# What is the "pH" of a solution of magnesium hydroxide given that K_(sp) = 1.8 * 10^(-11) ?

Feb 28, 2016

$\text{pH} = 10.52$

#### Explanation:

In order to find the pH of a solution, you must determine the concentration of hydronium ions, ${\text{H"_3"O}}^{+}$, either directly or indirectly.

When you're dealing with a Bronsted - Lowry acid, you will be solving for the concentration of hydronium ions directly.

When you're dealing with a Bronsted - Lowry base, like you are here, you will be solving for the concentration of hydronium ions indirectly, i.e. by solving for the concentration of hydroxide ions, ${\text{OH}}^{-}$.

Magnesium hydroxide, "Mg"("OH")_2, us insoluble in aqueous solution. This means that when you place magnesium hydroxide in water, an equilibrium will be established between the undissolved solid and the dissolved ions.

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

The solubility product constant, ${K}_{s p}$, for this equilibrium looks like this

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

What you need to do here is use the ${K}_{s p}$ to find the molar solubility, $s$, of magnesium hydroxide in aqueous solution at ${25}^{\circ} \text{C}$.

To do that, set up an ICE table

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

color(purple)("I")" " " " " "-" " " " " " " " " " " " " "0" " " " " " " " " " 0
color(purple)("C")" " " " " "-" " " " " " " " " " " "(+s)" " " " " "(+color(red)(2)s)
color(purple)("E")" " " " " "-" " " " " " " " " " " " " "s" " " " " " " " " "color(red)(2)s

Remember, the concentration of the solid is assumed to be constant.

Here ${K}_{s p}$ will be equal to

${K}_{s p} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}} = 4 {s}^{3}$

Rearrange to solve for $s$

$s = \sqrt[3]{{K}_{s p} / 4} = \sqrt[3]{\frac{1.8 \cdot {10}^{- 11}}{4}} = 1.65 \cdot {10}^{- 4}$

This means that the concentration of hydroxide ions in a saturated solution of magnesium hydroxide at ${25}^{\circ} \text{C}$ will be

["OH"^(-)] = color(red)(2) * 1.65 * 10^(-4)"M" = 3.30 * 10^(-4)"M"

Now, at ${25}^{\circ} \text{C}$, the concentration of hydronium ions and the concentration of hydroxide ions have the following relationship

color(blue)(K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-4))" ", where

${K}_{W}$ - the ion product for water's auto-ionization

Plug in your value to get

["H"_3"O"^(+)] = 10^(-14)/(3.30 * 10^(-4)) = 3.03 * 10^(-11)"M"

This means that the pH of the solution will be

color(blue)("pH" = - log(["H"_3"O"^(+)]))

$\text{pH} = - \log \left(3.03 \cdot {10}^{- 11}\right) = \textcolor{g r e e n}{10.52}$

Alternatively, you can use the equation

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

Here

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