The degree of dissociation of 0.1 M acetic acid is `1.4xx10^(-2)`. Find out the pKa?

Answer

To find the pKa of acetic acid given its concentration and degree of dissociation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Concentration of acetic acid (C) = 0.1 M
- Degree of dissociation (α) = 1.4 × 10^(-2)

2. **Calculate the Dissociation Constant (Ka)**:
The dissociation constant (Ka) can be calculated using the formula:
\[ K_a = C \cdot \alpha^2 \]
Substituting the values:
\[ K_a = 0.1 \cdot (1.4 \times 10^{-2})^2 \]
First, calculate \((1.4 \times 10^{-2})^2\):
\[ (1.4 \times 10^{-2})^2 = 1.96 \times 10^{-4} \]
Now, substitute this back into the equation for Ka:
\[ K_a = 0.1 \cdot 1.96 \times 10^{-4} = 1.96 \times 10^{-5} \]

3. **Calculate pKa**:
The pKa is calculated using the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value of Ka:
\[ pK_a = -\log(1.96 \times 10^{-5}) \]
Using logarithmic properties:
\[ pK_a = -\log(1.96) - \log(10^{-5}) = -\log(1.96) + 5 \]
Now, calculate \(-\log(1.96)\):
\[ -\log(1.96) \approx -0.293 \]
Therefore:
\[ pK_a = 5 - 0.293 = 4.707 \]

4. **Final Result**:
The pKa of acetic acid is approximately:
\[ pK_a \approx 4.707 \]