Question
Question: What is the percentage of pyridine \[{\text{(}}{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{...
What is the percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N + H) in a 0.10 M aqueous pyridine solution ( Kb for {{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$$$$ = 1.7 \times {10^{ - 9}})?
(A) 0.77%
(B) 1.6%
(C) 0.0060%
(D) 0.013%
Solution
The law of dilution gives us a relation between dissociation constant and the degree of dissociation. This relation can be explained as: the degree of dissociation of a weak electrolyte (α) is directly proportional to dilution constant Kb, and it is inversely proportional to the concentration C0.
Complete step by step solution:
For the reaction,
Dilution constant can be written as:
Kb=[AB][A+][B−]=(1−α)C0(αC0)(αC0)=1−αα2×C0
Where αis the degree of dissociation of a weak electrolyte. And C0 is the concentration.
For weak electrolyte, α<<0 so (1−α) can be neglected and the resulting equation is:
So, α=C0Kb
So, for pyridine on dilution with water results in pyridinium ion. In question, we are given that molarity of pyridine solution is 0.10M and Kb for {C_5}{H_5}N$$$$ = 1.7 \times {10^{ - 9}}.
α=C0Kb=0.101.7×10−9=1.30×10−4
So the degree of dissociation of pyridinium ion {{\alpha = }}$$$$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=Λ0ΛC.