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 8.2
M
(b) pH 10.8
M
(c) pH 12.3
M

Answer 1

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

Ksp [ Fe2+][OH-]^2   ,

a) pH = 8.2 , pOH = 14-8.2 = 5.8 , [OH-] = 10^ -5.8 = 1.585 x 10^ -6 M

Ksp = 7.9 x 10^ -16 = [Fe2+] ( 1.585x10^-6)^2

solibility = [Fe2+] = 3.14 x 10^ -4 M

b) pH = 10.8 , pOH= 3.2 , [OH-] = 10^-3.2 = 0.00063

Ksp = 7.9x10^-16 = [Fe2+] ( 0.00063^2)

[fe2+] = solubility = 1.99 x 10^ -9 M

c) pH = 12.3 , [OH-] = 1.7 , [OH-] = 10^ -1.7 = 0.0199526

Ksp = 7.9 x 10^ -16 = [Fe2+]( 0.0199526)^2

[Fe2+] = solubility = 1.984 x 10^ -12 M