Question

The solubility of hydroxide $A(OH{)}_2$ in a buffer of $pH=11$ is found to be $0.0416{\text{g L}}^{-1}$. Molar solubility (in ${\text{mol L}}^{-1}$) of $A(OH{)}_2$ in pure water would be:
$($Given: molecular weight of $A(OH{)}_2=104{\text{g mol}}^{-1}$ and $(100{)}^{\frac{1}{3}}=4.64)$

• A. $2.32\times {10}^{-3}$
• B. $4.64\times {10}^{-3}$
• C. $2.32\times {10}^{-4}$
• D. $4.64\times {10}^{-4}$

Answer 1

The correct option is D $4.64\times {10}^{-4}$

$A(OH{)}_2\rightleftharpoons A^{2+}+2O{H}^{-}S=\frac{0.0416{\text{g L}}^{-1}}{104{\text{g mol}}^{-1}}[A^{2+}]=4\times {10}^{-4}{\text{mol L}}^{-1}[O{H}^{-}]={10}^{-pOH}={10}^{-3}{\text{mol L}}^{-1}{K}_{sp}=[A^{2+}\left]\right[O{H}^{-}{]}^2=(4\times {10}^{-4}){\left({10}^{-3}\right)}^2=4\times {10}^{-10}{\text{mol}}^3{\text{L}}^{-3}{K}_{sp}=4s^3\text{for}A(OH{)}_2$
$S=3\sqrt{\frac{K_{sp}}{4}}=3\sqrt{\frac{4\times {10}^{-10}}{4}}=4.64\times {10}^{-4}$