Problem 54 - Chemistry and Chemical Reactivity

Question

(a) Which aqueous solution is expected to have the higher boiling point: \(0.10 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}\) or\(0.15 \mathrm{m}\) sugar? (b) For which aqueous solution is the vapor pressure of water higher: \(0.30 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3}\) or \(0.15 m \mathrm{Na}_{7} \mathrm{SO}_{4} ?\)

Step-by-Step Solution

Answer

(a) The \(0.10\, \mathrm{m}\, \mathrm{Na}_2\mathrm{SO}_4\) solution has a higher boiling point. (b) The \(0.30 \,\mathrm{m}\, \mathrm{NH}_4\mathrm{NO}_3\) solution has a higher vapor pressure.

Step 1: Understanding the Boiling Point Elevation

Boiling point elevation depends on the number of solute particles present in a solution. This is calculated using the formula: \[ \Delta T_b = i \cdot K_b \cdot m \]where \( \Delta T_b \) is the boiling point elevation, \( i \) is the van't Hoff factor (ionization factor), \( K_b \) is the ebullioscopic constant (a property of the solvent), and \( m \) is the molality.

Step 2: Calculate Van't Hoff Factor (i)

First, calculate the van't Hoff factor for each compound, which represents the number of particles the compound dissociates into per formula unit when dissolved:- \( \text{Na}_2\text{SO}_4 \) dissociates into 3 ions (2 Na\(^+\) and 1 SO\(_4^{2-}\)), so \( i = 3 \).- Sugar (e.g., glucose) does not dissociate in solution, so \( i = 1 \).

Step 3: Compare Boiling Point Elevations

Now use the boiling point elevation equation to find which solution has a higher boiling point:- For Na\(_2\)SO\(_4\): \( \Delta T_b = 3 \cdot K_b \cdot 0.10 \).- For sugar: \( \Delta T_b = 1 \cdot K_b \cdot 0.15 \).- Since \( 3 \cdot 0.10 = 0.30 \) and \( 1 \cdot 0.15 = 0.15 \), the Na\(_2\)SO\(_4\) solution has a higher boiling point because the value is greater.

Step 4: Understanding Vapor Pressure Lowering

Vapor pressure lowering is also connected to the number of solute particles in solution. The presence of solute lowers the vapor pressure of the solvent, more solute particles mean a greater decrease in vapor pressure.

Step 5: Calculate Total Particles for Vapor Pressure Comparison

Calculate the total number of particles for each solution:- \( \ ext{NH}_4\text{NO}_3\) dissociates into 2 ions (NH\(_4^+\) and NO\(_3^−\)), so \( i = 2 \): \( \text{Total concentration} = 2 \cdot 0.30 = 0.60 \).- \( \ ext{Na}_7\text{SO}_4\) dissociates into 8 ions (7 Na\(^+\) ions and 1 SO\(_4^{2-}\)), so \( i = 8 \): \( \text{Total concentration} = 8 \cdot 0.15 = 1.20 \).- Since \( 1.20 \) is greater than \( 0.60 \), the Na\(_7\)SO\(_4\) solution will lower the vapor pressure of water more effectively. Thus, \( \ ext{NH}_4\text{NO}_3\) solution will have a higher vapor pressure.

Key Concepts

Boiling Point Elevation

Boiling point elevation is a fascinating colligative property of solutions. It occurs when a solute is dissolved in a solvent, causing the boiling point of the solution to be higher than that of the pure solvent. This happens because the solute particles interfere with the escape of solvent molecules into the gas phase, requiring more energy (or a higher temperature) to boil.

The magnitude of boiling point elevation can be calculated using the formula \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the increase in boiling point, \( i \) is the van't Hoff factor, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.

One interesting application of boiling point elevation is in comparing different solutions, like in the exercise above, where a solution of \( 0.10 \mathrm{m} \ \mathrm{Na}_2\mathrm{SO}_4 \) was compared to \( 0.15 \mathrm{m} \) sugar. Since \( \mathrm{Na}_2\mathrm{SO}_4 \) dissociates into more ions, it increases the boiling point more significantly than sugar, which does not dissociate. This demonstrates how the number of solute particles (determined by the van't Hoff factor \( i \)) directly influences the boiling point elevation.

Vapor Pressure Lowering

Vapor pressure lowering is another important colligative property that plays a crucial role in how solutions behave. When a non-volatile solute is dissolved in a solvent, it lowers the solvent's vapor pressure. This occurs because solute particles take up space at the surface, making it harder for solvent molecules to escape into the vapor phase.

The reduction in vapor pressure depends on the number of solute particles present. More particles mean a more significant reduction, as there are fewer solvent molecules available to escape. For example, in our exercise, the solution with \( \mathrm{Na}_7\mathrm{SO}_4 \) significantly lowers the vapor pressure compared to \( \mathrm{NH}_4\mathrm{NO}_3 \). This is because \( \mathrm{Na}_7\mathrm{SO}_4 \) dissociates into eight particles per formula unit, whereas \( \mathrm{NH}_4\mathrm{NO}_3 \) dissociates into only two particles.

Therefore, the solution with \( \mathrm{Na}_7\mathrm{SO}_4 \) exerts a greater effect on vapor pressure lowering due to its higher particle concentration in the solution.

Van't Hoff Factor

The van't Hoff factor, denoted as \( i \), is a critical concept in understanding colligative properties. It represents the number of particles into which a solute dissociates in solution. This factor is especially important when comparing ionic and non-ionic compounds.

For instance, an ionic compound like \( \mathrm{Na}_2\mathrm{SO}_4 \) dissociates into three ions: two \( \mathrm{Na}^+ \) ions and one \( \mathrm{SO}_4^{2-} \) ion. Thus, its van't Hoff factor \( i \) is 3. In contrast, a non-ionic substance such as sugar does not dissociate in solution, so its van't Hoff factor is 1.

The van't Hoff factor influences both boiling point elevation and vapor pressure lowering. The more particles a solute generates, the greater its impact on these properties. This is why \( \mathrm{Na}_2\mathrm{SO}_4 \) has a more pronounced effect on boiling point elevation compared to sugar, due to its higher van't Hoff factor. Understanding this concept helps to predict the effects various solutes will have on different colligative properties.