For Cr(OH)_3,the K_{sp} = 1.6 times 10^{-30}. | vedclass

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(C) The dissociation of $Cr(OH)_3$ is given by: $Cr(OH)_3(s) \rightleftharpoons Cr^{3+}(aq) + 3OH^-(aq)$.Let the molar solubility be $S$.Then $[Cr^{3+

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$\sqrt[4]{1.6 	imes 10^{-30} / 27}$

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(C) The dissociation of $Cr(OH)_3$ is given by: $Cr(OH)_3(s) ightleftharpoons Cr^{3+}(aq) + 3OH^-(aq)$.Let the molar solubility be $S$.Then $[Cr^{3+}] = S$ and $[OH^-] = 3S$.The solubility product expression is $K_{sp} = [Cr^{3+}][OH^-]^3$.Substituting the values: $K_{sp} = (S)(3S)^3 = 27S^4$.Therefore,$S^4 = K_{sp} / 27$.$S = \sqrt[4]{K_{sp} / 27} = \sqrt[4]{1.6 	imes 10^{-30} / 27}$.

For $Cr(OH)_3$,the $K_{sp} = 1.6 \times 10^{-30}$. What is the molar solubility of this compound in water?

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