The solubility product constant for "Mg(OH)"_2 is 1.8 × 10^"-11". What would be the solubility of "Mg(OH)"_2 in 0.345 M "NaOH"?

The solubility product constant for ${\text{Mg(OH)}}_{2}$ is 1.8 × 10^"-11". What would be the solubility of ${\text{Mg(OH)}}_{2}$ in 0.345 M $\text{NaOH}$?

Jun 3, 2016

The solubility would be 1.5 × 10^"-10"color(white)(l) "mol/L".

Explanation:

Set up the chemical equation for the equilibrium:

$\textcolor{w h i t e}{m m m m m m} \text{Mg(OH)"_2"(s)" ⇌ "Mg"^(2+)"(aq)" + 2"OH"^"-""(aq)}$
$\text{E:/mol·L"^"-1} \textcolor{w h i t e}{m m m m m m m m m m} x \textcolor{w h i t e}{m m m m} 0.345 + 2 x$

Set up the solubility product expression:

${K}_{\text{sp" = ["Mg"^(2+)]["OH"^"-"]^2 = x(0.345 +2x)^2 = 1.8 × 10^"-11}}$

Check that $2 x \textcolor{w h i t e}{l} \text{<< 0.345}$

0.345/(1.8 × 10^"-11") = 1.9 × 10^10 > 400. ∴ $2 x \textcolor{w h i t e}{l} \text{<< 0.345}$,

Solve the ${K}_{\text{sp}}$ expression:

x(0.345)^2 = 1.8 × 10^"-11"

0.119x = 1.8 × 10^"-11"

x = (1.8 × 10^"-11")/0.119 = 1.5 × 10^"-10"

∴ Solubility of "Mg(OH)"_2 = 1.5 × 10^"-10" color(white)(l) "mol/L"