Question

The solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in water is approximately \(9.6 \mathrm{mg} / \mathrm{L}\) at a given temperature. (a) Calculate the \(K_{\mathrm{sp}}\) of magnesium hydroxide. (b) Calculate the hydroxide concentration needed to precipitate \(\mathrm{Mg}^{2+}\) ions such that no more than \(5.0 \mu \mathrm{g} \mathrm{Mg}^{2+}\) per liter remains in the solution.

Short answer

(a) \(K_{sp} = 1.79 \times 10^{-11}\) (b) \([\text{OH}^-] = 9.29 \times 10^{-3} \text{ mol/L}\) needed.

Step-by-step solution

Convert Solubility to Moles

The solubility of \(\text{Mg(OH)}_2\) is given as \(9.6 \text{ mg/L}\). Convert this mass to moles by using the molar mass of \(\text{Mg(OH)}_2\), which is \(58.32 \text{ g/mol}\). Calculate the moles per liter using the formula:\[\text{Moles of } \text{Mg(OH)}_2 = \frac{9.6 \text{ mg/L}}{58.32 \text{ g/mol} \times 1000 \text{ mg/g}} = 1.65 \times 10^{-4} \text{ mol/L}\]

Write the Dissolution Equation

\(\text{Mg(OH)}_2\) dissolves in water according to the equation:\[\text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-\]This indicates that 1 mole of \(\text{Mg(OH)}_2\) produces 1 mole of \(\text{Mg}^{2+}\) and 2 moles of \(\text{OH}^-\).

Express Ion Concentrations

Based on the dissolution equation, if the concentration of \(\text{Mg(OH)}_2\) is \(1.65 \times 10^{-4} \text{ mol/L}\):- The concentration of \(\text{Mg}^{2+}\) ions = \(1.65 \times 10^{-4} \text{ mol/L}\).- The concentration of \(\text{OH}^-\) ions = \(2 \times 1.65 \times 10^{-4} = 3.30 \times 10^{-4} \text{ mol/L}\).

Calculate the Ksp

The solubility product \(K_{sp}\) expression for \(\text{Mg(OH)}_2\) is given by:\[K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2\]Substitute the concentrations:\[K_{sp} = (1.65 \times 10^{-4})(3.30 \times 10^{-4})^2 = 1.79 \times 10^{-11}\]

Establish Ion Concentrations for Precipitation

To prevent more than \(5.0 \mu\text{g/L}\) of \(\text{Mg}^{2+}\) in solution, convert the concentration:\[5.0 \mu\text{g/L} = \frac{5.0 \times 10^{-6} \text{ g/L}}{24.305 \text{ g/mol}} = 2.06 \times 10^{-7} \text{ mol/L}\]Use this concentration to calculate \([\text{OH}^-]\).

Calculate Required Hydroxide Ion Concentration

Rearrange the \(K_{sp}\) expression to solve for \([\text{OH}^-]\):\[K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2\]Substitute \(K_{sp}\) and \([\text{Mg}^{2+}]\):\[1.79 \times 10^{-11} = (2.06 \times 10^{-7})[\text{OH}^-]^2\]Solve for \([\text{OH}^-]\):\[[\text{OH}^-] = \sqrt{\frac{1.79 \times 10^{-11}}{2.06 \times 10^{-7}}} = 9.29 \times 10^{-3} \text{ mol/L}\]

Conclusion

To maintain no more than \(5.0 \mu\text{g/L}\) of \(\text{Mg}^{2+}\) ions, the hydroxide concentration needs to be at least \(9.29 \times 10^{-3} \text{ mol/L}\).

Key concepts

Molar Solubility

Molar solubility refers to the number of moles of a solute that can be dissolved in a liter of solution before the solution becomes saturated. For magnesium hydroxide (\(\text{Mg(OH)}_2\)), the molar solubility is calculated based on its given solubility in water, which is 9.6 mg/L. To find the molar solubility, we first convert the mass from milligrams to grams, and then use the molar mass of magnesium hydroxide, 58.32 g/mol, to determine the number of moles.The conversion is performed as follows:- Convert 9.6 mg to grams: \(9.6 \text{ mg} = 9.6 \times 10^{-3} \text{ g}\)- Apply the conversion formula: \[\text{Moles of Mg(OH)}_2 = \frac{9.6 \times 10^{-3} \text{ g}}{58.32 \text{ g/mol}} = 1.65 \times 10^{-4} \text{ mol/L}\]Understanding molar solubility is crucial as it helps us to predict how much of a solute like \(\text{Mg(OH)}_2\) can dissolve in water before it starts forming a precipitate, which ties into the next concept.

Precipitation Reaction

A precipitation reaction occurs when ions in solution react to form an insoluble compound, which then falls out of the solution as a solid. This is what happens with magnesium hydroxide in water under certain conditions.When \(\text{Mg(OH)}_2\) is added to water, it dissociates into \(\text{Mg}^{2+}\) and \(\text{OH}^-\) ions:\[\text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-\]This dissociation is in a dynamic equilibrium, meaning that the rate of \(\text{Mg(OH)}_2\) dissolving equals the rate of it precipitating back out. When the product of the concentrations of \(\text{Mg}^{2+}\) and \(\text{OH}^-\) ions exceeds the solubility product constant, \(K_{sp}\), precipitation occurs, forming solid \(\text{Mg(OH)}_2\). Ensuring a clear understanding of precipitation reactions helps in predicting when a precipitate will form and is essential for controlling reactions in a laboratory setting.

Hydroxide Ion Concentration

The concentration of hydroxide ions (\(\text{OH}^-\)) in a solution is crucial for calculating when magnesium ions (\(\text{Mg}^{2+}\)) will begin to precipitate. From the dissolution equation:\[\text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-\]we see that two hydroxide ions are produced for every magnesium ion. If the concentration of \(\text{OH}^-\) is very high, the equilibrium shifts towards forming more \(\text{Mg(OH)}_2\), leading to precipitation. This balance can be manipulated to ensure only small amounts of \(\text{Mg}^{2+}\) remain in solution.In this exercise, to keep \(\text{Mg}^{2+}\) ions at or below 5.0 µg/L, the necessary hydroxide concentration can be calculated using the equation:\[K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2\]Solving for \([\text{OH}^-]\), we ensure that the \(\text{OH}^-\) concentration is high enough to precipitate excess \(\text{Mg}^{2+}\), cementing the necessity of precise hydroxide levels in maintaining solution balance.

Equilibrium Expressions

Equilibrium expressions are mathematical representations of the balance between the reactants and products in a chemical reaction that is at equilibrium. For magnesium hydroxide, the equilibrium expression allows us to relate the solubility of the compound to the concentrations of its dissociated ions.For the given reaction: \[\text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-\]the solubility product (\(K_{sp}\)) is expressed as:\[K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2\]This equation takes into account the stoichiometry of the reaction, where the concentration of \(\text{OH}^-\) ions is squared because two hydroxide ions are produced for every unit of Mg(OH)\(_2\) that dissolves. Equilibrium expressions are fundamental in predicting how changes in concentrations or conditions can affect the balance and lead to either dissolving or precipitating more of a solute.