pH After Dilution of Strong Acid

 

❓ Question

An aqueous solution of HCl has:

$$pH = 1$$

The solution is diluted by adding equal volume of water.

Find the new pH of the solution.
(Ignore dissociation of water.)

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🖼 Question Image

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✍️ Short Concept

pH is related to hydrogen ion concentration:

$$pH = -\log[H^+]$$

Dilution reduces concentration → pH increases.

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🔷 Step 1 — Find Initial $[H^+]$ 💯

Given:

$$pH = 1$$
$$[H^+] = 10^{-1} \, M$$

Since HCl is strong acid, it dissociates completely.

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🔷 Step 2 — Apply Dilution Rule

Equal volume of water added ⇒

Total volume becomes double.

So concentration becomes half.

$$[H^+]_{new} = \frac{10^{-1}}{2}$$
$$= 5 \times 10^{-2}$$

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🔷 Step 3 — Calculate New pH

$$pH = -\log(5 \times 10^{-2})$$
$$= -[\log5 + \log10^{-2}]$$
$$= -(0.699 - 2)$$
$$= 1.30$$

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🔷 Step 4 — Concept Shortcut

When concentration halves:

$$pH \text{ increases slightly}$$

Not by 1 unit — only about 0.3.

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✅ Final Answer

$$\boxed{pH \approx 1.30}$$

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⭐ Golden JEE Insight

For equal dilution:

$$[H^+] \rightarrow \frac{[H^+]}{2}$$

So:

$$pH \text{ increases by } \log 2 \approx 0.3$$

Very common JEE shortcut.