Question

If the ionic product of $M({OH)}_2$ is $5\times {10}^{-10}$ ,then the molar solubility of $M({OH)}_2$ in $0.1MNaOH$ is:

• A. $5\times {10}^{-12}M$
• B. $5\times {10}^{-8}M$
• C. $5\times {10}^{-10}M$
• D. $5\times {10}^{-9}M$

Answer 1

The correct option is B $5\times {10}^{-8}M$

$\left[O{H}^{-}\right]={10}^{-1}$

$M(OH{)}_2⟶M^{2+}+2O{H}^{-}$

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

$\Rightarrow 5\times {10}^{-10}=S\times {10}^{-2}$

$\Rightarrow S=5\times {10}^{-8}M$

Hence molar solubility= $5\times {10}^{-8}M$