The degree of dissociation of `0.1 N CH_3 COOH` is (given `K_a=1xx10^(-5)`) approximately

Options

A. `10^(-5)`
B. `10^(-4)`
C. `10^(-3)`
D. `10^(-2)`

Answer

To find the degree of dissociation (α) of 0.1 N acetic acid (CH₃COOH) given that the dissociation constant (Kₐ) is 1 × 10⁻⁵, we can use the Ostwald dilution law, which states:

\[ \alpha = \sqrt{\frac{K_a}{C}} \]

where:
- α = degree of dissociation
- Kₐ = dissociation constant
- C = concentration (normality in this case)

### Step-by-Step Solution:

1. **Identify the given values:**
- Kₐ = 1 × 10⁻⁵
- C = 0.1 N (normality)

2. **Convert normality to molarity:**
- For acetic acid, the normality is equal to the molarity since the n-factor is 1.
- Therefore, C = 0.1 M.

3. **Substitute the values into the formula:**
\[ \alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1 \times 10^{-5}}{0.1}} \]

4. **Calculate the concentration in the formula:**
- Convert 0.1 to scientific notation:
\[ 0.1 = 1 \times 10^{-1} \]
- Thus, we can rewrite the equation:
\[ \alpha = \sqrt{\frac{1 \times 10^{-5}}{1 \times 10^{-1}}} = \sqrt{1 \times 10^{-4}} \]

5. **Calculate the square root:**
\[ \alpha = \sqrt{10^{-4}} = 10^{-2} \]

6. **Final Result:**
- The degree of dissociation (α) is approximately \( 10^{-2} \) or 0.01.

### Conclusion:
The degree of dissociation of 0.1 N CH₃COOH is approximately \( 0.01 \) or \( 10^{-2} \).

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