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Solutions question

2025 · 8 Apr · Shift 2 · Q6
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Solutions question

2025 · 8 Apr · Shift 2 · Q6

JEE MainChemistrySolutionsMCQ+4 / −1
HA(aq)⇌H+(aq)+A−(aq)\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q)HA(aq)⇌H+(aq)+A−(aq) The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HAHAHA is 0.20 °C. The dissociation constant for the acid is Given: KfK_fKf​(H2OH_2OH2​O) = 1.8 K kg mol−1, molality ≡ molarity
  1. A
    1.90×10−31.90 \times 10^{-3}1.90×10−3
  2. B
    1.38×10−31.38 \times 10^{-3}1.38×10−3
  3. C
    1.1×10−21.1 \times 10^{-2}1.1×10−2
  4. D
    1.89×10−11.89 \times 10^{-1}1.89×10−1
View written solutionFree

Correct answer: B

  1. Use freezing point depression formula

For a solute that may dissociate, ΔTf=iKfm\Delta T_f = i K_f mΔTf​=iKf​m where:

  • ΔTf=0.20∘C\Delta T_f = 0.20^\circ CΔTf​=0.20∘C
  • Kf=1.8 K kg mol−1K_f = 1.8\, \mathrm{K\,kg\,mol^{-1}}Kf​=1.8Kkgmol−1
  • m=0.1m = 0.1m=0.1

So, i=ΔTfKfm=0.201.8×0.1=0.200.18=1.111≈1.11i = \frac{\Delta T_f}{K_f m} = \frac{0.20}{1.8 \times 0.1} = \frac{0.20}{0.18} = 1.111\approx 1.11i=Kf​mΔTf​​=1.8×0.10.20​=0.180.20​=1.111≈1.11

  1. Relate van’t Hoff factor to degree of dissociation

For the weak monobasic acid: HA⇌H++A−\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}HA⇌H++A−

If degree of dissociation is α\alphaα, then starting from 1 mole of HA\mathrm{HA}HA:

  • undissociated HA=1−α\mathrm{HA} = 1-\alphaHA=1−α
  • H+=α\mathrm{H^+} = \alphaH+=α
  • A−=α\mathrm{A^-} = \alphaA−=α

Total particles: 1−α+α+α=1+α1-\alpha + \alpha + \alpha = 1+\alpha1−α+α+α=1+α Thus, i=1+αi = 1+\alphai=1+α

Hence, α=i−1=1.111−1=0.111\alpha = i-1 = 1.111-1 = 0.111α=i−1=1.111−1=0.111

  1. Use Ostwald’s dilution law

For a weak acid of concentration CCC, Ka=Cα21−αK_a = \frac{C\alpha^2}{1-\alpha}Ka​=1−αCα2​

Given molality ≡\equiv≡ molarity, so C=0.1C = 0.1C=0.1

Substitute α=0.111\alpha = 0.111α=0.111: Ka=0.1×(0.111)21−0.111K_a = \frac{0.1\times (0.111)^2}{1-0.111}Ka​=1−0.1110.1×(0.111)2​

Now, (0.111)2=0.012321(0.111)^2 = 0.012321(0.111)2=0.012321 1−0.111=0.8891-0.111 = 0.8891−0.111=0.889

Therefore, Ka=0.1×0.0123210.889=0.00123210.889≈1.386×10−3K_a = \frac{0.1\times 0.012321}{0.889} = \frac{0.0012321}{0.889} \approx 1.386\times 10^{-3}Ka​=0.8890.1×0.012321​=0.8890.0012321​≈1.386×10−3

So, Ka≈1.38×10−3K_a \approx 1.38 \times 10^{-3}Ka​≈1.38×10−3

  1. Match with options

This corresponds to Option B.

  1. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.

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