Question

The molar conductivities at infinite dilution for sodium acetate, $HCl$ and $NaCl$ are $91Sc{m}^{2}mo{l}^{-1}$, $425.9Sc{m}^{2}mo{l}^{-1}$ and $126.4Sc{m}^{2}mo{l}^{-1}$ respectively. The molar conductivity of acetic acid at infinite dilution is

• A. $390.5Sc{m}^{2}mo{l}^{-1}$
• B. $530.9Sc{m}^{2}mo{l}^{-1}$
• C. $300.5Sc{m}^{2}mo{l}^{-1}$
• D. $930.5Sc{m}^{2}mo{l}^{-1}$

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, ${\Lambda }_{m(NaAc)}^{\circ }=91Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_{m(HCl)}^{\circ }=425.9Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_{m_{(NaCl)}^{\circ }}=126.4Sc{m}^{2}mo{l}^{-1}$

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

${\Lambda }_{m(HAC)}^{\circ }={\lambda }_{H^{+}}^{\circ }+{\lambda }_{A{C}^{-}}^{\circ }$

$={\lambda }_{H^{+}}^{\circ }+{\lambda }_{C{r}^{-}}^{\circ }+{\lambda }_{A{C}^{-}}^{\circ }+{\lambda }_{N{a}^{+}}^{\circ }-{\lambda }_{C{l}^{-}}^{\circ }-{\lambda }_{N{a}^{+}}^{\circ }$

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

$=(425.9+91.0-126.4)Sc{m}^{2}mo{l}^{-1}$

$=390.5Sc{m}^{2}mo{l}^{-1}$