1. [17.2] Buffers

2. Buffer: a solution that resists a change in pH The best buffer has large and equal amounts of proton donors (weak acid to neutralize OH - ) and proton acceptors (salt to neutralize H + ) Weak acid & its salt Weak base & its salt ½ titration point (save this for tomorrow)

3. Buffers: Solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.

4. Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH − to make F − and water.

5. Buffers If acid is added, the F − reacts to form HF and water.

6. Example 1 Calculate the pH of a solution of 1.00 M HNO 2 and 1.00 M NaNO 2. (K a = 4.0 x 10 -4 ) HNO 2 ↔ H + + NO 2 - adding acids Adding base OH - I1.00Negligible1.00 C-x+x E1.00 – xx1.00 + x

7. Ka = [H + ][NO 2 - ] [HNO 2 ] 4.0 x 10 -4 = (x)(1.00 +x) (1.00 –x) x = [H+] = 4.0 x 10 -4 M pH =

8. Example 2 A buffered solution is made by adding 75.0 g sodium acetate to 500.0 mL of 0.64 M acetic acid. What is the final pH? (assume no volume change; K a = 1.8 x 10 -5 ) HC 2 H 3 O 2 ↔ H + + C 2 H 3 O 2 - I0.64 M ≈ 0 75.0 g x 1 mole/82.03 g = 0.914 mole /0.500 L = 1.83 M C-x+x E0.64 – xx1.83 + x

9. Ka = [H + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] 1.8 x 10 -5 = (x)(1.83 +x) (0.64–x) x = [H+] = 6.3 x 10 -6 M pH = Or use Hendersen-Hasselbach equation pH = pK a + log[base] [acid] pH = -log(1.8 x 10 -5 ) + log (1.83/0.64) = 4.74 + 0.46 = 5.20

10. pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a pK a close to the desired pH. The best buffers can absorb most additions without changing pH significantly Large amount of acceptor = large amount of donor

11. When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.

12. Addition of Strong Acid or Base to a Buffer 1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

13. Example 3 Calculate pH after 0.010 mole HCl is added to 250.0 mL of 0.050 M NH 3 & 0.15 M NH 4 Cl (K a = 5.6 x 10 -10 ) NH 3 + H +  NH 4 + I0.050 M0.010/0.250 L =.040 M 0.15 M C-0.040M +0.040 M E0.010 M00.19 M A buffer still exists after H+ reacts completely. Use Hendersen- Hasselbach to calculate pH.

14. pH = pK a + log[base] [acid] pH = -log(5.6 x 10 -10 ) + log (0.010/0.19) = 7.97 In choosing buffers: pH = pKa + log ([A-]/[HA]) desired pH Ratio = 1 for most effective buffer!

15. Example 4 Which acid in table 14.2 (Zumdahl) would be (a) the best choice for making a pH = 7.00 buffer? pH = 7.00 best buffer would have K a = 1.0 x 10 -7

16. Buffer HOCl K a = 3.5 x 10 -8 pK a = 7.46 (b) How would you make this buffer? To make buffer, we need to calculate [base]/[acid] ratio. 7.00 = 7.46 + log [B]/[A] 7.00 = 7.46 + (-0.46) [OCl - ]/[HOCl] = 10 -0.46 =0.35 Any OCl - /HOCl buffer in a concentration ratio of 0.35 : 1 will have a pH of 7.00 One possibility is [NaOCl] = 0.35 M & [HOCl] = 1.0 M