SOLUBILITY OF SOLUTION

Question

The Ksp of Mg(OH)2 is 6 x 10^-12. How much 2x10^-4 M HCL (in mL) is needed to neutralize a 100 mL saturated solution of Mg(OH)2?

Answer 1

First, we must find the moles of OH- in solution so that we can then determine the moles of H+ needed to neutralize that.

Mg(OH)₂(s) <==> Mg²⁺(aq) + 2OH^-(aq)

Ksp = [Mg²⁺][OH^-]² = 6x10^-12

Let x = [Mg²⁺] and then [OH^-] = 2x

6x10^-12 = (x)(2x)² = 4x³

x = 1.14x10^-4 M

[OH-] = 2 * 1.14x10^-4 M = 2.28x10^-4 M

moles OH^- = 2.28x10^-4 moles/L x 0.100 L = 2.28x10^-5 moles

H⁺ + OH^- ==> H2O

moles HCl need = 2.28x10^-5 moles (1:1 mole ratio)

volume of HCl needed: (x L)(2x10^-4 mol/L) = 2.3x10^-5 moles

x = 0.115 L = 115 mls HCl needed