The 𝐾a of a monoprotic weak acid is 0.00623. What is the percent ionization of a 0.131 M solution of this acid?

Answer 1

The percent of ionization of this weak monoprotic acid is approximated by [H⁺] / [molarity of acid] x 100.

K_a = [H⁺]² / 0.131 M

[H⁺] = √ K_a x 0.131 M = 0.0286

%Ionization = 0.0286 / 0.131 x100 = 21.8%

Answer 2

Let HA represent the weak acid.

HA ==> H⁺ + A^-

The % ionization is [H⁺] / [HA] (x100%)

First, find [H⁺]

Ka = [H⁺][A^-] / [HA]

0.00623 = (x)(x) / 0.131 - x

x² = 0.000816 - 0.00623x

x² + 0.00623x - 0.000816 = 0 ... use the quadratic formula to solve for x

x = 0.0256 M = [H⁺]

% ionization = 0.0256 / 0.131 (x100%) = 19.5%