The 𝐾a of a monoprotic weak acid is 0.00304. What is the percent ionization of a 0.154 M solution of this acid?

Answer 1

Let HA be the weak acid.

HA ==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA]

0.00304 = (x)(x) / 0.154

x² = 4.68x10^-4

x = 0.0216 M = [H⁺] = [A^-]

Percent ionization = 0.0216 / 0.154 (x100%) = 14%