1. 1 Chapter Seventeen Additional Aspects of Aqueous Equilibria

2. 2 The Common Ion Effect and Buffer Solutions Common ion effect - solutions in which the same ion is produced by two different compounds Buffer solutions - resist changes in pH when acids or bases are added to them –due to common ion effect Two common kinds of buffer solutions 1solutions of a weak acid plus a soluble ionic salt of the weak acid 2solutions of a weak base plus a soluble ionic salt of the weak base

3. 3 The Common Ion Effect and Buffer Solutions Weak Acids plus Salts of Weak Acids –acetic acid CH 3 COOH –sodium acetate NaCH 3 COO

4. 4 The Common Ion Effect and Buffer Solutions Example: Calculate the concentration of H + and the pH of a solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate.

5. 5 The Common Ion Effect and Buffer Solutions Example: Calculate the concentration of H + and the pH of a solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate.

6. 6 The Common Ion Effect and Buffer Solutions Substitute these quantities into the ionization expression.

7. 7 The Common Ion Effect and Buffer Solutions Apply the simplifying assumption

8. 8 The Common Ion Effect and Buffer Solutions Compare the acidity of a pure acetic acid solution and the buffer we just described.

9. 9 The Common Ion Effect and Buffer Solutions Compare the acidity of a pure acetic acid solution and the buffer we just described. [H + ] is 89 times greater in pure acetic acid than in buffer solution.

10. 10 The Common Ion Effect and Buffer Solutions General expression for the ionization of a weak monoprotic acid.

11. 11 The Common Ion Effect and Buffer Solutions Its ionization constant expression is

12. 12 The Common Ion Effect and Buffer Solutions Solve the expression for [H + ]

13. 13 The Common Ion Effect and Buffer Solutions Making the assumption that the concentrations of the weak acid and the salt are reasonable. The expression reduces to

14. 14 The Common Ion Effect and Buffer Solutions The above relationship is valid for buffers containing a weak monoprotic acid and a soluble, ionic salt. The relationship changes if the salt’s cation is not univalent to

15. 15 The Common Ion Effect and Buffer Solutions Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation

16. 16 Weak Bases plus Salts of Weak Bases Buffers that contain a weak base plus the salt of a weak base - for example - ammonia plus ammonium nitrate.

17. 17 Weak Bases plus Salts of Weak Bases Buffers that contain a weak base plus the salt of a weak base - for example - ammonia plus ammonium nitrate.

18. 18 Weak Bases plus Salts of Weak Bases Example: Calculate the concentration of OH- and the pH of the solution that is 0.15 M in aqueous ammonia, NH 3, and 0.30 M in ammonium nitrate, NH 4 NO 3.

19. 19 Weak Bases plus Salts of Weak Bases Substitute these values into the ionization expression for ammonia and solve algebraically.

20. 20 Weak Bases plus Salts of Weak Bases Let’s compare the aqueous ammonia concentration to that of the buffer described above.

21. 21 Weak Bases plus Salts of Weak Bases Let’s compare the aqueous ammonia concentration to that of the buffer described above. The [OH - ] in aqueous ammonia is 180 times greater than in the buffer.

22. 22 Weak Bases plus Salts of Weak Bases Derive a general relationship for buffer solutions that contain a weak base plus a salt of a weak base. –Ionization equation

23. 23 Weak Bases plus Salts of Weak Bases Ionization expression –general form

24. 24 Weak Bases plus Salts of Weak Bases For salts that have univalent ions: For salts that have divalent or trivalent ions

25. 25 Weak Bases plus Salts of Weak Bases Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation

26. 26 Buffering Action Buffer solutions resist changes in pH. Example: If 0.020 mole of HCl is added to 1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the gaseous HCl. 1 Calculate the pH of the original buffer solution

27. 27 Buffering Action

28. 28 Buffering Action 2 Now we calculate the concentration of all species after the addition of HCl. –HCl will react with some of the ammonia

29. 29 Buffering Action 3 Now that we have the concentrations of our salt and base, we can calculate the pH.

30. 30 Buffering Action 4 Calculate the change in pH.

31. 31 Preparation of Buffer Solutions Example: Calculate the concentration of H + and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions. Determine the amounts of acetic acid and sodium hydroxide (before reaction)

32. 32 Preparation of Buffer Solutions Example: Calculate the concentration of H + and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions. Determine the amounts of acetic acid and sodium hydroxide (before reaction)

33. 33 Preparation of Buffer Solutions Sodium hydroxide and acetic acid react in a 1:1 mole ratio.

34. 34 Preparation of Buffer Solutions After the two solutions are mixed, the total volume is 300 mL (100 + 200), and the concentrations are:

35. 35 Preparation of Buffer Solutions Substitution into the ionization constant expression (or Henderson-Hasselbach equation) gives

36. 36 Preparation of Buffer Solutions For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH. Example: Calculate the number of moles of solid ammonium chloride, NH 4 Cl, that must be used to prepare 1.00 L of a buffer solution that is 0.10 M in aqueous ammonia, and that has a pH of 9.15. –Because pH=9.15

37. 37 Preparation of Buffer Solutions For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH. Example:Calculate the number of moles of solid ammonium chloride, NH 4 Cl, that must be used to prepare 1.00 L of a buffer solution that is 0.10 M in aqueous ammonia, and that has a pH of 9.15. –Because pH=9.15

38. 38 Preparation of Buffer Solutions Appropriate equations and equilibria representations are:

39. 39 Preparation of Buffer Solutions Substitute into the ionization constant expression (or Henderson-Hasselbach equation) for aqueous ammonia

40. 40 Acid-Base Indicators Equivalence point - point at which chemically equivalent amounts of acid and base have reacted End point - point at which chemical indicator changes color

41. 41 Acid-Base Indicators Equilibrium constant expression for an indicator would be

42. 42 Acid-Base Indicators Rearrange this expression to get a feeling for range over which indicator changes color.

43. 43 Acid-Base Indicators Some Acid-Base Indicators

44. 44 Strong Acid/Strong Base Titration Curves Plot pH vs. Volume of acid or base added in titration. Consider the titration of 100.0 mL of 0.100 M perchloric acid with 0.100 M potassium hydroxide. –Plot pH vs. mL of KOH added –1:1 mole ratio

45. 45 Strong Acid/Strong Base Titration Curves Before titration starts the pH of the HClO 4 solution is 1.00. –Remember perchloric acid is a strong acid

46. 46 Strong Acid/Strong Base Titration Curves After 20.0 mL of 0.100 M KOH has been added the pH is 1.17.

47. 47 Strong Acid/Strong Base Titration Curves After 50.0 mL of 0.100 M KOH has been added the pH is 1.48.

48. 48 Strong Acid/Strong Base Titration Curves After 90.0 mL of 0.100 M KOH has been added the pH is 2.28.

49. 49 Strong Acid/Strong Base Titration Curves After 100.0 mL of 0.100 M KOH has been added the pH is 7.00.

50. 50 Strong Acid/Strong Base Titration Curves We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

51. 51 Weak Acid/Strong Base Titration Curves Consider the titration of 100.0 mL of 0.100 M acetic acid, CH 3 COOH, with 0.100 M KOH. –react in a 1:1 mole ratio

52. 52 Weak Acid/Strong Base Titration Curves Before the equivalence point is reached, both CH 3 COOH and K CH 3 COO are present in solution forming a buffer.

53. 53 Weak Acid/Strong Base Titration Curves 1 Determine the pH of the acetic acid solution before titration is begun.

54. 54 Weak Acid/Strong Base Titration Curves 1 Determine the pH of the acetic acid solution before titration is begun.

55. 55 Weak Acid/Strong Base Titration Curves 1 Determine the pH of the acetic acid solution before titration is begun.

56. 56 Weak Acid/Strong Base Titration Curves 2 After 20.0 mL of KOH solution has been added, the pH is:

57. 57 Weak Acid/Strong Base Titration Curves 2 After 20.0 mL of KOH solution has been added, the pH is:

58. 58 Weak Acid/Strong Base Titration Curves 2 After 20.0 mL of KOH solution has been added, the pH is:

59. 59 Weak Acid/Strong Base Titration Curves 2 After 20.0 mL of KOH solution has been added, the pH is: Similarly for all other cases before the equivalence point is reached.

60. 60 Weak Acid/Strong Base Titration Curves At the equivalence point, the solution is 0.500 M in KCH 3 COO, the salt of a strong base and a weak acid which hydrolyzes to give a basic solution. –Both processes make the solution basic Cannot have a pH=7.00 at equivalence point. Let us calculate the pH at the equivalence point.

61. 61 Weak Acid/Strong Base Titration Curves 1 Set up the equilibrium reaction:

62. 62 Weak Acid/Strong Base Titration Curves 2 Determine the concentration of the salt in solution.

63. 63 Weak Acid/Strong Base Titration Curves 3 Perform a hydrolysis calculation for the potassium acetate in solution.

64. 64 Weak Acid/Strong Base Titration Curves After the equivalence point is reached, the pH is determined by the excess KOH - as in Strong Acid/Strong Base example.

65. 65 Weak Acid/Strong Base Titration Curves After the equivalence point is reached, the pH is determined by the excess KOH - as in Strong Acid/Strong Base example.

66. 66 Weak Acid/Strong Base Titration Curves We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

67. 67 Strong Acid/Weak Base Titration Curves Titration curves for Strong Acid/Weak Bases look similar to Strong Base/Weak Acid but they are inverted.

68. 68 Weak Acid/Weak Base Titration Curves Titration curves have very short vertical sections. Solution is buffered both before and after the equivalence point. Visual indicators cannot be used.

69. 69 Solubility Product Constants Silver chloride, AgCl,is rather insoluble in water. Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.

70. 70 Solubility Product Constants The equilibrium constant expression for this dissolution is called a solubility product constant. K sp =solubility product constant Molar concentration of ions raised to their stoichiometric powers at equilibrium

71. 71 Solubility Product Constants Solubility product constant for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound. Consider the dissolution of silver sulfide in water.

72. 72 Solubility Product Constants Its solubility product expression is

73. 73 Solubility Product Constants The dissolution of solid calcium phosphate in water is represented as

74. 74 Solubility Product Constants Its solubility product constant expression is

75. 75 Solubility Product Constants In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as K sp has a fixed value for a given system at a given temperature

76. 76 Solubility Product Constants The same rules apply for compounds that have more than two kinds of ions. An example is calcium ammonium phosphate.

77. 77 Determination of Solubility Product Constants Example: One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25 o C. Calculate the molar solubility of, and K sp for, AgCl. Molar solubility can be calculated from the data:

78. 78 Determination of Solubility Product Constants The equation for the dissociation of silver chloride and its solubility product expression are

79. 79 Determination of Solubility Product Constants Substitution into the solubility product expression gives

80. 80 Uses of Solubility Product Constants We can use the solubility product constant to calculate the solubility of a compound at 25 o C. Example: Calculate the molar solubility of barium sulfate, BaSO 4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25 o C. K sp = 1.1 x 10 -10.

81. 81 Uses of Solubility Product Constants Example: Calculate the molar solubility of barium sulfate, BaSO 4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25 o C. K sp = 1.1 x 10 -10.

82. 82 Uses of Solubility Product Constants Substitute into solubility product expression and solve for x, giving the ion concentrations.

83. 83 Uses of Solubility Product Constants Now we can calculate the mass of BaSO 4 in 1.00 L of saturated solution.

84. 84 The Reaction Quotient in Precipitation Reactions Use solubility product constants to calculate the concentration of ions in a solution and whether or not a precipitate will form. Example: We mix 100 mL of 0.010 M potassium sulfate, K 2 SO 4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO 3 ) 2 solutions. Will a precipitate form?

85. 85 The Reaction Quotient in Precipitation Reactions Example: We mix 100 mL of 0.010 M potassium sulfate, K 2 SO 4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO 3 ) 2 solutions. Will a precipitate form?

86. 86 The Reaction Quotient in Precipitation Reactions Calculate the Q sp for PbSO 4. –Solution volumes are additive. –Concentrations of the important ions are:

87. 87 The Reaction Quotient in Precipitation Reactions Finally, we calculate Q sp for PbSO 4.

88. 88 Fractional Precipitation Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution. –Look at a solution that contains Cu +, Ag +, and Au + –We could precipitate them as chlorides

89. 89 Fractional Precipitation Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution. –Look at a solution that contains Cu +, Ag +, and Au + –We could precipitate them as chlorides

90. 90 Fractional Precipitation Example: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu +, Ag +, and Au + ions, which compound precipitates first? Calculate the concentration of Cl - required to initiate precipitation of each of these metal I chlorides.

91. 91 Fractional Precipitation Example: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu +, Ag +, and Au + ions, which compound precipitates first? Calculate the concentration of Cl - required to initiate precipitation of each of these metal I chlorides.

92. 92 Fractional Precipitation Example: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu +, Ag +, and Au + ions, which compound precipitates first? Calculate the concentration of Cl - required to initiate precipitation of each of these metal I chlorides.

93. 93 Fractional Precipitation Repeat the calculation for silver chloride.

94. 94 Fractional Precipitation For copper (I) chloride to precipitate.

95. 95 Fractional Precipitation We have calculated the [Cl - ] required to precipitate AuCl, [Cl - ] >2.0 x 10 -11 M to precipitate AgCl, [Cl - ] >1.8 x 10 -8 M to precipitate CuCl, [Cl - ] >1.9 x 10 -5 M We can calculate the amount of Au + precipitated before Ag + begins to precipitate, as well as the amounts of Au + and Ag + precipitated before Cu + begins to precipitate.

96. 96 Fractional Precipitation Example: Calculate the percent of Au + ions that precipitate before AgCl begins to precipitate. Use the [Cl - ] from before to determine the [Au + ] remaining in solution just before AgCl begins to precipitate.

97. 97 Fractional Precipitation Example: Calculate the percent of Au + ions that precipitate before AgCl begins to precipitate. Use the [Cl - ] from before to determine the [Au + ] remaining in solution just before AgCl begins to precipitate.

98. 98 Fractional Precipitation The percent of Au + ions unprecipitated just before AgCl precipitates is

99. 99 Fractional Precipitation The percent of Au + ions unprecipitated just before AgCl precipitates is Therefore, 99.99989% of the Au + ions precipitates before AgCl begins to precipitate.

100. 100 Fractional Precipitation Similar calculations for the concentration of Ag + ions unprecipitated before CuCl begins to precipitate gives

101. 101 Fractional Precipitation The percent of Au + ions unprecipitated just before AgCl precipitates is

102. 102 Fractional Precipitation The percent of Au + ions unprecipitated just before AgCl precipitates is Thus, 99.905% of the Ag + ions precipitates before CuCl begins to precipitate.

103. 103 Factors that Affect Solubility CaF 2 (s)  Ca +2 (aq)+2F - (aq) Addition of a Common ion (F- from NaF) –Solubility decreases –Equilibrium shifts to left Changes in pH (H+ reacts with F-) –Solubility increases (with increasing pH) –Equilibrium shifts to right

104. 104 Factors that Affect Solubility Ag + (aq) + 2NH 3 (aq)  Ag(NH 3 ) 2 + (aq) complex ion Formation of a complex ion –Lewis Acid base chemistry Calculate K f Formation Constant

105. 105 Factors that Affect Solubility Ag + (aq) + 2NH 3 (aq)  Ag(NH 3 ) 2 + (aq) complex ion AgCl(s)  Ag + (aq) + Cl - (aq) –In formation of complex ion –Removed Ag + from the equilibrium –Equilibrium shifts to right –Favors dissolving AgCl

106. 106 Synthesis Question Bufferin is a commercially prepared medicine that is literally a buffered aspirin. How could you buffer aspirin? Hint - what is aspirin?

107. 107 Synthesis Question Aspirin is acetyl salicylic acid. So to buffer it all that would have to be added is the salt of acetyl salicylic acid.

108. 108 Group Question Blood is slightly basic, having a pH of 7.35 to 7.45. What chemical species causes our blood to be basic? How does our body regulate the pH of blood?

109. 109 Synthesis Question Most kidney stones are made of calcium oxalate, Ca(O 2 CCO 2 ). Patients who have their first kidney stones are given an extremely simple solution to stop further stone formation. They are told to drink six to eight glasses of water a day. How does this stop kidney stone formation?

110. 110 Synthesis Question

111. 111 Group Question The cavities that we get in our teeth are a result of the dissolving of the material our teeth are made of, calcium hydroxy apatite. How does using a fluoride based toothpaste decrease the occurrence of cavities?