Question

The solubility product of $BaCr{O}_4$ is $2.4\times {10}^{-10}{M}^2$. The maximum concentration of $Ba(N{O}_3{)}_2$ possible without precipitation in a $6\times {10}^{-4}M{K}_{2}Cr{O}_4$ solution is

• A. $4\times {10}^{-7}M$
• B. $1.2\times {10}^{10}M$
• C. $6\times {10}^{-4}M$
• D. $3\times {10}^{-4}M$

Answer 1

The correct option is A $4\times {10}^{-7}M$

Given
$K_{sp}=\left[B{a}^{+2}\right]\left[Cr{O}_4^{2-}\right]=2.4\times {10}^{-10}{M}^2$
In $6\times {10}^{-4}M{K}_{2}Cr{O}_4$ the $\left[Cr{O}_4^{-2}\right]=6\times {10}^{-4}M$

So $\left[B{a}^{+2}\right]\times 6\times {10}^{-4}=2.4\times {10}^{-10}$
Hence $\left[B{a}^{+2}\right]=4\times {10}^{-7}M$