The K_{sp} of Ag_2CrO_4, AgCl, AgBr and AgI a | vedclass

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(B) When $AgNO_3$ is added to a solution containing equal concentrations of anions $(Cl^-, Br^-, I^-, CrO_4^{2-})$,the salt that requires the highest

Vedclass · Accepted Answer

$Ag_2CrO_4$

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(B) When $AgNO_3$ is added to a solution containing equal concentrations of anions $(Cl^-, Br^-, I^-, CrO_4^{2-})$,the salt that requires the highest concentration of $Ag^+$ to exceed its $K_{sp}$ will precipitate last.For $AgCl, AgBr, AgI$ (type $AB$ salts),$[Ag^+] = \frac{K_{sp}}{[Anion]}$. Since $[Anion]$ is equal,the salt with the highest $K_{sp}$ precipitates last.For $Ag_2CrO_4$ (type $A_2B$ salt),$K_{sp} = [Ag^+]^2 [CrO_4^{2-}]$,so $[Ag^+] = \sqrt{\frac{K_{sp}}{[CrO_4^{2-}]}}$.Comparing the required $[Ag^+]$:$AgCl: [Ag^+] = \frac{1.8 	imes 10^{-10}}{C}$$AgBr: [Ag^+] = \frac{5.0 	imes 10^{-13}}{C}$$AgI: [Ag^+] = \frac{8.3 	imes 10^{-17}}{C}$$Ag_2CrO_4: [Ag^+] = \sqrt{\frac{1.1 	imes 10^{-12}}{C}}$Comparing these values,the $[Ag^+]$ required for $Ag_2CrO_4$ is significantly higher than the others,meaning $Ag_2CrO_4$ will precipitate last.

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