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MCQEasyJEE 2025Ionic Equilibria & pH

JEE Chemistry 2025 Question with Solution

An aqueous solution of HCl with pH 1.01.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would be: (Given log 2 = 0.30)\text{(Given log 2 = 0.30)}

  • A

    reduce to 0.50.5

  • B

    increase to 1.31.3

  • C

    remain same

  • D

    increase to 22

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: An aqueous solution of HCl has pH 1.01.0. Equal volume of water is added, so the total volume becomes double.

Find: The new pH of the diluted solution.

For HCl, assuming complete dissociation,

pH=−log⁡[H+]\text{pH} = -\log[H^+]

Since initial pH =1.0= 1.0,

[H+]=10−1=0.1 M[H^+] = 10^{-1} = 0.1 \, \text{M}

After adding an equal volume of water, the concentration becomes half:

[H+]new=0.12=0.05 M[H^+]_{\text{new}} = \frac{0.1}{2} = 0.05 \, \text{M}

Now calculate the new pH:

pHnew=−log⁡(0.05)\text{pH}_{\text{new}} = -\log(0.05)

Using

0.05=5×10−20.05 = 5 \times 10^{-2}

so

log⁡(0.05)=log⁡5+log⁡10−2\log(0.05) = \log 5 + \log 10^{-2}

Also, from log⁡2=0.30\log 2 = 0.30,

log⁡5=log⁡(102)=1−0.30=0.70\log 5 = \log \left(\frac{10}{2}\right) = 1 - 0.30 = 0.70

Therefore,

log⁡(0.05)=0.70−2=−1.30\log(0.05) = 0.70 - 2 = -1.30

Hence,

pHnew=−(−1.30)=1.30\text{pH}_{\text{new}} = -(-1.30) = 1.30

Therefore, the pH increases to 1.31.3 and the correct option is B.

Direct Dilution Insight

Given: Initial pH =1= 1 for aqueous HCl.

Find: pH after doubling the volume.

If the volume is doubled, the hydrogen ion concentration is halved:

[H+]new=[H+]2[H^+]_{\text{new}} = \frac{[H^+]}{2}

So the pH increases by

log⁡2=0.30\log 2 = 0.30

Hence,

pHnew=1.0+0.30=1.3\text{pH}_{\text{new}} = 1.0 + 0.30 = 1.3

This works because pH depends logarithmically on [H+][H^+], so halving [H+][H^+] adds log⁡2\log 2 to the pH. Therefore, the correct option is B.

Common mistakes

  • Students often think dilution makes the acid strength vanish and jump directly to pH =2= 2. This is wrong because adding equal volume of water only halves [H+][H^+]; it does not reduce it by a factor of 1010. First halve the concentration, then recalculate pH.

  • A common mistake is using concentration change linearly in pH and writing pH = 1.5. This is wrong because pH is logarithmic, not arithmetic. Use pH=−log⁡[H+]\text{pH} = -\log[H^+] after finding the new concentration.

  • Some students forget that HCl is a strong acid and do not take initial [H+]=10−1 M[H^+] = 10^{-1} \, \text{M} from pH =1= 1. This leads to an incorrect setup. For strong acids here, hydrogen ion concentration is obtained directly from the given pH.

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