Solubility product (K_{sp}) of salts of type | vedclass

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(D) For $MX$: $K_{sp} = S^2 = 4 	imes 10^{-8} \implies S = 2 	imes 10^{-4} \, M$.For $MX_2$: $K_{sp} = 4S^3 = 3.2 	imes 10^{-14} \implies S^3 = 0.8

Vedclass · Accepted Answer

$MX > M_3X > MX_2$

Vedclass · Answer

(D) For $MX$: $K_{sp} = S^2 = 4 	imes 10^{-8} \implies S = 2 	imes 10^{-4} \, M$.For $MX_2$: $K_{sp} = 4S^3 = 3.2 	imes 10^{-14} \implies S^3 = 0.8 	imes 10^{-14} = 8 	imes 10^{-15} \implies S = 2 	imes 10^{-5} \, M$.For $M_3X$: $K_{sp} = 27S^4 = 2.7 	imes 10^{-15} \implies S^4 = 10^{-16} \implies S = 10^{-4} \, M$.Comparing the solubilities: $2 	imes 10^{-4} > 10^{-4} > 2 	imes 10^{-5}$.Thus,the order is $MX > M_3X > MX_2$.

Solubility product $(K_{sp})$ of salts of type $MX$,$MX_2$,and $M_3X$ at temperature $T$ are $4 \times 10^{-8}$,$3.2 \times 10^{-14}$,and $2.7 \times 10^{-15}$ respectively. The solubility of the salts at temperature $T$ are in the order:

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