Question

Consider the following solutions: 0.010\(m \mathrm{Na}_{3} \mathrm{PO}_{4}\) in water 0.020 \(m \mathrm{CaBr}_{2}\) in water 0.020 \(m \mathrm{KCl}\) in water 0.020 \(m \mathrm{HF}\) in water \((\mathrm{HF} \text { is a weak acid.) }\) a. Assuming complete dissociation of the soluble salts, which solution(s) would have the same boiling point as 0.040 \(\mathrm{m} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water? \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is a nonelectrolyte. b. Which solution would have the highest vapor pressure at \(28^{\circ} \mathrm{C} ?\) c. Which solution would have the largest freezing-point depression?

Short answer

a) 0.010 m Na3PO4 and 0.020 m KCl will have the same boiling point as 0.040 m C6H12O6, as they have equal concentrations of solute particles (0.040 m). b) 0.020 m HF has the highest vapor pressure at \(28^{\circ} \mathrm{C}\) as it has the lowest concentration of solute particles. c) 0.020 m CaBr2 would have the largest freezing-point depression, due to its highest concentration of solute particles (0.060 m).

Step-by-step solution

Determine the concentration of solute particles in each solution

In order to determine the total concentrations of solute particles in each of the given solutions, we need to consider how many ions will be generated upon dissociation of each solute. Here's the breakdown: - 0.010 m Na3PO4: \[\mathrm{Na}_{3} \mathrm{PO}_{4}(s) \rightarrow 3\mathrm{Na}^{+}(aq) + \mathrm{PO}_{4}^{3-}(aq)\] So, the concentration of solute particles in the solution is \(0.010 \times (3 + 1) = 0.040 \: m\). - 0.020 m CaBr2: \[\mathrm{CaBr}_{2}(s) \rightarrow \mathrm{Ca}^{2+}(aq) + 2\mathrm{Br}^{-}(aq)\] So, the concentration of solute particles in the solution is \(0.020 \times (1 + 2) = 0.060 \: m\). - 0.020 m KCl: \[\mathrm{KCl}(s) \rightarrow \mathrm{K}^{+}(aq) + \mathrm{Cl}^{-}(aq)\] So, the concentration of solute particles in the solution is \(0.020 \times (1 + 1) = 0.040 \: m\). - 0.020 m HF: Since HF is a weak acid, it does not completely dissociate. Therefore, we can consider its concentration to be 0.020 m. - 0.040 m C6H12O6: Since C6H12O6 is a nonelectrolyte, its concentration remains 0.040 m.

Compare the boiling points of the solutions

The boiling point elevation of a solution is directly proportional to the concentration of solute particles. In comparing the boiling points, we find: - 0.040 m C6H12O6 - 0.010 m Na3PO4 (0.040 m solute particles) - 0.020 m KCl (0.040 m solute particles) These three solutions will have the same boiling point due to their equal concentrations of solute particles. So, the answer to question (a) is: 0.010 m Na3PO4 and 0.020 m KCl will have the same boiling point as 0.040 m C6H12O6.

Compare the vapor pressures of the solutions

The vapor pressure lowering of a solution is directly proportional to the concentration of solute particles. The solution with the highest vapor pressure at \(28^{\circ} \mathrm{C}\) will be the one with the lowest concentration of solute particles: - 0.020 m HF Thus, the answer to question (b) is: 0.020 m HF has the highest vapor pressure at \(28^{\circ} \mathrm{C}\).

Compare the freezing-point depressions of the solutions

The freezing-point depression of a solution is directly proportional to the concentration of solute particles. The solution with the largest freezing-point depression will be the one with the highest concentration of solute particles: - 0.020 m CaBr2 (0.060 m solute particles) Thus, the answer to question (c) is: 0.020 m CaBr2 would have the largest freezing-point depression.

Key concepts

Boiling Point Elevation

Boiling point elevation is a fascinating colligative property that occurs when a solute is dissolved in a solvent, causing the boiling point of the resulting solution to be higher than that of the pure solvent. This phenomenon is directly related to the concentration of solute particles in the solution. Simply put, the more particles present, the greater the elevation of the boiling point.

In this specific scenario, solutions with equal concentrations of solute particles will experience the same boiling point elevation. For example, a solution of 0.010 m sodium phosphate (\(\text{Na}_{3}\text{PO}_{4}\)) and 0.020 m potassium chloride (\(\text{KCl}\)) both have solute particle concentrations of 0.040 m, which matches that of a 0.040 m glucose solution. Consequently, they all have the same boiling point.

This effect can be understood through the equation \(\Delta T_b = i \cdot K_b \cdot m\), where \(\Delta T_b\) is the boiling point elevation, \(i\) is the van 't Hoff factor representing ion quantity, \(K_b\) is the ebullioscopic constant, and \(m\) is the molality of the solution. The boiling point elevation helps in determining solution properties, making it valuable for various chemical and industrial applications.

Vapor Pressure Lowering

Vapor pressure lowering is another key colligative property related to the presence of solute particles in a solution. When a non-volatile solute is dissolved in a solvent, the vapor pressure of the resulting solution is lower than that of the pure solvent. This is because solute particles occupy space at the surface, reducing the number of solvent molecules able to escape into the vapor phase.

The extent of vapor pressure lowering is determined by the total concentration of solute particles in a solution. Solutions with higher concentrations of solute particles exhibit more significant vapor pressure reductions. For instance, in this problem, the 0.020 m hydrofluoric acid (\(\text{HF}\)) solution has the lowest concentration of solute particles, and thus, it retains the highest vapor pressure among the compared solutions.

The equation for vapor pressure lowering is given by Raoult's Law: \(P = X_{solvent} \cdot P^o_{solvent}\), where \(P\) is the vapor pressure of the solution, \(P^o_{solvent}\) is the vapor pressure of the pure solvent, and \(X_{solvent}\) is the mole fraction of the solvent.

Freezing Point Depression

Freezing point depression is a colligative property that describes the lowering of a solvent's freezing point when a solute is introduced. Simply put, the more solute particles dissolved, the lower the freezing point. This effect is particularly useful when trying to understand and manipulate the properties of solutions in various chemical fields.

In our context, freezing point depression is observed with the solution of calcium bromide (\(\text{CaBr}_2\)), because it results in a solute particle concentration of 0.060 m upon dissociation. This concentration is the highest among the solutions considered, which accounts for the largest decrease in freezing point.

The equation representing freezing point depression is \(\Delta T_f = i \cdot K_f \cdot m\), where \(\Delta T_f\) is the freezing point depression, \(i\) is the van 't Hoff factor, \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution.

Ion Dissociation in Solutions

Ion dissociation in solutions is a crucial concept for understanding colligative properties, as it directly affects the number of solute particles present in the solution. When ionic compounds dissolve in water, they separate into individual positive and negative ions.

This dissociation process determines the effective concentration of particles in a solution, impacting properties like boiling point elevation, freezing point depression, and vapor pressure lowering.

For instance, consider \(\text{CaBr}_2\). It dissociates into three ions: one calcium ion (\(\text{Ca}^{2+}\)) and two bromide ions (\(\text{Br}^-\)). This increases the total particle concentration, intensifying colligative effects. Additionally, weak electrolytes like \(\text{HF}\) do not fully dissociate, leading to fewer particles in the solution.

Improved understanding of ion dissociation helps to predict and control the behavior of solutions in real-world applications, like chemical reactions or manufacturing processes.