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Question: What is the percentage of pyridine \[{\text{(}}{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{...

What is the percentage of pyridine (C5H5N){\text{(}}{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N)}} that forms pyridinium ion (C5H5N + H){\text{(}}{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{{\text{N}}^{\text{ + }}}{\text{H)}} in a 0.10 M aqueous pyridine solution ( Kb{{\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%

Explanation

Solution

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 (α{{\alpha }}) is directly proportional to dilution constant Kb{{\text{K}}_b}, and it is inversely proportional to the concentration C0{{\text{C}}_{\text{0}}}.

Complete step by step solution:
For the reaction,
Dilution constant can be written as:
Kb=[A+][B][AB]=(αC0)(αC0)(1α)C0=α21α×C0{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 α{{\alpha }}is the degree of dissociation of a weak electrolyte. And C0{{\text{C}}_{\text{0}}} is the concentration.
For weak electrolyte, α<<0\alpha < < 0 so (1α)\left( {1 - \alpha } \right) can be neglected and the resulting equation is:
So, α=KbC0\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 Kb{{\text{K}}_{\text{b}}} for {C_5}{H_5}N$$$$ = 1.7 \times {10^{ - 9}}.
α=KbC0=1.7×1090.10=1.30×104\alpha = \sqrt {\dfrac{{{K_b}}}{{{C_0}}}} = \sqrt {\dfrac{{1.7 \times {{10}^{ - 9}}}}{{0.10}} = } 1.30 \times {10^{ - 4}}
So the degree of dissociation of pyridinium ion {{\alpha = }}$$$$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 α=ΛCΛ0\alpha = \dfrac{{{\Lambda _C}}}{{{\Lambda _0}}}.