The solubility of the ionic compound \(\mathrm{M}_{2} \mathrm{X}_{3},\) having a molar mass of \(288 \mathrm{g} / \mathrm{mol},\) is $3.60 \times 10^{-7} \mathrm{g} / \mathrm{L}\( . Calculate the \)K_{\mathrm{sp}}$ of the compound.

Given the solubility of the compound in grams per liter, we will convert it to moles per liter using the molar mass: Solubility in moles per liter = \(\dfrac{\text{Solubility in grams per liter}}{\text{Molar mass}}\) Plug in the given values: Solubility in moles per liter = \(\dfrac{3.60 \times 10^{-7} \mathrm{g}/\mathrm{L}}{288 \mathrm{g}/\mathrm{mol}}\)

Write the equation for the dissociation of the ionic compound \(\mathrm{M}_{2} \mathrm{X}_{3}\) in water. \(\mathrm{M}_{2} \mathrm{X}_{3} \rightleftharpoons 2\mathrm{M}^{+} + 3\mathrm{X}^{-}\)

Let the solubility in moles per liter be represented by \(s\). Based on the stoichiometry of the dissociation equation, the concentration of \(\mathrm{M}^{+}\) ions will be \(2s\) and the concentration of \(\mathrm{X}^{-}\) ions will be \(3s\). From step 1, we found the solubility in moles per liter : \(s = \dfrac{3.60 \times 10^{-7} \mathrm{g}/\mathrm{L}}{288 \mathrm{g}/\mathrm{mol}}\) Now calculate the concentrations of ions: \([\mathrm{M}^{+}] = 2s\) \([\mathrm{X}^{-}] = 3s\)

The \(K_{\mathrm{sp}}\) expression for the dissociation of \(\mathrm{M}_{2} \mathrm{X}_{3}\) can be written as: \(K_{\mathrm{sp}} = [\mathrm{M}^{+}]^{2} [\mathrm{X}^{-}]^{3}\) Plug in the concentrations of ions from step 3: \(K_{\mathrm{sp}} = (2s)^{2} (3s)^{3}\) Now, substitute the value of \(s\) we calculated in step 1, and solve for \(K_{\mathrm{sp}}\): \(K_{\mathrm{sp}} = \left(2 \times \dfrac{3.60 \times 10^{-7}}{288}\right)^{2} \left(3 \times \dfrac{3.60 \times 10^{-7}}{288}\right)^{3}\) Evaluate this expression to find the \(K_{\mathrm{sp}}\) value: \(K_{\mathrm{sp}} \approx 4.91 \times 10^{-24}\) The solubility product constant, \(K_{\mathrm{sp}}\), for the ionic compound \(\mathrm{M}_{2} \mathrm{X}_{3}\) is approximately \(4.91 \times 10^{-24}\).