Question

If a solution of HF (Ka = 6.8 x 10-4) has a pH of 3.65, calculate the initial concentration of hydrofluoric acid.

Answer 1

Let the concentration of HF be c

use:
pH = -log [H+]
3.65 = -log [H+]
[H+] = 2.239*10^-4 M
HF dissociates as:

HF          ----->     H+   + F-
c                 0         0
c-x               x         x


Ka = [H+][F-]/[HF]
Ka = x*x/(c-x)
6.8*10^-4 = 2.239*10^-4*2.239*10^-4/(c-2.239*10^-4)
c-2.239*10^-4 = 7.37*10^-5
c=2.976*10^-4 M
Answer: 3.0*10^-4 M