Question

The $K_{sp}$ of $A{g}_{2}Cr{O}_4$ is $1.1\times {10}^{-12}$ at 298K. The solubility (in mol/L) of $A{g}_{2}Cr{O}_4$ in 0.1M $AgN{O}_3$ solution is:

• A. $1.1\times {10}^{-11}$
• B. $1.1\times {10}^{-10}$
• C. $1.1\times {10}^{-12}$
• D. $1.1\times {10}^{-9}$

Answer 1

The correct option is B

$1.1\times {10}^{-10}$

$A{g}_{2}Cr{O}_4$ is a weak electrolyte and will clearly ionize to a small extent compared to the strong electrolyte $AgN{O}_3$, which will complete dissociate into $A{g}^{+}$ and $N{O}_3^{-}$ ions. So the majority concentration of - in fact, for all practical purposes - comes from $AgN{O}_3$. Hence, for the calculation,
$\left[A{g}^{+}\right]=0.1M$
$K_{sp}=\left[A{g}^{+}{]}^2\right[Cr{O}_4^{-2}]=(0.1{)}^2\left(s\right)$
Solving, $s=1.1\times {10}^{-10}$