Question

If molar conductivity at infinite dilution of NaCl, HCl, and $C{{H}_{3}}COONa$ respectively are a, b, and c $Sc{{m}^{2}}mo{{l}^{-1}}$ then the molar conductivity at infinite dilution for $C{{H}_{3}}COOH$ in $Sc{{m}^{2}}mo{{l}^{-1}}$ will be
(A) a+b-c
(B) b+c-a
(C) a+c-b
(D) a-b-c

Answer 1

Hint: To solve this question we first need to understand what is molar conductivity. Molar conductivity can be defined as the conductivity of an electrolyte solution divided by the electrolyte's molar concentration.

Complete answer:
Now, the total molar conductivity is the conductance produced by all of the ions contained in a solution of one mole of electrolyte.
According to Ostwald's law, weak electrolytes ionize and dissociate completely upon infinite dilution.
Also, Kohlrauch law states that upon infinite dilution, the contribution of an ion to the total molar conductance of an electrolyte is not dependent on the other ions present or their nature.
So, we can say that at infinite dilution, the sum of molar conductivities $\lambda $ of the anions and cations gives the total molar conductivity or the molar conductance $\Lambda $ of the solution.
\[\Lambda =\lambda _{+}^{{}^\circ }+\lambda _{-}^{{}^\circ }\]
The reaction between hydrochloric acid and sodium acetate results in the formation of acetic acid and sodium chloride.
The balanced chemical equation is as follows
\[C{{H}_{3}}COONa+HCl\to C{{H}_{3}}COOH+NaCl\]
Now, it is given to us that the molar conductivity ($\lambda $) of NaCl is \[{{\lambda }_{NaCl}}=a\text{ }Sc{{m}^{2}}mo{{l}^{-1}}\], molar conductivity of HCl is \[{{\lambda }_{HCl}}=b\text{ }Sc{{m}^{2}}mo{{l}^{-1}}\], and molar conductivity of $C{{H}_{3}}COONa$ is \[{{\lambda }_{C{{H}_{3}}COONa}}=c\text{ }Sc{{m}^{2}}mo{{l}^{-1}}\]. Let us assume that the molar conductivity of $C{{H}_{3}}COOH$ is x $Sc{{m}^{2}}mo{{l}^{-1}}$. So, the equation relating the molar conductivities of the molecules is
c + b = x + a
x = c + b - a

So, the molar conductivity for $C{{H}_{3}}COOH$ at infinite dilution will be option (B) b + c - a $Sc{{m}^{2}}mo{{l}^{-1}}$.

Note:
It should be noted that as the dilution is depleted, the molar conductivity of both weak and strong electrolytes increases. This is because of the increase in dissociation of the electrolytes into ions with the increase in dilution and hence increase in the concentration of the ions. With the increase in the number of ions, conductivity increases.