1. Acid-Base Titrations Calculations

2. – buret to hold the titrant – beaker to hold the analyte – pH meter to measure the pH

3. Acid-Base Titration Types – Strong acid-strong base titration – Weak acid-strong base titration – Polyprotic acid-strong base titration

4. Strong Acid-Strong Base Titrations

5. We divide the curve into four regions: 1. Initial pH 2. Initial pH to eq. point 3. Equivalence point 4. After eq. point

6. Strong Acid - Strong Base Titrations We divide the curve into four regions: 1. Initial pH The pH of the solution is determined by the concentration of the strong acid.

7. 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 Strong Acid - Strong Base Titrations

8. We divide the curve into four regions: At the equivalence point, [OH − ] = [H + ]. The pH = 7.00. 3. Equivalence point Strong Acid - Strong Base Titrations

9. 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 Strong Acid - Strong Base Titrations

10. Let’s see how this works in practice. Strong Acid - Strong Base Titrations

11. Sample Exercise 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

12. 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 Sample Exercise

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 This is between the initial point and the equivalence point.

14. 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. Sample Exercise

15. 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. Sample Exercise

16. 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 Sample Exercise

17. 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) Sample Exercise

18. 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 Sample Exercise

19. 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 Sample Exercise

20. 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) Sample Exercise

21. 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 Sample Exercise

22. 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 Sample Exercise

23. 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 Sample Exercise

24. 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 Sample Exercise

25. 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 Sample Exercise

26. 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 Sample Exercise

27. 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 Sample Exercise

28. 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 Sample Exercise

29. 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 Sample Exercise

30. 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 Sample Exercise

31. 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 Sample Exercise

32. 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) Sample Exercise

33. 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 Sample Exercise

34. 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 + ] Sample Exercise

35. 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 ) Sample Exercise

36. 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 Sample Exercise

37. 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 Sample Exercise

38. 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 Sample Exercise

39. 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 Sample Exercise

40. 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 Sample Exercise

41. 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 Sample Exercise

42. 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. Sample Exercise

43. 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. Sample Exercise

44. 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. Sample Exercise

45. 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 Sample Exercise

46. 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 Sample Exercise

47. 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 Sample Exercise

48. 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) Sample Exercise

49. 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 Sample Exercise

50. 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 Sample Exercise

51. 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 Sample Exercise

52. 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 Sample Exercise

53. 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 Sample Exercise

54. 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 Sample Exercise

55. 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 Sample Exercise

56. 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 Sample Exercise

57. 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 Sample Exercise

58. 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 Sample Exercise

59. 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) Sample Exercise

60. 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 Sample Exercise

61. 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 − ] Sample Exercise

62. 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 ) Sample Exercise

63. 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 Sample Exercise

64. 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 Sample Exercise

65. 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 Sample Exercise

66. 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 Sample Exercise

67. 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 Sample Exercise

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

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. 1.The initial pH is determined by the K a of the acid.

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. 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.

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. 3.At the eq. point, we have no HX, only X −. We need to use the Kb value to find pH.

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. 4.Beyond the eq. point, we use the excess base to calculate 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. Let’s see how this works in practice.

74. 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. Sample Exercise

75. 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. Sample Exercise

76. 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 Sample Exercise

77. 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 Sample Exercise

78. 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) Sample Exercise

79. 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 Sample Exercise

80. 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 Sample Exercise

81. 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 Sample Exercise

82. 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) Sample Exercise

83. 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 Sample Exercise

84. 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 Sample Exercise

85. 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 Sample Exercise

86. 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 Sample Exercise

87. 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 Sample Exercise

88. 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 Sample Exercise

89. 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 Sample Exercise

90. 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 Sample Exercise

91. 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) Sample Exercise

92. 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 Sample Exercise

93. 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 Sample Exercise

94. 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) Sample Exercise

95. 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 Sample Exercise

96. 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 Sample Exercise

97. 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 Sample Exercise

98. 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 ) Sample Exercise

99. 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 Sample Exercise

100. 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) Sample Exercise

101. 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]) Sample Exercise

102. 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) Sample Exercise

103. 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 Sample Exercise

104. 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 Sample Exercise

105. 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. Sample Exercise

106. 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. Sample Exercise

107. 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. Sample Exercise

108. 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. Sample Exercise

109. 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 Sample Exercise

110. 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) Sample Exercise

111. 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 Sample Exercise

112. 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 Sample Exercise

113. 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: Sample Exercise

114. 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 ) Sample Exercise

115. 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) Sample Exercise

116. 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 Sample Exercise

117. 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 Sample Exercise

118. 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. Sample Exercise

119. 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 − Sample Exercise

120. 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 Sample Exercise

121. 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 Sample Exercise

122. 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 Sample Exercise

123. 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 Sample Exercise

124. 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 ) Sample Exercise

125. 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 Sample Exercise

126. 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 Sample Exercise

127. 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] Sample Exercise

128. 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 ) Sample Exercise

129. 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 Sample Exercise

130. 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 Sample Exercise

131. 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 − ] Sample Exercise

132. 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 Sample Exercise

133. 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 − ] Sample Exercise

134. 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 ) Sample Exercise

135. 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 Sample Exercise

136. 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 Sample Exercise

137. 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 Sample Exercise

138. 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 Sample Exercise

139. 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 Sample Exercise

140. 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 Sample Exercise

141. 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

142. 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)

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