Question

The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅​H₅​N⁺H) in a 0.10 M aqueous pyridine solution (K_b​ for C₅H₅N = 1.7×10⁻⁹) is

• A. 0.00006
• B. 0.00013
• C. 0.0077
• D. 0.016

Answer 1

The Correct Option is B

Pyridine and its Base Dissociation 

The dissociation of pyridine in water is represented by the following equation:

\(\text{C}_5\text{H}_5\text{N} + \text{H}_2\text{O} \rightarrow \text{C}_5\text{H}_5\text{NH}^+ + \text{OH}^-\)

Pyridine or \( \text{C}_5\text{H}_5\text{N} \) is a weak base, and its base dissociation constant is given by:

\(K_b = C\alpha^2\)

The degree of dissociation (\( \alpha \)) is calculated using the formula:

\(\alpha = \sqrt{\frac{K_b}{C}}\)

Substituting the given values:

\(\alpha = \sqrt{\frac{1.7 \times 10^{-9}}{0.1}}\)

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

\(\alpha = 1.3 \times 10^{-4}\)

To find the percentage of pyridine that forms pyridinium ion, we use:

\(\alpha \times 100 = 1.3 \times 10^{-4} \times 100 = 0.013\%\)

The correct option is: B - 0.013%

Answer 2

Relation Between \( K_a \) and \( K_b \) 

The relationship between the acid dissociation constant (\( K_a \)) and the base dissociation constant (\( K_b \)) is given by:

\(K_a \times K_b = K_w\)

• \( K_a \) = acid's dissociation constant
• \( K_b \) = base's dissociation constant
• \( K_w \) = water's ionic product

Now, the expression for \( K_a \) is:

\(K_a = \frac{K_w}{K_b}\) = \( 1.7 \times 10^{-8} \)

Reaction of Pyridine

The reaction of pyridine with water is as follows:

\(\text{C}_5\text{H}_5\text{N} + \text{H}_2\text{O} \rightarrow \text{C}_5\text{H}_5\text{NH}^+ + \text{OH}^-\)

The equilibrium expression for the dissociation is:

\(1 - x \quad \quad \quad \quad \quad \quad x \quad \quad \quad \quad x\)

Given \( K_b = 1.7 \times 10^{-8} \), the expression becomes:

\(1.7 \times 10^{-8} = x^2\)

Solving for \( x \):

\(x = 1.3 \times 10^{-4}\)

The percentage of pyridine that forms pyridinium ion is:

\(\text{Percentage of pyridine} = 0.013\%\)

Conclusion:

Hence, the percentage of pyridine (\( \text{C}_5\text{H}_5\text{N} \)) that forms pyridinium ion (\( \text{C}_5\text{H}_5\text{NH}^+ \)) in a 0.10 M aqueous pyridine solution (where \( K_b \) for pyridine is \( 1.7 \times 10^{-9} \)) is \( 0.013\% \).