Question

Calculate the percentage of pyridine (C₅H₅N) that forms pyridinium ion, C₅H₅NH⁺, in a 0.10 M aqueous solution of pyridine (K_b = 1.7 × 10^-9).

Short answer

Percent dissociation, of pyridine to pyridinium ion in 0.1 M solution = 0.013%

Step-by-step solution

Calculation of Amount Dissociated

C₅H₅N reaction with water is represented as

$$\begin{gathered} {\mathrm{C}}_5{\mathrm{H}}_5{\mathrm{NH}}_{\left(\mathrm{aq}\right)}+{\mathrm{H}}_2\mathrm{O}\iff {\mathrm{C}}_5{\mathrm{H}}_5{{\mathrm{NH}}^{+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ 0.1\mathrm{\_}00 \\ 0.1-\mathrm{x}-\mathrm{x}\mathrm{x} \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{x}}^2}{0.1-\mathrm{x}}=1.7\times {10}^{-9} \\ {\mathrm{x}}^2=1.7\times {10}^{-9}\times 0.1(\mathrm{Taken},0.1-\times ~0.1>>\mathrm{x}) \\ \mathrm{x}=1.304\times {10}^{-5} \end{gathered}$$

Calculation of Percentage Dissociation

For calculating the percent dissociation, amount dissociated is divided by initial amount (0.1 M here).

$\mathrm{Percent}\mathrm{Dissociation}=\frac{\mathrm{Amount}\mathrm{dissociated}}{\mathrm{Total}\mathrm{amount}}\times 100=\frac{1.304\times {10}^{-5}}{0.1}\times 100=0.013\%$