When PH = ㏒(H^+)
[H+] = 10^-3.67
[H+] = 2.14x 10^-4
so according to the reaction equation:
HF ↔ H^+ + F^-
at equilibrium X-(2.14x10^-4) (2.14x10^-4) (2.14x10^-4)
by substitution in Ka formula:
Ka = [H+][F]-/[HF]
6.8x10^-4 = (2.14x10^-4)*(2.14x10^-4) /(X-2.4x10^-4)
X-2.4x10^-4 = (4.58x10^-8)* (6.8x10^-4)
∴X = 2.4x10^-4
∴[HF] = X- (2.14x10^-4)= (2.4x10^-4)-(2.14x10^-4) = 2.6 X 10^-5
