Question

Solubility of $AgCl$ in 0.1 M calcium chloride solution is:
$(K_{sp}\text{of}AgCl=1.8\times {10}^{–10})$

• A. $1.8\times {10}^{-9}M$
• B. $9\times {10}^{-10}M$
• C. $1.8\times {10}^{-8}M$
• D. $1.3\times {10}^{-5}M$

Answer 1

Answer: (B)

$CaC{l}_2\to \underset{0.1}{C{a}^{2+}}+\underset{0.2}{2C{l}^{-}}$
$AgCl\to \underset{S}{A{g}^{+}}+\underset{(S+0.2)}{C{l}^{-}}$
$\left[A{g}^{+}\right]\left[C{l}^{-}\right]=K_{sp}$
$s(s+0.2)=1.8\times {10}^{-10}$
$s^2+0.2s=1.8\times {10}^{-10}$
$s=\frac{1.8\times {10}^{-10}}{0.2}\left[s^2\text{is neglected}\right]$
$=9\times {10}^{-10}$