The degree of dissociation of acetic acid in a 0.1 N solution is `1.32 xx 10^(-2)?.` At what concentration of nitrous acid will its degree of dissociation be the same as that of acetic acid? `[K_a (HNO_2) =4 xx 10^(-4)]`

Answer

To solve the problem, we need to find the concentration of nitrous acid (HNO₂) that will give it the same degree of dissociation (α) as that of acetic acid (CH₃COOH) in a 0.1 N solution, where the degree of dissociation is given as \(1.32 \times 10^{-2}\).

### Step-by-Step Solution:

1. **Understand the relationship between degree of dissociation, concentration, and dissociation constant**:
The degree of dissociation (α) is defined as the fraction of the original acid that dissociates into ions. For a weak acid, the relationship can be expressed as:
\[ K_a = C \alpha^2 \]
where \(K_a\) is the dissociation constant, \(C\) is the concentration of the acid, and α is the degree of dissociation.

2. **Identify the known values**:
- For acetic acid:
- Degree of dissociation, \(α = 1.32 \times 10^{-2}\)
- Concentration, \(C = 0.1 \, N\) (which is equivalent to 0.1 M for acetic acid)
- For nitrous acid:
- \(K_a (HNO_2) = 4 \times 10^{-4}\)
- We need to find the concentration \(C\) for which \(α = 1.32 \times 10^{-2}\).

3. **Set up the equation for nitrous acid**:
Using the relationship \(K_a = C \alpha^2\) for nitrous acid:
\[ K_a (HNO_2) = C \cdot (1.32 \times 10^{-2})^2 \]
Substituting the known value of \(K_a\):
\[ 4 \times 10^{-4} = C \cdot (1.32 \times 10^{-2})^2 \]

4. **Calculate \(α^2\)**:
\[ (1.32 \times 10^{-2})^2 = 1.7424 \times 10^{-4} \]

5. **Rearrange the equation to solve for \(C\)**:
\[ C = \frac{4 \times 10^{-4}}{1.7424 \times 10^{-4}} \]

6. **Perform the calculation**:
\[ C = \frac{4}{1.7424} \approx 2.296 \, \text{M} \]

7. **Final result**:
The concentration of nitrous acid at which its degree of dissociation will be the same as that of acetic acid is approximately \(2.30 \, \text{M}\).

### Summary:
The concentration of nitrous acid required to have the same degree of dissociation as acetic acid in a 0.1 N solution is approximately \(2.30 \, \text{M}\).