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)

Precipitation occurs, if $Q_c>K_{sp}$. The maximum concentration of $Ba{\left(N{O}_3\right)}_3$ or $Ba2+$ without precipitation is obtained if

$Q_c=K_{sp}=\left[B{a}^{2+}\right]\left[Cr{O}_4^{2-}\right]$

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

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

$=0.4\times 10^{-6}M$

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