The ionisation constant of acetic acid is `1.74xx10^(-5)`. The degree of dissociation of acetic acid in its 0.05 m solution and its pH are respectively

Options

A. `1.86xx10^(-2), 4`
B. `1.24xx10^(-3), 4`
C. `1.24xx10^(-3), 3`
D. `1.86xx10^(-2), 3`

Answer

To solve the problem, we need to find the degree of dissociation (α) of acetic acid in a 0.05 M solution and its pH. The ionization constant (Ka) of acetic acid is given as \(1.74 \times 10^{-5}\).

### Step-by-Step Solution:

1. **Write the ionization equation for acetic acid:**
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

2. **Set up the initial concentrations:**
- Initial concentration of acetic acid, \([CH_3COOH] = 0.05 \, \text{M}\)
- Change in concentration due to dissociation:
- \([CH_3COOH] = 0.05 - 0.05\alpha\)
- \([CH_3COO^-] = 0 + 0.05\alpha\)
- \([H^+] = 0 + 0.05\alpha\)

3. **Write the expression for the ionization constant (Ka):**
\[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
Substituting the concentrations:
\[ 1.74 \times 10^{-5} = \frac{(0.05\alpha)(0.05\alpha)}{0.05 - 0.05\alpha} \]

4. **Assume α is small (α << 1), so \(0.05 - 0.05\alpha \approx 0.05\):**
\[ 1.74 \times 10^{-5} = \frac{(0.05\alpha)(0.05\alpha)}{0.05} \]
Simplifying:
\[ 1.74 \times 10^{-5} = 0.05\alpha^2 \]

5. **Solve for α:**
\[ \alpha^2 = \frac{1.74 \times 10^{-5}}{0.05} \]
\[ \alpha^2 = 3.48 \times 10^{-4} \]
\[ \alpha = \sqrt{3.48 \times 10^{-4}} \approx 0.0186 \]

6. **Calculate the concentration of H+:**
\[ [H^+] = 0.05 \times \alpha = 0.05 \times 0.0186 \approx 0.00093 \, \text{M} \]

7. **Calculate the pH:**
\[ \text{pH} = -\log[H^+] = -\log(0.00093) \approx 3.03 \]

### Final Results:
- Degree of dissociation (α) = 0.0186 (or 1.86%)
- pH = 3.03