Question
The pH of a solution obtained by mixing 5 mL of 0.1 M NH_4OH solution with 250 mL of 0.1 M NH_4Cl solution is _____ 10^ -2 . (Nearest integer) Given: pK_b(NH_4OH) = 4.74 2 = 0.30 3 = 0.48 5 = 0.70
JEE Main
Chemistry
Ionic Equilibrium
2026
JEE Main 2026 (04 April Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
Step-by-step solution
The given mixture forms a basic buffer solution. Millimoles of base NH_4OH = 5 0.1 = 0.5 mmol Millimoles of salt NH_4Cl = 250 0.1 = 25 mmol Using the Henderson-Hasselbalch equation for a basic buffer: pOH = pK_b + ( [ Salt ] [ Base ] ) pOH = 4.74 + ( 25 0.5 ) pOH = 4.74 + (50) We know that (50) = (5 10) = 5 + 10 = 0.70 + 1 = 1.70 pOH = 4.74 + 1.70 = 6.44 The pH of the solution is given by: pH = 14 - pOH pH = 14 - 6.44 = 7.56 pH = 756 10^ -2 Answer: 756
Related: Chemistry — Ionic Equilibrium · All PYQ Banks