Plotting $1 / \Lambda_{\mathrm{m}}$ against $\mathrm{c} \Lambda_{\mathrm{m}}$ 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 [$\Lambda_{\mathrm{m}}=$ molar conductivity $\Lambda_{\mathrm{m}}^{\circ}=$ limiting molar conductivity $\mathrm{c}=$ molar concentration $\mathrm{K}_{\mathrm{a}}=$ dissociation constant of HX]

(A) $\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ}$

(B) $\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ} / 2$

(C) $2 \mathrm{~K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ}$

(D) $1 /(\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ})$

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Redox Reactions and Electrochemistry

Plotting

1/ Λ_{m}

against

c Λ_{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

[

Λ_{m} =

molar conductivity

Λ_{m}^{\circ} =

limiting molar conductivity

c =

molar concentration

K_{a} =

dissociation constant of HX]

Answer: (A)

For a weak acid, the degree of dissociation

α = \frac{Λ _{m}}{Λ _{0}}

.

The dissociation constant is:

K_{a} = \frac{c α ^{2}}{1 - α} = \frac{c ( Λ _{m} / Λ _{0} ) ^{2}}{1 - ( Λ _{m} / Λ _{0} )}

Rearranging:

K_{a} (1 - \frac{Λ _{m}}{Λ _{0}}) = \frac{c Λ _{m}^{2}}{Λ _{0}^{2}}

Dividing by

Λ_{m}

\frac{K _{a}}{Λ _{m}} - \frac{K _{a}}{Λ _{0}} = \frac{c Λ _{m}}{Λ _{0}^{2}}

Rearranging:

\frac{1}{Λ _{m}} = \frac{c Λ _{m}}{K _{a} Λ _{0}^{2}} + \frac{1}{Λ _{0}}

This is of the form

y = S x + P

where:

Therefore:

\frac{P}{S} = \frac{1/ Λ _{0}}{1/ ( K _{a} Λ _{0}^{2} )} = K_{a} Λ_{0}

Answer: (A)

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option