At equilibrium, a 1.0 L reaction vessel contains 2.3 mol of Mg(OH)₂ and 0.170 mol of OH⁻. What is the equilibrium concentration of Mg²⁺

Question

. What is the equilibrium concentration of Mg²⁺ based on the reaction: Mg(OH)₂ (s) ⇌ Mg²⁺ (aq) + 2 OH⁻ (aq) Kc = 1.80 x 10⁻¹¹

Answer 1

I think this should be a Ksp problem and not a Kc problem. But, that is a minor point, and one can address the problem by writing the equilibrium expression and then using the values to solve for the equilibrium concentration of Mg²⁺.

Mg(OH)₂(s) <==> Mg²⁺(aq) + 2OH^-(aq) Ksp = 1.80x10^-11

Ksp = [Mg²⁺][OH^-]²

Let x = [Mg²⁺] and then [OH^-] = 2x

1.80x10^-11 = (x)(2x)²

1.80x10^-11 = 4x³

x³ = 4.5x10^-12

x = 1.65x10^-4 M = [Mg²⁺]