Problem 67 - Chemistry and Chemical Reactivity

Question

(a) Which solution is expected to have the higher boiling point: \(0.20 \mathrm{m}\) KBr or \(0.30 \mathrm{m}\) sugar? (b) Which aqueous solution has the lower freezing point: \(0.12 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3}\) or \(0.10 \mathrm{m} \mathrm{Na}_{2} \mathrm{CO}_{3} ?\)

Step-by-Step Solution

Answer

(a) 0.20 m KBr has the higher boiling point. (b) 0.12 m NH₄NO₃ has the lower freezing point.

Step 1: Define the boiling point elevation formula

Boiling point elevation is described by the formula \( \Delta T_b = i \cdot K_b \cdot m \), where \( i \) is the van't Hoff factor, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality. A higher \( \Delta T_b \) indicates a higher boiling point.

Step 2: Calculate boiling point elevation for KBr solution

KBr dissociates into two ions: K⁺ and Br⁻, so \( i = 2 \). The molality \( m = 0.20 \mathrm{m} \) for KBr. Substitute these into \( \Delta T_b = i \cdot K_b \cdot m \) to get \( \Delta T_{b,\mathrm{KBr}} = 2 \cdot K_b \cdot 0.20 = 0.40 \cdot K_b \).

Step 3: Calculate boiling point elevation for the sugar solution

Sugar does not dissociate in solution, so \( i = 1 \). The molality \( m = 0.30 \mathrm{m} \). Substitute into the formula: \( \Delta T_{b,\mathrm{sugar}} = 1 \cdot K_b \cdot 0.30 = 0.30 \cdot K_b \).

Step 4: Compare boiling point elevations

Since \( \Delta T_{b,\mathrm{KBr}} = 0.40 \cdot K_b \) is greater than \( \Delta T_{b,\mathrm{sugar}} = 0.30 \cdot K_b \), the KBr solution has a higher boiling point.

Step 5: Define the freezing point depression formula

Freezing point depression is described by \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality. A lower \( \Delta T_f \) results in a lower freezing point.

Step 6: Calculate freezing point depression for NH₄NO₃ solution

NH₄NO₃ dissociates into two ions: NH₄⁺ and NO₃⁻, so \( i = 2 \). The molality \( m = 0.12 \mathrm{m} \). Substitute into the formula: \( \Delta T_{f,\mathrm{NH}_4 ext{NO}_3} = 2 \cdot K_f \cdot 0.12 = 0.24 \cdot K_f \).

Step 7: Calculate freezing point depression for Na₂CO₃ solution

Na₂CO₃ dissociates into three ions: 2 Na⁺ and CO₃²⁻, so \( i = 3 \). The molality \( m = 0.10 \mathrm{m} \). Substitute into the formula: \( \Delta T_{f,\mathrm{Na}_2 ext{CO}_3} = 3 \cdot K_f \cdot 0.10 = 0.30 \cdot K_f \).

Step 8: Compare freezing point depressions

\( \Delta T_{f,\mathrm{NH}_4 ext{NO}_3} = 0.24 \cdot K_f \) is less than \( \Delta T_{f,\mathrm{Na}_2 ext{CO}_3} = 0.30 \cdot K_f \). Therefore, the NH₄NO₃ solution has a lower freezing point.

Key Concepts

Boiling Point Elevation

Boiling point elevation is a property of solutions. When you dissolve a solute in a solvent, the boiling point of the solution becomes higher than that of the pure solvent. This occurs because the presence of solute particles causes the solvent molecules to require more energy to escape into the vapor phase.
This effect is quantitatively described by the formula \( \Delta T_b = i \cdot K_b \cdot m \).

• \( \Delta T_b \) is the increase in boiling point.
• \( i \) is the van't Hoff factor, representing the number of particles the solute dissociates into.
• \( K_b \) is the ebullioscopic constant, specific to each solvent.
• \( m \) is the molality of the solution.

For instance, in the problem given, when comparing a \(0.20 \mathrm{m}\) KBr solution to a \(0.30 \mathrm{m}\) sugar solution, we observe that KBr dissociates into two ions, contributing to a higher van't Hoff factor of 2. This results in a greater boiling point elevation for KBr, demonstrating how the increase in boiling point depends not only on the concentration of the solute but also on its ability to dissociate.

Freezing Point Depression

Freezing point depression occurs when a solute is added to a solvent, causing the freezing point of the solution to be lower than that of the pure solvent. In simple terms, it takes a lower temperature for the solution to solidify.
This phenomenon is described by the formula \( \Delta T_f = i \cdot K_f \cdot m \).

• \( \Delta T_f \) is the decrease in freezing point.
• \( i \) is the van't Hoff factor, indicating the number of particles the solute breaks into.
• \( K_f \) is the cryoscopic constant.
• \( m \) is the solution's molality.

In the example provided, \(0.12 \mathrm{m}\) \(\mathrm{NH}_{4}\mathrm{NO}_{3}\) solution with a van't Hoff factor of 2 results in a smaller decrease in temperature compared to \(0.10 \mathrm{m}\) \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) with a factor of 3. This demonstrates how substances that dissociate into more ions tend to create a more significant reduction in freezing point, explaining why the \(\mathrm{NH}_{4}\mathrm{NO}_{3}\) solution ends up with a lower freezing point.

Van't Hoff Factor

The van't Hoff factor, denoted by \( i \), is a crucial part of colligative properties such as boiling point elevation and freezing point depression. The van't Hoff factor measures the effect of a solute on the colligative properties of a solution. More specifically, it accounts for the number of particles into which a solute dissociates in solution. The more particles that are produced, the greater the effect on colligative properties.
To determine \( i \):

• A non-dissociating solute, like sugar, has \( i = 1 \).
• Ionic compounds dissociate into their ions; for example, \(\mathrm{KBr}\) dissociates into \(\mathrm{K}^+\) and \(\mathrm{Br}^-\), thus \( i = 2 \).
• Compounds like \(\mathrm{Na}_2\mathrm{CO}_3\) split into three ions (2 \(\mathrm{Na}^+\) and \(\mathrm{CO}_3^{2-}\)), giving \( i = 3 \).

The van't Hoff factor directly influences the magnitude of both boiling point elevation and freezing point depression. Consequently, deeper understanding and accurate calculation of this factor are imperative for analyzing and predicting changes in solutions' properties.