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

To determine the percentage of pyridine that forms the pyridinium ion in a 0.10 M aqueous pyridine solution, we can use the basicity of pyridine and the equilibrium concepts. Here are the step-by-step calculations:

1. Write the equilibrium reaction:

   Pyridine (C_5H_5N) is a weak base. In water, it reacts to form the pyridinium ion:

   C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^−

2. Apply the base dissociation constant:

   The expression for the base dissociation constant K_b is given by:

   K_b = \frac{[C_5H_5NH^+][OH^-]}{[C_5H_5N]}

   Given, K_b = 1.7 \times 10^{-9} and initial concentration of C_5H_5N = 0.10 M.

3. Assume small dissociation:

   Let x be the change in concentration due to dissociation. At equilibrium:

   • [C_5H_5NH^+] = [OH^-] = x
   • [C_5H_5N] = 0.10 - x \approx 0.10 (since x is small)
4. Use the K_b expression:

   1.7 \times 10^{-9} = \frac{x \cdot x}{0.10}

   Solve for x:

   x^2 = 1.7 \times 10^{-10}

   x = \sqrt{1.7 \times 10^{-10}} = 1.3 \times 10^{-5}

5. Calculate percentage of pyridinium ion:

   The fraction of pyridine that forms pyridinium ion is given by \frac{x}{0.10}

   \text{Percentage} = \left( \frac{1.3 \times 10^{-5}}{0.10} \right) \times 100 = 0.00013\%

6. Conclusion:

   Therefore, the percentage of pyridine that forms the pyridinium ion is 0.00013%. Hence, the correct answer is 0.00013.