Calculate the pH and % dissociation of a 0.25 M solution of propanoic acid at 25 degrees Celsius. (Ka= 1.34 x 10^5)

Question

Please post steps to find solution, thanks!

Answer 1

Let propanoic acid be represented by HA, a weak acid.  The dissociation of a weak acid is given by

HA ==> H⁺ + A^- and

Ka = [H⁺][A^-]/[HA]

1.34x10^-5 = [x][x]/0.25 -x (note: assuming x is small relative to 0.25, we can neglect it and avoid using a quadratic)

1.34x10^-5 = x²/0.25

x² = 3.35x10^-6

x = 1.83x10^-3 M = [H⁺] = [A^-] (note: this value is indeed small compared to 0.25 so assumption was valid)

pH = -log [H⁺] = -log 1.83x10^-3 = 2.74

% dissociation = [H⁺]/[HA] (x100%) = 1.83x10^-3/0.25 (x100%) = 0.73% dissociated