The ionization constant of `HF,HCOOH` and `HCN` at `298 K` are `6.8xx10^(-4), 1.8xx10^(-4)` and `4.8xx10^(-9)` respectively. Calculate the ionization constant of the corresponding conjugate base.

Explanation

To calculate the ionization constant of the conjugate bases of HF, HCOOH, and HCN, we can use the relationship between the ionization constants of acids and their conjugate bases. The relationship is given by:

\[ K_a \times K_b = K_w \]

Where:
- \( K_a \) is the ionization constant of the acid,
- \( K_b \) is the ionization constant of the conjugate base,
- \( K_w \) is the ionization constant of water, which is \( 1.0 \times 10^{-14} \) at 298 K.

### Step-by-Step Solution

1. **Identify the given values:**
- For HF: \( K_a = 6.8 \times 10^{-4} \)
- For HCOOH: \( K_a = 1.8 \times 10^{-4} \)
- For HCN: \( K_a = 4.8 \times 10^{-9} \)
- \( K_w = 1.0 \times 10^{-14} \)

2. **Calculate \( K_b \) for HF:**
\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} \]
\[ K_b = 1.47 \times 10^{-11} \]

3. **Calculate \( K_b \) for HCOOH:**
\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} \]
\[ K_b = 5.56 \times 10^{-11} \]

4. **Calculate \( K_b \) for HCN:**
\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-9}} \]
\[ K_b = 2.08 \times 10^{-6} \]

### Final Results
- \( K_b \) for HF: \( 1.47 \times 10^{-11} \)
- \( K_b \) for HCOOH: \( 5.56 \times 10^{-11} \)
- \( K_b \) for HCN: \( 2.08 \times 10^{-6} \)