The factor by which the degree of ionisation of `200mL` of `0.1M` benzoic acid solution `(K_(a) = 4 xx 10^(-5))` changes on addition of `100mL` of `0.2M HCl` is:

Options

A. `0.02`
B. `0.03`
C. `33.33`
D. None

Explanation

To solve the problem step-by-step, we will follow these steps:

### Step 1: Calculate the initial degree of ionization (α) of benzoic acid.
Given:
- Concentration of benzoic acid (C) = 0.1 M
- Acid dissociation constant (K_a) = \(4 \times 10^{-5}\)

Using the formula for the degree of ionization of a weak acid:
\[ \alpha = \sqrt{\frac{K_a}{C}} \]

Substituting the values:
\[ \alpha = \sqrt{\frac{4 \times 10^{-5}}{0.1}} = \sqrt{4 \times 10^{-4}} = 2 \times 10^{-2} \]

### Step 2: Determine the effect of adding HCl on the degree of ionization.
When we add HCl, it provides additional H⁺ ions, which will affect the degree of ionization due to the common ion effect.

### Step 3: Calculate the new concentration of H⁺ ions after adding HCl.
Given:
- Volume of benzoic acid solution = 200 mL
- Concentration of HCl = 0.2 M
- Volume of HCl added = 100 mL

Total volume after mixing = 200 mL + 100 mL = 300 mL.

Using the formula for dilution (M1V1 = M2V2), we can find the new concentration of H⁺ ions from HCl:
\[ M_1 = 0.2 \, \text{M}, \quad V_1 = 100 \, \text{mL}, \quad V_2 = 300 \, \text{mL} \]
\[ M_2 = \frac{M_1 \cdot V_1}{V_2} = \frac{0.2 \times 100}{300} = \frac{20}{300} = \frac{1}{15} \approx 0.0667 \, \text{M} \]

### Step 4: Calculate the new degree of ionization (α') of benzoic acid after adding HCl.
Using the new concentration of H⁺ ions:
\[ \alpha' = \frac{K_a}{M} = \frac{4 \times 10^{-5}}{0.0667} \]

Calculating α':
\[ \alpha' = \frac{4 \times 10^{-5}}{0.0667} \approx 6 \times 10^{-4} \]

### Step 5: Calculate the factor by which the degree of ionization changes.
The factor of change in the degree of ionization is given by:
\[ \text{Factor} = \frac{\alpha'}{\alpha} = \frac{6 \times 10^{-4}}{2 \times 10^{-2}} = \frac{6}{2} \times 10^{-4 + 2} = 3 \times 10^{-2} = 0.03 \]

### Final Answer:
The factor by which the degree of ionization changes is **0.03**.

---