Problem 144 - A Complete Resource Book in Chemistry for JEE Main

Question

What would be the \(\mathrm{pH}\) of a \(0.1\) molal aqueous solution of a monoprotic acid 'HA', that freezes at \(-0.2046^{\circ} \mathrm{C}\) ? \(\left[\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86^{\circ} \mathrm{mol}^{-1} \mathrm{~kg} ;\right.\) assuming molality \(=\) molarity]

Step-by-Step Solution

Answer

The pH of the solution is 2.

Step 1: Understanding the Problem

We need to determine the pH of a 0.1 molal solution of a monoprotic acid that freezes at a specific temperature. The key to solving this is using the freezing point depression formula to find the degree of dissociation of the acid.

Step 2: Use Freezing Point Depression Formula

Apply the formula for freezing point depression: \[ \Delta T_f = i \cdot K_f \cdot m \]Where \( \Delta T_f = 0.2046^{\circ}C \), \( K_f = 1.86 \), and \( m = 0.1 \). Solve for \( i \) (van't Hoff factor), representing the extent of dissociation.

Step 3: Calculate Van’t Hoff Factor

Plug the values into the formula: \[ 0.2046 = i \cdot 1.86 \cdot 0.1 \]This simplifies to \[ i = \frac{0.2046}{0.186} \approx 1.1 \].This indicates that the acid partially dissociates.

Step 4: Determine Degree of Dissociation

Monoprotic acid, \( HA \), dissociates to \( H^+ \) and \( A^- \). The degree of dissociation, \( \alpha \), relates to \( i \): \[ i = 1 + \alpha \]Thus, \( \alpha = i - 1 = 1.1 - 1 = 0.1 \).

Step 5: Calculate Concentration of Hydrogen Ions

If \( \alpha = 0.1 \), then \[ [H^+] = \alpha \cdot m = 0.1 \cdot 0.1 = 0.01 \text{ mol/L} \].

Step 6: Find the pH

Use the formula: \[ \text{pH} = -\log[H^+] \]Substitute \[ [H^+] = 0.01 \], giving \[ \text{pH} = -\log(0.01) = 2 \].

Key Concepts

Freezing Point Depression

Freezing point depression is a fascinating colligative property of solutions. When a solute is dissolved in a solvent, it disrupts the formation of the solid structure needed to freeze, thus lowering the freezing point of the solution. This property helps us determine various characteristics of a solution, such as the degree of dissociation of an acid.

• The formula for freezing point depression is: \[ \Delta T_f = i \cdot K_f \cdot m \]
• Where \(\Delta T_f\) is the change in freezing point, \(i\) is the van't Hoff factor, \(K_f\) is the cryoscopic constant of the solvent, and \(m\) is the molality of the solution.

For the problem at hand, we use this formula to understand the extent to which our monoprotic acid, \(HA\), dissociates in water.

Degree of Dissociation

The degree of dissociation, often symbolized as \( \alpha \), describes how much an acid ionizes in solution. For a monoprotic acid like \( HA \), it dissociates into \( H^+ \) and \( A^- \) ions.

• When an acid dissociates, it increases the number of particles in the solution.
• This is where colligative properties become handy. By measuring the freezing point depression, we can determine \( \alpha \).
• The relationship between the degree of dissociation and the van't Hoff factor is given by \( i = 1 + \alpha \).

In this scenario, \( \alpha \) can be calculated by rearranging to \( \alpha = i - 1 \). Given that \( i \approx 1.1 \), we find \( \alpha = 0.1 \), indicating that 10% of the acid molecules dissociate in the solution.

Van't Hoff Factor

The van't Hoff factor, denoted as \( i \), is crucial for understanding how dissociation affects the physical properties of a solution. It shows how many particles a solute divides into when it dissolves.

• A completely dissociated solute has an \( i \) value greater than one, while a non-dissociating solute has \( i = 1 \).

For our monoprotic acid, the van't Hoff factor is calculated using the freezing point depression: \[ i = \frac{0.2046}{0.186} \approx 1.1 \]. This means that each \( HA \) molecule partially dissociates, increasing the number of particles in the solution slightly more than the original amount, but not to the extent that would occur with full dissociation.

Monoprotic Acid

Monoprotic acids are acids that can donate one proton per molecule in a reaction. An example is the generic acid \( HA \) used in our problem. These acids simplify many calculations because they dissociate in a straightforward way:

• \( HA \rightarrow H^+ + A^- \)

Knowing that the acid is monoprotic helps in finding how many ions are produced upon dissociation. Since we're dealing with a monoprotic acid, the van't Hoff factor and degree of dissociation directly inform us about the concentration of \( H^+ \) ions, which is essential for calculating the \( \mathrm{pH} \) of the solution.In our exercise, we dissipated any complex calculation by direct application of these principles, finding \( [H^+] = 0.01 \text{ mol/L} \), and subsequently, the \( \mathrm{pH} = 2 \). This demonstrates the beauty and simplicity inherent in monoprotic systems.