The ionization constants of $HF$,$HCOOH$,and $HCN$ at $298 \ K$ are $6.8 \times 10^{-4}$,$1.8 \times 10^{-4}$,and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of their corresponding conjugate bases.

The relationship between the ionization constant of an acid $(K_a)$ and its conjugate base $(K_b)$ is given by the equation: $K_a \times K_b = K_w$,where $K_w = 10^{-14}$ at $298 \ K$.
$1$. For $HF$ $(K_a = 6.8 \times 10^{-4})$,the conjugate base is $F^-$.
$K_b(F^-) = \frac{10^{-14}}{6.8 \times 10^{-4}} = 1.47 \times 10^{-11}$.
$2$. For $HCOOH$ $(K_a = 1.8 \times 10^{-4})$,the conjugate base is $HCOO^-$.
$K_b(HCOO^-) = \frac{10^{-14}}{1.8 \times 10^{-4}} = 5.56 \times 10^{-11}$.
$3$. For $HCN$ $(K_a = 4.8 \times 10^{-9})$,the conjugate base is $CN^-$.
$K_b(CN^-) = \frac{10^{-14}}{4.8 \times 10^{-9}} = 2.08 \times 10^{-6}$.