The $K_{sp}$ of $Ag_2CrO_4$, $AgCl$, $AgBr$ and $AgI$ are, respectively, $1.1 imes 10^{-12}$, $1.8 imes 10

Question

The $K_{sp}$ of $Ag_2CrO_4$, $AgCl$, $AgBr$ and $AgI$ are, respectively, $1.1 	imes 10^{-12}$, $1.8 	imes 10^{-10}$, $5.0 	imes 10^{-13}$, and $8.3 	imes 10^{-17}$. The salt that precipitates last if $AgNO_3$ solution is added to the solution containing equal moles of $NaCl$, $NaBr$, $NaI$ and $Na_2CrO_4$ is:

Accepted Answer

Let the concentration of $Cl^-$, $Br^-$, $I^-$, and $CrO_4^{2-}$ be $c$.
The concentration of $Ag^+$ required to precipitate each salt is calculated as follows:
For $AgCl$: $[Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8 	imes 10^{-10}}{c}$
For $AgBr$: $[Ag^+] = \frac{K_{sp}}{[Br^-]} = \frac{5.0 	imes 10^{-13}}{c}$
For $AgI$: $[Ag^+] = \frac{K_{sp}}{[I^-]} = \frac{8.3 	imes 10^{-17}}{c}$
For $Ag_2CrO_4$: $[Ag^+] = \sqrt{\frac{K_{sp}}{[CrO_4^{2-}]}} = \sqrt{\frac{1.1 	imes 10^{-12}}{c}}$
Assuming $c = 0.1	ext{ M}$ for comparison:
$[Ag^+]$ for $AgI = 8.3 	imes 10^{-16}	ext{ M}$
$[Ag^+]$ for $AgBr = 5.0 	imes 10^{-12}	ext{ M}$
$[Ag^+]$ for $AgCl = 1.8 	imes 10^{-9}	ext{ M}$
$[Ag^+]$ for $Ag_2CrO_4 = \sqrt{1.1 	imes 10^{-11}} \approx 3.32 	imes 10^{-6}	ext{ M}$
The salt that requires the highest concentration of $Ag^+$ to exceed its ionic product over $K_{sp}$ will precipitate last. Here, $Ag_2CrO_4$ requires the maximum $[Ag^+]$, so it will precipitate last.

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