Question

(a) When sufficient \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is added to a solution containing \(\mathrm{Mg}^{2+}, \mathrm{Mg}(\mathrm{OH})_{2}\) will precipitate. Explain by writing balanced equations of the reactions. (b) Will \(\mathrm{Mg}(\mathrm{OH})_{2}\) precipitate when \(2.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is added to \(1.00 \mathrm{~L}\) of a solution containing 25 ppm of \(\mathrm{Mg}^{2+}\) ?

Short answer

(a) The balanced chemical equations for the precipitation of Mg(OH)₂ are: 1. Na₂CO₃ + H₂O -> 2Na⁺ + CO₃²⁻ + 2OH⁻ 2. Mg²⁺ + 2OH⁻ -> Mg(OH)₂↓ (b) Yes, Mg(OH)₂ will precipitate since the available amount of OH⁻ ions (0.0378 mol) is greater than the required amount of OH⁻ ions (0.00206 mol) when 2.0 g of Na₂CO₃ is added to the 1 L Mg²⁺ solution containing 25 ppm of Mg²⁺.

Step-by-step solution

(a) Writing balanced chemical equations for Mg(OH)₂ precipitation

The process occurs in two steps: 1. The reaction between Na₂CO₃ and H₂O: \[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \longrightarrow 2\text{Na}^+ + \text{CO}_3^{2-} + 2\text{OH}^- \] 2. The precipitation of Mg(OH)₂: \[ \text{Mg}^{2+} + 2\text{OH}^- \longrightarrow \text{Mg}(\text{OH})_2\downarrow \] Now let's move on to part (b) of the problem.

(b) Determining if Mg(OH)₂ will precipitate with given amount of Na₂CO₃

First, we need to convert the concentration of Mg²⁺ from ppm to moles per liter: \[ \frac{25 \hspace{1mm} \text{mg}}{\text{L}} \times \frac{1 \text{mol}}{24,305 \text{mg}} = \frac{25}{24,305} \frac{\text{mol}}{\text{L}} \approx 0.00103 \frac{\text{mol}}{\text{L}}\] Next, we need to determine the moles of Na₂CO₃ added to 1 L of the solution: \[ 2.0 \hspace{1mm} \text{g} \times \frac{1 \text{mol}}{105.99 \hspace{1mm} \text{g}} \approx 0.0189 \hspace{1mm} \text{mol}\] Since 1 mole of Na₂CO₃ produces 2 moles of OH⁻, 1 L of the solution will have: \[ 0.0189 \hspace{1mm} \text{mol} \times 2 = 0.0378 \hspace{1mm} \text{mol} \hspace{1mm} \text{OH}^-\] Now we can see if the concentration of OH⁻ is enough to react with all Mg²⁺ ions. We know that it takes two OH⁻ ions to react with one Mg²⁺ ion. Therefore, we calculate the moles of OH⁻ needed for the reaction with all available Mg²⁺ ions: \[ 0.00103 \hspace{1mm} \text{mol} \times 2 \approx 0.00206 \hspace{1mm} \text{mol} \hspace{1mm} \text{OH}^-\] Since the available amount of OH⁻ ions (0.0378 mol) is greater than the required amount of OH⁻ ions (0.00206 mol), Mg(OH)₂ will precipitate from the solution when 2.0 g of Na₂CO₃ is added to it.

Key concepts

Balanced Chemical Equations

In chemistry, balanced chemical equations are essential because they keep track of the number of atoms involved in a chemical reaction. Each side of the equation must have the same number of atoms for each element. Balancing equations follows the Law of Conservation of Mass, which states that matter cannot be created or destroyed.

Take for example the reaction between sodium carbonate, \(\text{Na}_2\text{CO}_3\), and water, \(\text{H}_2\text{O}\), to form ions:

• \(\text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \longrightarrow 2\text{Na}^+ + \text{CO}_3^{2-} + 2\text{OH}^-\)

Here, the equation is balanced because there is an equal number of each type of atom on both sides of the equation: sodium, carbon, oxygen, and hydrogen.

Further, when magnesium ions \(\text{Mg}^{2+}\) react with hydroxide ions \(\text{OH}^-\) to form magnesium hydroxide, the equation is:

• \(\text{Mg}^{2+} + 2\text{OH}^- \longrightarrow \text{Mg(OH)}_2\downarrow\)

Notice the importance of the coefficients (such as the 2 in \(2\text{Na}^+\) and \(2\text{OH}^-\)), which ensure that the equation reflects the conservation of atoms.

Molar Concentration Conversion

Converting concentrations between different units is a critical skill in chemistry, especially when dealing with reactions in solution. Concentration describes how much solute is present in a given amount of solution.

In this exercise, we are given magnesium ion concentration in parts per million (ppm), a common measure for dilute solutions. To use it in stoichiometric calculations, we convert it to moles per liter (mol/L).

Here's how the conversion works:

• Magnesium's concentration in ppm is 25 ppm, equivalent to 25 mg/L, since 1 ppm equals 1 mg/L for water-based solutions.
• To convert mg to moles, use the molar mass of magnesium: \(\frac{25 \, \text{mg}}{24,305 \, \text{mg/mol}} \approx 0.00103 \, \text{mol/L}\).

This conversion allows us to proceed with calculations by comparing molar amounts, facilitating further chemical analysis.

Reaction Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves calculating how much of each substance is involved before, during, and after a reaction.

In this problem, we determine if magnesium hydroxide \(\text{Mg(OH)}_2\) will precipitate by comparing the number of moles of available \(\text{OH}^-\) ions (from \(\text{Na}_2\text{CO}_3\)) to the moles of \(\text{Mg}^{2+}\) ions.

• First, calculate the moles of \(\text{Na}_2\text{CO}_3\): \(2.0 \, \text{g} \times \frac{1 \, \text{mol}}{105.99 \, \text{g}} = 0.0189 \, \text{mol}\).
• Each mole of \(\text{Na}_2\text{CO}_3\) yields two moles of \(\text{OH}^-\) ions: \(0.0189 \, \text{mol} \times 2 = 0.0378 \, \text{mol OH}^-\).
• The reaction requires two \(\text{OH}^-\) ions per \(\text{Mg}^{2+}\) ion: \(0.00103 \, \text{mol Mg}^{2+} \times 2 = 0.00206 \, \text{mol OH}^-\).

Since we have more \(\text{OH}^-\) ions available (0.0378 mol) than needed (0.00206 mol), all the \(\text{Mg}^{2+}\) in solution will precipitate as \(\text{Mg(OH)}_2\).

Thus, stoichiometry helps predict the outcome of reactions by helping us calculate and ensure correct proportions of substances involved.