Q116 / 240 · KCET 2026 · Chemistry
A $0.15$ mole of pyridinium chloride has been added to $500\text{ cm}^3$ of $0.2$ M pyridine solution (a base). Assuming there is no change in volume upon mixing(Given: $pK_b$ of pyridine = 8.82), the pH of the resulting solution is
- Correct answerA$5$
- B$6$
- C$7$
- D$8$
Solution
Moles of pyridine (base) $= 0.2 \times 0.5 = 0.1$ mol
Moles of pyridinium chloride (salt) $= 0.15$ mol
Using the Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_b + \log\left(\dfrac{\text{Moles of salt}}{\text{Moles of base}}\right)$
$pOH = 8.82 + \log\left(\dfrac{0.15}{0.10}\right)$
$pOH = 8.82 + \log(1.5)$
$pOH = 8.82 + 0.176 \approx 9$
$pH = 14 - pOH = 14 - 9 = 5$
Answer: $5$
Moles of pyridinium chloride (salt) $= 0.15$ mol
Using the Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_b + \log\left(\dfrac{\text{Moles of salt}}{\text{Moles of base}}\right)$
$pOH = 8.82 + \log\left(\dfrac{0.15}{0.10}\right)$
$pOH = 8.82 + \log(1.5)$
$pOH = 8.82 + 0.176 \approx 9$
$pH = 14 - pOH = 14 - 9 = 5$
Answer: $5$
Hence, the correct answer is option: 'A'.