Question

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

(a) pH 7.4
______________M
(b) pH 10.8
______________M
(c) pH 13.3
______________M

Answer 1

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

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

so..

a)

pOH = 14-pH = 14-7.4 = 6.6

[OH-] = 10^-6.6

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

7.9*10^-16 = (S)(10^-6.6)^2

S = (7.9*10^-16) / ((10^-6.6)^2) = 0.0125206M

b)

pOH = 14-pH = 14-10.8= 3.2

[OH-] = 10^-3.2

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

7.9*10^-16 = (S)( 10^-3.2)^2

S = (7.9*10^-16) / ((10^-3.2)^2) = 1.98 *10^-9 M

c)

pOH = 14-pH = 14-13.3= 0.7

[OH-] = 10^-0.7

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

7.9*10^-16 = (S)( 10^-0.7)^2

S = (7.9*10^-16) / ((10^-0.7)^2) = 1.9843*10^-14 M