Question

Calculate ${\Lambda }_{HN{O}_3}^0$ and ${\Lambda }_{HOAc}^0$ using appropriate molar conductances of the electrolytes listed below at infinite dilution in $H_{2}O$ at ${25}^{\circ }C$.

Electrolyte	${\Lambda }^0\left(Sc{m}^{2}mo{l}^{-1}\right)$
$KCl$	$149.9$
$NaCl$	$126.5$
$HCl$	$426.2$
$KN{O}_3$	$145.0$
$NaOAc$	$91.0$

• A. ${\Lambda }_{HN{O}_3}^0=390.7Sc{m}^2{\text{mol}}^{-1}$
  ${\Lambda }_{HOAc}^0=517.2Sc{m}^2{\text{mol}}^{-1}$
• B. ${\Lambda }_{HN{O}_3}^0=517.2Sc{m}^2{\text{mol}}^{-1}$
  ${\Lambda }_{HOAc}^0=149.8Sc{m}^2{\text{mol}}^{-1}$
• C. ${\Lambda }_{HN{O}_3}^0=421.3Sc{m}^2{\text{mol}}^{-1}$
  ${\Lambda }_{HOAc}^0=390.7Sc{m}^2{\text{mol}}^{-1}$
• D. ${\Lambda }_{HN{O}_3}^0=149.8Sc{m}^2{\text{mol}}^{-1}$
  ${\Lambda }_{HOAc}^0=444.7Sc{m}^2{\text{mol}}^{-1}$

Answer 1

The correct option is C ${\Lambda }_{HN{O}_3}^0=421.3Sc{m}^2{\text{mol}}^{-1}$

${\Lambda }_{HOAc}^0=390.7Sc{m}^2{\text{mol}}^{-1}$
Kohlrausch law of independent migration of ions:
${\Lambda }_m^0={\lambda }_{+}^0+{\lambda }_{-}^0$

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

Using the given data,
${\Lambda }_{HN{O}_3}^0={\Lambda }_{KN{O}_3}^0+{\Lambda }_{HCl}^0-{\Lambda }_{KCl}^0$
${\Lambda }_{HN{O}_3}^0=145.0+426.2-149.9$
${\Lambda }_{HN{O}_3}^0=421.3Sc{m}^2{\text{mol}}^{-1}$


${\Lambda }_{HOAc}^0={\Lambda }_{NaOAC}^0+{\Lambda }_{HCl}^0-{\Lambda }_{NaCl}^0$
${\Lambda }_{HOAc}^0=91.0+426.2-126.5$
${\Lambda }_{HOAc}^0=390.7Sc{m}^2{\text{mol}}^{-1}$