Question

Molar conductances of ${\text{BaC}}{{\text{l}}_2}{\text{, }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ and ${\text{HCl}}$ at infinite dilution are ${{\text{X}}_1}{\text{, }}{{\text{X}}_2}$ and ${{\text{X}}_3}$ respectively. Molar conductance of ${\text{BaS}}{{\text{O}}_4}$ at infinite dilution is:
A) ${{\text{X}}_1}{\text{ + }}{{\text{X}}_2} - {{\text{X}}_3}$
B) ${{\text{X}}_1}{\text{ + }}{{\text{X}}_2} - 2{{\text{X}}_3}$
C) ($\dfrac{{{{\text{X}}_1}{\text{ + }}{{\text{X}}_2} - {{\text{X}}_3}}}{2}$
D) $\dfrac{{{{\text{X}}_1}{\text{ + }}{{\text{X}}_2} - 2{{\text{X}}_3}}}{2}$

Answer 1

According to the Kohlrausch’s law of independent migration of ions, the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. In other words, “We can represent the limiting molar conductivity of an electrolyte as the sum of the individual contributions of the cations and anions present in the electrolyte”.

Complete step-by-step solution:
Molar conductances of ${\text{BaC}}{{\text{l}}_2}{\text{, }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ and ${\text{HCl}}$ at infinite dilution are ${{\text{X}}_1}{\text{, }}{{\text{X}}_2}$ and ${{\text{X}}_3}$ respectively.
$\lambda _m^\infty {\text{BaC}}{{\text{l}}_2} = \lambda _{{\text{B}}{{\text{a}}^{2 + }}}^o + 2\lambda _{{\text{C}}{{\text{l}}^ - }}^o = {X_1}$ … …(1)
$\lambda _m^\infty {{\text{H}}_2}{\text{S}}{{\text{O}}_4} = 2\lambda _{{{\text{H}}^ + }}^o + \lambda _{{\text{SO}}_4^{2 - }}^o = {X_2}$ .. ….(2)
$\lambda _m^\infty {\text{HCl}} = \lambda _{{{\text{H}}^ + }}^o + \lambda _{{\text{C}}{{\text{l}}^ - }}^o = {X_3}$ … …(3)
$\lambda _m^\infty {\text{BaS}}{{\text{O}}_4} = \lambda _{{\text{B}}{{\text{a}}^{2 + }}}^o + \lambda _{{\text{SO}}_4^{2 - }}^o$ … …(4)
Add equations (1) and (2)
$\lambda _m^\infty {\text{BaC}}{{\text{l}}_2} + \lambda _m^\infty {{\text{H}}_2}{\text{S}}{{\text{O}}_4} = \lambda _{{\text{B}}{{\text{a}}^{2 + }}}^o + 2\lambda _{{\text{C}}{{\text{l}}^ - }}^o + 2\lambda _{{{\text{H}}^ + }}^o + \lambda _{{\text{SO}}_4^{2 - }}^o = {X_1} + {X_2}$ … …(5)
Multiply equation (3) with 2.
$2\lambda _m^\infty {\text{HCl}} = 2\lambda _{{{\text{H}}^ + }}^o + 2\lambda _{{\text{C}}{{\text{l}}^ - }}^o = 2{X_3}$ … …(6)
Subtract equation (6) from equation (5)
$\lambda _m^\infty {\text{BaC}}{{\text{l}}_2} + \lambda _m^\infty {{\text{H}}_2}{\text{S}}{{\text{O}}_4} - 2\lambda _m^\infty {\text{HCl}} = \lambda _{{\text{B}}{{\text{a}}^{2 + }}}^o + 2\lambda _{{\text{C}}{{\text{l}}^ - }}^o + 2\lambda _{{{\text{H}}^ + }}^o + \lambda _{{\text{SO}}_4^{2 - }}^o - 2\lambda _{{{\text{H}}^ + }}^o - 2\lambda _{{\text{C}}{{\text{l}}^ - }}^o = {X_1} + {X_2} - 2{X_3}$
$\lambda _m^\infty {\text{BaC}}{{\text{l}}_2} + \lambda _m^\infty {{\text{H}}_2}{\text{S}}{{\text{O}}_4} - 2\lambda _m^\infty {\text{HCl}} = \lambda _{{\text{B}}{{\text{a}}^{2 + }}}^o + \lambda _{{\text{SO}}_4^{2 - }}^o = {X_1} + {X_2} - 2{X_3}$ … …(7)
The equation (7) and the equation (4) are one and the same
Hence, molar conductance of ${\text{BaS}}{{\text{O}}_4}$ at infinite dilution is ${{\text{X}}_1}{\text{ + }}{{\text{X}}_2} - 2{{\text{X}}_3}$ .

Hence, the correct answer is option (B).

Note: We can call the molar conductivity as the conductance of the given volume of electrolyte having one mole of electrolyte. The solution is kept between two electrodes. The electrodes have unit area of cross-section and distance of unit length. We can call the molar conductive as limiting molar conductivity if the concentration of the electrolyte is nearly zero.