If the solubility product of CuS is 9 times 1 | vedclass

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Question

(C) The dissolution of $CuS$ in water is represented as: $CuS(s) \rightleftharpoons Cu^{2+}(aq) + S^{2-}(aq)$.Let the solubility of $CuS$ be $s \ mol/

Vedclass · Accepted Answer

$3 	imes 10^{-8} \ M$

Vedclass · Answer

(C) The dissolution of $CuS$ in water is represented as: $CuS(s) ightleftharpoons Cu^{2+}(aq) + S^{2-}(aq)$.Let the solubility of $CuS$ be $s \ mol/L$.Then,$[Cu^{2+}] = s$ and $[S^{2-}] = s$.The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [Cu^{2+}][S^{2-}] = s 	imes s = s^2$.Given $K_{sp} = 9 	imes 10^{-16}$.Therefore,$s^2 = 9 	imes 10^{-16}$.Taking the square root on both sides,$s = \sqrt{9 	imes 10^{-16}} = 3 	imes 10^{-8} \ M$.Thus,the maximum molarity of $CuS$ in an aqueous solution is $3 	imes 10^{-8} \ M$.

If the solubility product of $CuS$ is $9 \times 10^{-16}$,then what will be the maximum molarity of $CuS$ in an aqueous solution?

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