Question

A decinormal solution of NaCl has specific conductivity equal to 0.0092. If ionic conductances of $N{a^ + }$ and $C{l^ - }$ ions at the same temperature
are 43.0 and 65 $oh{m^{ - 1}}$ respectively, calculate the degree of dissociation of NaCl solution.

Answer 1

Hint: Degree of dissociation of an electrolyte $\left( \alpha \right)$ is the fraction of one mole of the electrolyte that has dissociated under the given conditions. The value of $\alpha$ depends upon: nature of electrolyte, nature of solvent, dilution and the temperature.

Complete step by step answer:
Given, Normality of solution of $NaCl = \dfrac{1}{{10}}N$

Specific conductivity $\left( K \right) = 0.0092$
Ionic conductances of $N{a^ + }$ and $C{l^ - }$ are 43.0 and 65 $oh{m^{ - 1}}$.
To find the equivalent conductance in terms of concentration\[\left( {\Lambda _{eq}^c} \right)\]
Equivalent conductance is the conducting power of all ions furnished by an equivalent of an electrolyte in any solution.
Equivalent conductance \[\left( {\Lambda _{eq}^c} \right) = \dfrac{{Conductivity\left( K \right) \times 1000}}{{Normality{\text{ of solution}}}}\]
= $\dfrac{{0.0092 \times 1000}}{{0.1}}$
= 92 $oh{m^{ - 1}}$.
To find equivalent conductance at infinite dilution.
Equivalent conductance at infinite dilution \[\left( {\Lambda _{eq}^\infty } \right)\] for $NaCl$ $ = \Lambda _{N{a^ + }}^\infty + \Lambda _{C{l^ - }}^\infty $

By putting the value of ionic conductance $\left( {\lambda _{N{a^ + }}^\infty } \right)$ and ionic conductance $\left( {\lambda _{C{l^ - }}^\infty } \right)$ at infinite dilution, we will get
Equivalent conductance at infinite dilution$\left( {\Lambda _{eq}^c} \right)$
= 43.0 + 65
= 108
To find the degree of dissociation $\left( \alpha \right)$
We know, degree of dissociation is the ratio of equivalent conductance at concentration to the equivalent conductances at infinite dilution i.e.
Degree of dissociation\[\left( \alpha \right) = \dfrac{{\Lambda _{eq}^c}}{{\Lambda _{eq}^\infty }}\]
By putting the value, we will get
Degree of dissociation $\left( \alpha \right) = \dfrac{{92}}{{108}}$ = 0.85

Note: One gram equivalent of an electrolyte carries the same charge in solution i.e. one faraday of charge on cation and one faraday of charge on anions. Since the total charge in solution is the same, equivalent conductivity is a useful parameter to compare the conducting power of electrolytes. Higher the value of equivalent conductivity of an electrolyte, higher is its conducting power.