Question

Magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2},\) found in milk of magnesia, has a solubility of \(7.05 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}\) at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Mg}(\mathrm{OH})_{2}\)

Short answer

Calculate the molar solubility using the solubility given and the molar mass of \(\mathrm{Mg}(\mathrm{OH})_{2}\), use the molar ratio between the dissolved ions and the original compound to find their concentrations, then use those concentrations in the expression for \(K_{sp}\) to find its value.

Step-by-step solution

Write the Dissolution Equation

Write the balanced chemical equation for the dissolution of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in water: \[\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\mathrm{OH}^{-}(aq)\].

Determine Molar Solubility

Convert the solubility from grams per liter to moles per liter (molarity) by dividing the solubility by the molar mass of \(\mathrm{Mg}(\mathrm{OH})_{2}\). The molar mass of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(24.31 \text{ g/mol for Mg} + 2 \times 17.01 \text{ g/mol for OH}\), which sums up to \(58.33 \text{ g/mol}\). Thus, the molar solubility (\(S\)) is \[S = \frac{7.05 \times 10^{-3} \text{ g/L}}{58.33 \text{ g/mol}}\].

Calculate the Ion Concentrations

For every mole of \(\mathrm{Mg}(\mathrm{OH})_{2}\) that dissolves, it produces one mole of \(\mathrm{Mg}^{2+}\) ions and two moles of \(\mathrm{OH}^{-}\) ions. So, the concentrations of the ions will be \([\mathrm{Mg}^{2+}] = S\) and \([\mathrm{OH}^{-}] = 2S\).

Express the Solubility Product Constant \(K_{sp}\)

The solubility product constant \(K_{sp}\) for \(\mathrm{Mg}(\mathrm{OH})_{2}\) can be expressed as \[K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^2\]. Substitute the concentrations with the expressions from step 3 into the equation for \(K_{sp}\).

Solve for \(K_{sp}\)

Plugging the values of ion concentrations in terms of \(S\) into the \(K_{sp}\) expression, we get \[K_{sp} = S(2S)^2 = 4S^3\]. Calculate the value of \(S\) from Step 2 and then cube it, and multiply by 4 to find the value of \(K_{sp}\).

Key concepts

Molar Solubility

Molar solubility refers to the number of moles of a substance that can be dissolved per liter of solution until the solution becomes saturated. It's essential when dealing with the solubility of sparingly soluble salts, like magnesium hydroxide, in the context of chemistry.

In the exercise, we translate the solubility from grams per liter to molarity, the unit for molar solubility, by dividing by the molar mass of \( \mathrm{Mg}(\mathrm{OH})_{2} \). Understanding molar solubility is crucial because it directly relates to the ion concentrations in a saturated solution, which we need to calculate the solubility product constant, or \( K_{sp} \).

Dissolution Equation

The dissolution equation represents the process by which a solid dissolves in a solvent to produce ions in a solution. It is written as a balanced chemical equation showing the reactant (the solid) on the left and the products (the ions in solution) on the right.

For our example of magnesium hydroxide \( \mathrm{Mg}(\mathrm{OH})_{2} \), the dissolution equation is: \[\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\mathrm{OH}^{-}(aq)\]. This equation indicates that one mole of \( \mathrm{Mg}(\mathrm{OH})_{2} \) will produce one mole of \( \mathrm{Mg}^{2+} \) ions and two moles of \( \mathrm{OH}^{-} \) ions when dissolved.

Ion Concentration

Ion concentration is the amount of ions present in a particular volume of solution, typically expressed in molarity (moles per liter). In the context of solubility, knowing the ion concentrations that result from a dissolved compound is essential.

For every mole of \( \mathrm{Mg}(\mathrm{OH})_{2} \) that dissolves, we can state the ion concentrations as \( [\mathrm{Mg}^{2+}] = S \) and \( [\mathrm{OH}^{-}] = 2S \), where \( S \) is the molar solubility. This direct relationship allows us to plug these concentrations into the expression for \( K_{sp} \) when calculating the solubility product constant.

Ksp Calculation

The solubility product constant, or \( K_{sp} \), represents the equilibrium constant for the dissolution process of a sparingly soluble salt. It's a crucial concept when understanding the solubility equilibrium of a compound.

Most generically expressed, \( K_{sp} \) for a salt is the product of the concentrations of its ions, each raised to the power corresponding to their stoichiometric coefficients in the balanced dissolution equation.

Using the ion concentrations deduced from the molar solubility, we calculate the \( K_{sp} \) for \( \mathrm{Mg}(\mathrm{OH})_{2} \) by plugging these into the equation \[K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^2\]. This \( K_{sp} \) value is a key indicator of the solubility of magnesium hydroxide and helps predict whether a precipitate will form in a given solution.