JEE Advanced 2023 Paper 1 Q05 Chemistry Electrochemistry Nernst Equation Hard

JEE Advanced 2023 Paper 1 · Q05 · Nernst Equation

Plotting $1/\Lambda_m$ against $c\Lambda_m$ for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with $y$-axis intercept of $P$ and slope of $S$. The ratio $P/S$ is [$\Lambda_m$ = molar conductivity, $\Lambda_m^\circ$ = limiting molar conductivity, $c$ = molar concentration, $K_a$ = dissociation constant of HX]

  1. A. $K_a \Lambda_m^\circ$
  2. B. $K_a \Lambda_m^\circ / 2$
  3. C. $2 K_a \Lambda_m^\circ$
  4. D. $1/(K_a \Lambda_m^\circ)$
Reveal answer + step-by-step solution

Correct answer:A

Solution

For a weak acid, $\alpha = \Lambda_m/\Lambda_m^\circ$ and $K_a = c\alpha^2/(1-\alpha) \approx c\alpha^2$ for small $\alpha$. Rearranging: $\dfrac{1}{\Lambda_m} = \dfrac{1}{\Lambda_m^\circ} + \dfrac{c\Lambda_m}{K_a (\Lambda_m^\circ)^2}$. So intercept $P = 1/\Lambda_m^\circ$ and slope $S = 1/(K_a (\Lambda_m^\circ)^2)$. Ratio $P/S = K_a \Lambda_m^\circ$.

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