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

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(A) To find which compound precipitates first,we calculate the $[Ag^{+}]$ required for each:$1. \ AgBr: [Ag^{+}] = \frac{K_{sp}}{[Br^{-}]} = \frac{5 \

Vedclass · Accepted Answer

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

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

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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