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Standard XII

Chemistry

Depression in Freezing Point

The measured ...

Question

The measured freezing point depression for a $0.1 m$ aqueous ${C H}_{3} C O O H$ solution is ${0.19}^{o} C$ . The acid dissociation constant $K_{a}$ at this concentration will be:
(Given $K_{f}$ the molal cryoscopic constant $= 1.86 K$ $k g {m o l}^{- 1}$ )

4.76 \times 10^{- 5}

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4 \times 10^{- 5}

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8 \times 10^{- 5}

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2 \times 10^{- 5}

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Solution

The correct option is C $4 \times 10^{- 5}$
$∵ Δ T_{f} = i \times K_{f} \times m$

$∴ i = \frac{Δ T_{f}}{K_{f} \times m} = \frac{0.19}{1.86 \times 0.1} = 1.02$

Again $α = \frac{i - 1}{n - 1} = \frac{1.02 - 1}{2 - 1}$

$= 0.02 = 2.0 \times 10^{- 2}$

For ${C H}_{3} C O O H ⇌ {C H}_{3} C O O^{-} + H^{+}$

$K_{a} = {C α}^{2} = 0.1 \times {(2.0 \times 10^{- 2})}^{2} = 4 \times 10^{- 5}$

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The measured freezing point depression for a $0.1 m$ aqueous ${C H}_{3} C O O H$ solution is ${0.19}^{o} C$ . The acid dissociation constant $K_{a}$ at this concentration will be:
(Given $K_{f}$ the molal cryoscopic constant $= 1.86 K$ $k g {m o l}^{- 1}$ )

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The measured freezing point depression for a 0.1m aqueous CH3COOH solution is 0.19oC. The acid dissociation constant Ka at this concentration will be:(Given Kf the molal cryoscopic constant =1.86K kgmol−1)

The correct option is C 4×10−5∵ΔTf=i×Kf×m∴i=ΔTfKf×m=0.191.86×0.1=1.02Again α=i−1n−1=1.02−12−1=0.02=2.0×10−2For CH3COOH⇌CH3COO−+H+Ka=Cα2=0.1×(2.0×10−2)2=4×10−5

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The measured freezing point depression for a $0.1 m$ aqueous ${C H}_{3} C O O H$ solution is ${0.19}^{o} C$ . The acid dissociation constant $K_{a}$ at this concentration will be:
(Given $K_{f}$ the molal cryoscopic constant $= 1.86 K$ $k g {m o l}^{- 1}$ )