The ionization of pyridine in water can be represented by the equation: $C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^-$ 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. $\alpha = \sqrt{\frac{K_b}{c}} = \sqrt{\frac{1.7 \times 10^{-9}}{0.10}} = \sqrt{1.7 \times 10^{-8}} \approx 1.30 \times 10^{-4}$ The percentage of pyridine that forms pyridinium ion is the percentage ionization: Percentage ionization = $\alpha \times 100 = 1.30 \times 10^{-4} \times 100 = 1.30 \times 10^{-2} = 0.013\%$.