1. Copyright © 2012, The McGraw-Hill Compaies, Inc. Permission required for reproduction or display. Chemistry Third Edition Julia Burdge Lecture PowerPoints Chapter 17 Acid-Base Equilibria and Solubility Equilibria

2. CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria 2 17.1The Common Ion Effect 17.2Buffer Solutions 17.3Acid-Base Titrations 17.4Solubility Equilibria 17.5Factors Affecting Solubility 17.6Separation of Ions Using Differences in Solubility

3. Topics 17.1The Common Ion Effect 3

4. 17.1The Common Ion Effect 4 Let’s examine how the properties of a solution change when a second solute is introduced. Consider a liter of solution containing 0.10 mole of acetic acid. [H 3 O + ] = [CH 3 COO ‒ ] = 1.34 × 10 ‒3 M and pH = 2.87

5. 17.1The Common Ion Effect 5 Now consider what happens when we add 0.050 mole of sodium acetate (CH 3 COONa) to the solution.

6. 17.1The Common Ion Effect 6 In general, when a compound containing an ion in common with a dissolved substance is added to a solution at equilibrium, the ionization equilibrium shifts to the left. This phenomenon is known as the common ion effect. The common ion can also be H 3 O + or OH –.

7. SAMPLE PROBLEM 17.1 7 Determine the pH at 25°C of a solution prepared by adding 0.050 mole of sodium acetate to 1.0 L of 0.10 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.)

8. SAMPLE PROBLEM 17.1 8 Setup We use the stated concentration of acetic acid, 0.10 M, and [H 3 O + ]  0 M as the initial concentrations in the ice table. Solution

9. SAMPLE PROBLEM 17.1 9 x = 3.6 × 10 ‒5 M. [H 3 O + ] = x, so pH = ‒log (3.6 × 10 ‒5 ) = 4.44 The common ion effect has shifted the ionization of acetic acid to the left.

10. Topics 17.2Buffer Solutions 10 Calculating the pH of a Buffer Preparing a Buffer Solution with as Specific pH

11. 17.2Buffer Solutions Calculating the pH of a Buffer 11 A solution that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) is a buffer solution or simply a buffer. A buffer’s ability to convert a strong acid to a weak acid minimizes the effect of the addition on the pH :

12. 17.2Buffer Solutions Calculating the pH of a Buffer 12 A buffer’s ability to convert a strong base to a weak base minimizes the effect of the addition on the pH :

13. 17.2Buffer Solutions Calculating the pH of a Buffer 13 Suppose that we have 1 L of the acetic acid-sodium acetate solution:

14. 17.2Buffer Solutions Calculating the pH of a Buffer 14

15. 17.1The Common Ion Effect Calculating the pH of a Buffer 15 Consider what happens when we add 0.10 mole of HCl to the buffer. (We assume that the addition of HCl causes no change in the volume of the solution.)

16. 17.1The Common Ion Effect Calculating the pH of a Buffer 16 A change of only 0.08 pH units.

17. 17.1The Common Ion Effect Calculating the pH of a Buffer 17 Had we added 0.10 mol of HCl to 1 L of pure water, the pH would have gone from 7.00 to 1.00!

18. 17.2Buffer Solutions Calculating the pH of a Buffer 18 In the determination of the pH of a buffer, we always neglect the small amount of weak acid that ionizes (x) because ionization is suppressed by the presence of a common ion. Similarly, we ignore the hydrolysis of the acetate ion because of the presence of acetic acid. This enables us to derive an expression for determining the pH of a buffer.

19. 17.2Buffer Solutions Calculating the pH of a Buffer 19

20. 17.2Buffer Solutions Calculating the pH of a Buffer 20 Henderson-Hasselbalch equation

21. SAMPLE PROBLEM 17.2 21 Starting with 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate, calculate the pH after the addition of 0.100 mole of NaOH. (Assume that the addition does not change the volume of the solution.) Strategy Added base will react with the acetic acid component of the buffer, converting OH ‒ to CH 3 COO ‒ : CH 3 COOH(aq) + OH ‒ (aq) → H 2 O(l) + CH 3 COO ‒ (aq)

22. SAMPLE PROBLEM 17.2 22 Setup Solution Thus, the pH of the buffer after addition of 0.10 mol of NaOH is 4.83.

23. 17.2Buffer Solutions Preparing a Buffer Solution with as Specific pH 23 A solution is only a buffer if it has the capacity to resist pH change when either an acid or a base is added. If the concentrations of a weak acid and conjugate base differ by more than a factor of 10, the solution does not have this capacity.

24. 17.2Buffer Solutions Preparing a Buffer Solution with as Specific pH 24 Therefore, we consider a solution a buffer, and can use to calculate its pH, only if the following condition is met: Consequently, the log term can only have values from –1 to 1, and the pH of a buffer cannot be more than one pH unit different from the pK a of the weak acid it contains.

25. 17.2Buffer Solutions Preparing a Buffer Solution with as Specific pH 25 This is known as the range of the buffer, where pH = pK a  1. This enables us to select the appropriate conjugate pair to prepare a buffer with a specific, desired pH.

26. 17.2Buffer Solutions Calculating the pH of a Buffer 26 To prepare a buffer with a specific, desired pH: 1.Choose a weak acid whose pK a is close to the desired pH.

27. 17.2Buffer Solutions Calculating the pH of a Buffer 27 2.Substitute the pH and pK a values into the following equation to obtain the necessary ratio of [conjugate base]/[weak acid].

28. SAMPLE PROBLEM 17.3 28 Select an appropriate weak acid from the table, and describe how you would prepare a buffer with a pH of 9.50.

29. SAMPLE PROBLEM 17.3 29 Setup Two of the acids listed in the table have pK a values in the desired range: hydrocyanic acid (HCN, pK a = 9.31) and phenol (C 6 H 5 OH, pK a = 9.89).

30. SAMPLE PROBLEM 17.3 30 Solution One way to achieve this would be to dissolve 0.41 mol of C 6 H 5 ONa and 1.00 mol of C 6 H 5 OH in 1 L of water.

31. Topics 17.3Acid‒Base Titrations 31 Strong Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Strong Acid-Weak base Titrations Acid-Base Indicators

32. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 32 The titrant is the solution that is added from the burette.

33. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 33 Titration curve (pH as a function of volume titrant added) of a strong acid-strong base titration.

34. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 34 Consider the addition of a 0.100 M NaOH solution (from a burette) to a vessel containing 25.0 mL of 0.100 M HCl.

35. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 35 Prior to the addition of base: Here, the pH of the acid is given by ‒log (0.100), or 1.00 Simply calculate the pH of a strong acid: ‒log [H 3 O + ]. Calculate the pH of the solution at every stage of titration.

36. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 36 Consider the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. Before the equivalence point: Note: Using millimoles instead of moles simplifies the calculations. Remember that millimoles = M × mL

37. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 37 Before the equivalence point: The total volume of the solution is 35.0 mL. The number of millimoles of NaOH in 10.0 mL is

38. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 38 The number of millimoles of HCl originally present is The amount of HCl left after partial neutralization is 2.50 ‒ 1.00, or 1.50 mmol.

39. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 39 Thus [H 3 O + ] = 0.0429 M, and the pH of the solution is pH = ‒log (0.04289) = 1.37 We have 1.50 mmol in 35.0 mL: Next, determine [H 3 O + ].

40. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 40 At the equivalence point: Consider the addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. This is a complete neutralization reaction. [H 3 O + ] = [OH ‒ ] = 1.00 × 10 ‒7 M and the pH of the solution is 7.000.

41. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 41 After the equivalence point: Consider the addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl. The total volume of the solution is now 60.0 mL. The number of millimoles of NaOH added is

42. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 42 There are 2.50 mmol of HCl in 25.0 mL of solution. 2.50 mmol of NaOH have been consumed, and there are 3.5 ‒ 2.5 or 1.00 mmol of NaOH remaining. The [NaOH] in 60.0 mL of solution is

43. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 43 [OH ‒ ] = 0.0167 M and pOH = ‒log (0.0167) = 1.78. The pH of the solution is 14.00 ‒ 1.78 or 12.22.

44. 17.3Acid‒Base Titrations Strong Acid-Strong Base Titrations 44

45. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 45 Titration curve of a weak acid-strong base titration: Strong-acid/strong-base curve, for comparison)

46. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 46 Consider the neutralization reaction between acetic acid (a weak acid) and sodium hydroxide (a strong base):

47. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 47 Prior to the addition of base: The pH is determined by the ionization of acetic acid. Use concentration (0.10 M) and K a (1.8 × 10 ‒5 ) to calculate the [H 3 O + ]:

48. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 48

49. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 49 Solving for x, gives [H 3 O + ] = 1.34 × 10 ‒3 M and pH = 2.87

50. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 50 Before the equivalence point: After the first addition of base, some of the acetic acid has been converted to acetate ion via the reaction

51. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 51 The solution is a buffer and we use the Henderson-Hasselbalch equation to calculate the pH. After the addition of 10.0 mL of base, the solution contains 1.5 mmol of acetic acid and 1.0 mmol of acetate ion

52. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 52 At the equivalence point: All the acetic acid has been neutralized and we are left with acetate ion in solution. pH is determined by the concentration and the K b of acetate ion.

53. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 53 The equivalence point occurs when 25.0 mL of base has been added, making the total volume 50.0 mL. The 2.5 mmol of acetic acid has all been converted to acetate ion.

54. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 54 The K b for acetate ion is 5.6 × 10 ‒10.

55. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 55 Solving for x, gives [OH ‒ ] = 5.3 × 10 ‒6 M, pOH = 5.28, and pH = 8.72

56. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 56 After the equivalence point: The curve for titration of a weak acid with a strong base is identical to the curve for titration of a strong acid with a strong base. Because all the acetic acid has been consumed, there is nothing in solution to consume the additional added OH ‒, and the pH levels off between 12 and 13.

57. 17.3Acid‒Base Titrations Weak Acid-Strong Base Titrations 57

58. SAMPLE PROBLEM 17.4 58 Calculate the pH in the titration of 50.0 mL of 0.120 M acetic acid by 0.240 M sodium hydroxide after the addition of (a) 10.0 mL of base, (b) 25.0 mL of base, and (c) 35.0 mL of base. Strategy The reaction is

59. SAMPLE PROBLEM 17.4 59 Setup a)The solution originally contains (0.120 mmol/mL)(50.0 mL) = 6.00 mmol of acetic acid.

60. SAMPLE PROBLEM 17.4 60 Solution a)Prior to the equivalence point, the solution is a buffer. We use the Henderson-Hasselbalch equation.

61. SAMPLE PROBLEM 17.4 61 b)After the addition of 25.0 mL of base, the titration is at the equivalence point. Calculate the pH using the concentration and the K b of acetate ion.

62. SAMPLE PROBLEM 17.4 62 The total volume 50.0 mL + 25.0 mL = 75.0 mL. The concentration of acetate ion is

63. SAMPLE PROBLEM 17.4 63 x = [OH ‒ ], so [OH ‒ ] = 6.7 × 10 ‒6 M pOH = –log (6.7 × 10 ‒6 ) = 5.17 and pH = 14.00 ‒ 5.17 = 8.83.

64. SAMPLE PROBLEM 17.4 64 A 35.0 mL amount of the base contains (0.240 mol/mL)(35.0 mL) = 8.40 mmol of OH ‒. c)After the addition of 35.0 mL of base, the titration is past the equivalence point and we solve for pH by determining the concentration of excess hydroxide ion. 8.40 ‒ 6.00 = 2.40 mmol of excess OH ‒

65. SAMPLE PROBLEM 17.4 65 The total volume is 50.0 + 35.0 = 85.0 mL. [OH ‒ ] = 2.40 mmol/85.0 mL = 0.0280 M pOH = ‒log (0.0280) = 1.553 pH = 14.000 ‒ 1.553 = 12.447

66. 17.3Acid‒Base Titrations Strong Acid-Weak base Titrations 66

67. 17.3Acid‒Base Titrations Strong Acid-Weak base Titrations 67 Consider the titration of HCl, a strong acid, with NH 3, a weak base: or The pH at the equivalence point is less than 7 because the ammonium ion acts as a weak Brønsted acid:

68. 17.3Acid‒Base Titrations Strong Acid-Weak base Titrations 68 Before the equivalence point: The pH is determined by the concentration and the K b of ammonia. Consider the titration of 25.0 mL of 0.10 M NH 3 with 0.10 M HCl.

69. 17.3Acid‒Base Titrations Strong Acid-Weak base Titrations 69

70. 17.3Acid‒Base Titrations Strong Acid-Weak base Titrations 70 The pH is calculated using the concentration and K a of the conjugate base of NH 3, the NH 4 + ion, and an equilibrium table. At the equivalence point:

71. 17.3Acid‒Base Titrations Strong Acid-Weak base Titrations 71 The curve for titration of a weak base with a strong acid is identical to the curve for titration of a strong base with a strong acid. Because all the ammonia has been consumed, there is nothing in solution to consume the additional added H 3 O +, and the pH levels off between 1 and 2. After the equivalence point:

72. SAMPLE PROBLEM 17.5 72 Calculate the pH at the equivalence point when 25.0 mL of 0.100 M NH 3 is titrated with 0.100 M HCl. Strategy The reaction between NH 3 and HCl is Setup The solution originally contains (0.100 mmol/mL)(25.0 mL) = 2.50 mmol NH 4 +. At the equivalence point, 2.50 mmol of HCl has been added.

73. SAMPLE PROBLEM 17.5 73 The volume of 0.100 M HCl that contains 2.50 mmol is It takes 25.0 mL of titrant to reach the equivalence point, so the total solution volume is 25.0 + 25.0 = 50.0 mL. At the equivalence point, all the NH 3 originally present has been converted to NH 4 +.

74. SAMPLE PROBLEM 17.5 74 Solution The [NH 4 + ] is (2.50 mmol)/(50.0 mL) = 0.0500 M.

75. SAMPLE PROBLEM 17.5 75 [H 3 O + ] = x = 5.3 × 10 ‒6 M At equilibrium, pH = ‒log (5.3 × 10 ‒6 ) = 5.28 The equilibrium expression is

76. 17.3Acid‒Base Titrations Acid-Base Indicators 76 The equivalence point in a titration can be determined using an acid-base indicator. An acid-base indicator is usually a weak organic acid or base for which the ionized and un-ionized forms are different colors.

77. 17.3Acid‒Base Titrations Acid-Base Indicators 77 HIn and its conjugate base, In ‒, must have distinctly different colors. Acidic solutions are the color of HIn. Basic solutions are the color of In ‒. Acidic Basic

78. 17.3Acid‒Base Titrations Acid-Base Indicators 78 The endpoint of a titration is the point at which the color of the indicator changes. There is a range of pH over which the color change occurs. We select an indicator whose color change occurs over a pH range that coincides with the steepest part of the titration curve.

79. 17.3Acid‒Base Titrations Acid-Base Indicators 79 The titration curves for hydrochloric acid and acetic acid— each being titrated with sodium hydroxide

80. 17.3Acid‒Base Titrations Acid-Base Indicators 80 Phenolphthalein or methyl red are suitable indicators for the titration of a strong acid with a strong base because both endpoints coincide with the steepest part of the HCl-NaOH titration curve. Methyl red is not a suitable indicator for use in the titration of acetic acid with sodium hydroxide. Its endpoint occurs significantly before the equivalence point.

81. 17.3Acid‒Base Titrations Acid-Base Indicators 81

82. SAMPLE PROBLEM 17.6 82 Which indicator (or indicators) would you use for the following acid-base titrations (a) strong acid-strong base(b) weak acid-strong base, and (c) strong acid-weak base? Setup The steep part of the curve spans a)a pH range of about 4 to 10. b)a pH range of about 7 to 10. c)a pH range of about 7 to 3.

83. SAMPLE PROBLEM 17.6 83

84. SAMPLE PROBLEM 17.6 84 Solution a)Most of the indicators, with the exceptions of thymol blue, bromophenol blue, and methyl orange, would work. b)Cresol red and phenolphthalein are suitable indicators. c)Bromophenol blue, methyl orange, methyl red, and chlorophenol blue are all suitable indicators.

85. Topics 17.4Solubility Equilibria 85 Solubility Product Expression and K sp Calculations Involving K sp and Solubility Predicting Precipitation Reactions

86. 17.4Solubility Equilibria Solubility Product Expression and K sp 86 The solubility of ionic compounds is important in industry, medicine, and everyday life. The compounds described as “insoluble” in Chapter 4 are actually very slightly soluble. The K sp is called the solubility product constant. It allows for quantitative predictions about how much of a given ionic compound will dissolve in water.

87. 17.4Solubility Equilibria Solubility Product Expression and K sp 87 K sp is equal to the concentrations of products over the concentrations of reactants, each raised to its coefficient from the balanced chemical equation. The smaller the K sp, the less soluble the compound.

88. 17.4Solubility Equilibria Calculations Involving K sp and Solubility 88 There are two ways to express the solubility of a substance: 1.The molar solubility: number of moles of solute in 1 L of a saturated solution (mol/L). 2.The solubility: number of grams of solute in 1 L of a saturated solution (g/L). Both of these refer to concentrations at a particular temperature (usually 25°C).

89. 17.4Solubility Equilibria Calculations Involving K sp and Solubility 89 To calculate molar solubility from K sp : The procedure is essentially identical to the procedure for solving weak acid or weak base equilibrium problems 1.Construct an equilibrium table. 2.Fill in what we know. 3.Figure out what we don’t know.

90. 17.4Solubility Equilibria Calculations Involving K sp and Solubility 90 The K sp of silver bromide (AgBr) is 7.7 × 10 ‒13. Let s be the molar solubility (in mol/L) of AgBr. At equilibrium: [Ag + ] = [Br ‒ ] = s

91. 17.4Solubility Equilibria Calculations Involving K sp and Solubility 91 The equilibrium expression is Therefore,

92. 17.4Solubility Equilibria Calculations Involving K sp and Solubility 92 The molar solubility of AgBr is 8.8 × 10 ‒7 M. Express this solubility in g/L by multiplying the molar solubility by the molar mass of AgBr:

93. SAMPLE PROBLEM 17.7 93 Calculate the solubility of copper(II) hydroxide [Cu(OH) 2 ] in g/L. Setup The equation for the dissociation of Cu(OH) 2 is K sp for Cu(OH) 2 = 2.2 × 10 ‒20. The molar mass of Cu(OH) 2 is 97.57 g/mol. K sp = [Cu 2+ ][OH ‒ ] 2

94. SAMPLE PROBLEM 17.7 94 Solution Therefore, Remember to raise an entire term to the appropriate power!

95. SAMPLE PROBLEM 17.7 95 The molar solubility of Cu(OH) 2 is 1.8 × 10 ‒7 M. Multiplying by its molar mass gives:

96. SAMPLE PROBLEM 17.8 96 The solubility of calcium sulfate (CaSO 4 ) is measured experimentally and found to be 0.67 g/L. Calculate the value of K sp for calcium sulfate. Setup The molar mass of CaSO 4 is 136.2 g/mol. The molar solubility of CaSO 4 is

97. SAMPLE PROBLEM 17.8 97 Solution This is a relatively large K sp value.

98. 17.4Solubility Equilibria Predicting Precipitation Reactions 98 We use the reaction quotient (Q) to predict when a precipitate will form. Q has the same form as K sp except the concentrations of ions are not equilibrium concentrations.

99. 17.4Solubility Equilibria Predicting Precipitation Reactions 99 If we mix a solution containing Ag + ions with one containing Cl ‒ ions, we write: where “i” denotes initial concentrations If Q ≤ K sp no precipitate will form If Q > K sp AgCl will precipitate

100. SAMPLE PROBLEM 17.9 100 Predict whether a precipitate will form when each of the following is added to 650 mL of 0.0080 M K 2 SO 4 : (a)250 mL of 0.0040 M BaCl 2 (b)175 mL of 0.15 M AgNO 3 (c)325 mL of 0.25 M Sr(NO 3 ) 2 (Assume volumes are additive.) Setup The compounds that might precipitate and their K sp values are a)BaSO 4, K sp = 1.1 × 10 ‒10 b)Ag 2 SO 4, K sp = 1.5 × 10 ‒5 c)SrSO 4, K sp = 3.8 × 10 ‒7

101. SAMPLE PROBLEM 17.9 101 Solution (a) Concentrations of the constituent ions of BaSO 4 are: Q = [Ba 2+ ][SO 4 2‒ ] = (0.0011)(0.0058) = 6.4 × 10 ‒6 Q > K sp (1.1 × 10 ‒10 ) BaSO 4 will precipitate.

102. SAMPLE PROBLEM 17.9 102 (b) Concentrations of the constituent ions of Ag 2 SO 4 are: Q = [Ag + ] 2 [SO 4 2‒ ] = (0.032) 2 (0.0063) = 6.5 × 10 ‒6 Q < K sp (1.5 × 10 ‒5 ) Ag 2 SO 4 will not precipitate.

103. SAMPLE PROBLEM 17.9 103 (c) Concentrations of the constituent ions of SrSO 4 are: Q = [Sr 2+ ][SO 4 2‒ ] = (0.083)(0.0053) = 4.4 × 10 ‒4 Q > K sp (3.8 × 10 ‒7 ) SrSO 4 will precipitate.

104. Topics 17.5Factors Affecting Solubility 104 The Common Ion Effect pH Complex Ion Formation

105. 17.5Factors Affecting Solubility The Common Ion Effect 105 In a saturated solution of AgCl, K sp = [Ag + ][Cl ‒ ] s = 1.3 × 10 ‒5 M In a solution in which AgCl is the only solute, [Ag + ] = [Cl ‒ ].

106. 17.5Factors Affecting Solubility The Common Ion Effect 106 Determine the solubility of AgCl in a solution already containing a solute that has an ion in common with AgCl. For example, consider dissolving AgCl in a 0.10 M solution of AgNO 3. [Ag + ] will be equal to 0.10 M plus the concentration contributed by AgCl. In this case, [Ag + ] ≠ [Cl ‒ ]

107. 17.5Factors Affecting Solubility The Common Ion Effect 107 The equilibrium expression is Because we expect s to be very small, s = 1.6 × 10 ‒9 M AgCl is significantly less soluble in 0.10 M AgNO 3 than in pure water—due to the common ion effect.

108. SAMPLE PROBLEM 17.10 108 Calculate the molar solubility of silver chloride in a solution that is 6.5 × 10 ‒3 M in silver nitrate. Setup The dissolution equilibrium and the equilibrium expression are Solution

109. SAMPLE PROBLEM 17.10 109 Because we expect s to be very small, The presence of AgNO 3 reduces the solubility of AgCl by a factor of ~ 500!

110. 17.5Factors Affecting Solubility pH 110 The solubility of a substance can also depend on the pH of the solution. Insoluble bases tend to dissolve in acidic solutions while insoluble acids tend to dissolve in basic solutions.

111. 17.5Factors Affecting Solubility pH 111 At equilibrium In a solution with a pH of less than 10.45, the solubility of Mg(OH) 2 would increase.

112. 17.5Factors Affecting Solubility pH 112 The effect of additional H 3 O + ions are summarized as follows: If pH > 10.45, [OH ‒ ] would be higher and the solubility of Mg(OH) 2 would decrease because of the common ion (OH ‒ ) effect.

113. 17.5Factors Affecting Solubility pH 113 The pH also influences the solubility of salts that contain a basic anion. For example, In an acidic medium, the high [H 3 O + ] will shift the following equilibrium to the left, consuming F ‒ :

114. 17.5Factors Affecting Solubility pH 114 As [F ‒ ] decreases, [Ba 2+ ]must increase to satisfy the equality K sp = [Ba 2+ ][F ‒ ] 2. Thus, more BaF 2 dissolves. The solubilities of salts containing anions that do not hydrolyze, such as Cl ‒, Br ‒, and NO 3 ‒, are unaffected by pH.

115. SAMPLE PROBLEM 17.11 115 Which of the following compounds will be more soluble in acidic solution than in water: (a) CuS, (b) AgCl, (c) PbSO 4 ? Setup a.S 2‒ is the conjugate base of the weak acid HS ‒. S 2‒ reacts with H 3 O + b.Cl ‒ is the conjugate base of the strong acid HCl. Cl ‒ does not react with H 3 O +.

116. SAMPLE PROBLEM 17.11 116 c.SO 4 2‒ is the conjugate base of the weak acid HSO 4 ‒. It reacts with H 3 O + as follows: A salt that produces an anion that reacts with H 3 O + will be more soluble in acid than in water. Solution CuS and PbSO 4 are more soluble in acid than in water. (AgCl is no more or less soluble in acid than in water.)

117. 17.5Factors Affecting Solubility Complex Ion Formation 117 A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Transition metals have a particular tendency to form complex ions. For example, a solution of cobalt(II) chloride (CoCl 2 ) is pink because of the presence of the Co(H 2 O) 6 2+ ions. When HCl is added, the solution turns blue because the complex ion CoCl 4 2‒ forms:

118. 17.5Factors Affecting Solubility Complex Ion Formation 118

119. 17.5Factors Affecting Solubility Complex Ion Formation 119 CuSO 4 (Center). After the addition of a few drops of concentrated NH 3, a light-blue precipitate of Cu(OH) 2 is formed. (Right) When more NH 3 is added, the Cu(OH) 2 precipitate dissolves to form the dark-blue complex ion Cu(NH 3 ) 4 2+.

120. 17.5Factors Affecting Solubility Complex Ion Formation 120 A measure of the tendency of a metal ion to form a particular complex ion is given by the formation constant (K f ) (also called the stability constant), which is the equilibrium constant for the complex ion formation. The larger K f is, the more stable the complex ion.

121. 17.5Factors Affecting Solubility Complex Ion Formation 121 The formation of the Cu(NH 3 ) 4 2+ ion can be expressed as The large value of K f indicates that the complex ion is very stable in solution.

122. 17.5Factors Affecting Solubility Complex Ion Formation 122

123. 17.5Factors Affecting Solubility Complex Ion Formation 123 The dissolution of silver chloride is represented by the equation The sum of this equation and the one representing the formation of Ag(NH 3 ) 2+ is

124. 17.5Factors Affecting Solubility Complex Ion Formation 124 The corresponding equilibrium constant is (1.6 × 10 ‒10 )(1.5 × 10 7 ) = 2.4 × 10 ‒3. This is significantly larger than the K sp value, indicating that much more AgCl will dissolve in the presence of aqueous ammonia than in pure water. In general, the effect of complex ion formation generally is to increase the solubility of a substance.

125. 17.5Factors Affecting Solubility Complex Ion Formation 125 Amphoteric hydroxides, can react with both acids and bases. Examples are Al(OH) 3, Pb(OH) 2, Cr(OH) 3, Zn(OH) 2, and Cd(OH) 2. For example,

126. 17.5Factors Affecting Solubility Complex Ion Formation 126 Solving an equilibrium problem involving complex ion formation is complicated by the magnitude of K f and stoichiometry. Consider the following reaction We cannot neglect x, and have to raise [NH 3 ]to the fourth power. This equation is not easily solved. Another approach is needed.

127. 17.5Factors Affecting Solubility Complex Ion Formation 127 1.Assume that all the copper(II) ion is consumed to form the complex ion (since K f is so large.) 2.Consider the equilibrium in terms of the reverse reaction; the dissociation of Cu(NH 3 ) 4 2+, for which the equilibrium constant is the reciprocal of K f.

128. 17.5Factors Affecting Solubility Complex Ion Formation 128 Because this K is so small, we can expect x to be insignificant. Note that [NH 3 ], which had been 3.0 M, has been diminished by 4 × 0.10 M due to the amount required to complex 0.10 mole of copper(II) ion.

129. 17.5Factors Affecting Solubility Complex Ion Formation 129 Neglecting x, x = 4.4 × 10 ‒18 M Because K f is so large, the amount of copper that remains uncomplexed is extremely small.

130. SAMPLE PROBLEM 17.12 130 In the presence of aqueous cyanide, cadmium(II) forms the complex ion Cd(CN) 4 2‒. Determine the molar concentration of free (uncomplexed) cadmium(II) ion in solution when 0.20 mole of Cd(NO 3 ) 2 is dissolved in a liter of 2.0 M sodium cyanide (NaCN). Setup K f for the complex ion Cd(CN) 4 2‒ is 7.1 × 10 16. The reverse process has an equilibrium constant of 1/K f = 1.4 × 10 ‒17

131. SAMPLE PROBLEM 17.12 131 The equilibrium expression for the dissociation is Stoichiometry indicates that four CN ‒ ions are required to react with one Cd 2+ ion. Therefore, [CN ‒ ] will be [2.0 M ‒ 4(0.20 M)] = 1.2 M. Solution

132. SAMPLE PROBLEM 17.12 132 Neglecting x, x = 1.4 × 10 ‒18 M Don’t forget to adjust the concentration of the complexing agent before entering it in the equilibrium table.

133. Topics 17.6Separation of Ions Using Differences in Solubility 133 Fractional Precipitation Qualitative Analysis of Metal Ions in Solution

134. 17.6Separation of Ions Using Differences in Solubility Fractional Precipitation 134 One way to separate ions is to convert them to insoluble salts. When a soluble compound such as silver nitrate is slowly added to a solution that contains Cl ‒, Br ‒, and I ‒ ions, AgI precipitates first, followed by AgBr, and then AgCl. This practice is known as fractional precipitation. The solubility of the silver halides decreases from AgCl to AgI.

135. SAMPLE PROBLEM 17.13 135 Silver nitrate is added slowly to a solution that is 0.020 M in Cl ‒ ions and 0.020 M in Br ‒ ions. Calculate the concentration of Ag + ions (in mol/L) required to initiate the precipitation of AgBr without precipitating AgCl. Strategy Use the equilibrium expressions to calculate the maximum [Ag + ] that can exist without exceeding K sp for each compound.

136. SAMPLE PROBLEM 17.13 136 Setup AgBr should precipitate first. Use the equilibrium expression for AgBr to determine the minimum [Ag + ] necessary to initiate precipitation of AgBr. Solve for [Ag + ] again, using the equilibrium expression for AgCl to determine the maximum [Ag + ] that can exist in the solution without initiating the precipitation of AgCl.

137. SAMPLE PROBLEM 17.13 137 Solution Solving the AgBr equilibrium expression for [Ag + ], For AgBr to precipitate from solution, [Ag + ] must exceed 3.9 × 10 ‒11 M. Solving the AgCl equilibrium expression for the [Ag + ],

138. SAMPLE PROBLEM 17.13 138 For AgCl not to precipitate from solution, [Ag + ] must stay below 8.0 × 10 ‒9 M. To precipitate Br ‒ ions without precipitating Cl ‒ from this solution, 3.9 × 10 ‒11 M < [Ag + ] < 8.0 × 10 ‒9 M

139. 17.6Separation of Ions Using Differences in Solubility Qualitative Analysis of Metal Ions in Solution 139 The principle of selective precipitation can be used to identify the types of ions present in a solution. This practice is called qualitative analysis. There are about 20 common cations that can be analyzed readily in aqueous solution. Analysis of these ions must be carried out systematically from group 1 through group 5.

140. 17.6Separation of Ions Using Differences in Solubility Qualitative Analysis of Metal Ions in Solution 140 Group 1 cations. When dilute HCl is added to the unknown solution, only the Ag +, Hg 2 2+, and Pb 2 + ions precipitate as insoluble chlorides. Group 2 cations. The chloride precipitates are removed by filtration. Hydrogen sulfide is added to the unknown acidic solution to produce metal sulfides:

141. 17.6Separation of Ions Using Differences in Solubility Qualitative Analysis of Metal Ions in Solution 141 Only the metal sulfides with the smallest K sp values precipitate under acidic conditions. These are Bi 2 S 3, CdS, CuS, and SnS. The solution is filtered to remove the insoluble sulfides. Group 3 cations. NaOH is added to the solution. In a basic solution, the more soluble sulfides (CoS, FeS, MnS, NiS, ZnS) precipitate. The Al 3+ and Cr 3+ ions precipitate as hydroxides. The solution is filtered again to remove the insoluble sulfides and hydroxides.

142. 17.6Separation of Ions Using Differences in Solubility Qualitative Analysis of Metal Ions in Solution 142 Group 4 cations. Sodium carbonate is added to the basic solution to precipitate Ba 2+, Ca 2+, and Sr 2+ ions as BaCO 3, CaCO 3, and SrCO 3. These precipitates are filtered. Group 5 cations. The only cations possibly remaining in solution are Na +, K +, and NH 4 +. NH 4 + ions can be determined by adding sodium hydroxide:

143. 17.6Separation of Ions Using Differences in Solubility Qualitative Analysis of Metal Ions in Solution 143 To confirm the presence of Na + and K + ions, a flame test is used in which a piece of platinum wire is dipped into the solution and held over a flame. Na + ions emit a yellow flame, whereas K + ions emit a violet flame. Ammonia gas is detected by either its odor or a litmus test.

144. 17.6Separation of Ions Using Differences in Solubility Qualitative Analysis of Metal Ions in Solution 144 A summary of the scheme for separating metal ions: