Question

The solubility product, $K_{sp}$ of $A{g}_2$ $Cr{O}_4$ is $3.2\times {10}^{-11}$. Its solubility in mol L${}^{-1}$ is:

• A. $4\times {10}^{-4}$
• B. $2\times {10}^{-4}$
• C. $2\times {10}^{-3}$
• D. $4\times {10}^{-6}$

Answer 1

Answer: (B)

$2\times {10}^{-4}$

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

Let the solubility be $S$.

Then, $K_{sp}=$$(2S{)}^2$ x $(1S{)}^1$

$K_{sp}=3.2\times {10}^{-11}=4S^3$

$S=\sqrt[3]{33.2\times {10}^{-11})/4}$ $=0.0002$ $=2\times {10}^{-4}$ $mol.L^{-1}$