Question

The $K_{sp}$ of $A{g}_{2}Cr{O}_4,AgCl,AgBr$ and $AgI$ are respectively, $1.1\times {10}^{-12}$, $1.8\times {10}^{-10}$, $5.0\times {10}^{-13}$ and $8.3\times {10}^{-17}$. Which of the following salts will precipitate last if $AgN{O}_3$ solution is added to the solution containing equal moles of $NaCl,NaBr,NaI$ and $N{a}_{2}Cr{O}_4$?

• A. $A{g}_{2}Cr{O}_4$
• B. $AgI$
• C. $AgCl$
• D. $AgBr$

Answer 1

Answer: (A)

$A{g}_{2}Cr{O}_4$

1. $A{g}_{2}Cr{O}_4\rightleftharpoons 2A{g}^{+}+{Cr{O}_4}^{2-}$

$Ksp=(2s{)}^2\times s=4s^2$

$Ksp=(1.1\times {10}^{-12})$

$S=3\sqrt{\frac{Ksp}{4}}=6.5\times {10}^{-5}$M

2. $AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$

$Ksp=S\times S$ ; $Ksp=1.8\times {10}^{-10}$

$S=\sqrt{Ksp}=1.34\times {10}^{-5}$M

3. $AgBr\rightleftharpoons A{g}^{+}+B{r}^{-}$

$Ksp=S\times S$ ; $Ksp=5\times {10}^{-13}$

$S=\sqrt{Ksp}=0.71\times {10}^{-6}$M

4. $AgI\rightleftharpoons A{g}^{+}+I^{-}$

$Ksp=S\times S$ ; $Ksp=8.3\times {10}^{-17}$

$S=\sqrt{Ksp}=0.9\times {10}^{-8}$M

$\therefore$ Solubility of $A{g}_{2}Cr{O}_4$ is highest, so it will precipitate last.