Question

The solubility of sparingly soluble salt $CdC{O}_3$ is $2\times {10}^{-6}$ $mol{L}^{-1}$ at ${25}^{o}C$. The $K_{sp}$ value for $CdC{O}_3$ is:

• A. $8\times {10}^{-12}mo{l}^2{L}^{-2}$
• B. $2\times {10}^{-12}mo{l}^2{L}^{-2}$
• C. $4\times {10}^{-12}mo{l}^2{L}^{-2}$
• D. $4\times {10}^{-15}mo{l}^2{L}^{-2}$

Answer 1

The correct option is C $4\times {10}^{-12}mo{l}^2{L}^{-2}$

Given, solubility, $s=2\times {10}^{-6}$ $mol{L}^{-1}$
Lets consider solid $CdC{O}_3$ is in contact with its saturated aqueous solution. The equilibrium is established between undissolved $CdC{O}_3$ and its ions. The equilibrium reaction is given by :
$CdC{O}_3\left(s\right)\rightleftharpoons C{d}^{2+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)t=0c00t=t_{eq}c-sss$
Solubility product: $K_{sp}=\left[C{d}^{2+}\right]\left[C{O}_3^{2-}\right]=s\times s$
$s^2=(2\times {10}^{-6}{)}^2{K}_{sp}=4\times {10}^{-12}mo{l}^2{L}^{-2}$