Silver ions are added to a solution with [Br^ | vedclass

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(A) To determine which compound precipitates first,we calculate the $[Ag^{+}]$ required for each:$1$. For $AgBr$: $K_{sp} = [Ag^{+}][Br^{-}] \implies

Vedclass · Accepted Answer

$AgBr \ (K_{sp} = 5 	imes 10^{-13})$

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(A) To determine which compound precipitates first,we calculate the $[Ag^{+}]$ required for each:$1$. For $AgBr$: $K_{sp} = [Ag^{+}][Br^{-}] \implies [Ag^{+}] = \frac{5 	imes 10^{-13}}{0.1} = 5 	imes 10^{-12} \ M$$2$. For $AgCl$: $K_{sp} = [Ag^{+}][Cl^{-}] \implies [Ag^{+}] = \frac{1.8 	imes 10^{-10}}{0.1} = 1.8 	imes 10^{-9} \ M$$3$. For $Ag_2CO_3$: $K_{sp} = [Ag^{+}]^2[CO_3^{2-}] \implies [Ag^{+}]^2 = \frac{8.1 	imes 10^{-12}}{0.1} = 8.1 	imes 10^{-11} \implies [Ag^{+}] \approx 9 	imes 10^{-6} \ M$$4$. For $Ag_3AsO_4$: $K_{sp} = [Ag^{+}]^3[AsO_4^{3-}] \implies [Ag^{+}]^3 = \frac{1 	imes 10^{-22}}{0.1} = 10^{-21} \implies [Ag^{+}] = 10^{-7} \ M$Comparing the values: $5 	imes 10^{-12} The lowest $[Ag^{+}]$ required is for $AgBr$.

Silver ions are added to a solution with $[Br^{-}] = [Cl^{-}] = [CO_3^{2-}] = [AsO_4^{3-}] = 0.1 \ M$. Which compound will precipitate with the lowest $[Ag^{+}]$ concentration?

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