Question

Calculate the molar solubility of Mg(OH)₂ in a solution that is basic with a pH of 12.68.

K_sp = [Mg²⁺][OH^–]² = 5.6 × 10^–12

Answer 1

pH = 12.68

pOH = 14 - 12.68 = 1.32

[OH-] = 10^-1.32

         = 0.04786 M

Mg(OH)2   ---------------> Mg+2 +   2 OH-

                                           S           0.0478

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

5.6 x 10^-12 = S x (0.04786)^2

S = 2.44 x 10^-9 M

molar solubiity = 2.4 x 10^-9 M