Question

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

• 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)

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

$K_{SP}=\left[{Mg}^{2+}\right]{\left[{OH}^{-}\right]}^{2}9\times {10}^{-12}=0.01\times {\left[{OH}^{-}\right]}^2\frac{{\left[{OH}^{-}\right]}^2}{{10}^{-2}}={\left[{OH}^{-}\right]}^{2}9\times {10}^{-18}={\left[{OH}^{-}\right]}^2\left[{OH}^{-}\right]=\sqrt{9\times {10}^{-10}}=3\times {10}^{-5}$