Question

The solubility of benzoic acid, is 0.34 \(\mathrm{g} / 100 \mathrm{mL}\) in water at \(25^{\circ} \mathrm{C}\) and 10.0 \(\mathrm{g} / 100 \mathrm{mL}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(25^{\circ} \mathrm{C} .\) Rationalize this solubility behavior. For a \(1.0-\mathrm{m}\) solution of benzoic acid in benzene, would the measured freezing point depression be equal to, greater than, or less than \(5.12^{\circ} \mathrm{C} ?\left(K_{\mathrm{f}}=5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} \text { for benzene.) }\right.\)

Short answer

The solubility of benzoic acid is higher in benzene than in water because benzoic acid is an organic compound and more soluble in nonpolar solvents like benzene. For a 1.0 m solution of benzoic acid in benzene, the measured freezing point depression would be equal to 5.12 °C, as calculated using the formula ΔT = Kf × m × i, with i=1 for benzoic acid in benzene.

Step-by-step solution

Analyze the nature of the substances and their interactions

Benzoic acid is an organic compound, while water is a polar compound due to the presence of the hydroxyl group (-OH). Benzene is also an organic compound, which makes it nonpolar. According to the principle "like dissolves like", polar compounds are more soluble in polar solvents and nonpolar compounds in nonpolar solvents. This explains why benzoic acid has higher solubility in benzene.

Calculate the expected freezing point depression

To determine whether the measured freezing point depression would be equal to, greater than, or less than \(5.12^{\circ} \mathrm{C}\), we will use the freezing point depression formula: ΔT = Kf × m × i where ΔT represents the freezing point depression, Kf is the cryoscopic constant for the solvent, m is the molality of the solution, and i is the dissociation factor. Since benzoic acid is a weak acid and does not dissociate into ions in a nonpolar solvent like benzene, i = 1.

Plug in the values and calculate the freezing point depression

Given a 1.0 m solution of benzoic acid in benzene, and using the Kf value provided for benzene, we can plug in the values into the formula: ΔT = (5.12 °C · kg/mol) × (1.0 mol/kg) × (1) ΔT = 5.12 °C

Compare the calculated freezing point depression to the given value of 5.12 °C

Since the calculated freezing point depression is exactly equal to the provided value of 5.12 °C, we conclude that the measured freezing point depression for a 1.0 m solution of benzoic acid in benzene would be equal to 5.12 °C.

Key concepts

Freezing Point Depression

Freezing point depression is an important concept in chemistry, especially when dealing with solutions. This phenomenon occurs when a solute is added to a solvent, causing the freezing point of the solvent to decrease. The reason behind this effect is that the presence of the solute disrupts the orderly crystal formation of the solvent molecules.
The greater the concentration of the solute particles, the more significant the disruption, leading to a lower freezing point. In our exercise, the formula for freezing point depression is \[\Delta T = K_f \times m \times i\]Where:

• \(\Delta T\) is the freezing point depression
• \(K_f\) is the cryoscopic constant of the solvent
• \(m\) is the molality of the solution
• \(i\) is the van 't Hoff factor, related to the degree of dissociation of the solute

In a nonpolar solvent like benzene, where the solute does not ionize, \(i\) typically equals 1.

Like Dissolves Like Principle

The phrase 'like dissolves like' is crucial in understanding solubility. It essentially means that polar solvents are better at dissolving polar solutes, whereas nonpolar solvents are more effective at dissolving nonpolar solutes.
This concept is derived from the molecular interactions within solutions. Polar molecules have positive and negative ends due to uneven charge distribution, thus they dissolve well in solvents that can also exhibit similar charge interactions.
Conversely, nonpolar molecules, which have more uniform charge distribution, dissolve better in nonpolar solvents where they experience similar types of molecular forces. In the exercise, benzoic acid dissolves better in benzene than in water due to this principle.

Polar and Nonpolar Interactions

Polar and nonpolar interactions describe the types of forces that occur between molecules, impacting solubility and other physical properties. Polar interactions involve attraction between molecules with partial positive and negative charges, akin to magnets. These interactions can lead to significant dissolving of polar substances in polar solvents.
Nonpolar interactions, on the other hand, are based on van der Waals forces, which are weaker than polar interactions but sufficient for nonpolar molecules. They rely on temporary charges induced in nonpolar molecules. In our exercise, the stronger polar interactions in water cannot effectively overcome the lack of charges in benzoic acid, making it less soluble. While benzene, being nonpolar, facilitates these weaker, but suitable interactions.

Cryoscopic Constant

The cryoscopic constant \(K_f\) is a property unique to each solvent, representing the freezing point depression per molal concentration of a solute. It quantifies how much the presence of a mole of solute particles lowers the freezing point of a given solvent.
Benzene has a cryoscopic constant of 5.12 °C·kg/mol, meaning that for each mole of solute added per kilogram of benzene, the freezing point decreases by 5.12 °C.
In the provided exercise, this constant helps us calculate the depression in the freezing point of benzene when benzoic acid is dissolved in it. By using this constant along with the molality of the solution, we can determine the extent of freezing point lowering accurately, showing its significance in freezing point depression calculations.