Question

Molar conductivities at infinite dilution of $NaCl,HCl$ and $C{H_3}COONa$ are 126.4, 425.9 and 91.0 S cm2 mol-1 respectively. Molar conductivity for $C{H_3}COOH$ will be:

Answer 1

Hint:Conductivity and molar conductivity depend on the electrolyte concentration. The limiting molar conductivity is molar conductivity when concentration approaches to zero. Kohlrausch law of independent migration of ions states that the limiting molar conductivity of an electrolyte is equal to the sum of individual anions and cations present in the electrolyte. We shall use ${\lambda _{C{H_3}COOH}} = {\lambda _{C{H_3}COONa}} + {\lambda _{HCl}} - {\lambda _{NaCl}}$ for the given question.

Complete step-by-step answer:The conductivity of a solution for different electrolytes at a constant temperature differs due to the charge and size of dissociated ions. Thus, it makes it important to determine a physical quantity related to conductivity.
${ \wedge _m} = \dfrac{K}{C}$
Where, $K$ is conductivity in S m-1 and $C$ is concentration in mol m-3 and ${ \wedge _m}$ is molar conductivity expressed in S m2 mol-1.
Both conductivity and molar conductivity depend on the concentration of the electrolyte. Conductivity is inversely proportional to the concentration of both strong and weak electrolyte.
The limiting molar conductivity is molar conductivity when concentration approaches to zero. The limiting molar conductivity is represented by ${ \wedge _m}^0$ . Kohlrausch law of independent migration of ions states that the limiting molar conductivity of an electrolyte is equal to the sum of individual anions and cations present in the electrolyte.
Thus, for the above given question we can calculate the molar conductivity of $C{H_3}COOH$ at infinite dilution by using the Kohlrausch law of independent migration of ions
The equation for the above reaction will be,
${\lambda _{C{H_3}COOH}} = {\lambda _{C{H_3}COONa}} + {\lambda _{HCl}} - {\lambda _{NaCl}}$
It is given that
${\lambda _{C{H_3}COONa}} = 91Sc{m^2}mo{l^{ - 1}}$

${\lambda _{HCl}} = 425.9Sc{m^2}mo{l^{ - 1}}$
${\lambda _{NaCl}} = 126.4Sc{m^2}mo{l^{ - 1}}$
Now substitute the value in the equation,
We get,
${\lambda _{C{H_3}COOH}} = (91 + 425.9 - 126.4)Sc{m^2}mo{l^{ - 1}}$
$ \Rightarrow {\lambda _{C{H_3}COOH}} = 390.5Sc{m^2}mo{l^{ - 1}}$
Thus, we conclude that molar conductivity of $C{H_3}COOH$ at infinite dilution is $390.5Sc{m^2}mo{l^{ - 1}}$ .


Note: Acetic acid is a weak electrolyte in which the dissociation degree is low at high concentrations and therefore, for weak electrolytes the change in dilution is due to increase in the degree of dissociation. Kohlrausch law of independent migration of ions makes it possible to calculate the dissociation constant for even weak electrolytes.