Question

At ${20}^{o}C$ the $A{g}^{+}$ ion concentration in a saturated solution of $A{g}_{2}Cr{O}_4$ is $1.5\times {10}^{-4}$ mole/L.At ${20}^{o}C$ the solubility product of $A{g}_{2}Cr{O}_4$ would be:

• A. $3.375\times {10}^{-12}$
• B. $1.6875\times {10}^{-10}$
• C. $1.6875\times {10}^{-12}$
• D. $1.6875\times {10}^{-11}$

Answer 1

Answer: (C)

$1.6875\times {10}^{-12}$

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

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

$=(1.5\times {10}^{-4}{)}^2\times \left(\frac{1.5\times {10}^{-4}}{2}\right)=1.6875\times {10}^{-12}$