What is the maximum number of moles of "Mg"("OH")_2, that can remain dissolved in a 5.0 L solution whose pH is 9.70?

Aug 5, 2016

"0.036 moles Mg"("OH")_2

Explanation:

!! CORRECT SOLUTION !!

$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
SIDE NOTE My first attempt at solving this problem was incorrect, to put it mildly. I will add the correct answer here, and leave the incorrect one below to serve as a 'How not to solve this problem' example.
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

The idea here is that magnesium hydroxide, "Mg"("OH")_2, is insoluble in aqueous solution. This means that when you dissolve this salt in water, the following dissociation equilibrium is established

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

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

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

The value listed here is

${K}_{s p} = 1.8 \cdot {10}^{- 11}$

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

Assuming that the solution is buffered at a pH of $9.70$, i.e. the concentration of hydroxide anions, ${\text{OH}}^{-}$, is constant, your first goal here will be figure out the concentration of magnesium cations, ${\text{Mg}}^{2 +}$.

Start by calculating the concentration of hydroxide anions. You know that for aqueous solutions at room temperature, you have

color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))" " and " "color(purple)(|bar(ul(color(white)(a/a)color(black)("poH" = - log(["OH"^(-)]))color(white)(a/a)|)))

In your case, you will have

$\text{pOH} = 14 - 9.70 = 4.30$

and

"pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")

Plug in your value to find

["OH"^(-)] = 10^(-4.30) = 5.01 * 10^(-5)"M"

Now, according to the expression you have for the solution's solubility product constant, the equilibrium concentration of magnesium cations will be equal to

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

Plug in your values to find

["Mg"^(2+)] = (1.8 * 10^(-11))/(5.01 * 10^(-5))^color(red)(2) = 7.17 * 10^(-3)"M"

This means that when you place some solid magnesium hydroxide in a buffered solution that has a pH equal to $9.70$, the salt will dissolve until the equilibrium concentration of magnesium cations is equal to $7.17 \cdot {10}^{- 3} \text{M}$.

Use the volume of the solution to find the number of moles of magnesium cations

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

n_("Mg"^(2+)) = 7.17 * 10^(-3)"mol" color(red)(cancel(color(black)("L"^(-1)))) * 5.0 color(red)(cancel(color(black)("L"))

n_("Mg"^(2+)) = "0.03585 moles Mg"^(2+)

Since magnesium hydroxide dissociates in a $1 : 1$ mole ratio to produce magnesium cations, you can say that the number of moles of magnesium hydroxide that dissolve in this solution will be

n_("Mg"("OH")_2) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.036 moles Mg"("OH")_2)color(white)(a/a)|)))

The answer is rounded to two sig figs.

$\textcolor{red}{\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{}}}$

!! INCORRECT SOLUTION !!

The problem wants you to use the pH of the solution to find the concentration of hydroxide anions, ${\text{OH}}^{-}$.

Since magnesium hydroxide, "Mg"("OH")_2, is insoluble in aqueous solution, its dissociation equilibrium will be affected by the pH of the solution. More specifically, a higher concentration of hydroxide anions will reduce the solubility of magnesium hydroxide.

Start by calculating the concentration of hydroxide anions. You know that for aqueous solutions at room temperature, you have

color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))" " and " "color(purple)(|bar(ul(color(white)(a/a)color(black)("poH" = - log(["OH"^(-)]))color(white)(a/a)|)))

In your case, you will have

$\text{pOH} = 14 - 9.70 = 4.30$

and

"pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")

Plug in your value to find

["OH"^(-)] = 10^(-4.30) = 5.01 * 10^(-5)"M"

Magnesium hydroxide's dissociation equilibrium looks like this

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

Notice that every mole of magnesium hydroxide that dissociates in aqueous solution produces

• one mole of magnesium cations, $1 \times {\text{Mg}}^{2 +}$
• two moles of hydroxide anions, $\textcolor{red}{2} \times {\text{OH}}^{-}$

This tells you that the equilibrium concentration of hydroxide anions will be twice as high as the equilibrium concentration of magnesium cations.

You can thus say that this solution will contain

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

["Mg"^(2+)] = 1/color(red)(2) xx 5.01 * 10^(-5)"M" = 2.5 * 10^(-5)"M"

Use the volume of the solution to calculate how many moles of magnesium cations would be present

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

n_("Mg"^(2+)) = 2.5 * 10^(-5)"mol" color(red)(cancel(color(black)("L"^(-1)))) * 5.0 color(red)(cancel(color(black)("L"))

n_("Mg"^(2+)) = 1.25 * 10^(-4)"moles Mg"^(2+)

Now, you already know that for every mole of magnesium hydroxide that dissociates you get $1$ mole of magnesium cations in solution.

This means that in a solution that has a pH equal to $9.70$ you can only hope to dissolve

n_("Mg"("OH")_2) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.3 * 10^(-4)"moles Mg"("OH")_2)color(white)(a/a)|)))

The answer is rounded to two sig figs.