Problem 79 - Chemistry An Atoms First Approach

Question

Consider the following solutions: \(0.010 m \mathrm{Na}_{3} \mathrm{PO}_{4}\) in water \(0.020 \mathrm{m} \mathrm{CaBr}_{2}\) in water \(0.020 \mathrm{m} \mathrm{KCl}\) in water \(0.020 \mathrm{m}\) HF in water (HF 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?

Step-by-Step Solution

Answer

a) The Na3PO4 and KCl solutions will have the same boiling point as the 0.040 m C6H12O6 solution. b) The HF solution will have the highest vapor pressure at 28°C. c) The CaBr2 solution will have the largest freezing-point depression.

Step 1: Calculate the molality for each solution

Since the given molality values are provided directly, we don't need to make any calculations for molality. We already have the molality for each of the solutions: - 0.010 m Na3PO4 - 0.020 m CaBr2 - 0.020 m KCl - 0.020 m HF - 0.040 m C6H12O6 (reference)

Step 2: Determine the Van't Hoff factor (i) for each solution

For non-electrolytes like C6H12O6, i=1. For electrolytes, i is equal to the number of ions produced on dissociation: - Na3PO4: 3 Na+ + PO4^3- → i = 4 - CaBr2: Ca^2+ + 2 Br- → i = 3 - KCl: K+ + Cl- → i = 2 - HF: weak acid, doesn't completely dissociate → i ≈ 1

Step 3: Calculate i×molality for each solution

Multiply molality and the number of ions produced(i) for each solution: - Na3PO4: 0.010 m × 4 = 0.040 - CaBr2: 0.020 m × 3 = 0.060 - KCl: 0.020 m × 2 = 0.040 - HF: 0.020 m × 1 ≈ 0.020 - C6H12O6: 0.040 m × 1 = 0.040 (reference) a) The solutions which have the same i×molality as the C6H12O6 solution will have the same boiling point. Both Na3PO4 and KCl solutions show the same i×molality value (0.040) as C6H12O6. b) The highest vapor pressure will be produced by the solution with the lowest molality. In this case, the HF solution has the lowest i×molality, so it will have the highest vapor pressure at 28°C. c) The largest freezing-point depression will occur in the solution with the highest i×molality. So, CaBr2 with i×molality value of 0.060 will have the largest freezing-point depression. Answers: a) The Na3PO4 and KCl solutions will have the same boiling point as the 0.040 m C6H12O6 solution. b) The HF solution will have the highest vapor pressure at 28°C. c) The CaBr2 solution will have the largest freezing-point depression.

Key Concepts

Boiling Point Elevation

Boiling point elevation is a key property of solutions that results from the introduction of a solute to a solvent, causing the boiling point of the resulting solution to be higher than that of the pure solvent. This occurs because the solute particles disrupt the solvent's ability to vaporize, which requires additional energy, thereby increasing the boiling point.
The extent of boiling point elevation is determined by the equation:\[\Delta T_b = i \cdot K_b \cdot m\]where \(\Delta T_b\) is the boiling point elevation, \(K_b\) is the ebullioscopic constant of the solvent, \(m\) is the molality of the solution, and \(i\) is the van 't Hoff factor, which represents the number of particles the solute breaks into upon dissolving.
In the exercise, two solutions have an i\(\times\)molality equal to that of the reference C\(_{6}\)H\(_{12}\)O\(_{6}\) solution: \(Na_3PO_4\) and \(KCl\). Thus, they will exhibit the same boiling point elevation as the reference solution.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. It is influenced by the presence of a solite; when a non-volatile solute is added, the vapor pressure of the solvent decreases. This happens because solute particles occupy space at the surface, reducing the number of solvent molecules available to escape into the vapor phase.
The fundamental principle behind this is Raoult's Law, given by:\[P_{solution} = X_{solvent} \cdot P_{solvent}^0\]where \(P_{solution}\) is the vapor pressure of the solution, \(X_{solvent}\) is the mole fraction of the solvent, and \(P_{solvent}^0\) is the vapor pressure of the pure solvent.
In the provided exercise scenario, the HF solution exhibits the lowest contribution to vapor pressure reduction due to its low i\(\times\)molality value, meaning it will have the highest vapor pressure at the given temperature of 28°C.

Freezing-Point Depression

Freezing-point depression is another colligative property that describes the reduction of a solution's freezing point compared to that of the pure solvent. This occurs because the solute disrupts the structure necessary for the solvent to solidify, requiring a greater removal of thermal energy (cooling).The formula to calculate freezing-point depression is:\[\Delta T_f = i \cdot K_f \cdot m\]where \(\Delta T_f\) is the freezing point depression, \(K_f\) is the cryoscopic constant of the solvent, \(m\) is the molality of the solute, and \(i\) is the van 't Hoff factor.
For this exercise, the solution of \(CaBr_2\), with the highest i\(\times\)molality value (0.060), will show the largest freezing-point depression. This indicates that the \(CaBr_2\) solution will require a greater decrease in temperature to solidify, compared to other listed solutions.