1. 16.2 Buffers: Solutions That Resist pH Change
Casey Schulkind

2. Buffers
Buffer: a solution that resists pH change by neutralizing added acid or base
Buffer Characteristics:
Resists pH change & can maintain a nearly constant pH
When additional acid is added to a buffer, the conjugate base reacts with the acid, neutralizing it
Contains significant amounts of both a weak acid and its conjugate base (or a weak base and its conjugate acid)
The weak acid neutralizes added base
The base neutralizes added acid

3. Buffer consisting of a weak base and its conjugate acid
Buffer containing a weak acid (which can neutralize added base) and its conjugate base (which can neutralize added acid)
Buffer consisting of a weak base and its conjugate acid

4. Practice Which solution is a buffer?
A solution that is M in HNO₂ and M in HCl
A solution that is M in HNO₃ and M in NaNO₃
A solution that is M in HNO₂ and M in NaCl
A solution that is M in HNO₂ and M in NaNO₂
(d) A solution that is M in HNO₂ and M in NaNO₂

5. Calculating the pH of a Buffer Solution
Common Ion Effect: the tendency of a common ion to decrease the solubility of an ionic compound or to decrease the ionization of a weak acid or weak base
To find the pH of a buffer solution containing common ions:
Work an equilibrium problem in which the initial concentrations include both the acid and its conjugate base

6. HC₂H₃O₂ (aq) + H₂O (l) ⇋ H₃O⁺ (aq) + C₂H₃O₂⁻ (aq)
Practice
Calculate the pH of a buffer solution that is M in HC₂H₃O₂ and M in NaC₂H₃O₂. Kₐ = 1.8 x 10⁻⁵
HC₂H₃O₂ (aq) + H₂O (l) ⇋ H₃O⁺ (aq) + C₂H₃O₂⁻ (aq)
[HC₂H₃O₂]
[H₃O⁺]
[C₂H₃O₂⁻] I
0.100 ~0 C -x +x E x x x

7. (continued) X is Small: Kₐ = [H₃O⁺][C₂H₃O₂⁻] / [HC₂H₃O₂]
1.8 x 10⁻⁵ = (x)(0.100+x) / (0.100-x)
1.8 x 10⁻⁵ = x
X is Small:
((1.8 x 10⁻⁵) / 0.100) x 100% = 0.018%
pH= -log[H₃O⁺]
pH = -log(1.8 x 10⁻⁵)
pH = 4.74
[H₃O⁺] = x = 1.8 x 10⁻⁵

8. The Henderson-Hasselbalch Equation
pH = pKₐ + log([base] / [acid])
Allows us to calculate the pH of a buffer solution from the initial concentrations of the buffer components
*can be used only if x is small and if you are given the concentrations of both acid and base

9. Practice
Calculate the pH of a buffer solution that is M in benzoic acid (C₇H₆O₂) and M in sodium benzoate (NaC7H5O2). Kₐ=6.5 x 10-5
pH = pKₐ + log([base] / [acid])
pH = -log(6.5 x 10-5) + log(0.150 / 0.050)
pH =
pH = 4.66

10. Calculating pH Changes in a Buffer Solution
Although a buffer resists pH change as acid or base is added to it, the pH does change by a small amount
Calculating the pH change requires breaking up the problem into two parts:
Stoichiometric calculation: to calculate how the addition changes the relative amounts of acid and conjugate base
Instead of an ICE table, set up a “before addition, addition, after addition” table to track stoichiometric changes that occur during the neutralization of the added acid.
Equilibrium calculation: calculate the pH based on the new amounts of acid and conjugate base
Set up an ICE table, using the “after addition” values as initial concentrations

11. Practice
A 1.0L buffer solution contains mol HC₂H₃O₂ and mol NaC₂H₃O₂. Kₐ = 1.8 x Because the initial amounts of acid and conjugate base are equal, pH = pKₐ = -log(1.8 x 10-5) = Calculate the new pH after adding mol of solid NaOH to the buffer.
Part 1: Stoichiometry
Part 2: Equilibrium
HC₂H₃O₂ + H2O⇌ H3O+ + C₂H₃O₂-
OH-
HC₂H₃O₂
C₂H₃O₂-
Before addition
~0.00 mol
0.100 mol
Addition
0.010 mol
---
After Addition
0.090 mol
0.110 mol
[HC₂H₃O₂]
[H3O+]
[C₂H₃O₂-] I
0.090 ~0
0.110 C -x +x E x x x

12. (continued) pH = -log(1.47 x 10-5) Kₐ = [H3O+][C₂H₃O₂-] / [HC₂H₃O₂]
Kₐ = x(0.110+x) / x
1.8 x 10-5 = 0.110x / 0.090
X = [H3O+] = 1.47 x 10-5
pH = -log(1.47 x 10-5)
pH = 4.83
X is Small:
(1.47 x 10-5 / 0.090) x 100% = 0.016%

13. Buffers Containing a Base and its Conjugate Acid
When a buffer is composed of a base and its conjugate acid, the conjugate acid is the ion
To calculate the pH of a solution using the Henderson-Hasselbalch equation, first calculate pKₐ for the conjugate acid of the weak base
*Kₐ x Kb = Kw
*pKₐ + pKb = 14