(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 \(\mathrm{Mg}^{2+} ?\)

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.

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.