Question Detail
Question
A 0.01 M solution of a weak monoprotic acid has a pH of 4.50. What is the Ka of the acid?
Correct Answer
Option A is the correct answer.
Detailed Explanation
[H⁺] = 10⁻⁴·⁵⁰ = 3.16 × 10⁻⁵ M. Ka = [H⁺]²/(C - [H⁺]) = (1.0×10⁻⁹)/(0.01 - 3.16×10⁻⁵) ≈ 1.0×10⁻⁹/0.01 = 1.0 × 10⁻⁷ M.
Hint
Ka = [H⁺]²/(initial concentration - [H⁺]) for a weak acid.