1. Acid-Base Titrations Section 17.3

2. Introduction Definition: – In an acid-base titration, a solution containing a known concentration of a base is slowly added to an acid. – An indicator is used to signal the equivalence point of the titration. This is the point at which stoichiometrically equivalent amounts of acid and base have been mixed. – A pH meter can also be used to find the equivalence point.

3. Introduction The typical titration apparatus includes: – a buret to hold the titrant – a beaker to hold the analyte – a pH meter to measure the pH

4. Introduction In this section, we will be looking at a series of titrations in detail to understand why acids and behave the way they do. – Strong acid-strong base titration – Weak acid-strong base titration – Polyprotic acid-strong base titration

5. Strong Acid-Strong Base Titrations The titration curve of a strong acid-strong base titration has the following shape.

6. Strong Acid-Strong Base Titrations The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: 1. Initial pH 2. Initial pH to eq. point 3. Equivalence point 4. After eq. point

7. Strong Acid-Strong Base Titrations The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: 1. Initial pH The pH of the solution is determined by the concentration of the strong acid.

8. Strong Acid-Strong Base Titrations The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: As base is added, pH increases slowly and then rapidly. The pH is determined by the concentration of the acid that is not neutralized. 2. Initial pH to eq. point

9. Strong Acid-Strong Base Titrations The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: At the equivalence point, [OH − ] = [H + ]. The pH = 7.00. 3. Equivalence point

10. Strong Acid-Strong Base Titrations The titration curve of a strong acid-strong base titration has the following shape. We divide the curve into four regions: As more base is added, pH increases rapidly and then slowly. pH is determined by the concentration of the excess base. 4. After eq. point

11. Strong Acid-Strong Base Titrations The titration curve of a strong acid-strong base titration has the following shape. Let’s see how this works in practice.

12. Sample Exercise 17.6 (page 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 b)51.0 mL

13. Sample Exercise 17.6 (page 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

14. Sample Exercise 17.6 (page 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 This is between the initial point and the equivalence point.

15. Sample Exercise 17.6 (page 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 This is between the initial point and the equivalence point. pH is determined by the amount of acid that has not been neutralized.

16. Sample Exercise 17.6 (page 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 Therefore, we need to determine the number of mols of acid remaining, n acid, and the total volume, V total, of the solution. (Remember, adding the NaOH increases the total volume.

17. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i

18. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i = (0.100 M)(0.0500 L)

19. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol

20. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = M base,added × V base,added

21. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = (0.100 M)(0.0490 L)

22. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol

23. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol n acid,remaining = n acid,i − n base,added

24. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol n acid,remaining = (5.00 − 4.90) × 10 −3 mol

25. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol n acid,remaining = 0.10 × 10 −3 mol

26. Sample Exercise 17.6 (page 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 n acid,i = M acid,i × V acid,i = 5.00 × 10 −3 mol n base,added = 4.90 × 10 −3 mol n acid,remaining = 1.0 × 10 −4 mol

27. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol

28. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = V acid + V base

29. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = V acid + V base = 0.0500 L + 0.0490 L

30. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = V acid + V base = 0.0990 L

31. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = 0.0990 L

32. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = 0.0990 L [H + ] = n acid,remaining /V total

33. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = 0.0990 L [H + ] = (1.0 × 10 −4 mol)/(0.0990 L)

34. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = 0.0990 L [H + ] = 1.0 × 10 −3 M

35. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = 0.0990 L [H + ] = 1.0 × 10 −3 M pH = −log[H + ]

36. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = 0.0990 L [H + ] = 1.0 × 10 −3 M pH = −log(1.0 × 10 −3 )

37. Sample Exercise 17.6 (page 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 n acid,remaining = 1.0 × 10 −4 mol V total = 0.0990 L [H + ] = 1.0 × 10 −3 M pH = 3.00

38. Sample Exercise 17.6 (page 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 pH = 3.00

39. Sample Exercise 17.6 (page 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

40. Sample Exercise 17.6 (page 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 b)51.0 mL

41. Sample Exercise 17.6 (page 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. b)51.0 mL

42. Sample Exercise 17.6 (page 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. b)51.0 mL

43. Sample Exercise 17.6 (page 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. b)51.0 mL This is beyond the equivalence point.

44. Sample Exercise 17.6 (page 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. b)51.0 mL This is beyond the equivalence point. All of the strong acid has been used up.

45. Sample Exercise 17.6 (page 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. b)51.0 mL This is beyond the equivalence point. All of the strong acid has been used up. The pH is determined by the excess base that has been added.

46. Sample Exercise 17.6 (page 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. b)51.0 mL We already determined n acid,i = 5.00 × 10 −3 mol

47. Sample Exercise 17.6 (page 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. b)51.0 mL n acid,i = 5.00 × 10 −3 mol

48. Sample Exercise 17.6 (page 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. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = M base,added × V base,added

49. Sample Exercise 17.6 (page 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. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = (0.100 M)(0.0510 L)

50. Sample Exercise 17.6 (page 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. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = 5.10 × 10 −3 mol

51. Sample Exercise 17.6 (page 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. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = 5.10 × 10 −3 mol n base,remaining = n base,added − n acid,i

52. Sample Exercise 17.6 (page 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. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = 5.10 × 10 −3 mol n base,remaining = (5.10 − 5.00) × 10 −3 mol

53. Sample Exercise 17.6 (page 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. b)51.0 mL n acid,i = 5.00 × 10 −3 mol n base,added = 5.10 × 10 −3 mol n base,remaining = 0.10 × 10 −3 mol

54. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol

55. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = V acid + V base

56. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = V acid + V base = 0.0500 L + 0.0510 L

57. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = V acid + V base = 0.1010 L

58. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L

59. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L [OH − ] = n base,remaining /V total

60. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L [OH − ] = (1.0 × 10 −4 mol)/(0.1010 L)

61. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L [OH − ] = 9.9 × 10 −4 M

62. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L [OH − ] = 9.9 × 10 −4 M pOH = −log[OH − ]

63. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L [OH − ] = 9.9 × 10 −4 M pOH = −log(9.9 × 10 −4 )

64. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L [OH − ] = 9.9 × 10 −4 M pOH = 3.00

65. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L [OH − ] = 9.9 × 10 −4 M pOH = 3.00 ⇒ pH = 14.00 − 3.00

66. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L [OH − ] = 9.9 × 10 −4 M pOH = 3.00 ⇒ pH = 14.00 − 3.00 = 11.00

67. Sample Exercise 17.6 (page 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. b)51.0 mL n base,remaining = 1.0 × 10 −4 mol V total = 0.1010 L [OH − ] = 9.9 × 10 −4 M pH = 11.00

68. Sample Exercise 17.6 (page 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. b)51.0 mL pH = 11.00

69. Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.

70. Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. 1.The initial pH is determined by the K a of the acid.

71. Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. 2.To determine the pH from the initial point to the eq. point, we first neutralize the weak acid and then use the Henderson- Hasselbach equation.

72. Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. 3.At the eq. point, we have no HX, only X −. We need to use the K b value to find pH.

73. Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. 4.Beyond the eq. point, we use the excess base to calculate pH.

74. Weak Acid-Strong Base Titrations The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base. Let’s see how this works in practice.

75. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH.

76. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction.

77. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O

78. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = M acid × V acid

79. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = M acid × V acid = (0.100 M)(0.0500 L)

80. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = M acid × V acid = 5.00 × 10 −3 mol

81. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol

82. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = M base × V base

83. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = M base × V base = (0.100 M)(0.0450 L)

84. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = M base × V base = 4.50 × 10 −3 mol

85. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. First, calculate the concentrations of materials before neutralization reaction. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = 4.50 × 10 −3 mol

86. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial,acid = 5.00 × 10 −3 mol n base = 4.50 × 10 −3 mol

87. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial 5.0 × 10 −3 mol4.5 × 10 −3 mol0.0 mol

88. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial 5.0 × 10 −3 mol4.5 × 10 −3 mol0.0 mol n change −4.5 × 10 −3 mol +4.5 × 10 −3 mol

89. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n initial 5.0 × 10 −3 mol4.5 × 10 −3 mol0.0 mol n change −4.5 × 10 −3 mol +4.5 × 10 −3 mol n final 5.0 × 10 −4 mol0.0 mol4.5 × 10 −3 mol

90. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n acid,final = 5.0 × 10 −4 mol

91. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n acid,final = 5.0 × 10 −4 mol [acid] = n acid,final /V total

92. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n acid,final = 5.0 × 10 −4 mol [acid] = (5.0 × 10 −4 mol)/(0.0950 L)

93. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O n acid,final = 5.0 × 10 −4 mol [acid] = 5.3 × 10 −3 M

94. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = n base,final /V total

95. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = (4.50 × 10 −3 mol)/(0.0950 L)

96. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M

97. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M

98. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M pK a = −logK a

99. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M pK a = −log(1.8 × 10 −5 )

100. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O [acid] = 5.3 × 10 −3 M [base] = 4.74 × 10 −2 M pK a = 4.74

101. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = pK a + log([base]/[acid)

102. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = 4.74 + log([4.74 × 10 −2 M]/[5.3 × 10 −3 M])

103. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = 4.74 + log(9.00)

104. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = 4.74 + 0.954

105. Sample Exercise 17.7 (page 735) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. Next, use the values to determine the concentrations after the neutralization. CH 3 COOH + OH − ➙ CH 3 COO − + H 2 O pH = 5.69

106. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH.

107. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. At the equivalence point, all of the weak acid is converted to its conjugate weak base.

108. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. At the equivalence point, all of the weak acid is converted to its conjugate weak base. This means that we will be using the base hydrolysis expression to find [OH − ], pOH, and pH.

109. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. The number of mols of acetate in solution at the equivalence point is equal to the number of mols of acetic acid at the start of the titration.

110. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. n base,eq. pt. = n initial,acid

111. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. n base,eq. pt. = n initial,acid = (0.100 M)(0.0500 L)

112. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. n base,eq. pt. = n initial,acid = 5.00 × 10 −3 mol

113. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol

114. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol The concentration of the acetate is given by:

115. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol The concentration of the acetate is given by: [base] = (n base,eq. pt. )/(V total )

116. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol The concentration of the acetate is given by: [base] = (5.00 × 10 −3 mol)/(0.100 L)

117. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. n base,eq. pt. = 5.00 × 10 −3 mol The concentration of the acetate is given by: [base] = 5.00 × 10 −2 M

118. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M

119. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M Next, we do an i-c-e table on the base hydrolysis.

120. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M CH 3 COO − + H 2 O ➙ CH 3 COOH + OH −

121. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M CH 3 COO − + H 2 O ➙ CH 3 COOH + OH − i5.00 × 10 −2 0.0

122. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M CH 3 COO − + H 2 O ➙ CH 3 COOH + OH − i5.00 × 10 −2 0.0 c−x+x

123. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M CH 3 COO − + H 2 O ➙ CH 3 COOH + OH − i5.00 × 10 −2 0.0 c−x+x e5.00 × 10 −2 − xxx

124. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x CH 3 COO − + H 2 O ➙ CH 3 COOH + OH − i5.00 × 10 −2 0.0 c−x+x e5.00 × 10 −2 − xxx

125. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = K w /K a = (1.0 × 10 −14 )/(1.8 × 10 −5 )

126. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = K w /K a = 5.6 × 10 −10

127. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10

128. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = ([acid][OH − ])/[base]

129. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = (x 2 )/(5.00 × 10 −2 )

130. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = (x 2 )/(5.00 × 10 −2 ) x 2 = (5.6 × 10 −10 )(5.00 × 10 −2 ) = 2.8 × 10 −11 x = (2.8 × 10 −11 ) ½

131. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = (x 2 )/(5.00 × 10 −2 ) x 2 = (5.6 × 10 −10 )(5.00 × 10 −2 ) = 2.8 × 10 −11 x = (2.8 × 10 −11 ) ½ = 5.3 × 10 −6

132. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [base] = 5.00 × 10 −2 M, [acid] = [OH − ] = x K b = 5.6 × 10 −10 = (x 2 )/(5.00 × 10 −2 ) x 2 = (5.6 × 10 −10 )(5.00 × 10 −2 ) = 2.8 × 10 −11 x = (2.8 × 10 −11 ) ½ = 5.3 × 10 −6 = [OH − ]

133. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [OH − ] = 5.3 × 10 −6 M

134. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = −log[OH − ]

135. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = −log[OH − ] = −log(5.3 × 10 −6 )

136. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = −log[OH − ] = −log(5.3 × 10 −6 ) = 5.28

137. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = 5.28

138. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = 5.28 pH = 14.00 – pOH

139. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = 5.28 pH = 14.00 – pOH = 14.00 − 5.28

140. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. [OH − ] = 5.3 × 10 −6 M pOH = 5.28 pH = 14.00 – pOH = 14.00 − 5.28 = 8.72

141. Sample Exercise 17.8 (page 735) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. pH = 8.72

142. The pH of the titration at the equivalence point depends on the type of acids and bases being titrated. – Strong acid-Strong base: eq. pt. = 7.0 – Weak acid-Strong base: eq. pt. > 7.0 – Strong acid-Weak base: eq. pt. < 7.0

143. Titrations of Polyprotic Acids When weak acid contain more than one ionizable H atom, as in phosporous acid, H 3 PO 3, reaction with OH − occurs in a series of steps. – H 3 PO 3 (aq) + H 2 O(l) ➙ H 2 PO 3 − (aq) + H 3 O + (aq) – H 2 PO 3 − (aq) + H 2 O(l) ➙ HPO 3 2− (aq) + H 3 O + (aq) – HPO 3 2− (aq) + H 2 O(l) ➙ PO 3 3− (aq) + H 3 O + (aq)

144. Titrations of Polyprotic Acids When the neutralization steps are sufficiently separated, the substance exhibits multiple equivalence points.