1. 1081

2. y = 1.0 x 10 -5 M [OH - ] = 1.0 x 10 -5 M 1082

3. y = 1.0 x 10 -5 M [OH - ] = 1.0 x 10 -5 M Hence pOH = 5.0 1083

4. y = 1.0 x 10 -5 M [OH - ] = 1.0 x 10 -5 M Hence pOH = 5.0 Since pOH + pH = 14.0, therefore pH = 9.0 1084

5. Acid-base titrations: The impact of hydrolysis 1085

6. Acid-base titrations: The impact of hydrolysis Salt hydrolysis has an important effect on the pH profile of acid-base titrations. 1086

7. Acid-base titrations: The impact of hydrolysis Salt hydrolysis has an important effect on the pH profile of acid-base titrations. The equivalence point may be above or below neutral conditions (i.e. pH = 7). 1087

8. Acid-base titrations: The impact of hydrolysis Salt hydrolysis has an important effect on the pH profile of acid-base titrations. The equivalence point may be above or below neutral conditions (i.e. pH = 7). For the titration of a strong acid and a strong base, the equivalence point should be at pH = 7. 1088

9. Acid-base titrations: The impact of hydrolysis Salt hydrolysis has an important effect on the pH profile of acid-base titrations. The equivalence point may be above or below neutral conditions (i.e. pH = 7). For the titration of a strong acid and a strong base, the equivalence point should be at pH = 7. Example: HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O 1089

10. 1090

11. 1091

12. 1092

13. 1093

14. Indicators 1094

15. Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. 1095

16. Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) 1096

17. Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O 1097

18. Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O The first drop of excess NaOH then reacts with the indicator that is present: 1098

19. Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O The first drop of excess NaOH then reacts with the indicator that is present: HIn (aq) + OH - (aq) H 2 O + In - (aq) 1099

20. Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O The first drop of excess NaOH then reacts with the indicator that is present: HIn (aq) + OH - (aq) H 2 O + In - (aq) Now HIn and In - have different colors, so we can detect that the acid-base reaction is complete. 1100

21. For the equilibrium: HIn (aq) H + (aq) + In - (aq) 1101

22. For the equilibrium: HIn (aq) H + (aq) + In - (aq) 1102

23. For the equilibrium: HIn (aq) H + (aq) + In - (aq) Midway in the transition of the indicator color change: [HIn] = [In - ], and hence K In = [H + ] (midway point). 1103

24. For the equilibrium: HIn (aq) H + (aq) + In - (aq) Midway in the transition of the indicator color change: [HIn] = [In - ], and hence K In = [H + ] (midway point). Take the log of both sides of this relationship, leads to pK In = pH (midway point). 1104

25. 1105

26. IONIC EQUILIBRIUM 1106

27. IONIC EQUILIBRIUM Buffers 1107

28. Buffers Buffer: A solution whose pH remains approximately constant despite the addition of small amounts of either acid or base. 1108

29. Buffers Buffer: A solution whose pH remains approximately constant despite the addition of small amounts of either acid or base. A buffer is a combination of species in solution that maintains an approximately constant pH by virtue of a pair of chemical reactions. 1109

30. Buffers Buffer: A solution whose pH remains approximately constant despite the addition of small amounts of either acid or base. A buffer is a combination of species in solution that maintains an approximately constant pH by virtue of a pair of chemical reactions. One reaction describes a reaction of a buffer component with added acid, the other reaction describes the reaction of a buffer component with added base. 1110

31. Example: acetic acid/sodium acetate buffer 1111

32. Example: acetic acid/sodium acetate buffer A solution containing these two substances has the ability to neutralize both added acid and added base. 1112

33. Example: acetic acid/sodium acetate buffer A solution containing these two substances has the ability to neutralize both added acid and added base. If base is added to the buffer, it will react with the acid component: CH 3 CO 2 H (aq) + OH - (aq) CH 3 CO 2 - (aq) + H 2 O 1113

34. Example: acetic acid/sodium acetate buffer A solution containing these two substances has the ability to neutralize both added acid and added base. If base is added to the buffer, it will react with the acid component: CH 3 CO 2 H (aq) + OH - (aq) CH 3 CO 2 - (aq) + H 2 O If acid is added to the buffer, it will react with the base component: CH 3 CO 2 - (aq) + H + (aq) CH 3 CO 2 H (aq) 1114

35. Example: acetic acid/sodium acetate buffer A solution containing these two substances has the ability to neutralize both added acid and added base. If base is added to the buffer, it will react with the acid component: CH 3 CO 2 H (aq) + OH - (aq) CH 3 CO 2 - (aq) + H 2 O If acid is added to the buffer, it will react with the base component: CH 3 CO 2 - (aq) + H + (aq) CH 3 CO 2 H (aq) Note that the Na + is not directly involved in the buffer chemistry. 1115

36. Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation 1116

37. Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) 1117

38. Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) 1118

39. Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) Now take the log of both sides of the preceding equation, to obtain 1119

40. Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) Now take the log of both sides of the preceding equation, to obtain 1120

41. That is, 1121

42. That is, 1122

43. That is, 1123

44. That is, This is the Henderson-Hasselbalch equation for the acetic acid system. 1124

45. We could repeat the previous approach for the weak acid HA to obtain: 1125

46. We could repeat the previous approach for the weak acid HA to obtain: 1126

47. We could repeat the previous approach for the weak acid HA to obtain: In a more general form it would be: 1127

48. We could repeat the previous approach for the weak acid HA to obtain: In a more general form it would be: Either of the preceding two equations are called the Henderson-Hasselbalch equation. 1128

49. Example: Calculate the pH of a buffer system containing 1.0 M CH 3 CO 2 H and 1.0 M NaCH 3 CO 2. What is the pH of the buffer after the addition of 0.10 moles of gaseous HCl to 1.00 liter of the buffer solution? The K a for acetic acid is 1.8 x 10 -5. 1129

50. Example: Calculate the pH of a buffer system containing 1.0 M CH 3 CO 2 H and 1.0 M NaCH 3 CO 2. What is the pH of the buffer after the addition of 0.10 moles of gaseous HCl to 1.00 liter of the buffer solution? The K a for acetic acid is 1.8 x 10 -5. Because acetic acid is a weak acid, we can ignore the small amount of dissociation and assume at equilibrium that [CH 3 CO 2 H] = 1.0 M 1130

51. Example: Calculate the pH of a buffer system containing 1.0 M CH 3 CO 2 H and 1.0 M NaCH 3 CO 2. What is the pH of the buffer after the addition of 0.10 moles of gaseous HCl to 1.00 liter of the buffer solution? The K a for acetic acid is 1.8 x 10 -5. Because acetic acid is a weak acid, we can ignore the small amount of dissociation and assume at equilibrium that [CH 3 CO 2 H] = 1.0 M It is also important to keep in mind that there is a lot of acetate ion present, and this will suppress the dissociation of the acetic acid (Le Châtelier’s Principle). 1131

52. A similar situation applies to the acetate ion, that is, we can ignore the hydrolysis of this ion. Also, the acetic acid present will suppress the hydrolysis of the acetate ion (Le Châtelier’s Principle), so that [CH 3 CO 2 - ] = 1.0 M 1132

53. A similar situation applies to the acetate ion, that is, we can ignore the hydrolysis of this ion. Also, the acetic acid present will suppress the hydrolysis of the acetate ion (Le Châtelier’s Principle), so that [CH 3 CO 2 - ] = 1.0 M Now K a = 1.8 x 10 -5 so that pK a = 4.7 1133

54. A similar situation applies to the acetate ion, that is, we can ignore the hydrolysis of this ion. Also, the acetic acid present will suppress the hydrolysis of the acetate ion (Le Châtelier’s Principle), so that [CH 3 CO 2 - ] = 1.0 M Now K a = 1.8 x 10 -5 so that pK a = 4.7 Hence, from the Henderson-Hasselbalch equation: = 4.7 1134

55. Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: 1135

56. Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: CH 3 CO 2 - + H + CH 3 CO 2 H 1136

57. Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: CH 3 CO 2 - + H + CH 3 CO 2 H 0.10 mols 0.10 mols 0.10 mols (from the HCl) 1137

58. Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: CH 3 CO 2 - + H + CH 3 CO 2 H 0.10 mols 0.10 mols 0.10 mols (from the HCl) At equilibrium, the concentrations of the buffer components are (keep in mind the total volume is 1.0 liter): 1138

59. Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: CH 3 CO 2 - + H + CH 3 CO 2 H 0.10 mols 0.10 mols 0.10 mols (from the HCl) At equilibrium, the concentrations of the buffer components are (keep in mind the total volume is 1.0 liter): [CH 3 CO 2 H] = 1.0 + 0.10 = 1.10 M [CH 3 CO 2 - ] = 1.0 - 0.10 = 0.90 M 1139

60. To calculate the new pH, use the Henderson- Hasselbalch equation: 1140