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Henderson equation; pH after adding NaOH to acetic acid

Short answer

A buffer resists pH changes. An acidic buffer (weak acid + its salt) follows pH = pKa + log([salt]/[acid]), derived from Ka. Adding 0.4 g NaOH to 500 mL 0.2 M acetic acid gives 0.01 mol salt and 0.09 mol acid; pH = 4.74 + log(0.01/0.09) = 3.79.

Define Buffer. Derive Henderson's equation for acidic buffer. Calculate the pH of the solution formed by adding 0.4 g of NaOH on 500 mL 0.2 M acetic acid. pKa for acetic acid = 4.74.

Buffer

A buffer is a solution that resists a change in pH on adding small amounts of acid or base. An acidic buffer contains a weak acid (HA) and its salt (A⁻) with a strong base.

Derivation of Henderson's equation (acidic buffer)

For the weak acid HA: HA ⇌ H⁺ + A⁻

Ka = [H⁺][A⁻]/[HA]
[H⁺] = Ka [HA]/[A⁻]

In the presence of the salt (fully ionised) the ionisation of HA is suppressed, so [HA] ≈ initial concentration of acid and [A⁻] ≈ concentration of salt:

[H⁺] = Ka [acid]/[salt]
Taking −log on both sides:
−log[H⁺] = −log Ka − log([acid]/[salt])
pH = pKa + log([salt]/[acid])

This is Henderson's equation.

Calculation

Moles NaOH = 0.4/40 = 0.01 mol
Moles CH₃COOH = 0.2 × 0.5 = 0.10 mol
CH₃COOH + NaOH → CH₃COONa + H₂O
Acid left = 0.10 − 0.01 = 0.09 mol
Salt formed = 0.01 mol

pH = 4.74 + log(0.01/0.09)
   = 4.74 − 0.954 = 3.79

Answer: pH ≈ 3.79

✓ Answer checked by Claude Opus 5.5 (high effort)

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