# Solubility of Mg(OH)_2 is 1.6 x 10^-4 "mol/L" at 298 K. What is its solubility product?

Jan 31, 2016

$1.6 \cdot {10}^{- 11}$

#### Explanation:

Magnesium hydroxide, "Mg"("OH")_2, is considered Insoluble in aqueous solution, which implies that the ions it forms will be in equilibrium with the solid.

Magnesium hydroxide dissociates only partially to form magnesium cations, ${\text{Mg}}^{2 +}$, and hydroxide anions, ${\text{OH}}^{-}$

${\text{Mg"("OH")_text(2(s]) rightleftharpoons "Mg"_text((aq])^(2+) + color(red)(2)"OH}}_{\textrm{\left(a q\right]}}^{-}$

For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.

In your case, a molar solubility of

$s = 1.6 \cdot {10}^{- 4}$

means that you can only dissolve $1.6 \cdot {10}^{- 4}$ moles of magnesium in a liter of water at that temperature.

Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces $1$ mole of magnesium cations and $\textcolor{red}{2}$ moles of hydroxide anions.

This tells you that if you successfully dissolve $1.6 \cdot {10}^{- 4}$ moles of magnesium hydroxide in one liter of solution, number of moles of each ion will be

${n}_{M {g}^{2 +}} = 1 \times 1.6 \cdot {10}^{- 4} = 1.6 \cdot {10}^{- 4} {\text{moles Mg}}^{2 +}$

and

${n}_{O {H}^{-}} = \textcolor{red}{2} \times 1.6 \cdot {10}^{- 4} = 3.2 \cdot {10}^{- 4} \text{moles}$

Since we're working with one liter of solution, you can cay that

["Mg"^(2+)] = 1.6 * 10^(-4)"M"

["OH"^(-)] = 3.2 * 10^(-4)"M"

By definition, the solubility product constant, ${K}_{s p}$, will be equal to

${K}_{s p} = {\left[{\text{Mg"^(2+)] * ["OH}}^{-}\right]}^{\textcolor{red}{2}}$

Plug in these values to get

${K}_{s p} = 1.6 \cdot {10}^{- 4} \cdot {\left(3.2 \cdot {10}^{- 4}\right)}^{\textcolor{red}{2}}$

${K}_{s p} = \textcolor{g r e e n}{1.6 \cdot {10}^{- 11}} \to$ rounded to two sig figs

The listed value for magnesium hydroxide's solubility product is $1.6 \cdot {10}^{- 11}$, so this is an excellent result.

http://www.wiredchemist.com/chemistry/data/solubility-product-constants