What is the solubility of Mg(OH)₂ at a pH of 12.10? (Ksp Mg(OH)₂ is 1.6 × 10⁻¹³)

Answer 1

Mg(OH)₂(s) <==> Mg²⁺(aq) + 2OH^-(aq) .. dissolution equilibrium

Ksp = [Mg²⁺] [OH^-]²

From pH = 12.10, we can find pOH and from pOH we can determine [OH-] to plug into the equation.

pOH = 14 - pH = 14 - 12.1 = 1.9

[OH-] = 1x10^-1.9 = 1.26x10^-2 M

1.6x10^-13 = [Mg²⁺][1.26x10^-2]²

[Mg²⁺] = 1.0x10^-9 M = solubility of Mg(OH)2 @ pH 12.1