Question

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

Ionic product of $M\left( OH _{2}=5 \times 10^{-10}\right.$

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

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

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

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