
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.
Answer
511.8k+ views
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.
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.
Recently Updated Pages
Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain why it is said like that Mock drill is use class 11 social science CBSE

Which of the following blood vessels in the circulatory class 11 biology CBSE

Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

What is the chemical name and formula of sindoor class 11 chemistry CBSE

How much is 23 kg in pounds class 11 chemistry CBSE

How do I get the molar mass of urea class 11 chemistry CBSE
