The degree of dissociation of acetic acid (0.1 mol `L^(-1)`) in water (`K_a` of acetic acid is `10^(-5)`) is

Options

A. 0.01
B. 0.5
C. 0.1
D. 1

Explanation

To find the degree of dissociation (α) of acetic acid in water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- Concentration of acetic acid (C) = 0.1 mol/L
- Acid dissociation constant (K_a) = \(10^{-5}\)

2. **Assume the degree of dissociation (α):**
- Since acetic acid is a weak acid, we assume that the degree of dissociation (α) is much less than 1. Therefore, we can approximate \(1 - α \approx 1\).

3. **Write the expression for K_a:**
- The dissociation of acetic acid (CH₃COOH) can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
- The expression for the acid dissociation constant (K_a) is:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]
- At equilibrium, the concentrations can be expressed as:
- \([\text{CH}_3\text{COO}^-] = Cα\)
- \([\text{H}^+] = Cα\)
- \([\text{CH}_3\text{COOH}] = C(1 - α) \approx C\) (since α is small)

4. **Substituting into the K_a expression:**
- Therefore, we can write:
\[ K_a = \frac{(Cα)(Cα)}{C} = Cα^2 \]
- Rearranging gives:
\[ K_a = Cα^2 \implies α^2 = \frac{K_a}{C} \]

5. **Calculate α:**
- Substitute the values of K_a and C:
\[ α^2 = \frac{10^{-5}}{0.1} = 10^{-4} \]
- Taking the square root:
\[ α = \sqrt{10^{-4}} = 10^{-2} = 0.01 \]

6. **Conclusion:**
- The degree of dissociation (α) of acetic acid is 0.01.

### Final Answer:
The degree of dissociation of acetic acid (0.1 mol/L) in water is **0.01**.