The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N+H) in a 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 × 10–9) is
C5H5N + H5O → C5H5N+H + OH–
$α = \sqrt{\dfrac{K_b}{c}} = \sqrt{\dfrac{1.7 \times 10^{-9}}{0.1}} = \sqrt{1.7 \times 10^{-8}}$
% α = 1.3 × 10–4 ×100
= 1.3 × 10–2 = 0.013.
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