Solubility product constants (K_{sp}) of silv | vedclass

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(C) To find the solubility $(S)$ of each salt,we use the $K_{sp}$ expressions:$1$. For $AgCl$ ($1:1$ type): $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} =

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$Ag_2CO_3$

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(C) To find the solubility $(S)$ of each salt,we use the $K_{sp}$ expressions:$1$. For $AgCl$ ($1:1$ type): $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} = \sqrt{2 	imes 10^{-10}} \approx 1.41 	imes 10^{-5} \ M$.$2$. For $AgBr$ ($1:1$ type): $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} = \sqrt{5 	imes 10^{-13}} \approx 7.07 	imes 10^{-7} \ M$.$3$. For $Ag_2CO_3$ ($2:1$ type): $K_{sp} = 4S^3 \implies S = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{8 	imes 10^{-12}}{4}} = \sqrt[3]{2 	imes 10^{-12}} \approx 1.26 	imes 10^{-4} \ M$.$4$. For $AgI$ ($1:1$ type): $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} = \sqrt{8 	imes 10^{-17}} \approx 8.94 	imes 10^{-9} \ M$.Comparing the values,$1.26 	imes 10^{-4} > 1.41 	imes 10^{-5} > 7.07 	imes 10^{-7} > 8.94 	imes 10^{-9}$. Thus,$Ag_2CO_3$ has the highest solubility.

Solubility product constants $(K_{sp})$ of silver salts are given as $AgCl = 2 \times 10^{-10}$,$AgBr = 5 \times 10^{-13}$,$Ag_2CO_3 = 8 \times 10^{-12}$,and $AgI = 8 \times 10^{-17}$. Which salt shows the highest solubility?

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