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Q.
Solubility product of $AgCl$ is $2.8 \times 10^{-10}$ at $25^{\circ} C .$ Calculate solubility of the salt in $0.1 \,M\, AgNO _{3}$ solution
Equilibrium
Solution:
In $0.1 \,M \,AgNO _{3}$
$AgNO _{3} \rightleftharpoons Ag ^{+}+ NO _{3}^{-}$
$AgCl \rightleftharpoons Ag ^{+}+ Cl ^{-}$
$K_{sp}=\left[ Ag ^{+}\right]\left[ Cl ^{-}\right]$
Now $\left[ Ag ^{+}\right]$ can be taken as $\left[ AgNO _{3}\right]$ while $[Cl ^{-}]$ is the solubility of $AgCl$
$\therefore Cl =\frac{K_{ sp }}{\left[ Ag ^{+}\right]}=\frac{2.8 \times 10^{-10}}{0.1}$
$\therefore $ Solubility of $AgCl =2.8 \times 10^{-9}$ mol liter.