The 𝐾a of a monoprotic weak acid is 0.00353. What is the percent ionization of a 0.108 M solution of this acid?

Answer 1

HA = weak monoprotic acid

HA ==> H⁺ + A^-

Ka = [H⁺][A^-]/[HA]

0.00353 = (x)(x) / 0.108 - x

x² = 0.0003812 - 0.00353x

x² + 0.00353x - 0.0003812 = 0

x = 0.0178 M = [H⁺] = [A^-]

% ionization = 0.0178 / 0.108 (x100%) = 16.5% ionized