Question

The solubility of $AgCl$ in $0.2M$ $NaCl$ solution is ($K_{sp}$ of $AgCl=1.20\times {10}^{-10}$)

• A. $6.0\times {10}^{-10}M$
• B. $1.2\times {10}^{-10}M$
• C. $0.2M$
• D. $0.2\times {10}^{-10}M$

Answer 1

The correct option is A $6.0\times {10}^{-10}M$

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

$\underset{0.02}{NaCl}\rightleftharpoons \underset{0.02}{N{a}^{+}}+\underset{0.02}{C{l}^{-}}$

$K_{sp}AgCl=1.20\times {10}^{-10}$

$K_{sp}AgCl=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=a\times [a+0.2]=a^2+0.2a$

$a^2$ is a very small so it is a neglected.

$K_{sp}AgCl=0.2a$

$1.20\times {10}^{-10}=0.2a$

$a=\frac{1.20\times {10}^{-10}}{0.20}=6\times {10}^{-10}mole$