Question

Arrange \(0.30 \mathrm{~m}\) solutions of the following solutes in order of increasing freezing point and boiling point. (a) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (c) \(\mathrm{Ba}(\mathrm{OH})_{2}\) (d) \(\mathrm{CaCr}_{2} \mathrm{O}_{7}\)

Short answer

Arrange the following solutions in order of increasing freezing points and boiling points: Fe(NO₃)₃, C₂H₅OH, Ba(OH)₂, and CaCr₂O₇. The order of increasing freezing points and boiling points for these solutions is: C₂H₅OH, Ba(OH)₂, CaCr₂O₇, and Fe(NO₃)₃.

Step-by-step solution

Determine the van't Hoff factor (i) for each solute.

The van't Hoff factor is the number of particles produced when the compound dissociates in water. Let's determine the van't Hoff factor for each solute: (a) Fe(NO₃)₃: Dissociates into Fe³⁺ and 3NO₃¯ ions, so i = 4. (b) C₂H₅OH: Doesn't dissociate as it's a non-electrolyte, so i = 1. (c) Ba(OH)₂: Dissociates into Ba²⁺ and 2OH⁻ ions, so i = 3. (d) CaCr₂O₇: Dissociates into Ca²⁺ and 2CrO₄²⁻ ions, so i = 3.

Apply colligative properties

In this problem, we are asked to arrange the solutions in order of increasing freezing and boiling points. To do this, we will apply the colligative properties for freezing point depression and boiling point elevation: Freezing point depression: ΔTf = Kf · m · i Boiling point elevation: ΔTb = Kb · m · i Where ΔTf and ΔTb represent the changes in freezing and boiling points, respectively, Kf and Kb are the solvent's constants, m is the molality of the solute, and i is the van't Hoff factor.

Compare freezing point depressions and boiling point elevations

Since all the solutions have the same molality (0.30 m) and we are comparing the solutions in the same solvent (water), the Kf and Kb values will be constant for all solutes. Therefore, we can compare the freezing point depressions and boiling point elevations based only on the van't Hoff factors: ΔTf and ΔTb will be smaller for smaller i values. Therefore, we can arrange the solutions in order of increasing freezing points and boiling points based on their van't Hoff factors: Freezing point: C₂H₅OH (i=1) < Ba(OH)₂ (i=3) = CaCr₂O₇ (i=3) < Fe(NO₃)₃ (i=4) Boiling point: C₂H₅OH (i=1) < Ba(OH)₂ (i=3) = CaCr₂O₇ (i=3) < Fe(NO₃)₃ (i=4) So, the solutions in order of increasing freezing point are: C₂H₅OH, Ba(OH)₂, CaCr₂O₇, and Fe(NO₃)₃. The solutions in order of increasing boiling point are also: C₂H₅OH, Ba(OH)₂, CaCr₂O₇, and Fe(NO₃)₃.

Key concepts

Freezing Point Depression

In chemistry, freezing point depression refers to the effect where the addition of a solute to a pure solvent reduces the temperature at which the liquid becomes a solid. It's a crucial concept in colligative properties, which are properties that depend on the ratio of the number of solute particles to the number of solvent molecules.

When a solute, like salt, is added to a solvent, such as water, it disrupts the interactions between the solvent molecules. This disruption requires a lower temperature to achieve the organized structure of a solid (crystal lattice), which is why the freezing point is depressed. The more solute particles present, the greater the effect.

One equation used to calculate the freezing point depression is \( \Delta T_f = K_f \cdot m \cdot i \),where \( \Delta T_f \) is the change in freezing point, \( K_f \) is the cryoscopic constant (a characteristic of the solvent), \( m \) is the molality of the solution, and \( i \) is the van't Hoff factor, which accounts for the number of particles the solute forms in solution.

To illustrate, let's consider the impact of a solution with a fixed molality on various solutes. A molecule that dissolves without ionizing (such as ethanol, \( C_2H_5OH \)) has a smaller effect on the freezing point compared to ionic compounds that dissociate into multiple particles (like \( Ba(OH)_2 \)). Thus, for equally concentrated solutions, the extent of freezing point depression follows the number of particles formed in solution.

• Non-electrolytes (do not dissociate): Smallest freezing point depression
• Electrolytes (dissociate into ions): Greater freezing point depression

Boiling Point Elevation

Boiling point elevation is another colligative property which describes how the presence of a solute that’s dissolved in a solvent can raise the solvent’s normal boiling point. Similar to freezing point depression, boiling point elevation is dependent on the number of solute particles compared to the solvent molecules.

Adding a non-volatile solute (one that does not vaporize easily) to a liquid solvent means that there are now fewer solvent molecules at the surface to escape into the gas phase. To boil, the liquid must have higher kinetic energy, which necessitates higher temperatures. Higher boiling points often signal greater intermolecular forces or greater amounts of solute.

The equation for boiling point elevation is \( \Delta T_b = K_b \cdot m \cdot i \),with \( \Delta T_b \) being the change in boiling point, \( K_b \) the ebullioscopic constant, \( m \) the molality, and \( i \) the van't Hoff factor. As with freezing point depression, the van't Hoff factor is crucial since it determines how many particles the solute creates when dissolved. Boiling point elevation is directly proportional to the product of the solute's molality and the van't Hoff factor.

When comparing the boiling point elevations of various solutions with the same molality, the solution with the highest van't Hoff factor causes the greatest boiling point elevation. For instance, an ionic compound with a van't Hoff factor greater than 1 will elevate the boiling point more than an equal concentration of a molecular compound (which usually has a van't Hoff factor of 1). This concept explains why, for example, adding table salt to water will make the water boil at a temperature higher than 100°C under normal atmospheric conditions.

van't Hoff Factor

The van't Hoff factor, denoted by \( i \), is fundamental to understanding colligative properties like freezing point depression and boiling point elevation. It represents the number of particles into which a substance dissociates or ionizes in solution. For example, a molecule that stays intact as a whole in solution, such as glucose, has a van't Hoff factor of 1. In contrast, an ionic compound like sodium chloride, which splits into sodium and chloride ions, has a van't Hoff factor of 2.

Determining the van't Hoff factor is a step-by-step process:

• Identify the solute (the substance being dissolved).
• Understand whether the solute is an electrolyte (which dissociates into ions) or a non-electrolyte (does not dissociate).
• Predict the number of particles the solute will form in solution. For ionic compounds, this typically equals the sum of the cations and anions produced upon dissolution.

In solutions with multiple solutes, the total van't Hoff factor is the sum of the factors for each solute. However, the real-world factor can be lower than the theoretical value due to ion pairing or incomplete dissociation, which need to be considered in precise calculations.

To reiterate the importance of the van't Hoff factor, let's summarize: It impacts the degree to which the boiling point increases and the freezing point decreases in a solution. The greater the van't Hoff factor, the more pronounced these colligative effects will be. In our original exercise, substances like \( Fe(NO_3)_3 \), which dissociate completely into multiple ions, significantly alter the boiling and freezing points of water in comparison to substances like ethanol, which does not dissociate and thus has only a moderate impact.