The ionization constant of $\mathrm{HF}, \mathrm{HCOOH}$ and $\mathrm{HCN}$ at $298 \mathrm{~K}$ are $6.8 \times 10^{-4}, 1.8 \times 10^{-4}$ and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.
It is known that,
$K_{b}=\frac{K_{w}}{K_{a}}$
Given,
$K_{a}$ of HF $=6.8 \times 10^{-4}$
Hence, $K_{b}$ of its conjugate base $\mathrm{F}^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{6.8 \times 10^{-4}}$
$=1.5 \times 10^{-11}$
Given,
$K_{a}$ of $\mathrm{HCOOH}=1.8 \times 10^{-4}$
Hence, $K_{b}$ of its conjugate base $\mathrm{HCOO}^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{1.8 \times 10^{-4}}$
$=5.6 \times 10^{-11}$
Given,
$K_{a}$ of HCN $=4.8 \times 10^{-9}$
Hence, $K_{b}$ of its conjugate base $\mathrm{CN}^{-}$
$=\frac{K_{w}}{K_{a}}$
$=\frac{10^{-14}}{4.8 \times 10^{-9}}$
$=2.08 \times 10^{-6}$