Problem 150 - A Complete Resource Book in Chemistry for JEE Main

Question

In a \(0.2\) molal aqueous solution of a weak acid HX, the degree of ionization is \(0.3\). Taking \(\mathrm{K}_{\mathrm{f}}\) for water as \(1.85 \mathrm{k} \mathrm{kg}\) melt, the freezing point of the solution will be nearest to (a) \(-0.480^{\circ} \mathrm{C}\) (b) \(-0.360^{\circ} \mathrm{C}\) (c) \(-0.260^{\circ} \mathrm{C}\) (d) \(+0.480^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The freezing point of the solution is closest to option (a) \(-0.480^{\circ} \text{C}\).

Step 1: Determine the van't Hoff factor (i)

For a weak acid HX, the acid dissociates as \( ext{HX} \rightleftharpoons ext{H}^+ + ext{X}^- \). Let the initial concentration of HX be \(0.2 \text{ m}\), and the degree of ionization be \(0.3\). The moles of ions produced are \(2 \times 0.3 \times 0.2 = 0.12\text{ m}\) for the ions \(H^+\) and \(X^-\). Thus, the total concentration of particles is \(0.2 - 0.2 \times 0.3 + 0.12 = 0.26 \text{ m}\). So, the van't Hoff factor \(i\) is \(\frac{0.26}{0.2} = 1.3\).

Step 2: Apply the Freezing Point Depression Formula

The freezing point depression \( \Delta T_f \) is calculated using the formula: \[ \Delta T_f = i \cdot K_f \cdot m \] where \( i = 1.3 \), \( K_f = 1.85 \text{ K kg mole}^{-1} \), and \( m = 0.2 \). Substitute these values into the formula to get: \[ \Delta T_f = 1.3 \times 1.85 \times 0.2 = 0.481 \text{ K} \].

Step 3: Calculate the New Freezing Point

Since the normal freezing point of water is \(0^{\circ} \text{C}\), the new freezing point of the solution will be \(0^{\circ} \text{C} - 0.481^{\circ} \text{C} = -0.481^{\circ} \text{C}\).

Step 4: Compare with the Given Options

Among the options provided, the calculated freezing point \(-0.481^{\circ} \text{C}\) is closest to option (a) \(-0.480^{\circ} \text{C}\).

Key Concepts

Van't Hoff Factor

The van't Hoff factor, often denoted as \( i \), is an important concept in colligative properties, such as freezing point depression. It accounts for the effect of solute particles in solution. Specifically, it tells us how many particles a compound dissociates into when dissolved in a solvent.
For non-electrolytes, which do not dissociate into ions, the van't Hoff factor is \( i = 1 \). However, electrolytes, like the weak acid HX in the problem, dissociate into multiple ions.
In the dissociation process of the weak acid, HX splits into \( \text{H}^+ \) and \( \text{X}^- \). Given the degree of ionization is 0.3, 30% of HX dissociates. This means that for each mole of HX, there are approximately 0.3 moles of \( \text{H}^+ \) and 0.3 moles of \( \text{X}^- \) produced. Hence, the total ionic concentration in solution becomes higher than the initial concentration of HX, requiring the van't Hoff factor's adjustment.
The van't Hoff factor is crucial because it scales the impact of solute particles on properties such as the freezing point. In our scenario, the calculated \( i \) was found to be 1.3, reflecting both the undissociated and dissociated portions of the weak acid.

Solution Concentration

Solution concentration refers to the amount of solute present in a given quantity of solvent or solution. It's a fundamental aspect in chemistry, especially when discussing reactions and properties. Molality is a common unit of concentration used in colligative properties as it is temperature-independent.
In the context of our exercise, we're dealing with a 0.2 molal (m) solution, reflecting that there are 0.2 moles of HX dissolved in 1 kilogram of water. Unlike molarity, molality considers mass over volume, making it particularly useful when studying freezing point depression, since temperature changes do not affect mass.
Concentration plays a pivotal role in determining how solutes like HX will behave in solution, impacting how it affects the properties of the solvent. Here, the given concentration is used to accurately compute the change in freezing point, as it affects factors like ionization and the van't Hoff factor.

Degree of Ionization

The degree of ionization is an indication of the extent to which a solute, like a weak acid, dissociates into ions in solution. It is typically expressed as a fraction or percentage.
For weak acids such as HX, which do not fully dissociate, the degree of ionization provides insight into the percentage of the original compound that forms ions. For example, a degree of ionization of 0.3, or 30%, indicates that only 30% of the weak acid's molecules dissociate into ions, specifically \( \text{H}^+ \) and \( \text{X}^- \).
Understanding the degree of ionization is important as it directly influences the number of particles in solution, affecting colligative properties such as freezing point depression. During calculation, it helps in determining the effective concentration of ions, which then contributes to calculating the van't Hoff factor. Therefore, the degree of ionization is integral to predicting how a weak acid like HX changes the freezing point of its solution.