The K_b of C₆H₅NH₂ (aniline) is 3.9 × 10⁻¹⁰. Determine pH of a buffer solution made up of 500 mL of 1.4 M C₆H₅NH₂ and 230 mL of 2.3 M C₆H₅NH₃⁺.

• A. 4.81
• B. 9.62
• C. 9.29
• D. 4.71

Answer: (D)

Verified step by step guidance

1. Start by understanding the components of the buffer solution. Aniline (C6H5NH2) is a weak base, and C6H5NH3+ is its conjugate acid. The buffer solution consists of both these components.
2. Use the Henderson-Hasselbalch equation to find the pH of the buffer solution: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \). First, calculate the pKa from the given Kb using the relation \( \text{pKa} = 14 - \text{pKb} \).
3. Calculate the concentration of C6H5NH2 and C6H5NH3+ in the buffer solution. Use the formula \( C = \frac{n}{V} \) where \( n \) is the number of moles and \( V \) is the total volume of the solution. Find the moles of each component using their initial concentrations and volumes, then divide by the total volume (500 mL + 230 mL).
4. Substitute the concentrations of the base (C6H5NH2) and the acid (C6H5NH3+) into the Henderson-Hasselbalch equation. Ensure the concentrations are in the same units (Molarity, M).
5. Calculate the logarithmic term \( \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \) and add it to the pKa value to find the pH of the buffer solution.