The percentage of pyridine (C_5H_5N) that for | vedclass

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(B) The dissociation of pyridine in water is given by: $C_5H_5N + H_2O \rightleftharpoons C_5H_5N^{+}H + OH^{-}$.For a weak base,the degree of dissoci

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$0.013$

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(B) The dissociation of pyridine in water is given by: $C_5H_5N + H_2O ightleftharpoons C_5H_5N^{+}H + OH^{-}$.For a weak base,the degree of dissociation $\alpha$ is calculated as $\alpha = \sqrt{\frac{K_b}{c}}$.Given $K_b = 1.7 	imes 10^{-9}$ and $c = 0.10 \ M$.$\alpha = \sqrt{\frac{1.7 	imes 10^{-9}}{0.10}} = \sqrt{1.7 	imes 10^{-8}} = 1.3 	imes 10^{-4}$.The percentage of dissociation is $\% \alpha = \alpha 	imes 100$.$\% \alpha = 1.3 	imes 10^{-4} 	imes 100 = 1.3 	imes 10^{-2} = 0.013 \%$.

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

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