Chemistry Math Problem

Question

Calculate the percentage of acetic acid that exists in ionized form when dissolved in water at 0.65 M total concentration. K_a(HOAc) = 1.76 x 10^-5.

Answer 1

HoAc <==> H⁺ + OAc^-

Ka = 1.76x10^-5 = [H⁺][OAc^-] / [HOAc] = (x)(x) / 0.65 -x

x² = 1.14x10^-5

x = 3.38 x10^-3

% ionization = 3.38x10^-3 / 0.65 (x100) = 0.52%

NOTE: Because the % ionization is so low, it was justified to neglect x in the denominator of the Ka expression above.