Question

Determine the OH- conc of a solution that is 0.180 M in F-

Answer 1

Ka of HF = 6.6*10^-4

SO,

Kb of F- = 10^-14 / Ka

      = (10^-14)/(6.6*10^-4)

      =1.52*10^-11

F- + H2O   ßà HF + OH-

0.180-X            X     X

Kb =x*x/(0.180-x)

since kb is small x will be small and can be ignored as compared to 0.180

1.52*10^-11 = x^2 /(0.180)

x = 1.65*10^-6 M

[OH-] = x = 1.65*10^-6 M

Answer: 1.65*10^-6 M