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:

$$\text{pH}=\text{p}{K}_a+\log \frac{[\text{base}]}{[\text{acid}]}$$

For the NH₄⁺/NH₃ system, $\text{p}K_a = 14 - \text{p}K_b(\text{NH}_3)$.
Use $\text{p}K_b(\text{NH}_3)=4.75$ ⇒ $\text{p}K_a(\text{NH}_4^+)=9.25$.

Initial moles (in 1 L):

• $n_{{\text{NH}}_3}=\mathrm{0.10\; \to }$$[{\rm base}]=0.10\ \mathrm{M}$
  

• $n_{\text{NH}_4^+}=0.10$ → $[{\rm acid}]=0.10\ \mathrm{M}$
  

So initial pH:

$${\text{pH}}_{\text{initial}}=9.25+\log \frac{0.10}{0.10}=9.25+0=\mathrm{9.25.}$$

Add 0.05 mol HCl: HCl reacts quantitatively with NH₃:

$${\text{NH}}_3+\text{HCl}\to {\text{NH}}_4^{+}+{\text{Cl}}^{-}\mathrm{.}$$

New mole numbers:

• $n_{\text{NH}_3,\text{new}} = 0.10 - 0.05 = 0.05$
  

• $n_{\text{NH}_4^+,\text{new}} = 0.10 + 0.05 = 0.15$
  

New concentrations (≈ moles per L since volume ≈1 L):
$[\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}]=0.05\mathrm{M},[\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{d}]=0.15\mathrm{M}\mathrm{.}$

New pH:

$${\text{pH}}_{\text{final}}=9.25+\log \text{\hspace{0.17em}⁣}\left(\frac{0.05}{0.15}\right)=9.25+\log \text{\hspace{0.17em}⁣}\left(\frac{1}{3}\right)=9.25-\log 3.$$

Numeric:

$$\log 3\approx 0.4771\Rightarrow {\text{pH}}_{\text{final}}\approx 9.25-0.4771=8.7729\approx \mathrm{8.77.}$$

Change in pH:

$$\mathrm{\Delta }\text{pH}={\text{pH}}_{\text{final}}-{\text{pH}}_{\text{initial}}=8.77-9.25=-\mathrm{0.48.}$$

So pH decreases by 0.48 units (about 0.5 pH unit).

---

🧮 Image Solution

---

✅ Conclusion & Video Solution

✅ Final Answer:

$$\boxed{{\displaystyle \text{pH decreases by }0.48\text{ units (from 9.25 to ≈8.77).}}}$$

Concept recap: HCl consumes base (NH₃) converting it to NH₄⁺; buffer resists large pH change — here the pH shifts by only ~0.48 units. The Henderson–Hasselbalch equation is the quickest route.