1. 1 Acid-Base Equilibria

2. 2 Solutions of a Weak Acid or Base The simplest acid-base equilibria are those in which a single acid or base solute reacts with water. –In this chapter, we will first look at solutions of weak acids and bases. –We must also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water.

3. 3 Acid-Ionization Equilibria Acid ionization (or acid dissociation) is the (See Animation: Acid Ionization Equilibrium) –When acetic acid is added to water it reacts as follows.

4. 4 Acid-Ionization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the –Consider the generic monoprotic acid, HA.

5. 5 Acid-Ionization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid- dissociation constant). –The corresponding equilibrium expression is:

6. 6 Acid-Ionization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant). –Since the concentration of water remains relatively constant, we rearrange the equation to get:

7. 7 Acid-Ionization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant). –Thus, K a, the acid-ionization constant, equals the constant [H 2 O]K c.

8. 8 Acid-Ionization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid- dissociation constant). –Table 17.1 lists acid-ionization constants for various weak acids.

9. 9 Experimental Determination of Ka The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions.

10. 10 A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.

11. 11 A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. Starting Change Equilibrium –Let x be the moles per liter of product formed.

12. 12 A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. –The equilibrium-constant expression is:

13. 13 A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. –Substituting the expressions for the equilibrium concentrations, we get

14. 14 A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. –We can obtain the value of x from the given pH.

15. 15 A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. –Substitute this value of x –Note that the concentration of unionized acid remains virtually unchanged.

16. 16 A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. –Substitute this value of x

17. 17 A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. –To obtain the degree of dissociation:

18. 18 Calculations With K a Once you know the value of K a, you can calculate the equilibrium concentrations of species HA, A-, and H 3 O + for solutions of different molarities.

19. 19 Calculations With K a Note that in our previous example, the degree of dissociation was so small that “x” was negligible compared to the concentration of nicotinic acid.

20. 20 Calculations With K a How do you know when you can use this simplifying assumption?

21. 21 Calculations With K a How do you know when you can use this simplifying assumption?

22. 22 A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C.

23. 23 A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C.

24. 24 A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C. –Note that

25. 25 A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C.

26. 26 A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C. Starting Change Equilibrium –These data are summarized below.

27. 27 –The equilibrium constant expression is A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C.

28. 28 –If we substitute the equilibrium concentrations and the K a into the equilibrium constant expression, we get A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C.

29. 29 A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C. –Rearranging the preceding equation to put it in the form ax 2 + bx + c = 0, we get –You can solve this equation exactly by using the quadratic formula.

30. 30 –Now substitute into the quadratic formula. A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C.

31. 31 –Now substitute into the quadratic formula. A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C.

32. 32 –Taking the upper sign, we get A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C. –Now we can calculate the pH.

33. 33 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. –Sulfuric acid, for example, can lose two protons in aqueous solution.

34. 34 Polyprotic Acids –For a weak diprotic acid like carbonic acid, H 2 CO 3, two simultaneous equilibria must be considered. Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

35. 35 –Each equilibrium has an associated acid-ionization constant. Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

36. 36 –Each equilibrium has an associated acid-ionization constant. Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

37. 37 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

38. 38 –However, reasonable assumptions can be made that simplify these calculations as we show in the next example. Polyprotic Acids –When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions. Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

39. Polyprotic Acids

40. 40 A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

41. 41 –The pH can be determined by simply solving the equilibrium problem posed by the first ionization. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

42. 42 –If we abbreviate the formula for ascorbic acid as H 2 Asc, then the first ionization is: A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

43. 43 A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12. Starting Change Equilibrium

44. 44 –The equilibrium constant expression is A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

45. 45 –Substituting into the equilibrium expression A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

46. 46 –Assuming that x is much smaller than 0.10, you get A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

47. 47 –The hydronium ion concentration is 0.0028 M, so A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

48. 48 –The ascorbate ion, Asc 2-, which we will call y, is produced only in the second ionization of H 2 Asc. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

49. 49 –Assume the starting concentrations for HAsc - and H 3 O + to be those from the first equilibrium. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

50. 50 A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12. Starting Change Equilibrium

51. 51 –The equilibrium constant expression is A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

52. 52 –Substituting into the equilibrium expression A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

53. 53 –Assuming y is much smaller than 0.0028, the equation simplifies to A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12.

54. 54 –Hence, A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x 10 -12. –The concentration of the ascorbate ion equals K a2.

55. 55 Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. –Ammonia, for example, ionizes in water as follows. –The corresponding equilibrium constant is:

56. 56 Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. –Ammonia, for example, ionizes in water as follows. –The concentration of water is nearly constant.

57. 57 Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. –In general, a weak base B with the base ionization has a base ionization constant equal to Table 17.2 lists ionization constants for some weak bases.

58. 58 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10-9. 1.Write the equation and make a table of concentrations. 2.Set up the equilibrium constant expression. 3.Solve for x = [OH - ]. –As before, we will follow the three steps in solving an equilibrium.

59. 59 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. –Pyridine ionizes by picking up a proton from water (as ammonia does). Starting Change Equilibrium

60. 60 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. –Note that

61. 61 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. –The equilibrium expression is

62. 62 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. –If we substitute the equilibrium concentrations and the K b into the equilibrium constant expression, we get

63. 63 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. –Using our simplifying assumption that the x in the denominator is negligible, we get

64. 64 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. –Solving for x we get

65. 65 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. –Solving for pOH

66. 66 Acid-Base Properties of a Salt Solution One of the successes of the Brønsted- Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. –Consider a solution of sodium cyanide, NaCN.

67. 67 One of the successes of the Brønsted- Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. Acid-Base Properties of a Salt Solution

68. 68 One of the successes of the Brønsted- Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. Acid-Base Properties of a Salt Solution

69. 69 One of the successes of the Brønsted- Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. –You can also see that OH - ion is a product, so you would expect –The reaction of the CN - ion with water is referred to as the hydrolysis of CN -. Acid-Base Properties of a Salt Solution

70. 70 The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. Acid-Base Properties of a Salt Solution

71. 71 Acid-Base Properties of a Salt Solution The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

72. 72 Acid-Base Properties of a Salt Solution The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

73. 73 Acid-Base Properties of a Salt Solution The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

74. 74 Predicting Whether a Salt is Acidic, Basic, or Neutral How can you predict whether a particular salt will be acidic, basic, or neutral? –The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs.

75. 75 Predicting Whether a Salt is Acidic, Basic, or Neutral How can you predict whether a particular salt will be acidic, basic, or neutral?

76. 76 Predicting Whether a Salt is Acidic, Basic, or Neutral How can you predict whether a particular salt will be acidic, basic, or neutral?

77. 77 Predicting Whether a Salt is Acidic, Basic, or Neutral To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt. –Consider potassium acetate, KC 2 H 3 O 2. The potassium ion is the cation of a strong base (KOH) and does not hydrolyze.

78. 78 Predicting Whether a Salt is Acidic, Basic, or Neutral To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt. –Consider potassium acetate, KC 2 H 3 O 2.

79. 79 Predicting Whether a Salt is Acidic, Basic, or Neutral These rules apply to normal salts (those in which the anion has no acidic hydrogen) 1.A salt of a strong base and a strong acid.

80. 80 Predicting Whether a Salt is Acidic, Basic, or Neutral These rules apply to normal salts (those in which the anion has no acidic hydrogen) 2.A salt of a strong base and a weak acid.

81. 81 Predicting Whether a Salt is Acidic, Basic, or Neutral These rules apply to normal salts (those in which the anion has no acidic hydrogen) 3.A salt of a weak base and a strong acid.

82. 82 Predicting Whether a Salt is Acidic, Basic, or Neutral These rules apply to normal salts (those in which the anion has no acidic hydrogen) 4.A salt of a weak base and a weak acid.

83. 83 The pH of a Salt Solution To calculate the pH of a salt solution would require the K a of the acidic cation or the K b of the basic anion. (see Figure 17.8)(see Figure 17.8) –The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species. –Thus the K b for CN - is related to the K a for HCN.

84. 84 The pH of a Salt Solution To see the relationship between K a and K b for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN -. KaKa KbKb –When these two reactions are added you get the ionization of water. KwKw

85. 85 The pH of a Salt Solution To see the relationship between K a and K b for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN -. KaKa KbKb –When two reactions are added, their equilibrium constants are multiplied. KwKw

86. 86 The pH of a Salt Solution To see the relationship between K a and K b for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN -. KaKa KbKb Therefore, KwKw

87. 87 The pH of a Salt Solution For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases. –The only difference is first obtaining the K a or K b for the ion that hydrolyzes.

88. 88 A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 °C? The K a for HCN is 4.9 x 10 -10.

89. 89 A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 °C? The K a for HCN is 4.9 x 10 -10. –The CN - ion is acting as a base, so first, we must calculate the K b for CN -.

90. 90 A Problem To Consider –Let x = [OH - ] = [HCN What is the pH of a 0.10 M NaCN solution at 25 °C? The K a for HCN is 4.9 x 10 -10.

91. 91 A Problem To Consider –This gives What is the pH of a 0.10 M NaCN solution at 25 °C? The K a for HCN is 4.9 x 10 -10.

92. 92 A Problem To Consider –Solving the equation, you find that What is the pH of a 0.10 M NaCN solution at 25 °C? The K a for HCN is 4.9 x 10 -10.

93. 93 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. –Consider a solution of acetic acid (HC 2 H 3 O 2 ), in which you have the following equilibrium.

94. 94 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

95. 95 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

96. 96 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

97. 97 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. –Consider the equilibrium below. Starting 0.025 00.018 Change -x +x Equilibrium 0.025-x x0.018+x

98. 98 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. –The equilibrium constant expression is:

99. 99 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. –Substituting into this equation gives:

100. 100 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4.

101. 101 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. –The equilibrium equation becomes

102. 102 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4. –Hence,

103. 103 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCH 2 O and 0.018 M in sodium formate, NaCH 2 O. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4.

104. Buffers

105. 105 Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

106. 106 Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

107. 107 Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

108. 108 Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

109. 109 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? –To illustrate, consider a buffer of a weak acid HA and its conjugate base A -. The acid ionization equilibrium is:

110. 110 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? –The acid ionization constant is: –By rearranging, you get an equation for the H 3 O + concentration.

111. 111 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? –Taking the negative logarithm of both sides of the equation we obtain:

112. 112 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? –More generally, you can write

113. 113 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH?

114. 114 Calculate the pH of a 44ml solution of.202 M acetic acid and 15.5 ml of.185 M NaOH solution. HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O 44.0 ml 15.5 ml.202 M.185 M The Henderson-Hasselbalch Equation

115. 115 The Henderson-Hasselbalch Equation Calculate the pH of a 44ml solution of.202 M acetic acid and 15.5 ml of.185 M NaOH solution. Calculating the new molarity in 59.5 ml HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O + 1.01 x 10 -1 M 4.82 x 10 -2 M x

116. Titration Curves

117. 117 Acid-Ionization Titration Curves An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid).

118. 118 Titration of a Strong Acid by a Strong Base

119. Figure 17.12: Curve for the titration of a strong acid by a strong base.

120. 120 Figure 17.12 shows a curve for the titration of HCl with NaOH. –At the equivalence point, the pH of the solution is Titration of a Strong Acid by a Strong Base

121. 121 Figure 17.12 shows a curve for the titration of HCl with NaOH. –To detect the equivalence point, you need Titration of a Strong Acid by a Strong Base

122. 122 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl. NaOH + HCl  NaCl + H 2 O

123. 123 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl. NaOH + HCl  NaCl + H 2 O –We get the amounts of reactants by multiplying the volume of each (in liters) by their respective molarities.

124. 124 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl. NaOH + HCl  NaCl + H 2 O

125. 125 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl.

126. 126 A Problem To Consider Calculate the pH of a solution in which 10.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.100 M HCl. –Hence,

127. 127 The titration of a weak acid by a strong base gives a somewhat different curve. Titration of a Weak Acid by a Strong Base

128. Figure 17.13: Curve for the titration of a weak acid by a strong base.

129. 129 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O –At the equivalence point,.

130. 130 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O –First, calculate the concentration of the acetate ion.

131. 131 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O 2.5 x 10 -3 mol

132. 132 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. –The total volume of the solution is 50.0 mL. Hence,

133. 133 Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O.025 M.025 M -.025 M -.025 M +.025 M 0 0.025 M A Problem To Consider

134. 134 Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. NaC 2 H 3 O 2  Na + + C 2 H 3 O 2 -.025 M.025 M.025 M A Problem To Consider

135. 135 A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The K a for acetic acid is 1.7 x 10 -5. The K b for the acetate ion is 5.9 x 10 -10

136. 136 The titration of a weak base with a strong acid is a reflection of our previous example. –Figure 17.14 shows the titration of NH 3 with HCl. –In this case, the pH declines slowly at first, then falls abruptly from about pH 7 to pH 3. –Methyl red, which changes color from yellow at pH 6 to red at pH 4.8, is a possible indicator. Titration of a Strong Acid by a Weak Base

137. Figure 17.14: Curve for the titration of a weak base by a strong acid.

138. 138 Just a Reminder…

139. 139 Building Equations with K values Weak acids and bases are assigned a K a or K b values based on the degree to which they ionize in water. Larger K values indicate a greater degree of ionization (strength). K a and K b, along with other K values that we will study later (Ksp, K D, K f ) are all manipulated in the same manner.

140. 140 When equations are added K values are multiplied. MnS ↔ Mn+2 + S-2K= 5.1 x 10 -15 S -2 + H 2 O ↔ HS- + OH-K= 1.0 x 10 -19 2H+ + HS- OH- ↔ H 2 S +H 2 O K= 1.0 x 10 -7 MnS + 2H+ ↔ Mn +2 + H 2 K= Building Equations with K values

141. 141 When equations are reversed the K values are reciprocated. Al(OH) 3 ↔ Al +3 + 3OH - K = 1.9 x 10-33 3OH- + Al +3 ↔ Al(OH) 3 K = 1/1.9 x 10 -33 = Building Equations with K values

142. 142 When equations are multiplied the K values are raised to the power. NH 3 + H 2 O ↔ NH 4 + + OH-K = 1.8 x 10-5 2NH 3 + 2H 2 O ↔ 2NH 4 + + 2OH- K = (1.8 x 10 -5 ) 2 = Building Equations with K values

143. 143 When equations are divided the root of the K values are taken. 2HPO 4 -2 ↔ 2H + + 2PO 4 3- K = 1.3 x 10 -25 HPO 4 -2 ↔ H + + PO 4 3- K = √1.3 x 10 -25 = Building Equations with K values

144. 144 Operational Skills Determining K a (or K b ) from the solution pH Calculating the concentration of a species in a weak acid solution using K a Calculating the concentration of a species in a weak base solution using K b Predicting whether a salt solution is acidic, basic, or neutral Obtaining K a from K b or K b from K a Calculating concentrations of species in a salt solution

145. 145 Operational Skills Calculating the common-ion effect on acid ionization Calculating the pH of a buffer from given volumes of solution Calculating the pH of a solution of a strong acid and a strong base Calculating the pH at the equivalence point in the titration of a weak cid with a strong base

146. 146 Animation: Acid Ionization Equilibrium Return to slide 3 (Click here to open QuickTime animation)

147. 147 Figure 17.3: Variation of percent ionization of a weak acid with concentration. Return to slide 19

148. 148 Figure 17.8: A pH meter reading NH 4 Cl. Photo courtesy of American Color. Return to slide 82

149. 149 Animation: Adding an Acid to a Buffer Return to slide 104 (Click here to open QuickTime animation)