Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question A basic buffer contains: NH 3 = 0.10  mol , NH 4 + = 0.10  mol Strong acid added: HCl = 0.05  mol Given: p K b ( NH 3 ) = 4.745 Find the change in pH after adding HCl. ✍️ Short Solution This is a weak base buffer (NH₃–NH₄Cl). When HCl is added, it reacts completely with NH₃, converting it into NH₄⁺. We then apply the Henderson–Hasselbalch equation for basic buffers to find the new pH. 🔹 Step 1 — Buffer System NH 3 + NH 4 + \text{NH}_3 + \text{NH}_4^+ Both initially = 0.10 mol → Perfect weak base buffer. 🔹 Step 2 — Henderson–Hasselbalch Equation (Basic Buffer) pOH = p K b + log ⁡ ( salt base ) \text{pOH} = pK_b + \log\left(\frac{\text{salt}}{\text{base}}\right) pH = 14 − pOH \text{pH} = 14 - \text{pOH} Given: p K b = 4.745 🔹 Step 3 — Reaction With Strong Acid HCl reacts fully with NH₃: NH 3 + HCl → NH 4 + \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ HCl added = 0.05 mol New moles: NH 3 = 0.10 − 0.05 = 0.05  mol \text...

  ❓ 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...