Question

Calculate the concentration of all species in a 0.14 M KF solution.

[K+], [F−], [HF], [OH−], [H3O+]

please give each value clearly

Answer 1

Actually, [F-] is not exactly equal to 0.14 because HF, the conjugate acid of F-, is a weak acid (pKa (HF) = 3.2), so some F- will react with H2O to form some HF and thereby reduce the F- concentration.

It may be easier to work this using pKb:

F- + H2O → HF + OH-
Kb = ([HF][OH-])/[F-]
pKa + pKb = 14
pKb = 14 - pKa = 14 - 3.2 = 10.8
Kb = 10^(-10.8) = 1.585 x 10^-11

Set up an ICE table:
Species...Initial ...Change...Equil...
F-............0.14 M....-x.......(0.14 -x)
H2O (not included in calculations)
HF.............0...........x.............
OH-..........0...........x..........x....

Plug these into the expression for Kb:

Kb = ([HF][OH-])/[F-]
1.585 x 10^-11 = (x)*(x)/(0.14 -x)
(1.585 x 10^-11)*(0.14 -x) = x^2
x^2 + (1.585 x 10^-11)x - 2.219 x 10^-22 = 0

Solution of this quadratic equation using the quadratic formula gives

x = 8.948 x 10^-12

So the equilibrium concentrations of the species are

[F-] = (0.14 - x) = (0.14 - 8.948 x 10^-12) M ≈ 0.14 M (to two sig figs)
[HF] = x = 8.948 x 10^-12 M
[OH-] = x = 8.948 x 10^-12 M

We then use [OH-] tp calculate [H3O+]:

[H3O+]*[OH-] = 10^-14
[H3O+] = (10^-14)/[OH-]
[H3O+] = (10^-14)/(8.948 x 10^-12 M)
[H3O+] = 1.1 x 10^-3 M