Question

Determine the [OH−] of a solution that is 0.180 M in F−.

Answer 1

Ka for HF = 6.6*10^-4

use:

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/6.6*10^-4

Kb = 1.515*10^-11

F- dissociates as

F- + H2O -----> HF + OH-

0.18 0 0

0.18-x x x

Kb = [HF][OH-]/[F-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.515*10^-11)*0.18) = 1.651*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.651*10^-6 M

[OH-] = x = 1.651*10^-6 M

Answer: 1.65*10^-6 M