1. To calculate the new pH, use the Henderson- Hasselbalch equation: 1141

2. To calculate the new pH, use the Henderson- Hasselbalch equation: = 4.6 1142

3. To calculate the new pH, use the Henderson- Hasselbalch equation: = 4.6 Note in this example that the [H + ] changed from [H + ] = 10 -pH = 10 -4.7 = 2.0 x 10 -5 M to [H + ] = 10 -pH = 10 -4.6 = 2.5 x 10 -5 M 1143

4. To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H 2 O. 1144

5. To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H 2 O. The initial [H + ] = 1.0 x 10 -7 M (from self-dissociation of water) 1145

6. To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H 2 O. The initial [H + ] = 1.0 x 10 -7 M (from self-dissociation of water) After the HCl is added, [H + ] = 1.0 x 10 -1 M, so there is a million fold increase in [H + ]! 1146

7. To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H 2 O. The initial [H + ] = 1.0 x 10 -7 M (from self-dissociation of water) After the HCl is added, [H + ] = 1.0 x 10 -1 M, so there is a million fold increase in [H + ]! Whereas for the buffer, [H + ] changes from 2.0 x 10 -5 M to 2.5 x 10 -5 M. 1147

8. Distribution Curve 1148

9. Distribution Curve Distribution curve (for a buffer): Gives the fraction of the acid component and the base component present in solution as a function of the pH. 1149

10. Distribution Curve Distribution curve (for a buffer): Gives the fraction of the acid component and the base component present in solution as a function of the pH. Example: Consider the acetic acid/acetate buffer. 1150

11. Distribution Curve Distribution curve (for a buffer): Gives the fraction of the acid component and the base component present in solution as a function of the pH. Example: Consider the acetic acid/acetate buffer. At low pH: H + + CH 3 CO 2 - CH 3 CO 2 H 1151

12. Distribution Curve Distribution curve (for a buffer): Gives the fraction of the acid component and the base component present in solution as a function of the pH. Example: Consider the acetic acid/acetate buffer. At low pH: H + + CH 3 CO 2 - CH 3 CO 2 H At high pH: OH - + CH 3 CO 2 H CH 3 CO 2 - 1152

13. Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows: 1153

14. Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows: 1154

15. Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows: For the acetic acid/acetate ion buffer (pK a =4.7) the buffer range is pH = 3.7 – 5.7. 1155

16. Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows: For the acetic acid/acetate ion buffer (pK a =4.7) the buffer range is pH = 3.7 – 5.7. Note that the buffer functions best at pH = 4.7, i.e. when [CH 3 CO 2 H] = [CH 3 CO 2 - ] 1156

17. 1157

18. 1158

19. Exercise: Comment on the following as possible buffer systems. (a) NaCl (aq) /HCl (aq) (b) NH 3(aq) /NH 4 Cl (aq) (c) H 2 PO 4 - (aq) / HPO 4 2- (aq) (d) NaHCO 3(aq) 1159

20. (a) NaCl (aq) /HCl (aq) To be a buffer, it needs to be able to react with both added H + and added OH -. If H + is added, the following reaction does not occur to any significant extent: H + (aq) + Cl - (aq) HCl (aq) (because HCl is a strong acid). That means that (a) cannot be a buffer system. 1160

21. (b) NH 3(aq) /NH 4 Cl (aq) Addition of H + : H + (aq) + NH 3(aq) NH 4 + (aq) Addition of OH - : OH - (aq) + NH 4 + (aq) NH 3(aq) + H 2 O Hence (b) is a buffer system. 1161

22. (b) NH 3(aq) /NH 4 Cl (aq) Addition of H + : H + (aq) + NH 3(aq) NH 4 + (aq) Addition of OH - : OH - (aq) + NH 4 + (aq) NH 3(aq) + H 2 O Hence (b) is a buffer system. (c) H 2 PO 4 - (aq) / HPO 4 2- (aq) Addition of H + : H + (aq) + HPO 4 2- (aq) H 2 PO 4 - (aq) Addition of OH - : OH - (aq) + H 2 PO 4 - (aq) HPO 4 2- (aq) + H 2 O Hence (c) is a buffer system. 1162

23. (d)NaHCO 3(aq) 1163

24. (d)NaHCO 3(aq) (The Na + cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.) 1164

25. (d)NaHCO 3(aq) (The Na + cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.) Addition of H + : H + (aq) + HCO 3 - (aq) H 2 CO 3(aq) 1165

26. (d)NaHCO 3(aq) (The Na + cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.) Addition of H + : H + (aq) + HCO 3 - (aq) H 2 CO 3(aq) Addition of OH - : OH - (aq) + HCO 3 - (aq) CO 3 2- (aq) + H 2 O 1166

27. (d)NaHCO 3(aq) (The Na + cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.) Addition of H + : H + (aq) + HCO 3 - (aq) H 2 CO 3(aq) Addition of OH - : OH - (aq) + HCO 3 - (aq) CO 3 2- (aq) + H 2 O Hence (d) is a buffer system. 1167

28. (d)NaHCO 3(aq) (The Na + cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.) Addition of H + : H + (aq) + HCO 3 - (aq) H 2 CO 3(aq) Addition of OH - : OH - (aq) + HCO 3 - (aq) CO 3 2- (aq) + H 2 O Hence (d) is a buffer system. Note that in this example, the HCO 3 - ion functions as both the acid and the base. 1168

29. (d)NaHCO 3(aq) (The Na + cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.) Addition of H + : H + (aq) + HCO 3 - (aq) H 2 CO 3(aq) Addition of OH - : OH - (aq) + HCO 3 - (aq) CO 3 2- (aq) + H 2 O Hence (d) is a buffer system. Note that in this example, the HCO 3 - ion functions as both the acid and the base. Various anions of multi-proton acids can function as buffer systems with a single species present. 1169

30. IONIC EQUILIBRIUM Solubility Products 1170

31. IONIC EQUILIBRIUM Solubility Products A saturated solution of an insoluble salt is a heterogeneous equilibrium. 1171

32. IONIC EQUILIBRIUM Solubility Products A saturated solution of an insoluble salt is a heterogeneous equilibrium. Example: In a saturated solution of AgCl solution, the following equilibrium is present: 1172

33. IONIC EQUILIBRIUM Solubility Products A saturated solution of an insoluble salt is a heterogeneous equilibrium. Example: In a saturated solution of AgCl solution, the following equilibrium is present: AgCl (s) Ag + (aq) + Cl - (aq) 1173

34. IONIC EQUILIBRIUM Solubility Products A saturated solution of an insoluble salt is a heterogeneous equilibrium. Example: In a saturated solution of AgCl solution, the following equilibrium is present: AgCl (s) Ag + (aq) + Cl - (aq) The equilibrium constant is: 1174

35. Now 1175

36. Now but [AgCl] is a constant, recall 1176

37. Now but [AgCl] is a constant, recall 1177

38. Now but [AgCl] is a constant, recall We set where the subscript sp stands for solubility product. 1178

39. Now but [AgCl] is a constant, recall We set where the subscript sp stands for solubility product. Hence, 1179

40. Examples: MgF 2(s) Mg 2+ (aq) + 2 F - (aq) 1180

41. Examples: MgF 2(s) Mg 2+ (aq) + 2 F - (aq) 1181

42. Examples: MgF 2(s) Mg 2+ (aq) + 2 F - (aq) Ca 3 (PO 4 ) 2(s) 3 Ca 2+ (aq) + 2 PO 4 3- (aq) 1182

43. Examples: MgF 2(s) Mg 2+ (aq) + 2 F - (aq) Ca 3 (PO 4 ) 2(s) 3 Ca 2+ (aq) + 2 PO 4 3- (aq) 1183

44. Examples: MgF 2(s) Mg 2+ (aq) + 2 F - (aq) Ca 3 (PO 4 ) 2(s) 3 Ca 2+ (aq) + 2 PO 4 3- (aq) Note that the pure solids do not occur in the expression for K sp. 1184

45. Very small values for K sp indicate very insoluble salts. For example, K sp = 1.6 x 10 -10 for AgCl at 25 o C. 1185

46. Very small values for K sp indicate very insoluble salts. For example, K sp = 1.6 x 10 -10 for AgCl at 25 o C. In a solution containing Ag + (aq) and Cl - (aq) at 25 o C, we have one of the following situations: 1186

47. Very small values for K sp indicate very insoluble salts. For example, K sp = 1.6 x 10 -10 for AgCl at 25 o C. In a solution containing Ag + (aq) and Cl - (aq) at 25 o C, we have one of the following situations: unsaturated solution 1187

48. Very small values for K sp indicate very insoluble salts. For example, K sp = 1.6 x 10 -10 for AgCl at 25 o C. In a solution containing Ag + (aq) and Cl - (aq) at 25 o C, we have one of the following situations: unsaturated solution saturated solution 1188

49. Very small values for K sp indicate very insoluble salts. For example, K sp = 1.6 x 10 -10 for AgCl at 25 o C. In a solution containing Ag + (aq) and Cl - (aq) at 25 o C, we have one of the following situations: unsaturated solution saturated solution supersaturated solution 1189

50. Very small values for K sp indicate very insoluble salts. For example, K sp = 1.6 x 10 -10 for AgCl at 25 o C. In a solution containing Ag + (aq) and Cl - (aq) at 25 o C, we have one of the following situations: unsaturated solution saturated solution supersaturated solution Some AgCl precipitate will form until the product of the ionic concentrations is equal to 1.6 x 10 -10. 1190

51. Sample problem 1: In a saturated solution of Ag 2 CO 3, the concentrations of the ions are [Ag + ] = 2.54 x 10 -4 M and [CO 3 2- ] = 1.27 x 10 -4 M. Calculate the K sp and the solubility of Ag 2 CO 3 in g/liter. 1191

52. Sample problem 1: In a saturated solution of Ag 2 CO 3, the concentrations of the ions are [Ag + ] = 2.54 x 10 -4 M and [CO 3 2- ] = 1.27 x 10 -4 M. Calculate the K sp and the solubility of Ag 2 CO 3 in g/liter. Ag 2 CO 3(s) 2 Ag + (aq) + CO 3 2- (aq) 1192

53. Sample problem 1: In a saturated solution of Ag 2 CO 3, the concentrations of the ions are [Ag + ] = 2.54 x 10 -4 M and [CO 3 2- ] = 1.27 x 10 -4 M. Calculate the K sp and the solubility of Ag 2 CO 3 in g/liter. Ag 2 CO 3(s) 2 Ag + (aq) + CO 3 2- (aq) = (2.54 x 10 -4 ) 2 (1.27 x 10 -4 ) = 8.19 x 10 -12 1193

54. Sample problem 1: In a saturated solution of Ag 2 CO 3, the concentrations of the ions are [Ag + ] = 2.54 x 10 -4 M and [CO 3 2- ] = 1.27 x 10 -4 M. Calculate the K sp and the solubility of Ag 2 CO 3 in g/liter. Ag 2 CO 3(s) 2 Ag + (aq) + CO 3 2- (aq) = (2.54 x 10 -4 ) 2 (1.27 x 10 -4 ) = 8.19 x 10 -12 The concentration of CO 3 2- (aq) is equal to the number of moles of Ag 2 CO 3(s) that have dissolved. 1194

55. molar mass of Ag 2 CO 3 = 275.8 g/mol The solubility of Ag 2 CO 3 = 1.27 x 10 -4 mol l -1 275.8 g mol -1 = 0.0350 g/l 1195

56. Sample problem 2: Calculate the solubility of PbF 2 in g/l given K sp = 4.1 x 10 -8. 1196

57. Sample problem 2: Calculate the solubility of PbF 2 in g/l given K sp = 4.1 x 10 -8. PbF 2(s) Pb 2+ (aq) + 2 F - (aq) 1197

58. Sample problem 2: Calculate the solubility of PbF 2 in g/l given K sp = 4.1 x 10 -8. PbF 2(s) Pb 2+ (aq) + 2 F - (aq) The abbreviated ICE table for this problem looks like Pb 2+ F - 1198

59. Sample problem 2: Calculate the solubility of PbF 2 in g/l given K sp = 4.1 x 10 -8. PbF 2(s) Pb 2+ (aq) + 2 F - (aq) The abbreviated ICE table for this problem looks like Pb 2+ F - y 2y 1199

60. Sample problem 2: Calculate the solubility of PbF 2 in g/l given K sp = 4.1 x 10 -8. PbF 2(s) Pb 2+ (aq) + 2 F - (aq) The abbreviated ICE table for this problem looks like Pb 2+ F - y 2y 1200