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Buffer action; pH change on adding KOH

Short answer

A buffer solution resists pH change. Adding KOH converts acetic acid to acetate: initially 10 mmol each, after adding 3 mmol KOH acid = 7 mmol, salt = 13 mmol. pH change = log(13/7) ≈ +0.27, from 4.74 to 5.01.

Define the term buffer and buffer action. If 3.0 millimoles of KOH is added to 100 ml of a solution which is 0.10 M in acetic acid and 0.10 M in sodium acetate. What will be the change in pH of the solution?

Buffer and buffer action

A buffer is a solution (a weak acid with its salt, or a weak base with its salt) that resists a change in pH on adding a small amount of strong acid or base. The property of resisting the change is called buffer action. In CH₃COOH / CH₃COONa the acetate ion neutralises added H⁺ (CH₃COO⁻ + H⁺ → CH₃COOH) and the acid neutralises added OH⁻ (CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O).

Calculation

Initial: acid = 0.10 M × 100 mL = 10 mmol ; salt = 10 mmol
Initial pH = pKa + log(10/10) = pKa = 4.74

KOH added = 3.0 mmol:  CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
Acid left = 10 − 3 = 7 mmol
Salt now  = 10 + 3 = 13 mmol

New pH = pKa + log(13/7) = 4.74 + 0.269 = 5.01
Change in pH = log(13/7) = +0.27

(The change does not depend on pKa.) Answer: pH rises by about 0.27 units (from 4.74 to 5.01, taking pKa = 4.74).

✓ Answer checked by Claude Opus 5.5 (high effort)

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