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 $C_{5} H_{5} N = 1.7 \times 10^{- 9})$ is

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0.0060 %

0.0130 %

0.77 %

1.6 %

answer is B.

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Detailed Solution

$\begin{array}{l} \underset{0.1 M}{C_{5} H_{5} N} + H_{2} O ⇌ \underset{}{C_{5} H_{5} \overset{+}{N} H} + \underset{}{{OH}^{-}} \end{array}$ ;

$α = \sqrt{\frac{K_{b}}{C}} = \sqrt{\frac{1.7 \times 10^{- 9}}{0.10}} = 1.30 \times 10^{- 4}$

$∴ Percentage of pyridine that forms pyridinium ion$

$= 1.30 \times 10^{- 4} \times 100 = 0.013 %$

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