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 7.7
M
(b) pH 11.0
M
(c) pH 13.7
M

Answer 1

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

Ksp = [Fe2+][OH-]^2 =7.9 x 10^-16

Molar solubility of Fe(OH)2 = [Fe2+], so find [OH-] in each case, and solve for [Fe2+]. In each case, let [Fe2+] = x.

pH + pOH = 14.0. So if pH = 7.7, pOH = 14 - 7 = 6.3

[OH-] = 10 ^-pOH =1 x 10^-6.3 = 5.01 *10^-7 M

If pH = 11, then pOH =14.0 - 11 = 3

[OH-] = 10 ^-pOH = 10^-3

And if pH = 13.7, then pOH = 14.0 - 12.7 = 0.3.

[OH-] = 10 ^-pOH =10^-0.3 =1.99

(a) For pH 7.

Ksp = (x)(5.01 *10^-7)^2 = 7.9 x 10^-16

x = 3.15 *10^-3 M

(b) pH =11

Ksp = (x)(10^-3)^2 = 7.9 x 10^-16

x =7.9 *10^-10 M

(c) pH = 13.7

Ksp = (x)(1.99)^2 = 7.9 x 10^-16

x =1.99 *10^-16 M