JEE Main · 2025 · Shift-IIhardIEQ-081

HA(aq) H+(aq) + A-(aq) The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20°C.…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

\ceHA(aq)⇌H+(aq)+A−(aq)\ce{HA(aq) \rightleftharpoons H+(aq) + A-(aq)}

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20°C.

The dissociation constant for the acid is

Given: Kf(\ceH2O)=1.8K_f(\ce{H2O}) = 1.8 kg mol−1^{-1}, molality ≡\equiv molarity

Options
  1. a

    1.38×10−31.38 \times 10^{-3}

    ✓
  2. b

    1.1×10−21.1 \times 10^{-2}

  3. c

    1.90×10−31.90 \times 10^{-3}

  4. d

    1.89×10−11.89 \times 10^{-1}

Correct Answera

1.38×10−31.38 \times 10^{-3}

Detailed Solution

Step 1: Find van't Hoff factor ii

ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m

i=ΔTfKf⋅m=0.201.8×0.1=0.200.18=1.11i = \tfrac{\Delta T_f}{K_f \cdot m} = \tfrac{0.20}{1.8 \times 0.1} = \tfrac{0.20}{0.18} = 1.11

Step 2: Find degree of dissociation α\alpha

For \ceHA⇌H++A−\ce{HA \rightleftharpoons H+ + A-}, n=2n = 2 (one molecule gives 2 ions)

i=1+(n−1)α=1+αi = 1 + (n-1)\alpha = 1 + \alpha

α=i−1=1.11−1=0.11\alpha = i - 1 = 1.11 - 1 = 0.11

Step 3: Calculate KaK_a

Ka=cα21−α=0.1×(0.11)21−0.11=0.1×0.01210.89=0.001210.89≈1.38×10−3K_a = \tfrac{c\alpha^2}{1 - \alpha} = \tfrac{0.1 \times (0.11)^2}{1 - 0.11} = \tfrac{0.1 \times 0.0121}{0.89} = \tfrac{0.00121}{0.89} \approx 1.38 \times 10^{-3}

Key Points to Remember:

  • ΔTf=iKfm\Delta T_f = i K_f m gives ii, then α=i−1\alpha = i - 1 for 1:1 electrolyte
  • Ka=cα21−αK_a = \tfrac{c\alpha^2}{1-\alpha} (Ostwald dilution law)

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