Question

What will be the solubility of $AgCl$ in a $0.1M$ $KCl$ solution?
$(K_{sp}AgCl=1.20\times {10}^{-10})$

• A. $0.1M$
• B. $1.2\times {10}^{-4}M$
• C. $1.2\times {10}^{-9}M$
• D. $1.2\times {10}^{-10}M$

Answer 1

Answer: (C)

$1.2\times {10}^{-9}M$

Let the solubility of $AgCl$ be $S$.

Then,
$Ag{Cl}^{}\rightleftharpoons {Ag}^{+}+{Cl}^{-}$
$S$ $S$ $S$
Potassium chloride is a strong electrolyte and is completely ionised. It shall provide ${Cl}^{-}$ ion concentration $=0.1M$

$\left[{Ag}^{+}\right]=S$
$\left[{Cl}^{-}\right]=(S+0.1)M$

$K_{sp}=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]=S\times (S+0.01)=S^2+0.1S$

Neglecting $S^3$ and $S^2$

$1.20\times {10}^{-10}=0.1S$

or $S=\frac{1.20\times {10}^{-10}}{0.1}=1.20\times {10}^{-9}mol{L}^{-1}$