Answer to Question #284769 in General Chemistry for John

Question

In a saturated solution of Mg(OH)2, the concentration of Mg2+ is 1.31 × 10–4 M. What is the solubility product for Mg(OH)2?

Answer 1

First write the equation for dissociation of magnesium hydroxide.

$Mg(OH)_2\longrightarrow Mg^{2+} +2OH^-\\ K_{SP}=[Mg^{2+}][OH^-]^2$

From this equation, one mole of magnesium hydroxide dissociates to produce a mole of magnesium ions and two moles of hydroxide ions. We can therefore use proportions to find the concentration of the hydroxide ions. And consequently the solubility product.

$1\equiv1.31×10^{-4}M\\2\equiv?\\ [OH]^-=\frac{2×1.31×10^{-4}}{1}=2.62×10^{-4}\\ [Mg^{2+}]=1.31×10^{-4}\\K_{SP}=(1.31×10^-4)×(2.62×10^-4)^2\\ K_{sp}=8.99×10^{-12}$