Question

Ionic conductance of \[{H^ + }\] and \[S{O_4}^{2 - }\] are \[x\] and \[y{\text{ S c}}{{\text{m}}^2}mo{l^{ - 1}}\]. Find the equivalent conductivity of \[{H_2}S{O_4}\].

Answer 1

Hint: Sulphuric acid is an ionic compound and its equivalent conductivity can be found by obtaining a relationship between its conductivity and that of its constituent ions. The stoichiometric numbers associated with the ions must be kept in mind.

Complete answer:
Kohlrausch's law of independent migration of ions talks about the disappearance of all interionic effects at infinite dilution allowing all the ions to move freely and independently of their counter ions (ions containing opposite charges). This law establishes a relationship between the conductivity of ions migrating independently in a solution and that of its compound.
Sulphuric acid when hydrolyzed gives two moles of protons and one mole of sulphate ions per mole.
\[{H_2}S{O_4}(aq) \to 2{H^ + }(aq) + S{O_4}^{2 - }(aq)\]
Thus equivalent conductivity of an electrolyte can be calculated by adding the products of conductivity of each contributing ion and its stoichiometric number.
\[{\text{Conductivity of electrolyte AB = a}}A + bB\]
Where, \[{\text{a}}\]is the stoichiometric number of A ions, \[b\] is the stoichiometric number of B ions, \[A\] is the conductivity of A ions migrating independently and \[B\] is the conductivity of B ions migrating independently.
The equivalent conductivity of sulphuric acid can be obtained as follows:
\[{\text{Conductivity of }}{H_2}S{O_4}{\text{ = 2}} \times {\text{(conductivity of }}{{\text{H}}^ + }) + 1 \times ({\text{conductivity of S}}{{\text{O}}_4}^{2 - })\]
The given conductivities of individual ions can be substituted from the question. And therefore the equivalent conductivity of sulphuric acid is:
\[{\text{Conductivity of }}{H_2}S{O_4}{\text{ = 2}}x + y{\text{ Sc}}{{\text{m}}^2}mo{l^{ - 1}}\]

Note:
Kohlrausch's law of independent migration of ions works at infinite dilution only because Oswald's law assumes that an electrolyte dissociates completely into its constituent ions only when the solution achieves infinite dilution.