Question

What will be the molar conductance of oxalic acid from following data:
$N{a_2}{C_2}{O_4} = 400{\Omega ^{ - 1}}c{m^2}e{q^{ - 1}}$, ${H_2}S{O_4} = 700{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}}$ and $N{a_2}S{O_4} = 450{\Omega ^{ - 1}}c{m^2}e{q^{ - 1}}$

Answer 1

Hint: Conductance is defined as the reciprocal of the resistance i.e. it is defined as the ability to conduct heat and electricity.
Conductivity: It is defined as the measure of the ability to conduct heat and electricity.
Molar conductivity: It is defined as the conductivity divided by its molar concentration.

Complete step-by-step answer:Let us first talk about conductance, molar conductance and limiting molar conductance.
Conductance is defined as the reciprocal of the resistance i.e. it is defined as the ability to conduct heat and electricity.
Conductivity: It is defined as the measure of the ability to conduct heat and electricity.
Molar conductivity: It is defined as the conductivity divided by its molar concentration. It is represented by ${\Delta _m}$. As name indicates, molar conductivity changes with concentration of the solution.
Limiting molar conductivity: It is defined as the molar conductivity at infinite dilution i.e. zero concentration. It is represented as ${\lambda ^\infty }$.
In the question we are given with the data as: $N{a_2}{C_2}{O_4} = 400{\Omega ^{ - 1}}c{m^2}e{q^{ - 1}} = {\lambda _{eq}}$, ${H_2}S{O_4} = 700{\Omega ^{ - 1}}c{m^2}mo{l^{ - 1}} = {\lambda _m}$ and $N{a_2}S{O_4} = 450{\Omega ^{ - 1}}c{m^2}e{q^{ - 1}} = {\lambda _{eq}}$.
And the relation between ${\lambda _{eq}}$ and ${\lambda _m}$ is as follows: ${\lambda _{eq}} = \dfrac{{{\lambda _m}}}{2}$. So ${\lambda _{eq}}{H_2}S{O_4} = \dfrac{{700}}{2} = 350{\Omega ^{ - 1}}c{m^2}e{q^{ - 1}}$.
Kohlrausch’s law: It states that the molar conductivity at infinite dilution is the sum of the molar conductivity of its ion.
For example as in the question we are given with molar conductivity of $N{a_2}{C_2}{O_4},{H_2}S{O_4}$ and $N{a_2}S{O_4}$. We can write their equation according to Kohlrausch’s law as:
${\lambda _{eq}}^\infty N{a_2}{C_2}{O_4} = {\lambda _{eq}}^\infty N{a^ + } + {\lambda _{eq}}^\infty {C_2}{O_4}^{2 - }$ ----(1) where ${\lambda _{eq}}^\infty N{a_2}{C_2}{O_4}$ is equivalent molar conductivity of $N{a_2}{C_2}{O_4}$, ${\lambda _{eq}}^\infty N{a^ + }$ is the equivalent molar conductivity of sodium ion $N{a^ + }$ and ${\lambda _{eq}}^\infty {C_2}{O_4}^{2 - }$ is the equivalent molar conductivity of ${C_2}{O_4}^{2 - }$ at infinite dilution.
${\lambda _{eq}}^\infty {H_2}S{O_4} = {\lambda _{eq}}^\infty {H^ + } + {\lambda _{eq}}^\infty S{O_4}^{2 - }$ -----(2), ${\lambda _{eq}}^\infty {H_2}S{O_4}$ is equivalent molar conductivity of ${H_2}S{O_4}$, ${\lambda _{eq}}^\infty {H^ + }$ is the equivalent molar conductivity of hydrogen ion ${H^ + }$ and ${\lambda _{eq}}^\infty S{O_4}^{2 - }$ is the equivalent molar conductivity of $S{O_4}^{2 - }$ at infinite dilution.
${\lambda _{eq}}^\infty N{a_2}S{O_4} = {\lambda _{eq}}^\infty N{a^ + } + {\lambda _{eq}}^\infty S{O_4}^{2 - }$ -----(3), where ${\lambda _{eq}}^\infty N{a_2}S{O_4}$ is equivalent molar conductivity of $N{a_2}S{O_4}$, ${\lambda _{eq}}^\infty N{a^ + }$ is the equivalent molar conductivity of sodium ion $N{a^ + }$ and ${\lambda _{eq}}^\infty S{O_4}^{2 - }$ is the equivalent molar conductivity of $S{O_4}^{2 - }$ at infinite dilution.
Now we want to calculate the molar conductance of oxalic acid. So we have to calculate
${\lambda _{eq}}^\infty {H_2}{C_2}{O_4} = {\lambda _{eq}}^\infty {H^ + } + {\lambda _{eq}}^\infty {C_2}{O_4}^{2 - }$, ${\lambda _{eq}}^\infty {H_2}{C_2}{O_4}$ is equivalent molar conductivity of ${H_2}{C_2}{O_4}$, ${\lambda _{eq}}^\infty {H^ + }$ is the equivalent molar conductivity of hydrogen ion ${H^ + }$ and ${\lambda _{eq}}^\infty {C_2}{O_4}^{2 - }$ is the equivalent molar conductivity of ${C_2}{O_4}^{2 - }$ at infinite dilution.
So if we add (1) and (2) and subtract (3) from then we will get:
${\lambda _{eq}}^\infty {H^ + } + {\lambda _{eq}}^\infty {C_2}{O_4}^{2 - } = 400 + 350 - 450 = 300{\Omega ^{ - 1}}c{m^2}e{q^{ - 1}}$.
But in the equation we have to calculate molar conductance hence we have to multiply the equivalent conductance by two. So molar conductance will be $2 \times 300{\Omega ^{ - 1}}c{m^2}e{q^{ - 1}}$.

Note: For weak electrolytes, if dilution is increasing then the molar conductivity will be increasing as electrolyte will dissociate into more ions on increasing the dilution. But for strong electrolytes the electrolyte is already completely ionised so there is no so much effect of the dilution on strong electrolytes.