When 1.88 g of AgBr_{(s)} is added to a 10^{ | vedclass

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(D) The solubility equilibrium for $AgBr$ is: $AgBr_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Br^{-}_{(aq)}$.First,calculate the solubility product con

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$5 	imes 10^{-11} \ M$

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(D) The solubility equilibrium for $AgBr$ is: $AgBr_{(s)} ightleftharpoons Ag^{+}_{(aq)} + Br^{-}_{(aq)}$.First,calculate the solubility product constant $(K_{sp})$ using the given concentrations in $KBr$ solution:$K_{sp} = [Ag^{+}][Br^{-}] = (5 	imes 10^{-10} \ M) 	imes (10^{-3} \ M) = 5 	imes 10^{-13}$.When $AgBr_{(s)}$ is added to $10^{-2} \ M$ $AgNO_3$ solution,the common ion effect occurs,and the concentration of $Ag^{+}$ becomes $10^{-2} \ M$.Using the $K_{sp}$ value:$K_{sp} = [Ag^{+}][Br^{-}]$$5 	imes 10^{-13} = (10^{-2} \ M) 	imes [Br^{-}]$$[Br^{-}] = \frac{5 	imes 10^{-13}}{10^{-2}} = 5 	imes 10^{-11} \ M$.

When $1.88 \ g$ of $AgBr_{(s)}$ is added to a $10^{-3} \ M$ aqueous solution of $KBr$,the concentration of $Ag^{+}$ is $5 \times 10^{-10} \ M$. If the same amount of $AgBr_{(s)}$ is added to a $10^{-2} \ M$ aqueous solution of $AgNO_3$,the concentration of $Br^{-}$ is

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