The percentage of pyridine () that forms pyridinium ion () in a aqueous pyridine solution ( for ) is — Ionic Equilibrium Chemistry Question
Question
The percentage of pyridine ($C_5H_5N$) that forms pyridinium ion ($C_5H_5N^+H$) in a $0.10 \text{ M}$ aqueous pyridine solution ($K_b$ for $C_5H_5N = 1.7 \times 10^{-9}$) is:
Answer: A
💡 Solution & Explanation
For a weak base, the degree of dissociation $\alpha = \sqrt{K_b / C}$. Substituting the given values: $\alpha = \sqrt{1.7 \times 10^{-9} / 0.10} = \sqrt{1.7 \times 10^{-8}} \approx 1.3 \times 10^{-4}$. To convert this to a percentage, multiply by 100: $1.3 \times 10^{-4} \times 100 = 1.3 \times 10^{-2}\% = 0.013\%$.
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