# Given the value of K_(sp) = 2.9 xx 10^(-12), what would be the solubility of "Mg"("OH")_2 in a 0.345 M "NaOH" solution?

## Given the value of Ksp 2.9x10^-12, what would be the solubility of Mg(OH)2 in a 0.345 M NaOH solution?

Jun 3, 2016

$2.44 \cdot {10}^{- 11} \text{M}$

#### Explanation:

Your strategy here will be to use an ICE table to find the solubility of magnesium hydroxide, "Mg"("OH")_2, in a solution that contains the soluble sodium hydroxide, $\text{NaOH}$.

Sodium hydroxide dissociates in a $1 : 1$ mole ratio to form hydroxide anions

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

Since every mole of sodium hydroxide produces one mole of hydroxide anions, your solution will contain

["OH"^(-)] = ["NaOH"] = "0.345 M"

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

color(purple)("I")color(white)(aaaaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0.345)
color(purple)("C")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaacolor(black)((+s))aaaaaaaacolor(black)((+color(red)(2)s))
$\textcolor{p u r p \le}{\text{E}} \textcolor{w h i t e}{a a a a a a \textcolor{b l a c k}{-} a a a a a a a a a a a a a a \textcolor{b l a c k}{s} a a a a a a a a a \textcolor{b l a c k}{0.345 + \textcolor{red}{2} s}}$

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

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

In your case, this will be equal to

$2.9 \cdot {10}^{- 12} = s \cdot {\left(0.345 + \textcolor{red}{2} s\right)}^{\textcolor{red}{2}}$

$2.9 \cdot {10}^{- 12} = s \cdot \left(0.119025 + 1.38 s + 4 {s}^{2}\right)$

Rearrange to get

$4 {s}^{3} + 1.38 {s}^{2} + 0.119025 s - 2.9 \cdot {10}^{- 12} = 0$

This cubic equation will produce one real solution

$s = 2.44 \cdot {10}^{- 11}$

This represents the molar solubility of magnesium hydroxide in a solution that contains $\text{0.345 M}$ hydroxide anions. You can thus say that you have

"molar solubility"_("in 0.345M OH"^(-)) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.44 * 10^(-11)"M")color(white)(a/a)|)))

I'll leave the answer rounded to three sig figs.

SIDE NOTE You should be able to predict that the solubility of magnesium hydroxide in this solution is significantly lower than the solubility of the salt in pure water.

In this case, you would have

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

color(purple)("I")color(white)(aaaaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaacolor(black)((+s))aaaaaaaacolor(black)((+color(red)(2)s))
$\textcolor{p u r p \le}{\text{E}} \textcolor{w h i t e}{a a a a a a \textcolor{b l a c k}{-} a a a a a a a a a a a a a a \textcolor{b l a c k}{s} a a a a a a a a a a a \textcolor{b l a c k}{\textcolor{red}{2} s}}$

This time, the solubility product constant would be

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

$2.9 \cdot {10}^{- 12} = 4 {s}^{3} \implies s = \sqrt{\frac{2.9 \cdot {10}^{- 12}}{4}} = 9.0 \cdot {10}^{- 5}$

The solubility of the salt decreases in a solution that contains hydroxide anions because of the common-ion effect.