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. $5\times {10}^{-9}M$
• E. $5\times {10}^{-16}M$

Answer 1

Answer: (B)

$5\times {10}^{-8}M$

Ionic product of $M(OH{)}_2=5\times {10}^{-10}$

or, $\left[M^{+}\right][O{H}^{-}{]}^2=5\times {10}^{-10}$

Given: $\left[O{H}^{-}\right]={10}^{-1}mol/L$

So, $M^{2+}=\frac{5\times {10}^{-10}}{[{10}^{-1}{]}^2}$

$=\frac{5\times {10}^{-10}}{{10}^{[}-2]}=5\times {10}^{-8}M$

Hence, option B is correct.