Question

The solubility of $ AgCl $ is $ 1 \times 10^{-5}\, mol/L $ . Its solubility in $ 0.1 $ molar sodium chloride solution is

• A. $ 1 \times 10^{-10} $
• B. $ 1 \times 10^{-5} $
• C. $ 1 \times 10^{-9} $
• D. $ 1 \times 10^{-4} $

Answer 1

Answer: (C)

$K_{s p}$ of $AgCl =(\text { solubility of } AgCl )^{2}$

$=\left(1 \times 10^{-5}\right)^{2}=1 \times 10^{-10}$

Suppose its solubility in $0.1 M NaCl$ is $x mol / L$

$AgCl \rightleftharpoons \underset{x}{Ag ^{+}}+ \underset{x}{Cl ^{-}}$

$NaCl \rightleftharpoons \underset{0.1M}{Na ^{+}}+\underset{0.1 M }{ Cl ^{-}}$

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

$K_{s p}$ of $AgCl =\left[ Ag ^{+}\right]\left[ Cl ^{-}\right]$

$=x \times(x+0.1)$

$1 \times 10^{-10}=x^{2}+0.1 x$

Higher power of $x$ are neglected

$1 \times 10^{-10} =0.1 x$

$x =1 \times 10^{-9} M$