Chemistry - Ionic Equilibrium Question with Solution | TestHub

Here$\mathrm{A}{\mathrm{g}}_3\mathrm{P}{\mathrm{O}}_4$is most soluble as it has highest${\mathrm{K}}_{\mathrm{s}\mathrm{p}}$
$\underset{\mathrm{S}}{\mathrm{A}{\mathrm{g}}_3\mathrm{P}{\mathrm{O}}_4}\to \underset{3\mathrm{S}}{3\mathrm{A}{\mathrm{g}}^{+}}+\underset{\mathrm{S}}{\mathrm{P}{\mathrm{O}}_4^{3-}}$
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}=\left[\mathrm{A}{\mathrm{g}}^{+}{]}^3\right[\mathrm{P}{\mathrm{O}}_4^{3-}]$
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}=27{\mathrm{S}}^4$
$\mathrm{S}={\left(\frac{{\mathrm{K}}_{\mathrm{s}\mathrm{p}}}{27}\right)}^{1/4}$