Question

A 0.15 mole of pyridinium chloride has been added to 500 cm\(^3\) of 0.2M pyridine solution (a base). Assuming there is no change in volume upon mixing, the pH of the resulting solution is
(Note: K\(_b\) for pyridine is 1.5 \(\times\) 10\(^{-9}\))

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

For any buffer solution problem, first identify if it's an acidic buffer (weak acid + its salt) or a basic buffer (weak base + its salt). Then, apply the appropriate Henderson-Hasselbalch equation:

• \textbf{Acidic Buffer:} pH = pK\(_a\) + log([Salt]/[Acid])
• \textbf{Basic Buffer:} pOH = pK\(_b\) + log([Salt]/[Base])

Always remember to convert pOH to pH if the question asks for pH of a basic buffer.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The mixture of a weak base (pyridine) and its conjugate acid salt (pyridinium chloride) forms a basic buffer solution. We need to calculate its pH.
Step 2: Key Formula or Approach:
Use the Henderson-Hasselbalch equation for basic buffers:
$pOH = pK_{b} + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right)$
Followed by the relation at $25^\circ C$: $pH = 14 - pOH$.
Step 3: Detailed Explanation:
1. Determine Concentrations:
Volume of solution $= 500 \text{ cm}^{3} = 0.5 \text{ L}$.
Concentration of Base (Pyridine) $= 0.2$ M.
Concentration of Salt (Pyridinium chloride) $= \frac{\text{Moles}}{\text{Volume}} = \frac{0.15 \text{ mol}}{0.5 \text{ L}} = 0.3$ M.

2. Calculate pK$_{b$:}
$K_b = 1.5 \times 10^{-9}$.
$pK_{b} = -\log(1.5 \times 10^{-9}) = 9 - \log(1.5)$.
We leave $\log(1.5)$ as it will cancel out perfectly in the next step.

3. Calculate pOH:
\[ pOH = pK_{b} + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
\[ pOH = [9 - \log(1.5)] + \log\left(\frac{0.3}{0.2}\right) \]
\[ pOH = 9 - \log(1.5) + \log(1.5) \]
The $\log(1.5)$ terms completely cancel out.
\[ pOH = 9 \]

4. Calculate pH:
\[ pH = 14 - pOH = 14 - 9 =
5. \]
Step 4: Final Answer:
The pH of the resulting buffer solution is
5.