1. Chapter 16 Aqueous Ionic Equilibrium

2. pH / pOH Calculations Strong acids Strong bases Weak acids Weak bases Salts

3. NaAc(aq) → Na + (aq) + Ac − (aq) Calculate pH for a mixture of HAc and Ac − HAc(aq) ⇌ H + (aq) + Ac − (aq) acid Ac − (aq) + H 2 O(l) ⇌ HAc(aq) + OH − (aq) base

4. NH 4 Cl(aq) → NH 4 + (aq) + Cl − (aq) Calculate pH for a mixture of NH 3 and NH 4 + NH 4 + (aq) ⇌ H + (aq) + NH 3 (aq) acid NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + OH − (aq) base

5. Calculate the pH of a solution that contains 0.100 M HAc and 0.100 M NaAc. K a for HAc is 1.8 x 10 −5. Example 16.1, page 716 pH = 4.74

6. weak acid + its conjugate base = buffer solution weak base + its conjugate acid = buffer solution Adding H + or OH −, pH does not change too much

7. Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution described in Example 16.1. Compare this pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L of water. Assume there is no volume change after solid NaOH is added. Example 16.3, page 722

8. Henderson-Hasselbalch equation for buffer solutions pH for buffer solutions: ICE → exact answer pH for buffer solutions: approximation

10. Calculate the pH of a solution that contains 0.50 mol/L HAc and 0.50 mol/L NaAc. K a for HAc is 1.8 x 10 −5. pH = 4.74 Example 16.1, page 716. revisited

11. Calculate the pH of a buffer solution that is 0.050 M in benzoic acid (HC 7 H 5 O 2 ) and 0.150 M in sodium benzoate (NaC 7 H 5 O 2 ). For benzoic acid, K a = 6.5  10 –5. Example 16.2, page 718.

12. Use the Henderson–Hasselbalch equation to calculate the pH of a buffer solution that is 0.50 M in NH 3 and 0.20 M in NH 4 Cl. For ammonia, pK b = 4.75. Example 16.4, page 725.

13. Calculate the change in pH that occurs when 0.010 mol gaseous HCl is added to 1.0 L of each of the following solutions: Solution A: 5.00 mol/L HAc and 5.00 mol/L NaAc Solution B: 0.050 mol/L HAc and 0.050 mol/L NaAc K a for HAc is 1.8 x 10 −5.

14. HAc Ac − HAc Ac − H+H+ HAc Ac − H+H+ Solution A HAc Ac − Solution B

15. Buffer capacity To maximize buffer capacity: 1) High [HA] and [A − ] 2) [HA] = [A − ] (example page 725-726) pH = pK a

16. Titration

17. Titrate strong acid with strong base

19. Calculate the pH when the following quantities of 0.100 mol/L NaOH solution have been added to 50.0 mL of 0.100 mol/L HCl solution a)0 mL;b) 49.0 mL; c) 50.0 mL;d) 51.0 mL; e) 60.0 mL. a)1.000;b) 3.00; c) 7.00;d) 11.00; e) 11.96

23. Titrate weak acid with strong base

24. Calculate the pH when the following quantities of 0.100 mol/L NaOH solution have been added to 50.0 mL of 0.100 mol/L HAc solution. K a of HAc is 1.8 x 10 −5. a)0 mL;b) 49.0 mL;c) 50.0 mL; d) 51.0 mL;e) 60.0 mL. A very similar example: page 733 -- 738 a)2.87;b) 6.43; c) 8.72;d) 11.00; e) 12.00

25. Chapter 17

27. The pH Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids with Various Ka Values with 0.10 M NaOH

28. Solubility Equilibrium

29. CaF 2 (s)Ca 2+ (aq) + 2F − (aq) H2OH2O Solubility equilibrium is established Aqueous solution of CaF 2 is saturated K sp = [Ca 2+ ][F − ] 2 solubility product constant “insoluble” salts

31. Write down the dissociation reactions and the expression of K sp

32. Solid silver chromate is added to pure water at 25 °C. Equilibrium is achieved between undissolved Ag 2 CrO 4 (s) and its aqueous solution. Silver ion concentration is 1.3 x 10 −4 mol/L. Calculate K sp for this compound.

33. CaF 2 (s)Ca 2+ (aq) + 2F − (aq) H2OH2O Solubility equilibrium is established Aqueous solution of CaF 2 is saturated Solubility: the concentration of a saturated solution. Unit: mol/L or ∙ ∙ ∙ solubility ≠ solubility product constant K sp Solubility ↔ K sp

34. value of K c or K p equilibrium concentrations or pressures

35. Copper(I) bromide has a measured solubility of 2.0 x 10 −4 mol/L at 25 °C. Calculate its K sp. K sp = 4.0 x 10 −8

36. Bismuth sulfide (Bi 2 S 3 ) has a measured solubility of 1.0 x 10 −15 mol/L at 25 °C. Calculate its K sp. K sp = 1.1 x 10 −73 Try example 16.9 on page 746

37. The K sp for copper(II) iodate, Cu(IO 3 ) 2, is 1.4 x 10 −7 at 25 °C. Calculate its solubility at 25 °C. 3.3 x 10 −3 M Try example 16.8 on page 745

38. Calculate the solubility of CaF 2 (K sp = 1.46 x 10 −10 ) in a 0.100 mol/L NaF solution. Example 16.10, page 747 CaF 2 (s)Ca 2+ (aq) + 2F − (aq) H2OH2O Increase [Ca 2+ ] or [F − ] → equilibrium shifts to left → solubility ↓ common ion effect

39. For Mg(OH) 2 K sp = 1.8 x 10 −11. What is the pH of a saturated solution of Mg(OH) 2 ? What is its solubility?

40. Suppose that solid Mg(OH) 2 is equilibrated with a solution buffered at a more acidic pH of 9.00. K sp of Mg(OH) 2 is 1.8 x 10 −11 What are the [Mg 2+ ] and solubility?

42. CaF 2 (s)Ca 2+ (aq) + 2F − (aq) H2OH2O Q = [Ca 2+ ][F − ] 2 If Q > K sp, precipitate will form If Q < K sp, precipitate will not form

43. A solution is prepared by adding 750.0 mL of 4.00 x 10 −3 mol/L Ce(NO 3 ) 3 to 300.0 mL of 2.00 x 10 −2 mol/L KIO 3. Will Ce(IO 3 ) 3 (K sp = 1.9 x 10 −10 ) precipitate from this solution? Try example 16.12 on page 750

44. 1 K a for acetic acid (CH 3 COOH) is 1.8 x 10 -5 while K a for hypochlorous (HClO) ion is 3.0 x 10 -8. A. Which acid is the stronger acid? B. Which is the stronger conjugate base? Acetate ion (CH 3 COO - ) or chlorous (ClO - ) ion? C. Calculate k b values for CH 3 COO - and ClO -. 2.a. Calculate the pH of a 1.50 L solution containing 0.750 mole of HCN and 0.62 mole of KCN. Ka = 4.0 x 10 -1 0 b. If 0.015 mole of KOH was added, calculate the pH of the solution. c. If 0.015 mole of HBr was added, calculate the pH of the solution.