Question

Calculate the molar solubility of Fe(OH)2 in a buffer solution where the pH has been fixed at the indicated values. Ksp = 7.9 x 10-16.

(a) pH 7.3 _______M

(b) pH 11.4 _______M

(c) pH 13.8 ________M

Answer 1

a)

use:

pH = -log [H+]

7.3 = -log [H+]

[H+] = 5.012*10^-8 M

use:

[OH-] = Kw/[H+]

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

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

[OH-] = (1.0*10^-14)/(5.012*10^-8)

[OH-] = 2.0*10^-7 M

At equilibrium:

Fe(OH)2 <----> Fe2+ + 2 OH-

   s 2*10^-7 + 2s

Ksp = [Fe2+][OH-]^2

7.9*10^-16=(s)*(2*10^-7+ 2s)^2

Since Ksp is small, s can be ignored as compared to 2*10^-7

Above expression thus becomes:

7.9*10^-16=(s)*(2*10^-7)^2

7.9*10^-16= (s) * 4*10^-14

s = 1.975*10^-2 M

Answer: 2.0*10^-2 M

b)

use:

pH = -log [H+]

11.4 = -log [H+]

[H+] = 3.981*10^-12 M

use:

[OH-] = Kw/[H+]

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

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

[OH-] = (1.0*10^-14)/(3.981*10^-12)

[OH-] = 2.512*10^-3 M

At equilibrium:

Fe(OH)2 <----> Fe2+ + 2 OH-

   s 2.512*10^-3 + 2s

Ksp = [Fe2+][OH-]^2

7.9*10^-16=(s)*(2.512*10^-3+ 2s)^2

Since Ksp is small, s can be ignored as compared to 2.512*10^-3

Above expression thus becomes:

7.9*10^-16=(s)*(2.512*10^-3)^2

7.9*10^-16= (s) * 6.31*10^-6

s = 1.252*10^-10 M

Answer: 1.3*10^-10 M

c)

use:

pH = -log [H+]

13.8 = -log [H+]

[H+] = 1.585*10^-14 M

use:

[OH-] = Kw/[H+]

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

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

[OH-] = (1.0*10^-14)/(1.585*10^-14)

[OH-] = 0.631 M

At equilibrium:

Fe(OH)2 <----> Fe2+ + 2 OH-

   s 0.631 + 2s

Ksp = [Fe2+][OH-]^2

7.9*10^-16=(s)*(0.631+ 2s)^2

Since Ksp is small, s can be ignored as compared to 0.631

Above expression thus becomes:

7.9*10^-16=(s)*(0.631)^2

7.9*10^-16= (s) * 0.3982

s = 1.984*10^-15 M

Answer: 2.0*10^-15 M