Question

The solubility product of $A{g}_{2}Cr{O}_4$ is $32\times {10}^{-12}$. What is the concentration of $Cr{O}_4^{-}$ ions in that solution?

• A. $2\times {10}^{-4}m/s$
• B. $16\times {10}^{-4}m/s$
• C. $8\times {10}^{-4}m/s$
• D. $8\times {10}^{-8}m/s$

Answer 1

The correct option is A $2\times {10}^{-4}m/s$

Let $SM$ be the solubility of $A{g}_{2}Cr{O}_4$.

$\left[A{g}^{+}\right]=2S;\left[Cr{O}_4^{2-}\right]=S$

$K_{sp}=\left[A{g}^{+}{]}^2\right[Cr{O}_4^{2-}]=(2S{)}^{2}S=4S^3$

$32\times {10}^{-12}=4S^3$

$S=2\times {10}^{-4}M$

Hence, the concentration of chromate ions $\left(CrO{4}^{2-}\right)$ in the solution is $2\times {10}^{-4}$.