If the freezing point of `0.1 M HA(aq)` solution is `-0.2046^(@)C` then `pH` of solution is
`(` If `K_(f)` water `=1.86mol^(-1)kg^(-1))`

Options

A. 1
B. 1.3
C. 1.7
D. 2

Explanation

To solve the problem, we need to determine the pH of a 0.1 M solution of HA given that its freezing point is -0.2046°C and the freezing point depression constant (Kf) for water is 1.86 mol/kg.

### Step-by-Step Solution:

1. **Calculate the Depression in Freezing Point (ΔTf)**:
\[ \Delta T_f = T_f(\text{water}) - T_f(\text{solution}) \]
Given that the freezing point of water is 0°C and the freezing point of the solution is -0.2046°C:
\[ \Delta T_f = 0 - (-0.2046) = 0.2046 \, °C \]

**Hint**: Remember that the depression in freezing point is the difference between the freezing point of pure solvent and the solution.

2. **Use the Freezing Point Depression Formula**:
The formula for freezing point depression is:
\[ \Delta T_f = K_f \cdot i \cdot m \]
Where:
- \( K_f \) = 1.86 mol/kg (freezing point depression constant)
- \( m \) = molality of the solution (which is approximately equal to molarity for dilute solutions)
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)

Plugging in the known values:
\[ 0.2046 = 1.86 \cdot i \cdot 0.1 \]

**Hint**: Ensure you understand the relationship between molality and molarity for dilute solutions.

3. **Solve for the van 't Hoff Factor (i)**:
Rearranging the equation to find \( i \):
\[ i = \frac{0.2046}{1.86 \cdot 0.1} = \frac{0.2046}{0.186} \approx 1.1 \]

**Hint**: The van 't Hoff factor indicates how many particles are produced from the dissociation of the solute.

4. **Relate i to Degree of Dissociation (α)**:
For a weak acid HA that dissociates into H⁺ and A⁻:
\[ i = 1 + (n - 1) \cdot \alpha \]
Here, \( n = 2 \) (since HA dissociates into two ions: H⁺ and A⁻):
\[ 1.1 = 1 + (2 - 1) \cdot \alpha \]
Simplifying gives:
\[ 1.1 = 1 + \alpha \implies \alpha = 0.1 \]

**Hint**: The degree of dissociation (α) tells you the fraction of the original solute that has dissociated.

5. **Calculate the Concentration of H⁺ Ions**:
The concentration of H⁺ ions in the solution can be calculated as:
\[ [H^+] = C \cdot \alpha \]
Where \( C = 0.1 \, M \):
\[ [H^+] = 0.1 \cdot 0.1 = 0.01 \, M \]

**Hint**: The concentration of H⁺ ions is directly proportional to the degree of dissociation.

6. **Calculate the pH**:
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Substituting the value of [H⁺]:
\[ pH = -\log(0.01) = -\log(10^{-2}) = 2 \]

**Hint**: The pH scale is logarithmic, so a change in concentration by a factor of 10 results in a change of 1 unit in pH.

### Final Answer:
The pH of the solution is **2**.