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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
A
0.77%
B
1.6%
C
0.0060%
D
0.013%
Explanation
$\alpha = \sqrt{K_b/C}$.
Detailed Solution
Pyridine is a weak base: $K_b = C\alpha^2$
$\alpha = \sqrt{\frac{1.7\times10^{-9}}{0.1}} = 1.30\times10^{-4}$
$\%\alpha = 1.30\times10^{-4}\times100 = 0.013\%$
$\alpha = \sqrt{\frac{1.7\times10^{-9}}{0.1}} = 1.30\times10^{-4}$
$\%\alpha = 1.30\times10^{-4}\times100 = 0.013\%$