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 7.6

(b) pH 10.0

(c) pH 13.8

Answer 1

(a) pH = 7.6

pH + pOH = 14

pOH = 6.4

[OH-] = 10^-pOH = 10^-6.4 = 3.98 x 10^-7 M

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

                                                        S            3.98 x 10^-7

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

7.9 x 10^-16 = S x (3.98 x 10^-7)^2

S = 4.987 x 10^-3 M

molar solubility = 4.987 x 10^-3 M

(b) pH = 10.0

pOH = 4

[OH-] = 1 x 10^-4 M

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

7.9 x 10^-16 = S x (1 x 10^-4)^2

S = 7.9 x 10^-8 M

molar solubility = 7.9 x 10^-8 M

(c) pH 13.8

[OH-] = 0.63 M

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

7.9 x 10^-16 = S x (0.63)^2

S = 1.99 x 10^-15 M

molar solubility = 1.99 x 10^-15 M