Henderson’s Equation of Acidic Buffer-Limitation, and Solved Examples

What is Henderson’s Equation of Acidic Buffer?

The Henderson-Hasselbalch equation describes the relationship between the pH, pKa, and concentrations of the weak acid and conjugate base that make up a buffered solution. It is derived from the acid dissociation constant expression.

The equation is:

pH = pKa + log([A^–]/[HA])

or

pH=pKa+log(salt/acid)

Where:

pH = the pH of the buffered solution

pKa = the negative log of the acid dissociation constant of the weak acid

[A^–] = the molar concentration of the conjugate base (anion) of the weak acid

[HA] = the molar concentration of the weak acid

Derivation of Henderson’s Equation of Acidic Buffer

We’ll start with the acid dissociation equilibrium of a weak acid HA:

HA + H₂O ⇌ H₃O⁺ + A^–

The acid dissociation constant Ka for this reaction is:

Ka = [H₃O⁺][A^–]/[HA]

Take the negative log of both sides:

-log(Ka) = -log([H₃O⁺][A^–]/[HA])

pKa = pH + log([A^–]/[HA])

Rearrange:

pH = pKa + log([A^–]/[HA])

This is the Henderson-Hasselbalch equation.

The key steps are:

1. Writing out the dissociation equilibrium of the weak acid HA.
2. Expressing Ka in terms of the equilibrium concentrations.
3. Taking the negative log of both sides to convert Ka to pKa.
4. Substituting -log[H3O+] = pH
5. Rearranging to isolate pH.

The derivation relies on the equilibrium constant expression for Ka and taking logs to convert between Ka and pKa.

Limitations of the Henderson-Hasselbalch Equation

• It only applies to buffered solutions containing a weak acid and its conjugate base. It does not apply to buffers made from a weak base and its conjugate acid.
• The weak acid and conjugate base must be in equilibrium – no other major disturbances to the equilibrium like additions of strong acid/base.
• It assumes ideal behavior – the activity coefficients of the weak acid and conjugate base are taken as 1. This may not be valid at very high or very low concentrations.
• It assumes the pKa value is constant and not affected by changes in temperature, ionic strength, etc. In reality, pKa values can shift.
• It assumes the solution is dilute – typically valid for concentrations < 0.1 M. At higher concentrations of buffer components, deviations from ideality are more likely.
• It assumes the pH is far enough from the pKa that [A^–] and [HA] do not become comparable to the concentration of water (55 M). Extremely high or low pH conditions violate this.
• It assumes measurements of [A^–], [HA], pH, and pKa are all at the same temperature. Temperature changes can shift pH and pKa.

Solved Example

• A buffer solution contains 0.20 M of acetic acid (CH₃COOH) and 0.15 M of sodium acetate (CH₃COO^–Na⁺)
• The pKa of acetic acid is 4.76

We are asked to calculate the pH of this buffered solution.

Solution:

1. Write out the form of the Henderson-Hasselbalch equation:

pH = pKa + log([A^–]/[HA])

2. Plug in the known values:

pKa = 4.76 [A^–] = 0.15 M (concentration of acetate) [HA] = 0.20 M (concentration of acetic acid)

3. Calculate the log term:

log([A^–]/[HA]) = log(0.15/0.20) = -0.11

4. Plug all values into the equation:

pH = 4.76 + (-0.11) = 4.65

Therefore, the pH of this acetic acid/acetate buffer solution is 4.65.