For two weak acids A and B, the ratio of their percent ionization is 4 : 9. The ratio of their Ka would be–

Options

A. `4:9`
B. `2:3`
C. `16:81`
D. `3:2`

Answer

To solve the problem, we need to find the ratio of the dissociation constants (Ka) of two weak acids A and B given the ratio of their percent ionization.

### Step-by-Step Solution:

1. **Understanding Percent Ionization**:
The percent ionization of a weak acid is defined as the fraction of the acid that dissociates into ions. For weak acids A and B, let the percent ionization be represented as αA and αB respectively. According to the problem, we have:
\[ \frac{\alpha_A}{\alpha_B} = \frac{4}{9} \]

2. **Relation Between Ionization and Ka**:
For a weak acid HA, the degree of ionization (α) is related to its dissociation constant (Ka) and the initial concentration (C) by the formula:
\[ \alpha \propto \sqrt{K_a} \]
This implies:
\[ \alpha_A \propto \sqrt{K_{aA}} \quad \text{and} \quad \alpha_B \propto \sqrt{K_{aB}} \]

3. **Setting Up the Ratio**:
From the proportionality, we can express the ratio of the percent ionizations in terms of Ka:
\[ \frac{\alpha_A}{\alpha_B} = \frac{\sqrt{K_{aA}}}{\sqrt{K_{aB}}} \]
Substituting the given ratio:
\[ \frac{4}{9} = \frac{\sqrt{K_{aA}}}{\sqrt{K_{aB}}} \]

4. **Squaring Both Sides**:
To eliminate the square roots, we square both sides of the equation:
\[ \left(\frac{4}{9}\right)^2 = \frac{K_{aA}}{K_{aB}} \]
This simplifies to:
\[ \frac{16}{81} = \frac{K_{aA}}{K_{aB}} \]

5. **Final Ratio of Ka**:
Therefore, the ratio of the dissociation constants (Ka) for acids A and B is:
\[ K_{aA} : K_{aB} = 16 : 81 \]

### Conclusion:
The ratio of their Ka values is \( \boxed{16 : 81} \).