Question

The degree of hydrolysis of $0.1MNaCN$ solution is $4\%$. What will be the solubility of $Al(OH{)}_3$ in this solution?
$\left[K_{sp}ofAl\right(OH{)}_3=6.4\times {10}^{-20}]$

• A. $0.04{\text{mol L}}^{-1}$
• B. ${10}^{-15}{\text{mol L}}^{-1}$
• C. ${10}^{-12}{\text{mol L}}^{-1}$
• D. $1.6\times {10}^{-4}{\text{mol L}}^{-1}$

Answer 1

The correct option is C ${10}^{-12}{\text{mol L}}^{-1}$

$C{N}^{-}+H_{2}O\rightleftharpoons HCN+O{H}^{-}$
$t=00.100t=eq.0.1-0.1h0.1h0.1h$

Here, the $O{H}^{-}$ ions produced from $Al(OH{)}_3$ will be negligible compared to the $O{H}^{-}$ ions produced from hydrolysis of $NaCN.$
So, $\left[O{H}^{-}\right]=0.1\times 0.04=4\times {10}^{-3}M$
$K_{sp}=\left[A{l}^{3+}\right][O{H}^{-}{]}^3$
$6.4\times {10}^{-20}=S\times [4\times {10}^{-3}{]}^3$
$S={10}^{-12}{\text{mol L}}^{-1}$