Question

What will be the solubility of $Cd(CN{)}_2$ in $0.05\text{M}NaCN$ solution if the $K_{sp}$ is $1.0\times {10}^{-8}{\text{mol}}^3{\text{L}}^{-3}$?
$Cd(CN{)}_2\rightleftharpoons C{d}^{2+}+2C{N}^{-}NaCN\to N{a}^{+}+C{N}^{-}$

• A. $1.0\times {10}^{-6}$
• B. $1.0\times {10}^{-8}$
• C. $2\times {10}^{-8}$
• D. $4\times {10}^{-6}$

Answer 1

The correct option is D $4\times {10}^{-6}$

$Cd(CN{)}_2\rightleftharpoons C{d}^{2+}+2C{N}^{-}S2SNaCN\to N{a}^{+}+C{N}^{-}0.05M{K}_{sp}=S\times (0.05{)}^2\text{Here S is negligible than 0.05.}\text{So,}\left[C{N}^{-}\right]\approx 0.05S=\frac{1\times {10}^{-8}}{0.05\times 0.05}=4\times {10}^{-6}$