Question

What will be the solubility of $AgCl$ in $0.05$M $NaCl$ aqueous solution if solubility product of $AgCl$ is $1.5\times {10}^{-10}mo{l}^2{L}^{-2}$?

• A. $3\times {10}^{-9}mol{L}^{-1}$
• B. $0.05mol{L}^{-1}$
• C. $1.5\times {10}^{-5}mol{L}^{-1}$
• D. $3\times {10}^{9}mol{L}^{-1}$

Answer 1

The correct option is A $3\times {10}^{-9}mol{L}^{-1}$

$NaCl\to N{a}^{+}+C{l}^{-}$

$\left[C{l}^{-}\right]=0.05M$

$AgCl\to \underset{s}{A{g}^{+}}+\underset{s}{C{l}^{-}}$

$\left[A{g}^{+}\right]\left[C{l}^{-}\right]=Ksp$

$\left(5\right)(5+0.05)=1.5\times {10}^{-10}$

$5+0.05\simeq 0.05$

$5\times 0.05=1.5\times {10}^{-10}$

$5=\frac{1.5\times {10}^{-10}}{0.05}=3\times {10}^{-9}mo{l}^{-1}$