A 0.15 mole of pyridinium chloride has been added to 500 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

Step-by-step solution

Moles of pyridine (base) = 0.2 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 + ( Moles of salt Moles of base ) pOH = 8.82 + ( 0.15 0.10 ) pOH = 8.82 + (1.5) pOH = 8.82 + 0.176 9 pH = 14 - pOH = 14 - 9 = 5 Answer: 5