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NEET CHEMISTRYMedium

The molar conductivity of 0.007 M acetic acid is 20 S cm2^2 mol−1^{-1}. What is the dissociation constant of acetic acid? Choose the correct option. [ΛH+∘=350 S cm2 mol−1,ΛCH3COO−∘=50 S cm2 mol−1]\left[ \Lambda_{H^+}^\circ = 350 \text{ S cm}^2 \text{ mol}^{-1}, \Lambda_{CH_3COO^-}^\circ = 50 \text{ S cm}^2 \text{ mol}^{-1} \right]

A

1.75 \times 10^{-4} \text{ mol L}^{-1}

B

2.50 \times 10^{-4} \text{ mol L}^{-1}

C

1.75 \times 10^{-5} \text{ mol L}^{-1}

D

2.50 \times 10^{-5} \text{ mol L}^{-1}

Step-by-Step Solution

Given Λm=20 S cm2 mol−1\Lambda_m = 20 \text{ S cm}^2 \text{ mol}^{-1}. ΛCH3COOH∘=ΛCH3COO−∘+ΛH+∘=50+350=400 S cm2 mol−1\Lambda_{CH_3COOH}^\circ = \Lambda_{CH_3COO^-}^\circ + \Lambda_{H^+}^\circ = 50 + 350 = 400 \text{ S cm}^2 \text{ mol}^{-1}. Degree of dissociation α=ΛmΛm∘=20400=120\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{20}{400} = \frac{1}{20}. Dissociation constant Ka=Cα21−α≈Cα2=7×10−3×(120)2=7×10−3×1400=1.75×10−5 mol L−1K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2 = 7 \times 10^{-3} \times (\frac{1}{20})^2 = 7 \times 10^{-3} \times \frac{1}{400} = 1.75 \times 10^{-5} \text{ mol L}^{-1}.

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