Problem 58 - 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 \(0.15 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4} ?\)

Step-by-Step Solution

Answer

(a) Na2SO4 solution has a higher boiling point; (b) Na2SO4 solution has a higher vapor pressure.

Step 1: Understand Boiling Point Elevation

Boiling point elevation 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 can be 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, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.

Step 2: Determine the van’t Hoff Factor (i)

The van't Hoff factor \( i \) is the number of particles the solute splits into or forms when dissolved. For Na2SO4, which dissociates into 2 Na+ and 1 SO42-, \( i = 3 \). Sugar does not dissociate in solution, so \( i = 1 \).

Step 3: Calculate Boiling Point Elevation for Each Solution

For a \(0.10 \text{ m} \; \text{Na}_2\text{SO}_4\) solution, \( \Delta T_b = 3 \times K_b \times 0.10 \). For a \(0.15 \text{ m} \) sugar solution, \( \Delta T_b = 1 \times K_b \times 0.15 \). Since 0.30 \( K_b \) is greater than 0.15 \( K_b \), the Na2SO4 solution will have a higher boiling point.

Step 4: Understand Vapor Pressure Lowering

The vapor pressure of a solvent decreases when a solute is dissolved in it, causing a phenomenon known as vapor pressure lowering. This can be observed through Raoult's Law: \( P_{solution} = P_{pure} \cdot X_{solvent} \), where \( X_{solvent} \) is the mole fraction of the solvent.

Step 5: Calculate Effective Concentration for Vapor Pressure

For the \(0.30 \text{ m} \; \text{NH}_4\text{NO}_3\) solution, \( i = 2 \) (NH4NO3 dissociates into NH4+ and NO3-), making its effective molality \(0.60 \text{ m}\). For the \(0.15 \text{ m} \; \text{Na}_2\text{SO}_4\) solution, the effective molality is \(0.45 \text{ m}\). The solution with higher effective concentration will lower the vapor pressure more.

Step 6: Compare Vapor Pressures

Given the calculations, the solution of \(0.15 \text{ m} \; \text{Na}_2\text{SO}_4\) has a lower effective concentration (0.45). The vapor pressure of its solution is therefore higher than that of the \(0.30 \text{ m} \; \text{NH}_4\text{NO}_3\) solution.

Key Concepts

Boiling Point Elevation

When a solute is added to a solvent, the boiling point of the solution increases. This phenomenon is known as boiling point elevation. It happens because the solute particles interfere with the liquid molecules, making it harder for them to escape into the vapor phase. Boiling point elevation can be quantified using the following formula:
- \( \Delta T_b = i \cdot K_b \cdot m \)
Here:

• \( \Delta T_b \) is the change in boiling point.
• \( i \) is the van't Hoff factor, which indicates the number of particles the solute produces.
• \( K_b \) is the ebullioscopic constant for the solvent.
• \( m \) is the molality of the solution.

To understand this with an example, let's compare two solutions: a 0.10 m Na\(_2\)SO\(_4\) solution and a 0.15 m sugar solution. Na\(_2\)SO\(_4\) dissociates into three ions (2 Na\(^+\) and one SO\(_4^{-2}\)), giving it a van't Hoff factor \(i = 3\). The sugar, on the other hand, does not dissociate and thus has \(i = 1\). This means the boiling point elevation for Na\(_2\)SO\(_4\) will be greater, evidenced by the formula \(3 \times K_b \times 0.10\), compared to \(1 \times K_b \times 0.15\) for sugar. Therefore, the Na\(_2\)SO\(_4\) solution will have a higher boiling point than the sugar solution.

Vapor Pressure Lowering

When a solute is mixed with a solvent, the solvent's vapor pressure decreases. This is known as vapor pressure lowering. The presence of solute particles reduces the number of solvent molecules that can escape into the vapor phase, which in turn lowers the vapor pressure.
The change in vapor pressure can be explained through Raoult's Law:
- \( P_{solution} = P_{pure} \cdot X_{solvent} \)
Where:

• \( P_{solution} \) is the vapor pressure of the solution.
• \( P_{pure} \) is the vapor pressure of the pure solvent.
• \( X_{solvent} \) is the mole fraction of the solvent in the solution.

Now, let's compare a 0.30 m NH\(_4\)NO\(_3\) solution with a 0.15 m Na\(_2\)SO\(_4\) solution to understand this concept more practically. NH\(_4\)NO\(_3\) dissociates into two ions (NH\(_4^+\) and NO\(_3^-\)), giving it a van't Hoff factor of \(i = 2\). For Na\(_2\)SO\(_4\), with \(i = 3\), the effective molality becomes 0.45 m compared to NH\(_4\)NO\(_3\)'s 0.60 m. The higher the effective molality, the more vapor pressure is lowered. Thus, even though 0.15 m \( \text{Na}_2\text{SO}_4 \) seems like a lower concentration, due to its lower effective concentration (0.45 m), it leads to higher vapor pressure than the NH\(_4\)NO\(_3\) solution with an effective concentration of 0.60 m.

van't Hoff Factor

The van't Hoff factor \(i\) is a crucial element when discussing colligative properties. This factor denotes the number of particles a solute separates into when dissolved in a solution. Understanding \(i\) helps predict the colligative properties like boiling point elevation and vapor pressure lowering.
To clarify, let's delve into how various solutes dissociate:

• For Na\(_2\)SO\(_4\), which dissociates into 3 ions in solution: 2 Na\(^+\) and 1 SO\(_4^{-2}\), \(i = 3\).
• Sugar molecules do not dissociate upon dissolution, hence \(i = 1\).

The significance of \(i\) comes into play in determining colligative properties. For instance, in boiling point elevation, \(i\) multiplies the molality of the solution to reflect how many solute particles are actually in the solution, influencing the boiling point significantly more than just solute concentration alone.
For example, even if two solutions have the same molality, the one with the higher van't Hoff factor will exhibit a more significant change to its colligative properties. In summary, the van't Hoff factor provides an understanding of how effective a solute is in a solution, depending on its dissociation nature.