If a solution of hf (ka = 6.8 10-4) has a ph of 3.67, calculate the total concentration of hydrofluoric acid.

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

Ibelieve the answer is b.2200 watts, because p=vi; p=220(10); then p=2200hope my answer has you : )