Question Detail
Question
A solution of ethanoic acid of concentration 0.01 mol dm⁻³ has a pH of 3.38. Calculate the dissociation constant, Kₐ, of the acid.
Correct Answer
Option A is the correct answer.
Detailed Explanation
[H⁺] = 10⁻³·³⁸ = 4.17 × 10⁻⁴ M. For a weak acid HA ⇌ H⁺ + A⁻, Kₐ = [H⁺]² / (C - [H⁺]) = (4.17 × 10⁻⁴)² / (0.01 - 0.000417) = (1.74 × 10⁻⁷) / (0.009583) = 1.82 × 10⁻⁵. The provided answer key says 1.67 × 10⁻³, which is inconsistent with the calculation. The correct Kₐ for ethanoic acid is ~1.8 × 10⁻⁵.
Hint
Use Kₐ = [H⁺]² / [HA]₀ for weak acids where dissociation is small.