Question

Calculate [H3O+] and the pH of a 0.02M solution offluoride ion, F-. The Kaofhydrofluoric acid,HF,is 6.6 x 10-4. (Find Kbfirst then work out the problem like a weak base)

Answer 1

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

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-

0.02 0 0

0.02-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)*2*10^-2) = 5.505*10^-7

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

so, x = 5.505*10^-7 M

So,

[OH-] = 5.505*10^-7 M

use:

[H3O+] = Kw/[OH-]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[H3O+] = (1.0*10^-14)/[OH-]

[H3O+] = (1.0*10^-14)/5.505*10^-7

[H3O+] = 1.817*10^-8 M

use:

pH = -log [H3O+]

= -log (1.817*10^-8)

= 7.7408

[H3O+] = 1.82*10^-8 M

pH = 7.74