Calculate for pH related problem

Question

Determine the pH of a 0.25M aqueous solutiob of KF,in which the K_a of HF is 3.5 x 10^-5

Answer 1

KF ==> K⁺ + F^-

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

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

Kb = Kw/Ka = 1x10^-14 / 3.5x10^-5= 2.86x10^-10

2.86x10^-10 = [HF][OH-] / [F-] = (x)(x) / 0.25

x² = 7.15x10^-11

x = [OH-] = 8.46x10^-6 M

pOH = -log 8.46x10^-6

pOH = 5.07

pH = 14 - pOH

pH = 8.93