The degree of dissociation of `0.1N CH_(3)COOH` is `(K_(a)= 1xx10^(-5))`

Options

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

Explanation

To find the degree of dissociation (α) of 0.1N acetic acid (CH₃COOH) given that the dissociation constant (Kₐ) is \(1 \times 10^{-5}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dissociation of Acetic Acid:**
The dissociation of acetic acid can be represented as:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
Here, one mole of acetic acid dissociates into one mole of acetate ion and one mole of hydrogen ion.

2. **Determine the Concentration (C):**
Given that the normality (N) of acetic acid is 0.1N and the n-factor for acetic acid is 1 (since it produces one ion of H⁺), the molarity (C) is also:
\[ C = 0.1 \, \text{M} \]

3. **Using the Expression for Kₐ:**
The dissociation constant \(K_a\) can be expressed using the degree of dissociation (α):
\[ K_a = C \alpha^2 \]
Substituting the values we have:
\[ 1 \times 10^{-5} = 0.1 \alpha^2 \]

4. **Rearranging to Find α:**
Rearranging the equation to solve for α:
\[ \alpha^2 = \frac{1 \times 10^{-5}}{0.1} \]
\[ \alpha^2 = 1 \times 10^{-4} \]

5. **Taking the Square Root:**
Now, taking the square root to find α:
\[ \alpha = \sqrt{1 \times 10^{-4}} = 10^{-2} \]

6. **Final Result:**
Therefore, the degree of dissociation (α) of 0.1N acetic acid is:
\[ \alpha = 0.01 \, \text{or} \, 1\% \]