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NEET CHEMISTRYMedium

The molar conductance of M/32 solution of a weak monobasic acid is 8.0 ohm1 cm28.0 \text{ ohm}^{-1} \text{ cm}^2 and at infinite dilution is 400 ohm1 cm2400 \text{ ohm}^{-1} \text{ cm}^2. The dissociation constant of this acid is:

A

1.25×1051.25 \times 10^{-5}

B

1.25×1061.25 \times 10^{-6}

C

6.25×1046.25 \times 10^{-4}

D

1.25×1041.25 \times 10^{-4}

Step-by-Step Solution

  1. Calculate Degree of Dissociation (α\alpha): For a weak electrolyte, the degree of dissociation α\alpha is the ratio of molar conductivity at a specific concentration (Λm\Lambda_m) to the limiting molar conductivity (Λm\Lambda_m^\circ). α=ΛmΛm\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} Given: Λm=8.0 S cm2 mol1\Lambda_m = 8.0 \text{ S cm}^2 \text{ mol}^{-1} and Λm=400 S cm2 mol1\Lambda_m^\circ = 400 \text{ S cm}^2 \text{ mol}^{-1}. α=8.0400=0.02\alpha = \frac{8.0}{400} = 0.02

  2. Calculate Dissociation Constant (KaK_a): The dissociation constant for a weak acid is given by: Ka=cα21αK_a = \frac{c\alpha^2}{1 - \alpha} Given: Concentration c=M/32=1/32 mol L10.03125 Mc = \text{M}/32 = 1/32 \text{ mol L}^{-1} \approx 0.03125 \text{ M}. Since α(0.02)\alpha (0.02) is very small compared to 1, we can approximate 1α11 - \alpha \approx 1. Kacα2K_a \approx c\alpha^2 Ka=132×(0.02)2K_a = \frac{1}{32} \times (0.02)^2 Ka=132×4×104K_a = \frac{1}{32} \times 4 \times 10^{-4} Ka=18×104=0.125×104K_a = \frac{1}{8} \times 10^{-4} = 0.125 \times 10^{-4} Ka=1.25×105K_a = 1.25 \times 10^{-5}

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