1. Chapter 16 Lecture 1 Buffers I.The Common Ion Effect A.Addition of a salt containing the conjugate base (acid) to a solution of a weak acid (base) decreases its ionization B.Example: A solution of 1.0 M HF and 1.0 M NaF 1)HF H + + F - K a = 7.2 x 10 -4 2)NaF Na + + F - 3)F - is the common ion of these two equations 4)By Le Chatelier’s Principle, the excess F - forces the HF equilibrium to the left. Since less H + is produced, the solution is less acidic that HF alone. C.Example: A solution of 1.0 M NH 3 and 1.0 M NH 4 Cl 1)NH 3 + H 2 O NH 4 + + OH - K b = 1.8 x 10 -5 2)NH 4 Cl NH 4 + + Cl - 3)The common ion NH 4 + forces the equilibrium to the left, and the solution is less basic that NH 3 alone

2. D.Polyprotic Acids are Effected by the Common Ion Effect 1.H 3 PO 4 H + + H 2 PO 4 - 2.H 2 PO 4 - H + + HPO 4 2- 3.H + from the first ionization inhibits the second ionization E.Solving Common Ion Problems 1.The only difference from previous weak acid or weak base problems, is that now, [A - ] 0 or [HB + ] 0 ≠ 0 2.Example: Find [H + ] and % Diss. of 1.0 M HF + 1.0 M NaF a.In Ch. 15, we found [H + ] and % Diss. of 1.0 M HF = 0.027 and 2.7% b.This time, we have F - common ion = [A - ] 0 II.Buffers A.A Buffer is a solution containing a weak acid and the salt of its conjugate base (or a weak base and the salt of its conjugate acid) which resist changes to its pH. 1)Example: Blood stays at pH = 7.4, even when various reactions in the body produce much H + and OH - 2)The buffer system in blood is H 2 CO 3 /HCO 3 -

4. 3)Different buffers maintain different pH’s a)The NH 3 /NH 4 + buffer maintains a pH = 8.0 b)Example: What is pH of 0.5 M HC 2 H 3 O 2 /0.5 M NaC 2 H 3 O 2 ? (K a = 1.8 x 10 -5 ) 4)What happens when we add acid or base to a buffered system? Example: Add 0.01 mol of NaOH to 1 L of the acetic acid buffer in the last Example. What will be the new pH? a)NaOH will completely dissociate to OH - b)OH - will deprotonate HC 2 H 3 O 2 to completion because OH - is a strong base c)Carry out a new equilibrium problem after considering the stoichiometry of these reactions. HC 2 H 3 O 2 H + + C 2 H 3 O 2 - Initial 0.50 0.5 After OH- 0.490 0.51 Change -x +x +x Equililibrium 0.49 – x x 0.51 + x x = 1.7 x 10 -5 pH = 4.76

5. d)Before adding the NaOH, the pH = 4.74. Change is only 0.02 units!! e)What happens if we add 0.01 mol NaOH to 1 L of water? i.Before, pH = 7.00 ii.After, pH = 12.00 (change of 5.00 units) 5)Hints for Buffer Problems a)Buffer problems are just common ion problems b)Do the stoichiometry of OH - /H + addition first c)Then do the equilibrium problem with the new concentrations B.How a Buffer Works 1)Buffers have a large concentration of both HA and A - 2)When we add OH -, it is used up by reaction with HA and replaced in solution with A - 3)When we add H +, it is used up by reaction with A - and replaced in solution by HA 4)HA H + + A -

6. pH depends on [HA]/[A - ] i.When we add OH-, [HA] decreases and [A - ] increases ii.But, if [HA] and [A - ] are large, their ratio won’t change much iii.If the ratio doesn’t change much, the pH doesn’t change much Since ratio changes little, the pH changes little 5)When we add H +, the strong base A - uses it up, again only with a small change in the ratio of [HA]/A - ] 6)Example: Find the pH of 0.10 M HF (K a = 7.2 x 10 -4 ) and 0.30 M NaF pH = 3.62 In the last Example:

7. 7)Henderson-Hasselbalch Equation for Buffers: a)We can use this simple equation instead of doing the longer equilibrium problem approach for most buffer problems b)Any concentration of buffer with the same [A - ]/[HA] ratio will have the same pH c)The Henderson-Hasselbalch Equation assumes that [A - ] and [HA] are the same as [A - ] 0 and [HA] 0. If we keep the buffer concentration high, this is a valid approximation. d)Example: Find pH of a buffer made of 0.75 M lactic acid (K a = 1.4 x 10 -4 ) and 0.25 M Sodium Lactate.

8. C.Buffers made up of a weak base and the salt of its conjugate acid. 1.We still have an A - and HA, they are now the weak base and its conjugate acid, respectively. We just call them B and BH + in the case of a base. B + H 2 O BH + + OH - 2.We can still use the Henderson-Hasselbalch Equation, as long as we recall that the protonated species is on the bottom. a)Example: Find the pH of 0.25 M NH 3 (K b = 1.8 x 10 -5 ) and 0.4 M NH 4 Cl. i. ii.K a = K W /K b = 5.6 x 10 -10, so pK a = 9.26 b)Example: Find pH of above solution after we add 0.1 mol of H + to 1 L of it. NH 3 + H + NH 4 + Initial 0.25 0.1 0.40 After H + 0.15 00.50

9. D.Buffer Capacity = amount of H + /OH - a buffer can absorb with only a small pH change 1.Remember, the pH of a buffer depends only on the ratio [A - ]/[HA] 2.Buffer capacity depends on the size of [A - ] and [HA] 3.Example: Find  pH for 0.01 mol of HCl added to: a)5 M HC 2 H 3 O 2 / 5 M NaC 2 H 3 O 2 (K a = 1.8 x 10 -5 ) b)0.05 M HC 2 H 3 O 2 / 0.05 M NaC 2 H 3 O 2 i.pH = pK a + log(1) = pK a + 0 = 4.74 for both ii.Large Concentration: C 2 H 3 O 2 - +H + HC 2 H 3 O 2 Initial 5 0.01 5 After H + 4.99 0 5.01 pH = 4.74 + log(4.99/5.01) = 4.74 iii.Small Concentration: Initial 0.05 0.01 0.05 After H + 0.04 0 0.06 pH = 4.74 + log(0.04/0.06) = 4.56

10. 4)What are optimal buffering conditions? a)Try to avoid large changes in [A - ]/[HA] ratio b)Best case is when [A - ] = [HA] i.Add 0.01 mol of H + to 1 L of a weak acid buffer where [A-] = [HA] = 1 M. Change in ratio is from 1.00  0.98 ii. Add 0.01 mol of H + to 1 L of a weak acid buffer where [A-] = 1 M and [HA] = 0.01 M. Change in ratio is from 100  49.5 c)Choose a buffer whose pK a value is near the pH you want to maintain. i.pH = pK a + log([A - ]/HA]) ii.If [A - ] = [HA], then pH = pK a + log(1) = pK a + 0 = pK a iii.Buffers most effective if: d)Example: Choose the best buffer for pH = 4.30 i.chloroacetic acid (K a = 1.35 x 10 -3 ) propanoic acid (K a = 1.3 x 10 -5 ) benzoic acid (K a = 6.4 x 10 -5 ) hypochlorous acid (K a = 3.5 x 10 -8 ) ii.Chloroacetic: pK a = 2.87 iii.Propanoic: pK a = 4.89 iv.Benzoic: pK a = 4.19 v.Hypochlorous: pK a = 7.46 Benzoic Acid is Best Buffer