What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from table 7.7 ($4.3\text{ x 1}{{\text{0}}^{-10}}$ ). Calculate the degree of ionization of aniline in the solution. Also, calculate the ionization constant of the conjugate acid of aniline.

Hint: pH of the solution is equal to the negative logarithm of the concentration of the hydrogen ions. The degree of ionization is equal to the square root of ionization constant divided with concentration. The ionization constant of the conjugate acid of aniline is calculated by dividing the ionic product of water by the ionization constant of aniline.

Complete answer:
When aniline is dissociated in water it forms aniline ion and hydroxyl ion.
Let c is the initial concentration of $[{{C}_{6}}{{H}_{5}}N{{H}_{3}}^{+}]$ and x is the degree of ionization.
So, the reaction will be and the concentration after the equilibrium:
$\begin{align}
  & PhN{{H}_{2}}\text{ + }{{H}_{2}}O\text{ }\rightleftharpoons \text{ }PhN{{H}_{3}}^{+}\text{ + }O{{H}^{-}} \\
 & \text{ c(1-}x)\text{ c}x\text{ c}x \\
\end{align}$
So, the ionization constant is equal to the ratio of product of concentration of products to the ratio of product of concentration of reactants. Hence,
${{K}_{b}}=\dfrac{[PhN{{H}_{3}}^{+}][O{{H}^{-}}]}{[PhN{{H}_{2}}]}=\dfrac{(cx)(cx)}{c(1-x)}=\dfrac{c{{x}^{2}}}{1-x}$
Since the concentration very small, therefore, 1-x = 1
Hence, ${{K}_{b}}=c{{x}^{2}}$
So, the degree of ionization of aniline can be calculated with:
$x=\sqrt{\dfrac{{{K}_{b}}}{c}}$
The ionization constant is equal to $4.3\text{ x 1}{{\text{0}}^{-10}}$and concentration is 0.001 M
So, putting these values,
$x=\sqrt{\dfrac{{{K}_{b}}}{c}}=\sqrt{\dfrac{4.3\text{ x 1}{{\text{0}}^{-10}}}{0.001}}=6.56\text{ x }{{10}^{-4}}$
So, the concentration of hydroxyl ions is:
$[O{{H}^{-}}]=cx=0.001\text{ x }6.56\text{ x 1}{{\text{0}}^{-4}}=6.56\text{ x 1}{{\text{0}}^{-7}}M$
The concentration of hydrogen ions can be calculated by dividing the ionic product of water with concentration of hydroxyl ions.
$[{{H}^{+}}]=\dfrac{{{K}_{w}}}{[O{{H}^{-}}]}=\dfrac{{{10}^{-14}}}{6.56\text{ x 1}{{\text{0}}^{-7}}}=1.52\text{ x 1}{{\text{0}}^{-8}}$
Now, the pH of the solution is equal to the negative logarithm of the concentration of the hydrogen ions.
$pH=-\log [{{H}^{+}}]=-\log 1.52\text{ x 1}{{\text{0}}^{-8}}=7.818$
Hence, the pH of the solution is 7.818.
The ionization constant of the conjugate acid of aniline is calculated by dividing the ionic product of water by the ionization constant of aniline.
${{K}_{a}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}=\dfrac{{{10}^{-14}}}{4.3\text{ x 1}{{\text{0}}^{-10}}}=2.32\text{ x 1}{{\text{0}}^{-5}}$
So, the ionization constant of the conjugate acid of aniline is $2.32\text{ x 1}{{\text{0}}^{-5}}$

Note:
All the formulas must be taken right. Sometimes the value of the ionic product of water is not mentioned in the question, so its value must be taken ${{10}^{-14}}$ for the solution.