K_{sp} for Cr(OH)_3 is 1.6 times 10^{-30}. Wh | vedclass

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(A) The dissociation of $Cr(OH)_3$ is represented as: $Cr(OH)_{3(s)} \rightleftharpoons Cr^{3+}_{(aq)} + 3OH^-_{(aq)}$Let the molar solubility be $s \

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

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(A) The dissociation of $Cr(OH)_3$ is represented as: $Cr(OH)_{3(s)} ightleftharpoons Cr^{3+}_{(aq)} + 3OH^-_{(aq)}$Let the molar solubility be $s \ mol/L$.At equilibrium,the concentration of $Cr^{3+}$ is $s$ and $OH^-$ is $3s$.The solubility product expression is: $K_{sp} = [Cr^{3+}][OH^-]^3$Substituting the values: $K_{sp} = (s)(3s)^3 = 27s^4$Given $K_{sp} = 1.6 	imes 10^{-30}$,we have: $27s^4 = 1.6 	imes 10^{-30}$Solving for $s$: $s^4 = \frac{1.6 	imes 10^{-30}}{27}$$s = \sqrt[4]{\frac{1.6 	imes 10^{-30}}{27}}$

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

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