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

<p>To find the molar solubility of $\mathrm{Cr}(\mathrm{OH})_3$ in water, we start with the dissolution equilibrium:</p> <p>$ \mathrm{Cr}(\mathrm{OH})_{3(\mathrm{~s})} \rightleftharpoons \mathop {\mathrm{Cr^{3+}_{(aq)}}}\limits_s + \mathop {\mathrm{3OH^-_{(aq)}}}\limits_{3s} $</p> <p>At equilibrium, the solubility product $\mathrm{K}_{\mathrm{sp}}$ is given by:</p> <p>$ \mathrm{K}_{\mathrm{sp}} = (\mathrm{s}) \cdot (3 \mathrm{~s})^3 = 27 \mathrm{~s}^4 $</p> <p>Substituting the given $\mathrm{K}_{\mathrm{sp}}$ value:</p> <p>$ 27 \mathrm{~s}^4 = 1.6 \times 10^{-30} $</p> <p>Solving for $\mathrm{s}$, we divide both sides by 27:</p> <p>$ \mathrm{s} = \left(\frac{1.6 \times 10^{-30}}{27}\right)^{1/4} $</p> <p>Thus, the molar solubility of $\mathrm{Cr}(\mathrm{OH})_3$ is:</p> <p>$ \left(\frac{1.6}{27} \times 10^{-30}\right)^{1/4} $</p>