The solubility product constants (K_{sp}) of | vedclass

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(D) For $CuS$ ($AB$ type): $K_{sp} = s^2 \implies s = \sqrt{K_{sp}} = \sqrt{10^{-37}} \approx 10^{-18.5} \ M$.For $Ag_2S$ ($A_2B$ type): $K_{sp} = 4s^

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$Ag_2S > CuS > HgS$

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(D) For $CuS$ ($AB$ type): $K_{sp} = s^2 \implies s = \sqrt{K_{sp}} = \sqrt{10^{-37}} \approx 10^{-18.5} \ M$.For $Ag_2S$ ($A_2B$ type): $K_{sp} = 4s^3 \implies s = \sqrt[3]{K_{sp}/4} = \sqrt[3]{10^{-44}/4} \approx 10^{-14.9} \ M$.For $HgS$ ($AB$ type): $K_{sp} = s^2 \implies s = \sqrt{K_{sp}} = \sqrt{10^{-54}} = 10^{-27} \ M$.Comparing the values: $10^{-14.9} > 10^{-18.5} > 10^{-27}$.Thus,the order of solubility is $Ag_2S > CuS > HgS$.

The solubility product constants $(K_{sp})$ of $CuS$,$Ag_2S$,and $HgS$ are $10^{-37}$,$10^{-44}$,and $10^{-54}$ respectively. The order of their solubility is:

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