An acid `HA` is `40%` dissociated in an aqueous solution. The hydronium ion concentration of its `0.2M` solution would be

Options

A. `0.08 M`
B. `0.4M`
C. `0.2 M`
D. None

Explanation

To find the hydronium ion concentration of a 0.2 M solution of the acid HA that is 40% dissociated, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data**:
- The acid HA is 40% dissociated.
- The initial concentration (C) of the acid solution is 0.2 M.

2. **Convert the percentage of dissociation to a fraction (alpha)**:
- Since the acid is 40% dissociated, we can express this as:
\[ \alpha = \frac{40}{100} = 0.4 \]

3. **Set up the dissociation equation**:
- The dissociation of the acid HA can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
- Initially, we have:
- Concentration of HA = C = 0.2 M
- Concentration of H⁺ = 0 M
- Concentration of A⁻ = 0 M

4. **Calculate the concentrations at equilibrium**:
- At equilibrium, the concentration of HA will be:
\[ [HA] = C(1 - \alpha) = 0.2(1 - 0.4) = 0.2 \times 0.6 = 0.12 \text{ M} \]
- The concentration of H⁺ (or hydronium ion) will be:
\[ [H^+] = C \alpha = 0.2 \times 0.4 = 0.08 \text{ M} \]
- The concentration of A⁻ will also be:
\[ [A^-] = C \alpha = 0.2 \times 0.4 = 0.08 \text{ M} \]

5. **Conclusion**:
- Therefore, the hydronium ion concentration in the solution is:
\[ [H^+] = 0.08 \text{ M} \]

### Final Answer:
The hydronium ion concentration of the 0.2 M solution of acid HA is **0.08 M**.