Question

The concentration of the ${Ag}^{+}$ ion in a saturated solution of ${Ag}_{2}Cr{O}_4$ at ${20}^{o}C$ is $1.5\times {10}^{-4}mol{L}^{-1}$. The solubility product constant of ${Ag}_{2}Cr{O}_4$ at ${20}^{o}C$ is:

• A. $3.4\times {10}^{-12}$
• B. $1.7\times {10}^{-10}$
• C. $1.7\times {10}^{-8}$
• D. $1.7\times {10}^{-12}$
  

Answer 1

The correct option is D $1.7\times {10}^{-12}$

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

now, $\left[A{g}^{+}\right]=1.5\times {10}^{-4}M$

$\therefore \left[Cr{O}_4^{2-}\right]=\frac{1}{2}\times \left[A{g}^{+}\right]=\frac{1.5\times {10}^{-4}}{2}=0.75\times {10}^{-4}M$

Solubility product of $A{g}_{2}Cr{O}_4,K_{sp}=\left[\right(A{g}^{+}){]}^2\times [Cr{O}_4^{2-}]$

$=(1.5\times {10}^{-4}{)}^2\times (0.75\times {10}^{-4})$

$=1.7\times {10}^{-12}$