In 0.10 M aqueous solution of pyridine (C₅H₅N), find the percentage of pyridine that forms pyridinium ion (C₅H₅N⁺H) (Kᵦ for C₅H₅N = 1.7 x 10⁻⁹ )

The correct option is 0.013%.

To find the percentage of pyridine that forms the pyridinium ion, we can analyze the ionization of pyridine in water. Pyridine (C₅H₅N) behaves as a weak base and reacts with water as follows:

C₅H₅N + H₂O ⇌ C₅H₅NH⁺ + OH^-

Let $C$ be the initial concentration of pyridine and $\alpha$ be the degree of ionization. The equilibrium concentrations of the species are:

[C₅H₅N] = $C(1-\alpha )$

[C₅H₅NH⁺] = [OH^-] = $C\alpha$

The base dissociation constant ($K_b$) expression is:

$$K_b=\frac{\left[C_5{H}_5{\mathrm{NH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[C_5{H}_{5}N\right]}$$

Substituting the equilibrium concentrations into the expression:

$$K_b=\frac{\left(C\alpha \right)\left(C\alpha \right)}{C(1-\alpha )}=\frac{C{\alpha }^2}{1-\alpha }$$

Given that pyridine is a very weak base with a small $K_b$ value of $1.7\times {10}^{-9}$, the degree of ionization $\alpha$ is extremely small compared to 1 ($\alpha \ll 1$). Therefore, we can approximate $1-\alpha \approx 1$.

Using this approximation, the equation simplifies to:

$$K_b\approx C{\alpha }^2$$

Solving for $\alpha$:

$$\alpha =\sqrt{\frac{K_b}{C}}$$

Substitute the given values ($C=0.10\text{M}$ and $K_b=1.7\times {10}^{-9}$) into the equation:

$$\alpha =\sqrt{\frac{1.7\times {10}^{-9}}{0.10}}$$

$$\alpha =\sqrt{1.7\times {10}^{-8}}$$

$$\alpha \approx 1.30\times {10}^{-4}$$

The percentage of pyridine that forms pyridinium ion is given by:

$$\text{Percentage of ionization}=\alpha \times 100\%$$

$$\text{Percentage of ionization}=(1.30\times {10}^{-4})\times 100\%$$

$$\text{Percentage of ionization}=0.013\%$$