Question

calculate the pH of 0.25M solution of propanoic acid​

Answer 1

Answer:

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