The dissociation constants of two acids `HA_(1)` and `HA_(2)` are `3.0xx10^(-4)` and `1.8xx10^(-5)` respectively. The relative strengths of the acids will be approximately

Options

A. `h_(1) gt h_(2)`
B. `h_(1) lt h_(2)`
C. `h_(1) = h_(2)`
D. None of these

Answer

To determine the relative strengths of the two acids \( HA_1 \) and \( HA_2 \) based on their dissociation constants \( K_{a1} \) and \( K_{a2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dissociation Constants**:
- For acid \( HA_1 \), the dissociation constant \( K_{a1} = 3.0 \times 10^{-4} \).
- For acid \( HA_2 \), the dissociation constant \( K_{a2} = 1.8 \times 10^{-5} \).

2. **Use the Formula for Relative Strength**:
The relative strength of the acids can be calculated using the formula:
\[ \text{Relative Strength} = \sqrt{\frac{K_{a1}}{K_{a2}}} \]

3. **Substitute the Values**:
\[ \text{Relative Strength} = \sqrt{\frac{3.0 \times 10^{-4}}{1.8 \times 10^{-5}}} \]

4. **Simplify the Fraction**:
- First, simplify the constants:
\[ \frac{3.0}{1.8} = \frac{30}{18} = \frac{5}{3} \]
- Now simplify the powers of ten:
\[ \frac{10^{-4}}{10^{-5}} = 10^{(-4) - (-5)} = 10^{1} = 10 \]
- Combine these results:
\[ \frac{K_{a1}}{K_{a2}} = \frac{5}{3} \times 10 \]

5. **Calculate the Final Value**:
\[ \frac{K_{a1}}{K_{a2}} = \frac{50}{3} \]
- Now take the square root:
\[ \text{Relative Strength} = \sqrt{\frac{50}{3}} \approx \sqrt{16.67} \approx 4.08 \]

6. **Interpret the Result**:
- The relative strength of \( HA_1 \) to \( HA_2 \) is approximately \( 4.08:1 \).
- This indicates that \( HA_1 \) is stronger than \( HA_2 \).

### Conclusion:
Thus, the relative strengths of the acids \( HA_1 \) and \( HA_2 \) are approximately \( 4:1 \), meaning \( HA_1 \) is stronger than \( HA_2 \).