What is the $pH$ of $0.001 \,M$ aniline solution? The ionization constant of aniline can be taken from 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, $\left( CH _{3}\right)_{2} NH$ $5.4 \times 10^{-4}$
Triethylamine, $\left( C _{2} H _{5}\right)_{3} N$ $6.45 \times 10^{-5}$
Ammonia, $NH _{3}$ or $NH _{4} OH$ $1.77 \times 10^{-5}$
Quinine, ( $A$ plant product) $1.10 \times 10^{-6}$
Pyridine, $C _{5} H _{5} N$ $1.77 \times 10^{-9}$
Aniline, $C _{6} H _{5} NH _{2}$ $4.27 \times 10^{-10}$
Urea, $CO \left( NH _{2}\right)_{2}$ $1.3 \times 10^{-14}$

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

$c=0.001 \,M\, pH$ $=?$

$a=?$

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

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

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

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

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

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

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

$=6.187$

$pH =7.813$

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 $2.34 \times 10^{-5}$

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