What is the ionization constant of an acid if the hydronium ion concentration of a 0.40 M solution is `1.40xx10^(-4)` M?
(a)`1.96xx10^(-8)`
(b)`1.22xx10^(-9)`
(c)`4.90xx10^(-8)`
(d)`1.40xx10^(-6)`

Options

A. `1.96xx10^(-8)`
B. `1.22xx10^(-9)`
C. `4.90xx10^(-8)`
D. `1.40xx10^(-6)`

Answer

To find the ionization constant (Ka) of an acid given the hydronium ion concentration and the concentration of the acid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Concentration of the acid (C) = 0.40 M
- Hydronium ion concentration ([H₃O⁺]) = 1.40 × 10⁻⁴ M

2. **Define the Degree of Dissociation (α)**:
- The degree of dissociation (α) is defined as the fraction of the acid that dissociates into ions.

3. **Use the Relationship Between [H₃O⁺], C, and α**:
- The concentration of hydronium ions can be expressed as:
\[ [H₃O⁺] = C \cdot α \]
- Rearranging this gives:
\[ α = \frac{[H₃O⁺]}{C} \]

4. **Substitute the Known Values**:
- Substitute the values into the equation:
\[ α = \frac{1.40 \times 10^{-4}}{0.40} \]

5. **Calculate α**:
- Performing the calculation:
\[ α = 3.5 \times 10^{-4} \]

6. **Use the Expression for Ka**:
- The ionization constant (Ka) can be expressed as:
\[ K_a = \frac{C \cdot α^2}{1 - α} \]
- Since α is small (less than 0.05), we can approximate:
\[ K_a \approx C \cdot α^2 \]

7. **Substitute Values into the Ka Expression**:
- Substitute C and α into the expression:
\[ K_a \approx 0.40 \cdot (3.5 \times 10^{-4})^2 \]

8. **Calculate Ka**:
- First, calculate \( (3.5 \times 10^{-4})^2 \):
\[ (3.5 \times 10^{-4})^2 = 1.225 \times 10^{-7} \]
- Now calculate \( K_a \):
\[ K_a \approx 0.40 \cdot 1.225 \times 10^{-7} = 4.90 \times 10^{-8} \]

9. **Final Answer**:
- The ionization constant (Ka) of the acid is:
\[ K_a = 4.90 \times 10^{-8} \]

### Conclusion:
The correct answer is (c) \( 4.90 \times 10^{-8} \).