Buffer pH Change After Adding 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.)

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Buffer solution
👉 Henderson equation
👉 Effect of strong acid on buffer.

Main idea:

For basic buffer:

$$\boxed{ pOH=pK_b+\log\frac{[salt]}{[base]} }$$

Find initial pH and final pH after adding HCl.

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🔷 Step 1 — Initial pH of Buffer 💯

Initially:

$$NH_3=0.10\text{ mol}$$
$$NH_4Cl=0.10\text{ mol}$$

Thus:

$$\frac{salt}{base}=1$$

So:

$$pOH=pK_b+\log1$$
$$=4.745$$

Hence:

$$pH=14-4.745$$
$$=9.255$$

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🔷 Step 2 — Reaction with HCl

Reaction:

$$NH_3+HCl\rightarrow NH_4Cl$$

HCl added:

$$0.05\text{ mol}$$

Thus:

Base decreases:

$$NH_3=0.10-0.05$$
$$=0.05$$

Salt increases:

$$NH_4Cl=0.10+0.05$$
$$=0.15$$

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🔷 Step 3 — Final pOH

Using Henderson equation:

$$pOH = 4.745+\log\frac{0.15}{0.05}$$
$$= 4.745+\log3$$

Given:

$$\log3=0.477$$

Thus:

$$pOH=4.745+0.477$$
$$=5.222$$

Hence:

$$pH=14-5.222$$
$$=8.778$$

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🔷 Step 4 — Change in pH

$$\Delta pH = 9.255-8.778$$
$$=0.477$$

Now express as:

$$x\times10^{-2}$$
$$0.477=47.7\times10^{-2}$$

Nearest integer:

$$\boxed{ 48 }$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Direct pH formula use kar dena

❌ HCl addition ke baad mole changes ignore kar dena

Remember:

Strong acid added to basic buffer:

$$\boxed{ \text{Base decreases, salt increases} }$$

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✅ Final Answer

$$\boxed{ 48 }$$

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