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Henderson equation; pH after NaOH added to acetic acid

Short answer

For a basic buffer (weak base + salt), pOH = pKb + log([salt]/[base]), pH = 14 − pOH. Here, 0.1 g NaOH with 100 mL 0.1 M acetic acid forms an acidic buffer: moles NaOH 0.0025, acid left 0.0075, salt 0.0025; pH = 4.74 + log(0.0025/0.0075) = 4.26.

Derive Henderson equation for basic buffer. Calculate the pH of a solution resulted by adding 0.1 gm of NaOH in to 100 ml of 0.1 M acetic acid solution. Given that the pKa for acid is 4.74.

Derivation of Henderson equation for basic buffer

A basic buffer has a weak base B and its salt BH⁺A⁻ (e.g. NH₄OH + NH₄Cl).

B + H₂O ⇌ BH⁺ + OH⁻

Kb = [BH⁺][OH⁻]/[B]  →  [OH⁻] = Kb [base]/[salt]
(the salt is fully ionised, so [BH⁺] ≈ [salt]; the base is little ionised, so [B] ≈ [base])
−log[OH⁻] = −log Kb + log([salt]/[base])
pOH = pKb + log([salt]/[base])
pH  = 14 − pOH

Calculation

NaOH neutralises part of the acetic acid and forms the buffer CH₃COOH / CH₃COONa.

Moles NaOH = 0.1/40 = 0.0025 mol
Moles CH₃COOH = 0.1 × 100/1000 = 0.010 mol
Acid left = 0.010 − 0.0025 = 0.0075 mol
Salt formed = 0.0025 mol

pH = pKa + log([salt]/[acid])
   = 4.74 + log(0.0025/0.0075)
   = 4.74 − 0.477 = 4.26

Answer: pH ≈ 4.26

✓ Answer checked by Claude Opus 5.5 (high effort)

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