Question ab157

Dec 27, 2015

["Mg"^(2+)] = 2.0 * 10^(-3)"M"

Explanation:

The clue to what actually happens here lies with the molar solubility and pH of a magnesium hydroxide solution dissolved in pure water.

As you know, magnesium hydroxide, "Mg"("OH")_2, will not dissociate completely to form magnesium cations, ${\text{Mg}}^{2 +}$, and hydroxide anions, ${\text{OH}}^{-}$.

Instead, an equilibrium will be established between the undissolved salt and the aforementioned ions

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

For magnesium hydroxide dissolved in pure water, you will have

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

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

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

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

This will give you

$s = \sqrt[3]{{K}_{s p} / 4} = \sqrt[3]{\frac{2.0}{4} \cdot {10}^{- 11}} = 1.7 \cdot {10}^{- 4}$

The equilibrium concentration of the hydroxide anions in pure water will be

["OH"^(-)] = 2 * 1.7 * 10^(-4)"M" = 3.4 * 10^(-4)"M"

This corresponds to a pOH of

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

$\text{pOH} = - \log \left(3.4 \cdot {10}^{- 4}\right) = 3.47$

The pH of this solution will thus be

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

$\text{pOH" = 14 - "pH} = 14 - 3.47 = 10.53$

Now, you know that your solution has a pH of $10$. This should immediately lead you to believe that an acid was added to the solution of magnesium hydroxide in pure water.

According to Le Chatelier's Principle, you can expect the equilibrium to shift to the right, since the presumed neutralization reaction consumed some of the hydroxide anions.

As a result, more salt will dissolve in this solution.

A solution of $\text{pH} = 10$ will have

$\text{pOH} = 14 - 10 = 4$

This is equivalent to

$\textcolor{b l u e}{\left[\text{OH"^(-)] = 10^(-"pOH}\right)}$

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

This goes to show that an acid was present in the solution, since the concentration of hydroxide anions is smaller than it would have been for the magnesium hydroxide in pure water solution.

So, if the equilibrium concentration of hydroxide anions is ${10}^{- 4} \text{M}$, you can say that

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

Therefore,

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

$\left[\text{Mg"^(2+)] = (2.0 * 10^(-11))/(10^(-4))^2 = color(green)(2.0 * 10^(-3)"M}\right)$

Compare this to the equilibrium concentration of magnesium cations in the pure water solution, which is equal to the previously calculated $s$

["Mg"^(2+)] = 1.7 * 10^(-4)"M" -> magnesium hydroxide in pure water

["Mg"^(2+)] = 2.0 * 10^(-3)"M" -># magnesium hydroxide in your solution

You can thus conclude that the solubility of the salt increased when compared with the pure water solution. More salt dissolved as a result of the decrease in the concentration of hydroxide anions.

Here is a video that explains this topic in details, in addition to a very similar exercise starting at minute 5:04.
Solubility Equilibria | Common Ion Effect.