Problem 77 - Chemistry: The Central Science

Question

(a) Explain why \(\mathrm{Mg}(\mathrm{OH})_{2}\) precipitates when \(\mathrm{CO}_{3}{\underline{\phantom{xx}}}^{2-}\) 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 \(\mathrm{Mg}^{2+}\) ?

Step-by-Step Solution

Answer

Mg(OH)₂ precipitates when CO₃²⁻ is added to a solution containing Mg²⁺ ions because it forms an insoluble compound that precipitates out of the solution. After calculating the ion concentrations of CO₃²⁻ and Mg²⁺ in the given solution and comparing the ion product (IP) with the solubility product constant (Ksp) for Mg(OH)₂, we find that IP > Ksp. This indicates that the solution is supersaturated concerning Mg(OH)₂, and therefore, Mg(OH)₂ will precipitate when 4.0 g of Na₂CO₃ is added to 1.00 L of a solution containing 125 ppm of Mg²⁺ ions.

Step 1: Discuss the chemical reaction

When CO₃²⁻ ions are added to a solution containing Mg²⁺ ions, the following chemical reaction occurs: \(Mg^{2+} (aq) + 2OH^- (aq) \rightarrow Mg(OH)_{2} (s)\) This reaction forms Mg(OH)₂, which is insoluble in water and precipitates out of the solution.

Step 2: Calculate ion concentrations

We'll calculate the concentrations of CO₃²⁻ and Mg²⁺ ions in the solution after adding 4.0 g of Na₂CO₃ to 1.00 L of the solution. First, let's find the concentration of CO₃²⁻ ions: - We have 4.0 g of Na₂CO₃. - The molar mass of Na₂CO₃ is \(2\times(22.99)+12.01+3\times(16.00) = 105.99~g/mol\). - Therefore, the moles of Na₂CO₃ = \(\frac{4.0~g}{105.99~g/mol} = 0.0377~mol\). - The concentration of CO₃²⁻ ions is \(\frac{0.0377~mol}{1.00~L} = 0.0377~M\). Now, let's calculate the concentration of Mg²⁺ ions: - We have 125 ppm of Mg²⁺ ions, which means there are 125 mg of Mg²⁺ ions per 1 L of the solution. - The molar mass of Mg is 24.30 g/mol. - Therefore, the moles of Mg²⁺ = \(\frac{125~mg}{24.30~g/mol} \times \frac{1~g}{1000~mg} = 0.00514~mol\). - The concentration of Mg²⁺ ions is \(\frac{0.00514~mol}{1.00~L} = 0.00514~M\).

Step 3: Determine if Mg(OH)₂ will precipitate

Next, let's determine the ion product and compare it with the solubility product constant, \(K_{sp}\), of Mg(OH)₂. The ion product, IP, is given by the expression: \[IP = [Mg^{2+}][OH^-]^2\] In this case, the hydroxide ions come from the dissociation of CO₃²⁻ ions. We know that for each CO₃²⁻ ion that dissolves, one OH⁻ ion is formed. So, the concentration of OH⁻ ions is equal to the concentration of CO₃²⁻ ions. Hence, \[[OH⁻] = [CO₃^{2-}] = 0.0377~M\] Now, calculate the ion product: \[IP = (0.00514)(0.0377)^2 = 7.13 \times 10^{-6}\] The solubility product constant, \(K_{sp}\), for Mg(OH)₂ is given as \(K_{sp} = 5.61 \times 10^{-12}\). Comparing the ion product and the solubility product constant, we see that IP > \(K_{sp}\). That means, the solution is supersaturated with respect to Mg(OH)₂, and Mg(OH)₂ will precipitate.

Key Concepts

Solubility Product Constant

The solubility product constant, represented as \( K_{sp} \), is a valuable tool in chemistry used to determine the solubility of a compound in a solution. It describes the equilibrium point for the dissolution of a sparingly soluble compound. For the compound magnesium hydroxide, \( \text{Mg(OH)}_2 \), this constant captures how much of the compound can dissolve in water to form its ions Mg²⁺ and OH⁻. The expression for the solubility product constant for magnesium hydroxide is:

• \( K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \)

This equation indicates that the solubility of a compound is dependent on the concentrations of its ions in solution. A situation where the ion product exceeds \( K_{sp} \) results in precipitation, as the solution becomes supersaturated. Hence, by comparing the calculated ion product to the \( K_{sp} \), we can predict whether a precipitate will form. This is the key method used to determine the formation of magnesium hydroxide precipitate in chemical solutions.

Magnesium Hydroxide

Magnesium hydroxide, \( \text{Mg(OH)}_2 \), is a white solid known for its low solubility in water. When a solution containing magnesium ions \( \text{(Mg}^{2+}) \) meets a source of hydroxide ions \( \text{(OH}^-) \), the conditions are ripe for forming magnesium hydroxide. This compound appears as a precipitate because it does not dissolve easily at room temperature.
Magnesium hydroxide is often encountered in antacids and laxatives and is a simple but essential material in the real world. Its formation fascinates chemists as it considers ion interactions in a reaction. Understanding these concepts ensures that the manipulation of solutions in laboratory settings is accurate, which could be critical in industrial applications and academic pursuits.

Ion Concentration Calculation

Calculating ion concentrations in a solution involves stoichiometry and a grasp of molecular masses and conversions. When you add a compound like sodium carbonate, \( \text{Na}_2\text{CO}_3 \), into a solution, its dissociation provides concentrations of specific ions—in this case, \( \text{CO}_3^{2-} \) and a resultant impact on \( \text{OH}^- \) concentration.

• The process starts by determining the moles of \( \text{Na}_2\text{CO}_3 \) by dividing the mass of the substance by its molar mass.
• The concentration of ions is then calculated by dividing the moles by the volume of the solution in liters.

This thorough calculation helps predict whether a reaction between ions will result in a precipitate. For magnesium hydroxide, understanding the ion concentrations post-reaction is crucial. This guides us in assessing whether conditions allow for precipitation based on the comparison with the \( K_{sp} \). Accurate ion concentration calculation ensures precise interpretations of chemical behavior in solutions.