The 𝐾a of a monoprotic weak acid is 0.00506. What is the percent ionization of a 0.114 M solution of this acid?

Answer 1

HA <==> H⁺ + A^- .. ionization of a monoprotic acid

Ka = [H⁺][A^-] / [HA]

0.00506 = (x)(x) / 0.114 - x

x² = 6.38x10^-4 - 0.00506x

x² + 0.00506x - 6.38x10^-4 = 0

x = 0.0229 M = [H⁺] = [A^-]

% ionization = 0.0229 M / 0.114 M (x100%) = 20.1% ionized