What is the percent ionization of a 0.183 M solution of this acid?

Question

The K_a of a monoprotic weak acid is 0.00282. What is the percent ionization of a  0.183M solution of this acid?

Answer 1

HA <==> H⁺ + A^-

Ka = [H⁺][A^-]/[HA]

0.00282 = (x)(x)/0.182 - x

x² = 0.000513 - 0.00282x

x² + 0.00282x - 0.000513 = 0

x = 0.0213 = [H⁺] = [A^-]

Percent ionization = 0.0213/0.183 (x100%) = 11.6%

NOTE: you should use the quadratic equation and not the assumption that x is insignificant as it is not.

Answer 2

K_a = [H+][A-]/[HA]

At equilibrium: [HA] = 0.183-x

Set up K_a expression: x²/0.183 = 0.00282 (make small x approximation)

x = 0.0227 = [H+]

Percent Ionization = (0.0227/0.183)*100 = 12.4%