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 \(K_{f}\) 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 nearest to \(-0.480^{\circ} \mathrm{C}\), option (a).

Step 1: Understand the Problem

We have a 0.2 molal aqueous solution of a weak acid, HX, with a degree of ionization of 0.3. We need to find the freezing point depression using the given van't Hoff factor and cryoscopic constant (\(K_f\)) for water.

Step 2: Calculate the Van't Hoff Factor

The van't Hoff factor \(i\) accounts for the degree of ionization in the solution. For a weak acid \(HX\), which partially ionizes to \(H^+\) and \(X^-\), the degree of ionization \(\alpha\) is 0.3. Thus, \(i = 1 + \alpha = 1 + 0.3 = 1.3\).

Step 3: 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 \(i\) is the van't Hoff factor, \(K_f\) is the cryoscopic constant (1.85 °C kg/mol), and \(m\) is the molality (0.2 mol/kg). Plugging in the values, we have \(\Delta T_f = 1.3 \times 1.85 \times 0.2 = 0.481\) °C.

Step 4: Find the New Freezing Point

The freezing point of pure water is 0 °C. Since the freezing point is depressed by \(\Delta T_f\), the new freezing point is \(0 - 0.481\) °C, which is approximately \(-0.480\) °C.

Key Concepts

van't Hoff factor

The van't Hoff factor is a key concept in understanding how solutions behave when solutes are dissolved. It reflects how many particles a solute forms in solution, influencing properties like boiling point and freezing point. When a solute dissolves, it can break into ions. For example, a salt like sodium chloride (NaCl) separates into two ions, Na⁺ and Cl⁻. The van't Hoff factor, represented by the symbol \( i \), is the number of particles that result from this dissolution.

In a solution of a weak acid like HX, the van't Hoff factor isn't simply the number of particles but also considers how much the acid ionizes. Since weak acids do not fully ionize, the van't Hoff factor integrates the degree of ionization, \( \alpha \). In this exercise, the degree of ionization for HX is given as 0.3, meaning 30% ionizes into its respective ions (\( H^+ \) and \( X^- \)). Therefore, the equation for the van't Hoff factor becomes:

• \( i = 1 + \alpha \)

Placing the values for the exercise:

• \( i = 1 + 0.3 = 1.3 \)

This adjustment accounts for both the unionized and ionized forms of the acid. This factor is critical for calculating properties such as freezing point depression.

degree of ionization

The degree of ionization refers to how much of a solute ionizes in a solvent. It is a key concept when studying weak acids and bases. The degree of ionization, often denoted as \( \alpha \), is defined as the fraction of solute molecules that dissociate into ions in a solution.

For example, if you dissolve a weak acid like HX in water, not every molecule will break into \( H^+ \) and \( X^- \) ions. Instead, a percentage will, which is determined by the nature of the acid and the solution conditions.This percentage is expressed as \( \alpha = \frac{n_{ionized}}{n_{total}} \), where \( n_{ionized} \) is the number of molecules that have ionized, and \( n_{total} \) is the total number of molecules originally added.

• In the given exercise, \( \alpha = 0.3 \) indicates that 30% of the weak acid HX is ionized in the solution.

This level of ionization impacts how the solution behaves, especially in colligative properties such as freezing point and boiling point.

• If \( \alpha \) were 1, it would mean complete ionization.

However, for weak acids, \( \alpha \) is typically less than 1, affecting how we calculate things like the van't Hoff factor.

cryoscopic constant

The cryoscopic constant, denoted by \( K_f \), is an essential parameter in calculating freezing point depression. It reflects how a solute affects the freezing point of a solvent. Each solvent has a unique cryoscopic constant that is determined experimentally and is usually given in degrees Celsius per molal (°C kg/mol).

For water, the cryoscopic constant is often given as 1.85 °C kg/mol. This constant tells us how much the freezing point is lowered per mole of solute particles added. It's a critical component of the freezing point depression formula, which is:

• \( \Delta T_f = i \times K_f \times m \)

Here, \( \Delta T_f \) is the change in the freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality of the solution.

Using this formula allows us to calculate how much the freezing point of a solution decreases when a solute is added. This concept is especially important in scenarios where precise control of temperature is needed, like in antifreeze solutions or in laboratories conducting chemical reactions at specific temperatures.