1. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Chapter 17 Additional Aspects of Aqueous Equilibria Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Aqueous Equilibria Overview Application of equilibrium: common ion effect buffers (application of common ion effect to acid-base situations) titrations, especially acid-base solubility product (Ksp) complex ions © 2009, Prentice-Hall, Inc.

3. Aqueous Equilibria © 2009, Prentice-Hall, Inc. The Common-Ion Effect Consider a solution of acetic acid: If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. CH 3 COOH (aq) + H 2 O (l) H 3 O + (aq) + CH 3 COO − (aq)

4. Aqueous Equilibria © 2009, Prentice-Hall, Inc. The Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.” Use the same principle to solve common ion problems – include the initial concentration of the common ion in the RICE table.

5. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.1 (p. 720) What is the pH of a solution made by adding 0.30 mol of acetic acid (HC 2 H 3 O 2 ) and 0.30 mol of sodium acetate (NaC 2 H 3 O 2 ) to enough water to make 1.0 L of solution? (4.74)

6. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.1 (p. 720) Dissociation of sodium acetate: NaCH 3 COO (aq)  Na + (aq) + CH 3 COO - (aq) totally dissociates no effect on pH +0.30 M

7. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.1 (p. 720) What is the pH of a 0.30 M solution of acetic acid? (i.e. without the added sodium acetate)

8. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.1 (p. 720) Calculate the pH of a solution containing 0.085 M nitrous acid (HNO 2 ; K a = 4.5 x 10 -4 ) and 0.10 M potassium nitrite (KNO 2 ). (3.42)

9. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.2 (p. 722) Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. (Note: common ion is H 3 O + ) K a for HF is 6.8  10 −4. [H 3 O + ] [F − ] [HF] K a = = 6.8  10 -4

10. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Because HCl, a strong acid, is also present, the initial [H 3 O + ] is not 0, but rather 0.10 M. [HF], M[H 3 O + ], M[F − ], M Initially0.200.100 Change−x−x+x+x+x+x At Equilibrium 0.20 − x  0.200.10 + x  0.10 x HF (aq) + H 2 O (l) H 3 O + (aq) + F − (aq) The Common-Ion Effect HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) totally dissociates 0.10 M no effect

11. Aqueous Equilibria © 2009, Prentice-Hall, Inc. = x 1.4  10 −3 = x (0.10) (x) (0.20) 6.8  10 −4 = (0.20) (6.8  10 −4 ) (0.10) The Common-Ion Effect

12. Aqueous Equilibria © 2009, Prentice-Hall, Inc. The Common-Ion Effect Therefore, [F − ] = x = 1.4  10 −3 [H 3 O + ] = 0.10 + x = 0.10 + 1.4  10 −3 = 0.10 M So,pH = −log (0.10) pH = 1.00

13. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice 17.2 Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCOOH; K a = 1.8 x 10 -4 ) and 0.10 M in HNO 3. ([HCOO-] = 9.0 x 10 -5 M; pH = 1.00) (for HW)

14. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffers Buffers are solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.

15. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffers Consists of a mixture of a weak acid and its conjugate base  Contains both an acidic species (to neutralize OH - ) and a basic species (to neutralize H + )

16. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffers The most effective buffers contain equal amounts of the weak acid and its conjugate base Add the salt of the conjugate base to make [HX] and [X - ] equal  moves the equilibrium to the left

17. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffers If a small amount of OH - is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH − to make F − and water. The [HX]/[X - ] ratio remains more or less constant  no significant change in pH.

18. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffers Similarly, if acid is added, the F − reacts with it to form HF and water.

19. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H + ] [A − ] [HA] K a = HA + H 2 OH 3 O + (H + ) + A −

20. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffer Calculations Rearranging slightly, this becomes [A − ] [HA] K a = [H + ] Taking the negative log of both sides, we get [A − ] [HA] −log K a = −log [H + ] + − log pKapKa pH acid base

21. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffer Calculations So pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation.

22. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Buffer Capacity and pH Range Buffer capacity = the amount of acid or base that can be neutralized before there is a significant change in pH depends on [buffer components]  [conjugate acid-base pair]   buffer capacity

23. Aqueous Equilibria © 2009, Prentice-Hall, Inc. pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a pK a close to the desired pH.

24. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.3 (p. 725) What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4  10 −4. (3.77) Note: You may solve this problem using either the standard equilibrium method, with RICE table, or the Henderson-Hasselbalch Equation.

25. Aqueous Equilibria Sample Exercise 17.3 © 2009, Prentice-Hall, Inc. pH = ? [HC 2 H 3 O 2 ] = 0.12 M [C 2 H 3 O 2 - ] = 0.10 M K a = 1.4 x 10 -4 from NaC 3 H 5 O 3

26. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Henderson–Hasselbalch Equation pH = pK a + log [base] [acid] pH = −log (1.4  10 −4 ) + log (0.10) (0.12) pH pH = 3.77 pH = 3.85 + (−0.08) pH

27. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.4 (p. 726) How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH is 9.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution.) (0.36 mol)

28. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.4 Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (HC 7 H 5 O 2 ) to produce a pH of 4.00. (0.13 M)

29. Aqueous Equilibria © 2009, Prentice-Hall, Inc. When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction. 1 st RICE table or stoich calculation 2 nd RICE table or H-H calculation

30. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Addition of Strong Acid or Base to a Buffer 1.Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2.Use the Henderson–Hasselbalch equation to determine the new pH of the solution (or use a RICE table)

31. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Calculating pH Changes in Buffers – Sample Exercise 17.5 (p. 728) A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. (We have done this calculation before.) a) Calculate the pH of this solution after 0.020 mol of NaOH is added. (4.80) b) Calculate the pH of a 0.020 M NaOH solution for comparison. (no buffer). (12.30)

32. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Calculating pH Changes in Buffers Using the H-H equation: Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 − pH = pK a = −log (1.8  10 −5 ) = 4.74 i.e. K a = [H + ][C 2 H 3 O 2 − ] => [H + ] = (1.8 x 10 -5 )(0.300) [HC 2 H 3 O 2 ] (0.300)

33. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Calculating pH Changes in Buffers Stoich: The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC 2 H 3 O 2 (aq) + OH − (aq)  C 2 H 3 O 2 − (aq) + H 2 O (l) HC 2 H 3 O 2 OH − C2H3O2−C2H3O2− Before reaction0.300 mol0.020 mol0.300 mol After reaction0.280 mol0.000 mol0.320 mol

34. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0.280) pH = 4.74 + 0.06pH pH = 4.80

35. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Calculating pH Changes in Buffers Or use a second RICE table

36. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.5 (cont’d) and, for comparison, b)calculate the pH that would result if 0.020 mol of NaOH were added to 1.00 L of pure water (neglect any volume changes). Note: strong base,  simple calculation. (12.30)

37. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.5 Determine a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl, and(4.68) b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.00 L of pure water.(1.70)

38. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).

39. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

40. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration Curve = a plot of pH versus volume during a titration Titration labs – early April –Choose appropriate indicators –Standardize an NaOH solution

41. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration of a Strong Acid with a Strong Base Four regions of the titration curve: 1.Initial pH – given by the strong acid solution (pH < 7) 2.Between initial pH and equivalence point (pH still < 7) 3.At the equivalence point: pH = 7.00 4.After the equivalence point – given by the amount of excess base (pH > 7)

42. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

43. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration of a Strong Acid with a Strong Base Just before (and after) the equivalence point, the pH increases rapidly.

44. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

45. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

46. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Seven steps to solving titration problems: Analyze the problem, then: 1.Determine # moles of each reactant. 2.Determine # moles of all species after reaction. 3.Calculate new volume after reaction. 4.Determine molarities of all species (combine Steps 2 & 3). 5.Equilibrium calculation – RICE table. * 6.Equilibrium calculation, substituting results from Step 5 into K a. * 7.[H + ]  pH. May mean [OH - ]  pOH  14.00 – pOH = pH. *not required for strong acid-strong base titrations

47. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.6 (p. 731) Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution: a) 49.0 mL( 3.00) b) 51.0 mL(11.00)

48. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.6 Calculate the pH when the following quantities of 0.10 M HNO 3 have been added to 25.0 mL of 0.10 M KOH solution: a) 24.9 mL(10.30) B) 25.1 mL( 3.70)

49. Aqueous Equilibria © 2009, Prentice-Hall, Inc. How to perform and analyze a titration curve e.g. strong base added to strong acid Equivalence point is at pH = 7.00 Choose an indicator that changes color near pH = 7.00, e.g. phenolphthalein (colorless in acid, fuschia in base) End point = the volume of the titrant at the point when the indicator changes color Titration error = difference between equivalence point and end point

50. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Strong Base-Strong Acid Titration Note that the pH begins high then decreases as acid is added. After the equivalence point, the pH is given by the strong acid in excess.  pH < 7

51. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. At the equivalence point the pH is >7. Phenolphthalein is commonly used as an indicator in these titrations.

52. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Weak Acid-Strong Base Titration Four regions of the titration curve: 1. Initial pH: from equilibrium calculation of weak acid 2. Buffer: neutralization  weak acid and its conjugate base: pH does not change very much

53. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Weak Acid – Strong Base Titration 3. equivalence point: pH depends on [conjugate base] and equilibrium calculation: pH > 7 4. after the equivalence point: pH depends on strong base: pH > 7

54. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

55. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.7 (p. 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH solution is added to 50.0 mL of 0.100 M HC 2 H 3 O 2. (K a = 1.8 x 10 -5 ) (2.0 x 10 -6 M)

56. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.7

57. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.7 a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (HC 7 H 5 O 2, K a = 6.3 x 10 -5 ).(4.20) b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH 3.(9.26)

58. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.8 (p. 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M HC 2 H 3 O 2 with 0.100 M NaOH. (8.72)

59. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.8 Calculate the pH at the equivalence point when a)40.0 mL of 0.025 M benzoic acid (HC 7 H 5 O 2, K a = 6.3 x 10 -5 ) is titrated with 0.050 M NaOH(8.21) b) 40.0 mL of 0.100 M NH 3 is titrated with 0.100 HCl(5.28)

60. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

61. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Strong acid/Weak acid Titrations pH begins at 7, gradual  as base is added pH  dramatically near equivalence point pH at equivalence point = 7.00 Shape of curve after equivalence point is due to [base] Initial pH  is steeper than for strong acid pH levels off (buffer effect) pH at equivalence point > 7.00 Shape of curve after equivalence point is due to [base]

62. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.

63. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.

64. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Solubility Products K sp = equilibrium constant for the equilibrium between an ionic solid solute and its saturated aqueous solution. Called the solubility-constant or the solubility product K sp = [cation] m [anion] n where m, n = stoichiometric coefficients

65. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2− (aq)

66. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2− ] where the equilibrium constant, K sp, is called the solubility product. (BaSO 4 falls out of K because it is a solid.)

67. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.9 (p. 738) Write the expression for the solubility- product constant for CaF 2, and look up the corresponding K sp value in Appendix D in your textbook.

68. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.9 Give the solubility-product constant expressions and the values of the solubility-product constants (from Appendix D) for the following compounds: a) barium carbonate b)silver sulfate

69. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Solubility Products K sp is not the same as solubility. Solubility = the maximum amount of solute that can be dissolved in a standard volume of solvent –often expressed as grams of solid that will dissolve per liter of solution (g/L). Molar solubility = the number of moles of solute that dissolve to form a liter of saturated solution.

70. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Solubility Products We can use solubility to find K sp and vice versa.

71. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Solubility  K sp 1.Solubility (g/vol)  molar solubility (M) 2.Molar solubility  molar [ions] at equilibrium. 3. Use equilibrium [ions] in the K sp expression.

72. Aqueous Equilibria © 2009, Prentice-Hall, Inc. K sp  Solubility 1.Write K sp expression. 2.Let x = molar solubility of the salt. 3.Use stoichiometry of the reaction to find the exponents in the K sp expression. 4. Substitute into K sp and solve for x. e.g. K sp Ag 2 SO 4 = 1.5 x 10 -5 = [Ag + ] 2 [SO 4 2- ] = [2x] 2 [x]

73. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.10 (p. 739) Solid silver chromate is added to pure water at 25 o C. Some of the solid remains undissolved at the bottom of the flask. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 x 10 -4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving the Ag + or CrO 4 2- ions in the solution, calculate K sp for this compound. (1.1 x 10 -12 )

74. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.10 A saturated solution of Mg(OH) 2 in contact with undissolved solid is prepared at 25 o C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH - ions in the solution, calculate K sp for this compound. (1.6 x 10 -12 )

75. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.11 (p. 740) The K sp for CaF 2 is 3.9 x 10 -11 at 25 o C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. (1.6 x 10 -2 g CaF 2 /L soln)

76. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.11 (cont’d)

77. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.11 The K sp for LaF 3 is 2 x 10 -19. What is the solubility of LaF 3 in water in moles per liter? (9 x 10 -6 mol/L)

78. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Factors That Affect Solubility 1. The presence of a common ion. 2.The pH of the solution. 3.The presence or absence of complexing agents.

79. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Factors Affecting Solubility 1. The Common-Ion Effect Solubility  when a common ion is added (Le Châtelier’s principle): e.g. CaF 2 : CaF 2(s)  Ca 2+ (aq) + 2F – (aq) Add more F – (i.e., + NaF), the equilibrium shifts to offset the increase.  CaF 2(s) is formed and precipitation occurs. As NaF is added to the system, the solubility of CaF 2 

80. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.12 (p. 742) Calculate the molar solubility of CaF 2 at 25 o C in a solution that is a) 0.010 M Ca(NO 3 ) 2 (3.1 x 10 -5 mol CaF 2 /L in 0.010 M Ca(NO 3 ) 2 ) b) 0.010 M in NaF (3.9 x 10 -7 mol CaF 2 /L in 0.010 M NaF)

81. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.12 (p. 742)

82. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.12 The value for K sp for manganese (II) hydroxide, Mn(OH) 2, is 1.6 x 10 -13. Calculate the molar solubility of Mn(OH) 2 in a solution that contains 0.020 M NaOH. (4.0 x 10 -10 M)

83. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Factors Affecting Solubility 2. pH –If a substance has a basic anion, it will be more soluble in an acidic solution. –Substances with acidic cations are more soluble in basic solutions.

84. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Factors Affecting Solubility Add more info from packet here??

85. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.13 (p. 745) Which of the following substances will be more soluble in acidic solution than in basic solution: a) Ni(OH) 2(s) b) CaCO 3(s) c) BaF 2(s) d) AgCl (s) (a-c)

86. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.13 Write the net ionic equation for the reaction of the following copper (II) compounds with acid: a) CuS b) Cu(N 3 ) 2

87. Aqueous Equilibria © 2009, Prentice-Hall, Inc. MOMA Demo Mg(OH) 2  Mg 2+ (aq) + 2OH - (aq) Solubility of Mg(OH) 2 = 9 x10 -4 g/100. mL What is its molar solubility? What is the pH of a saturated Mg(OH) 2 solution? What happens when HCl is added? Mg(OH) 2 + 2 HCl  MgCl 2(a q) + H 2 O, i.e. OH - ions are removed from the solution, thus allowing more Mg(OH) 2 to dissolve. As the Mg(OH) 2 dissolves, it immediately reacts with the HCl. After all the added HCl has been used up, the pH gradually increases as more Mg(OH) 2 is added, up to its maximum solubility.

88. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Complex Ions Add more from packet??

89. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Formation of Complex Ions e.g. AgCl has a very low solubility: K sp = 1.8 x 10 -10 Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) K f = the formation constant, for the complex ion Overall reaction: AgCl (s) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) + Cl – (aq) As the small amount of AgCl dissolves, the Ag + is taken up by the NH 3 allowing more Ag + to form.  By Le Châtelier’s principle, the forward reaction (the dissolving of AgCl) is favored.

90. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Factors Affecting Solubility 3. Complex Ions –Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

91. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Factors Affecting Solubility Complex Ions –The formation of these complex ions increases the solubility of these salts.

92. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Complex Ions Common ions that form complexes: NH 3 CN - OH - SCN - halogens Clue in Reactions Question = An excess of concentrated …. is added to …. Ch 5 & 13 The Ultimate Equations Handbook

93. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.14 (p. 748) Calculate the concentration of Ag + present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO 3 to give an equilibrium concentration of [NH 3 ] = 0.20 M. Neglect the small volume change that occurs when NH 3 is added. ([Ag + ] = 1.5 x 10 -8 M)

94. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.14 (cont’d)

95. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.14 Calculate [Cr 3+ ] in equilibrium with Cr(OH) 4 - when 0.010 mol of Cr(NO 3 ) 3 is dissolved in a liter of solution buffered at pH 10.0. ([Cr 3+ ] = 1 x 10 -16 M)

96. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Factors Affecting Solubility Amphoterism –Substances that are capable of acting either as an acid or a base are amphoteric. –Compare with amphiprotic, which relates more generally to any species that can either gain or lose a proton.

97. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Factors Affecting Solubility Amphoterism –Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. –Examples of such cations are Al 3+, Cr 3+, Zn 2+, and Sn 2+.

98. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Amphoterism The hydroxides generally form complex ions with several hydroxide ligands attached to the metal: Al(OH) 3(s) + OH – (aq)  Al(OH) 4 – (aq) Hydrated metal ions act as weak acids. As strong base is added, protons are removed: Al(H 2 O) 6 3+ (aq) + OH – (aq)  Al(H 2 O) 5 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 5 (OH) 2 + (aq) + OH – (aq)  Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) etc. Addition of an acid reverses these reactions

99. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Precipitation and Separation of Ions Consider the following: BaSO 4(s)  Ba 2+ (aq) + SO 4 2– (aq) At any instant in time, Q = [Ba 2+ ][ SO 4 2– ] - If Q < K sp, more solid can dissolve until Q = K sp. - If Q = K sp, the system is at equilibrium and the solution is saturated. - If Q > K sp, the salt will precipitate until Q = K sp.

100. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.15 (p.751) Will a precipitate form when 0.10 L of 8.0 x 10 -3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0 x 10 -3 M Na 2 SO 4 ? (yes)

101. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.15 Will a precipitate form when 0.050 L of 2.0 x 10 -2 M NaF is mixed with 0.010 L of 1.0 x 10 -2 M Ca(NO 3 ) 2 ? (yes)

102. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.

103. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Exercise 17.16 (p. 751) A solution contains 1.0 x 10 -2 M Ag + and 2.0 x 10 -2 M Pb 2+. When Cl - is added to the solution, both AgCl (K sp = 1.8 x 10 -10 ) and PbCl 2 (K sp = 1.7 x 10 -5 ) precipitate from the solution. What concentration of Cl - is necessary to begin the precipitation of each salt? Which salt precipitates first? (> 2.9 x 10 -2 M for PbCl 2 ; > 1.8 x 10 -8 M for AgCl, AgCl precipitates first)

104. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Practice Exercise 17.16 A solution consists of 0.050 M Mg 2+ and 0.020 M Cu 2+. Which ion will precipitate first as OH - is added to the solution? What concentration of OH - is necessary to begin the precipitation of each cation? (K sp = 1.8 x 10 -11 for Mg(OH) 2 and K sp = 4.8 x 10 -20 for Cu(OH) 2 ) (Cu(OH) 2 precipitates first, when [OH - ] > 1.5 x 10 -9 M; Mg(OH) 2 precipitates when [OH - ] > 1.9 x 10 -5 M)

105. Aqueous Equilibria © 2009, Prentice-Hall, Inc. Sample Integrative Exercise 17 (p. 755) A sample of 1.25 L of HCl gas at 21 o C and 0.950 atm is bubbled through 0.500 L of 0.150 M NH 3 solution. Calculate the pH of the resulting solution, assuming that all of the HCl dissolves and that the volume of the solution remains 0.500 L. (pH = 8.97)