Question

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.

Short answer

The solubility product constant, \(K_{sp}\), for the ionic compound \(M_{2}X_{3}\) is approximately \(1.24 \times 10^{-43}\).

Step-by-step solution

Convert solubility from g/L to mol/L

First, we need to convert the given solubility, which is expressed in g/L, to a molar solubility expressed in mol/L. To do this, we will divide the given solubility in g/L by the given molar mass of the compound. \(Molar\: solubility = \frac{Solubility\: (g/L)}{Molar\: mass\: (g/mol)}\)

Calculate ion concentration in the saturated solution

We will use the molar solubility to find the concentration of ions in a saturated solution of M₂X₃. Each M₂X₃ unit dissociates into 2 M⁺ ions and 3 X⁻ ions. So, if 's' is the molar solubility, the resulting concentrations will be: MM: 2s XX: 3s

Calculate Ksp

Now that we have the ion concentrations in the saturated solution, we can calculate the Ksp using the following expression: \(K_{sp} = [M^+]^2 [X^-]^3\) Plugging in the ion concentrations (2s for M⁺ and 3s for X⁻) into the formula and simplifying: \(K_{sp} = (2s)^2 (3s)^3\) Now we can substitute the molar solubility from Step 1 and compute the Ksp.

Substitute the molar solubility and simplify

Substitute the molar solubility found in Step 1 into the Ksp expression and simplify: \( K_{sp} = (2 \times (\frac{3.60 \times 10^{-7}}{288}))^2 (3 \times (\frac{3.60 \times 10^{-7}}{288}))^3 \) After substituting and solving for Ksp: \(K_{sp} \approx 1.24 \times 10^{-43}\) The solubility product constant, Ksp, for the ionic compound M₂X₃ is approximately 1.24 x 10⁻⁴³.