
What is the percentage of pyridine ${{C}_{5}}{{H}_{5}}N$ that forms pyridinium ion ${{C}_{5}}{{H}_{5}}{{N}^{+}}H$ in a 0.10 M aqueous pyridine solution (${{K}_{b}}$ for ${{C}_{5}}{{H}_{5}}N$= $1.7\times {{10}^{-9}}$)?
A. 0.77%
B. 1.6%
C. 0.0060%
D. 0.013%
Answer
539.4k+ views
Hint: Pyridine is a basic heterocyclic compound having the chemical formula ${{C}_{5}}{{H}_{5}}N$ and its structure is related with benzene where one methine group $(=CH)$ is replaced with nitrogen atom. It is highly flammable in nature.
Step by step solution: ${{K}_{a}}$ value defines the acid dissociation constant or we can say that an equilibrium constant for the dissociation of acids. Strong acids completely dissociate in water while weak acids do not. ${{K}_{a}}$ value shows the strength of the acid whereas ${{K}_{b}}$ is base dissociation constant and in water both weak and strong bases attains an equilibrium value that value is known as ${{K}_{b}}$ value. There is another term ${{K}_{w}}$ defined by water dissociation constant having fixed value of ${{10}^{-14}}$ and the relation between these terms is given by
${{K}_{a}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}$; ${{K}_{b}}$ for ${{C}_{5}}{{H}_{5}}N$= $1.7\times {{10}^{-9}}$(Given)
${{K}_{a}}=\dfrac{{{10}^{-14}}}{1.7\times {{10}^{-9}}}=5.88\times {{10}^{-6}}$
The reaction of pyridine can be as follows:
${{C}_{5}}{{H}_{5}}N+{{H}_{2}}O\rightleftarrows {{C}_{5}}{{H}_{5}}NH+{{C}_{5}}{{H}_{5}}N{{H}^{+}}+O{{H}^{-}}$
${{K}_{a}}=\dfrac{{{x}^{2}}}{0.1-x}$; x is very less value i.e. less than 0.1 so it can be neglected and the value of ${{K}_{a}}$ is $5.88\times {{10}^{-6}}$ as we find earlier.
${{x}^{2}}=5.88\times 0.1\times {{10}^{-6}}$
$x=7.6\times {{10}^{-4}}$
Therefore percentage of pyridine can be calculated by:
$\dfrac{7.6\times {{10}^{-4}}}{0.1}\times 100=0.77%$
This formula can be explained on the basis of
$\dfrac{dissociated\text{ ions}}{concentration\ \text{of solution}}\times 100$
From this we can conclude that option A is correct.
Note:
Pyridine is a weakly alkaline and water miscible compound having unpleasant fish-like smell and colorless in nature but when it gets old its impure sample becomes yellow in color. The main application of pyridine is used as pesticides.
Step by step solution: ${{K}_{a}}$ value defines the acid dissociation constant or we can say that an equilibrium constant for the dissociation of acids. Strong acids completely dissociate in water while weak acids do not. ${{K}_{a}}$ value shows the strength of the acid whereas ${{K}_{b}}$ is base dissociation constant and in water both weak and strong bases attains an equilibrium value that value is known as ${{K}_{b}}$ value. There is another term ${{K}_{w}}$ defined by water dissociation constant having fixed value of ${{10}^{-14}}$ and the relation between these terms is given by
${{K}_{a}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}$; ${{K}_{b}}$ for ${{C}_{5}}{{H}_{5}}N$= $1.7\times {{10}^{-9}}$(Given)
${{K}_{a}}=\dfrac{{{10}^{-14}}}{1.7\times {{10}^{-9}}}=5.88\times {{10}^{-6}}$
The reaction of pyridine can be as follows:
${{C}_{5}}{{H}_{5}}N+{{H}_{2}}O\rightleftarrows {{C}_{5}}{{H}_{5}}NH+{{C}_{5}}{{H}_{5}}N{{H}^{+}}+O{{H}^{-}}$
| Initial concentration | 0.1 | 0 | 0 | 0 |
| Final concentration | 0.1-x | x | x |
${{K}_{a}}=\dfrac{{{x}^{2}}}{0.1-x}$; x is very less value i.e. less than 0.1 so it can be neglected and the value of ${{K}_{a}}$ is $5.88\times {{10}^{-6}}$ as we find earlier.
${{x}^{2}}=5.88\times 0.1\times {{10}^{-6}}$
$x=7.6\times {{10}^{-4}}$
Therefore percentage of pyridine can be calculated by:
$\dfrac{7.6\times {{10}^{-4}}}{0.1}\times 100=0.77%$
This formula can be explained on the basis of
$\dfrac{dissociated\text{ ions}}{concentration\ \text{of solution}}\times 100$
From this we can conclude that option A is correct.
Note:
Pyridine is a weakly alkaline and water miscible compound having unpleasant fish-like smell and colorless in nature but when it gets old its impure sample becomes yellow in color. The main application of pyridine is used as pesticides.
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