Question

(a) Explain why Mg \(\mathrm{Mg}(\mathrm{OH})_{2}\) precipitates when \(\mathrm{CO}_{3}^{2-}\) ion is added to a solution containing \(\mathrm{Mg}^{2+} .\) (b) Will Mg \((\mathrm{OH})_{2}\) precipitate when 4.0 \(\mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is added to 1.00 \(\mathrm{L}\) of a solution containing 125 \(\mathrm{ppm}\) of \(\mathrm{Mg}^{2+} ?\)

Short answer

(a) When CO₃²⁻ ion is added to a solution containing Mg²⁺, it forms soluble MgCO₃, decreasing the concentration of Mg²⁺ ions in the solution and shifting the equilibrium of Mg(OH)₂ dissolution reaction, causing more Mg(OH)₂ to precipitate. (b) After calculating the ion product ([Mg²⁺][OH⁻]²) and comparing it to the solubility product (Ksp) of Mg(OH)₂, we find that the ion product (5.45 × 10⁻¹²) is very close to but still less than the Ksp (5.61 × 10⁻¹²), so Mg(OH)₂ precipitation will not occur.

Step-by-step solution

Part (a): Explain why Mg(OH)₂ precipitates when CO₃²⁻ is added to a solution containing Mg²⁺

Adding CO₃²⁻ ion to a solution containing Mg²⁺ ion results in the formation of MgCO₃, which is soluble in water. Once Mg²⁺ ions react with CO₃²⁻ ions, an equilibrium is established between Mg(OH)₂ and MgCO₃, as well as water and OH⁻ ions. The chemical reactions can be represented as follows: Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq) Mg²⁺(aq) + CO₃²⁻(aq) ⇌ MgCO₃(aq) When CO₃²⁻ ions are added to the solution, some Mg²⁺ ions will react with CO₃²⁻ ions to form soluble MgCO₃. This decreases the concentration of Mg²⁺ ions in the solution, and as a result, the equilibrium shifts leftward in the first reaction (Le Chatelier's principle), causing more Mg(OH)₂ to precipitate.

Part (b): Will Mg(OH)₂ precipitate when 4.0 g of Na₂CO₃ is added to 1.00 L of a solution containing 125 ppm of Mg²⁺?

To determine if Mg(OH)₂ will precipitate, we need to compare the ion product ([Mg²⁺][OH⁻]²) with the solubility product (Ksp) for Mg(OH)₂. If the ion product is greater than Ksp, then precipitation will occur; otherwise, it will not. Step 1: Calculate the concentrations of Mg²⁺ and CO₃²⁻ in the solution Mg²⁺ concentration = 125 ppm = 125 mg/L To convert this to mol/L, divide by the molar mass of Mg²⁺, which is 24.3 g/mol: Mg²⁺ concentration = \( \dfrac{125 \times 10^{-3}}{24.3} \) mol/L ≈ 0.00514 mol/L Na₂CO₃ mass = 4.0 g To find the CO₃²⁻ concentration, divide by the molar mass of Na₂CO₃ (105.99 g/mol) and by the volume of the solution in liters (1 L): CO₃²⁻ concentration = \( \dfrac{4.0}{105.99} \) mol/L ≈ 0.0377 mol/L Step 2: Calculate the concentration of OH⁻ produced by the reaction between CO₃²⁻ and Mg²⁺ Mg²⁺ + CO₃²⁻ ⇌ MgCO₃ + 2OH⁻ Since 1 mol of Mg²⁺ reacts with 1 mol of CO₃²⁻, the limiting reactant is Mg²⁺ (0.00514 mol/L), and the resulting concentration of OH⁻ is twice the limiting reactant (2 x 0.00514 mol/L) ≈ 0.0103 mol/L. Step 3: Calculate the ion product and compare it to the Ksp of Mg(OH)₂ Ksp (Mg(OH)₂) = 5.61 × 10⁻¹² Ion product = [Mg²⁺][OH⁻]² = (0.00514)(0.0103)² ≈ 5.45 × 10⁻¹² Since the ion product is very close to but still less than the Ksp, Mg(OH)₂ precipitation will not occur.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept that helps us understand how a chemical system at equilibrium responds to changes in concentration, pressure, volume, or temperature. When a change is imposed on a dynamic equilibrium system, the system will shift its position to counteract the effect of the change, thus re-establishing equilibrium.
This principle is particularly evident in chemical reactions involving equilibrium states. When addition of external ions or reactants occurs, the system adjusts by shifting the equilibrium position to reduce the effect of the added components.
For example, in the case of adding \(\text{CO}_3^{2-}\) ions to a solution containing \(\text{Mg}^{2+}\), the formation of \(\text{MgCO}_3\) occurs, which is more soluble. As \(\text{Mg}^{2+}\) from the solution gets consumed, the equilibrium in the reaction \(\text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-\) shifts to the left, producing more \(\text{Mg(OH)}_2\), hence causing the precipitation of magnesium hydroxide. This adjustment is a direct application of Le Chatelier's Principle in balancing the chemical reaction.

Solubility Product Constant

The Solubility Product Constant, or \(K_{sp}\), of a compound is a key concept in predicting whether a precipitate will form when two solutions are mixed. It is specific to sparingly soluble ionic compounds and represents the product of the concentrations of the ions in a saturated solution, raised to the power of their respective coefficients in the balanced chemical equation.
For example, for magnesium hydroxide, \(\text{Mg(OH)}_2\), the \(K_{sp}\) expression is given by \([\text{Mg}^{2+}][\text{OH}^-]^2\). The value of \(K_{sp}\) can predict if a compound will precipitate under certain conditions. If the ion product of the solutions exceeds the \(K_{sp}\), the solution is supersaturated, and precipitation will occur.
In the context of the exercise provided, we compare the calculated ion product of the solution to the given \(K_{sp}\) for magnesium hydroxide, which allows us to determine whether a precipitate will form in the solution when additional ions are introduced.

Precipitation Reaction

Precipitation reactions occur when ions in solution combine to form an insoluble compound, which separates as a solid. This process is pivotal in chemistry, especially in qualitative analysis to identify ions in a solution and in various industrial applications.
For a precipitation reaction to occur, the product of the reacting ion concentrations (ion product) in the solution must exceed the compound's solubility product constant, \(K_{sp}\).
Consider the reaction in the given problem where \(\text{Mg(OH)}_2\) precipitates when \(\text{CO}_3^{2-}\) ions are added to a solution of \(\text{Mg}^{2+}\). Here, the interaction between \(\text{Mg}^{2+}\) and \(\text{OH}^-\) ions leads to the formation of solid \(\text{Mg(OH)}_2\). This approach helps predict the conditions under which precipitation happens, serving as a practical application of ionic equilibrium in chemical reactions.

Ion Product

The Ion Product is a crucial concept in determining whether a solution will undergo precipitation. It is calculated from the concentrations of the ions in solution, similar to the solubility product constant (\(K_{sp}\)). However, while \(K_{sp}\) is a constant specific to a particular substance at a given temperature, the ion product varies with the concentrations of dissolved ions.
When the ion product exceeds \(K_{sp}\), the solution is supersaturated, prompting a shift towards the formation of a solid precipitate as the reaction seeks equilibrium. If the ion product is less than \(K_{sp}\), the solution remains unsaturated, and no precipitation occurs.
In the example from the exercise, determining the ion product by calculating \([\text{Mg}^{2+}][\text{OH}^-]^2\) shows whether \(\text{Mg(OH)}_2\) will precipitate when \(\text{Na}_2\text{CO}_3\) is added. Since the calculated ion product was slightly less than the \(K_{sp}\), no precipitation occurs, highlighting the importance of understanding ion product in predicting chemical behavior in solutions.