A 0.1 M solution of ethanoic acid is 1.34% ionized. What is the value of K_a for ethanoic acid? A) K_a = 5.3*10^-11 B) K_a = 1.82* 10^-5 C) K_a = 1.34 *10^-6 D) K_a = 2.67 * 10^-2 E) K_a = 3.01 * 10^-4

Question

Quizplus · Accepted Answer

The degree of ionization is 1.34%, which means [H⁺] = 0.00134 M. Using the formula Kₐ = [H⁺]² / [CH₃COOH_initial - H⁺], we find Kₐ ≈ 1.82 × 10⁻⁵.

A 01 M Solution of Ethanoic Acid Is 1