Question

(a) Why does a \(0.10 m\) aqueous solution of NaCl have a higher boiling point than a \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?\) (b) Calculate the boiling point of each solution. (c) The experimental boiling point of the \(\mathrm{NaCl}\) solution is lower than that calculated, assuming that \(\mathrm{NaCl}\) is completely dissociated in solution. Why is this the case?

Short answer

The boiling point of a 0.10 m NaCl solution is higher than a 0.10 m C₆H₁₂O₆ solution because NaCl dissociates into two ions (Na⁺ and Cl⁻) in water, while C₆H₁₂O₆ doesn't dissociate and remains as a single molecule. The boiling points of the NaCl and C₆H₁₂O₆ solutions are 100.1024 °C and 100.0512 °C, respectively. The experimental boiling point of the NaCl solution is lower than the calculated value because not all NaCl molecules dissociate completely, leading to a lower boiling point elevation.

Step-by-step solution

(a) Understanding the boiling point elevation due to solutes

Boiling point elevation occurs when a non-volatile solute is added to a solvent, causing the boiling point of the solution to be higher than that of the pure solvent. In this case, both NaCl and C₆H₁₂O₆ are non-volatile solutes being added to water. The main reason why the boiling point of a 0.10 molarity (m) NaCl solution is higher than a 0.10 m C₆H₁₂O₆ solution is due to the number of particles in the solution. NaCl is an ionic compound and dissociates into two ions (Na⁺ and Cl⁻) when dissolved in water, while C₆H₁₂O₆, a sugar molecule, does not dissociate and remains as a single molecule in solution.

(b) Calculating the boiling points of NaCl and C₆H₁₂O₆ solutions

We can use the formula for boiling point elevation to calculate the boiling points of the NaCl and C₆H₁₂O₆ solutions: ΔT = i × Kb × m Where ΔT is the boiling point elevation, i is the van't Hoff factor (number of particles produced per formula unit in the solution), Kb is the molal boiling point constant of water (0.512 °C/m), and m is the molality. For NaCl: i = 2 (since NaCl dissociates into two ions) For C₆H₁₂O₆: i = 1 (since it doesn't dissociate) Now we can calculate the boiling point elevation for each solution: ΔT_NaCl = 2 × 0.512 × 0.10 = 0.1024 °C ΔT_C₆H₁₂O₆ = 1 × 0.512 × 0.10 = 0.0512 °C The boiling point of pure water is 100 °C, so we can find the boiling points of both solutions by adding the calculated boiling point elevation to the boiling point of water: Boiling point of NaCl solution = 100 °C + 0.1024 °C = 100.1024 °C Boiling point of C₆H₁₂O₆ solution = 100 °C + 0.0512 °C = 100.0512 °C

(c) Explaining the discrepancy between experimental NaCl boiling point and calculated boiling point

The calculated boiling point of the NaCl solution assumes complete dissociation of NaCl into Na⁺ and Cl⁻ ions. In reality, however, not all NaCl molecules dissociate completely in solution. Some Na⁺ and Cl⁻ ions may associate with each other, forming ion pairs. These ion pairs effectively reduce the number of particles in the solution, leading to a lower boiling point elevation than expected from the calculation. Consequently, the experimental boiling point of the NaCl solution is lower than the calculated boiling point.

Key concepts

Colligative Properties

Colligative properties are fascinating aspects of chemistry that only depend on the number of solute particles in a solution, not the nature of the solute itself. This means that characteristics like boiling point elevation, freezing point depression, and osmotic pressure are affected by how many particles are present, regardless of their specific type. This property is crucial when considering solutions like saltwater, where adding more particles changes how the solution behaves.
For instance, in a salt solution compared to pure water, the introduction of salt affects the boiling point. After adding the solute, the solution requires more energy (heat) to convert into vapor, resulting in a higher boiling point. This is why the boiling point of a saline solution is higher than that of pure water. Understanding colligative properties helps us predict and control these changes in solution behavior for various practical applications.

Van't Hoff Factor

The van't Hoff factor, denoted as \(i\), is a critical concept when discussing colligative properties. It represents the number of particles formed in a solution from a single solute molecule. For instance, when sodium chloride (NaCl) dissolves in water, it dissociates into two ions: Na⁺ and Cl⁻. Therefore, the van't Hoff factor for NaCl is 2.
In contrast, glucose (C₆H₁₂O₆) does not dissociate into separate particles when it dissolves. It stays as a single intact molecule, giving it a van’t Hoff factor of 1. This difference significantly impacts the boiling point elevation between solutions of NaCl and C₆H₁₂O₆, even if they have the same concentration. The NaCl solution generates more particles, enhancing the colligative effect and increasing the boiling point more than the glucose solution.

Solute Dissociation

Solute dissociation refers to how a solute breaks down into its individual ions when dissolved in a solvent. This process is pivotal in determining the van't Hoff factor and thus the colligative properties. Sodium chloride is a classic example of a solute that dissociates. In water, it separates into two ions, Na⁺ and Cl⁻, contributing to the higher boiling point elevation compared to non-dissociating solutes like glucose.
This breakdown directly impacts how solutions behave under various conditions. However, solute dissociation is not always complete. Factors like concentration and intermolecular forces can cause some ions to reassociate or form ion pairs, reducing the number of free particles. This incomplete dissociation can lead to experimental boiling points being lower than theoretical predictions, as seen with NaCl solutions in reality.