Chemistry - Ionic Equilibrium Question with Solution | TestHub

For,${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\rightleftharpoons {\text{2Ag}}^{\text{+}}{\text{+CrO}}_{\text{4}}^{\text{2-}}$
solubility product
${\text{K}}_{\text{sp}}\text{=}{\left(\text{25}\right)}^{\text{2}}{\text{×5=4s}}^{\text{3}}$
${\text{K}}_{\text{sp}}=\left(1.1\times {10}^{-12}\right)$
$\text{S}=3\sqrt{\frac{{\text{K}}_{\text{sp}}}{4}}=0.65\times {10}^{-4}$
For,$\text{AgCl}\rightleftharpoons {\text{Ag}}^{\text{+}}{\text{+Cl}}^{\text{-}}$
${\text{K}}_{\text{s}\text{p}}=\text{S}\times \text{S}$$\left({\text{K}}_{\text{s}\text{p}}=1.8\times {10}^{-10}\right)$
$\text{S=}\sqrt{{\text{K}}_{\text{sp}}}=1.34\times {10}^{-5}$
For,$\text{AgBr}\rightleftharpoons {\text{Ag}}^{\text{+}}{\text{+Br}}^{\text{-}}$
${\text{K}}_{\text{sp}}\text{=S×S}\left({\text{K}}_{\text{sp}}{\text{=5×10}}^{\text{-13}}\right)$
$\text{S=}\sqrt{K_{sp}}=0.71\times {10}^{-6}$
For,$\text{AgI}\rightleftharpoons {\text{Ag}}^{\text{+}}{\text{+I}}^{\text{-}}$
${\text{K}}_{\text{sp}}\text{=S×S}\left({\text{K}}_{\text{sp}}\text{=8}{\text{.3×10}}^{\text{-17}}\right)$
$\text{S=}\sqrt{{\text{K}}_{\text{s}\text{p}}}=0.9\times {10}^{-8}$
therefore, the solubility of$\text{A}{\text{g}}_2\text{C}\text{r}{\text{O}}_4$is highest so it is precipitate last.