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)

$1\times {10}^{-9}$

$K_{sp}$ of $AgCl={\left(solubilityofAgCl\right)}^2$

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

Suppose its solubility in $0.1M$ $NaCl$ is $xmol/L$

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

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

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

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

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

The higher power of $x$ is neglected

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

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