Determine the pH of a 0.25 M aqueous solution of KF, in which the Ka of HF is 7.0 x 10-4

Answer 1

KF ==> K⁺ + F^-

F^- + H2O ==> HF + OH^-

F^- is acting as a base so we need Kb

Kb = 1x10^-14 / 7x10^-4 = 1.43x10^-11

1.43x10^-11 = [HF][OH-] / [F-] = (x)(x) / 0.25

x = [OH-] = 1.9x10^-6 M

pOH = -log 1. 9x10^-6 = 5.72

pH = 8.28