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In aqueous solution of $AgCl(s)$, $AgBr(s)$ and $AgI(s)$. Find the ratio $[Cl^- ]$ : $[Br- ]$ : $[I^-]$ at equilibrium if $K_{sp}\, (AgCl)=10^{-10}$, $K_{sp}\, (AgBr) = 10^{-13}$ , $K_{sp}\, (AgI) = 10^{-16}$
${10^6}:10:{10^2}$
$1:\frac{1}{{10}}:\frac{1}{{100}}$
${10^6}:{10^3}:1$
${10^3}:1:{10^6}$
Solution
$\mathrm{AgCl} \rightarrow \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \ldots(1) \mathrm{Ksp}=10^{-10}$
$\mathrm{s}_{1}+\mathrm{s}_{2}+\mathrm{s}_{3} \quad \mathrm{s}_{1}$
$\mathrm{AgBr} \rightarrow \mathrm{Ag}^{+}+\mathrm{Br}^{-} \ldots(2)$
$\mathrm{Ksp}=10^{-13}$
$\mathrm{s}_{1}+\mathrm{s}_{2}+\mathrm{s}_{3} \quad \mathrm{S}_{2}$
$\mathrm{AgI} \rightarrow \mathrm{Ag}^{+}+\mathrm{I}^{-} \ldots(3)$
$\mathrm{Ksp}=10^{-16}$
$\mathrm{s}_{1}+\mathrm{s}_{2}+\mathrm{s}_{\mathrm{g}} \quad \mathrm{S}_{3}$
$\frac{10^{-10}}{10^{-13}}=\frac{s_{1}}{s_{2}} \quad \frac{10^{-13}}{10^{-16}}=\frac{s_{2}}{s_{3}}$
$\left[\mathrm{Cl}^{-}\right]:\left[\mathrm{Br}^{-}\right]:\left[\mathrm{I}^{-}\right]=\mathrm{s}_{1}: \mathrm{s}_{2}: \mathrm{s}_{3}$
$10^{3} s_{2}: s_{2}: \frac{1}{10^{3}} s_{2}=10^{6}: 10^{3}: 1$
