Question

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

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

Answer 1

Answer: (A)

Given, concentration of $NaCl=0.2M$
$K_{sp}(AgCl)=1.20\times 10^{-10}$
Let the solubility of $AgCl$ in $NaCl=x$
$AgCl\to A{g}^{+}+C{l}^{-}$

$\therefore \left[A{g}^{+}\right]=x$ and $\left[C{l}^{-}\right]=(x+0.2)$
$\therefore K_{sp}(AgCl)=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$=x(x+0.2)$
$=x^2+0.2x$
$\therefore K_{sp}=0.2x\left(x^2<<1\right)$
or $1.2\times 10^{-10}=0.2x$
$\therefore x=6\times 10^{-10}$