Question

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

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

Answer 1

The correct option is A $2\times {10}^{-4}M$

$A{g}_{2}Cr{O}_4\to 2A{g}^{+}+Cr{O}_4^{2-}$

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

$S={\left(\frac{k_{sp}}{4}\right)}^{1/3}\Rightarrow {\left(\frac{32\times {10}^{-12}}{4}\right)}^{1/3}$

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