At what concentration of the solution will the degree of dissociation of nitrous acid be 0.2? `K_a" for "HNO_2" is "4 xx 10^(-4). `

Answer

To solve the problem, we need to determine the concentration of a nitrous acid (HNO₂) solution at which the degree of dissociation (α) is 0.2, given that the acid dissociation constant (Kₐ) for HNO₂ is 4 × 10^(-4).

### Step-by-Step Solution:

1. **Understanding Degree of Dissociation**:
The degree of dissociation (α) is defined as the fraction of the total number of moles of the acid that dissociates into ions. In this case, α = 0.2 means that 20% of the acid dissociates.

2. **Setting Up the Dissociation Equation**:
The dissociation of nitrous acid can be represented as:
\[ HNO_2 \rightleftharpoons H^+ + NO_2^- \]
If we start with an initial concentration of HNO₂ as \( C \), then at equilibrium, the concentration of dissociated ions will be:
- Concentration of \( H^+ \) = \( \alpha C \)
- Concentration of \( NO_2^- \) = \( \alpha C \)
- Concentration of undissociated \( HNO_2 \) = \( C - \alpha C = C(1 - \alpha) \)

3. **Substituting the Values**:
Given that α = 0.2, we can substitute this into our expressions:
- Concentration of \( H^+ \) = \( 0.2C \)
- Concentration of \( NO_2^- \) = \( 0.2C \)
- Concentration of \( HNO_2 \) = \( C(1 - 0.2) = 0.8C \)

4. **Writing the Expression for Kₐ**:
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][NO_2^-]}{[HNO_2]} \]
Substituting the equilibrium concentrations into this expression, we get:
\[ K_a = \frac{(0.2C)(0.2C)}{0.8C} \]

5. **Simplifying the Equation**:
This simplifies to:
\[ K_a = \frac{0.04C^2}{0.8C} = \frac{0.04C}{0.8} = 0.05C \]

6. **Solving for C**:
Now, we can set this equal to the given \( K_a \):
\[ 0.05C = 4 \times 10^{-4} \]
Solving for \( C \):
\[ C = \frac{4 \times 10^{-4}}{0.05} = \frac{4 \times 10^{-4}}{5 \times 10^{-2}} = \frac{4}{5} \times 10^{-2} = 0.08 \text{ M} \]

### Final Answer:
The concentration of the solution at which the degree of dissociation of nitrous acid is 0.2 is **0.08 M**.