Question

Consider the following solubility data for various chromates at 25${}^o$C.
$K_{sp}$
$A{g}_{2}Cr{O}_4$ $9.0\times {10}^{-12}$
$BaCr{O}_4$ $2.0\times {10}^{-10}$
$PbCr{O}_4$ $1.8\times {10}^{-14}$
The chromate that is the most soluble in water at 25${}^o$C on a molar basis is

• A. $A{g}_{2}Cr{O}_4$
• B. $BaCr{O}_4$
• C. $PbCr{O}_4$
• D. impossible to determine
• E. none of these

Answer 1

The correct option is A $A{g}_{2}Cr{O}_4$

Let S represent the molar solubility.
For $A{g}_{2}Cr{O}_4$, $4S^3=K_{SP}=9.0\times {10}^{-12}$
$S=1.31\times {10}^{-4}M$
For $BaCr{O}_4$, $S^2=K_{SP}=2.0\times {10}^{-10}$.
$S=1.4\times {10}^{-5}M$
For $PbCr{O}_4$, $S^2=K_{SP}=1.8\times {10}^{-14}$.
$S=1.35\times {10}^{-7}M$
Hence, $A{g}_{2}Cr{O}_4$ is most soluble in water at ${25}^{o}C$.