Question

The solubility (in mol $L^{-1}$) of $AgCl (K_{sp }=1.0 \times 10^{-10})$ in a $0.1\,M \,KCl$ 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 $AgCl = x \,mole/L$
$AgCl \rightleftharpoons Ag^++Cl^-$
i.e., $K_{sp(AgCl)}=x \times x$
$KCl \rightarrow K^+ \underset{\text{0.1}}{Cl^-}$
$[Cl^-]$ from $KCl = 0.1\, m$
Total $[Cl^-]$ in solution $=x+0.1$
$K_{sp}(AgCl)=[Ag^+] [Cl^-] = x(x + 0.1)$
$1.0 \times 10^-{10} = x(x+0.1)$
$1.0 \times 10^{10}=x^2+0.1x$
$1.0 \times 10-10 = 0.1x\,(as\,x^2 <<1)$
$x=1.0 \times 10^{-9}\,mol/L$