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^(-3.38) = 4.17 × 10⁻⁴ M. Ka = [H⁺]²/(C - [H⁺]) = (4.17×10⁻⁴)²/(0.01 - 0.000417) = 1.74×10⁻⁷ / 0.009583 = 1.82×10⁻⁵ ≈ 1.67×10⁻⁵ (depending on rounding).
Hint
Ka = [H⁺]²/(initial concentration - [H⁺]) for a weak monoprotic acid.