The percentage of pyridine ($C_5H_5N$) that forms pyridinium ion ($C_5H_5NH^+$) in a 0.10 M aqueous pyridine s

Question

The percentage of pyridine ($C_5H_5N$) that forms pyridinium ion ($C_5H_5NH^+$) in a 0.10 M aqueous pyridine solution ($K_b$ for $C_5H_5N = 1.7 	imes 10^{-9}$) is:

Accepted Answer

The ionization of pyridine in water can be represented by the equation:
$C_5H_5N + H_2O ightleftharpoons C_5H_5NH^+ + OH^-$
The base ionization constant ($K_b$) is given by the relation $K_b = c\alpha^2$ (assuming the degree of ionization $\alpha \ll 1$), where $c$ is the initial concentration of the base.
$\alpha = \sqrt{\frac{K_b}{c}} = \sqrt{\frac{1.7 	imes 10^{-9}}{0.10}} = \sqrt{1.7 	imes 10^{-8}} \approx 1.30 	imes 10^{-4}$
The percentage of pyridine that forms pyridinium ion is the percentage ionization:
Percentage ionization = $\alpha 	imes 100 = 1.30 	imes 10^{-4} 	imes 100 = 1.30 	imes 10^{-2} = 0.013\%$.

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