An aqueous solution of `0.01 M CH_(3) COOH` has van't Hoff factor `1.01` . If `pH = - "log"[H^(+)]`, pH of `0.01` M `CH_(3)COOH` solution would be :

Answer

To find the pH of a `0.01 M CH₃COOH` solution with a van't Hoff factor of `1.01`, we can follow these steps:

### Step 1: Understand the dissociation of acetic acid
Acetic acid (CH₃COOH) dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Define initial concentrations
Let:
- Initial concentration of acetic acid, \( C = 0.01 \, \text{M} \)
- Degree of dissociation, \( \alpha \)

At time \( t = 0 \):
- Concentration of CH₃COOH = \( C \)
- Concentration of CH₃COO⁻ = 0
- Concentration of H⁺ = 0

### Step 3: Write expressions for concentrations at equilibrium
At equilibrium, the concentrations will be:
- Concentration of CH₃COOH = \( C(1 - \alpha) \)
- Concentration of CH₃COO⁻ = \( C\alpha \)
- Concentration of H⁺ = \( C\alpha \)

### Step 4: Use the van't Hoff factor to find \( \alpha \)
The van't Hoff factor \( i \) is given by:
\[ i = 1 + \alpha \]

Given \( i = 1.01 \):
\[ 1.01 = 1 + \alpha \]
\[ \alpha = 0.01 \]

### Step 5: Calculate the concentration of H⁺ ions
Using the value of \( \alpha \):
\[ \text{Concentration of H}^+ = C\alpha = 0.01 \times 0.01 = 0.0001 \, \text{M} = 10^{-4} \, \text{M} \]

### Step 6: Calculate pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the concentration of H⁺:
\[ \text{pH} = -\log(10^{-4}) = 4 \]

### Final Answer
The pH of the `0.01 M CH₃COOH` solution is **4**.

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