Q39 / 180 · COMEDK 2026 (9 May Shift 1) · Chemistry

Identify the correct mathematical expression which represents the variation in molar conductivity of a weak acid having concentration C and ionisation constant $K_a$
($\lambda_m^\infty$ = molar conductivity at infinite dilution, $\lambda_m$ = molar conductivity at concentration C)

  • Correct answer
    A
    $K_a = \lambda_m^2 C / \lambda_m^\infty(\lambda_m^\infty - \lambda_m)$
  • B
    $K_a = \lambda_m \lambda_m^\infty - (\lambda_m^\infty)^2 + \lambda_m^2 C$
  • C
    $\lambda_m + \lambda_m^\infty + K_a C^{\frac{1}{2}} = 0$
  • D
    $K_a = \lambda_m^2 C / \lambda_m^\infty(\lambda_m^\infty + \lambda_m)$

Solution

The degree of dissociation $\alpha$ for a weak electrolyte is given by:

$\alpha = \dfrac{\lambda_m}{\lambda_m^\infty}$

For a weak acid, the ionisation constant $K_a$ is given by Ostwald's dilution law:

$K_a = \dfrac{C \alpha^2}{1 - \alpha}$

Substituting the expression for $\alpha$ into the equation for $K_a$:

$K_a = \dfrac{C \left( \dfrac{\lambda_m}{\lambda_m^\infty} \right)^2}{1 - \dfrac{\lambda_m}{\lambda_m^\infty}}$

$K_a = \dfrac{C \dfrac{\lambda_m^2}{(\lambda_m^\infty)^2}}{\dfrac{\lambda_m^\infty - \lambda_m}{\lambda_m^\infty}}$

$K_a = \dfrac{C \lambda_m^2}{\lambda_m^\infty (\lambda_m^\infty - \lambda_m)}$

Answer: $K_a = \lambda_m^2 C / \lambda_m^\infty(\lambda_m^\infty - \lambda_m)$

Hence, the correct answer is option: 'A'.

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