Acid-Base Buffers and the Henderson-Hasselbalch Equation

Acid-Base Buffers

Definition and Composition of Buffers

A buffer solution is a mixture that resists significant changes in pH when small amounts of acid or base are added. Buffers are typically composed of either:

• A weak acid and its conjugate base

• A weak base and its conjugate acid

For example, a buffer can be made from acetic acid (CH3COOH) and sodium acetate (CH3COONa), or from ammonia (NH3) and ammonium chloride (NH4Cl).

How Buffers Work

Buffers function by neutralizing added acids (H+) or bases (OH-) through equilibrium reactions involving the weak acid/base and its conjugate. This minimizes changes in the concentration of H+ ions, thus stabilizing the pH.

• When acid (H+) is added, the conjugate base component of the buffer reacts to form the weak acid.

• When base (OH-) is added, the weak acid component reacts to form water and the conjugate base.

Example: In an acetic acid/acetate buffer:

• Added H+: $\mathrm{CH_3COO^-} + \mathrm{H^+} \rightarrow \mathrm{CH_3COOH}$

• Added OH-: $\mathrm{CH_3COOH} + \mathrm{OH^-} \rightarrow \mathrm{CH_3COO^-} + \mathrm{H_2O}$

Buffer Capacity

Buffer capacity refers to the amount of acid or base a buffer can neutralize before the pH changes significantly. It depends on the concentrations of the buffer components; higher concentrations provide greater capacity.

Buffer Calculations and the Henderson-Hasselbalch Equation

Equilibrium Expression for Weak Acids

The ionization of a weak acid (HA) in water is represented as:

$\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}$

The acid dissociation constant ($K_a$) is:

$K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a convenient way to calculate the pH of a buffer solution:

$\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\mathrm{A^-}]}{[\mathrm{HA}]} \right)$

• $\mathrm{p}K_a = -\log K_a$

• $[\mathrm{A^-}]$ = concentration of conjugate base

• $[\mathrm{HA}]$ = concentration of weak acid

This equation assumes that the concentrations of acid and conjugate base do not change significantly upon addition of small amounts of acid or base.

Special Cases

• If $[\mathrm{A^-}] = [\mathrm{HA}]$, then $\mathrm{pH} = \mathrm{p}K_a$

• If $[\mathrm{A^-}] > [\mathrm{HA}]$, then $\mathrm{pH} > \mathrm{p}K_a$

• If $[\mathrm{A^-}] < [\mathrm{HA}]$, then $\mathrm{pH} < \mathrm{p}K_a$

Buffer Action: Examples and Calculations

Example: Calculating pH of a Buffer Solution

Given a buffer containing 0.10 M acetic acid and 0.10 M sodium acetate, with $K_a = 1.8 \times 10^{-5}$:

• $\mathrm{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74$

• $\mathrm{pH} = 4.74 + \log\left(\frac{0.10}{0.10}\right) = 4.74$

Adding Strong Base: If 0.010 mol of NaOH is added to 1.0 L of this buffer:

• NaOH reacts with acetic acid: $\mathrm{CH_3COOH} + \mathrm{OH^-} \rightarrow \mathrm{CH_3COO^-} + \mathrm{H_2O}$

• New concentrations: $[\mathrm{CH_3COOH}] = 0.09$ M, $[\mathrm{CH_3COO^-}] = 0.11$ M

• $\mathrm{pH} = 4.74 + \log\left(\frac{0.11}{0.09}\right) = 4.86$

The pH changes only slightly, demonstrating buffer action.

Example: Calculating pH After Addition of Strong Acid

If 0.010 mol of HCl is added to the same buffer:

• HCl reacts with acetate: $\mathrm{CH_3COO^-} + \mathrm{H^+} \rightarrow \mathrm{CH_3COOH}$

• New concentrations: $[\mathrm{CH_3COOH}] = 0.11$ M, $[\mathrm{CH_3COO^-}] = 0.09$ M

• $\mathrm{pH} = 4.74 + \log\left(\frac{0.09}{0.11}\right) = 4.62$

Again, the pH changes only slightly.

Table: Buffer Action Comparison

The following table summarizes the effect of adding strong acid or base to a buffered versus an unbuffered solution:

Solution Type	Initial pH	pH after adding acid	pH after adding base
Buffered (acetic acid/acetate)	4.74	4.62	4.86
Unbuffered (water)	7.00	2.00	12.00

Additional info: Table values for unbuffered solution are typical for strong acid/base addition to pure water.

Summary of Key Equations

• $K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$

• $\mathrm{p}K_a = -\log K_a$

• $\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\mathrm{A^-}]}{[\mathrm{HA}]} \right)$

Applications of Buffers

• Biological systems (e.g., blood plasma buffer)

• Industrial processes (e.g., fermentation, pharmaceuticals)

• Laboratory experiments requiring stable pH