = 4, then [H+] = 10–4 mol/l): φ1 = 0 + 0.059 log 10 − 4 = 0 + 0.059·(–4) = –0.24 V. 1 The electrode potential in the 2nd solution is (if pH = 10, then [H+] = 10–10 mol/l) φ2 = 0 + 0.059 log 10− 10 = 0 + 0.059·(–10) = –0.59 V. 1 Calculate the electromotive force: EMF = –0.24 – (–0.59) = 0.35 V. Numerical problem 6. The electrode potential of zinc electrode Zn|ZnSO4 (0.2 M) is –0.79 V. Calculate the percent of dissociation of ZnSO4 in this solution. Steps to Solution: According to the Nernst equation the electrode potential at some moment of time is: 2+ 2+ φZn2+|Zno = φoZn2+|Zno + R ⋅ T ln [ Zn ] = φoZn2+|Zno + 0.059 log [ Zn ] 0 n⋅F n [ Zn ] 2+ o 1 Zn + 2e → Zn Two electrons are transferred from Zn2+ ion to zinc metal in the balanced equation for this reaction, so n is 2 for this electrode. The standard potential of zinc electrode is –0.76 V. Therefore, the electrode potential is: 0.059 φZn2+/Zno = –0.76 + log[ Zn 2 + ] ⋅ α 2 Calculate the percent of ZnSO4 dissociation: 0.059 –0.79 = –0.76 + log 0.2α 2 0.03 log 0.2α = –0.03 log 0.2α = –1 0.2α = 10–1 α = 0.1/0.2 = 0.5 or 50 %. с) Problems to Solve 33 1. Calculate the electrode potential value for the aluminium half-cell lowered into 0.2М Al2(SO4)3 solution at 25 оС, if its standard potential (ϕо) = –1.663 V, the activity coefficient of aluminium ions in the solution f (Al+3) = 0.255. Answer:–1.674 V 2. Give the schematic representation of the galvanic cell consisting of silver and zink 1st kind half-cells. Calculate the EMF for the cell if both metals activities are 1 mol/l in respective solutions. Answer: 1.50 V 3. Give the schematic representation of the galvanic cell consisting of calomel electrode and redox half-cell PtFe3+, Fe2+. Calculate the EMF for the cell at standard conditions. Answer: 0.52 V 4. Laboratory Activities and Experiments Section 4.1. Practical Skills and Suggested Learning Activities − to measure the electromotive force (EMF) for Weston cell; − to provide the student with the opportunity to construct a number of voltaic cells; − to measure the EMF produced by each cell; − to determine the 1st kind electrode potentials and redox potentials. 4.2. Experimental Guidelines 4.2.1. The electromotive force measurement for Weston cell with multimeter or pH-meter A Weston cell is an example of a cell that can be made to definite specifications, has a definite EMF, is long lived, and produces an EMF that changes little with temperature. Such cells are used as standard cells in potentiometry circuits to determine the EMF of another cell. Weston cell is an H type cell. One electrode consists of Cd amalgam covered with crystals of CdSO4⋅8/3H2O. Another electrode contains Hg with solid Hg2SO4 and covered with crystals of CdSO4⋅8/3H2O. The whole cell is filled with a saturated solution of CdSO4. The cell is represented as follows: (-) Cd(Hg)CdSO4⋅8/3H2OCdSO4(sat’d)Hg2SO4Hg (+) The cadmium electrode is treated as if it were the anode, the electrode at which oxidization occurs: Cd(Hg) + SO42- + 8/3H2O → CdSO4⋅8/3H2O + 2e−. The mercury electrode is the cathode, at which reduction occurs: Hg2SO4 + 2e− → 2Hg + SO42-. The overall reaction: Cd(Hg) + Hg2SO4 + 8/3H2O → CdSO4⋅8/3H2O + 2Hg. The EMF of this cell is 1.01807 V at 25 oC. To connect Weston cell with pH-meter and measure its electromotive force. 34 Compare it with the known value. 4.2.2. To measure the electromotive force produced by the galvanic cell Disconnect Weston cell. Construct the galvanic cell of two 1st kind electrodes as it’s shown on the Figure: Lower the salt bridge into position so that one end of the U-tube is immersed in the copper (II) chloride solution and the other end is in the zinc sulfate. Clamp the salt bridge into position and determine the voltage of the cell with pH-meter. Compare the EMF of the cell with the standard EMF calculated with the standard electrode potentials. (–) (+) Zn C KCl (–) ZnCl KCl CuCl2 Cu(+ . Figure 1. A galvanic cell constructed of two 1st kind electrodes 4.2.3. The 1st kind electrode potential determining Construct the galvanic cell consisting of the 1st kind electrode with the unknown potential and saturated silver-silver chloride half-cell as a reference electrode with the known potential (ϕ = +0.22 V). Write the schematical representation of the cell, for example: (–) Mе | Men+ | KCl | KCl, AgCl | Ag(+) Determine the voltage of the cell with pH-meter. As EMF = ϕ+ – ϕ– it is possibly to calculate the potential of the 1st kind electrode ϕ x : − ϕ x− = – EMF + ϕsilver -silver chloride Compare the calculated potential value of the 1st kind electrode with its standard potential. Make conclusions. Comment on any cases in which the standard potential does not agree with the experimentally observed value. 4.2.4. The redox electrode potential determining Construct the galvanic cell consisting of redox electrode with the unknown potential and saturated silver-silver chloride half-cell as a reference electrode with the known potential (ϕ = +0.22 V). Write the schematical representation of the cell, 35 for example: Ag | AgCl, KCl | KCl | I2, 2KI | С(graphite)(+) . Determine the voltage of the cell with pH-meter, calculate the redox potential as: ϕ(I |2I − ) = EMF + 0.22 . (–) 2 Compare the calculated redox potential value with the standard potential. Make conclusions. Comment on any cases in which the standard potential does not agree with the experimentally observed value. 5. Conclusions and Interpretations. Lesson Summary Topic 4 The Reduction-Oxidation Potentials Measuring. Potentiometry Determining of pH for Solutions and Biological Liquids. Potentiometry Titration 1. Objectives A very large part of chemistry is concerned, either directly or indirectly, with determining the concentrations of ions in solution. Any method that can accomplish such measurements using relatively simple physical techniques is bound to be widely exploited. Cell potentials are fairly easy to measure, and although the Nernst equation relates them to ionic activities rather than to concentrations, the difference between them becomes negligible in solutions where the total ionic concentration is less than about 10–3 M. Potentiometry as the method of determining pH of solutions has a few advantages as compared with another methods: it is more exact and accurate (allows to measure pH with the accuracy of 0.03 – 0.05 points), allows to measure pH of multi-component systems and colour solutions. This method is widely used in biology, medicine, pharmacy. It can be also used for measuring pH of different objects of environments and biological liquids. The large importance has the determining of pH for the researches of biochemical and physiological processes. 2. Learning Targets: – to learn the skills of the potentiometer (pH-meter) applying for the potentiometry experiments; – to learn the techniques of pH potentiometry measurements for biological liquids (blood plasma, gastric juice, urine etc.) using a glass electrode; – to perform a potentiometry titration of acids, bases and their mixtures. 3. Self Study Section 3.1. Syllabus Content Potentiometry. The technique of potentiometry determining of pH and ions activity in analyte solutions. Indicator electrodes (hydrogen, glass electrode) and their using in electrometric methods based on electromotive force (EMF) 36 measurement of a galvanic cell. The technique of potentiometry titration. The proper choice of the indicator electrode. The integral and differential curves of potentiometry titration plotting and the equivalence point determining. The analyte solution concentration determining. 3.2. Overview Potentiometry as the method of determining pH of solutions has a few advantages as compared with another methods: it is more exact and accurate (allows to measure pH with the accuracy of 0.03 – 0.05 points), allows to measure pH of multi-component systems and colour solutions. This method is widely used in biology, medicine, pharmacy. It can be also used for measuring pH of different objects of environments and biological liquids. The large importance has the determining of pH for the researches of biochemical and physiological processes. For the pH determining by potentiometry titration the electrochemical cell should consist of the reference electrode (silver/silver chloride or calomel electrode usually) and the indicator electrode, which is immersed in the analyte solution: reference electrode | salt bridge | analyte solution | indicator electrode. A reference is an electrode that has the half-cell potential known, constant, and completely insensitive to the composition of the solution under study. In conjunction with this reference is the indicator (or working) electrode, whose response depends upon the analyte concentration (φ depends on [H+]). The following galvanic cells are the most commonly used for the potentiometry determining of pH. Hydrogen-silver/silver chloride galvanic cell. In this cell hydrogen electrode acts as an indicator electrode while silver/silver chloride acts as a reference one. The schematical representation of the cell is given below: (–) (Pt) H2 2H3O+ KCl, AgCl Ag (+). At 298 К EMF of hydrogen-silver/silver chloride galvanic cell is: EMF = ϕ ( +) Ag,AgCl| KCl − ϕ (−) 2H + |H 2 = ϕ o Ag,AgCl|KCl + 0.059pH , then pH = EMF − ϕ o Ag, AgCl|KCl EMF − 0.22 , = 0.059 0.059 (1) 0 where ϕ Ag, is the potential of saturated silver/silver chloride electrode, which AgCl |KCl equals +0.22 V at 298 К. The saturated calomel electrode may be used instead of the saturated silver/silver chloride electrode (ϕcalomel = +0.25 V), then the рН value may be calculated by the equation: рН = Е − ϕ calom E − 0.25 . = 0.059 0.059 The hydrogen electrode applying for pH measuring has a few limitations one of which is the complexity of its manufacturing. It is also not appropriate to measure the pH of biological liquids because some organic substances, such as proteins, are 37 able to precipitate on the platinum plate surface, resulting in pH values obtained to be false. Glass electrode is often used as an indicator electrode being widely used in biomedical researches, as its potential depends on activity of hydrogen ions in solution. Glass-silver/silver chloride galvanic cell. The glass electrode in this cell acts as an indicator while silver/silver chloride is a reference electrode. The schematical representation of the cell is given below: (–) Ag AgCl, HCl glass membrane analyte solution with рНх KCl, AgCl Ag(+) The electromotive force of this cell may be calculated by the equation: (+) (-) . EMF = ϕ Ag, AgCl| KCl − ϕ glass Taking into account that Nernst equation for glass electrode is: 0 ϕ glass = ϕ glass − 0.059pH , (+) 0 EMF = ϕ Ag, AgCl|KCl − ϕ glass + 0.059pH If we assume ϕ(+)Ag,AgCl|KCl – ϕ 0glass = ϕ 0/, then EMF = ϕ 0/ + 0.059 pH, so EMF − ϕ 0 / . (2) рH = 0.059 The standard potential of a glass electrode depends on the grade of glass the electrode membrane is made of. Moreover it varies with time. That is why the calibration of the cell including glass electrode is commonly carried out. The cell calibration includes electromotive force measurements of the cell with standard buffer solutions of known рН values. Then the calibration curve should be plotted as the pH value versus the pH of buffer solutions. For the unknown pH value of a solution determining the EMF of the cell should be firstly measured. EMF value is then used for determination of the solution рН value by using plot on which find the EMF value (Ex value on Figure 1), draw horizontal line to the intersection with calibration line and find рН value which corresponds to the measured EMF value of the solution (pHx ). Figure1. Calibration curve of the cell including glass electrode and its using for pH determining pH measurements are now probably the most numerous type of quantitative chemical assay performed, and nearly all of these measurements are determined potentiometrically (electrochemically) using a pH meter, and a glass indicator and calomel or silver/silver chloride reference electrode. In many situations, accurate determination of an ion concentration by direct 38 measurement of a cell potential is impossible due to the presence of other ions and a lack of information about activity coefficients. In such cases it is often possible to determine the ion indirectly by titration with some other ion. The potentiometry titration is based on the fixing the equivalence point using the results of measurements of a galvanic cell potential during the titration process. Near the equivalence point there will be the significant change of the indicator electrode potential. The potentiometry titration is the electrochemical method based on the equivalent volume of a titrant determining by a major change in electromotive force (EMF) of the proper galvanic cell. For potentiometry titration the electrochemical cell should consist of the reference electrode (silver/silver chloride or calomel electrode usually) and the indicator electrode, which is immersed in the analyte solution. For example, for acid-base titration the indicator electrode potential should vary under the concentration changing of H+ ions. So for acid-base titration quingydrone, hydrogen and glass electrodes can be used as indicator ones. So the indicator electrode potential should vary under the concentration changing of the ions which take part or are forming in the process of titration while the reference electrode potential should remain constant. The potentiometry titration may be carried out using the reactions of neutralization, reduction-oxidation, precipitation, etc. The galvanic cell electrodes during the titration should be immersed in the analyte solution while the titrant being added with a burette in small portions (0.51.0 ml). After each portion adding the EMF should be measured using the potentiometer (pH meter). The titration data are used then for the integral curve plotting in the coordinates of determined EMF values as a function of the titrant volume (Fig. 2). ∆E ∆V EEV , ?mV 200 100 2000 0 1000 - 10 0 equivalence point mV/ml equivalence point 8 00 . - 20 0 6 00 - 30 0 4 00 - 40 0 2 00 - 50 0 Volume of titrant, ml 0 2 4 6 8 0 2 4 6 Volume of titrant, ml Figure 2. The curves of potentiometry titration of a solution of sulfuric acid with an alkali solution. a) Integral curve; b) Differential curve. Lowering the perpendicular from the middle of the jump (point of equivalence) it is possible to determine the equivalent volume of a titrant as the intersection with x-axis. The equivalence point may be also determined by the differential curve of titration which is plotted as ∆EMF/∆V – V (Fig. 2). 39 The perpendicular, dropped from the top of the curve on x-axis indicates the equivalent volume of a titrant. The analyte solution concentration may be calculated by the equation: Cx⋅V = Ct⋅Vt , so: Cx = C t ⋅ Vt , V where Сx, Сt are the concentrations of the analyte solution and the titrant, respectively; V, Vt are the volumes of the analyte solution and the titrant, respectively. 3.3. References 1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p. 2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995 p. 3. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. – Saunders College Publishing. – 1990. – 806 p. 4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001. – 1067 p. 5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGraw-Hill. – 1998. – 561 p. 6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The Central Science. – Prentice Hall. – 2000. – 1017 p. 7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular Science. – Mosby. – 1994. – 977 p. 8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. – 1997. – 1031 p. 9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991. – 625 p. 3.4. Self Assessment Exercises а) Review Questions 1. Explain the matter of the potentiometry method of pH determining. 2. Give the examples of indicator electrodes and reference electrodes used for the potentiometry determining of pH. 3. Construction and uses of glass electrode in the potentiometry measuring of pH. 4. Explain the potentiometry titration technique. 5. Plot the integral and differential curves for the potentiometry titration of a strong acid with an alkali. 6. Explain the titrant equivalent volume determining using the curves of titration. b) Types of Numerical Problems and Their Solving Strategies Numerical problem 1. Calculate the value of the quinhydrone electrode potential if it is immersed in solution with pH = 4.5. 40 Steps to Solution: The electrode potential at some moment of time is: φ = φ0 – 0.059pH. The standard electrode potential of quinhydrone electrode φ0 = 0.704 V. Therefore: φ = 0.704 – 0.059·4.5 = 0.44 V. Numerical problem 2. Calculate the pH of solution, in which the quinhydrone electrode potential is 0.20 V at the temperature of 298 K. Steps to Solution: The electrode potential at some moment of time is: φ = φ0 – 059pH. Therefore, the pH of the solution is: o pH = ϕ − ϕ . 0.059 The standard electrode potential of quinhydrone electrode φ0 = 0.704 V. Calculate the pH of solution: pH = 0.704 − 0.2 = 8.54. 0.059 Numerical problem 3. Calculate the pH of solution, if the electromotive force of the galvanic cell consisting of quinone-hydroquinone and silver-silver chloride electrodes is 0.08 V at the temperature of 298 K. Steps to Solution: Electromotive force of the galvanic cell is: EMF = φcathode – φanode = φQ/H – φAg,AgCl|KCl = φ0Q/H – 0.059pH – φAg,AgCl|KCl. The standard electrode potential of silver-silver chloride electrode is 0.222 V. The standard electrode potential of quinhydrone electrode φ0 = 0.704 V. Therefore, EMF = 0.704 – 0.222 – 0.059 pH = 0.482 – 0.059 pH Calculate the pH of the solution is: 0.482 − EMF 0.482 − 0.08 pH = = = 6.8. 0.059 0.059 Numerical problem 4. Calculate the electrode potential for the Pt | Fe3+, Fe2+ electrode, if the concentrations of Fe3+ and Fe2+ ions in the solution are 0.1 and 0.05 mol/l respectively. Steps to Solution: According to the Nernst equation the electrode potential at some moment of time is: 3+ 3+ 0.059 φ(Fe3+/Fe2+) = φo(Fe3+/Fe2+) + R ⋅ T ln [Fe ] = φo(Fe3+/Fe2+) + log [Fe ] 2 + n n ⋅ F [Fe ] [Fe 2 + ] 41 Fe3+ + 1e → Fe2+ One electron is transferred from Fe2+ ion to Fe3+ in the balanced equation for this reaction, so n is 1 for this electrode. The standard electrode potential φo(Fe3+/Fe2+) = 0,77 V. Therefore, the electrode potential is: 0.059 0.1 φ(Fe3+/Fe2+) = 0.77 + log = 0.77 + 0.059·lg 2 = 0.77 + 0.02 = 0.79 V. 1 0.05 Numerical problem 5. Calculate the pH value and hydrogen ions concentration in a gastric juice into which hydrogen and saturated calomel electrodes are immersed and the galvanic cell EMF is measured as 0.32 V at 25 0С. The standard potential value for the saturated calomel electrode is 0.25 V. Steps to Solution: Write down the schematical representation of the galvanic cell and the equation for its EMF calculation: (–} (Pt) Н2, 2Н+| KCl, gastric juice | KCl | KCl, Hg2Cl2 | Hg (+} EMF = ϕ+– ϕ– = 0.25 + 0.059 рН Calculate the pH value of gastric juice: рН = EMF − ϕ calomel = 0.32 − 0.18 = 1.19. 0.059 0.059 Calculate the hydrogen ions [H+] concentration in gastric juice: pH = –log [Н+] = 1.19; [Н+] = alog(–1.19) = 6.46·10–2 mol/l. с) Problems to Solve 1. Calculate the pH value of the blood if the EMF of hydrogen-silver/silver chloride galvanic cell value is 0.65 V at 25 оС. The electrode potential silver/silver chloride half-cell is 0.22 V. Answer: рН = 7.29 2. Calculate the pH and pOH values of gastric juice into which hydrogen electrode is immersed. Another electrode in the galvanic cell is saturated calomel electrode. The EMF value of this cell is 0.315 V at 25 оС and saturated calomel electrode potential is 0.247 V. Answer: рН = 1.15, рОН = 12.85 3. The potentiometry titration of HCl solution with 1 M NaOH solution was carried out. Determine the mass of HCl in the analyte solution using the following experimental data: VNaOH , ml 0 0.50 0.90 0.99 1.0 1.01 1.10 ϕ of hydrogen 398 416 457 516 683 890 929 electrode, mV Answer: 0.0365 g 42 4. Laboratory Activities and Experiments Section 4.1. Practical Skills and Suggested Learning Activities – to determine pH of a biological liquid of human’s organism; – to determine a strong acid or an alkali solution concentration with the potentiometry titration method. 4.2. Experimental Guidelines 4.2.1. The pH value determining for a biological liquid of human’s organism 1. Compose a galvanic cell consisting of glass indicator electrode and saturated silver-silver chloride reference electrode. Connect the galvanic cell to pHmeter-millivoltmeter. 2. Press the "POWER" button on the device. 3. Set the temperature of the solution by pressing the "MODE" button to set the unit of measurement " оС " and, turning the knob "SET TEMPERATURE MANUALLY". 4. Switch to pH measurement by pressing the "MODE" to set units "pH". 5. Immerse electrodes into the analyte solution, which was poured into a beaker and carry out the pH measurement. 6. After the measurement electrodes should be rinsed with distilled water and immersed into distilled water or 0.1 M HCl solution. 4.2.2. Determining the concentration of a strong acid or an alkali solution with the potentiometry titration method 1. Use a volumetric pipette to transfer 10.00 ml of the hydrochloric acid to a 100 ml beaker and, using a graduated cylinder, add about 20 ml of distilled water to the acid. 2. Place the beaker on a magnetic stirrer, place a small stirrer bar in the solution, and set the stirrer in motion—not too fast, otherwise some of the acid may splash from the beaker. 3. Compose a galvanic cell consisting of glass indicator electrode and silversilver chloride reference electrode: (–) Ag | AgCl, KCl | glass membrane | analyte solution | KCl, AgCl | Ag (+) 4. Fill your burette with 0.1 M sodium hydroxide solution. 5. Connect the galvanic cell to pH-meter-millivoltmeter. Record the pH of the solution to two places past the decimal. 6. Position your burette so that you can begin to add sodium hydroxide to the beaker of hydrochloric acid. Add the sodium hydroxide approximately 0.5 ml at a time until a total of 9.0 ml has been added. Record the pH of the solution after the addition of each 0.5 ml aliquot. Continue adding the sodium hydroxide, but now in approximately 0.2 ml aliquots, until a total of 11.0 ml has been added. As before, record the pH and total volume of sodium hydroxide after each aliquot has been added. Finally, add more sodium hydroxide in approximately 0.5 ml aliquots until the total volume of 43 sodium hydroxide added is 18.0 ml. Again the pH of the solution should be recorded after each addition. The above procedure may be summarized as follows: Table 1 The experimental data of potentiometry titration The volume of titrant EMF (or ϕ), ∆EMF (∆ϕ), ∆V, ∆EMF /∆V, V, ml mV mV ml mV/ml 0 1 2 so on 7. Plot a graph of EMF (y-axis) against total volume of sodium hydroxide added (x-axis). Determine the equivalence point for the titration between hydrochloric acid and sodium hydroxide. In a titration curve, the equivalence point is at the mid-point of the steepest part of the curve. 8. Plot also the differential curve of titration as ∆EMF /∆V (y-axis) against total volume of sodium hydroxide added (x-axis). Determine the equivalence point for the titration between hydrochloric acid and sodium hydroxide. In a titration curve, the equivalence point is in the top of the curve. 9. From the volume and concentration of NaOH solution used and the volume of HCl solution used for titration, calculate the concentration of HCl solution. 5. Conclusions and Interpretations. Lesson Summary Topic 5 The Surface Tension and Surface-Active Substances. Adsorption on the Movable Interfaces 1. Objectives Surface tension has significant impacts on the survival as well as the processes needed for day to day existence of living beings. . For example: phospholipids are a key component in cell membranes, which act as a protective surface against the environment. Most household detergents contain sodium dodecyl sulfate which reduces the surface tension of water. If local water reaches specific concentration of detergent, water striders will break the surface tension and sink. Surface tension is also essential for the transfer of energy from wind to water to create waves. Waves are necessary for rapid oxygen diffusion in lakes and seas. Understanding surface tension and what affects it is vitally important for researchers and engineers alike because new solutions for treating water have to maintain water's delicate and complex balance. 44 Surfactants have many commercial uses and depending on the use can be called by many names, including: -wetting agents, emulsifying agents, solubilizing agents, and detergents or soap. Surfactants are also widely used in pharmaceuticals. They are commonly added to drug suspensions to hinder caking of medications during storage, for reconstitution of powdered forms of medication into water at later use, or as an additive to tablets to aid in the penetration of moisture into the tablet for ready disintegration upon administration. 2. Learning Targets: – to understand the concept of surface tension, and that liquids tend to minimize their area; − to demonstrate that surface tension is a physical property of liquids; − to evaluate the surface activity of substances on the basis of their structure; − to analyze the structural features of the surface layer of adsorbed surfactant molecules, to explain the principles of biological membranes structure; − to analyze the adsorption equations, to distinguish a monomolecular and polymolecular adsorption; − to interpret the physical-chemical properties of proteins, which are structural components of all tissues. 3. Self Study Section 3.1. Syllabus Content Surface phenomena and their significance for biology and medicine. Surface tension of liquids and solutions. A surface tension isotherm. Surfactants and surfaceinactive substances. Surface activity. Traube’s rule. Adsorption on a liquid-gas and liquid-liquid interfaces. Gibbs’ equation. Surfactants molecules orientation in the interfacial layer. The structure of biological membranes. 3.2. Overview The molecular basis for surface tension may be explained by considering the attractive forces that molecules in a liquid exert on one another. If the liquid is not acted upon by external forces, a liquid sample forms a sphere, which has the minimum surface area for a given volume. Nearly spherical drops of water are a familiar sight, for example, when the external forces are negligible. To break or change the shape or area of a surface it is necessary to do work. This work counterbalances the resistance the surface is offering to the change. The work per unit area is called the surface tension. The surface tension is the ratio between the excess of free surface Gibbs energy and the surface area of the interface: σ = GS , [J/m2 or N/m] S where GS – free surface Gibbs energy; S – the surface area of the interface. Solutes can have different effects on surface tension depending on their structure. Surface active substances are absorbed at the interface and decrease the 45 surface tension of water (σsolution < σwater): salts of carbonic acids R-COONa (C11 < R < C18) (sodium oleate С17Н33СООNa); sulphur-acids salts R-C6H4-SO3–Na+ (R > C12). Surface inactive substances increase the surface tension of water (σsolution > σwater) and are concentrated in the bulk of solution (non-organic acids, bases, salts) and some strongly polar organic compounds (glycerine, amino acids etc.). The Gibbs adsorption equation enables the extent of adsorption at a liquid surface to be estimated from surface tension data: 2 С ∆σ C (σ 1 − σ ) , [mol/m ] A=− 2 ⋅ =− ⋅ RT ∆C RT (C1 − C 2 ) where A – adsorption, mol/m2; C – general concentration of solution, mol/m3; R=8,314 J/(mol·K) – universal gas constant; ∆σ/∆c – change of surface tension with changing of substance concentration. In case of surface active substances surface tension decreases at increasing of ∆σ solute concentration < 0 and adsorption is positive A > 0. ∆C In case of surface inactive substances surface tension increases at increasing of ∆σ solute concentration > 0 and adsorption is negative A < 0. ∆C 3.3. References 1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p. 2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995 p. 3. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. – Saunders College Publishing. – 1990. – 806 p. 4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001. – 1067 p. 5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGraw-Hill. – 1998. – 561 p. 6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The Central Science. – Prentice Hall. – 2000. – 1017 p. 7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular Science. – Mosby. – 1994. – 977 p. 8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. – 1997. – 1031 p. 9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991. – 625 p. 3.4. Self Assessment Exercises а) Review Questions 1. What is free surface energy? Explain the reasons of its origin. 46 2. 3. 4. 5. What is surface tension? The surface tension units. Why does surface tension try to minimize surface area? Factors affecting the surface tension. Define surface active substances (surfactants) and surface inactive substances. Give examples. 6. Illustrate the surface tension and adsorption magnitudes for surfactants with the surface tension and adsorption isotherms. 7. Explain the properties of surfactants on the basis of their molecules structure. 8. Illustrate the surfactants molecules orientation in the interfacial layers of waterthe air and benzene-water. 9. Define the surface activity, its units. Different variables affection the surface activity. 10. State and explain Traube’s rule. 11. What is the specific surface area of the adsorbent and how it may be calculated? 12. What experimental data are needed for the surfactants molecules sizes (surface area and the length) and the monomolecular layer thickness calculating. 13. Explain the structure of biological membranes. b) Types of Numerical Problems and Their Solving Strategies Numerical problem 1. Calculate the adsorption quantity (in mol/m2) at 10 °С for the solution which contains 50 mol/L of pelargonic acid С8Н17СООН. The surface tension of water is 74.22·10–3 J/m2 and the surface tension of the researched solution is 57.0·10–3 J/m2 at this temperature. Steps to Solution: 1. According to the Gibbs equation adsorption on the solution-gas interface is: A=− С ∆σ C (σ sol − σ H 2 O ) ⋅ =− ⋅ RT ∆C RT (Csol − CH 2 O ) 2. Calculate the adsorption quantity of solution: A=− 33 50 mol / L (57 ⋅ 10−3 − 74.22 ⋅ 10−3 ) J / m 2 − 17.22 ⋅ 10 −3 ⋅ =− = 7.32 ⋅ 10−6 mol/m2 8.314 J / mol ⋅ K ⋅ 283 K (50 − 0) mol / L 2352.86 . σ2<σ1, ∆σ < 0, A > 0, so pelargonic acid is surface active substance. ∆C Numerical problem 2. Calculate the free surface energy (in kJ) for the pelargonic acid С8Н17СООН solution at the temperature of 10°С if the surface tension of this solution is 57.0·10–3 J/m2. The surface area of the interface is 50 m2. Steps to Solution: 1. The free surface energy could be calculated from the formula for surface tension: → GS = σ ⋅ S GS σ= S 47 2. Calculate the free surface energy: GS = 57·10–3 · 50 = 2.85 J = 2.85·10–3 kJ Numerical problem 3. Calculate the specific surface area of arsenic (III) suldide As2S3 if its particles average diameter is 1·10–7 m, and its density is 3.43·103 kg/m3. Steps to Solution: 1. Calculate the particles average radius: r= d 10 −7 = = 5 ⋅ 10 −8 m. 2 2 2. The specific surface area of the adsorbent is the total area of all its particles (Sspecific=4πr2) in the unit volume ( V = 4 πr 3 ) or in the unit mass (m). Accepting 3 the spherical shape of the dispersed phase particles, it may be written: Sspecific = S 4 πr 2 3 = = [m -1 ] 4 3 r V πr 3 or S specific = 3 m2 S S 4 πr 2 = = = m Vρ 4 3 rρ kg πr ⋅ ρ 3 3. Calculate the specific surface area: S specific = –1 3 5 ⋅ 10 −8 = 6.7 ⋅ 10 7 m or S specific = m2 = 1.75 ⋅ 10 7 ⋅ 3.43 ⋅ 10 kg 3 5 ⋅ 10 −8 . 3 с) Problems to Solve 1. Calculate the adsorption quantity (in mol/m2) for the 20% solution of СaCl2 at the temperature of 293 К, if the surface tension of this solution is 80·10–3 J/m2 and the surface tension of water is 73·10–3 J/m2. The density of solution is 1.177 g/ml. Answer: –2.86·10–6 mol/m2 2. Calculate the adsorption quantity (in mol/m2) at 15 оС for the aqueous solution which contains 49 g/l of acetone if the surface tension of this solution is 59.4·10– 3 J/m2. The surface tension of water is 73.49·10–3 J/m2 at this temperature. The density of solution is 0.990 g/ml. Answer: 5.8·10–6 mol/m2 3. For aqueous solutions with the concentration of carbonic acid 0.005 and 0.01 mol/l the surface tension quantities at 273 К are 65.8·10–3 and 60.05·10–3 J/m2, respectively. Calculate the adsorption quantity (in mol/m2). Answer: 3.82·10–6 mol/m2 4. Calculate the specific surface area of kaolin suspension (its density is 1.8·10–3 kg/m3), if its particles are suppose to be of a spherical shape with the average diameter of 0.5·10–6 m. Answer: 4.8·103 m2/kg 48 4. Laboratory Activities and Experiments Section 4.1. Practical Skills and Suggested Learning Activities – To demonstrate that surfactants can reduce the surface tension of a liquid; – To get practical skills of the surface tension experimental determination on the liquid-gas interface with the maximal bubble pressure method; – To learn how Gibbs’ equation should be applied for the liquid-gas interfacial layer adsorption calculations. 4.2. Experimental Guidelines 4.2.1. Analysis of alcohol concentration influence on the surface tension of solutions Bubble pressure method consists in the following: 1. In vessel 2 distilled water is poured, preliminary the capillary has to be washed out by the distilled water. 2. Vessel 2 is covered with plug with inserted capillary 1. The end of the capillary should only become moist by the liquid meniscus. 3. Then vessel 2 is set in the thermostat, and connected with manometer (pressure gauge) 3 and aspirator 4. 4. The tap is turned on and the water is poured out gradually. Hereby in the vessel 2 rarefaction occurs, and through the liquid the bubble of ear slips. The bubble has to tear off of the capillary 1 evenly, almost every 10 s. 3 1 45 h 0 0 00 0 2 3 45 0 00 0 5 43 20 2 Fig. 1. Scheme of the Rebinder’s Bubble pressure method device for the surface tension evaluation 5. In vessel 2 distilled water is poured, preliminary the capillary has to be washed out by the distilled water. 6. Vessel 2 is covered with plug with inserted capillary 1. The end of the capillary should only become moist by the liquid meniscus. 7. Then vessel 2 is set in the thermostat, and connected with manometer (pressure gauge) 3 and aspirator 4. 49 8. The tap is turned on and the water is poured out gradually. Hereby in the vessel 2 rarefaction occurs, and through the liquid the bubble of ear slips. The bubble has to tear off of the capillary 1 evenly, almost every 10 s. Pressure P equals the difference between external pressure (atmospheric pressure) and the pressure in vessel 2 and is proportional to surface tension σ = Kh, K is constant for given capillary. As there are tabular data for surface tension of water at different temperatures, so constant K is measured using water. 9. Calculation of constant K: K = σH2O/h Calculation of surface tension for other liquids and solutions: σ = Kh, or σ = σH2O · hsol/hH2O 10. In the bubble pressure method the surface tension of 4 solutions with different concentration has to be defined. Obtained experimental data are put in the table 1 below. 11. According to experimental data students draw a diagram of surface tension isotherm, where ordinate axis is σ (J/m2) and abscissa axis is C (mole/l) 12. According Gibbs equation adsorption of alcohol is calculated. Students draw a diagram of Gibbs adsorption isotherm: ordinate axis is G (mole/l), abscissa axis is C (mole/l). Table 1 Experimental data for the surface tension and adsorption evaluation and the surfactant solutions concentrations affecting their values Alcohol Difference between Surface tension Adsorption A, concentration the levels Substance С, mole/m2 σ, J/m2 h, mm mol/l 1. Water 0.00 2. Alcohol 0.025 3. Alcohol 0.050 4. Alcohol 0.1 5. Alcohol 0.2 4.2.2. The hydrocarbon radical length in a surfactant molecule affecting the surface activity studying 1. Determine and calculate the capillary constant K as it is described in the previous experiment (4.2.1). 50 2. Determine the surface tension of alcohols solutions of the same homologous series (СН3ОН, С2Н5ОН, С3Н7ОН, С4Н9ОН, С5Н11ОН) with the concentration of 0.1 mol/l. 3. Plot the correlation of the solutions surface tensions (J/m2) versus the number of carbon atoms in the hydrocarbon radical. 4. Calculate the surface tension increasing ∆σ with the hydrocarbon radical length increasing. Make the conclusions about the Traube’s rule experimental verification. 5. Conclusions and Interpretations. Lesson Summary Topic 6 Adsorption on the Immovable Interfaces. The Adsorptive Ability of Activated Charcoal Studying. Ions-Exchange Adsorption and Chromatographic Methods of Analysis 1. Objectives Gas adsorption is of practical consequence to engineers and chemists in many ways. It can provide a convenient, cheap and reusable method for fluid purification and purification. More significantly, perhaps, the phenomenon of surface adsorption has been used to modify the rates of product yields of chemical reactions through heterogeneous catalysis. There are many environmental applications of adsorption in practice and many others are being developed. Activated carbons and clays are frequently used for the removal of organic contaminants, such as phenol and aniline. Moreover, the adsorption on inexpensive and efficient solid supports has been considered a simple and economical viable method for the removal of dyes from water and wastewater. The most frequently used procedures in pharmaceutical processing for solid dosage formulations are mixing, granulation, and compaction, as well as storage of finished dosage forms. The effects of adsorption on these procedures have been utilized widely in the pharmaceutical industry. Enterosorbents are very effective for adsorbing bacterial enterotoxins and endotoxins, so they are the common nonspecific treatment for intoxications. Kaolinpectin formulations are popular for symptomatic adsorption therapy. 2. Learning Targets: – to interpret the adsorption of substances from solutions on solid surfaces regularities; – to explain the physical-chemical foundations of the adsorption therapy methods; – to distinguish the selective and ion exchange adsorption of electrolytes; – to interpret the chromatographic analysis methods and their application in medical and biological researches. 51 3. Self Study Section 3.1. Syllabus Content Adsorption at the solid-gas interface. Langmuir equation. Adsorption from solution at the solid-liquid interface. Physical sorption (or physisorption) and chemical sorption (or chemisorption). General rules for the solutes, vapours and gases adsorption. Freundlich equation. Physico-chemical basis of adsorption therapy (hemosorbtion, plazmosorbtsiya, limfosorbtsiya, enterosorption, Application therapy). Immunosorbents. Adsorption of electrolytes: specific (selective) and ion exchange. Fajans-Peneth precipitation and adsorption rule. Naturally occurring ion exchangers and synthetically produced organic resins. Adsorption and ion exchange significance or the vital process in plants and living organisms. Chromatography. Chromatographic methods of analysis classification based on the phases states of matter, the technique and the separation mechanism. Adsorption, ion exchange and distribution chromatography. Chromatography applications in biology and medicine. 3.2. Overview When a gas or vapour is brought into contact with a solid, part of it is taken up by the solid. The molecules that disappear from the gas either enter the inside of the solid, or remain on the outside attached to the surface. The former phenomenon is termed absorption (or dissolution) and the latter adsorption. When the phenomena occur simultaneously, the process is termed sorption. The solid that takes up the gas is called the adsorbent, and the gas or vapour taken up on the surface is called the adsorbate. Molecules and atoms can attach themselves onto surfaces in two ways. In physisorption (physical adsorption), there is a weak van der Waals attraction of the adsorbate to the surface. The attraction to the surface is weak but long ranged and the energy released upon accommodation to the surface is of the same order of magnitude as an enthalpy of condensation (on the order of 20 kJ/mol). During the process of physisorption, the chemical identity of the adsorbate remains intact, i.e. no breakage of the covalent structure of the adsorbate takes place. Physisorption, to be a spontaneous thermodynamic process, must have a negative ∆G. Because translational degrees of freedom of the gas phase adsorbate are lost upon deposition onto the substrate ∆S is negative for the process. Since ∆G = ∆H – T∆S, ∆H for physisorption must be exothermic. In chemisorption (chemical adsorption), the adsorbate sticks to the solid by the formation of a chemical bond with the surface. This interaction is much stronger than physisorption, and, in general, chemisorption has more stringent requirements for the compatibility of adsorbate and surface site than physisorption. The chemisorption may be stronger than the bonds internal to the free adsorbate which can result in the dissociation of the adsorbate upon adsorption (dissociative adsorption). In 52 some cases ∆S for dissociative adsorption can be greater than zero, which means endothermic chemisorption, although uncommon, is possible. The adsorption quantity on immovable interfaces may be defined as the ratio of an adsorbate moles n by the adsorbent surface area S (m2), or by its mass m (kg): A= n , [mol/m 2 ]; S A= n , [mol/kg] m Adsorption is usually described through isotherms, that is, functions which connect the amount of adsorbate on the adsorbent, with its pressure (if gas) or concentration (if liquid). Langmuire adsorption theory is based on four assumptions: 1. The surface of the adsorbent is uniform, that is, all the adsorption sites are equivalent. 2. Adsorbed molecules do not interact. 3. All adsorption occurs through the same mechanism. 4. At the maximum adsorption, only a monolayer is formed: molecules of adsorbate do not deposit on other, already adsorbed, molecules of adsorbate, only on the free surface of the adsorbent. The Langmuir adsorption isotherm for gases adsorbed on solids: A = A∞ Kp 1 + Kp where A is the substance amount of adsorbate adsorbed per gram (or kg) of the adsorbent, mol/g or mol/kg; A∞ is the number of adsorption centres (maximum boundary adsorption – maximal substance amount of adsorbate per gram (or kg) of the adsorbent), mol/g or mol/kg; p – pressure of a gas; K – proportionality adsorption constant, l/mol. For liquids (adsorbate) adsorbed on solids (adsorbent), the Langmuir isotherm can be expressed by: KC A = A∞ 1 + KC where C – concentration of solution (of adsorbate in liquid), mol/l. In practice, activated carbon (also called activated charcoal or activated coal) is used as an adsorbent for the adsorption of mainly organic compounds along with some larger molecular weight inorganic compounds such as iodine and mercury. It is a material with an exceptionally high surface area. Just one gram of activated carbon has a surface area of approximately 500 m². The three main physical carbon types are granular, powder and extruded (pellet). All three types of activated carbon can have properties tailored to the application. Activated carbon is frequently used in everyday life, in: industry, food production, medicine, pharmacy, military, etc. In pharmacy, activated charcoal is considered to be the most effective single agent available as an emergency decontaminant in the gastrointestinal tract. It is used after a person swallows or absorbs almost any toxic drug or chemical. Empirical Freundlich isotherm: 53 x x = kp1 / n or A = = kC 1 / n m m where x is the number of moles of adsorbate, m – mass of adsorbent, K and n are temperature-dependent parameters. K and n depend on the nature of adsorbate and adsorbent. K is also known as the specific adsorption (if p = 1, than A = K). In logarithmic form this equation may be written as: or log x = logk + 1 logC x 1 log = log k + log p m n m n Ion exchange is a reversible chemical reaction where an ion from solution is exchanged for a similarly charged ion attached to an immobile solid particle. Ion exchange is the process through which ions in solution are transferred to a solid matrix (ion exchanger, or ion exchange resin) which, in turn releases ions of a different type but of the same polarity. In other words the ions in solutions are replaced by different ions originally present in the solid. Since ion exchange occurs between a solution and the internal surface of a solid it can be viewed as a special type of sorption process. Although ion exchange is similar to sorption since a substance is captured by a solid in both processes, there is a characteristic difference between them: ion exchange is a stoichiometric process in contrast to sorption. It means that in the ionexchange process, for every ion that is removed, another ion of the same sign is released into the solution. In contrast, in sorption, no replacement of the solute takes place. So ion exchange can be seen as a reversible reaction involving chemically equivalent quantities Conventional ion exchange resins consist of a cross-linked polymer matrix with a relatively uniform distribution of ion-active sites throughout the structure. Ion exchangers are either naturally occurring inorganic zeolites or synthetically produced organic resins. An organic ion exchange resin is composed of high-molecularweight polyelectrolytes that can exchange their mobile ions for ions of similar charge from the surrounding medium. Each resin has a distinct number of mobile ion sites that set the maximum quantity of exchanges per unit of resin. The synthetic organic resins are the predominant type used today because their characteristics can be tailored to specific applications. Ionisable groups attached to the resin bead determine the functional capability of the resin. Industrial water treatment resins are classified into four basic categories: − Strong Acid Cation (SAC); − Weak Acid Cation (WAC); − Strong Base Anion (SBA); − Weak Base Anion (WBA). SAC resins can neutralize strong bases and convert neutral salts into their corresponding acids. SAC resins derive their functionality from sulfonic acid groups (HSO3¯ ). When used in demineralization, SAC resins remove nearly all raw water cations, replacing them with hydrogen ions. A= 54 SBA resins can neutralize strong acids and convert neutral salts into their corresponding bases. These resins are utilized in most softening and full demineralization applications. SBA resins derive their functionality from quaternary ammonium functional groups. When in the hydroxide form, SBA resins remove all commonly encountered anions. WAC and WBA resins are able to neutralize strong bases and acids, respectively. These resins are used for dealkalization, partial demineralization, or (in combination with strong resins) full demineralization. Weak acid cation exchange resins derive their exchange activity from a carboxylic group (-COOH). When operated in the hydrogen form, WAC resins remove cations that are associated with alkalinity, producing carbonic acid. Weak base resin functionality originates in primary (R-NH2), secondary (RNHR'), or tertiary (R-NR'2) amine groups. WBA resins readily remove sulfuric, nitric, and hydrochloric acids. Weak acid cation resins are used primarily for softening and dealkalization of high-hardness, high-alkalinity waters, frequently in conjunction with SAC sodium cycle polishing systems. In full demineralization systems, the use of WAC and SAC resins in combination provides the economy of the more efficient WAC resin along with the full exchange capabilities of the SAC resin. 3.3. References 1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p. 2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995 p. 3. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. – Saunders College Publishing. – 1990. – 806 p. 4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001. – 1067 p. 5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGraw-Hill. – 1998. – 561 p. 6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The Central Science. – Prentice Hall. – 2000. – 1017 p. 7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular Science. – Mosby. – 1994. – 977 p. 8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. – 1997. – 1031 p. 9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991. – 625 p. 3.4. Self Assessment Exercises а) Review Questions 1. Adsorption on a solid-gas interface. Different variables affection the adsorption magnitude. 55 2. Langmuir adsorption theory, its basic statements. Langmuir equation and Langmuir adsorption isotherm. The equation constants defining. 3. Adsorption on a solid-gas interface. Different variables affection the adsorption magnitude. The rule of polarities aligning. Freundlich equation and adsorption isotherm. The equation constants defining. 4. Nature of adsorption forces. Define the sorption processes: adsorption, desorption, physisorption, chemisorption, physicosorption, capillary condensation. 5. Polymolecular adsorption theory of Polany and BET-theory. The equation of an isotherm of polymolecular adsorption, its analysis. 6. Fundamentals of adsorption therapy (haemosorbtion, enterosorption, Application therapy). Immunosorbents. 7. Adsorption from solutions. Molecular and ionic adsorption. Ion-exchange adsorption. Its features. 8. State the Fajans-Peneth adsorption rule. Illustrate if for BaSO4 – K2SO4 system. 9. Define the ion exchangers and give their classification 10. Write down the equations of the processes occurring during water demineralization with ion exchange. 11. Explain chromatographic methods of analysis and give their classifications. b) Types of Numerical Problems and Their Solving Strategies Numerical problem 1. When 2.8 g of oxygen is adsorbed with activated charcoal at 68 K 1.36 kJ of heat release. Calculate the adsorption heat effect (in kJ/mol) of oxygen with charcoal. Steps to Solution: The adsorption heat effect is the amount of heat which flows out of or into a reacting system under the 1 mole substance adsorption with an adsorbent. 1.36 kJ of heat is given to flow out while 2.8 g of oxygen been adsorbed, then X kJ of heat will flow out for 1 mole (32 g) oxygen adsorption: X= 32 ⋅ 1.36 = 15.54 kJ/mol. 2.8 Numerical problem 2. Calculate the adsorption quantity of acetic acid on activated charcoal, if the equilibrium concentration of the acid is 3.76 mmol/l, К and n constants in Freundlich equation are 2.82 and 1.21, respectively. Steps to Solution: Calculate the adsorption quantity according to Freundlich equation: A = kC 1 / n = 2.82 ⋅ 3.76 1 / 1.21= 2.82 ⋅ 3.760.826 = 2.82 ⋅ 2.986 = 8.42 . 56 Numerical problem 3. Calculate the equilibrium concentration of acetic acid, if adsorption quantity equals 0.012 mol/g, К and n constants in Freundlich equation are 0.25 and 3.1, respectively. Steps to Solution: 1. According to Freundlich adsorption isotherm equation: 1 1 1 x A = = К ⋅ Сeqn ; = = 0.32 ; 0,012 = 0,25 ⋅ Сeq0.32 , m n 3.1 therefore: Сeq0.32 = 0,048. 2. Calculate the equilibrium concentration of acetic acid: C= 0.32 0.048 = 8 ⋅10−5 mmol/g . Alternatively the equilibrium concentration may be calculated by taking logarithm of this equation: log Сeq0.32 = log 0,048, 0.32 log Ceq. = log 0.048, log Ceq = log 0.048 = −1.3188 = − 4.1213 0.32 0.32 Сeq. = alog(–4.1213) = alog (–5 + 0.8787) = 7.57⋅10–5 ≈ 8⋅10–5 mmol/g. Numerical problem 4. To 5 numbered flasks, each containing 100 ml of acetic acid solutions of various concentrations at 20 °C 3 g of activated charcoal were added. The acid solutions concentrations before and after the adsorption were evaluated by 25 ml each solution titration with 0.1 M NaOH solution with phenolphthalein. Calculate the adsorption magnitudes (A) and acetic acid equilibrium concentrations (Ceq) for each solution according to the experimental data listed in the Table below: The volumes of NaOH solution consumed for 25 ml acids Flask No. solution titration, ml 1 2 3 4 5 before the adsorption 2.75 3.90 5.60 11.50 22.20 after the adsorption 0.60 1.05 1.83 5.10 12.50 Steps to Solution: 1. The following values should be calculated firstly: 1) Со – the acid concentration before the adsorption (mmol/l ), 2) Сeq – the acid concentration after the adsorption (mol/l). For example, before the adsorption the acid concentration in solution No 1 may be calculated using the equation of the equivalents law: СacidVacid = СalkaliValkali; 57 C0 = C alkali ⋅ Valkali 0.1 ⋅ 2.75 = = 0.0111 mol/l = 11.1 mmol/l; Vacid 25 2. After the adsorption for the same solution: Ceq = 0.1 ⋅ 0.6 = 0.0024 mol/l = 2.4 mmol/l. 25 3. The initial acid concentrtions and its concentrations after the adsorption for other solutions may be calculated in the same manner. 4. The adsorption magnitude A= x/m (mmol per 1.0 g of the adsorbent) should be calculated according to the equation: x (C0 − Ceq ) ⋅ V A= = m m where: х – the change of the adsorptive moles number owing to adsorption, mmol; m – the mass of the adsorbent, g; Со – the initial cid concentration, mol/l. Сeq – the cid concentration after the adsorption, mol/l; V – the volume of the acid solution in which the adsorption was studied, l. For solution No 1: A= (C 0 − Ceq ) ⋅ V (0.0111 − 0.0024 ) ⋅ 0.1 = = m 3 0.29 mmol/g. 5. The same calculations should be done for other solutions. Numerical problem 5. Define the surface charge sign of phosphorite in the natural spring water which contain different soluble salts like Са(НСО3)2. Steps to Solution: Water insoluble phosphorite Са3(РО4)2 is a polar adsorbent of trigonal symmetry structure in the nodes of which Са2+ and РО43– ions are alternating. According to Fajans-Peneth adsorption rule Са2+ cations from water are attracting electrostatically to РО43– anions on the phosphorite surface and are finishing building its crystal lattice. That is why the phosphorite surface will gain the positive charge. с) Problems to Solve 1. Calculate the equilibrium concentration of acetic acid if it was found that 1 g of activated charcoal can adsorb 3.76·10–3 moles of the acid. K and n constants in Freundlich equation are 2.82 and 2.44, respectively. Answer: 10–7 mol/l 2. Define К and 1/n constant in Freundlich equation for oxalic acid adsorption with activated charcoal (m = 1 g) at К using the following experimental data: С, mmol/l A, mmol/g 280 1.26 440 1.58 660 2.00 58 1260 2.80 2950 4.00 What does Freundlich equation look like? Answer: log K = –1.2; К = 0.063; tgφ = 1/n = 0.53 3. Calculate the boundary adsorption magnitude for the Langmuir adsorption isotherm if the solute adsorption magnitude was determined as 6·10–3 mol/m2 at the equilibrium solute concentration of 0.03 mol/l. The constant K for the Langmuir equation is 0.8. Answer: 0.166 mol/m2. 4. Illustrate schematically the selective adsorption phenomena and define the surface charge signs: − for the chalk particles immersed into Ca(NO3)2 solution; − for calcium silicate CaSiO3 particles immersed into soluble glass ( Na2SiO3) solution. Answer: In the first case the adsorbent surface would gain a positive charge, in the second case – a negative charge 4. Laboratory Activities and Experiments Section 4.1. Practical Skills and Suggested Learning Activities – to determine and interpret gas adsorption isotherms; – to determine the magnitude of a solute adsorption in terms of the Langmuir adsorption isotherm; – to study the adsorption ability of activated charcoal; – to separate colored salts from their mixture. 4.2. Experimental Guidelines 4.2.1. The acetic acid adsorption on activated charcoal determining 1. Prepare aqueous solutions of acetic acid into numbered flasks using pipettes and measuring cylinders from 0.4 M solution following the scheme given in the Table 1. The total volume of each solution is 50 ml. Use flasks fitted with stoppers. Table 1. Scheme for acetic acid dilution Flask No. 1 2 3 4 Volume of 0.4 M acetic acid solution V1 (ml) 50 25 12.5 6.25 Volume of distilled water V2 (ml) – 25 37.5 43.75 Total volume (ml) 50 50 50 50 2. Transfer 10 ml of the solution from each flask into numbered titrimetric flask, so final volume of acetic acid solution in the numbered flasks is VA=40 ml per flask. 3. Determine the actual concentration of acetic acid in flasks by titration in this way: a) For titration, modify the volume in each titrimetric flask. Take away defined volume of the solution, to obtain in each flask the volume as given in the Table 2. 59 Table 2. Volumes of the acetic acid solutions used for titration before and after the adsorption Titrimetric flask No. 2 3 4 1 Volume V (ml) 10 10 10 10 b) Add 2-3 drops of phenolphthalein and titrate by NaOH. c) Once the endpoint has been reached, read the burette. The volume of the base Vio(ml) that was required to reach the endpoint write down to the Table 3. d) Calculate the actual concentration of acetic acid Cio in the flasks No. 1 – 4, respectively, and write it down to the Table 3: V o ⋅ СT , [mol/l] (1) Ci o = i V 4. 5. 6. 7. 8. where Vio is the volume of the titrant (NaOH), CT is the concentration of the titrant, V is the volume of the analyte (acetic acid according to Table 2), i=1–4 is the number of flask. Using practical balance and glazed paper, weigh 4 portions of activated charcoal, each portion 1 g. The accuracy of weighing must be 0.01 g. Put activated charcoal into numbered flasks with stoppers (1 portion per flask). Plug up the flasks, and shake them. Wait for 20 minutes, the process of adsorption is in progress. Mix the mixtures for several times by flasks shaking within this period. (Remark: The process of adsorption is a function of time too. It is important to put charcoal into flasks at the same time, to provide adsorption for the same period in each flask). Filter the mixtures into clean and dry flasks. To avoid disturbing effect of adsorption of acetic acid into filtering paper, remove away the first portion of filtration, app. 5 ml. Determine the final concentration of acetic acid Ci in each of the flasks after adsorption. From each solution, transfer the asked volume into clean and dry titrimetric flask, again following Table 2. Calculate the concentration of acetic acid after adsorption (Ci), using the Eq. 1 and data form Table 3 after adsorption. Repeat points 3a-3d, and from the consumed base Vi (ml) determine the concentration of acetic acid Ci after adsorption. Write them down to the Table 3. Table 3. Experimental data for the adsorption Flask No. Vio (ml) Cio (mol/l) Vi (ml) 1 2 3 4 60 Ci (mol/l) A= х m mmol/g 9. Finishing experiment, wash carefully used flasks, pipettes, etc. 10. Determine of the adsorption A as the substance amount of acetic acid adsorbed per gram of the charcoal m (mol/g) in individual flask: A= o х (С i − С i ) ⋅ V = m m (2) where Cio is initial concentration of acetic acid, mmol/l; C i is equilibrium concentration of acetic acid, mmol/l; V is volume of acetic acid taken for the adsorption (the volume of the liquid phase in the mixture charcoal – acetic acid), ml; m is mass of adsorbent, (1 g).; i=1–4 is the number of flask. Eq. 2 supposes that V is the same for i=1–4, and also the mass of the charcoal (g). Write down the obtained values of Ai to the Table 3. 11. Plot the dependence (isotherm) A = X – f(С) (T = const) for the adsorption of m acetic acid on activated charcoal. Make a conclusion about the equilibrium concentration of the acid affecting its adsorption. A= X m Ci Fig. 6.1 Freundlich adsorption isotherm 5. Conclusions and Interpretations. Lesson Summary Topic 7 Lyophobic Sols Preparation and Their Properties Studying 1. Objectives Colloids and colloidal systems are essential to life. Most of the substances, we come across in our daily life, are colloids. The meals we eat, the clothes we wear, the wooden furniture we use, the houses we live in, the newspapers we read, are largely composed of colloids. Following are the interesting and noteworthy examples of colloids: blue colour of the sky, fog, mist and rain, food articles: milk, butter, halwa, ice creams, fruit juices, etc., are all colloids in one form or the other. Colloids function in every body cell, in the blood, and in all body fluids, especially the intercellular fluids. Blood is a colloidal solution of an albuminoidal substance. The styptic action of alum and ferric chloride solution is due to coagulation of blood forming a clot which stops further bleeding. All life processes take place in a 61 colloidal system, and that is true both of the normal fluids and secretions of the organism, and of the bacterial toxins, as well as, in large measure, of the reactions, which confer immunity. Atmospheric soils are mainly colloidal in nature. On account of colloidal nature, soils adsorb moisture and nourishing materials. Dispersion refers to the spatial distribution of organisms, and is a fundamental component of a species' ecology and life history. Patterns of dispersion influence other aspects of a species' behavior and ecology. Dispersion patterns themselves are affected by the distribution of resources (including sunlight, nutrients, prey species, etc.) in the environment, as well as through the reciprocal influence of species' behavioral characteristics. 2. Learning Targets: – to get practical skills in lyophobic sols preparation; – to learn their properties; – to learn the structure of micelle; – to study capillary analysis bases and its application for the determining the charge sign of colloidal drugs; – to learn the theory of light dispersion and Rayleigh’s equation; − electrical properties of dispersions, introduction to electrokinetic phenomena; − to learn the micelle structure in isoelectric point; − to get practical skills in zeta-potential magnitude measurements and to know how to use it for lyophobic sols stability characteristics; – to learn electrophoresis applications in medical and biological researches. 3. Self Study Section 3.1. Syllabus Content The living organism as a disperse systems combination. Classification of disperse systems according to the aggregative state, interphase interaction, dispersion. Lyophilic and lyophobic dispersions. A structure of micelle. Structure of a double electric layer (DEL). The overall performance and history of development the ideas about DEL structure. Electrokinetial potential of a colloidal particle. Purification of colloidal solutions. Dialysis, electro-dialysis, ultrafiltration, compensatory dialysis. Haemodialysis and “artificial kidney” device. Molecular - kinetic properties of dispersions. Thermal molecular motion and Brownian motion, diffusion, and osmotic pressure. Optical properties of dispersions. Electrical properties of dispersions. Introduction to electrokinetic phenomena. Electrophoresis and electroosmosis. Helmholz-Smoluchowsky equation. The electrophoresis uses in clinical and laboratory practice researches 3.2. Overview Colloidal solutions are intermediate between true solutions and suspensions. The size of the colloidal particles range from 10-9 to 10-7 m. A colloidal system consists of two phases – the dispersed phase and the dispersion medium. Colloidal systems 62 are classified in three ways depending upon physical states of the dispersed phase and dispersion medium; nature of interaction between the dispersed phase and dispersion medium, and nature of particles of dispersed phase. The colloidal systems show interesting optical, mechanical and electrical properties. Colloidal sols can be formed by dispersion methods (e.g. by mechanical subdivision of larger particles or by dissolution in the case of lyophilic sols) or by condensation methods (from supersaturated solutions or supercooled vapours, or as the product of chemical reactions) or by a combination of these two (e.g. in an electrical discharge). Dispersion can be done mechanically, in a colloid mill that grinds the substance into small, equal particles. Another method is with an electric arc. Metal electrodes are used, at a current of 5-10A and voltage of 30-40V. When a condensation method is applied, molecules (or ions) are deposited on nuclei, which may be of the same chemical species as the colloid (homogeneous nucleation) or different (heterogeneous nucleation). Peptization may be defined as the process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence a small amount of electrolyte. The electrolyte used for this purpose is called peptizing agent. This method is applied, generally, to convert a freshly prepared precipitate into a colloidal sol. During peptization, the precipitate adsorbs one of the ions of the electrolyte on its surface. This causes the development of positive or negative charge on precipitates, which ultimately break up into smaller particles of the size of a colloid. According to the micellar theory of colloids solution structure, sol consists of micelles and intermicellar liquid. A micelle is a colloidal-sized particle formed by the association of molecules, each of which has a hydrophobic end and a hydrophilic end. There are 3 main parts in the micelle structure: nucleus, adsorptive and diffuse layers. A nucleating agent is a material either added to or present in the system, which induces either homogeneous or heterogeneous nucleation. Colloidal particles always carry an electric charge. The nature of this charge is the same on all the particles in a given colloidal solution and may be either positive or negative. The charge on the sol particles is due to preferential (selective) adsorption of ions from solution and/or due to electrical double layer formation. The sol particles acquire positive or negative charge by selective adsorption of cations or anions. When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle usually takes place according to the Fajans-Peneth adsorption rule. Having acquired a positive or a negative charge by selective adsorption on the surface of a colloidal particle, this layer attracts counter ions from the medium forming adsorptive and diffusive layers of counterions. According to modern views, the first layer of counterions is firmly held and is termed the adsorptive layer while the second layer is mobile which is termed the diffusive layer. Since separation of charge is a seat of potential, the charges of 63 opposite signs on the adsorptive and diffusive parts of the double layer results in a difference in potential between these layers. This potential difference between the adsorptive layer and the diffusive layer of opposite charges is called the electrokinetic potential, or zeta potential. The magnitude of zeta potential is crucial in determining the stability of a colloidal suspension. When all the particles have a large negative or large positive charge they will repel each other, and so the suspension will be stable. A sol is supposed to be stable when its zeta-potential magnitude is in the range of 30–90 mV. If the zeta potential is low the tendency for flocculation is increased. Another important consideration when discussing zeta potentials is pH; in fact, quoting a zeta potential without an accompanying pH is almost meaningless. This is due to the fact that, for suspensions of most materials, a plot of zeta potential versus pH exhibits an isoelectric point, a particular value of solution pH where the net charge on the particles is zero. At this point the suspension is highly unstable, and flocculation is at its most likely. An important consequence of the existence of electrical charges on the surface of particles is that they interact with an applied electric field. These effects are collectively defined as electrokinetic effects. There are four distinct effects depending on the way in which the motion is induced. These are: Electrophoresis: the movement of a charged particle relative to the liquid it is suspended in under the influence of an applied electric field. Electroosmosis: the movement of a liquid relative to a stationary charged surface under the influence of an electric field. Streaming potential: the electric field generated when a liquid is forced to flow past a stationary charged surface. Sedimentation potential: the electric field generated when charged particles sediment. The Helmholts-Smoluchowski equation for zeta-potential calculation by electrophoresis method: ξ= Slη τEε 0ε where S – linear motion on a sol boundary, m; l – distance between the electrodes in the device for electrophoresis, m; η – viscosity of the dispersion medium, Pa·s; τ – time of electrophoresis, sec.; E – electric field voltage V, ε – dielectric constant of medium (=81 for water), ε0 = 8.85·10–12 Fh/m – dielectric constant of vacuum. Systems containing colloid particles are in some properties different from systems which are homogeneous or composed of macroscopic phases. When the beam of light is passing through a dispersed system the following phenomena may be observed: transmission, absorption, refraction, reflection and scattering of light. The prevailing of any of these processes depends on the ratio of the wavelengths of the light and the sizes of the suspended particles. 64 The molecular and ionic solutions are optically transparent as the beam of light is just transmitting through them. Any medium is able to absorb selectively the light waves with the certain wavelengths. The optical properties of colloids depend on the size and structure of suspended particles and the dispersed phase concentration. The diffraction of light will be mainly observed for colloids as their particles sizes are equal about half of the initial light wavelength (λ/2). The diffraction courses the opalescence of colloids – the lusterless luminarity of soles usually with a touch of blue. If light passes through a system containing colloid particles, part of the light is scattered and consequently the ray passage through the environment can be observed (the Tyndall effect). Rayleigh equation shows the scattered light intensity affection with different factors: 2 n 2 − n02 υV particle I scat = 24π 3 I 0 21 n1 + 2n02 λ4 where Io and Iscat are the intensities of the initial and scattered light beam, respectively; υ is the number of particles of the volume Vparticle in unit volume; λ is the wavelength of the incident light, n0 is the refraction index or the dispersion medium, and n1 is refraction index of the dispersed phase. From this equation analysis it is obvious that: − in order for scattered light to originate, the diffraction index of the dispersion ratio and that of the dispersion environment have to be different; − light scattering is proportional to the number of particles in the system; − the scattered light is proportional to the quadrate of the particles volume; − the scattering is inversely proportional to the fourth power of the wavelength of the incident light. 3.3. References 1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p. 2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995 p. 3. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. – Saunders College Publishing. – 1990. – 806 p. 4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001. – 1067 p. 5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGraw-Hill. – 1998. – 561 p. 6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The Central Science. – Prentice Hall. – 2000. – 1017 p. 7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular Science. – Mosby. – 1994. – 977 p. 8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. – 65 1997. – 1031 p. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991. – 625 p. 3.4. Self Assessment Exercises а) Review Questions 1. Define the disperse systems and give their classifications according to different criteria. 2. Show the divergences between colloids and true solutions. 3. Name and explain the main methods of sols preparation. The difference between the peptization method and the dispersion and condensation methods. 4. Name the parts of the micelle structure. State the Fajans-Peneth adsorption rule. Define the potential-defining ions, nucleus, adsorptive and diffuse layers of a micelle. 5. Consider the examples of AgCl, As2S3, PbS lyophobic sols preparation with positively and negatively charged granules. Write down the schemes of their micelles structure. 6. The prevailing of light scattering or absorption and reflection depending on the sizes of the suspended particles. The Tyndall effect. 7. Write down the Rayleigh’s equation and make its analysis. Which variables do affect the scattered light intensity? 8. Research techniques of colloid systems, based on the phenomenon of dispersion of light. Nephelometry. Turbidimetry. Ultramicroscopy. Electronic microscopy. 9. Name the main methods of colloidal solutions purification. Their medicinal applications. The principle of the "artificial kidney" device operation. 10. Give the definitions of electrophoresis and electroosmosis. 11. Draw the schematical representation of the device for electrophoresis and electroosmosis. 12. Define the electrokinetical (zeta-) potential and explain the mechanism of its appearance. Explain its role in determining the stability of colloids. 13. Draw the micelle structure for Fe(OH)3 sol prepared in the presence of Fe(NO3)3 as a nucleating agent. Show the location of zeta-potential occurrence. Draw the structure of the same micelle in isoelectric state. 14. Application of electrophoresis in medical and biological researches. b) Types of Numerical Problems and Their Solving Strategies 9. Numerical problem 1. 12 ml of 0.01 М of FeCl3 solution were added to 10 ml of 0.05 N K4[Fe(CN)6] solution to obtain sol of Berlin blue. Write down the formula of micelle of this sol. What is the charge of the granule? Steps to Solution: 1. Calculate the normal concentration of 0.01 M FeCl3 solution: СN = СМ⋅3 = 0.01⋅3 = 0.03 mol-eq/l. 66 2. Write down the equation of Berlin blue formation: 12 KCl. 4 FeCl3 + 3 K4[Fe(CN)6] → Fe4([Fe(CN)6])3↓ + 3. Calculate which compound is in excess and is stabilizer: Number of mmol-eq (FeCl3) = 12⋅0.02 = 0.24 mmol-eq; Number of mmol-eq (K4[Fe(CN)6]) = 10⋅0.05 = 0.5 mmol-eq. 4. In solution there is the excess of K4[Fe(CN)6], which is the stabilizer. Therefore, hydrophobic Fe4([Fe(CN)6])3 particles will adsorb [Fe(CN)6]4– ions (FajansPeneth adsorption rule), which determine the negative charge of the granule. К+ ions will be counterions. 5. The scheme of structure of Berlin blue micelle in this case is: {[(mFe4([Fe(CN)6])3) ⋅ n[Fe(CN)6]4– ]4n– ⋅ 4(n – x) K+}4x– ⋅ 4x K+ (mFe4([Fe(CN)6])3) – aggregate; [(mFe4([Fe(CN)6])3) ⋅ n [Fe(CN)6]4– ]4n– – nucleus; {[(mFe4([Fe(CN)6])3) ⋅ n[Fe(CN)6]4– ]4n– ⋅ 4(n – x) K+}4x– – granule; n[Fe(CN)6]4– – potential determining ions; 4(n – x) K+ – adsorptive layer; 4xK+ – diffuse layer. Numerical problem 2. What is the minimum volume of 0.1 % AgNO3 solution which should be added to 50 ml of 0.15 % KI solution for preparing of sol with positively charged granules? Steps to Solution: The reaction of AgI formation is: AgNO3 + KI → AgI + KNO3 Sol with positively charged granules will form if AgNO3 is the stabilizer of the sol. The scheme of micelle is: {[(mAgI)nAg+]n+(n–x)NO3–}x+xNO3– In this case the AgNO3 solution must be in excess. According to the equivalent law: CM(KI)·V(KI) = CM(AgNO3)·V(AgNO3) The volume of AgNO3 solution is: V ( AgNO 3 ) = С M (KI) ⋅ V ( KI) C M (AgNO 3 ) Calculate the molarity of 0.15% KI solution: С M ( KI) = m(KI) 0.15 g = = 0.009 mol/l . M (KI) ⋅ V 166 g/mol ⋅ 0.1 l Calculate the molarity of 0.1% AgNO3 solution: CM (AgNO3 ) = m(AgNO3 ) 0.1 g = = 0.006 mol/l . M (AgNO3 ) ⋅ V 170 g/mol ⋅ 0.1 l Calculate the equivalent volume of AgNO3 solution: 67 0.009 mol/l ⋅ 50 ml = 75 ml . 0.006 mol/l Answer: the minimum volume of AgNO3 solution is 76 ml. V = Numerical problem 3. Calculate the diameter of an aerosol particles, if 120 of such particles were found in the volume of 4·10–11 m3. The mass concentration of sol is 1·10–4 kg/m3, and the dispersed phase density is 2.2·103 kg/m3. Steps to Solution: The quantity of particles in the unit of volume equals: n ν= . V The volume of 1 particle is equal to the ratio between the mass of 1 particle and the density of dispersed phase: m C С ⋅V V0 = = = d ν ⋅d n⋅d For the spherical particles the volume of 1 particle is: 4 V0 = πr 3 3 Then, С ⋅ V0 4 = πr n⋅d 3 → r = 3 3 ⋅ C ⋅ V0 4π ⋅ n ⋅ ρ Calculate the radius of the spherical particles: r =3 3 ⋅1 ⋅10− 4 ⋅ 4 ⋅10−11 4 ⋅ 3.14 ⋅120 ⋅ 2.2 ⋅103 = 1.535 10-7 m Calculate the diameter of the spherical particles: D = 2r = 1.535⋅10–7 ⋅ 2 = 3.07⋅10–7 m. Answer: the diameter of sol particles is 3.07⋅10–7 m. Numerical problem 4. Calculate the value of ξ-potential (in mV) for sol of collargolum using the following electrophoresis data: linear motion on a sol boundary S=10 mm, distance between the electrodes l=20 cm, electric field voltage E=300 V, time of electrophoresis τ=5 min, viscosity of the medium η=10–3 Pa·s, dielectric constant of medium ε=81 and dielectric constant of vacuum ε0=8.85·10–12 Fh/m. Steps to Solution: According to the Helmholts-Smoluchowski equation zeta-potential is: 68 ξ= Slη 10 ⋅ 10 −3 m ⋅ 20 ⋅ 10 −2 m ⋅ 10 −3 Pa ⋅ s = = 0.031 V = 31 mV . τEε 0ε 5 ⋅ 60s ⋅ 300V ⋅ 81 ⋅ 8.85 ⋅ 10−12 Fh/m Answer: zeta-potential of sol particles is 31 мV. c) Problems to Solve 1. For preparing of silver chloride sol 85 ml of 0.005 М silver nitrate solution were added to 15 ml of 0.025 М potassium chloride solution. Write down the formula of obtained sol micelle. What is the charge of its granule? Answer: positive 2. What volumes of 0.029 % NaCl solution (the density is 1 g/cm3) and 0.001 N AgNO3 solutions should be mixed to prepare uncharged particles of AgCl sol? Draw the structure of the micelle in isoelectric state. Answer: V(NaCl) : V(AgNO3) = 1 : 4.96 3. At ultramicroscopy investigation of gold hydrosol in volume of V0=1.6⋅10–11 m3 70 particles (n) were found. Sol concentration С=7⋅10–6 kg/m3, gold density d=19.3⋅103 kg/m3. Determine the radius of dispersed phase particles (in m) assuming their shape as spherical. Answer: 2.705⋅10–8 m 4. AgI sol was obtained at slowly adding 15 ml of 0.2 % AgNO3 solution (the density is 1 g/ml) to 20 ml of 0.01 М KI solution. Write down the formula of micelle of obtained sol. Determine the direction of its particles movement in the applied electric field. 5. What volume (in ml) of 0.005 М ZnCl2 solution should be added to 20 ml of 0.015 М (NH4)2S solution to prepare ZnS sol with positively charged granules? Write down the formula of micelle of obtained sol. Answer: 60 ml 6. Calculate zeta-potential of protein particles if under the electrophoresis its particles have been moved at a distance of 22.5 mm during 20 min at the applied voltage of 200 V. Distance between the electrodes is 15 cm. Viscosity of the solution η = 10–3 Pа⋅sec and constants ε = 81 і ε0 = 8.85⋅10–12 Fh/m. Answer: 19.6 mV 4. Laboratory Activities and Experiments Section 4.1. Practical Skills and Suggested Learning Activities – to prepare lyophobic sols with physical condensation method (a solvent replacement); – to prepare lyophobic sols with chemical condensation method under the double exchange reactions, hydrolysis and redox reactions; – to prepare sols with the methods of direct and indirect peptization; – to determine the sign of particles charge by the capillary analysis method; – to determine the charge sign of a sol granules and zeta potential magnitude for the particles of a drug. 69 4.2. Experimental Guidelines 4.2.1. Preparation of sols by physical condensation (by replacing solvent) а) Preparation of colophony sol Pour 5 ml of distilled water into the test tube and add the solution of colophony in ethyl alcohol by drops while stirring. Record your observation. Pay attention to the opalescence - a characteristic feature of colloidal solutions. What method does a sol form with? b) Preparation of sulfur sol Pour 5 ml of distilled water into the test tube and add 0.5 ml of saturated solution of sulfur in ethyl alcohol while stirring. What is observed? 4.2.2. Preparation of sols by method of chemical condensation а) Preparation of silver iodide sol by double-exchange reaction To 100 ml of 0.002 M KI solution add while stirring 1 ml of 0.01 M of AgNO3 solution. Note the color of sol and write the micelle structure (in the excess of KI). What is the charge of the granule? Write the structure of AgI micelle, obtained at the excess of AgNO3. b) Preparation of Berlin blue sols with different charged granules In two test tubes prepare Berlin blue sols from 0.005 M FeCl3solution and 0.005 M K4[Fe(CN)6] solution in the following ratios: 1) 3 ml of FeCl3 and 1 ml of K4[Fe(CN)6]; 2) 3 ml of K4[Fe(CN)6] and 1 ml of FeCl3. Write the equations of reactions and the structures of the micelles for both sols. What factor is affecting the charge of the granules? c) Preparation of iron hydroxide (III) sol by the reaction of hydrolysis 50 ml of distilled water should be heated to boiling in the flask. Add 5% FeCl3 solution into the boiling water by drops. What is observed? Write the equations of reactions and the structure of obtained sol micelle. d) Preparation of copper hexacyano(II)ferrate sol by double-exchange reaction To 5 ml of 1% CuSO4 solution add a few drops of 0.01% K4[Fe(CN)6] solution. Write the equation of the reaction and the structure of the obtained sol micelle. Write the micelle structure of the sol obtained in the K4[Fe(CN)6] excess. e) Preparation of silicic acid sol To 5 ml of 5% Na2SiO3 solution add 0.5 ml of 0.1 M HCl solution while stirring. What is the structure of the obtained sol micelle? f) Preparation of metallic silver sol by reduction reaction To 10 ml of 0.001 M AgNO3 solution add 2-3 drops of 1% K2CO3 solution and 1 ml of freshly prepared solution of tannin (aldehyde), and heat. Yellowish-brown sol of metallic silver is formed: AgNO3 + K2CO3 = AgOK + KNO3 + CO2 stabilizer 70 2AgOK + R–COH + H2O = 2Ag↓ + R–COOH + 2KOH Write the structure of obtained sol micelle and explain the method of its preparation. g) Preparation of manganese dioxide sol by redox reaction To 25 ml of 0.1 M KMnO4 solution in the flask add 10% H2O2 solution from the burette by small portions until the solution droplet placed on filter paper with a glass stick would not give pink spot. Write the structure of sol micelle. h) Preparation of lead sulfide sol To 3 ml of 1% Pb(CH3COO)2 solution add 3 ml of 5% Na2S solution by drops. Record your observations, the equation of reaction and the structure of the obtained sol micelle. What is the structure of the micelle of PbS sol, obtained at the excess of Pb(CH3COO)2? 4.2.3. Preparation of sols by peptization а) Preparation of Fe(OH)3 Prepare the precipitate of Fe(OH)3 by the reaction of FeCl3 with NH4OH. For this purpose, 1 ml of saturated FeCl3 solution should be diluted with water to 20 ml. To the obtained solution add 5% NH4OH solution slowly till bleaching of the liquid above the precipitate. Write the equation of the reaction. Note the color of the precipitate. Pour away the liquid above the precipitate and rinse several times with distilled water (decantation). Washed precipitate divide into 3 test tubes and add: into І – 10 ml of distilled water; into ІІ – 10 ml of 2 % FeCl3 solution; into ІІІ – 10 ml of 0.02 М HCl solution. Record your observations after 10 minutes. Explain what happened with the precipitate in the second and third tubes? What are the types and mechanism of peptization? Write the structure of the prepared sols micelles. b) Preparation of Berlin blue sol. To 5 ml of 2% FeCl3 solution add 1 ml of a saturated K4[Fe(CN)6] solution. The precipitate should be filtered and washed with distilled water. After washing of the precipitate on the filter with 0.1 M solution of oxalate acid (Н2С2O4) a blue sol of Berlin blue is filtered. Write the equation of the reaction and the structure of the sol micelle, considering that precipitate is peptized by oxalate acid due to С2O42– ions adsorption. 4.2.4. Determination of the charge sign of colloidal particles of drugs In colored sols the sign of particles charge can be determined by method of capillary analysis when capillaries surface of filter paper is used as charged surface. It is based on the fact that cellulose capillary walls of filter paper are negatively charged (because of dissociation of cellulose hydroxyl groups), and water that moistens the paper – positively. Place the drop of investigated sol on the piece of 71 filter paper. After the drop absorption sol with positively charged particles is adsorbed on the paper and gives stain colored at the center and colorless on the edges; sol with negatively charged particles does not absorbe by the paper and forms an evenly colored stain. 4.2.5. Determination of the sign and value of zeta-potential of a medicine (hydrophobic sol directed by the teacher: protargol or collargolum) by macroelectrophoresis method 1) Into the knee A of the device for electrophoresis (see Figure 1), which is fixed in tripod, pour in intermediate fluid for a given sol to level ~ 3-5 mm above the upper edge K. Tap should be opened to let the air come out through it. 2) Turn off the tap K and place the electrodes E into both knees, but do not clamp the stoppers to let the fluid leak out. 3) Pour a given sol through the funnel В up to the top. 4) Use wire to clean the bottom end of the tube under the funnel В to remove any air bubbles. 5) Open the tap K slightly to let sol from the funnel leak slowly to the bottom of Ushaped tube and form a clear boundary between sol and intermediate fluid, which is lightweight than sol, in both knees. 6) Add some more sol into the funnel and withdraw separation boundary between sol and intermediate fluid in front of the digital divisions on the scale. 7) Turn off the tap K and record the positions of sol boundaries in both knees respectively to scales (they may not coincide). 8) Turn on the current. Write the pointer of voltmeter will show the voltage between electrodes, note the start time of electrophoresis. 9) Record the voltage value E by direct indications of voltmeter or calculate it, if the scale of the device has implicit graduation. Note, what is the voltage (300 V, 450 V or other) supplied by rectifier on the electrophoresis and where the pointer of the voltmeter stopped. S Figure 1. Scheme of the device for electrophoresis:А intermediate fluid; S - sol; В - funnel; Е and Е - electrodes; К - tap. 10) Stop the electrophoresis, turn off the current in 5-10 minutes (when the boundary of sol in the knee lower to 5-7 mm - Scale Ш). Record the end time of electrophoresis and determine the duration of current flow τ. 11) By the difference between the positions of sol boundaries determine the 72 displacement of sol particles (S). Note the direction (to which of the electrodes sol boundary shifted during the electrophoresis). Make a conclusion about the sign of the charge on the granules of investigated colloidal solution. 12) After switching off the current measure the distance l between the electrodes using thread and ruler (by the course of the current in the U-shaped tube). 13) Pour away the solution from the U-shaped tube, rinse it with water. 14) Briefly describe the procedure of the work done and calculate the results. 15) Convert all measured values into SI system, substitute them in the formula of Helmholtz-Smoluchowski equation to calculate ζ-potential: η ⋅l ⋅ S ζ= τ ⋅ D ⋅ E ⋅ε . where besides the already mentioned values l (m), S (m), τ (s), E (V) take the viscosity η = 10-3 Pa⋅s and D = 81 – dielectric constant for water, and absolute dielectric permittivity constant for vacuum ε = 8.85⋅10-12 F/m. 16) Recalculate the obtained value of ζ in volts to millivolts (mV). 17) Make a conclusion about the sign of the granules charge of investigated sol and indicate whether it is stable, comparing the obtained value of zetapotential with the critical value (30 mV). 5. Conclusions and Interpretations. Lesson Summary Topic 8 The Stability of Colloidal Systems. Coagulation and Colloidal Protection 1. Objectives The long-term colloidal stability of dispersion will be of great importance in a number of industries such as pharmaceutical, ceramic, paints and pigments. The term “stability” can have different connotations to different applications. When applied to colloids, a stable colloidal system is one in which the particles resist flocculation or aggregation and exhibits a long shelf-life. This will depend upon the balance of the repulsive and attractive forces that exist between particles as they approach one another. If all the particles have a mutual repulsion then the dispersion will remain stable. However, if the particles have little or no repulsive force then some instability mechanism will eventually take place e.g. flocculation, aggregation etc. Coagulation processes are widely used in the industry, for example in the electrical precipitation of smoke. They are also applied for the purification of drinking water: the water obtained from natural sources often contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes. Most of the medicines are colloidal in nature. For example, argyrol is a silver sol 73 used as an eye lotion. Colloidal antimony is used in curing kalaazar. Colloidal gold is used for intramuscular injection. Milk of magnesia, an emulsion, is used for stomach disorders. Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated. So the colloidal protection in drugs manufacturing is of a great importance. 2. Learning Targets: – to get practical skills in critical concentration of sols coagulation experimental determining; – to learn how is it possible to calculate the coagulating ability of electrolytes with different valences ions; – to know the protective action evaluation of polymers or hydrophilic colloids towards hydrophobic sols and to determine experimentally the protective number. 3. Self Study Section 3.1. Syllabus Content Kinetic (sedimentation) and aggregative stability of disperse systems. The reasons of colloidal stability. Coagulation. The mechanism of electrolytes coagulating action. Coagulation threshold or critical concentration of coagulation. Schulze-Hardy rule. Mutual coagulation of sols. Coagulation proceedings for the potable water and wastewater purification. 3.2. Overview Disperse systems stability is the ability to keep during certain time composition and main properties, such as dispersion, concentration, even distribution of dispersed phase particles in dispersion medium and way of particles interaction unchangeable. A stable colloidal system is one in which the particles resist flocculation or aggregation and exhibits a long shelf-life. This will depend upon the balance of the repulsive and attractive forces that exist between particles as they approach one another. If all the particles have a mutual repulsion then the dispersion will remain stable. However, if the particles have little or no repulsive force then some instability mechanism will eventually take place e.g. flocculation, aggregation etc. Kinetic (sedimentation) stability of a disperse system is the conservation of particles been distributed throughout the whole volume of a system. The main conditions for kinetic stability are high dispersion and Brownian movement (motion). This stability increases with temperature increasing and decreasing of particles size. Aggregative stability of disperse systems is the ability of a system to counteract adhesion (aggregation) of particles and keep certain degree of dispersion. The stability of the lyophobic sols is due to the presence of charged on colloidal particles. If, somehow, the charge is removed, the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. 74 To coagulate lyophilic sols, it is necessary to remove charge as well as the solvent layer. This is indicated by the fact that after the addition of an electrolyte if we also add a dehydrating agent, the coagulation sets in at once due to the removal of water layer. Thus the stability of the colloidal solutions is mainly due to three reasons: − Brownian motion; − presence of similar charge on colloidal particles; − solvation of colloidal particles. The process of settling of colloidal particles is called coagulation or precipitation of the sol. Coagulation is the process of decreasing of system dispersivity due to the dispersion phase particles enlargement. The coagulation of the lyophobic sols can be carried out in the following ways: − by electrophoresis: the colloidal particles move towards oppositely charged electrodes, get discharged and precipitated; − by mixing two oppositely charged sols: oppositely charged sols when mixed in almost equal proportions, neutralise their charges and get partially or completely precipitated. This type of coagulation is called mutual coagulation; − by boiling: when a sol is boiled, the adsorbed layer is disturbed due to increased collisions with the molecules of dispersion medium. This reduces the charge on the particles and ultimately lead to settling down in the form of a precipitate; − by persistent dialysis: on prolonged dialysis, traces of the electrolyte present in the sol are removed almost completely and the colloids become unstable and ultimately coagulate; − by addition of electrolytes: When excess of an electrolyte is added, the colloidal particles are precipitated. The reason is that colloids interact with ions carrying charge opposite to that present on themselves. This causes neutralisation leading to their coagulation. The ion responsible for neutralisation of charge on the particles is called the coagulating ion. A negative ion causes the precipitation of positively charged sol and vice versa. The most important coagulation factor for soles is electrolytes influence. Only certain quantity of electrolyte causes coagulation. Minimal concentration of electrolyte in mmol that may cause 1 l of sole coagulation is called coagulation threshold (Cthr) or critical concentration of coagulation (Ccr). The beginning of coagulation first signs are system color changing, appearing of turbidity. C cr = Vel ⋅ C el , Vsole + Vel where Vel – volume of the electrolyte (l), that causes coagulation; Cel – concentration of electrolyte (mol/l), Vsole – volume of colloidal solution (l). The value that is inverse to coagulation threshold is called coagulation capacity: Vc = 1/Cthr. Coagulation capacity is the volume of sol, for whose coagulation 1 mmol of the 75 electrolyte is needed. Critical concentration of coagulation value is mainly determined by the valence rather than the type of the ions with opposite charge sing to the particles. Ions with the same charge sing than the particles are of secondary importance. According to the Schulze-Hardy rule the coagulating ability of ion increases with increasing of its charge, therefore the lowest critical concentration of coagulation have electrolytes with multi-charged coagulating ion: С thr (Cl–) > С thr (SO42–) > С thr (PO43–). Сthr(K+) > Сthr (Ca2+) > Сthr (Al3+) Coagulation threshold is inversely proportional to ion charge (z) raised to the power of 6: Сthr = 1/z6. That’s why the magnitudes of coagulation thresholds for one-, two- and threevalence ions correlate as: 1 1 1 C1 : C 2 : C 3 = 6 : 6 : 6 = 1 : 0.0156 : 0.00137 = 730 : 11.4 : 1 . 1 2 3 For inorganic ions with the same charges their coagulating ability increases with decreasing their degree of hydratation and increasing of the ion radius. Lyotropic, or Hofmeister Series: coagulating ability Li+ < Na+ < K+ < Rb+ < Cs+ coagulating ability CI– < Br– < I– < CNS– degree of hydratation degree of hydratation According to the kind of the repulsion force two mechanisms of the colloidal stability protection take place: − electrostatic stabilization of colloids; − stabilization of colloids with polymers. Polymeric stabilization of colloids involves polymeric molecules added to the dispersion medium in order to prevent the aggregation of the colloidal particles. The protective action of different polymers is expressed in terms of the protective number. Protective number is the number of milligrams of dry protective substance (polymer or hydrophilic colloid) which just prevents the precipitation of 10 ml of certain sol on the addition of 1 ml of 10% NaCl solution. 3.3. References 1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p. 2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995 p. 3. Rodney J. Sime. Physical Chemistry. Methods. Techniques. Experiments. – Saunders College Publishing. – 1990. – 806 p. 4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001. 76 – 1067 p. 5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGrawHill. – 1998. – 561 p. 6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The Central Science. – Prentice Hall. – 2000. – 1017 p. 7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular Science. – Mosby. – 1994. – 977 p. 8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. – 1997. – 1031 p. 9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991. – 625 p. 3.4. Self Assessment Exercises а) Review Questions 1. Define the reasons of colloidal stability. Kinetic (sedimentation) and aggregative stability of disperse systems. 2. Give the definition of the coagulation process. List the factors which may cause the coagulation of sols. 3. Define the coagulation threshold (critical concentration of coagulation). 4. State Schulze-Hardy rule. Explain it on some examples. 5. Explain the mutual coagulation of sols, give a few examples. 6. Explain the protective action of polymers or hydrophilic colloids towards hydrophobic sols. Define the term of the protective number. 7. Describe the protective action significance and applications in biology, medicine, pharmacy. b) Types of Numerical Problems and Their Solving Strategies Numerical problem 1. Calculate the critical concentration of coagulation for AgI sol, if 0.4 ml of 0.05 M BaCl2 solution should be added for 2 ml of sole coagulation. Steps to Solution: 1. Calculate the number of BaCl2 moles in 0.4 ml of 0.05 М solution: n = 0.05 mol/l · 0.4·10–3 l = 0.02·10–3 mol = 0.02 mmol. 2. Calculate the volume of solution after coagulation: V = 2 + 0.4 = 2.4 ml= 2.4·10–3 l. 3. Calculate the critical concentration of coagulation – the minimal number of BaCl2 mmol which is necessary for coagulation of 1 l of sol: Ccr = n 0.02 mmol = = 8.3 mmol/l V 2.4 ⋅ 10 − 3 l Answer: the critical concentration of coagulation is 8.3 mmol/l. 77 Numerical problem 2. How many millilitres of 1 M NaCl solution should be added to 500 ml of Fe(OH)3 sol to cause its coagulation if the critical concentration of coagulation is 104·10–3 mol/l? Steps to Solution: From the equation for critical concentration of coagulation: n V ⋅C Ccr = el = el el V Vsol + Vel → Vel С 104 ⋅ 10 −3 mol/l = cr = = 104 ⋅ 10 −3 . Vsol + Vel С el 1 mol/l Calculate the volume of NaCl solution solving the following equation: Vel = 104 ⋅ 10 −3 ⋅ (Vsol + Vel ) , Vel − 104 ⋅ 10 −3Vel = 104 ⋅ 10−3 ⋅ Vsol , 52 0.896Vel = 52 , Vel = = 58 ml . 0.896 Answer: the volume of 1 M NaCl solution is 58 mmol/l. Numerical problem 3. How many times the coagulation threshold of As2S3 sol will decrease if 0.5 М NaCl solution firstly been used for 10 ml of sol coagulation (1.2 ml were used for the coagulation) would be substituted with 0.036 М MgCl2 solution (0.4 ml were needed) or 0.001 М AlCl3 solution (0.1 ml were needed) for the same volume of sol (10 ml) coagulation. Steps to Solution: 1. Calculate the number of mmoles of each electrolyte needed for 10 ml of Ag2S3 sol coagulation: n1(NaCl) = С(NaCl)⋅V(NaCl) = 0.5 ⋅ 1.2 = 0.6 mmols NaCl; n2(MgCl2) = С(MgCl2)⋅V(MgCl2) = 0.036 ⋅ 0.4 = 0.0144 mmols MgCl2; n3(AlCl3) = С(AlCl3)⋅V(AlCl3) = 0.01 ⋅ 0.1 = 0.001 mmols AlCl3. 2. Calculate the total volumes of obtained solutions containing sol and coagulating electrolyte (V): V1 = Vsol + V(NaCl) = 10 + 1.2 = 11.2 ml = 11.2⋅10–3 l; V2 = Vsol + V(MgCl2) = 10 + 0.4 = 10.4 ml = 10.45⋅10–3 l; V3 = Vsol + V(AlCl3) = 10 + 0.1 = 10.1 ml = 10.1⋅10–3 l. 3. Calculate the coagulation threshold values for each electrolyte: n 0.6 Сcr1 = 1 = = 53.6 mmol/l NaCl; V1 11.2 ⋅ 10 − 3 Сcr2 = Сcr3 = n2 V2 n3 V3 = 0.0144 = 1.4 mmol/l MgCl2; 10.4 ⋅ 10 −3 = 0.001 10.1 ⋅ 10 − 3 = 0.1 mmol/l AlCl3. 4. How many times the coagulation threshold values of As2S3 sol will be lower for 78 MgCl2 and AlCl3 than for NaCl: Ccr1 Ccr 2 Ccr1 Ccr 2 = 53.6 = 38.3 times for MgCl ; 2 1.4 = 53.6 = 536 times for AlCl3. 0 .1 Numerical problem 4. A sol of AgCl was prepared by adding of 10 ml of 0.03 % NaCl solution to 25 ml of 0.002 M AgNO3 solution. Which electrolyte through the listed ones will have the lowest value of the critical concentration of coagulation for this sol: KBr, Ba(NO3)2, K2CrO4, MgSO4, AlCl3, K3PO4? Steps to Solution: 1. Calculate the molar concentration of 0.03 % NaCl solution, accepting the density being equal 1 g/ml and М(NaCl) = 58.5 g/mol using the following equation: ω ⋅ d ⋅ 10 0.03 ⋅ 1 ⋅ 10 = 0.00513 mmol/ml. СM = M ( NaCl) = 58.5 2. Calculate the volume of mixed solutions: V = V1 + V2 = 10 + 25 = 35 ml. 3. Calculate the number of mmoles of each reactant: n1(NaCl) = С1⋅V1 = 0.00513 ⋅ 10 = 0.0513 mmol; n2 (AgNO3) = С2⋅V2 = 0.001 ⋅ 25 = 0.025 mmol. 4. According to the equation of reaction: NaCl + AgNO3 → AgCl + NaNO3 0.0513 0.025 NaCl is in excess, so it is the stabilizer of the sol AgCl. The formula of the micelle is: {[m(AgCl) ⋅ nCl–]n– ⋅ (n – x) Na+}x– ⋅ xNa+. + 2+ 5. Cations K , Ba , Mg2+, Al3+ will cause the coagulation of this sol. According to the Schulze-Hardy rule the coagulating ability of ion increases with increasing of its charge. Therefore, the lowest value of critical concentration of coagulation has AlCl3. KBr, K2CrO4 and K3PO4 solutions of the same normality have the least coagulating ability. Numerical problem 5. Calculate the protective number and the coagulation threshold for Al2S3 sol with negatively charged granules if 3 ml of 0.5 % dextrin aqueous solution were added to 5.5 ml of the sol and the first signs of coagulation were observed with 1.5 ml of 2.5 M NaCl solution adding. 79 Steps to Solution: 1. Calculate the mass of dextrin in 3 ml of 0.5 % diluted solution (accepting its density being 1 g/ml): 3 ⋅ 0 .5 = 0.015 g = 15.0 mg of dextrin. 100 2. Calculate the mass of dextrin needed for its protective action toward 10 ml of sol instead of 5.5 ml: for 5.5 ml of sol protection 15 mg of dextrin are needed for 10 ml – х mg 15 ⋅ 10 х= = 27.3 mg. 5 .5 3. Find out the mass of NaCl in 1.5 ml (0.0015 l) of its 2.5 M solution, the molar mass of NaCl is 58.5 g/mol: m(NaCl) = CΜ⋅V(NaCl)⋅M(NaCl) = 2.5 ⋅ 0.0015 ⋅ 58.5 = 0.219 g NaCl. 4. Calculate the protective number that is the mass of dextrin (in mg) needed for just preventing the precipitation of 10 ml of sol on the addition of 1 ml of 10% NaCl solution (or ≈ 0.1 g of NaCl): for 0.219 g of NaCl 23.3 mg of dextrin re needed, for 0.1 g – х mg 27.3 ⋅ 0.1 x= = 12.5 (mg of dextrin). 0.219 5. Calculate the coagulation threshold for given sol: C ⋅V1 Сcr = , V where: С – the initial concentration of the coagulating electrolyte, mol/l; V1 – the volume of electrolyte needed for the coagulation, ml; V – the total volume of solution, ml . 2.5 ⋅ 1.5 C ⋅ V1 Сcr = = = 0.375 mol/l. Vsol + Vdextrin + V1 5.5 + 3.0 + 1.5 c) Problems to Solve 1. How many times the coagulation threshold of AgI sol will decrease if 1 М КNO3 solution firstly been used for 10 ml of sol coagulation (1.5 ml were used for the coagulation) would be substituted with 0.1 М Сa(NO3)2 solution (0.5 ml were needed) or 0.01 М Al(NO3)3 solution (0.2 ml were needed) for the same volume of sol (10 ml) coagulation. Answer: the coagulation threshold would decrease 27.4 times and 665.2 times, respectively 2. The coagulation of Al2S3 sol with negatively charged granules was carried out by the following electrolytes adding: KNO3, MgCl2 і AlCl3. Their critical concentrations of coagulation are 50.0, 0.72, and 0.093, respectively. Calculate the coagulation capacities ration for the cations with the different valences. 80 Answer: 1 : 69.4 : 537.6 3. Calculate the protective number for Fe(OH)3 sol if 5 mg of powder starch were added to 2 ml of the sol and the first signs of coagulation were observed with 0.2 ml of 10 % NaCl solution adding. Answer: 125 mg 4. Laboratory Activities and Experiments Section 4.1. Practical Skills and Suggested Learning Activities – to improve the skills of micelles structure composition for sols with different charges of granules; – to get practical skills in the critical threshold calculations; – to state and interprete Schulze-Hardy rule. 4.2. Experimental Guidelines 4.2.1. The coagulation threshold determining for iron (III) hydroxide sol Iron (III) sol Fe(OH)3 with positively charged granules was prepared in advance under the reaction of FeCl3 hydrolysis. Compose the micelle structure for this sol. For Fe(OH)3 sol coagulation electrolytes with the different valence anions are used, for example KCl, K2SO4, K3PO4 or K3[Fe(CN)6] their concentrations in mol/l are: 2.5; 0.1; 0.01, respectively. In three test tubes pour with the pipette 5 ml of the Fe(OH)3 sol each. The respective electrolyte solution should be added from burettes drop be drop into each test-tube until the first signs of coagulation would be observed: the turbidity appearance, color change (darkening), precipitation. Record the experimental data into the table below. Table 1 The experimental data obtained while coagulation threshold determining Test tube No Electrolyte added for coagulation The electrolyte Electrolyte solution used solution Coagulating ion for the concentration and its valence coagulation V, С, mol/l ml Coagulation threshold Сcr = C el ⋅ Vel Vsol +V el 1. 2. 3. Make the conclusions which of electrolytes and ions used have the highest coagulating ability. Explain why. Calculate the coagulation capacities ration for the anions with the different valences. 4.2.2. The protective action of a polymer towards hydrophobic sol coagulation studying In four test tubes pour with the pipette 2 ml of Fe(OH)3 sol with positively charged granules each. Study the coagulation of sol in two test tubes causing it by 81 two electrolytes adding with the anions of the different valence (two and three, respectively). The respective electrolyte solution should be added from burettes drop be drop into each test-tube until the first signs of coagulation would be observed. To the test tubes No 3 and 4 add firstly 1 ml of 0.5 % polymer solution (gelatine, starch etc.) and then study their coagulation in the same way as for the test tubes No 1 and 2. Compare the volumes of expended electrolytes solutions. Record the experimental data into the table below. Table 2 The experimental data for the polymer protective action studying Electrolyte Concentratio n of the electrolyte solution С, mol/l Used for coagulation V, ml without a in the polymer presence of polymer Coagulation threshold Ccr, mmol/l without a in the polymer presence of polymer Calculate the coagulation threshold values for each test tube by the equation: Сcr = Cel ⋅ Vel Vsol + Vpolymer + Vel , where: Сel – concentration of the electrolyte, mol/l; Vsol, Vpolymer, Vel – the volumes of a sol, polymer and electrolyte, respectively, l. Explain the obtained results and make the conclusion about the protective action of polymer towards sol coagulation. 4.2.3. Mutual coagulation of sols Pour 2 ml of Fe(OH)3 sol with positively charged granules into the test tube (compose the micelle structure for this sol if FeCl3 was used as a nucleating agent). Add 2 ml of Berlin blue sol with negatively charged granules. Compose the micelle structure for the obtained sol with K4[Fe(CN)6] being used as stabilizer. What process was observed after the two sols were mixed? Record and explain your observation. 5. Conclusions and Interpretations. Lesson Summary Topic 9 High molecular compounds. The determination of the swelling degree of gels and the influence of different factors on it. The determination of isoelectric point of proteins 1. Objectives Many polymers occur in nature, such as silk, cellulose, natural rubber, and proteins. In addition, a large number of polymers have been synthesized in the laborato82 ry, leading to such commercially important products as plastics, synthetic fibers, and synthetic rubber. The variation in the form of macromolecules is largely responsible for molecular diversity. Much of the variation that occurs both within an organism and among organisms can ultimately be traced to differences in macromolecules. There are four basic kinds of biological macromolecules. They are carbohydrates, lipids, proteins and nucleic acids. These polymers are composed of different monomers and serve different functionsЖ − сarbohydrates – composed of sugar monomers and necessary for energy storage; − lipids – include fats, phospholipids and steroids. Lipids help to store energy, cushion and protect organs, insulate the body and form cell membranes; − proteins – composed of amino acid monomers and have a wide variety of functions including molecular transport and muscle movement; − nucleic acids – include DNA and RNA that are so important in genes and life processes. Nucleic acids contain instructions for protein synthesis In fact, messenger RNA is what makes possible proteins, peptides, and enzymes. Natural and synthetic polymers are widely used in medicine and pharmacy. Many biomaterials, especially heart valve replacements and blood vessels, are made of polymers like Dacron, Teflon and polyurethane. Polymers are used for orthopedic and reconstructive surgery, and for tissue engineering.. They are used for drug delivery and as excipients in pharmaceutical tablets and hard gelatin capsules: − fillers ( starches); − disintegrants (starches, crosslinked polyvinyl pyrrolidone, sodium starch glycolate, sodium carboxymethyl cellulose); − binders (polyvinylpyrrolidone, cellulose derivatives, polyethylene glycol); − lubricants (polyethylene glycol; − antiadherents (starches); − film coatings (cellulose derivates, methylmethacrylate copolymers, methacrylic acid copolymers, polyvinyl acetate phthalate, polyethylene glycols). 2. Learning Targets: – to learn the main methods of polymers preparation; – to learn the structure of natural and synthesized polymers; – to get practical skills in the degree of swelling measuring; – to learn the electrolyted presence and narute affection the degree of polymers and living tissues swelling; – to learn the main properties of polyelectrolytes; – to get practical skills in the experimental determining of polyelectrolytes isoelectrical point; – to learn the technique of polyelectrolytes isoelectrical point determining by their solutions viscosity change and their precipitation. 83 3. Self Study Section 3.1. Syllabus Content Macromolecular compounds as the basis of living organisms. Globular and fibrillar structure of proteins. Macromolecular solutions features and their similarities and differences with true and colloidal solutions. Swelling and dissolution of polymers. The mechanism of swelling. Swelling affecting with pH, temperature and electrolytes nature. The role of swelling in the organism physiology. Gels creation in polymers solutions. The mechanism of gels formation. The influence of pH, temperature and electrolytes presence on the gels formation rate. Thixotropy. Syneresis. Diffusion in gels. Salting out effect of biopolymers. Coacervation and phase separation and its role in biological systems. Anomalous viscosity of polymers solutions. The viscosity of the blood. Donnan membrane equilibrium. Isoelectric state of proteins. Isoelectric point and its determining methods. Ionic state of biopolymers in aqueous solutions. 3.2. Overview A polymer molecule consists of the same repeating units, called monomers, or of different but resembling units. Figure 1 shows an example of a vinyl polymer, an industrially important class of polymer. In the repeating unit, X is one of the monofunctional units such as H, CH3, Cl, and C6H5 (phenyl). The respective polymers would be called polyethylene, polypropylene, polyvinyl chloride), and polystyrene. A double bond in a vinyl monomer CH2=CHX opens to form a covalent bond to the adjacent monomer. Repeating this polymerization step, a polymer molecule is formed that consists of n repeating units. We call n the degree of polymerization (DP). Usually, n is very large. It is not uncommon to find polymers with n in the range of 104–105. H H Figure 1. Vinyl polymer C C H X n Fig. 2 shows three architectures of a polymer molecule: a linear chain (a), a branched chain (b), and a cross-linked polymer (c). A bead represents a monomer here. A vinyl polymer is a typical linear polymer. A branched chain has branches, long and short. A cross-linked polymer forms a network encompassing the entire system. In fact, there can be just one supermolecule in a container. In the branched chain, in contrast, the branching does not lead to a supermolecule. A cross-linked polymer can only be swollen in a solvent. It cannot be dissolved. We will learn linear chain polymers in detail and about branched polymers to a lesser extent. 84 Figure 2. Architecture of polymer chain: a linear chain (a), a branched chain (b), and a cross-linked polymer (c). Some polymer molecules consist of more than one kind of monomers. An A–B copolymer has two constituent monomers, A and B. When the monomer sequence is random, i.e., the probability of a given monomer to be A does not depend on its neighbor, then the copolymer is called a random copolymer. Two main reactions are used for polymers preparation: − Polymerization: monomers' double-bonds open up to form continuous chain − Condensation: elimination of smaller molecule when functional groups react. Solutions of a high-molecular-weight polymer, even at low concentrations, can flow only slowly. Addition of a small amount of the polymer to the fluid can make it viscous, thereby preventing unwanted turbulence in the flow. When a polymer is added to a given solvent, attraction as well as dispersion forces begin acting between its segments, according to their polarity, chemical characteristics, and solubility parameter. If the polymer-solvent interactions are higher than the polymer-polymer attraction forces, the chain segment start to absorb solvent molecules, increasing the volume of the polymer matrix, and loosening out from their coiled shape. We say the segments are now "solvated" instead of "aggregated", as they were in the solid state. The whole "solvation-unfolding-swelling" process takes a long time, and given it is influenced only by the polymer-solvent interactions, stirring plays no role in this case. However, it is desirable to start with fine powdered material, in order to expose more of their area for polymer-solvent interactions. When crystalline, hydrogen bonded or highly crosslinked substances are involved, where polymer-polymer interactions are strong enough, the process does stop at this first stage, giving a swollen gel as a result. If on the contrary, the polymer-solvent interactions are still strongly enough, the "solvation-unfolding-swelling" process will continue until all segments are solvated. Thus, the whole loosen coil will diffuse out of the swollen polymer, dispersing into a solution. At this stage, the disintegration of the swollen mass can be favored by stirring, which increases the rate of dissolution. The degree of swelling is thus used as a common measure of the degree of crosslinking. The degree of swelling is measured as follows: 1. Measure the dry weight and/or volume of the polymer before swelling. 2. Swell the polymer in a solvent till equilibrium swelling would be achieved. 3. Measure the weight of the swollen polymer and/or volume of the polymer after equilibrium swelling. 85 4. The degree of swelling is measured using the following two swell ratios: Mass swell ratio i = m − m0 ⋅ 100%, m0 Volume swell ratio i= V − V0 V0 ⋅ 100%, where m0 and V0 are the initial mass nd volume of a polymer sample, m and V are its mass and volume after the swelling. The swell ratio is influenced with various factors: the nature of the polymer and the solvent, temperature, addition of electrolytes and pH. The latter variable is of the greatest importance. Electrolytes affecting is based on the anions hydratation ability. Some anions have the low ability for hydratation and can enhance the polymers swelling while other anions being well-hydrated would reduce the swelling. Lyotropic series of anions: CNS– > I– > KBr– > NO3– > Cl– > CH3COO– > SO42– > C2O42– no influence on swelling swelling reducing swelling enhancing Proteins are polyelectrolytes whose macromolecules contain acid and basic functional groups in the amino acids side chains. Amino acids in aqueous solutions are in the form of bipolar ions: H2N–R–COOH ⇄ +H3N–R–COO–. In acidic medium, when the ionization of carboxylic groups decreases, the protein molecule shows basic properties and gains positive charge: + H3N–R–COO– + H+ ⇄ +H3N–R–COOH. In basic medium vice versa the ionization of amino groups decreases, the protein molecule shows acidic properties and gains negative charge: + H3N–R–COO– + OH– ⇄ H2N–R–COO– + H2O. So proteins are amphoteric molecules while changes in pH value of the medium influence whether protein will be positively or negatively charged. Proteins net charge is the sum of all positive and all negative charges of amino acid side chains and of amino and carboxyl groups. Isoelectric point (pI) is characteristic pH value at which proteins net charge is zero. Proteins are positively charged at pH values of solution below their pI value and negatively charged when pH value of solution is higher then their pI value. When there is no electrical field, ampholytes are randomly distributed. By applying electrical field, ampholites start to move to one of the electrodes. For example, acidic ampholitic molecules with the lowest pI values will be most negatively charged and thus will migrate to anode until their net charge reaches zero. They will concentrate in that region. Since each ampholyte is a good buffer pH value of the medium is the sum of pI values of ampholites. After concentration of ampholytes is finished, steady state is reached and there is constant gradient of pH throughout the gel. If proteins’ pI values are inside the range of pH covered by ampholites which are used to construct pH gradient, when applied on gel, they will 86 migrate because of their charge until they reach zone with pH value identical to their pI (because at that pH value they have no net charge). This means that proteins with positive net charge will migrate to cathode and will become less and less charged until they reach their pI. Proteins with net negative charges will under the influence of an electrical field migrate across pH gradient to anode; again, until their pI value is reached. Because of this, in the focusing step, concentration of proteins happens according to their pI values. The particular behavior shown by polymer molecules, explains the high viscosity of polymer solutions. Solvent and low molecular weight solutes have comparable molecular size, and the solute does not swell when dissolving. Since molecular mobility is not restricted, and therefore intermolecular friction does not increase drastically, the viscosity of the solvent and the solution are similar. But the molecular size of polymer solutes is much bigger than that of the solvent. In the dissolution process such molecules swell appreciably, restricting their mobility, and consequently the intermolecular friction increases. The solution in these cases, becomes highly viscous. 3.3. References 1. V.O. Kalibabchuk, V.I. Halynska, L.I. Hryshchenko et al. Medical Chemistry. – AUS MEDICINE Publishing. – 2010. – 224 p. 2. Raymond Chang. Chemistry (6th Edition). – WCB/McGraw-Hill. – 1998. – 995 p. 3. Rodney J. Sime Physical Chemistry. Methods. Techniques. Experiments. – Saunders College Publishing. – 1990. – 806 p. 4. John McMurry, Robert C. Fay. Chemistry (3rd Edition). – Prentice Hall. – 2001. – 1067 p. 5. David E. Goldberg. Fundamentals of Chemistry (2nd Edition). – WCB/McGrawHill. – 1998. – 561 p. 6. Theodore L. Brown, H.Eugene LeMay, Bruce E. Bursten. Chemistry. The Central Science. – Prentice Hall. – 2000. – 1017 p. 7. John Olmsted III, Gregory M. Williams. Chemistry. The Molecular Science. – Mosby. – 1994. – 977 p. 8. Steven S. Zumdahl. Chemistry (4th Edition). – Houghton Mifflin Company. – 1997. – 1031 p. 9. Gary L. Miessler, Donald A. Tarr. Inorganic Chemistry. – Prentice Hall. – 1991. – 625 p. 3.4. Self Assessment Exercises а) Review Questions 1. Explain the terms polymer and monomer. 2. What are natural and synthetic polymers? Give two examples of each type. 3. How can you differentiate between addition and condensation polymerisation? 4. Discrete polymer’s dissolution steps. 87 5. Swelling and dissolution of polymers. Swelling mechanism and stages. Types of swelling. 6. Swell ratio and its affection with defferent variables. 7. Explain the nature of electrolytes affecting the swelling ratio. 8. Common and distinctive features of polymers solutions, true solutions and lyiphobic sols. 9. Explain the mechanism of gels formation in polymers solutions. What factors do influence gels formation? 10. Polyelectrolytes. Amphoteric properties of proteins. The charge sign of a ptotein macromolecule as a function of pH value of the medium. 11. Isoelectric point of proteins and its determining methods. 12. Properties of polymers solutions (viscosity, osmotic pressure, thixotropy). 13. Explain the salting out effect of biopolymers. Define coacervation as the process of phase separation and its role in biological systems. b) Types of Numerical Problems and Their Solving Strategies Numerical problem 1. A natural rubber sample with the volume of 1.094⋅10–4 m3 (V0) was placed into carbon disulfide. After 48 hours exposure at 293 K the sample volume becomes 9.204⋅10–4 m3 (V). Calculate the volume swell ratio of the polymer (%). Steps to Solution: The volume swell ratio of a polymer may be calculated by the equation: α= α= 9.204 ⋅ 10 −4 − 1.094 ⋅ 10 −4 1.094 ⋅ 10 −4 V − V0 ∆V ⋅ 100 % = ⋅ 100 % V0 V0 ⋅ 100 % = 8.11⋅ 10 −4 1.094 ⋅ 10 −4 ⋅ 100 % = 7.4132 ⋅ 100 % = 741.32 % Numerical problem 2. Calculate the molecular weight of ethylcellulose (in kg/mol) if its solution in aniline with the concentration of 4.9 kg/m3 has the osmotic pressure magnitude as 250.9 Pa. The second virial coefficient b2 is 0.52 Pа⋅m6/kg2. Steps to Solution: Osmotic pressure is a colligative property, which means that it is proportional to the concentration of solute. The van’t Hoff equation is often presented in introductory chemistry for calculating osmotic pressure from the moles of solute (nsolute) that occupy a given volume (V) and the absolute temperature (T) of the solution: π= nsolute ⋅ R ⋅ T (1) V Since all non-volatile, non-electrolytic solutions approach ideal behavior in the dilute limit, equation (1) is actually a limiting law, and should be written in the form: n ⋅ R ⋅ T (2) lim π = solute nsolute →0 V 88 It is more convenient to express concentration in terms of the 'grams' of solute per liter (or kg/m3), whereby we can make the following substitution in equation (1): nsolute C , = V M where M is the molecular weight of the solute. In this fashion, and with minor rearrangement, equation (2) can be written as: C ⋅ R ⋅ T (3) lim π = nsolute →0 M Since equation (3) is only exact in the dilute limit, we can recognize this relationship as the first term in a more general power series expansion in C: C ⋅ R ⋅T π= + b2 C 2 + b3C 3 + ... (4) M where b2 and b3 are called the second and third virial coefficients, respectively. These coefficients are empirically determined constants for a given solute-solvent system, and also depend on temperature. According to statistical mechanical solution theory, b2 represents the interaction of a single solute particle with the solvent, and higher order virial coefficients are associated with correspondingly larger number solute particle cluster interactions with the solvent. For the given problem solving we’ll use the equation (4) just with the second virial coefficient: π= C ⋅ R ⋅T + b2 C 2 M where: π is the osmotic pressure of solution, Pa; С– the polymer solution concentration, kg/m3; R = 8.314 J/mol⋅К; T– temperature, К; M – the molecular weight of the polymer. CRT M 4.9 ⋅ 8.314 ⋅ 313 π − b2 C 2 = M = CRT π − b2 C 2 = 250.9 − 0.52 ⋅ 4.9 2 = 53.5. Answer: the molecular weight of ethylcellulose is 53.5. Numerical problem 3. Blood serum albumin (рНpI = 4.64) was placed into the buffer solution with the hydrogen ions concentration [H+] = 10–6 mol/l. Towards which electrode the albumen perticles will migrate under electrophoresis? Steps to Solution: 1. Calculate the pH value of the buffer solution in which [H+] = 10–6 mol/l: pH = –log [H+] = –log10–6 = 6. 2. As pH of the buffer is higher than pHpI, then the albumen particles would gains negative charge: + NH3– R–COO– + OH– → NH3OH–R–COO–. 89 That is why the albumen particles will migrate in the direction of positively charged electrode – anode during the electrophoresis. Answer: albumen particles will migrate towards anode. Numerical problem 4. The efflux time for the egg albumin solution with the density of 1200 kg/m3 at 293 K from Ostwald viscometer is 27 sec. while the efflux time for the distilled water is 15 sec. Calculate the kinematic viscosity of the albumin solution if the density of water is 1000 kg/m3, water viscosity being equal 1.005⋅10–3 Pa⋅sec. Steps to Solution: The viscosity ηx of a solution (density dx) may be obtain the from the measurement of the efflux time τx, if the efflux time τ0 of a liquid with a known viscosity η0 and density d0 is measured for the same viscometer: d ⋅τ η =η x x , x 0 d0 ⋅ τ 0 where: η0 and ηx – kinematic viscosities of water and solution, respectively, Pa⋅sec; τ0 and τx – efflux times for water and solution, sec.; d0 and dx – the densities of water and solution, kg/m3. η x = 1.005 ⋅ 10 −3 ⋅ 27 ⋅ 1200 = 2.17 ⋅ 10 −3 Pa⋅sec. 15 ⋅ 1000 Answer: the kinematic viscosity of the albumin solution is 2.2⋅10–3 Pa⋅sec. c) Problems to Solve 1. Gelatine with рНpI of 4.7 is placed into the solution with the hydrogen ions concentration [H+] 1000 times greater than in water. What charge gelatine particles will gain in this solution? Answer: positive charge 2. The efflux time for gelatine solution with the density of 1050 kg/m3 at 298 K from Ostwald viscometer is 25 sec. while the efflux time for the distilled water is 10 sec. Calculate the kinematic viscosity of the gelatine solution if the density of water is 977.04 kg/m3, water viscosity being equal 0.8937⋅10–3 Pa⋅sec. Answer: 2.6 Pa⋅sec. 3. Calculate the efflux time of nitro cellulose solution from Ostwald viscometer. The density of the solution is 1120 kg/m3, its kinematic viscosity being 1.52⋅10–3 Pa⋅sec. The efflux time for water of the same volume is 12 sec. The kinematic viscosity of water is 1.11⋅10–3 Pa⋅sec, its density is 998.94 kg/m3. Answer: 14.7 sec. 4. Laboratory Activities and Experiments Section 4.1. Practical Skills and Suggested Learning Activities – to measure experimentally the swelling degree of a polymer sample; – to learn the electrolytes nature affection the polymer swell ratio; 90 – to determine gelatine pI and pHpI by swelling; – to learn the pH value affection the viscosity of a protein solution. To determine gelatine pI and pHpI as the minimum of viscosity; – to determine gelatine pI and pHpI as its maximal precipitation. 4.2. Experimental Guidelines 4.2.1. Swell ratio of gels determining with the mass-volumetric method Fill the vials of the device for swelling experiments (Fig.3) with pure water so that when you turn the device enclosed vial down the water level reaches about mid graduated tube. Record the level (V0). Figure 3. The device for the swell ratio of gels determining with the mass-volumetric method Then introduce a piece of dry gelatin (or other gel) with the mass m tied to filament into the opened vial, shut the stopper, carefully turn the device into a horizontal position and mark the swelling starting time. Every 15 minutes within 1 hour again pour the liquid into the enclosed vial and determine the loss of water volume recording its level on the graded tube (V). Calculate the swelling degree of gel (і). Table 1 Experimental data of swell ratio determining Time from No of The initial level The loss of water the, τ , record of water, V0, ml volume V, ml minutes 1 2 3 4 5 The solvelt volume been swollen with the gel sample during the time τ ∆V=(V0–V), ml Swell ratio і =∆V/m, ml/g 0 15 30 45 60 Draw the plot of swell ratio as a function of time. Make the conclusion about the type of swelling more typical for this polymer at a room temperature. 4.2.2. Determining the electrolytes nature affection swell ratio of gelatine Measure 10-15 ml of the following electrolytes 0.1 M solutions: K2SO4, CH3COOK, KCl, KI, KCNS. Pour each into one of five marked vials. Weigh five gelatine slabs, record the masses. Place the slabs into the vials and leave for 1 hour. 91 Then take out the gelatine slabs, dry gently with filter paper and reweigh. Record the experimental data into the table and calculate mass swell ratio for each electrolyte solution. Table 2 Experimental data of electrolytes nature affection swell ratio of gelatine No of vial Gelatine mass, g Electrolyte before the swelling, m0 after the swelling, m The solvelt mass been swollen with the gel sample, g ∆m = m – m0 Mass swell ratio і = ∆m/m0 Make the conclusion about the nature of anions affecting gelatine swelling. Write down the Lyotropic series of anions and explain which anions can enhance the polymers swelling while other anions reduce the swelling. 4.2.3. Gelatine pI and pHpI by swelling determining Pipette 10 ml of one of the acetate buffer solutions with different pH each into one of five marked vials. Weigh five gelatine slabs, record the masses (mo, g). Place the slabs into the vials and leave for 1 hour. Then pour out the solutions, dry gelatine slabs gently with filter paper and reweigh (m, g). Record the experimental data into the table and calculate mass swell ratio for each buffer solution. Table 3 Experimental data of pH affection swell ratio of gelatine No of vial Gelatine mass, g before the after the swelling, m0 swelling, m The solvelt mass been swollen with the gel sample, g ∆m = m – m0 Mass swell ratio і = ∆m/m0 1 2 3 4 5 Draw the plot of swell ratio і (Y-axis) as a function of pH (x-axis). Determine pI and pHpI for gelatine using the plot. Make the conclusion about the swell ratio magnitude of a polymer in its isoelectric point. 4.2.4. The pH value affection the viscosity of a protein solution. Gelatine pI and pHpI as the minimum of viscosity determining Pipette 10 ml of one of the acetate buffer solutions with different pH: 3.7; 3.9; 4.2; 4.7; 5.2 and pour each into one of five marked test tubes. Add 5 ml of 3 % gelatine solution into each tst tube and mix. Determine the viscosity of solutions with Ostwald viscometer. 92 Figure 2. Capillary Ostwald viscometer 1 – container for measuring the volume of liquid flowing through the capillary; 2 – capillary; 3 – container for the liquid collecting. To measure the efflux time of solvent (τ0) and gelatine solution (τх) from capillary Ostwald viscometer using stopwatch. Firstly the efflux time for water should be fixed, then the efflux time for each solution. The time measurements should be performed three times for each case and the average value for each solution should be recorded. Calculate the relative viscosity for each solution by the equation: τ η x = η0 x , τ0 where:η0– the relative viscosity of water at given temperature (1⋅10–3 Pa⋅sec at 298 K); ηх – the relative viscosity of the analyte solution. Record the measurements into the table. Table 4 Experimental data for the relative viscosity determining No of the test tube pH of solution Efflux time τ, sec. Relative viscosity 1 3.7 2 3.9 3 4.2 4 4.7 5 5.2 Н2О Plot the relative viscosity values against pH. What pH corresponds to the isoelectric point? Explain the effect of pH on the viscosity of protein solutions. 4. Conclusions and Interpretations. Lesson Summary 93 Appendixes Appendix A Standard Thermodynamical Functions or Some Substances at 298 К G ∆ S , , kJ/mol , J/(mol·К) kJ/mol 8 9 02 8 9 02 H ∆ 8 9 02 Substances 0 28.3 0 –1676.0 50.9 –1582.0 0 5.7 0 ССl4 (l) –135.4 214.4 –64.6 СО (g) –110.5 197.5 –137.1 CO2 (g) –393.5 213.7 –394.4 СаСО3 (s) –1207.0 88.7 –1127.7 СаF2 (s) –1214.6 68.9 –1161.9 Ca3N2 (s) –431.8 105.0 –368.6 СаО (s) –635.5 39.7 –604.2 Са(ОН)2 (s) –986.6 76.1 –896.8 0 222.9 0 Сl2O (g) 76.6 266.2 94.2 СlO2 (g) 105.0 257.0 122.3 Сl2O7 (l) 251.0 – – Cr2O3 (s) –1440.6 81.2 –1050.0 CuO (s) –162.0 42.6 –129.9 Al (s) Al2O3 (s) С (graphite) Cl2 (g) Fe (s) FeO (s) 0 27.2 0 –264.8 60.8 –244.3 0 130.5 0 –36.3 198.6 –53.3 HCN (g) 135.0 113.1 125.5 HCl (g) –92.3 186.8 –95.2 HF (g) –272.8 H2 (g) HBr (g) –270.7 178.7 HI (g) 26.6 206.5 1.8 HN3 (l) 294.0 328.0 238.8 H2O (g) –241.8 188.7 –228.6 H2O (l) –285.8 70.1 –237.3 H2S (g) –21.0 205.7 –33.8 KCl (s) –435.9 82.6 –408.0 KClO3 (s) –391.2 143.0 –289.9 MgCl2 (s) –641.1 94 89.9 –591.6 G ∆ S , , kJ/mol , J/(mol·К) kJ/mol 8 9 02 8 9 02 H ∆ 8 9 02 Substances Mg3N2 (s) –461.1 87.9 –400.9 MgO (s) –601.8 26.9 –569.6 N2 (g) 0 191.5 0 HN3 (l) 294.0 328.0 238.8 –46.2 192.6 –16.7 NH4NO2 (s) –256.0 – – NH4NO3 (s) NH3 (g) –365.4 151.0 –183.8 N2O (g) 82.0 219.9 104.1 NO (g) 90.3 210.6 86.6 N2O3 (g) 83.3 307.0 140.5 NO2 (g) 33.5 240.2 51.5 N2O4 (g) 9.6 303.8 98.4 N2O5 (g) –42.7 178.0 114.1 NiO (s) –239.7 38.0 –211.6 О3 (g) 142.3 237.7 163.4 0 205.0 0 –1492.0 114.5 –1348.8 PbO (s) –219.3 66.1 –189.1 PbO2 (s) –276.6 74.9 –218.3 0 31.9 0 SO2 (g) –296.9 248.1 –300.2 SO3 (g) –395.8 256.7 –371.2 O2 (g) P2O5 (s) S (s) 34.7 204.6 57.2 –910.9 41.8 –856.7 SnO (s) –286.0 56.5 –256.9 SnO2 (s) –580.8 52.3 –519.3 SiH4 (g) SiO2 (quartz) 0 30.6 0 TiCl4 (l) –804.2 252.4 –737.4 TiO2 (s) –943.9 50.3 –888.6 WO3 (s) –842.7 75.9 –763.9 ZnO (s) –350.6 43.6 –320.7 Ti (s) 95 Appendix B Standard Thermodynamical Functions or Some Organic Substances at 298 К , kJ/mol , J/(mol·К) kJ/mol G S , 8 9 02 ∆ 8 9 02 H 8 9 02 ∆ Formule and State СН4 (g) –74.9 186.2 –50.8 С2Н2 (g) 226.8 200.8 209.2 С2Н4 (g) 52.3 219.4 68.1 С2Н6 (g) –89.7 229.5 –32.9 49.0 124.5 172.8 СН3ОН (l) –238.7 126.8 –166.3 С2Н5ОН (l) –277.6 160.7 –174.8 С3Н8О3 (l) –669.1 204.6 –479.4 СН3СООН (l) –484.4 159.9 –389.6 СО(NH2)2 (s) СО(NH2)2 (l) –333.0 –317.7 104.7 175.7 –196.9 –202.7 С6Н12О6 (s) С6Н12О6 (l) –1273.0 –1263.1 212.1 264.0 –910.5 –914.5 С12Н22О11 (s) С12Н22О11 (l) –2220.9 –2215.8 360.2 403.8 –1544.3 –1551.4 С12Н22О11 (l) –2232.4 394.1 –1564.9 С6Н6 (l) Appendix C Surface Tension of Water at Different Temperatures t, °C 15 16 17 18 19 20 σ ×10 , –3 Н 2О t, °C 2 J/m 73.49 73.34 73.19 73.05 72.90 72.75 21 22 23 24 25 26 96 σ ×10 , –3 Н2О 2 J/m 72.59 72.44 72.28 72.13 71.97 71.80 Appendix D The main Half-Reactions and Standard Red-Ox Potentials Values Half-Reaction Oxidized Form nе + H2O2 + 2H + PbO2 + 4H – + MnO4 + 8H – + ClO +2H – + – + ClO3 + 6H ClO4 + 8H Cl2 0 Cr2O7 2– + + 14H – + 2NO3 + 12 H 0 + O2 + 4H Br2 0 – 0 Reduced Form е ,V 2e – 2H2O 2e – Pb 5e – Mn 2e – – 1.49 6e – – 1.45 8e – Cl + 4H2O 2e – 2Cl 6e – 2Cr 10e – + 2H2O 1.69 2+ + 4H2O 1.51 Cl + H2O Cl +3H2O – 1.39 – 1.36 3+ + 7H2O 0 1.35 N2 + 6H2O 1.24 1.23 4e – 2H2O 2e – 2Br – 1.78 2+ – 1.07 – + e NO + H2O 1.00 – + 3e – NO + 2H2O 0.96 NO3 +2H – + 2e – NO2 + H2O – + e – e – NO2 + 2H NO3 + 4H NO3 +2H Fe 3+ H2O2 3e – MnO2 + 4OH – e + O2 + 2H SO4 S4O6 SO4 + + 8H 2– 2– Fe 0.77 – 0 2– 0.78 2+ – + MnO4 + 2H2O І2 0.84 NO2 + H2O 2e 0 O2 + 2H MnO4 – + + 2H – 0.68 – MnO4 2– 0.57 0.54 2e – 2І 4e – 4OH – 0.40 6e – 0 S + 4H2O 0.36 2e – 2S2O3 2e – 2– – 0.54 2– SO3 + H2O 97 0.22 0.20 Appendix E Solubility Product Conctant Ksp Values for Feebly Soluble Electrolytes at 25 °С Electrolyte AgBr AgCl Ag2CrO4 AgI Ag2S Ag2SO4 BaCO3 Ksp 6·10 1.8·10 4·10 –10 –12 1.1·10 6·10 Electrolyte –13 –16 –50 Ksp Fe(OH)3 3.7·10 –40 FePO4 1.3·10 –22 –18 FeS 5·10 HgS 1.6·10 MgCO3 2.1·10 –52 2·10 Mg(OH)2 1.3·10 –11 5·10 –9 –10 BaCrO4 1.6·10 –10 BaSO4 1.1·10 –10 MnS 2.5·10 PbBr2 9.1·10 PbCl2 2·10 –5 PbCrO4 1.8·10 –14 –14 6·10 CaCO3 5·10 –9 PbCO3 7.5·10 2·10 –9 PbI2 8.0·10 –11 CaF2 4·10 CaSO4 6.3·10 Ca3(PO4)2 1·10 –6 –39 Ba3(PO4)2 CaC2O4 –5 –5 –5 –29 PbS 2.5·10 PbSO4 1.6·10 SrCO3 98 1.1·10 –9 –27 –8 –10