Problem 166 - Introductory Chemistry Atoms First

Question

The freezing point of a solution prepared by dissolving \(0.200\) mole of \(\mathrm{HF}(g)\) in \(2.00 \mathrm{~kg}\) of water is \(-0.19{ }^{\circ} \mathrm{C}\). Is HF primarily intact in solution, existing as \(\mathrm{HF}(a q)\), or has it dissociated to \(\mathrm{H}^{+}(a q)\) and \(\mathrm{F}^{-}(a q)\) ions? Does this mean \(\mathrm{HF}\) is a weak or a strong acid?

Step-by-Step Solution

Answer

The calculated molality of the HF solution is 0.1\(\frac{moles}{kg}\), and the experimental molality using the freezing point depression formula is approximately 0.102\(\frac{moles}{kg}\). Since these values are very close to each other, it indicates that HF does not fully dissociate in the solution and primarily remains intact as \(\mathrm{HF}(aq)\). Therefore, \(\mathrm{HF}\) is a weak acid, as it does not dissociate completely in the solution.

Step 1: Calculate molality of HF in the solution

First, we need to calculate the molality of the HF solution using the formula: \[molality = \frac{moles \ of \ solute}{mass \ of \ solvent \ (in \ kg)}\] Substitute the given values into the formula: \[molality = \frac{0.200 \ moles}{2.00 \ kg}\] Calculate the molality: \[molality = 0.1 \ \frac{moles}{kg}\]

Step 2: Calculate the experimental molality

Now, we can calculate the experimental molality using the freezing point depression formula: \[ΔT_f = K_f × molality × i\] Where ΔT_f = Freezing point depression (0.19 °C in this case) K_f = Freezing point depression constant for water (1.86 °C kg/mol) molality = Experimental molality of the solution i = van't Hoff factor (number of ions formed during dissociation) First, let's assume HF remains primarily intact in the solution (does not dissociate). In this case, the van't Hoff factor (i) will be 1. Rearrange the formula to find molality: \[molality = \frac{ΔT_f}{K_f × i}\] Substitute the values: \[molality = \frac{0.19°C}{1.86 \ °C\ kg/mol × 1}\] Calculate the molality: \[molality \approx 0.102 \ \frac{moles}{kg}\]

Step 3: Compare the calculated and experimental molality

Now, compare the calculated molality from step 1 (0.1\(\frac{moles}{kg}\)) and the experimental molality from step 2 (0.102\(\frac{moles}{kg}\)) to determine if HF primarily remains intact or dissociates in solution. The calculated molality (0.1\(\frac{moles}{kg}\)) and the experimental molality (0.102\(\frac{moles}{kg}\)) are very close to each other, which indicates that HF does not fully dissociate in the solution and primarily remains intact as \(\mathrm{HF}(aq)\).

Step 4: Conclude whether HF is a weak or strong acid

Since we have seen that HF does not fully dissociate in the solution and primarily remains intact as \(\mathrm{HF}(aq)\), we can conclude that \(\mathrm{HF}\) is a weak acid. A strong acid would fully dissociate into ions in the solution, while a weak acid does not dissociate completely.

Key Concepts

Colligative Properties

Colligative properties are characteristics of solutions that depend on the number, not the identity, of dissolved solute particles. When a solute is added to a solvent, properties such as vapor pressure, boiling point, and freezing point can be altered. One notable example is freezing point depression, a colligative property which occurs when the addition of a solute lowers the freezing point of the solution compared to that of the pure solvent. This happens because solute particles disrupt the crystalline structure formation of the ice, requiring a lower temperature to freeze.

For the problem at hand, the freezing point depression is used to determine whether hydrogen fluoride (HF) behaves primarily as an intact molecule or as dissociated ions when dissolved in water. By understanding the degree to which the solute affects the freezing point, one can infer the extent of ionization and subsequently the acid strength of HF. The comparison of the calculated molality with the experimental one reveals that the change in freezing point is consistent with HF not fully dissociating into ions, hence indicating its property as a weak electrolyte.

Molality

Molality is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of solution, molality is temperature-independent as it relies on mass. This makes it particularly useful in colligative properties calculations, where temperature changes are involved.

In the given exercise, we first calculate the molality of HF in solution, which helps us to understand the concentration of the solute. The molality, in turn, plays a critical role in calculating the freezing point depression of the solution. By calculating the difference between the experimental and the theoretical molality, the extent of the acid dissociation can be determined, leading to conclusions about whether HF is a weak or strong acid.

Acid Dissociation

Acid dissociation is a chemical process where an acid donates a proton (H⁺) to water, producing a hydronium ion (H₃O⁺) and an anion. The extent of dissociation is a measure of an acid's strength. A strong acid completely dissociates into its ions in aqueous solution, while a weak acid only dissociates partially.

In the context of our problem, the near-equal values for the calculated and experimental molality suggest that HF only partially dissociates, confirming it as a weak acid. If HF were a strong acid, we would observe a higher experimental molality due to an increased number of particles from the full dissociation, affecting the freezing point depression significantly.

van't Hoff Factor

The van't Hoff factor (\( i \)) quantifies the number of particles a compound forms when it dissolves in solution. For non-electrolytes and compounds that remain intact in solution, the van't Hoff factor is 1. However, for substances that dissociate into multiple particles, such as salts or acids in water, the factor will be equal to the number of ions produced per formula unit of solute.

Assuming HF does not dissociate in the solution, its van't Hoff factor is 1. Our exercise uses this value to calculate the molality based on the freezing point depression data. Since the calculated molality closely matches the experimental molality, this suggests that the van't Hoff factor for HF in solution is indeed near 1. This reinforces the idea that HF is a weak acid, as it does not fully dissociate to contribute a larger number of ions to the solution.