Question

The molar conductivities of $NaOAc$ and $HCl$ at infinite dilution in water at ${25}^{\circ }C$ are $91.0$ and $426.2Sc{m}^{2}mo{l}^{-1}$ respectively. To calculate the limiting molar conductivity of $HOAc$ at same temperature, the additional value required is:

• A. ${\Lambda }_{H_{2}O}^0$
• B. ${\Lambda }_{KCl}^0$
• C. ${\Lambda }_{NaOH}^0$
• D. ${\Lambda }_{NaCl}^0$

Answer 1

The correct option is D ${\Lambda }_{NaCl}^0$

Kohlrausch law:
At infinite dilution, when all the interionic effects disappear and dissociation of electrolyte is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it is associated.

Kohlrausch law of independent migration of ions:
${\Lambda }_m^0={\lambda }_{+}^0+{\lambda }_{-}^0$

${\lambda }_{+}^0\to \text{Limiting molar conductivities of cation}$
${\lambda }_{-}^0\to \text{Limiting molar conductivities of anion}$

In general,
${\Lambda }_m^0=v_{+}{\lambda }_{+}^0+v_{-}{\lambda }_{-}^0$

$v_{+}$ and $v_{-}$ are the number of cations and anions after dissociation of an electrolyte.
${\Lambda }_{HOAc}^0={\lambda }_{H^{+}}^0+{\lambda }_{Ac{O}^{-}}^0$

${\Lambda }_{NaOAc}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{Ac{O}^{-}}^0$

${\Lambda }_{HCl}^0={\lambda }_{H^{+}}^0+{\lambda }_{C{l}^{-}}^0$
Thus,
${\Lambda }_{HOAc}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{Ac{O}^{-}}^0+{\lambda }_{H^{+}}^0+{\lambda }_{C{l}^{-}}^0-[{\lambda }_{N{a}^{+}}^0+{\lambda }_{C{l}^{-}}^0]$
$\therefore$
${\Lambda }_{HOAc}^0={\Lambda }_{NaOAc}^0+{\Lambda }_{HCl}^0-{\Lambda }_{NaCl}^0$
Hence, the additional value required is ${\Lambda }_{NaCl}^0$