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Ionic Equilibrium question

2013 · Shift 0 · Q2
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Ionic Equilibrium question

2013 · Shift 0 · Q2

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?
  1. A
    0.1 L
  2. B
    0.9 L
  3. C
    2.0 L
  4. D
    9.0 L
View written solutionFree

Correct answer: D

  1. Find the initial hydrogen ion concentration

Given initial pH =1=1=1,

[H+]1=10−1 M[H^+]_1 = 10^{-1} \text{ M}[H+]1​=10−1 M

Since HCl is a strong acid, it dissociates completely, so this is the concentration of HCl.

  1. Find the initial moles of H+H^+H+ in 1 L

Initial volume V1=1 LV_1 = 1 \text{ L}V1​=1 L.

moles of H+=[H+]1×V1=10−1×1=0.1 mol\text{moles of } H^+ = [H^+]_1 \times V_1 = 10^{-1} \times 1 = 0.1 \text{ mol}moles of H+=[H+]1​×V1​=10−1×1=0.1 mol
  1. Find the final hydrogen ion concentration

Required final pH =2=2=2,

[H+]2=10−2 M[H^+]_2 = 10^{-2} \text{ M}[H+]2​=10−2 M
  1. Use dilution: moles remain constant

After adding water, moles of H+H^+H+ remain 0.10.10.1 mol. If final volume is V2V_2V2​ litres, then

[H+]2=0.1V2[H^+]_2 = \frac{0.1}{V_2}[H+]2​=V2​0.1​

Set this equal to 10−210^{-2}10−2:

0.1V2=10−2\frac{0.1}{V_2} = 10^{-2}V2​0.1​=10−2 V2=0.110−2=10 LV_2 = \frac{0.1}{10^{-2}} = 10 \text{ L}V2​=10−20.1​=10 L
  1. Calculate water added

Initial volume =1=1=1 L, final volume =10=10=10 L.

water added=10−1=9 L\text{water added} = 10 - 1 = 9 \text{ L}water added=10−1=9 L
  1. Check options
  • A: 0.10.10.1 L ❌
  • B: 0.90.90.9 L ❌
  • C: 2.02.02.0 L ❌
  • D: 9.09.09.0 L ✅

Therefore, the correct answer is D.

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