# Calculate the pH at which Mg(OH)_2 begins to precipitate from a solution containing 0.1 M Mg^(2+) ions? K_(sp) for Mg(OH)_2 = 1.0 x 10^-11.

Jan 31, 2016

$\text{pH} = 9$

#### Explanation:

The idea here is that you need to use magnesium hydroxide's solubility product constant to determine what concentration of hydroxide anions, ${\text{OH}}^{-}$, would cause the solid to precipitate out of solution.

As you know, the dissociation equilibrium for magnesium hydroxide looks like this

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

The solubility product constant, ${K}_{s p}$, will be equal to

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

Rearrange to find the concentration of the hydroxide anions

["OH"^(-)] = sqrt(K_(sp)/(["Mg"^(2+)]))

Plug in your values to get

["OH"^(-)] = sqrt((1.0 * 10^(-11))/0.1) = sqrt(1.0 * 10^(-10)) = 10^(-5)"M"

As you know, you can use the concentration of hydroxide anions to find the solution's pOH

color(blue)("pOH" = - log(["OH"]^(-)))

$\text{pOH} = - \log \left({10}^{- 5}\right) = 5$

Finally, use the relationship that exists between pOH and pH at room temperature

$\textcolor{b l u e}{\text{pOH " + " pH} = 14}$

to find the pH of the solution

$\text{pH} = 14 - 5 = \textcolor{g r e e n}{9}$

So, for pH values that are below $9$, the solution will be unsaturated. Once the pH of the solution becomes equal to $9$, the solution becomes saturated and the magnesium hydroxide starts to precipitate.