Question

The molar conductivities at infinite dilution for sodium acetate, $HCl$ and $NaCl$ are $91\, S \,cm^2\, mol^{-1}$, $425.9 \,S \,cm^2\, mol^{-1}$ and $126.4\, S \,cm^2\, mol^{-1}$ respectively. The molar conductivity of acetic acid at infinite dilution is

• A. $390.5\, S \,cm^2\, mol^{-1}$
• B. $530.9\, S \,cm^2\, mol^{-1}$
• C. $300.5\, S \,cm^2\, mol^{-1}$
• D. $930.5\, S \,cm^2\, mol^{-1}$

Answer 1

Answer: (A)

Key ldea Kohlrausch law of independent migration of ions can be used to calculate $\Lambda_{ m }^{\circ}$ for weak electrolyte such as acetic acid.

Given, $\Lambda_{ m ( NaAc )}^{\circ} =91 S cm ^{2} mol ^{-1}$

$\Lambda_{ m ( HCl )}^{\circ} =425.9 S cm ^{2} mol ^{-1}$

$\Lambda_{ m _{( NaCl )}^{\circ}} =126.4 Scm ^{2} mol ^{-1}$

Using Kohlrausch law, to find the molar conductivity of acetic acid.

$\Lambda_{ m ( HAC )}^{\circ} =\lambda_{ H ^{+}}^{\circ}+\lambda_{ AC ^{-}}^{\circ} $

$=\lambda_{ H ^{+}}^{\circ}+\lambda_{ Cr ^{-}}^{\circ}+\lambda_{ AC ^{-}}^{\circ}+\lambda_{ Na ^{+}}^{\circ}-\lambda_{ Cl ^{-}}^{\circ}-\lambda_{ Na ^{+}}^{\circ}$

$=\Lambda_{ m ( HCl )}^{\circ}+\Lambda_{ m ( NaAC )}^{\circ}-\Lambda_{ m ( NaCl )}^{\circ}$

$=(425.9+91.0-126.4) S cm ^{2} mol ^{-1}$

$=390.5 Scm ^{2} mol ^{-1}$