# Question f4fb3

Jul 10, 2016

${K}_{s p} = 3.2 \cdot {10}^{- 11}$

#### Explanation:

Magnesium hydroxide, "Mg"("OH")_2, is only sparingly soluble in aqueous solution, which means that it does not dissociate completely to form magnesium cations, ${\text{Mg}}^{2 +}$, and hydroxide anions, ${\text{OH}}^{-}$.

Instead, an equilibrium is established between the undissolved solid and the dissociated ions.

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

Notice that every mole of magnesium hydroxide that dissociates produces $1$ mole of magnesium cations and $\textcolor{red}{2}$ moles of hydroxide anions.

In other words, the saturated solution will contain twice as many moles of hydroxide anions than on magnesium cations. This means that the equilibrium concentration of the magnesium cations will be half that of the hydroxide anions

$\left[{\text{Mg"^(2+)] = 1/color(red)(2) xx ["OH}}^{-}\right] \to$ equilibrium concentrations

In your case, you will have

["Mg"^(2+)] = 1/2 * 4.0 * 10^(-4)"M" = 2.0 * 10^(-4)"M"

Now, the solubility product constant, ${K}_{s p}$, for magnesium hydroxide is defined as

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

Plug in your values to find

K_(sp) = 2.0 * 10^(-4)"M" * (4.0 * 10^(-4)"M")^color(red)(2)#

${K}_{s p} = 3.2 \cdot {10}^{- 11} {\text{M}}^{3}$

The units are usually omitted from the expression of the solubility product constant, which means that your answer will be

${K}_{s p} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{3.2 \cdot {10}^{- 11}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.