If the solubility product of CuS is 6 times 1 | vedclass

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(A) The solubility product of $CuS$ is given as $K_{sp} = 6 	imes 10^{-16}$.Let $s$ be the solubility of $CuS$ in $mol \ L^{-1}$.The dissociation equ

Vedclass · Accepted Answer

$2.45 	imes 10^{-8} \ M$

Vedclass · Answer

(A) The solubility product of $CuS$ is given as $K_{sp} = 6 	imes 10^{-16}$.Let $s$ be the solubility of $CuS$ in $mol \ L^{-1}$.The dissociation equilibrium is: $CuS(s) \leftrightarrow Cu^{2+}(aq) + S^{2-}(aq)$.The solubility product expression is $K_{sp} = [Cu^{2+}][S^{2-}] = s 	imes s = s^2$.Equating the values: $s^2 = 6 	imes 10^{-16}$.Taking the square root: $s = \sqrt{6 	imes 10^{-16}} = 2.45 	imes 10^{-8} \ mol \ L^{-1}$.Thus,the maximum molarity of $CuS$ in an aqueous solution is $2.45 	imes 10^{-8} \ M$.

If the solubility product of $CuS$ is $6 \times 10^{-16}$,calculate the maximum molarity of $CuS$ in aqueous solution.

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