
What is the percentage of pyridine \[{\text{(}}{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N)}}\] that forms pyridinium ion \[{\text{(}}{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{{\text{N}}^{\text{ + }}}{\text{H)}}\] in a 0.10 M aqueous pyridine solution ( \[{{\text{K}}_{\text{b}}}\] for \[{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}\]\[ = 1.7 \times {10^{ - 9}}\])?
(A) 0.77%
(B) 1.6%
(C) 0.0060%
(D) 0.013%
Answer
143.1k+ views
Hint: The law of dilution gives us a relation between dissociation constant and the degree of dissociation. This relation can be explained as: the degree of dissociation of a weak electrolyte (\[{{\alpha }}\]) is directly proportional to dilution constant \[{{\text{K}}_b}\], and it is inversely proportional to the concentration \[{{\text{C}}_{\text{0}}}\].
Complete step by step solution:
For the reaction,
Dilution constant can be written as:
\[{K_b} = \dfrac{{\left[ {{A^ + }} \right]\left[ {{B^ - }} \right]}}{{\left[ {AB} \right]}} = \dfrac{{\left( {\alpha {C_0}} \right)\left( {\alpha {C_0}} \right)}}{{\left( {1 - \alpha } \right){C_0}}} = \dfrac{{{\alpha ^2}}}{{1 - \alpha }} \times {C_0}\]
Where \[{{\alpha }}\]is the degree of dissociation of a weak electrolyte. And \[{{\text{C}}_{\text{0}}}\] is the concentration.
For weak electrolyte, \[\alpha < < 0\] so \[\left( {1 - \alpha } \right)\] can be neglected and the resulting equation is:
So, \[\alpha = \sqrt {\dfrac{{{K_b}}}{{{C_0}}}} \]
So, for pyridine on dilution with water results in pyridinium ion. In question, we are given that molarity of pyridine solution is 0.10M and \[{{\text{K}}_{\text{b}}}\] for \[{C_5}{H_5}N\]\[ = 1.7 \times {10^{ - 9}}\].
\[\alpha = \sqrt {\dfrac{{{K_b}}}{{{C_0}}}} = \sqrt {\dfrac{{1.7 \times {{10}^{ - 9}}}}{{0.10}} = } 1.30 \times {10^{ - 4}}\]
So the degree of dissociation of pyridinium ion \[{{\alpha = }}\]\[1.30 \times {10^{ - 4}}\].
Therefore, percentage of pyridine that forms pyridinium ion is \[{{1}}{{.30 \times 1}}{{\text{0}}^{{\text{ - 4}}}}{{ \times 100 = 0}}{\text{.013% }}\].
Hence the correct option is (D).
Note: The Degree of dissociation of any solute within a solvent is basically the ratio of molar conductivity at C concentration and limiting molar conductivity at zero concentration or infinite dilution. This can be mathematically represented as \[\alpha = \dfrac{{{\Lambda _C}}}{{{\Lambda _0}}}\].
Complete step by step solution:
For the reaction,
Dilution constant can be written as:
\[{K_b} = \dfrac{{\left[ {{A^ + }} \right]\left[ {{B^ - }} \right]}}{{\left[ {AB} \right]}} = \dfrac{{\left( {\alpha {C_0}} \right)\left( {\alpha {C_0}} \right)}}{{\left( {1 - \alpha } \right){C_0}}} = \dfrac{{{\alpha ^2}}}{{1 - \alpha }} \times {C_0}\]
Where \[{{\alpha }}\]is the degree of dissociation of a weak electrolyte. And \[{{\text{C}}_{\text{0}}}\] is the concentration.
For weak electrolyte, \[\alpha < < 0\] so \[\left( {1 - \alpha } \right)\] can be neglected and the resulting equation is:
So, \[\alpha = \sqrt {\dfrac{{{K_b}}}{{{C_0}}}} \]
So, for pyridine on dilution with water results in pyridinium ion. In question, we are given that molarity of pyridine solution is 0.10M and \[{{\text{K}}_{\text{b}}}\] for \[{C_5}{H_5}N\]\[ = 1.7 \times {10^{ - 9}}\].
\[\alpha = \sqrt {\dfrac{{{K_b}}}{{{C_0}}}} = \sqrt {\dfrac{{1.7 \times {{10}^{ - 9}}}}{{0.10}} = } 1.30 \times {10^{ - 4}}\]
So the degree of dissociation of pyridinium ion \[{{\alpha = }}\]\[1.30 \times {10^{ - 4}}\].
Therefore, percentage of pyridine that forms pyridinium ion is \[{{1}}{{.30 \times 1}}{{\text{0}}^{{\text{ - 4}}}}{{ \times 100 = 0}}{\text{.013% }}\].
Hence the correct option is (D).
Note: The Degree of dissociation of any solute within a solvent is basically the ratio of molar conductivity at C concentration and limiting molar conductivity at zero concentration or infinite dilution. This can be mathematically represented as \[\alpha = \dfrac{{{\Lambda _C}}}{{{\Lambda _0}}}\].
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Hydrocarbons Class 11 Notes: CBSE Chemistry Chapter 9

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry
