242 Chemistry Term 1 STPM CHAPTER 6 Example 6.14 Consider the following reaction: A + B C + D The equilibrium constants at 77 °C and 277 °C are 3.98 and 6000 respectively. Calculate the enthalpy change of reaction. Solution Using the equation: ln Kc = ln A – ∆H RT ln K2 K1 = ∆H R 1 T1 – 1 T2 ln 6000 3.98 = ∆H R 1 350 – 1 550 ∴ ∆H = +58 533 J = +58.5 kJ Quick Check 6.3 1 Which of the following systems would be affected by changes in pressure? (a) CH3 COOH(l) + C2 H5 OH(l) CH3 COOC2 H5 (l) + H2 O(l) (b) CaCO3 (s) CaO(s) + CO2 (g) (c) Ag(s) + Fe3+(aq) Ag+(aq) + Fe2+(aq) (d) CO(g) + H2 O(g) CO2 (g) + H2 (g) (e) PCl3 (g) + Cl2 (g) PCl5 (s) (f) 2H2 S(g) 2S(s) + 2H2 (g) 2 In which of the following reactions, the percentage of products decreases with increasing pressure? (a) Mg(OH)2 (s) MgO(s) + H2 O(g) (d) NH4 Cl(s) NH3 (g) + HCl(g) (b) 2SO2 (g) + O2 (g) 2SO3 (g) (e) SO2 (g) + Cl2 (g) SO2 Cl2 (g) (c) 2NO(g) + Br2 (g) 2NOBr(g) 3 The decomposition of ammonium nitrate is an endothermic reaction: NH4 NO3 (s) → N2 O(g) + 2H2 O(g) How would the equilibrium amount of N2 O change if (a) the pressure is decreased. (c) a catalyst is introduced. (b) the temperature is increased. 4 Consider the following system in equilibrium: CH4 (g) + 2H2 O(g) 4H2 (g) + CO2 (g) ∆H = +195 kJ Predict how the equilibrium composition of H2 and the value of Kc would change if (a) the volume of the vessel is increased. (b) the pressure is increased. (c) a catalyst is added. (d) the temperature is decreased.
243 Chemistry Term 1 STPM CHAPTER 6 6.8 Industrial Processes 1 Many important industrial reactions are reversible. As a result, the conditions of reactions must be carefully chosen so that the yield is reasonably high from the given amount of reactants. At the same time, the rate of reaction should be fast and cost of production low by avoiding using very high temperatures or pressures. 2 Here we will look at three important industrial processes: The Haber process, the Contact process and the Ostwald process and examine how the conditions are chosen to maximise yield and minimise cost. The Haber Process 1 The Haber process is used to manufacture ammonia from its elements, nitrogen and hydrogen. 2 Ammonia is an important compound used mainly to manufacture fertilisers, explosives, detergents, polymers and nitric acid. 3 The reaction in the Haber process is: N2 (g) + 3H2 (g) 2NH3 (g) ∆H = –92 kJ mol–1 4 The forward reaction is accompanied by a decrease in the number of gaseous particles. Hence, by Le Chatelier’s Principle, it is favoured by high pressures. However, too high a pressure would increase the cost of production as the pipes and containers must be strong enough to withstand high pressure. In practice, a pressure of 200 – 1000 atm is employed. 5 As the forward reaction is exothermic, it is favoured by low temperatures. However, at low temperatures the rate of reaction will be low and we have to wait a long time for the system to achieve equilibrium. 6 As a result, a compromise temperature of 450 °C to 550 °C is normally used. The temperature is not too low that the rate becomes too slow, and not too high that the yield is low. 7 The graph shows the percentage yield of ammonia at different temperatures and pressures. 70 60 50 40 30 20 10 0 100 400 °C 450 °C 500 °C 550 °C 350 °C 200 300 400 Pressure/atm Percentage yield High pressures incur higher cost. 450 °C is a compromise between rate and yield. Temperature: 450 °C Pressure: 200 – 1000 atm Catalyst: Iron powder 2015/P1/Q17(b) 2017/P1/Q17(a)
244 Chemistry Term 1 STPM CHAPTER 6 8 Iron is added as a catalyst to further increase the rate of reaction and shorten the time taken for the system to achieve equilibrium. The Contact Process 1 The Contact process is the industrial process in the manufacture of sulphuric acid. The process involves four stages. 2 Liquid sulphur is burned in a stream of hot air to produce sulphur dioxide: S(l) + O2 (g) → SO2 (g) 3 The sulphur dioxide is oxidised by oxygen in the air to sulphur trioxide: 2SO2 (g) + O2 (g) 2SO3 (g) ∆H = –197 kJ mol–1 4 The sulphur trioxide is absorbed into concentrated sulphuric acid to produce oleum (fuming sulphuric acid) SO3 (g) + H2 SO4 (aq) → H2 S2 O7 (aq) 5 Diluting the oleum with water followed by distillation yields 98% concentrated sulphuric acid. H2 S2 O7 (l) + H2 O → 2H2 SO4 (aq) 6 The second stage of the process is reversible. The forward reaction is favoured by high pressures as it occurs with a decrease in the number of gaseous particles. However, at 1 atm, the percentage conversion of SO2 to SO3 is already almost 95%. Hence, higher pressure, which would incur higher cost is not used. 7 The forward reaction is exothermic. It is favoured by low temperatures. However, low temperatures would mean slower rate of reaction. Thus a compromise temperature of 450 °C to 550 °C is usually employed. 8 Vanadium pentoxide is added as a catalyst to further increase the rate of reaction. The Ostwald Process 1 The Ostwald process (developed by Wilhelm Ostwald) is a chemical process for making nitric acid. 2 The major use of nitric acid is in the manufacture of nitrate fertilisers. It is also used to manufacture organic nitrogencompounds such as dynamite and TNT (trinitrotoluene) as well as organic dyes. 3 Ammonia (from the Haber process) and large amounts of purified air are passed rapidly over a platinum/rhodium catalyst kept at about 900 °C. 4 Ammonia is catalytically oxidised by oxygen (in the air supply) to form nitrogen monoxide. 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g) ∆H = –905.2 kJ At 1 atm, the percentage conversion of SO2 to SO3 is almost 95%. Temperature: 450 °C Pressure: 1 – 2 atm Catalyst: V2 O5 450 Temperature/°C 95 % of conversion to SO3 2018/P1/Q12
245 Chemistry Term 1 STPM CHAPTER 6 The reaction is highly sufficient exothermic to maintain the temperature of the catalyst at 900 °C. 5 The mixture is then cooled to room temperature and mixed with more air to convert the nitrogen monoxide to nitrogen dioxide. 2NO(g) + O2 (g) → 2NO2 (g) ∆H = –114 kJ mol-1 6 The nitrogen dioxide produced is then bubbled through water where it disproportionates to nitric acid and nitrogen monoxide. 3NO2 (g) + H2 O(l) → 2HNO3 (aq) + NO(g) ∆H = –117 kJ mol–1 The nitric acid produced has a concentration of about 55%. The unreacted nitrogen monoxide is recycled. 7 The aqueous nitric acid is then subjected to fractional distillation to give an azeotrope consisting of 68% nitric acid and 32% water. 8 Pure nitric acid can be obtained by distilling the azeotrope with phosphorus pentoxide, P4 O10. 9 The first stage of the process is reversible. Since the forward reaction is exothermic' it is favoured by low temperatures. The forward reaction is accompanied by an increase in the number of moles of gas, it is favoured by low pressures. 10 The conditions for the first step which contribute to an overall yield of 96% are: Temperature: About 900 °C Pressure: Between 4 to 10 atm SUMMARY SUMMARY 1 A reversible reaction is one that can occur in both directions. 2 Dynamic equilibrium is a situation where the rates of the forward and reverse reactions are the same, and the amount of substances do not change with time. 3 The equilibrium law states that when a reversible reaction has achieved dynamic equilibrium, the ratio of the molar concentrations of the products to the molar concentration of the reactants, both raised to their respective stoichiometric coefficient, is a constant at constant temperature. 4 For a general reversible reaction: aA + bB cC + dD [C]c [D]d [A]a [B]b = Kc (Where [ ] = equilibrium molar concentration) 5 The relationship between Kc and K p is given by: Kp = Kc (RT)∆n 6 The relationship between Kc and rate constants is given by: k1 k–1 = Kc 7 Le Chatelier’s Principle states that when a system in dynamic equilibrium is disturbed, the position of equilibrium will change to remove the effect of the disturbance so that equilibrium is reestablished. 8 Kc and K p depend only on temperature. 9 The relationship between equilibrium constant and temperature is given by the equation: Kc = Ae –∆H RT
246 Chemistry Term 1 STPM CHAPTER 6 1 The equilibrium constant, Kp , for the decomposition of magnesium nitrate: 2Mg(NO3 ) 2 (s) 2MgO(s) + 4NO2 (g) + O2 (g) is 2.3 10–18 atm5 at 50 °C and 7.6 10–11 atm5 at 85 °C. It can be concluded that A a catalyst is present. B Kp will decrease with increasing pressure. C the forward reaction is endothermic. D The amount of magnesium nitrate present at equilibrium will affect the value of Kp . 2 In the reaction: N2 (g) + 3H2 (g) 2NH3 (g); ∆H = negative Which of the following factors would affect the value of the equilibrium constant? A Increasing the pressure B Addition of more nitrogen C Decreasing the temperature D Adding a suitable catalyst 3 Consider the following system in equilibrium: CaCO3 (s) → CaO(s) + CO2 (g) The amount of CO2 will be increase by A adding more CaCO3 (s). B removing some of the CaO(s). C increasing the pressure. D increasing the volume of the system. 4 Which of the following is true when a system approaches dynamic equilibrium? A The rate of the forward reaction is decreasing. B The rate of the forward reaction is increasing. C The rate of the forward and reverse reactions approaches zero. D The activation energy approaches zero. 5 Consider the following equilibrium system: 2NH3 (g) N2 (g) + 3H2 (g); ∆H = +ve If the temperature is increased at constant pressure, will the volume of the mixture increase or decrease and why? Objective Questions A The volume will increase because the equilibrium shifts to the right. B The volume will increase because the equilibrium shifts to the right, and also because of thermal expansion. C The volume will decrease because the equilibrium shifts to the left. D There is no change in the volume because any thermal expansion would be balanced by a shift of equilibrium towards the left. 6 Solid silver chloride is soluble in aqueous ammonia via the following equilibrium: AgCl(s) + 2NH3 (aq) [Ag(NH3 )2 ]+(aq) + Cl– (aq) Which of the following would most likely cause the reappearance of silver chloride? A Adding more ammonia B Adding excess of dilute hydrochloric acid C Adding ammonium nitrate D Warming the mixture 7 The graphs below refer to the production of a gaseous compound Q. Pressure % of Q Temperature % of Q Which of the following statements is true about the production of Q? A The reaction is exothermic. B The percentage of Q would increase if a suitable catalyst is used. C The reaction is accompanied by an increase in volume. D The activation energy is low. 8 In Haber process, nitrogen and hydrogen react to form ammonia. N2 (g) + 3H2 (g) 2NH3 (g) STPM PRACTICE 6
247 Chemistry Term 1 STPM CHAPTER 6 Which of the following is not true of the process? A It requires high pressure B Iron is used as the catalyst to increase the yield of ammonia C The forward reaction is favoured by low temperature D The ammonia produced is liquefied and drained out of the reacting vessel 9 Which of the following would affect the equilibrium constant of a reaction? A Temperature B Catalyst C Concentration D All of the above 10 % of product Pressure Which of the following system would produce the graph shown above? A N2 O4 (g) 2NO2 (g) B 2SO2 (g) + O2 (g) 2SO3 (g) C 2H2 O2 (aq) 2H2 O(l) + O2 (g) D CaCO3 (s) CaO(s) + CO2 (g) 11 Which of the following graphs correctly represent the variation of the equilibrium constant, Kc , against temperature for an endothermic reaction? A T Kc C T Kc B T Kc D T Kc 12 Hydrogen iodide decomposes into its elements at very high temperature according to the equation 2HI(g) H2 (g) + I2 (g) In one experiment at 2.0 atm, it was found that 30% of the hydrogen iodide had decomposed. Which of the following statements is correct regarding the system in equilibrium? A The total pressure at equilibrium is 8 atm. B The equilibrium constant is independent of temperature. C The equilibrium constant is independent of pressure. D The percentage dissociation increases with increasing pressure. 13 For the reaction, N2 (g) + 3H2 (g) 2NH3 (g) what is the effect of adding neon to the reacting vessel at constant volume? A The yield of NH3 increases. B The yield of NH3 decreases. C The position of equilibrium remains unchanged. D The equilibrium constant decreases. 14 Consider the following reversible reaction. X(g) + Y(g) k1 k2 2Z(g) ∆H = –237 kJ mol–1 Which statement is true? A The forward reaction is favoured by low pressure B When the system is in equilibrium, –237 kJ of heat is released C At equilibrium, k1 = k2 D At equilibrium, k1 [X][Y] = k2 [Z]2 15 The Contact process for the manufacture of sulphuric acid involves the following equilibrium. 2SO2 (g) + O2 (g) 2SO3 (g) ∆H = Negative Which is not true when the reaction is at equilibrium? A k1 = k–1 B k1 [SO2 ]2 [O2 ] = k–1[SO3 ]2 C The total pressure of the system remains constant D ∆H = 0
248 Chemistry Term 1 STPM CHAPTER 6 16 Nitrosyl bromide, NOBr, decomposes according to the equation below. 2NOBr(g) : 2NO(g) + Br2 (g) ∆H = +23 kJ The reaction is second order with respect to nitrosyl bromide. Which statement is not true of the reaction? A The rate constant increases with the addition of a suitable catalyst B The rate law is rate = k[NOBr]2 C The activation energy is higher than 23 kJ D The activation energy for the reversible reaction is 23 kJ 17 An equilibrium system is subjected to the following changes separately. I Increasing temperature at constant pressure II Decreasing pressure at constant temperature In which system will the amount of product increases when subjected to the above changes? A 2Mg(NO3 )2 (s) 2MgO(s) + 4NO2 (g) + O2 (g) ∆H = positive B PCl3 (g) + Cl2 (g) PCl5 (g) ∆H = negative C 2HI(g) H2 (g) + I2 (g) ∆H = positive D CH3 COOC2 H5 (l) + H2 O(l) CH3 COOH(aq) + C2 H5 OH(aq) ∆H = zero 18 Consider the following reaction in equilibrium: A + B k1 k–1 C ∆H = negative Which statement(s) is/are true of the system when the temperature is increased? I k1 and k–1 increases. II There is a net reverse reaction. III Kc decreases. A I only B III only C II and III D I, II and III 19 The equilibrium constant Kp for the decomposition of nitrogen monoxide: 2NO(g) N2 (g) + O2 (g) varies with temperature T as shown in the graph below: Kp 1 T Which statement(s) is/are true of the above reaction? I The forward reaction is exothermic. II Pressure has no effect on the degree of decomposition of NO. III The value of Kp equals to the value of Kc . A I only B III only C I and II D I, II and III 20 For the equilibrium system: X(g) + Y(g) Z(g) the percentage yields of Z at pressure p are 35% at 780 K and 63% at 1200 K. Which statement is true of the equilibrium? A The percentage of Z can be increased by the addition of a suitable catalyst. B The forward reaction is accompanied by an absorption of heat. C The values of Kc and Kp are the same. D The rate is slower at 1200 K compared to 780 K. 21 Consider the following reaction: 2SO2 (g) + O2 (g) 2SO3 (g) ∆H = negative At equilibrium, the amount of sulphur trioxide can be increased by A adding a suitable catalyst B increasing the temperature C reducing the volume of the vessel D adding argon at constant volume 22 The equilibrium constant for a reversible reaction is 4.0 and 4.0 x 103 at 80 °C and 280 °C respectively. What is the enthalpy change for the reaction? A –64.6 kJ mol–1 B –55.8 kJ mol–1 C +55.8 kJ mol–1 D +64.6 kJ mol–1
249 Chemistry Term 1 STPM CHAPTER 6 23 Benzene (boiling point = 80°C) and water (boiling point = 100°C) forms an azeotrope with a composition by mass of 91% benzene and a boiling point of 69°C. Which of the following statement is not correct? A Benzene is more volatile than water. B If a mixture containing 50% by mass of benzene is fractionally distilled, the first distillate collected is the azeotrope. C The vapour pressure/composition curve shows a maximum. D A mixture of water and benzene shows positive deviation from Raoult’s law. 24 25.0 cm3 of 0.10 mol dm-3 ammonia is titrated against hydrochloric acid. Which statement is true regarding this titration? A The volume of hydrochloric acid needed is 25.0 cm3 . B The pH at mid-point titration is equal to the pKb for ammonia. C The solution at mid-point titration is a buffer solution with maximum buffer capacity. D The pH at the end-point is larger than 7. 25 Which of the following mixtures, that contains equimolar quantities of each component, will not give a buffer solution? A Methanamine and hydrochloric acid B Ammonia and ammonium sulphate C Ammonium chloride and hydrochloric acid D Ethanedioic acid and potassium hydroxide 26 What is the maximum mass of silver chloride precipitated when 25.0 cm3 of aqueous sodium chloride (0.10 mol dm–3) is added to 25.0 cm3 of aqueous silver nitrate (0.10 mol dm-3)? [Ksp of AgCl = 1.0 × 10–10 mol2 dm-6; AgCl = 143.5] A 3.2 × 10–4 g C 7.1 × 10–3 g B 6.5 × 10–4 g D 1.3 × 10–2 g Structured and Essay Questions 1 The molar mass of a mixture of a sample of N2 O4 is 60.0 g mol–1 at 323 K and 100 × 103 kPa. N2 O4 (g) 2NO2 (g) (b) Calculate the degree of dissociation of N2 O4 at 323 K and 100 kPa. (b) Calculate the equilibrium constant, Kp , for the dissociation of N2 O4 . (c) Calculate the degree of dissociation of N2 O4 at 323 K and 1000 kPa. Is your answer in agreement with Le Chatelier’s Principle? 2 (a) State the equilibrium law. (b) Name the factors that will affect the equilibrium position of a reversible reaction. Which of these factor(s) also affect(s) the value of the equilibrium constant? (c) Consider the reaction: N2 (g) + 3H2 (g) 2NH3 (g) A mixture of nitrogen and hydrogen in the mole ratio of 1 : 3 was allowed to reach equilibrium at 400 °C and 3.0 104 kPa. The percentage conversion of nitrogen into ammonia was 60%. Calculate Kp at the same temperature. 3 The following reaction is one of the steps in the Contact Process for the manufacture of sulphuric acid. 2SO2 (g) + O2 (g) 2SO3 (g); ∆H = –196 kJ (a) State the conditions of temperature and pressure necessary to ensure a high yield. (b) The temperature used in most sulphuric acid plants is a compromise between cost and rate. Suggest two reasons why a temperature of 450 °C is used. (c) (i) Name the catalyst used. (ii) What effect does the catalyst have on the equilibrium system?
250 Chemistry Term 1 STPM CHAPTER 6 (d) Sketch the energy profile for the reaction, showing both the catalysed and uncatalysed route. (e) At 450 °C, the equilibrium partial pressures of the mixture are: P(O2 ) = 0.70 atm P(SO2 ) = 0.10 atm P(SO3 ) = 0.80 atm (i) Calculate the equilibrium constant, Kp at 500 °C. (ii) What were the initial partial pressure of SO2 and O2 ? 4 Consider the following reaction: H2 (g) + I2 (g) 2HI(g); ∆H = –288 kJ (a) Sketch the energy profile for the reaction. Indicate on your diagram: (i) the enthalpy change of reaction, (ii) the activation energy of the forward reaction, (iii) the activation energy of the reverse reaction. (b) For the reaction: H2 (g) + Cl2 (g) 2HCl(g) Would you expect the activation energy of the forward reaction to be higher or lower than that involving hydrogen and iodine? Explain your answer. 5 The thermal decomposition of phosphorous(V) chloride can be represented by the following equilibrium: PCl5 (g) PCl3 (g) + Cl2 (g) At a total pressure of 200 kPa, the percentage dissociation of PCl5 is 50% at 200 °C and 85% at 350 °C. (a) Write an expression for the equilibrium constant, Kc and Kp for the system. (b) Calculate the value of Kp at 200 °C and 350 °C. (c) Is the dissociation of PCl5 an endothermic or exothermic process? (d) Predict how the percentage of dissociation would be affected by (i) an increase in the total pressure, (ii) a decrease in temperature, (iii) addition of a catalyst. 6 The variation of Kc with temperature of a reversible reaction is given in the table below: Temperature/°C 77 234 277 Kc 3.89 1700 6030 Plot a suitable graph to calculate the enthalpy change, ∆H, of the reaction. 7 In the Contact process, the following equilibrium is established. 2SO2 (g) + O2 (g) 2SO3 (g) ∆H = –390 kJ When a 2 : 1 mol ratio of SO2 and O2 at a total pressure of 5 atm was passed over vanadium pentoxide at 500 °C, the percentage conversion of SO2 to SO3 is 90%. (a) Calculate the equilibrium constant Kp for the reaction. (b) Explain the effect of an increase in temperature on (i) the yield of SO3 , (ii) the rate of attainment of the equilibrium.
251 Chemistry Term 1 STPM CHAPTER 6 8 Consider the following reaction: X(g) + 2Y(g) Z(g) At 700 K, equilibrium was achieved when the partial pressures of X, Y and Z are 0.4 atm, 0.9 atm and 1.2 atm respectively. (a) What do you understand by the term partial pressure? (b) Write an expression for the equilibirum constant, Kp , for the reaction. (c) Calculate the value of Kp at the same temperature. (d) Explain the effect on the position of equilibrium if argon gas is added to the mixture at constant pressure. 9 (a) The plot below shows the variation of pV —–— nRT for one mole of an ideal gas and one mole of carbon dioxide. 1 Carbon dioxide Ideal gas p pV nRT (i) State the ideal gas equation and explain the shape of the plot for the ideal gas. (ii) Explain what causes carbon dioxide to show positive deviation at high pressures. (b) The variation of the density (d) of a gas with pressure (p) under constant temperature (T) is given by the expression below: d = p1 M —–– RT 2 [Where M = Molar mass, R = 8.31 J mol–1K–1] (i) Sketch the variation of d —–p against p for an ideal gas at constant temperature, and explain the shape of your plot. (ii) The density of a gas is 0.57 g dm–3 at 35 °C and 101 kPa. Calculate the relative molecular mass of the gas. 10 (a) The equilibrium constant for a reversible reaction is 2.4 mol dm–3 at 727 °C and 1.6 at 1027 °C. (i) Is the forward reaction endothermic or exothermic? Explain your answer. (ii) Calculate the enthalpy change of the reaction. (iii) State and explain the effect on the equilibrium constant and the enthalpy change of reaction of adding a catalyst to the system at 727 °C. (b) The Haber process involves the following equilibrium: N2 (g) + 3H2 (g) 2NH3 (g) ∆H = negative The conditions used by most of the ammonia plants are: Temperature: 450 °C Pressure: 500 atm Catalyst: Finely divided iron Explain in terms of kinetics and yield the choice of conditions.
CHAPTER IONIC EQUILIBRIA AND 7 SOLUBILITY EQUILIBRIA Concept Map Students should be able to: Ionic equilibria • use Arrhenius, Brønsted-Lowry and Lewis theories to explain acids and bases; • identify conjugate acids and bases; • explain qualitatively the different properties of strong and weak electrolytes; • explain and calculate the terms pH, pOH, Ka, pKa, Kb, pKb, Kw and pKw from given data; • explain changes in pH during acid-base titrations; • explain the choice of suitable indicators for acidbase titrations; • define buffer solutions; • calculate the pH of buffer solutions from given data; • explain the use of buffer solutions and their importance in biological systems such as the role of H2CO3/ HCO3 – in controlling pH in blood. Solubility equilibria • define solubility product, Ksp; • calculate Ksp from given concentrations and vice versa; • describe the common ion effect, including buffer solutions; • predict the possibility of precipitation from solutions of known concentrations; • apply the concept of solubility equilibria to describe industrial procedure for water softening. • Salt hydrolysis • Salts formed between – Strong acid and weak base – Weak acid and strong base – Weak acid and weak base – Calculating pH of salt solutions • Buffer solution – Workings – Calculating pH – Preparation – Capacity Electrolytes • Asid and bases • Definitions – Arrhenius’ – Brønsted- Lowry’s – Lewis’ • Relative strength of Brønsted-Lowry asids and bases • Dissociation – Degree of – Constants of polyprotic acid Ostwald's Dilution Law pH and pOH Scale • Relationship between pH and pOH • Calculating pH of weak acids • Acid-base titration – Acid-based indicators – Choice of indicator – Buffer action – Important buffer in human blood • Sparingly soluble salts – Solubility – Solubility product Ksp and precipitation – Common ion effect – Selective precipitation – Precipitation and complex formation – Solubility of weak acids in strong acid Ionic Equilibria and Solubility Equilibria Learning earning Outcomes
253 Chemistry Term 1 STPM CHAPTER 7 7.1 Electrolytes 1 According to Arrhenius ionic theory, strong electrolytes dissociate completely when they dissolve in water to produce a high concentration of ions. The degree of dissociation is 1 or 100%. They have high conductivity. Degree of dissociation = Amount dissociated Original amount 2 On the other hand, weak electrolytes undergo only partial dissociation in water to produce a low concentration of ions. Their degree of dissociation is very much less than 1. They have low conductivity. 3 The degree of dissociation of a weak electrolyte, at constant temperature, is dependant of its concentration. Ostwald's Dilution Law 1 What is the relationship between the degree of dissociation of a weak electrolyte and its molar concentration? 2 Consider a weak electrolyte AB of concentration C mol dm–3 with a degree of dissociation of α. AB A+(aq) + B– (aq) Concentration [AB]/mol dm–3 [A+]/mol dm–3 [B– ]/mol dm–3 Initial C 0 0 At equilibrium C(1 – α) Cα Cα 3 Applying the equilibrium law: [A+][B– ] [AB] = Kc By substitution: Kc = (Ca)2 C(1 – α) Assuming that a is small, then (1 – α) ≈ 1 Kc = (Ca)2 C Or α = Kc C 4 The final expression is known as Ostwald’s dilution law which states that the degree of dissociation of a weak electrolyte is inversely proportional to the square root of its concentration. Electrolyte is anything that conducts electricity either in the molten state or in aqueous solution. Arrhenius was awarded the Nobel Prize for his work on electrolytes. The degree of dissociation of a 0.10 M CH3 COOH is only 1.6% at room temperature.
254 Chemistry Term 1 STPM CHAPTER 7 5 The degree of dissociation of a weak electrolyte increases with decreasing concentration of the electrolyte. Theoretically, weak electrolyte would undergo complete dissociation (α = 1 or 100%) at infinite dilution (or when the concentration approaches zero). 7.2 Acids and Bases 1 Acids are substances that have the following properties: (a) They have sour taste. (b) They turn blue litmus paper red. (c) They react with bases to form salts and water only. (d) They react with electropositive metals to liberate hydrogen gas. (e) They react with carbonates and bicarbonates to liberate carbon dioxide gas. 2 Bases are substances with the following properties: (a) They have bitter taste. (b) They turn red litmus paper blue. (c) They react with acids to form salts and water only. (d) They react with ammonium salts to produce ammonia gas. 3 There are three important definitions for acids and bases. They are the (a) Arrhenius definition (b) Brønsted-Lowry definition (c) Lewis definition Arrhenius’ Definition 1 In 1884, Svabte Arrhenius, a Swedish chemist, defines acids as compounds that dissociate to produce H+ ions when dissolved in water. For example: HCl(aq) → H+(aq) + Cl– (aq) HNO3 (aq) → H+(aq) + NO3 – (aq) H3 PO4 (aq) 3H+(aq) + PO4 3–(aq) H2 SO4 (aq) → 2H+(aq) + SO4 2–(aq) CH3 COOH(aq) CH3 COO– (aq) + H+(aq) 2 Bases are compounds that dissociate to produce OH– ions when dissolved in water. At infinite dilution (where concentration ≈ 0), all weak electrolytes dissociate 100%. General properties of acids General properties of bases Exam Tips Exam Tips Water must be present in Arrhenius's definition. The molecule dissociates to produce H+. 0 Concentration Degree of dissociation 1 2010/P1/Q44 2010/P2/Q1 2015/P1/Q20(a)
255 Chemistry Term 1 STPM CHAPTER 7 Examples are: NaOH(s) + aq → Na+(aq) + OH– (aq) Ca(OH)2 (s) + aq Ca2+(aq) + 2OH– (aq) Al(OH)3 (s) + aq Al3+(aq) + 3OH– (aq) 3 Strong acids and strong bases dissociate completely into ions in water. Weak acids and weak bases undergo only partial dissociation in water and the ions exist in equilibrium with the undissociated molecules. 4 Take note that Arrhenius acids and bases are limited to compounds containing H+ and OH– in their molecules. Furthermore, water must be present for the compounds to act as acids or bases. 5 However, Arrhenius’ definition cannot explain the acid/base properties of certain compounds that do not contain OH– or H+ in their structures. 6 For example, ammonia is a base. However, the ammonia molecule, NH3 , does not contain any –OH group. 7 Similarly, aqueous aluminium chloride is acidic. However, aluminium chloride, AlCl3 , does not contain any H atoms. Brønsted-Lowry’s Definition 1 To overcome the shortcomings of Arrhenius’ definition, in 1923 J N Brønsted and T M Lowry put forward their definition of acids and bases according to the proton transfer theory. 2 According to the Brønsted-Lowry’s definition: An acid is a proton donor. A base is a proton acceptor. 3 For example, when hydrogen chloride dissolves in water to form hydrochloric acid, the following reaction occurs. H+ HCl(aq) + H2 O(l) → H3 O+(aq) + Cl– (aq) 4 In the above example, the HCl molecule donates a proton to H2 O molecule to form H3 O+ and itself gets converted to Cl– . 5 In the reaction, HCl acts as a Brønsted-Lowry acid (proton donor), and H2 O acts as a Brønsted-Lowry base (proton acceptor). 6 Many aqueous cations with high charge are also Brønsted-Lowry acids. These highly-charged cations exist as hexaaqua ions with the general formula of [M(H2 O)6 ]n+. The hexaaqua cations act as Brønsted-Lowry acids by donating protons to water molecules. For example, [Al(H2 O)6 ]3+(aq) + H2 O(l) [Al(H2 O)5 (OH)]2+(aq) + H3 O+(aq) This explains the acidity of aqueous aluminium chloride mentioned above. The molecule dissociates to produce OH– . NH3 has no –OH group in its molecule. AlCl3 does not have H atom in its molecule. Proton transfer theory HCl is a proton donor. H2 O is a proton acceptor. Al(H2 O)6 3+ is a proton donor.
256 Chemistry Term 1 STPM CHAPTER 7 7 An example of a Brønsted-Lowry base is ammonia. H+ NH3 (aq) + H2 O(l) NH4 +(aq) + OH– (aq) In the example, ammonia receives a proton from H2 O. Hence, NH3 is a Brønsted-Lowry base and H2 O is a Brønsted-Lowry acid. 8 From the two examples given, water is both a Brønsted-Lowry acid as well as a Brønsted-Lowry base. 9 Brønsted-Lowry’s definition can even be extended to non-aqueous system. For example, the gas phase reaction between hydrogen chloride and ammonia. H+ HCl(g) + NH3 (g) NH4 +Cl– (s) In the above example, HCl donates a proton to NH3 . Hence, HCl is the acid. Ammonia in turn accepts a proton from HCl. Hence, NH3 is the base. 10 However, Brønsted-Lowry’s definition also has its shortcomings. For example, boric acid, H3 BO3 or B(OH)3 , reacts with water according to the equation: B(OH)3 (aq) + H2 O(l) B(OH)4 – (aq) + H+(aq) 11 In this reaction, boric acid neither dissociates to produce H+ ions nor donates a proton to water. Instead, it receives an OH– from water. Hence, it is not an acid by either Lewis’s definition or Brønsted-Lowry’s definition. Lewis’ Definition 1 To overcome the problems faced by Arrhenius’ definition and Brønsted-Lowry’s definition, an American chemist by the name of G. N. Lewis further extends the concept of acids and bases and proposes his electron-pair theory in 1938. 2 According to Lewis, (a) an acid is an electro-pair acceptor, (b) a base is an electron-pair donor. 3 A Lewis acid accepts a lone-pair of electron from a Lewis base to form a coordinate bond. 4 An example of a Lewis acid-base reaction is the reaction between copper(II) aqueous ions and ammonia to form a dark blue complex ion. Cu2+(aq) + 4NH3 (aq) Cu(NH3 ) 4 ] 2+(aq) 5 In the above reaction, the copper(II) ion made used of its empty orbitals to accept an electron pair each from four ammonia molecules to form four coordinate bonds. Non-aqueous system H3 BO3 does not act as a proton donor: H3 BO3 H+ + H2 BO3 – H3 N H3 N NH3 NH3 Cu2+ • • • • • • • • Electron-pair theory 2008/P1/Q10 INFO Lewis Acids and Bases
257 Chemistry Term 1 STPM CHAPTER 7 6 Cu2+ is a Lewis acid because it is an electron-pair acceptor while NH3 is a Lewis base because it is an electron-pair donor. 7 Let us refer to our previous problem concerning boric acid. When boric acid, an octet deficient compound, dissolves in water, the boron atom accepts a lone pair electrons from the water molecule to form a co-ordinate bond as shown below: H OH +H+ OH O OH OHB H • • – OH OH HO B Hence the boric acid is a Lewis acid and the water is the Lewis base. 8 Other examples of Lewis acid-base reactions are: (a) Reaction between boron trifluoride and ammonia BF3 + NH3 → BF3 • NH3 + F F • F NB • H H N H H H H F F F B (b) Reaction between aluminium chloride and ammonia + Cl Cl • • Cl Al N H H N H H H H Cl Cl Cl Al Info Chem BF3 : Lewis acid NH3 : Lewis base Info Chem OH | H HO—B←: + O | H OH ↓ OH – | HO—B←OH + H+ | OH Info Chem AlCl3 : Lewis acid NH3 : Lewis base Quick Check 7.1 Which of the following substances acts as Lewis acid in the following reactions? (a) Cu2+ + 4Cl– [CuCl4 ]2– (d) NH3 + H+ → NH4 + (b) CaO + SO3 → CaSO4 (e) H+ + OH– → H2 O (c) AlCl3 + Cl2 → [AlCl4 ]– + Cl+ 7.3 Conjugate Acid-base Pair 1 Consider the following equilibrium: H+ H+ NH3 (aq) + H2 O(l) NH4 +(aq) + OH– (aq) Base Acid Acid Base
258 Chemistry Term 1 STPM CHAPTER 7 Structure of H2 ClO4 +: O O O O H + Cl H 2 In the forward reaction, NH3 is a base and H2 O is an acid. However, in the reverse reaction, NH4 + is an acid and OH– is a base. 3 We see that when ammonia (a base) accepts a proton, it gets converted into NH4 +, an acid. Similarly, when H2 O (an acid) donates a proton to NH3 , it gets converted into OH– , a base. 4 Ammonia and the ammonium ions are known as a conjugate acid-base pair. The NH4 + ion is the conjugate acid of NH3 (which is a base). Water and the hydroxide ion is also a conjugate acid-base pair. OH– is the conjugate base of H2 O (which is an acid). 5 A conjugate acid is produced when a base accepts a proton. A conjugate base is formed when an acid loses a proton. 6 Examples of conjugate acid-base pairs are: CH3 COOH + H2 O CH3 COO– + H3 O+ Acid Base Conjugate base Conjugate acid H2 SO4 + HNO3 HSO4 – + H2 NO3 + Acid Base Conjugate base Conjugate acid H2 SO4 + HClO4 HSO4 – + H2 ClO4 + Acid Base Conjugate base Conjugate acid 7 Generally, HA + B HB+ + A– Acid Base Conjugate acid Conjugate base Example 7.1 Identify the Brønsted-Lowry acids and bases and their conjugate acids and bases in the following equations: (a) CH3 NH2 + HCl CH3 NH3 + + Cl– (b) HCO3 – + H2 O H2 CO3 + OH– (c) [Al(H2 O)6 ] 3+ + H2 O [Al(H2 O)5 OH]2+ + H3 O+ (d) CH3 COOH + C6 H5 O– → CH3 COO– + C6 H5 OH (e) H2 PO4 – + H2 O HPO4 2– + H3 O+ Solution Acid Base Conjugate base Conjugate acid (a) HCl CH3 NH2 Cl– CH3 NH3 + (b) H2 O HCO3 – OH– H2 CO3 (c) [Al(H2 O)6 ] 3+ H2 O [Al(H2 O)5 OH]2+ H3 O+ (d) CH3 COOH C6 H5 O– CH3 COO– C6 H5 OH (e) H2 PO4 – H2 O HPO4 2– H3 O+ Info Chem Base + H+ → Conjugate acid Acid → H+ + Conjugate base O O O H + N H Structure of H2 NO3 +:
259 Chemistry Term 1 STPM CHAPTER 7 Quick Check 7.2 1 Identify the Brønsted-Lowry acids and bases and their conjugate acids and bases in the following equations: (a) NH4 + + H2 O NH3 + H3 O+ (b) CN– + H2 O HCN + OH– (c) [Cu(H2 O)6 ] 2+ + H2 O [Cu(H2 O)5 (OH)]+ + H3 O+ (d) 2HCOOH HCOO– + HCOOH2 + (e) C2 H5 O– + H3 O+ C2 H5 OH + H2 O (f) H2 O + NH2 – NH3 + OH– 2 Give the formula of the conjugate acids of the following bases. (a) C2 H5 NH2 (c) HSO4 – (e) H2 O (b) HCO3 – (d) NH2 OH (f) CH3 CONH2 3 Give the formula of the conjugate bases of the following acids. (a) HCl (c) C6 H5 COOH (b) H2 O (d) HSO3 – 4 The hydrogencarbonate ion, HCO3 – , is both a Brønsted-Lowry acid and base. Write two equations showing how the HCO3 – ion reacts with water to illustrate its amphoteric nature. 7.4 Relative Strength of Brønsted-Lowry Acids and Bases The relative strength of Brønsted-Lowry’s acids and bases can be compared based on their (a) degree of dissociation, or (b) acid-base dissociation constant Degree of Dissociation 1 The degree of dissociation (α) is given by: α = Amount dissociated Original amount 2 A stronger acid (or base) will have a higher degree of dissociation compared to a weaker acid (or base). 3 However, the degree of dissociation of a weak acid (or weak base) is dependent on its concentration. The table below lists the degree of dissociation of ethanoic acid at different concentrations. [CH3 COOH]/mol dm–3 1.0 0.10 0.010 Degree of dissociation (%) 0.42 1.34 4.24 4 Hence, when comparing the strength of weak acids (or weak bases) we must ensure that the concentrations of the solutions under investigation are the same. 2009/P1/Q12 2017/P1/Q13 2016/P1/Q13 2018/P1/Q13, Q14 Degree of dissociation depends on concentration.
260 Chemistry Term 1 STPM CHAPTER 7 5 The table below lists the degree of dissociation of four acids all with concentrations of 0.10 mol dm–3. Acid H2 SO4 HNO2 CH3 COOH HSO4 – Degree of dissociation (%) ≈ 100 6.6 1.34 29.0 Based on the above table, we can arrange the four acids in order of increasing strength: CH3 COOH HNO2 HSO4 – H2 SO4 Dissociation Constant 1 Consider a weak monobasic acid HA: HA(aq) + H2 O(l) H3 O+(aq) + A– (aq) Or: HA(aq) H+ + A– 2 Applying the equilibrium law: [H+][A– ] [HA] = constant, Ka The constant Ka is known as the dissociation constant of the acid. The higher the magnitude of Ka , the stronger the acid. 3 However, Ka values for weak acids are usually very small, hence, another quantity known as pKa is also used. pKa = – log Ka 4 The acid dissociation constant, like the equilibrium constant, is dependent on temperature. The values used are usually quoted for 25 °C or 298 K. 5 The acid dissociation constants, Ka and pKa , of some acids are given in the table below: Acid Ka /mol dm–3 pKa CH3 COOH 1.8 10–5 4.74 HCOOH 1.8 10–4 3.74 C6 H5 COOH 6.3 10–5 4.20 HF 7.2 10–4 3.14 HCN 4.0 10–10 9.40 HClO 5.0 10–8 7.30 6 For a weak base, B, that dissociates in water according to the equation: B(aq) + H2 O(l) HB+(aq) + OH– At equilibrium, [HB+][OH– ] [B] = Kb And, pKb = – log Kb The acid dissociation is another form of equilibrium constant. Acid dissociation constants are quoted at 298 K. Exam Tips Exam Tips The higher the value of Ka , the lower the pKa .
261 Chemistry Term 1 STPM CHAPTER 7 7 The Kb and pKb values for some bases are given in the table below: Base Kb/mol dm–3 pKb NH3 1.74 10–5 4.76 C2 H5 NH2 4.7 10–4 3.33 CH3 NH2 4.2 10–4 3.38 NH2 OH 9.1 10–9 8.04 Example 7.2 The acid dissociation constant of chloroethanoic acid is 1.2 10–2 mol dm–3. (a) Write an equation to represent the dissociation of chloroethanoic acid in water. (b) Write an expression for Ka . (c) Calculate pKa for the acid. Solution (a) CH2 ClCOOH + H2 O CH2 ClCOO– + H3 O+ (b) Ka = [CH2 ClCOO– ][H3 O+] [CH2 ClCOOH] (c) pKa = – log (1.2 10–2) = 1.92 Info Chem The value of Ka and Kb is independant of the concentration of the solutions. Exam Tips Exam Tips [H2 O] is not included in the expression for Ka because it is the solvent and is present in a large amount that its concentration can be assumed to be constant. Quick Check 7.3 1 Calculate the pKa values for the following acids. Acid Ka /mol dm–3 Iodoethanoic acid 6.9 10–4 Nitrous acid 7.2 10–4 Cynic acid 3.6 10–4 Boric acid 5.8 10–10 2 The degree of dissociation of a 0.10 mol dm–3 ethanoic acid is 0.0134 at 298 K. (a) Write an equation to represent the dissociation of ethanoic acid in water. (b) Write an expression for Ka . (c) Calculate the value of Ka . (d) What is the pKa for ethanoic acid?
262 Chemistry Term 1 STPM CHAPTER 7 Weak polyprotic acids dissociate in stages. Exam Tips Exam Tips Ka = Ka, 1 × Ka, 2 3 The pKb of hydrazine, N2 H4 , is 5.77 at 298 K. (a) Calculate the Kb for hydrazine. (b) Write an equation to represent the reaction between hydrazine and water. (c) Calculate the degree of dissociation of 0.25 mol dm–3 solution of hydrazine. 4 The acid dissociation constant for HSO3 – ion is 6.2 10–8 mol dm–3 at 298 K. (a) Write an equation to represent the dissociation of HSO3 – in water. (b) Write an expression for Ka of HSO3 – . (c) Calculate its pKa value. Dissociation Constant of Polyprotic Acids 1 A polyprotic acid is an acid that can donate more than one H+ ion per molecule to other substances. 2 Examples of polyprotic acids are sulphuric acid, H2 SO4 ; sulphurous acid, H2 SO3 ; carbonic acid, H2 CO3 ; ethanedioic acid, H2 C2 O4 ; phosphoric acid, H3 PO4 . 3 For weak polyprotic acids, the acid molecules dissociate in stages. Each stage has its own dissociation constant. 4 For example, sulphurous acid dissociates in two stages: Stage one: H2 SO3 (aq) H+(aq) + HSO3 – (aq) Ka, 1 = [H+][HSO3 ] – [H2 SO3 ] = 1.30 10–2 mol dm–3 Stage two: HSO3 – (aq) H+(aq) + SO3 2–(aq) Ka, 2 = [H+][HSO3 ] 2– [HSO3 – ] = 6.30 10–8 mol dm–3 5 The acid dissociation constant for the combined process: H2 SO3 (aq) 2H+(aq) + SO3 2–(aq) Ka = [H+] 2 [HSO3 ] 2– [HSO3 – ] = Ka, 1 Ka, 2 = (1.30 10–2)(6.30 10–8) = 8.19 10–10 mol2 dm–6 Quick Check 7.4 1 The acid dissociation constants, Ka, 1 and Ka, 2 of carbonic acid, H2 CO3 , are 4.3 10–7 and 4.8 10–11 mol dm–3 respectively. (a) Write two equations to represent each stage of the dissociation of carbonic acid. (b) Write expressions for Ka, 1 and Ka, 2. (c) Calculate the acid dissociation constant for the following reaction: H2 CO3 2H+ + CO3 2–
263 Chemistry Term 1 STPM CHAPTER 7 2 The acid dissociation constants, Ka, 1, Ka, 2 and Ka, 3 of phosphoric acid, H3 PO4 are 7.9 10–3, 6.3 10–8 and 4.0 10–13 mol dm–3 respectively. (a) Write balanced equations to represent the three stages of dissociation of phosphoric acid. (b) Write expressions for Ka, 1, Ka, 2 and Ka, 3. (c) Calculate the acid dissociation constant for the dissociation of phosphoric acid as follows: H3 PO4 3H+ + PO4 3– The pH and pOH Scale 1 Strong acids dissociate completely in aqueous solution to produce high concentration of H+ ions. Weak acids dissociate partially in aqueous solution to produce low concentration of H+ ions. HCl(aq) + H2 O(l) 100% H3 O+(aq) + Cl– (aq) CH3 COOH(aq) + H2 O(l) CH3 COO– (aq) + H3 O+(aq) 2 The concentration or molarity of the H+ ions in an aqueous solution of an acid can be stated in terms of pH, where: pH = – log [H3 O+] where [ ] = mol dm–3 3 For example, a solution where the [H3 O+] = 0.25 mol dm–3, the pH is given by: pH = – log(0.25) = 0.60 4 Conversely, if we are given the pH of a solution (for example measured using a pH meter) is 3.4. Then the H3 O+ concentration can be calculated as follows: pH = 3.4 – log[H3 O+] = 3.4 ∴ [H3 O+] = 3.98 10–4 mol dm–3 5 The lower the pH value, the higher is the H3 O+ concentration and vice versa. 6 Similarly, the amount of OH– in a basic solution is expressed in terms of pOH where, pOH = – log[OH– ] 7 A solution where [OH– ] is 0.65 mol dm–3, the pOH is given by pOH = – log (0.65) = 0.19 Another unit to represent the H+ concentration of an acid. Quick Check 7.5 1 Calculate the pH or pOH of the following aqueous solutions: (a) [H+] = 3.4 10–3 mol dm–3 (b) [H+] = 0.80 mol dm–3 (c) [H+] = 2.4 mol dm–3 (d) [OH– ] = 5.6 10–2 mol dm–3 (e) [OH– ] = 2.8 mol dm–3
264 Chemistry Term 1 STPM CHAPTER 7 Relationship between [H+] and [OH– ]. Relationship between pH and pOH 2 An aqueous solution of sulphuric acid has a concentration of 0.25 mol dm–3. (a) Write an equation to show the dissociation of sulphuric acid. (b) What is the H+ concentration in the solution? (c) Calculate the pH of the solution. 3 Assuming that barium hydroxide, Ba(OH)2 , is a strong base. Calculate (a) the OH– concentration, and (b) pOH of 0.50 M Ba(OH)2 . 4 A 0.10 mol dm–3 ethanoic acid has a degree of dissociation of 0.0134. (a) Calculate the H+ concentration in the solution. (b) Calculate the pH. 5 The ‘degree of dissociation’ of aqueous ammonia of concentration 1.5 mol dm–3 is 0.34%. (a) Write a balanced equation to represent the ‘dissociation’ of ammonia in water. (b) Calculate the OH– concentration and pOH of the solution. Relationship between pH and pOH 1 It is rather inconvenient to have two different scales to represent the acidity and alkalinity of acids and bases. 2 It would be better if we can combine the two scales into one. 3 To do that, we need to know the relationship between pH and pOH in an aqueous solution. 4 Pure water undergoes autoionisation as shown below: 2H2 O(l) H3 O+(aq) + OH– (aq) Or: H2 O(l) H+(aq) + OH– (aq) Applying the equilibrium law: [H+][OH– ] [H2 O] = Kc 5 Since water is present in very large excess, the concentration of H2 O before and after dissociation is almost the same. So rearrange: [H+][OH– ] = Kc [H2 O] 6 Since [H2 O] is a constant, the product of Kc [H2 O] would give another constant. The new constant is given a symbol of Kw, known as the ionic product of water. [H+][OH– ] = Kw 7 At 25 °C, the concentration of H+ and OH– (obtained from conductivity measurement) is 1.0 10–7 mol dm–3 respectively. Hence, the ionic product of water at 25 °C is [H+][OH– ] = (1.0 10–7) 2 = 1.0 10–14 mol2 dm–6 8 Taking negative logarithm: (– log [H+]) + (– log [OH– ]) = – log (1.0 10–14) pH + pOH = 14.0 The above relationship holds for all aqueous solutions. Info Chem In all aqueous solutions, [H+][OH– ] = 1.0 × 10–14 mol2 dm–6
265 Chemistry Term 1 STPM CHAPTER 7 9 Thus, a solution with pOH of 4.5 will have a pH of = 14.0 – 4.5 = 9.5 10 Since pH and pOH are interconvertable, it is not necessary to have two scales. Instead the pH scale was chosen to represent the acidity or basicity of a solution. 11 For pure water at 298 K [H+] = 1.0 10–7 mol dm–3 ∴ pH = 7.0 12 In acidic solutions: [H+] 1.0 10–7 mol dm–3, and pH 7.0 In alkaline solution: [H+] 1.0 10–7 mol dm–3, and pH 7.0 In neutral solution: [H+] = 1.0 10–7 mol dm–3, and pH = 7.0 13 The pH scale: Strongly acidic Weakly acidic Weakly basic Strongly basic 0 7 14 Calculating pH of Weak Acids 1 The pH of a 0.10 mol dm–3 HCl can be calculated as follows: HCl(aq) → H+(aq) + Cl– (aq) The concentration of H+ = 0.10 mol dm–3 ∴ pH = – log (0.10) = 1.0 The calculation is straight forward because HCl is a strong acid that undergoes complete dissociation in water. 2 However, the situation is different when we consider a weak acid such as ethanoic acid. What is the [H+] of a 0.10 mol dm–3 ethanoic acid? CH3 COOH(aq) CH3 COO– (aq) + H+(aq) Obviously, the [H+] is less than 0.10 mol dm–3. 3 Consider an ethanoic acid of concentration C mol dm–3 with a degree of dissociation of α. CH3 COOH(aq) CH3 COO– (aq) + H+(aq) CH3 COOH CH3 COO– H+ Initial/mol dm–3 C 0 0 Final/mol dm–3 C(1 – α) Cα Cα 4 Applying the equilibrium law: Ka = [CH3 COO– ][H+] [CH3 COOH] = (Cα)2 C(1 – α) 2008/P1/Q13 Info Chem The [OH– ] in a 0.10 mol dm–3 HCl is 1.0 × 10–13 mol2 dm–6.
266 Chemistry Term 1 STPM CHAPTER 7 Assuming α is very small: (1 – α) ≈ α ∴ Ka = (Cα)2 C And α = Ka C Since, [H+] = Cα ∴ [H+] = C Ka C [H+] = Ka C Similarly, for a weak base: [OH– ] = Kb C ABBB ABBB Formula to calculate the [H+] of weak acids Formula to calculate the [OH– ] of weak bases Quick Check 7.6 1 Calculate the pH of a 0.025 mol dm–3 solution of an acid HA which has an acid dissociation constant of 5.6 10–5 mol dm–3. 2 Calculate the pH of a 0.10 mol dm–3 ammonia solution. [Kb for NH3 = 1.74 10–5 mol dm–3] 3 Calculate the pH and degree of dissociation of an acid with Ka = 3.4 10–6 mol dm–3 in (a) 0.20 mol dm–3 solution, and (b) 0.0020 mol dm–3 solution. (c) Comment on your answers. 4 The Kb of a base, BOH is 6.2 10–8 mol dm–3. (a) Calculate the molar concentration of B+ and OH– in a 1.0 10–3 M solution. (b) What is the pH of the above solution? 5 75 cm3 of water is added to 150 cm3 of 0.10 M ethanoic acid. (a) (i) What is the pH of 0.10 M ethanoic acid? (ii) Calculate the degree of dissociation of 0.10 M CH3 COOH. (b) What is (i) the pH, and (ii) the degree of dissociation, of the final mixture? [Ka , CH3 COOH = 1.8 10–5 mol dm–3] 6 Given the acid dissociation constant of chloric acid, HOCl, is 3.7 10–8 mol dm–3. What is the degree of dissociation of a 0.200 M chloric acid (a) in water, and (b) in 0.10 M HCl? 7 A 0.12 M of a weak monoprotic base has a degree of dissociation of 2.6%. Calculate, (a) the [OH– ], (c) the pH of the solution. (b) the [H+], and
267 Chemistry Term 1 STPM CHAPTER 7 Salt Hydrolysis 1 A salt is defined as the product formed when an acid is neutralised by a base. Acid + base → salt + water Examples are: HCl(aq) + NaOH(aq) → NaCl(aq) + H2 O(l) Ba(OH)2 (aq) + H2 SO4 (aq) → BaSO4 (s) + 2H2 O(l) CH3 COOH(aq) + NaOH(aq) → CH3 COONa(aq) + H2 O(l) 2 Most salts dissolve in water to give a neutral solution (pH = 7.0). That means the salt does not react chemically with water. 3 However, there are salts which when dissolved in water give either an acidic solution (pH 7) or an alkaline solution (pH 7). 4 Such salts are said to undergo hydrolysis. Salt + water Hydrolysis acidic or alkaline solution 5 Whether a salt will be hydrolysed by water or not depends on the type of acid and base that the salt is derived from. 6 We shall look at four types of salts in this section: (a) Salts formed between a strong acid and a weak base. (b) Salts formed between a weak acid and a strong base. (c) Salts formed between a strong acid and a strong base. (d) Salts formed between a weak acid and a weak base. Salts Formed between a Strong Acid and a Weak Base 1 An example of a salt formed between a strong acid and a weak base is ammonium chloride: NH3 + HCl → NH4 Cl Weak base Strong acid An aqueous solution of ammonium chloride is acidic (pH 7). 2 When ammonium chloride dissolves in water, it dissociates completely to form NH4 + and Cl– ions. NH4 Cl(s) + aq → NH4 +(aq) + Cl– (aq) 3 The ammonium ion, which is a conjugate acid of NH3 , reacts with water by donating a proton, H+, to water molecule to form H3 O+ ion and undissociated NH3 molecule (a weak base). NH4 +(aq) + H2 O(l) NH3 (l) + H3 O+(aq) 4 On the other hand, Cl– does not react with water: Cl– (aq) + H2 O(l) HCl(aq) + OH– (aq) This reaction does not occur because HCl is a strong acid. It will dissociate completely in water to regenerate Cl– ions. 5 The production of H3 O+ (from the hydrolysis of the NH4 + ion) makes the solution acidic. This is known as cationic hydrolysis. Exam Tips Exam Tips Anions of strong acids (Cl– , Br– , I– , NO3 – and SO4 2–) are not hydrolysed by water.
268 Chemistry Term 1 STPM CHAPTER 7 Example 7.3 Aqueous solution of ethylammonium sulphate, (C2 H5 NH3 +)2 SO4 has a pH of less than 7. Explain why it is so. Solution Aqueous solution of ethylammonium sulphate contains the C2 H5 NH3 + and SO4 2– ions. (C2 H5 NH3 +)2 SO4 → 2C2 H5 NH3 + + SO4 2– The ethylammonium ion acts as a Brønsted-Lowry acid and donates a proton to water molecule to produce H3 O+: C2 H5 NH3 +(aq) + H2 O(l) C2 H5 NH2 (aq) + H3 O+(aq) However, there is no reaction between SO4 2– ions and H2 O molecules. Example 7.4 Explain why aqueous copper(II) sulphate has a pH 7. Solution In aqueous solution, copper(II) ion exists in the form of hexaaquacomplex, [Cu(H2 O)6 ]2+. The haxaaqua-complex reacts with water to produce H3 O+ ion, making the solution acidic. [Cu(H2 O)6 ] 2+(aq) + H2 O(l) [Cu(H2 O)5 OH]+ (aq) + H3 O+ (aq) Salts Formed between a Weak Acid and a Strong Base 1 An example is sodium ethanoate, which is formed by the reaction between sodium hydroxide and ethanoic acid. CH3 COOH(aq) + NaOH(aq) → CH3 COONa(aq) + H2 O Weak acid Strong base An aqueous solution of sodium ethanoate has a pH 7. 2 In aqueous solution, sodium ethanoate undergoes complete dissociation to form ions: CH3 COONa(s) + aq → CH3 COO– (aq) + Na+(aq) 3 The ethanoate ion, a conjugate base, reacts with water to produce OH– and undissociated CH3 COOH molecules. CH3 COO– (aq) + H2 O(l) CH3 COOH(aq) + OH– (aq) 4 On the other hand, Na+ does not react with water. Na+(aq) + H2 O(l) NaOH(aq) + H+(aq) This reaction does not occur because the NaOH produced is a strong base and will dissociate completely in water to regenerate Na+ ions. Exam Tips Exam Tips Cations of strong bases (Na+, K+) are not hydrolysed by water.
269 Chemistry Term 1 STPM CHAPTER 7 5 Production of OH– (from the hydrolysis of CH2 COO– ) makes the solution basic. This is known as anionic hydrolysis. Example 7.5 Explain why an aqueous solution of sodium ethanedioate, Na2 C2 O4 , is basic. Solution In aqueous solution, sodium ethanedioate dissociates completely according to the equation: Na2 C2 O4 (aq) → 2Na+(aq) + C2 O4 2–(aq) The ethanedioate ion reacts with water to form undissociated H2 C2 O4 molecules (a weak acid) and OH– : C2 O4 2–(aq) + 2H2 O(l) H2 C2 O4 (aq) + 2OH– On the other hand, Na+ does not react with water. Salts Formed between a Strong Acid and a Strong Base 1 An example is sodium chloride which is formed by the reaction between sodium hydroxide and hydrochloric acid: NaOH(aq) + HCl(aq) → NaCl(aq) + H2 O(l) Strong base Strong acid An aqueous solution of sodium chloride is neutral (pH = 7). 2 Sodium chloride dissociates completely in water: NaCl(s) + aqueous → Na+(aq) + Cl– (aq) 3 Na+ does not react with water: Na+(aq) + H2 O(l) NaOH(aq) + H+(aq) Cl– does not react with water: Cl– (aq) + H2 O(l) HCl(aq) + OH– (aq) 4 Hence, the solution is neutral since there is no cationic hydrolysis or anionic hydrolysis to produce an excess of H+ or OH– . Salts Formed between a Weak Acid and a Weak Base 1 An example of a salt formed between a weak base and a weak acid is ammonium ethanoate. CH3 COOH(aq) + NH3 (aq) → CH3 COONH4 (aq) Weak acid Weak base 2 In water, ammonium ethanoate dissociates completely into the ammonium ions and ethanoate ions: CH3 COONH4 (s) → CH3 COO– (aq) + NH4 +(aq)
270 Chemistry Term 1 STPM CHAPTER 7 3 The ethanoate ion (a conjugate base) undergoes anionic hydrolysis to produce OH– ions: CH3 COO– (aq) + H2 O(l) CH3 COOH(aq) + OH– This would make the solution basic. 4 The ammonium ion (a conjugate acid) undergoes cationic hydrolysis to produce H+ ions: NH4 +(aq) + H2 O(l) NH3 (aq) + H3 O+(aq) This would make the solution acidic. 5 Thus, an aqueous solution of ammonium ethanoate may be neutral, slightly acidic or slightly basic depending on the relative amount of H3 O+ and OH– ions formed. This in turn depends on the extent of the cationic hydrolysis and the anionic hydrolysis. 6 The extent of hydrolysis depends on the relative values of acid dissociation constant (Ka ) of the cation, NH4 +, and, the base dissociation constant (Kb ) of the anion, CH3 COO– . 7 If Ka Kb , the solution will be acidic because the cation will be hydrolysed to a greater extent than the anion. At equilibrium, more H3 O+ will be produced. If Ka Kb , the solution will be basic because the anion will be hydrolysed to a greater extent than the cation. At equilibrium, more OH– will be produced. If Ka ≈ Kb , the solution will be neutral because both the cation and anion will be hydrolysed to the same extent. 8 In the case of aqueous ammonium ethanoate, the solution is neutral because the acid dissociation constant, Ka , for NH4 + and the base dissociation constant, Kb , for CH3 COO– are almost the same (about 5.4 10–10 mol dm–3). 9 Let us now look at an aqueous solution of ammonium fluoride which is formed between a weak base, NH3 and a weak acid, HF. NH4 F(s) + aq → NH4 +(aq) + F– (aq) NH4 + undergoes hydrolysis as follows: NH4 + + H2 O → NH3 + H3 O+ F– undergoes hydrolysis as follows: F– + H2 O HF + OH– However, the NH4 + ion (Ka = 5.4 10–10) is hydrolysed more extensively than the F– ion (Kb = 1.4 10–11). Thus, there is a surplus of H+ being produced and this makes aqueous ammonium fluoride acidic.
271 Chemistry Term 1 STPM CHAPTER 7 Summary Salt Aqueous solution Type of hydrolysis Strong acid and strong base Neutral No hydrolysis Strong acid and weak base Acidic Cationic hydrolysis Weak acid and strong base Basic Anionic hydrolysis Weak acid and weak base Acidic if Ka Kb Basic if Ka Kb Neutral if Ka ≈ Kb Cationic hydrolysis Anionic hydrolysis No hydrolysis Calculating the pH of Salt Solutions 1 An aqueous solution of ammonium chloride is acidic due to the hydrolysis of the ammonium ion. NH4 +(aq) + H2 O(l) NH3 (aq) + H3 O+(aq) or: NH4 +(aq) NH3 (aq) + H+(aq) 2 The [H+] in the solution can be calculated using the formula: [H+] = Ka C where Ka is the acid dissociation of NH4 +. 3 The acid dissociation constant of NH4 + can be obtained from the base dissociation constant of NH3 , as NH4 + is the conjugate acid of NH3 as shown below. 4 Consider a weak acid HA which dissociates partially in water: HA + H2 O H3 O+ + A– At equilibrium, Ka = [H3 O+][A– ] [HA] 5 Consider the conjugate base, A– , which reacts with water as shown: A– + H2 O HA + OH– At equilibrium, Kb = [HA][OH– ] [A– ] 6 Multiplying Ka and Kb : Ka Kb = [H3 O+][A– ] [HA] [HA][OH– ] [A– ] = [H3 O+][OH– ] = Kw 7 Hence, for any acid-base conjugate pair: Ka Kb = 1.0 10–14 at 25 °C or: pKa + pKb = 14
272 Chemistry Term 1 STPM CHAPTER 7 Example 7.6 Calculate the pH of 0.25 mol dm–3 ammonium chloride. (Kb for NH3 = 1.85 10–5 mol dm–3) Solution The equation of hydrolysis is: NH4 + + H2 O NH3 + H3 O+ To calculate the [H3 O+ ] we need to know the Ka value for the NH4 + ion. Ka for NH4 + = Kw Kb = 1.0 10–14 1.8 10–5 = 5.56 × 10–10 mol dm–3 Using the formula: [H3 O+] = Ka C = (5.56 10–10)(0.25) = 1.18 10–6 mol dm–3 ∴ pH = – log (1.18 10–6) = 4.93 ABBB ABBBBBBBBBBBBB Example 7.7 Calculate the pH of 0.50 mol dm–3 CH3 COONa. (Ka for CH3 COOH = 1.85 10–5 mol dm–3) Solution Equation of hydrolysis: CH3 COO– (aq) + H2 O(l) CH3 COOH(aq) + OH– (aq) Kb for CH3 COO– = Kw Ka = 1.0 10–14 1.85 10–5 = = 5.41 10–10 Using the formula: [H3 O+] = Kb C = (5.41 10–10)(0.50) = 1.64 10–5 mol dm–3 pOH = – log (1.64 10–5) = 4.78 ∴ pH = 14 – 4.78 = 9.22 ABBB ABBBBBBBBBBBBB Quick Check 7.7 1 Determine whether the following aqueous solution would be neutral, acidic or basic. (a) (NH4 )2 SO4 (e) FeCl3 (b) KNO3 (f) C2 H5 ONa (c) NH4 Br (g) Na2 S (d) NaOCl
273 Chemistry Term 1 STPM CHAPTER 7 7.5 Acid-base Titration 1 The aim of acid-base titrations is to determine the volume of two solutions that will exactly react with one another. 2 A known volume of a solution is placed in a conical flask. The other solution is added from a burette until there is enough of it to completely react with the solution originally in the flask. 3 However, most acids, bases and their salts are colourless. Therefore an indicator is necessary to pin-point the ‘end point’ of the titration. An acid-base indicator would change colour at the end-point of a titration. Acid-base Indicators 1 An acid-base indicator is a substance which changes colour according to the pH of the solution. Sometimes it is also known as a pH indicator. 2 Acid-base indicators are usually weak organic acids or weak organic bases that dissociate partially in water with the condition that the undissociated molecules have a different colour from their ions. 3 For example, the undissociated molecule of phenolphthalein is colourless and its anion is pink. HO C OH O C O HO C O O– C O + H+ Colourless Pink 4 In the presence of an acid, the equilibrium will shift towards the left-hand side and the indicator is colourless. 5 In the presence of a base, the equilibrium will shift to the right-hand side and the colour of the indicator changes to pink. 6 However, indicators change colour gradually over a range of pH value, and not at any particular pH. For example, the pH range of phenolphthalein is between 8.3 and 10.0. Pink 14 10 8.3 0 Colourless pH Working of phenolphthalein 2014/P1/Q13 2016/P1/Q14
274 Chemistry Term 1 STPM CHAPTER 7 7 Similarly for methyl orange: N+ H CH3 N N=N + H+ CH (Yellow) 3 SO3 Na – + CH3 N=N CH3 SO3 Na – + (Red) 8 The pH range of an acid-base indicator is usually about pKIn ± 1, where pKIn is the indicator dissociation constant of the indicator. HIn + H2 O H3 O+ + In– KIn = [H3 O+][In– ] [HIn] 9 For example, the KIn for bromothymol blue is 1.0 10–7 mol dm–3 and the pH range of bromothymol blue is calculated as follows: pKIn = – log(1.0 10–7) = 7 ∴ pH range = 7.0 ± 1 = 6.0 to 8.0 10 The table below lists the pH range of some common acid-base indicators. Indicator pH range Colour change Thymol blue 1.2 – 2.8 Red – Yellow Bromophenol blue 3.0 – 4.6 Yellow – Purple Methyl orange 3.1 – 4.4 Red – Yellow Methyl red 4.2 – 6.3 Red – Yellow Chlorophenol blue 4.8 – 6.4 Yellow – Red Bromothymol blue 6.0 – 7.6 Yellow – Blue Phenol red 6.8 – 8.4 Yellow – Red Cresol red 7.2 – 8.8 Yellow – Red Phenolphthalein 8.3 – 10.0 Colourless – Pink Example 7.8 An indicator has a Ka of 4.0 10–4 mol dm–3. Calculate the pH range over which the indicator changes colour. Solution Ka = 4.0 10–4 pKa = – log (4.0 10–4) = 3.40 ∴ pH range = pKa ± 1 = 2.40 to 4.40 Working of methyl orange
275 Chemistry Term 1 STPM CHAPTER 7 Example 7.9 An indicator has a Kb of 1.0 10–5. Over what pH range will the indicator change colour? Solution Kb = 1.0 10–5 pKb = – log (1.0 10–5) = 5.0 ∴ pOH range = pKb ± 1 = 4.0 to 6.0 Hence, pH range = 8.0 to 10.0 Choice of Indicator in Acid-base Titration 1 There are two points of interest in an acid-base titration: The equivalent point and the end point. 2 The equivalent point is when the amount of acid and base present exactly neutralises one another. Neither acid nor base is in excess. The end point is reached when the indicator changes colour. 3 Indicators should be chosen such that the end-point is as close to the equivalent point as possible. 4 For example, the titration between 25.0 cm3 of 0.10 M HCl and 0.15 M NaOH. The equation of reaction is: NaOH + HCl → NaCl + H2 O The number of moles of HCl in 25.0 cm3 = 25 1000 0.10 = 2.50 10–3 mol Hence, the number of moles of NaOH required to neutralise the acid = 2.50 10–3. Let the volume of NaOH required = V cm3 V 1000 0.15 = 2.50 10–3 V = 16.70 cm3 Hence, the equivalent point is reached when 16.70 cm3 of NaOH has been added. 5 Now, if the indicator changes colour before the equivalent point, then the volume recorded would be less than 16.70 cm3 . On the other hand, if the indicator changed colour after the equivalent point, the volume recorded would be more than 16.70 cm3 . 6 Therefore, for accurate results we have to use an indicator that changes colour at the equivalent point of the titration. 7 The choice of indicator depends on the type of acids or bases used. Definition of equivalent point Definition of end point The indicator must change colour only at the equivalent point, not before or after.
276 Chemistry Term 1 STPM CHAPTER 7 The Titration (or pH) Graph 1 The change in pH during an acid-base titration can be followed by measuring the pH of the mixture using a pH meter. 2 The pH change is then plotted against the volume of base (or acid) added from the burette. 3 The curves obtained are called a titration curve or pH curve. 4 These titration curves allow us to choose the most suitable indicator for the particular titration. 5 There are four types of titration curves depending on the type of acids or bases involved. Titration Involving a Strong Acid and a Strong Base 1 An example of a strong acid-strong base titration is the titration between 25.0 cm3 of 0.10 M HCl with 0.10 M NaOH from a burette. 2 The pH graph is as shown below: V pH 7 25 3 Before the addition of NaOH, the pH of the solution is 1.0. When NaOH is added, the pH of the solution increases gradually as more and more HCl is neutralised and the solution becomes less acidic. 4 There is a sharp increase in pH (from 4 to 11) slightly before and after the equivalent point. 5 At the equivalent point (when 25.0 cm3 of NaOH has been added), the solution contains only aqueous sodium chloride: NaOH(aq) + HCl(aq) → NaCl(aq) + H2 O(l) Since sodium chloride does not undergo hydrolysis, the pH at the equivalent point is 7.0. 6 Beyond the equivalent point, the pH increases gradually as more and more NaOH is added. 7 Any indictors whose pH range falls between 4 and 11 can be used. The suitable indicators are: Indicator pH range Colour change Methyl red 4.2 – 6.3 Red – Yellow Chlorophenol blue 4.8 – 6.4 Yellow – Red Bromothymol blue 6.0 – 7.6 Yellow – Blue Phenol red 6.8 – 8.4 Yellow – Red Cresol red 7.2 – 8.8 Yellow – Red Phenolphthalein 8.3 – 10.0 Colourless – Pink Alkali pH meter Electrode Acid
277 Chemistry Term 1 STPM CHAPTER 7 Titration between a Strong Base and a Weak Acid 1 Let us consider the titration between 25.0 cm3 of 0.10 M CH3 COOH and 0.10 M NaOH CH3 COOH + NaOH → CH3 COONa + H2 O The equation can be written as CH3 COOH + OH– → CH3 COO– + H2 O 2 The pH curve for the titration is as shown in the diagram. 3 Before the addition of NaOH, the pH of the solution is about 2.9 (ethanoic acid is a weak acid). When NaOH is added, the pH of the solution increases gradually as more and more CH3 COOH is neutralised and the solution becomes less acidic. 4 There is a sharp increase in pH (from 7 to 11) slightly before and after the equivalent point. 5 At the equivalent point (when 25.0 cm3 of NaOH has been added), the solution contains only aqueous sodium ethanoate. OH– (aq) + CH3 COOH(aq) → CH3 COO– (aq) + H2 O(l) However, the pH at the equivalent point is not 7.0 but about 10.0. This is because the ethanoate ion undergoes hydrolysis to produce OH– , thus making the solution slightly basic. CH3 COO– (aq) + H2 O(l) CH3 COOH(aq) + OH– (aq) 6 Beyond the equivalent point, the pH increases gradually as more and more NaOH is added. 7 Any indicators whose pH range is between 7 and 10 can be used. The suitable indicators are: Indicator pH range Colour change Cresol red 7.2 – 8.8 Yellow – Red Phenolphthalein 8.3 – 10.0 Colourless – Pink 8 Methyl orange is not suitable because it changes colour before the equivalent point. Titration between a Strong Acid and a Weak Base 1 Let us consider the titration between 25.0 cm3 of 0.10 M HCl and 0.10 M NH3 . HCl + NH3 → NH4 Cl The equation can be written as H+ + NH3 → NH4 + 2 The pH curve for the titration is as shown in the diagram. 3 Before the addition of NH3 , the pH of the solution is 1.0. When NH3 is added, the pH of the solution increases gradually as more and more HCl is neutralised and the solution becomes less acidic. 4 There is a sharp increase in pH (from 3 to 7) slightly before and after the equivalent point. V pH 7 25 pH 7 25 V 2014/P1/Q13
278 Chemistry Term 1 STPM CHAPTER 7 5 At the equivalent point (when 25.0 cm3 of NH3 has been added), the solution contains only aqueous ammonium chloride. H+(aq) + NH3 (aq) → NH4 +(aq) However, the pH at the equivalent point is not 7.0 but about 5.0. This is because the ammonium ion undergoes hydrolysis to produce H3 O+, thus making the solution slightly acidic. NH4 +(aq) + H2 O(l) NH3 (aq) + H3 O+(aq) 6 Beyond the equivalent point, the pH increases gradually as more and more NH3 is added. 7 Any indictors whose pH range is between 4 to 7 can be used. The suitable indicators are: Indicator pH range Colour change Methyl orange 3.1 – 4.4 Red – Yellow Methyl red 4.2 – 6.3 Red – Yellow Chlorophenol blue 4.8 – 6.4 Yellow – Red Titration between a Weak Base and a Weak Acid 1 An example is the titration between 25.0 cm3 of 0.10 M CH3 COOH with 0.10 M NH3 . CH3 COOH(aq) + NH3 (aq) → CH3 COONH4 (aq) 2 The pH curve has the following shape. 3 There is no sharp increase in pH at the equivalent point of the titration (after adding 25.0 cm3 NH3 ). 4 There are no suitable indicators for such titration. The end-point has to be determined by other suitable methods. Example 7.10 Sketch the pH curve for the following titrations: (a) 25.0 cm3 of 0.10 M NaOH against 0.10 M HCl (b) 25.0 cm3 of 0.10 M NH3 against 0.10 M HCl (c) 25.0 cm3 of 0.10 M NaOH against 0.10 M CH3 COOH Solution (a) Volume of HCI pH 7 25 (c) Volume of CH3 COOH pH 7 25 (b) Volume of HCI pH 7 25 pH V 7
279 Chemistry Term 1 STPM CHAPTER 7 Titration Involving Weak Polyprotic Acids 1 The pH graph involving weak polyprotic acids or polyprotic bases consists of more than one equivalent point (or inflexion points). These points correspond to the formation of acidic salts and normal salts. 2 Take the example of the reaction between 25.0 cm3 of 0.10 M oxalic acid (ethanedioic acid), H2 C2 O4 , and 0.10 M sodium hydroxide. 3 The pH graph is shown below: Volume of NaOH pH 25 50 Methyl orange Phenolphthalein 4 The first equivalent point (when 25.0 cm3 of NaOH has been added) corresponds to the formation of sodium hydrogen oxalate: NaOH + H2 C2 O4 → NaHC2 O4 + H2 O The pH rises from about 3.5 to 7. 5 The second equivalent point occurs when an additional 25.0 cm3 NaOH has been added, and this corresponds to the formation of sodium oxalate from sodium hydrogen oxalate: NaHC2 O4 + NaOH → Na2 C2 O4 + H2 O 6 Hence, it can been seen from the graph that methyl orange will detect the first end-point while phenolphthalein would detect the second end-point. Quick Check 7.8 1 25.0 cm3 of chloroethanoic acid required 31.50 cm3 of 0.050 M sodium hydroxide for reaction. [Ka of chloroethanoic acid = 1.2 10–2 mol dm–3] (a) Write an equation for the reaction between chloroethanoic acid and sodium hydroxide. (b) Calculate the molar concentration of chloroethanoic acid. (c) What is the pH of 25.0 cm3 of chloroethanoic acid? 2 25.0 cm3 of a monoprotic acid, HXO4 requires 35.00 cm3 of a solution containing 5.72 g dm–3 of NaOH for complete reaction using phenolphthalein as indicator. (a) Calculate the molar concentration of the NaOH solution. (b) Write a balanced equation for the reaction between NaOH and HXO4 . (c) Calculate the molar concentration of HXO4 . (d) If the concentration of the acid is 20.1 g dm–3, calculate the relative molecular mass of HXO4 . (e) Determine the relative atomic mass of X and suggest an identity for X. NaHC2 O4 is known as an acidic salt. NaHC2 O4 → Na+ + H+ + C2 O4 2–
280 Chemistry Term 1 STPM CHAPTER 7 7.6 Buffer Solution 1 A buffer solution is a solution whose pH does not change significantly when a little acid or base is added to it. 2 When 1 cm3 of 1 M hydrochloric acid is added to 1 dm3 of pure water, the pH changes from 7 to 3. If such a change was to occur in a biological system, the organism would perish. For example, if the pH of human blood was to change by as little as 0.4 unit, the person would lapse into coma. 3 Most biological reactions (such as those involving enzymes) are very sensitive to changes in pH. Hence, the reacting system has to be carefully buffered. 4 Buffers are also important in the food industries. Certain processed food has to be buffered so that they do not cause undue change in pH when consumed. 5 So, how does a buffer solution resists a change in pH? To do so, it must fulfill the following three conditions. (a) There must be a large amount of acid present to react with the bases added. (b) There must be a large amount of base present to react with the acids added. (c) The acid and base component of the buffer must not neutralise one another. 6 These conditions can be achieved if the solution contains a mixture of a weak acid and its conjugate base, or a weak base with its conjugate acid. 7 There are two types of buffer solutions: the acidic buffer (pH 7) and the alkaline buffer (pH 7). The Working of Buffer Solutions Acidic Buffer Solution 1 An acidic buffer is prepared by mixing a weak acid and its conjugate base. 2 An example is an aqueous mixture of ethanoic acid and sodium ethanoate (which supplies the conjugate base, CH3 COO– ). 3 Ethanoic acid, a weak acid, dissociates partially in water: CH3 COOH(aq) + H2 O(l) CH3 COO– (aq) + H3 O+(aq) On the other hand, sodium ethanoate dissociates completely in water: CH3 COONa(aq) → CH3 COO– (aq) + Na+(aq) 4 The presence of CH3 COO– ions from CH3 COONa further depresses the dissociation of CH3 COOH. Buffers are very important in biological system. pH of human blood = 7.4 Buffers in food industries A buffer must contain a significant amount of acid and base to react with any added OH– or H+. The acid and base components in the buffer must not neutralise one another. Weak acid + conjugate base → acidic buffer There is no reaction between ethanoic acid and sodium ethanoate. The dissociation of CH3 COOH is supressed by CH3 COONa. INFO Buffer Solution 2008/P1/Q14 2010/P1/Q13 2015/P1/Q20(b) 2009/P1/Q13 2014/P1/Q14 2013/P1/Q11
281 Chemistry Term 1 STPM CHAPTER 7 5 As a result, there is a large amount of undissociated CH3 COOH molecules (the weak acid) and CH3 COO– ions (the conjugate base) in the mixture. 6 When a little acid (H3 O+) is added to the mixture, the H3 O+ ions will be ‘mopped up’ by the CH3 COO– ion to form undissociated CH3 COOH molecules. H3 O+ (added) + CH3 COO– → CH3 COOH + H2 O 7 When a little base is added, the OH– ions will react with undissociated CH3 COOH molecules to form CH3 COO– ions. OH– (added) + CH3 COOH → CH3 COO– + H2 O 8 In both cases, the added H3 O+ and the OH– added are effectively removed. The pH of the solution is not much affected. Example 7.11 A mixture of carbonic acid, H2 CO3 and sodium hydrogencarbonate, NaHCO3 acts as a buffer solution. (a) Write an equation for the equilibrium that exists in the mixture. (b) Using balanced equations, show how the system maintains its pH on the addition of a little acid or base. Solution (a) H2 CO3 (aq) + H2 O(l) H3 O+(aq) + HCO3 – (aq) (b) H3 O+ (added) + HCO3 – → H2 CO3 + H2 O OH– (added) + H2 CO3 → HCO3 – + H2 O Basic Buffer Solution 1 A basic buffer is prepared by mixing a weak base and its conjugate acid. 2 An example is an aqueous mixture of ammonia and ammonium chloride (which supplies the conjugate acid, NH4 +). 3 Ammonia, a weak base, ‘dissociates’ partially in water: NH3 (aq) + H2 O(l) OH– (aq) + NH4 +(aq) On the other hand, ammonium chloride, dissociates completely in water: NH4 Cl(aq) → NH4 +(aq) + Cl– (aq) 4 The presence of NH4 + ions from NH4 Cl further depresses the ‘dissociation’ of NH3 . 5 As a result, there is a large reservoir of undissociated NH3 molecules (the weak base) and NH4 + ions (the conjugate acid) in the mixture. Removal of added base Weak base + conjugate acid → alkaline buffer There is no reaction between ammonia and ammonium chloride. The ‘dissociation’ of NH3 is supressed by the presence of NH4 Cl. Removal of added acid Info Chem This is one of the buffer systems that helps to regulate the pH of blood.
282 Chemistry Term 1 STPM CHAPTER 7 6 When a little acid (H3 O+ ) is added to the mixture, the H3 O+ ions will be ‘mopped up’ by the undissociated NH3 molecules to form NH4 +. H3 O+ (added) + NH3 → NH4 + + H2 O 7 When a little base is added, the OH– ions will react with NH4 + molecules to form undissociated NH3 molecules. OH– (added) + NH4 + → NH3 (aq) + H2 O 8 In both cases, the added H3 O+ and the OH– added are effectively removed. Hence, the pH of the solution is not much affected. Calculating the pH of a Buffer Solution The Acidic Buffer 1 Consider an aqueous mixture of ethanoic acid and sodium ethanoate. CH3 COOH + H2 O CH3 COO– + H3 O+ (I) CH3 COONa → CH3 COO– + Na+ 2 Applying the equilibrium law to reaction (I): [CH3 COO– ][H3 O+] [CH3 COOH] = Ka (The concentration in the above expression refers to the equilibrium concentration of the species. The [CH3 COO– ] comes from the dissociation of CH3 COOH as well as from CH3 COONa). 3 Rearranging the equation: [H3 O+] = Ka [CH3 COOH] [CH3 COO– ] 4 Since the degree of dissociation of CH3 COOH is very small in the presence of CH3 COONa, the equilibrium concentration of CH3 COOH can be approximated to be the same as the original concentration of the acid. 5 Similarly, the contribution of CH3 COO– from CH3 COOH is so little that it can be ignored. The equilibrium concentration of CH3 COO– can be approximated to be equal to the original concentration of the salt. [CH3 COOH]equilibrium ≈ [Acid]original [CH3 COO– ]equilibrium ≈ [Salt]original 6 Substituting into the above expression: [H3 O+] = Ka [Acid] [Salt] Taking – log throughout: – log [H3 O+] = – log Ka – log [Acid] [Salt] ∴ pH = pKa – log [Acid] [Salt] Removal of added acid Removal of added base Formula to calculate the pH of an acidic buffer
283 Chemistry Term 1 STPM CHAPTER 7 7 It should be noted that the pH of a buffer solution does not depend on the actual amount of acid or salt (the conjugate base) present. It is governed only by their ratio. However, they will have different buffer capacity. Example 7.12 Calculate the pH of a buffer solution by mixing (a) 0.25 mol of CH3 COOH and 1.0 mol of CH3 COONa in 1 dm3 of water. (b) 0.50 mol of CH3 COOH and 2.0 mol of CH3 COONa in 1 dm3 of water [Ka of CH3 COOH = 1.85 10–5 mol dm–3] Solution (a) Using the formula: pH = pKa – log [Acid] [Salt] pH = – log (1.85 10–5) – log 0.25 1.0 = 4.73 – (–0.60) = 5.33 (b) The pH = 5.33 The ratio of [Acid] [Salt] is the same in both cases. The Basic Buffer 1 Consider an aqueous mixture of ammonia and ammonium chloride. NH3 (aq) + H2 O(l) NH4 +(aq) + OH– (I) NH4 Cl → NH4 + + Cl– 2 Applying the equilibrium law to reaction (I): [OH– ][NH4 +] [NH3 ] = Kb The concentration in the above expression refers to the equilibrium concentration of the species. 3 Rearranging the equation: [OH– ] = Kb [NH3 ] [NH4 +] 4 Since the degree of ‘dissociation’ of ammonia is very small in the presence of ammonium chloride, the equilibrium concentration of NH3 can be approximated to be the same as the original concentration of the base. The pH is determined by the [Acid] [Salt] ratio. In both cases: [Acid] [Salt] = 0.25
284 Chemistry Term 1 STPM CHAPTER 7 5 Similarly, the contribution of NH4 + from NH3 is so little that it can be ignored. Hence, the equilibrium concentration of NH4 + can be approximated to be equal to the original concentration of the salt. [NH3 ] equilibrium ≈ [Base]original [NH4 +] equilibrium ≈ [Salt]original 6 Substituting into the above expression: [OH– ] = Kb [Base] [Salt] Taking – log throughout: – log [OH– ] = – log Kb – log [Base] [Salt] ∴ pOH = pKa – log [Base] [Salt] Example 7.13 Calculate the pH of a solution containing 0.40 mol dm–3 NH3 and 0.25 mol dm–3 NH4 Cl. [Kb for NH3 = 1.8 × 10–5 mol dm–3] Using the formula: pOH = pKb – log [Base] [Salt] = 4.73 – log 0.40 0.25 = 4.53 ∴ pH = 14 – 4.53 = 9.47 Example 7.14 A buffer is prepared by dissolving 0.25 mol CH3 COONa in 200 cm3 of 0.60 mol dm–3 aqueous CH3 COOH. (a) Calculate the pH of the buffer. (b) Calculate the pH of the solution after 1 cm3 of 1.0 M HCl is added to it. (c) Calculate the pH of the solution after 1 cm3 of 1.0 M NaOH is added to it. Solution (a) pH = pKa – log [Acid] [Salt] Concentration of CH3 COONa = 0.25 1000 200 = 1.25 mol dm–3 pH = – log (1.85 10–5) – log 0.60 1.25 = 4.73 – (– 0.32) = 5.05 Formula to calculate the pOH of an alkaline buffer
285 Chemistry Term 1 STPM CHAPTER 7 (b) Number of moles of H+ added = 1 1000 1 = 1.0 10–3 mol The added H+ will be removed by the following reaction: H+ (added) + CH3 COO– → CH3 COOH H+ (added) CH3 COO– CH3 COOH Original/mol 0 0.25 0.6 200 1000 = 0.12 Addition/mol 1.0 10–3 – – Final/mol 0 0.25 – (1.0 10–3) = 0.249 0.12 + (1.0 10–3) = 0.121 The final concentration of CH3 COO– = 0.249 1000 201 = 1.24 mol dm–3 The final concentration of CH3 COOH = 0.121 1000 201 = 0.602 mol dm–3 Hence, the final pH is given by: pH = 4.73 – log 0.602 1.24 = 5.04 The change is only by 0.01 unit. (c) Number of moles of OH– added = 1 1000 1 = 1.0 10–3 mol The added OH– will be removed by the following reaction: OH– (added) + CH3 COOH → CH3 COO– OH– (added) CH3 COOH CH3 COO– Original/mol 0 0.6 200 1000 = 0.12 0.25 Addition/mol 1.0 10–3 – – Final/mol 0 0.12 – (1.0 10–3) = 0.119 0.25 + (1.0 10–3) = 0.251 The final concentration of CH3 COO– = 0.251 1000 201 = 1.25 mol dm–3 The final concentration of CH3 COOH = 0.119 1000 201 = 0.592 mol dm–3 Hence, the final pH is given by: pH = 4.73 – log 0.592 1.24 = 5.05
286 Chemistry Term 1 STPM CHAPTER 7 Quick Check 7.9 1 Which of the following mixtures would form a buffer solution? (a) KCl and HCl (d) C2 H5 NH2 and C6 H5 NH3 +Cl– (b) H2 SO4 and NaHSO4 (e) H2 C2 O4 and CH3 COOH (c) CH3 COONa and CH3 COOK (f) NaOCN and HOCN 2 Calculate the pH of the following solutions. (a) 500 cm3 solution containing 0.45 mol of CH3 COOH and 0.67 mol of CH3 COONa. [Ka of CH3 COOH = 1.8 10–5 mol dm–3] (b) 8.20 g of sodium ethanoate in 450 cm3 of 0.15 M ethanoic acid. (c) A mixture containing 250 cm3 of 0.45 M potassium methanoate, HCOOK and 150 cm3 of 0.25 M methanoic acid. [Ka for HCOOH = 1.80 10–4 mol dm–3] (d) A mixture containing 0.10 M boric acid, H3 BO3 and 0.35 M sodium dihydrogen borate, NaH2 BO3 . [Ka of H3 BO3 = 7.30 10–10 mol dm–3] (e) A mixture formed by mixing 50.0 cm3 of 0.020 M NaOH and 50.0 cm3 of 0.035 M CH3 COOH. [Ka of CH3 COOH = 1.8 10–5 mol dm–3] (f) 35.0 cm3 of 0.10 M HCl is added to a 50.0 cm3 of a solution containing 0.15 M CH3 COONa and 0.20 M CH3 COOH. [Ka of CH3 COOH = 1.8 10–5 mol dm–3] Preparing an Acidic Buffer 1 For example, we are to prepare a buffer solution of pH = 4. This is an acidic buffer. 2 First, choose a weak acid whose pKa value is as close to 4 as possible. In this case, we would choose methanoic acid (pKa = 3.8). 3 Then, use the acidic buffer formula to calculate the ratio of [Acid] [Salt] : pH = pKa – log [Acid] [Salt] 4.0 = 3.8 – log [Acid] [Salt] – log [Acid] [Salt] = 0.2 ∴ [Acid] [Salt] = 0.63 4 The buffer solution can then be prepared using either one of the following methods: (a) Dissolve 0.63 mol of methanoic acid and 1.0 mol of sodium methanoate in water and make up to 1 dm3 . (b) Prepare, separately, a 1.0 M solution of methanoic acid and 1.0 M solution of sodium methanoate. Then mix the two solutions in the ratio of 1 : 0.63 by volume. Calculate the [acid] [salt] ratio.
287 Chemistry Term 1 STPM CHAPTER 7 Preparing a Basic Buffer Solution 1 For example, we are to prepare a buffer solution of pH = 9.8 (or pOH = 4.2). This is a basic buffer. 2 Choose a weak base whose pKb is as close to 4.2 as possible. Ammonia (with pKb = 4.73) is chosen for this case. 3 Using the basic buffer formula: pOH = pKb – log [Base] [Salt] 4.2 = 4.73 – log [Base] [Salt] – log [Base] [Salt] = – 0.53 ∴ [Base] [Salt] = 0.30 4 Either mix 0.30 mol of NH3 and 1.0 mol of NH4 Cl and make up to 1 dm3 with water, or mix a solution of 1.0 M NH3 and 1.0 M NH4 Cl in the ratio of 0.3 : 1 by volume. Example 7.15 Calculate the mass of sodium ethanoate needed to add to 500 cm3 of 0.10 mol dm–3 CH3 COOH to produce a buffer of pH = 4.5. [Ka for CH3 COOH = 1.85 10–5 mol dm–3]. Solution Using the formula: pH = pKa – log [CH3 COOH] [CH3 COONa] pKa for CH3 COOH = – log(1.85 10–5) = 4.74 Let the concentration of CH3 COONa needed be a mol dm–3. ∴ 4.5 = 4.74 – log 0.1 a a = 0.058 mol dm–3 = 0.058 82 g dm–3 = 4.76 g dm–3 ∴ Mass of CH3 COONa required = 500 1000 4.76 g = 2.38 g Buffer Capacity 1 Buffers do not have infinite ability to resist pH changes. When too much acid is added such that all the weak base components (or the conjugate base) are used up, the solution can no longer function as a buffer against additional acid. Buffer capacity is the amount of acid or base that needs to be added to a buffer solution for its pH to change by one unit. 2013/P1/Q14
288 Chemistry Term 1 STPM CHAPTER 7 2 Similarly, when too much base is added such that all the acid components (or conjugate acids) are used up, the solution can no longer function as a buffer against additional base. 3 Consider the following mixtures: A: 1.0 mol CH3 COOH and 1.0 mol CH3 COONa in 1 dm3 solution. B: 0.10 mol CH3 COOH and 0.10 mol CH3 COONa in 1 dm3 solution. Both the mixtures have the same pH, but mixture A has a higher buffer capacity because it contains more CH3 COOH and CH3 COO– and can ‘absorb’ more H+ or OH– that is added to it. 4 Consider a mixture containing 3.0 mol CH3 COOH and 0.10 mol CH3 COONa in 1 dm3 solution. The mixture is not a good buffer because, although it has high buffer action against addition of OH– , it has a very low buffer action against addition of H+. 5 Hence, a good buffer should have a large amount of the acid and base components in the ratio of 1 : 1. 6 A solution has maximum buffer capacity when the concentration of the acid and its conjugate base, or the base and its conjugate acid are the same. 7 Looking back at the ethanoic acid and ethanoate system: pH = pKa – log [Acid] [Salt] when [Acid] = [Salt], [Acid] [Salt] = 1, and pH = pKa 8 A buffer solution would have maximum buffer capacity when the pH of the buffer solution is the same as the pKa of the acid, or when the pOH of the buffer solution is equal to the pKb of the base. Buffer Action in Acid-base Titration 1 Consider the titration between 25.0 cm3 of 0.10 M ethanoic acid and 0.10 M sodium hydroxide. The pH graph is as follows: Volume of NaOH pH 12.5 25 A B 4.74 7 Maximum buffer capacity The region ‘AB’ is a buffer system. [CH3 COOH] [CH3 COONa] = 1 2017/P1/Q17(b) 2016/P1/Q20(b)(i)(ii)
289 Chemistry Term 1 STPM CHAPTER 7 2 The starting pH of the solution in the conical flask is about 2.9. On addition of a little sodium hydroxide, some of the ethanoic acid will react to form sodium ethanoate: CH3 COOH(aq) + NaOH(aq) → CH3 COONa(aq) + H2 O(l) The flask now contains a mixture of ethanoic acid and ethanoate ion (the conjugate base of CH3 COOH), and acts as a buffer solution. 3 The pH along the line AB does not change much because of the buffer action of the mixture. 4 The solution has a maximum capacity when 12.5 cm3 of sodium hydroxide has been added (Midway titration). At this point, half of the ethanoic acid has converted to sodium ethanoate: CH3 COOH(aq) + NaOH(aq) → CH3 COONa(aq) + H2 O(l) Initial/mol 0.0025 0 0 Addition/mol 0.00125 Midway titration/ml 0.00125 0 0.00125 5 At this point, the ratio of [acid] [salt] = 1, and the pH of the solution is equal to the pKa of ethanoic acid (which is 4.74). Example 7.16 The pH graph below refers to the titration between 25.0 cm3 of a monoprotic weak base and hydrochloric acid of 0.15 mol dm–3. 2 0 4 10 20 30 40 Volume of HCI /cm3 6 8 10 12 24 12 14 pH (a) Calculate the molar concentration of the base. (b) What is the Kb of the base?
290 Chemistry Term 1 STPM CHAPTER 7 Solution (a) Volume of HCl required = 24.00 cm3 (M 25.0)base = (0.15 24.00)HCl Molarity of the base = 0.144 mol dm–3 (b) From the graph, pH at mid-point of the titration (after the addition of 12.00 cm3 HCl) = 10.0 ∴ pOH = 14 – 10 = 4.0 Hence, pKb of the base = 4.0 Kb = 1.0 10–4 mol dm–3 7.7 Important Buffers in Human Blood 1 As mentioned before, if the pH of the human blood (pH = 7.4) is changed by as little as 0.4 unit, it could prove fatal. 2 There are three important buffer systems in the human blood: (a) carbonic acid/hydrogen carbonate system (b) the phosphate system (c) the amino acid systems 3 The carbonic/hydrogencarbonate system involves the following equilibrium: H2 CO3 (aq) H+(aq) + HCO3 – (aq) On addition of a little acid, it is destroyed by the following reaction: H+ (added) + HCO3 – (aq) → H2 CO3 (aq) On addition of a little base, it is destroyed by the following reaction: OH– (added) + H2 CO3 (aq) → HCO3 – (aq) + H2 O(l) 4 The phosphate system involves the following equilibrium: H2 PO4 – (aq) HPO4 2–(aq) + H+(aq) On addition of a little acid, it is destroyed by the following reaction: H+ (added) + HPO4 2–(aq) → H2 PO4 – (aq) On addition of a little base, it is destroyed by the following reaction: OH– (added) + H2 PO4 – (aq) → H2 O(l) + HPO4 2–(aq) 5 Amino acids are compounds with the basic amine group, and the acidic carboxyl group in its molecule. They can be represented by the general formula of: H2 N—R—COOH On the addition of a little acid, it is destroyed by the following reaction: H+(aq) + H2 N—R—COOH → H3 +N—R—COOH On the addition of a little base, it is destroyed by the following reaction: OH– (added) + H2 N—R—COOH → H2 N—R—COO– + H2 O Exam Tips Exam Tips The –NH2 group is basic. The –COOH group is acidic. The carbonic acid formed will decompose to carbon dioxide which is expel from the lungs: H2 CO3 (aq) → H2 O(l) + CO2 (g) The amino acids are all α-amino acids.
291 Chemistry Term 1 STPM CHAPTER 7 Quick Check 7.10 1 0.20 g of sodium hydroxide is added to 1 dm3 of a solution containing 0.15 mol of ethanoic acid and 0.19 mol of sodium ethanoate. (a) Calculate the pH of the solution before the addition of sodium hydroxide. (b) Calculate the change in pH after the addition of sodium hydroxide. (c) Comment on your answer. 2 A buffer solution was prepared by dissolving 0.020 mol of ethanoic acid and 0.018 mol of sodium ethanoate in enough water to make a 1 dm3 solution. (a) Calculate the pH of the buffer solution. (b) Calculate the change in pH when (i) 1.0 10–4 mol of sodium hydroxide, and (ii) 1.0 10–4 mol of hydrochloric acid, is added to 100 cm3 of the buffer solution. 7.8 Solubility Equilibria: Sparingly Soluble Salts Solubility Product, K sp 1 Not all ionic solids are soluble in water. Examples of insoluble salts are silver chloride, barium sulphate, calcium carbonate, lead iodide and many more. 2 However, no salts are completely insoluble in water. It is just that the solubility of such salts is very low at room conditions, and thus considered insoluble. 3 The solubility of a salt in water can be expressed in terms of mol dm–3 or g dm–3. Other units used are (g per 100 cm3 ) or (g per 100 g of solvent). 4 The solubility of a salt in water is affected by temperature. For most works, the temperature is fixed at 25 °C or 298 K. 5 A typical solubility curve is shown in the diagram. 6 For example, the solubility of silver chloride in water at 298 K is 1.0 10–5 mol dm–3 or 0.00144 g dm–3. On the other hand, the solubility of tin(II) sulphide, SnS, is only 1.0 10–26 mol dm–3 or 0.000 000 000 015 1 g dm–3 (or 1.51 10–11 g dm–3)! 7 When silver chloride is added, a little at a time, to water at room temperature, a saturated solution is eventually formed where no more silver chloride would dissolve no matter how much solid is added to the solution as long as the temperature remains constant. 8 In the saturated solution, the rate of dissolution of silver chloride is the same as the rate of deposition of silver chloride through the combination of Ag+(aq) and Cl– (aq). Tamperature Solubility Ag+(aq) + CI– (aq) AgCl(s) 2009/P1/Q14 2010/P1/Q14, Q15 2013/P1/Q15 2011/P1/Q14 2015/P1/Q16(b) 2017/P1/Q14