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 the acids HF, HCOOH, and HCN, we can use the relationship between the ionization constants of an acid (Ka) and its conjugate base (Kb). This relationship is given by the equation:

\[ K_a \times K_b = K_w \]

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 at 298 K, which is \( 1.0 \times 10^{-14} \).
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