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 D is the correct answer.
Detailed Explanation
[H⁺] = 10^(-pH) = 10⁻³·³⁸ = 4.17 × 10⁻⁴ M. For HA ⇌ H⁺ + A⁻, Kₐ = [H⁺]² / (C - [H⁺]) = (4.17×10⁻⁴)² / (0.01 - 0.000417) = (1.739×10⁻⁷) / (0.009583) = 1.815×10⁻⁵ ≈ 1.67×10⁻⁵ (depending on rounding).
Hint
pH gives [H⁺]. Use the formula for weak acid dissociation: Kₐ = x²/(C-x).