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, $CO \left( NH _{2}\right)_{2}$
$1.3 \times 10^{-14}$
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, $CO \left( NH _{2}\right)_{2}$ | $1.3 \times 10^{-14}$ |
$K_{b}=4.27 \times 10^{-10}$
$c=0.001 \,M\, pH$ $=?$
$a=?$
$k_{b}=c \alpha^{2}$
$4.27 \times 10^{-10}=0.001 \times \alpha^{2}$
$4270 \times 10^{-10}=\alpha^{2}$
$65.34 \times 10^{-5}=\alpha=6.53 \times 10^{-4}$
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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