# The ionization constant of HF, HCOOH and HCN at 298K

Question:

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.

Solution:

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 $\mathrm{CN}^{-}$

$=\frac{K_{w}}{K_{a}}$

$=\frac{10^{-14}}{4.8 \times 10^{-9}}$

$=2.08 \times 10^{-6}$