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)

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

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

Suppose its solubility in $0.1MNaCl$ is $xmol/L$

$AgCl\rightleftharpoons \underset{x}{A{g}^{+}}+\underset{x}{C{l}^{-}}$

$NaCl\rightleftharpoons \underset{0.1M}{N{a}^{+}}+\underset{0.1M}{C{l}^{-}}$

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

$K_{sp}$ of $AgCl=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

$=x\times (x+0.1)$

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

Higher power of $x$ are neglected

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

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