Chapter 15 Applications of Aqueous Equilibria

1. Chapter 15Applications of Aqueous Equilibria REACTIONS AND EQUILIBRIA INVOLVING ACIDS, BASES, AND SALTS

2. Common Ions Common ion effect - The addition of an ion already present (common) in a system causes equilibrium to shift away from the common ion.

3. Example 1 • Calculate the pH of 0.30 moles of acetic acid dissolved in 1.0 L of water. Ka = 1.8 X 10-5

4. Presence of the acetate ions initially causes the equilibrium position to be shifted to the left. In other words, less acid ionizes than normal and thus fewer hydrogen ions are formed and pH increases. Calculate the pH of a 0.30 moles of acetic acid and 0.30 moles of sodium acetate dissolved together in 1.0 liter of water. Ka = 1.8 X 10-5

5. Example 2 • Calculate the pH of a 0.075 M solution of C5H5N made in water (Kb = 1.7 X 10-9) and again made in a 0.0850 M solution of C5H5NCl.

6. Example 3 Calculate the pH and percent ionization in a 0.085 M solution of lactic acid (HLac) made in water (Ka = 1.4 X 10-4) and again made in a 0.50 M solution of sodium lactate.

7. Buffered Solutions Solutions that resist changes in pH when either OH- or H+ ions are added. Example: NH3/NH4+ buffer system HC2H3O2 / C2H3O2- buffer system

8. Usually contain a weak acid and its salt or a weak base and its salt. Pure water has no buffering capacity---acids and bases added to water directly affect the pH of the solution.

9. Systems that work • weak acid + salt of weak acid • weak base + salt of weak base • weak acid + ½ # of moles of strong base • weak base + ½ # of moles of strong acid • weak acid + weak base

10. How does it work? • Since a buffer consists of both an acid or base and its conjugate….an acid and a base are present in all buffer solutions. • If a small amount of strong acid is added to the buffer, there is a base component ready and waiting to neutralize the ‘invader’.

11. Buffer capacity The amount of acid or base that can be absorbed by a buffer system without a significant change in pH.

12. One way to calculate the pH of a buffer system is with the Henderson- Hasselbach equation. pH = pKa + log [base] [acid] pH = pKa + log [A-] [HA] Remember conjugate acid/base pairs!

13. The pH of a Buffered Solution A buffered solution contains 0.50M acetic acid (HC2H3O2, Ka = 1.8 X 10-5) and 0.50 M sodium acetate(NaC2H3O2). Calculate the pH. pH = 4.74

14. Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution in the example before. (pH = 4.74)

15. Compare this pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L of water.

16. Or you can use the McCormick equation…. [H+] = Ka [acid] [base]

17. If acid is added to the buffer, simply add acid to the numerator AND subtract the same quantity from the base since it was self-sacrificing and neutralized the acid. If base is added, simply add the base to the denominator and subtract from the numerator. When equal concentrations of Acid and Base are present [which occurs at the ½ equivalence point of a titration] the ratio of acid to base equals ONE and the pH = pKa.

18. Now, let’s try it. Same problem as before. Calculate the pH of 1.0 L of a buffer solution composed of 0.50 M acetic acid and 0.50 M sodium acetate before and after adding 0.010 mole solid NaOH. Ka = 1.8 X 10-5 before: [H+] = Ka [acid] = 1.8 X 10-5 [.50] = 1.8 X 10-5 [base] [.50] pH = 4.74 After: [H+] = 1.8 X 10-5 [.50-0.01] = 1.7 X 10-5 [.50 + 0.01] pH = 4.76

19. Preparing buffer solutions • Use 0.10 M to 1.0 M solutions of reagents & choose an acid whose Ka is near the [H3O+] concentration. • pKa should be as close as possible to the pH

20. A chemist needs a solution buffered at pH 4.30 and can choose from the following list of acids and their soluble salt: • chloroacetic acid Ka = 1.35 X 10-3 • propanoic acid Ka = 1.3 X 10-5 • benzoic acid Ka = 6.4 X 10-5 • hypochlorous acid Ka = 3.5 X 10-8