JEE Main Concept: NH₃–NH₄Cl Buffer Reaction with Strong Acid 💧
❓ Question One litre buffer solution was prepared by adding 0.10 mol each of NH₃ and NH₄Cl in deionised water. What is the change in pH on addition of 0.05 mol of HCl to the above solution? (Treat the solution volume change on addition as negligible — 1 L final volume.) 🖼️ Question Image ✍️ Short Solution (idea) This is a classical buffer problem . Use the Henderson–Hasselbalch equation for the NH₄⁺/NH₃ buffer: pH = p K a + log [ base ] [ acid ] For the NH₄⁺/NH₃ system, p K a = 14 − p K b ( NH 3 ) \text{p}K_a = 14 - \text{p}K_b(\text{NH}_3) . Use p K b ( NH 3 ) = 4.75 \text{p}K_b(\text{NH}_3)=4.75 ⇒ p K a ( NH 4 + ) = 9.25 \text{p}K_a(\text{NH}_4^+)=9.25 . Initial moles (in 1 L): n NH 3 = 0.10 → [ b a s e ] = 0.10 M [{\rm base}]=0.10\ \mathrm{M} n NH 4 + = 0.10 n_{\text{NH}_4^+}=0.10 → [ a c i d ] = 0.10 M [{\rm acid}]=0.10\ \mathrm{M} So initial pH: pH initial = 9.25 + log 0.10 0.10 = 9.25 + 0 = 9.25. Add 0.05 mo...