Question

The solubility product constant $Ksp$ of $Mg(OH{)}_2$ is $9.0\times {10}^{-12}$. If a solution is $0.010M$ with respect to $M{g}^{2+}$ ion. What is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg(OH{)}_2$?

• A. $1.5\times {10}^{-7}M$
• B. $3.0\times {10}^{-7}M$
• C. $1.5\times {10}^{-5}M$
• D. $3.0\times {10}^{-5}M$

Answer 1

Answer: (D)

$3.0\times {10}^{-5}M$

$Ksp\left(Mg\right(OH{)}_2)=9.0\times {10}^{-12}$

$\left(Mg\right(OH{)}_2\leftrightharpoons M{g}^{2+}+2[OH{]}^{-}$

$Ksp=\left[M{g}^{2+}\right][O{H}^{-}{]}^2$

$9\times {10}^{-12}=\left({10}^{-12}\right)(O{H}^{-}{)}^2$

$[O{H}^{-}{]}^2=3^2\times ({10}^{-5}{)}^2$

$\left[O{H}^{-}\right]=3.0\times {10}^{-5}M$

Maximum Hydroxide-ion concentration.