Question

The molarity of $M{g}^{2+}$ ions in a saturated solution of $M{g}_3{\left(P{O}_4\right)}_2$ whose solubility product is $1.08\times {10}^{-13}{M}^5$ is

• A. $1.0\times {10}^{-3}M$
• B. $2.0\times {10}^{-3}M$
• C. $3.0\times {10}^{-3}M$
• D. $4.0\times {10}^{-3}M$

Answer 1

The correct option is C $3.0\times {10}^{-3}M$

Let x moles per liter be the solubility of $M{g}_3(P{O}_4{)}_2$.
The dissociation equilibrium of $M{g}_3(P{O}_4{)}_2$ is $\underset{x}{M{g}_3(P{O}_4{)}_2}\rightleftharpoons \underset{3x}{3M{g}^{2+}}+\underset{2x}{2P{O}_4^{3-}}$
The solubility product is $1.08\times {10}^{-13}{M}^5$.
The expression for the solubility product is $K_{sp}=[M{g}^{2+}{]}^3+[P{O}_4^{3-}{]}^2$.
Substitute values in the above expression.
$1.08\times {10}^{-13}{M}^5=\left(3x{)}^3\right(2x{)}^2=108x^5$.
Hence, $x=9.88\times {10}^{-4}$.
$\left[M{g}^{2+}\right]=3x=3\times 9.88\times {10}^{-4}=3.0\times {10}^{-3}M$.