What is the $pH$ of $0.001 \,M$ aniline solution? The ionization constant of aniline can be taken from the table. Calculate the degree of ionization of aniline in the solution. Also,calculate the ionization constant of the conjugate acid of aniline.

Base	$K_{b}$
Dimethylamine,$(CH_{3})_{2}NH$	$5.4 \times 10^{-4}$
Triethylamine,$(C_{2}H_{5})_{3}N$	$6.45 \times 10^{-5}$
Ammonia,$NH_{3}$	$1.77 \times 10^{-5}$
Quinine	$1.10 \times 10^{-6}$
Pyridine,$C_{5}H_{5}N$	$1.77 \times 10^{-9}$
Aniline,$C_{6}H_{5}NH_{2}$	$4.27 \times 10^{-10}$
Urea,$CO(NH_{2})_{2}$	$1.3 \times 10^{-14}$

(A) Given: $c = 0.001 \,M = 10^{-3} \,M$,$K_{b} = 4.27 \times 10^{-10}$.
$1$. Degree of ionization $(\alpha)$:
For a weak base,$\alpha = \sqrt{\frac{K_{b}}{c}} = \sqrt{\frac{4.27 \times 10^{-10}}{10^{-3}}} = \sqrt{4.27 \times 10^{-7}} = \sqrt{42.7 \times 10^{-8}} \approx 6.53 \times 10^{-4}$.
$2$. $pH$ of the solution:
$[OH^-] = c \alpha = 10^{-3} \times 6.53 \times 10^{-4} = 6.53 \times 10^{-7} \,M$.
$pOH = -\log[OH^-] = -\log(6.53 \times 10^{-7}) = 7 - \log(6.53) \approx 7 - 0.815 = 6.185$.
$pH = 14 - pOH = 14 - 6.185 = 7.815$.
$3$. Ionization constant of conjugate acid $(K_{a})$:
$K_{a} \times K_{b} = K_{w} = 10^{-14}$.
$K_{a} = \frac{10^{-14}}{4.27 \times 10^{-10}} \approx 2.34 \times 10^{-5}$.