Chapter 16 Lecture 1 Buffers

1. Chapter 16 Lecture 1 Buffers • The Common Ion Effect • Addition of a salt containing the conjugate base (acid) to a solution of a weak acid (base) decreases its ionization • Example: A solution of 1.0 M HF and 1.0 M NaF • HF H+ + F- Ka = 7.2 x 10-4 • NaF Na+ + F- • F- is the common ion of these two equations • 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. • Example: A solution of 1.0 M NH3 and 1.0 M NH4Cl • NH3 + H2O NH4+ + OH- Kb = 1.8 x 10-5 • NH4Cl NH4+ + Cl- • The common ion NH4+ forces the equilibrium to the left, and the solution is less basic that NH3 alone

2. Polyprotic Acids are Effected by the Common Ion Effect • H3PO4 H+ + H2PO4- • H2PO4- H+ + HPO42- • H+ from the first ionization inhibits the second ionization • Solving Common Ion Problems • The only difference from previous weak acid or weak base problems, is that now, [A-]0 or [HB+]0≠ 0 • Example: Find [H+] and % Diss. of 1.0 M HF + 1.0 M NaF • In Ch. 15, we found [H+] and % Diss. of 1.0 M HF = 0.027 and 2.7% • This time, we have F- common ion = [A-]0 • Buffers • 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. • Example: Blood stays at pH = 7.4, even when various reactions in the body produce much H+ and OH- • The buffer system in blood is H2CO3/HCO3-

4. Different buffers maintain different pH’s • The NH3/NH4+ buffer maintains a pH = 8.0 • Example: What is pH of 0.5 M HC2H3O2/0.5 M NaC2H3O2? (Ka = 1.8 x 10-5) • 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? • NaOH will completely dissociate to OH- • OH- will deprotonate HC2H3O2 to completion because OH- is a strong base • Carry out a new equilibrium problem after considering the stoichiometry of these reactions. HC2H3O2 H+ + C2H3O2- Initial 0.5 0 0.5 After OH- 0.49 0 0.51 Change -x +x +x Equililibrium 0.49 – x x 0.51 + x x = 1.7 x 10-5 pH = 4.76

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

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

7. Henderson-Hasselbalch Equation for Buffers: • We can use this simple equation instead of doing the longer equilibrium problem approach for most buffer problems • Any concentration of buffer with the same [A-]/[HA] ratio will have the same pH • 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. • Example: Find pH of a buffer made of 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M Sodium Lactate.

8. Buffers made up of a weak base and the salt of its conjugate acid. • 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 + H2O BH+ + OH- • We can still use the Henderson-Hasselbalch Equation, as long as we recall that the protonated species is on the bottom. • Example: Find the pH of 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.4 M NH4Cl. • Ka = KW/Kb = 5.6 x 10-10, so pKa = 9.26 • Example: Find pH of above solution after we add 0.1 mol of H+ to 1 L of it. NH3 + H+ NH4+ Initial 0.25 0.1 0.40 After H+ 0.15 0 0.50

9. Buffer Capacity = amount of H+/OH- a buffer can absorb with only a small pH change • Remember, the pH of a buffer depends only on the ratio [A-]/[HA] • Buffer capacity depends on the size of [A-] and [HA] • Example: Find DpH for 0.01 mol of HCl added to: • 5 M HC2H3O2/ 5 M NaC2H3O2 (Ka = 1.8 x 10-5) • 0.05 M HC2H3O2/ 0.05 M NaC2H3O2 • pH = pKa + log(1) = pKa + 0 = 4.74 for both • Large Concentration: C2H3O2- + H+ HC2H3O2 Initial 5 0.01 5 After H+ 4.99 0 5.01 pH = 4.74 + log(4.99/5.01) = 4.74 • 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. Benzoic Acid is Best Buffer • What are optimal buffering conditions? • Try to avoid large changes in [A-]/[HA] ratio • Best case is when [A-] = [HA] • 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 pKa value is near the pH you want to maintain. • pH = pKa + log([A-]/HA]) • If [A-] = [HA], then pH = pKa + log(1) = pKa + 0 = pKa • Buffers most effective if: • Example: Choose the best buffer for pH = 4.30 • 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) • Chloroacetic: pKa = 2.87 • Propanoic: pKa = 4.89 • Benzoic: pKa = 4.19 • Hypochlorous: pKa = 7.46