1. Chapter 17 Buffers

2. Buffered solutions l A solution that resists a change in pH. l Buffers are: –A solution that contains a weak acid- weak base conjugate pair. –Often prepared by mixing a weak acid, or a weak base, with a salt of that acid or base.

3. Buffered solutions l The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. l We can make a buffer of any pH by varying the concentrations of these solutions.

4. What is the pH of a buffer that is 0.12M in lactic acid, HC 3 H 5 O 3, and 0.10M in sodium lactate? For lactic acid, K = 1.4 x 10 -4 I0.12 M00.10 M C-x+x E0.12 – x+x0.10 + x HC 3 H 5 O 3 ↔ H + + C 3 H 5 O 3 -1

5. [ H +1 ] [C 3 H 5 O 3 -1 ] Ka = [HC 3 H 5 O 3 ] 1.4 x 10 -4 = (x) (0.10 – x) (0.12 – x) Because of the small K a and the presence of the common ion, it is expected that x will be small relative to 0.12 or 0.10M. x = 1.7 x 10 -4 M therefore [H +1 ] = 1.7 x 10 -4 M pH = -log (1.7 x 10 -4 M) = 3.77

6. Use Henderson-Hasselbach Instead [base] pH = pK a + log [acid] [ 0.10] pH = -log (1.4 x 10 -4 ) + log [0.12] pH = 3.85 + (-0.08) = 3.77

7. Practice Problem Calculate the pH of a buffer composed of 0.12M benzoic acid (HC 7 H 5 O 2 ) and 0.20M sodium benzoate. K a = 6.3 x 10 -5 ans: 4.42 Remember the Aqueous Equilibrium Constants are located in Appendix D of your textbook.

8. Buffered Solutions l Buffers resist changes in pH because they contain both an acidic species to neutralize OH -1 ions and a basic one to neutralize H +1 ions. l It is important that the acidic and basic species of the buffer do not consume each other through a neutralization reaction.

9. Buffered Solutions l HA H + + A - l K a = [H + ] [A - ] [HA] l Buffers most effectively resist a change in pH in either direction when the concentrations of HA and A - are about the same. When [HA] equals [A - ] then [H + ] equals K a. l Scientists usually try to select a buffer whose acid form has a pK a close to the desired pH.

10. General equation l Ka = [H + ] [A - ] [HA] l so [H + ] = K a [HA] [A - ] l The [H + ] depends on the ratio [HA]/[A - ] l Take the negative log of both sides l pH = -log(K a [HA]/[A - ]) l pH = -log(K a )-log([HA]/[A - ]) l pH = pK a + log([A - ]/[HA])

11. This is called the Henderson- Hasselbach equation l pH = pK a + log([A - ]/[HA]) l pH = pK a + log(base/acid)

12. Try an Acid l Calculate the pH of the following mixture: l Prob. #1: 0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) l Answer: l pH = -log(1.4x10 -4 ) + log[0.25/0.75] l pH = 3.38

13. Now Try a Base l Calculate the pH of the following mixture: l 0.25 M NH 3 and 0.40 M NH 4 Cl l (K b = 1.8 x 10 -5 ) l Answer: l NH 3 NH 4 +1 + OH -1 l (NH 3 is B, NH 4 +1 is HB +1 ) l pOH = pK b + log[ HB +1 / B ] l pOH = -log(1.8x10 -5 ) + log[0.4/0.25] l pOH – 4.94 now convert to pH

14. Buffering Capacity l This is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree.

15. Buffering Capacity l The buffering capacity, that is, the effectiveness of the buffer solution, depends on the amount of acid and conjugate base from which the buffer is made. l The larger the amount the greater the buffering capacity. l In general, a buffer system can be represented as salt-acid or conjugate base-acid.

16. **Buffer capacity** l The pH of a buffered solution is determined by the ratio [A - ]/[HA]. l As long as this doesn’t change much the pH won’t change much. l The more concentrated these two are the more H + and OH - the solution will be able to absorb. l Larger concentrations means bigger buffer capacity.

17. Buffer Capacity l Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following: l 5.00 M HAc and 5.00 M NaAc l 0.050 M HAc and 0.050 M NaAc l Ka= 1.8x10 -5

18. Buffer capacity l The best buffers have a ratio [A - ]/[HA] = 1 l This is most resistant to change l True when [A - ] = [HA] l Make pH = pKa (since log1=0)

19. Addition of Strong Acids or Bases to Buffers l See pg. 648 B&L for explanation and examples.

20. Adding a strong acid or base l Do the stoichiometry first. l A strong base will take protons from the weak acid reducing [HA] 0 l A strong acid will add its proton to the anion of the salt reducing [A - ] 0 l Then do the equilibrium problem.

21. B & L pg 649

25. l Try the Buffer worksheet from the “Chang” chemistry text.

26. Prove they’re buffers l What would the pH be if 0.020 mol of HCl is added to 1.0 L of both of the preceding solutions. l What would the pH be if 0.050 mol of solid NaOH is added to each of the proceeding.

27. l Thanks to Mr. Green