Question

The solubility product of silver chloride is $1.5625\times {10}^{-10}$ at ${25}^{o}C$. Find its solubility in $g{L}^{-1}$.

Answer 1

Let the solubility of $AgCl$ be $S$ $mol{litre}^{-1}$
$AgCl\rightleftharpoons {Ag}^{+}+{Cl}^{-}$
Hence, $S^2=1.5625\times {10}^{-10}$
or $S=1.25\times {10}^{-5}mol{L}^{-1}$
molecular mass of $AgCl=(108+35.5)=143.5$
So, Solubility in $g{litre}^{-1}=Molmass\times S=1.79\times {10}^{-3}g{L}^{-1}$