The depression in freezing point of `0.01 m` aqueous `CH_(3)CooH` solution is `0.02046^(@)`, `1 m` urea solution freezes at `-1.86^(@)C`. Assuming molality equal to molarity, `pH` of `CH_(3)COOH` solution is

Options

A. 2
B. 3
C. 3.2
D. 4.2

Explanation

To find the pH of a 0.01 m aqueous acetic acid (CH₃COOH) solution, we can follow these steps:

### Step 1: Understand the given data
We are given:
- Depression in freezing point (ΔTf) = 0.02046°C
- Freezing point depression constant (Kf) for water = 1.86°C/m
- The molality of the acetic acid solution = 0.01 m

### Step 2: Use the freezing point depression formula
The freezing point depression can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f \) = depression in freezing point
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_f \) = freezing point depression constant
- \( m \) = molality of the solution

### Step 3: Determine the van 't Hoff factor (i)
For acetic acid (CH₃COOH), it dissociates into CH₃COO⁻ and H⁺ ions:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
Thus, the van 't Hoff factor \( i \) can be expressed as:
\[ i = 1 + \alpha \]
Where \( \alpha \) is the degree of dissociation.

### Step 4: Substitute values into the freezing point depression formula
Substituting the known values into the formula:
\[ 0.02046 = (1 + \alpha) \cdot 1.86 \cdot 0.01 \]
This simplifies to:
\[ 0.02046 = (1 + \alpha) \cdot 0.0186 \]

### Step 5: Solve for \( 1 + \alpha \)
Rearranging the equation gives:
\[ 1 + \alpha = \frac{0.02046}{0.0186} \]
Calculating this:
\[ 1 + \alpha \approx 1.1 \]
Thus,
\[ \alpha \approx 0.1 \]

### Step 6: Calculate the concentration of H⁺ ions
The concentration of H⁺ ions can be determined using:
\[ [\text{H}^+] = \alpha \cdot \text{initial concentration of acetic acid} \]
Given that the initial concentration is 0.01 m:
\[ [\text{H}^+] = 0.1 \cdot 0.01 = 0.001 \, \text{M} \]

### Step 7: Calculate the pH
The pH can be calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the value of \( [\text{H}^+] \):
\[ \text{pH} = -\log(0.001) = 3 \]

### Final Answer
The pH of the 0.01 m aqueous acetic acid solution is **3**.