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ExamsJEE AdvancedChemistry

The molar conductivities at 25 degrees C are given: lambdaₘ(HCl) = 426 S*cm²*mol⁻¹, lambdaₘ(NaCl) = 126 S*cm²*mol⁻¹, lambdaₘ(sodium crotonate) = 83 S*cm²*mol⁻¹. The conductivity of 0.001 M crotonic acid solution is 3.83 * 10⁻⁵ S*cm⁻¹. Find the ionisation constant Ka of crotonic acid.

  1. 10⁻⁵
  2. 1.11 * 10⁻⁵
  3. 1.11 * 10⁻⁴
  4. 0.01

Correct answer: 1.11 * 10⁻⁵

Solution

Using Kohlrausch's law, lambdaₘ_inf(crotonic acid) = 383 S*cm²*mol⁻¹. The measured molar conductivity is 3.83*10⁻⁵ S*cm⁻¹ / 0.001 mol/L = 38.3 S*cm²*mol⁻¹, giving alpha = 38.3/383 = 0.1. Then Ka = (0.001 * 0.1²)/(1-0.1) = 1.11*10⁻⁵.

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