Question

A 100.0 mL sample of 0.20 M HF is titrated with 0.10M KOH. Determine the pH of the solution before teh addition of any KOH. The Ka of HF is 3.5x10^-4

Answer 1

HF ------------------------> H+   + F-

0.20                               0          0 ------------> initial

0.20-x                            x           x ---------------> equilibrium

Ka = [H+][F-]/[HF]

Ka = x^2 / 0.20-x

3.5 x 10^-4 = x^2 / 0.20-x

x^2 + 3.5 x 10^-4 x - 7 x 10^-5 = 0

by solving this

x = 8.19 x 10^-3

[H+] = 8.19 x 10^-3 M

pH = -log [H+]

pH = -log (8.19 x 10^-3 )

pH = 2.09