Question

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

Short answer

(a) When \(CO_3^{2-}\) is added to a solution containing \(Mg^{2+}\), magnesium carbonate (\(MgCO_3\)) precipitates out of the solution. The presence of \(MgCO_3\) can then lead to the precipitation of magnesium hydroxide (\(Mg(OH)_2\)) in the presence of hydroxide ions. (b) The calculated solubility product constant (\(K_{sp}\)) for the given solution is \(8.69\times10^{10}\), which is greater than the known \(K_{sp}\) of \(Mg(OH)_2\) (\(1.5\times10^{-11}\)). Thus, magnesium hydroxide will precipitate in the given solution.

Step-by-step solution

Understand the reaction and precipitation concept

When the carbonate ion (\(CO_3^{2-}\)) is added to a solution containing magnesium ion (\(Mg^{2+}\)), the following chemical reaction occurs: \[Mg^{2+}(aq) + CO_3^{2-}(aq) \rightarrow MgCO_3(s)\] Magnesium carbonate (\(MgCO_3\)), which is insoluble in water, precipitates out of solution. The presence of \(MgCO_3\) can then lead to the precipitation of magnesium hydroxide (\(Mg(OH)_2\)): \[MgCO_3(s) + 2OH^-(aq) \rightarrow Mg(OH)_2(s) + CO_3^{2-}(aq)\]

Calculate the amount of carbonate ion added

We are given that 4.0 g of sodium carbonate (\(Na_2CO_3\)) is added to 1.00 L of the solution. Before calculating the concentration of carbonate ions, we need to find out the moles of \(Na_2CO_3\) added. Moles of \(Na_2CO_3\) = \(\frac{mass}{molar\,mass}\) The molar mass of \(Na_2CO_3 = (2\times22.99) + 12.01 + (3\times16.00) = 105.99\,g/mol\) Moles of \(Na_2CO_3\) = \(\frac{4.0 g}{105.99 g/mol} = 3.77\times10^{-2}\,mol\) As 1 mole of \(Na_2CO_3\) releases 1 mole of carbonate ion, the moles of carbonate ions are the same. So, the concentration of carbonate ions in the solution is: \[CO_3^{2-}\,Concentration = \frac{3.77\times10^{-2}\,mol}{1.00 L} = 3.77\times10^{-2} M\]

Calculate the concentration of magnesium ion in solution

The given solution contains 125 ppm (\(mg/L\)) of \(Mg^{2+}\) ions. Now let's convert this concentration from ppm to molarity. Molar mass of \(Mg = 24.31 g/mol\) \(1\,mg = 0.001\,g\) So, 125 ppm of magnesium ions \(=\frac{125\times0.001\,g}{24.31\,g/mol}=5.14\times10^{-3}\,mol\) We have 5.14×10⁻³ moles of \(Mg^{2+}\) ions in 1.00 L, so the concentration of \(Mg^{2+}\) ion in the solution is: \[Mg^{2+}\,Concentration = \frac{5.14\times10^{-3}\,mol}{1.00 L} = 5.14\times10^{-3} M\]

Compare the concentration values with solubility product constant

Now, we need to find out if the precipitation of \(Mg(OH)_2\) takes place by comparing the concentration values with the solubility product constant (\(K_{sp}\)) of \(Mg(OH)_2\). Using the reaction formula from Step 1: \[K_{sp}=[Mg^{2+}][OH^-]^2\] The solubility product constant (\(K_{sp})\) for \(Mg(OH)_2\) is 1.5×10⁻¹¹. Since the carbonate ion added results in a higher concentration of magnesium ions in solution, we can assume that precipitation occurs. Thus, we have: \(K_{sp} = (5.14\times10^{-3} M)(2\times(3.77\times10^{-2} M))^2\) By calculating, we get: \(K_{sp} = 8.69\times10^{10}\) As \(K_{sp}\) calculated is greater than the known \(K_{sp}\) (\(1.5\times10^{-11}\)), magnesium hydroxide will precipitate in the given solution.

Key concepts

Solubility Product Constant

The Solubility Product Constant, represented as \(K_{sp}\), is essential in predicting whether a precipitation reaction will occur. It is a specific type of equilibrium constant applied to the solubility of ionic compounds. This constant represents the maximum extent to which a solid can dissociate in solution. The formula for the solubility product constant of a compound \(A_xB_y\) is given as:\[ K_{sp} = [A^{m+}]^x[B^{n-}]^y \]where \([A^{m+}]\) and \([B^{n-}]\) are the molar concentrations of the ions, and \(x\) and \(y\) are the coefficients of the ions from the compound's dissociation equation. Key Concepts:

• The smaller the \(K_{sp}\), the less soluble the compound is.
• A solution is saturated when the ion product equals \(K_{sp}\).
• When the ion product exceeds \(K_{sp}\), precipitation occurs to restore equilibrium.

In the exercise, we calculated \(K_{sp}\) for magnesium hydroxide \(Mg(OH)_2\) using the concentrations of magnesium \(Mg^{2+}\) and hydroxide ions \(OH^-\) derived from the given conditions. Concluding that with a higher \(K_{sp}\) value than usual, precipitation of \(Mg(OH)_2\) occurred.

Molarity Calculation

Molarity is a crucial concept in understanding chemical reactions, as it involves measuring the concentration of solutes in a solution. Molarity is expressed in moles per liter (\(M\)) and is calculated using the formula:\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]Steps to Calculate Molarity:

• Determine the moles of solute. This often requires dividing the mass of the solute by its molar mass.
• Measure the volume of the solution in liters.
• Divide the moles of solute by the volume of solution to find molarity.

In the exercise, to determine if \(Mg(OH)_2\) would precipitate, we needed to calculate the molarity of both \(Mg^{2+}\) and \(CO_3^{2-}\). For \(Na_2CO_3\), we first found the moles using the molar mass and then divided by the 1.00 L of solution. Similarly, \(Mg^{2+}\) concentration was calculated from its given ppm, converting it to molarity by considering its molar mass.

Chemical Equilibrium

Chemical equilibrium is a dynamic state where the concentrations of reactants and products are stable, suggesting that the reaction rates in both forward and backward directions are equal. This state is essential, especially in precipitation reactions.Key Points about Chemical Equilibrium:

• In equilibrium, the macroscopic properties remain constant over time.
• The Equilibrium Constant \(K\) quantifies the relative concentrations of reactants and products at equilibrium.
• Changes in concentration, temperature, or pressure can shift equilibrium, following Le Chatelier's Principle.

For the reaction in the exercise, we examine equilibrium to decide if \(Mg(OH)_2\) precipitates in the solution. The presence of excess \(CO_3^{2-}\) ions shifts the reaction toward the formation of more \(CO_3^{2-}\) and fewer \(OH^-\) ions, influencing \(Mg(OH)_2\) precipitation. The equilibrium favors the formation of magnesium hydroxide precipitate whenever the ion product exceeds its \(K_{sp}\), as derived from step comparisons.