Chemistry - Ionic Equilibrium Question with Solution | TestHub

The given compound will dissociate as

$\begin{array}{ccccc}\text{Cr}{\left(\text{OH}\right)}_3\left(\mathrm{s}\right)& \to & {\text{Cr}}^{+3}\left(\mathrm{aq}\right)& +& 3{\text{OH}}^{-}\left(\mathrm{aq}\right)\\ & & \mathrm{s}& & 3\mathrm{s}\end{array}$

Writing the expression for the solubility product,

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Cr}}^{3+}\right]{\left[{\mathrm{OH}}^{-}\right]}^3=\mathrm{s}\times {\left(3\mathrm{s}\right)}^3=27{\mathrm{s}}^4$

$$\begin{gathered} \mathrm{s}=\sqrt[4]{\frac{{\mathrm{K}}_{\mathrm{sp}}}{27}} \\ \mathrm{s}=\sqrt[4]{\frac{2.7\times {10}^{-31}}{27}} \\ \mathrm{s}=1.0\times {10}^{-8} \mathrm{mol} {\mathrm{L}}^{-1} \end{gathered}$$