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
a. 2
b. 3
c. 4
d. 5

Options

A. 2
B. 3
C. 4
D. 5

Answer

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

### Step 1: Understand the Given Data
We have the following information:
- Depression in freezing point (ΔTf) of 0.01 m CH₃COOH solution = 0.02046°C
- Freezing point of 1 m urea solution = -1.86°C
- Assume molality (m) is equal to molarity (M).

### Step 2: Calculate the Freezing Point Depression for Urea
The freezing point depression for the urea solution can be calculated as follows:
- The freezing point of pure water (Tf0) is 0°C.
- Therefore, the depression in freezing point for urea (ΔTf2) is:
\[ \Delta Tf2 = Tf0 - Tf_{urea} = 0 - (-1.86) = 1.86°C \]

### Step 3: Apply the Freezing Point Depression Formula
The freezing point depression formula is given by:
\[ \Delta Tf = K_f \cdot m \cdot i \]
Where:
- \( K_f \) is the cryoscopic constant,
- \( m \) is the molality,
- \( i \) is the Van't Hoff factor.

For acetic acid (CH₃COOH) and urea, we can set up the equations:
1. For acetic acid:
\[ \Delta Tf1 = K_f \cdot m_1 \cdot i_1 \]
2. For urea:
\[ \Delta Tf2 = K_f \cdot m_2 \cdot i_2 \]

### Step 4: Set Up the Ratio of the Two Solutions
Dividing the two equations:
\[ \frac{\Delta Tf1}{\Delta Tf2} = \frac{m_1 \cdot i_1}{m_2 \cdot i_2} \]
Substituting the known values:
- \( \Delta Tf1 = 0.02046 \)
- \( \Delta Tf2 = 1.86 \)
- \( m_1 = 0.01 \) (for acetic acid)
- \( m_2 = 1 \) (for urea)
- \( i_2 = 1 \) (for urea)

This gives:
\[ \frac{0.02046}{1.86} = \frac{0.01 \cdot i_1}{1} \]

### Step 5: Solve for the Van't Hoff Factor (i1)
Rearranging gives:
\[ i_1 = \frac{0.02046 \cdot 1}{1.86 \cdot 0.01} \]
Calculating this:
\[ i_1 = \frac{0.02046}{0.0186} \approx 1.1 \]

### Step 6: Write the Dissociation Reaction for Acetic Acid
The dissociation of acetic acid in water can be represented as:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
Let the initial concentration of acetic acid be \( C = 0.01 \) m. If \( x \) is the amount that dissociates, at equilibrium we have:
- Concentration of \( CH_3COOH = 0.01 - x \)
- Concentration of \( CH_3COO^- = x \)
- Concentration of \( H^+ = x \)

### Step 7: Relate the Van't Hoff Factor to the Concentrations
The Van't Hoff factor \( i \) is related to the concentrations:
\[ i = 1 + \frac{x}{C} \]
Substituting \( i_1 = 1.1 \):
\[ 1.1 = 1 + \frac{x}{0.01} \]
Solving for \( x \):
\[ 0.1 = \frac{x}{0.01} \implies x = 0.001 \]

### Step 8: Calculate the Concentration of H⁺ Ions
The concentration of \( H^+ \) ions is:
\[ [H^+] = x = 0.001 \, \text{M} \]

### Step 9: Calculate the pH
Using the formula for pH:
\[ pH = -\log[H^+] \]
Substituting the value:
\[ pH = -\log(0.001) = 3 \]

### Conclusion
The pH of the 0.01 m aqueous acetic acid solution is 3.