Question

The solubility (in mol $L^{-1}$) of $AgCl(K_{sp}=1.0\times 10^{-10})$ in a $0.1MKCl$ solution will be

• A. $1.0\times 10^{-9}$
• B. $1.0\times 10^{-10}$
• C. $1.0\times 10^{-5}$
• D. $1.0\times 10^{-11}$

Answer 1

Answer: (A)

Let solubility of $AgCl=xmole/L$
$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$
i.e., $K_{sp(AgCl)}=x\times x$
$KCl\to K^{+}\underset{\text{0.1}}{C{l}^{-}}$
$[C{l}^{-}]$ from $KCl=0.1m$
Total $[C{l}^{-}]$ in solution $=x+0.1$
$K_{sp}(AgCl)=[A{g}^{+}][C{l}^{-}]=x(x+0.1)$
$1.0\times 10^{-}10=x(x+0.1)$
$1.0\times 10^{10}=x^2+0.1x$
$1.0\times 10-10=0.1x(as{x}^2<<1)$
$x=1.0\times 10^{-9}mol/L$