What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7.

Solution:

$K_{b}=4.27 \times 10^{-10}$

$c=0.001 \mathrm{M}$

$\mathrm{pH}=?$

$\mathrm{a}=?$

$k_{b}=c \alpha^{2}$

$4.27 \times 10^{-10}=0.001 \times \alpha^{2}$

$4270 \times 10^{-10}=\alpha^{2}$

$65.34 \times 10^{-5}=\alpha=6.53 \times 10^{-4}$

Then, [anion] $=c \alpha=.001 \times 65.34 \times 10^{-5}$

$=.065 \times 10^{-5}$

$\mathrm{pOH}=-\log \left(.065 \times 10^{-5}\right)$

$=6.187$

$\mathrm{pH}=7.813$

Now,

$K_{a} \times K_{b}=K_{w}$

$\therefore 4.27 \times 10^{-10} \times K_{a}=K_{w}$

$K_{a}=\frac{10^{-14}}{4.27 \times 10^{-10}}$

$=2.34 \times 10^{-5}$

Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times 10^{-5}$.