The percentage of pyridine that forms pyridinium ion in a 0.10 M aqueous pyridine solution ( for ) is [2016]

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Solution

Q.
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 [2016]

1 0.0060%

2 0.013%

3 0.77%

4 1.6%

Ans.
(2)

$\underset{0.10 M}{C_{5} H_{5} N} + H_{2} O ⇌ C_{5} H_{5} \overset{+}{N} H + {OH}^{-}$

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

$∴$ $Percentage of pyridine that forms pyridinium ion = 1.30 \times 10^{- 4} \times 100 = 0.013 %$

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