A 0.01 M solution of acetic acid is 1.34 % ionised (degree of dissociation = 0.0134) at 298 K. What is the ionization constant of acetic acid ?

Explanation

To find the ionization constant (K_a) of acetic acid from the given information, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Concentration of acetic acid (C) = 0.01 M
- Degree of ionization (α) = 0.0134 (which is 1.34% ionized)

2. **Write the Ionization Equation:**
- The ionization of acetic acid (CH₃COOH) can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

3. **Calculate the Concentration of Ions:**
- At equilibrium, the concentration of the ions (CH₃COO⁻ and H⁺) formed will be:
\[ [\text{CH}_3\text{COO}^-] = [\text{H}^+] = C \cdot \alpha = 0.01 \, \text{M} \cdot 0.0134 = 0.000134 \, \text{M} \]

4. **Calculate the Concentration of Undissociated Acetic Acid:**
- The concentration of undissociated acetic acid at equilibrium will be:
\[ [\text{CH}_3\text{COOH}] = C(1 - \alpha) = 0.01 \, \text{M} \cdot (1 - 0.0134) = 0.01 \, \text{M} \cdot 0.9866 = 0.009866 \, \text{M} \]

5. **Write the Expression for the Ionization Constant (K_a):**
- The expression for the ionization constant (K_a) is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]

6. **Substitute the Values into the K_a Expression:**
- Substitute the concentrations we calculated:
\[ K_a = \frac{(0.000134)(0.000134)}{0.009866} \]

7. **Calculate K_a:**
- Performing the calculation:
\[ K_a = \frac{0.000000017956}{0.009866} \approx 1.82 \times 10^{-6} \]

8. **Final Result:**
- The ionization constant of acetic acid (K_a) is approximately:
\[ K_a \approx 1.8 \times 10^{-6} \]