Question

The solubility product of $AgCl$ is $4.0\times {10}^{-10}$ at $298K$ . The solubility of $AgCl$ in $0.04MCaC{l}_2$ will be:

• A. $5.0\times {10}^{-9}M$
• B. $2.0\times {10}^{-5}M$
• C. $2.2\times {10}^{-4}M$
• D. $1.0\times {10}^{-4}M$

Answer 1

Answer: (A)

In $CaC{l}_2,\left[C{l}^{-}\right]=2\times 0.004=0.008mol{L}^{-1}$
$\left[C{l}^{-}\right]$from $AgCl$ is too small and is neglected
$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$4\times 10^{-10}=\left[A{g}^{+}\right]\times 0.08$
$\left[A{g}^{+}\right]=\frac{4\times 10^{-10}}{0.08}$
$=\frac{4}{8}\times 10^{-8}$
$=5.0\times 10^{-9}M$