Chemistry - Ionic Equilibrium Question with Solution | TestHub

The question asks which silver compound will precipitate first when silver ions are added to a solution containing equal concentrations of bromide, chloride, carbonate, and arsenate ions. The compound that precipitates first will be the one that requires the lowest concentration of $A{g}^{+}$ to reach its solubility product, $K_{sp}$.

To determine which compound precipitates first, we need to calculate the $[A{g}^{+}]$ required for each compound to reach its $K_{sp}$. We are given that $[B{r}^{-}]=[C{l}^{-}]=[C{O}_3^{2-}]=[As{O}_4^{3-}]=0.1M$.

A. $AgBr(s)\rightleftharpoons A{g}^{+}(aq)+B{r}^{-}(aq)$, $K_{sp}=[A{g}^{+}][B{r}^{-}]=5\times 10^{-13}$

$[A{g}^{+}]=\frac{K_{sp}}{[B{r}^{-}]}=\frac{5\times 10^{-13}}{0.1}=5\times 10^{-12}M$

B. $AgCl(s)\rightleftharpoons A{g}^{+}(aq)+C{l}^{-}(aq)$, $K_{sp}=[A{g}^{+}][C{l}^{-}]=1.8\times 10^{-10}$

$[A{g}^{+}]=\frac{K_{sp}}{[C{l}^{-}]}=\frac{1.8\times 10^{-10}}{0.1}=1.8\times 10^{-9}M$

C. $A{g}_{2}C{O}_3(s)\rightleftharpoons 2A{g}^{+}(aq)+C{O}_3^{2-}(aq)$, $K_{sp}=[A{g}^{+}{]}^2[C{O}_3^{2-}]=8.1\times 10^{-12}$

$[A{g}^{+}{]}^2=\frac{K_{sp}}{[C{O}_3^{2-}]}=\frac{8.1\times 10^{-12}}{0.1}=8.1\times 10^{-11}$

$[A{g}^{+}]=\sqrt{8.1\times 10^{-11}}=9\times 10^{-6}M$

D. $A{g}_{3}As{O}_4(s)\rightleftharpoons 3A{g}^{+}(aq)+As{O}_4^{3-}(aq)$, $K_{sp}=[A{g}^{+}{]}^3[As{O}_4^{3-}]=1\times 10^{-22}$

$[A{g}^{+}{]}^3=\frac{K_{sp}}{[As{O}_4^{3-}]}=\frac{1\times 10^{-22}}{0.1}=1\times 10^{-21}$

$[A{g}^{+}]=\sqrt[3]{1\times 10^{-21}}=1\times 10^{-7}M$

Comparing the $[A{g}^{+}]$ required for each compound to precipitate:

$AgBr$: $5\times 10^{-12}M$

$AgCl$: $1.8\times 10^{-9}M$

$A{g}_{2}C{O}_3$: $9\times 10^{-6}M$

$A{g}_{3}As{O}_4$: $1\times 10^{-7}M$

The lowest $[A{g}^{+}]$ is required for $AgBr$ to precipitate.

FINAL ANSWER: A