Question

Calculate the minimum concentration of $C{O}_3^{2-}$ required to precipitate $NiC{O}_3$ in a solution containing $2.0\times {10}^{-5}M$ of $N{i}^{2+}?$
Given : ($K_{sp}$ for $NiC{O}_3=6.6\times {10}^{-10}$)

• A. $3.3\times {10}^{-9}M$
• B. $3.3\times {10}^{-5}M$
• C. $6.6\times {10}^{-5}M$
• D. $12.2\times {10}^{-5}M$

Answer 1

The correct option is B $3.3\times {10}^{-5}M$

$Q_{sp}=\text{ionic product of}NiC{O}_3.$
$K_{sp}=\text{solubility product of}NiC{O}_3.$
For precipitation, Ionic product $>K_{sp}$
i.e. $Q_{sp}>K_{sp}$

$NiC{O}_3\left(s\right)\rightleftharpoons N{i}^{2+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)\left]\right[N{i}^{2+}\left]\right[C{O}_3^{2-}]>K_{sp}[C{O}_3^{2-}]>\frac{K_{sp}}{\left[N{i}^{2+}\right]}[C{O}_3^{2-}]>\frac{6.6\times {10}^{-10}}{2\times {10}^{-5}}\therefore [C{O}_3^{2-}]>3.3\times {10}^{-5}M$
So, the $[C{O}_3^{2-}{]}_{min}=3.3\times {10}^{-5}M$