Chapter 16 Aqueous Ionic Equilibrium

1. Chapter 16 Aqueous Ionic Equilibrium

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

3. HAc(aq) ⇌ H+(aq) + Ac−(aq) acid NaAc(aq) → Na+(aq) + Ac−(aq) Ac−(aq) + H2O(l) ⇌ HAc(aq) + OH−(aq) base Calculate pH for a mixture of HAc and Ac−

4. NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq) base NH4Cl(aq) → NH4+(aq) + Cl−(aq) NH4+(aq) ⇌ H+(aq) + NH3(aq) acid Calculate pH for a mixture of NH3 and NH4+

5. Example 16.1, page 716 Calculate the pH of a solution that contains 0.100 M HAc and 0.100 M NaAc. Ka for HAc is 1.8 x 10−5. 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. Example 16.3, page 722 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.

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

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

11. Example 16.2, page 718. Calculate the pH of a buffer solution that is 0.050 M in benzoic acid (HC7H5O2) and 0.150 M in sodium benzoate (NaC7H5O2). For benzoic acid, Ka = 6.5  10–5.

12. Example 16.4, page 725. Use the Henderson–Hasselbalch equation to calculate the pH of a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pKb = 4.75.

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 Ka for HAc is 1.8 x 10−5.

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

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

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 • 0 mL; b) 49.0 mL; • c) 50.0 mL; d) 51.0 mL; • e) 60.0 mL. • 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. Ka of HAc • is 1.8 x 10−5. • 0 mL; b) 49.0 mL; c) 50.0 mL; • d) 51.0 mL; e) 60.0 mL. • 2.87; b) 6.43; • c) 8.72; d) 11.00; • e) 12.00 A very similar example: page 733 -- 738

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. CaF2 (s) Ca2+ (aq) + 2F− (aq) H2O Solubility equilibrium is established Aqueous solution of CaF2 is saturated Ksp = [Ca2+][F−]2 solubility product constant “insoluble” salts

31. Write down the dissociation reactions and the expression of Ksp

32. Solid silver chromate is added to pure water at 25 °C. Equilibrium is achieved between undissolved Ag2CrO4(s) and its aqueous solution. Silver ion concentration is 1.3 x 10−4 mol/L. Calculate Ksp for this compound.

33. CaF2 (s) Ca2+ (aq) + 2F− (aq) Solubility equilibrium is established H2O Aqueous solution of CaF2 is saturated Solubility: the concentration of a saturated solution. Unit: mol/L or ∙ ∙ ∙ solubility ≠ solubility product constant Ksp Solubility ↔ Ksp

34. value of Kc or Kp 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 Ksp. Ksp = 4.0 x 10−8

36. Bismuth sulfide (Bi2S3) has a measured solubility of 1.0 x 10−15 mol/L at 25 °C. Calculate its Ksp. Ksp = 1.1 x 10−73 Try example 16.9 on page 746

37. The Ksp for copper(II) iodate, Cu(IO3)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. CaF2 (s) Ca2+ (aq) + 2F− (aq) H2O Increase [Ca2+] or [F−] → equilibrium shifts to left → solubility ↓ common ion effect Example 16.10, page 747 Calculate the solubility of CaF2 (Ksp = 1.46 x 10−10) in a 0.100 mol/L NaF solution.

39. For Mg(OH)2 Ksp = 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. Ksp of Mg(OH)2 is 1.8 x 10−11 What are the [Mg2+] and solubility?

42. CaF2 (s) Ca2+ (aq) + 2F− (aq) H2O Q = [Ca2+][F−]2 If Q > Ksp, precipitate will form If Q < Ksp, precipitate will not form

43. A solution is prepared by adding 750.0 mL of 4.00 x 10−3 mol/L Ce(NO3)3 to 300.0 mL of 2.00 x 10−2 mol/L KIO3. Will Ce(IO3)3 (Ksp = 1.9 x 10−10) precipitate from this solution? Try example 16.12 on page 750

44. 1 Ka for acetic acid (CH3COOH) is 1.8 x 10-5 while Ka 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 (CH3COO-) or chlorous (ClO-) ion? C. Calculate kb values for CH3COO- 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 10-10 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.