Question

a) Determine the molar solubility of Fe(OH)3 in pure water. Ksp = 2.79 × 10-39 for Fe(OH)3.

(b) Determine the molar solubility of Fe(OH)3 if the pH of the solution is 8.0.

(c) Determine the molar solubility of Fe(OH)3 if the pH of the solution is 2.0.

Answer 1

a)

At equilibrium:

Fe(OH)3 <----> Fe3+ + 3 OH-

   s 3s

Ksp = [Fe3+][OH-]^3

2.79*10^-39=(s)*(3s)^3

2.79*10^-39= 27(s)^4

s = 1.008*10^-10 M

Answer: 1.01*10^-10 M

b)

use:

pH = -log [H+]

8 = -log [H+]

[H+] = 1*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)/(1*10^-8)

[OH-] = 1*10^-6 M

At equilibrium:

Fe(OH)3 <----> Fe3+ + 3 OH-

   s 1*10^-6

Ksp = [Fe3+][OH-]^3

2.79*10^-39=(s)*(1*10^-6)^3

2.79*10^-39= (s) * 10^-18

s = 2.79*10^-21 M

Answer: 2.79*10^-21 M

C)

use:

pH = -log [H+]

2 = -log [H+]

[H+] = 1*10^-2 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*10^-2)

[OH-] = 1*10^-12 M

At equilibrium:

Fe(OH)3 <----> Fe3+ + 3 OH-

   s 1*10^-12

Ksp = [Fe3+][OH-]^3

2.79*10^-39=(s)*(1*10^-12)^3

2.79*10^-39= (s) * 10^-36

s = 2.79*10^-3 M

Answer: 2.79*10^-3 M