Question

The solubility product of $BaCr{O}_4$ is $2.4\times {10}^{-10}{M}^2$. The maximum concentration of $Ba{\left(N{O}_3\right)}_2$ 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

Answer: (A)

$4\times {10}^{-7}M$

Precipitation will occur when $Q_{sp}>K_{sp}$

The solubility product of $BaCr{O}_4$ is $2.4\times {10}^{-10}$

Here, $\left[Cr{O}_4^{2-}\right]=6\times {10}^{-4}$

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