Question

Determine the solubility of AgCl in 0.1 M $BaC{l}_2$. [$K_{sp}forAgCl=1\times {10}^{-10}$]

Answer 1

$\left[C{l}^{-}\right]={10}^{-1}$

$AgCl⟶A{g}^{+}+C{l}^{-}$

$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

$\therefore$ $1\times {10}^{-10}=S\times {10}^{-1}S={10}^{-9}$

$\therefore$ Solubility in $BaC{l}_2=1\times {10}^{-9}$