Question

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of$\text{ A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}\text{ }$. Given that $\text{ A}{{\text{l}}^{\text{3+}}}\text{ }$ and $\text{ SO}_{4}^{2-}\text{ }$ are the equivalent conductance at infinite dilution of the respective ions?

A) $\text{ }\Lambda _{\text{ A}{{\text{l}}^{\text{3+}}}\text{ }}^{0}+\text{ }\Lambda _{\text{ SO}_{4}^{2-}\text{ }}^{0}$

B)$\text{ (}\Lambda _{\text{ A}{{\text{l}}^{\text{3+}}}\text{ }}^{0}+\text{ }\Lambda _{\text{ SO}_{4}^{2-}\text{ }}^{0})\text{ }\times \text{ 6 }$

C) $\text{ }\frac{1}{3}\Lambda _{\text{ A}{{\text{l}}^{\text{3+}}}\text{ }}^{0}+\text{ }\frac{1}{2}\Lambda _{\text{ SO}_{4}^{2-}\text{ }}^{0}$

D) $\text{ 2}\Lambda _{\text{ A}{{\text{l}}^{\text{3+}}}\text{ }}^{0}+\text{ 3}\Lambda _{\text{ SO}_{4}^{2-}\text{ }}^{0}$

Answer 1

The conductance of a solution which contains the one equivalent weight of electrolyte is known as the equivalent conductance. It is denoted by the $\text{ }\Lambda \text{ }$. At infinite dilution the conductance of the electrolyte is equal to the sum of the conductance of the ions (cation and anions) in the solution. $\text{ }\\!\\!\Lambda\\!\\!\text{ }_{\text{m}}^{\text{0}}\text{ = }\\!\\!\lambda\\!\\!\text{ }_{\text{+}}^{\text{0}}\text{ + }\\!\\!\lambda\\!\\!\text{ }_{-}^{0}\text{ }$

Where, $\text{ }\\!\\!\Lambda\\!\\!\text{ }_{\text{m}}^{\text{0}}$ is equivalent conductance of electrolyte at infinite dilution, $\text{ }\\!\\!\lambda\\!\\!\text{ }_{\text{+}}^{\text{0}}$ and $\text{ }\\!\\!\lambda\\!\\!\text{ }_{-}^{0}\text{ }$ is the conductance of cation and anion.

Complete step by step answer:

The electricity passes through the solution of electrolytes due to the migration of ions when the potential difference is applied between the two electrodes.

The ease of a solution to conduct electricity is known as the conductance of the solution.

The conductance of the volume of a solution that contains the one equivalent weight of an electrolyte which is placed between the electrodes 1 meter apart and has a cross-sectional area equal to $\text{ 1 }{{\text{m}}^{\text{2}}}\text{ }$ is known as the equivalent conductance.

Let suppose that $\text{ 1 }{{\text{m}}^{3}}\text{ }$the volume of solution contains the 1 gram equivalent of the solution then conductance of solution will be equal to equivalent conductance, $\text{ }\Lambda \text{ }$ .

At infinite dilution, when electrolyte undergoes complete dissociation, each ion makes a definite contribution towards the equivalent conductance or towards the molar conductance of electrolyte irrespective of the nature of the other ion.

The conductance of a solution of an electrolyte is given by the sum of the contributions of the two ions. For example, let an electrolyte’ then conductance at the infinite dilution is:

$\text{ }\\!\\!\Lambda\\!\\!\text{ }_{\text{m}}^{\text{0}}\text{ = }\\!\\!\lambda\\!\\!\text{ }_{\text{+}}^{\text{0}}\text{ + }\\!\\!\lambda\\!\\!\text{ }_{-}^{0}\text{ }$

Where, $\text{ }\\!\\!\Lambda\\!\\!\text{ }_{\text{m}}^{\text{0}}$ is equivalent conductance of electrolyte at infinite dilution, $\text{ }\\!\\!\lambda\\!\\!\text{ }_{\text{+}}^{\text{0}}$ and $\text{ }\\!\\!\lambda\\!\\!\text{ }_{-}^{0}\text{ }$is the conductance of cation and anion.

We have given the $\text{ A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}\text{ }$electrolyte. It dissociates into its corresponding $\text{ A}{{\text{l}}^{\text{3+}}}\text{ }$ and $\text{ SO}_{4}^{2-}\text{ }$ as shown below,

$\text{ A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}\text{ }\to \text{ 2 A}{{\text{l}}^{\text{3+}}}\text{ + 3 SO}_{4}^{2-}\text{ }$

At infinite dilution, the equivalent conductance $\text{ A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}\text{ }$would be equal to the conductance of ions in the solution. It is given as follows,

$\text{ }\Lambda _{\text{ A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}\text{ }}^{\infty }=\text{ 2 }\Lambda _{\text{ A}{{\text{l}}^{\text{3+}}}\text{ }}^{0}+\text{ 3 }\Lambda _{\text{ SO}_{4}^{2-}\text{ }}^{0}$

Since we know that the equivalent conductance is given only concerning ions of the solution (irrespective of the number of ions) thus equivalent conductance at infinite dilution is as:

$\text{ }\Lambda _{\text{ A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}\text{ }}^{\infty }=\text{ }\Lambda _{\text{ A}{{\text{l}}^{\text{3+}}}\text{ }}^{0}+\text{ }\Lambda _{\text{ SO}_{4}^{2-}\text{ }}^{0}$

So, the correct answer is “Option A”.

Note: The conductivity per unit concentration is expressed as equivalent conductance or molar conductance. The relation between the equivalent conductance and the molar conductance is as shown below,

$\text{ }{{\Lambda }_{\text{eq}}}\text{ = }\left( \frac{1}{{{\text{n}}^{+}}\text{ + }{{\text{Z}}^{\text{+}}}} \right)\times {{\Lambda }_{\text{m}}}\text{ }$

Where, ${{\text{n}}^{+}}$ is the number of cations, and ${{\text{Z}}^{\text{+}}}$is the charge on the cation. Through this relation, the equivalent conductance and the molar conductance of electrolyte can be determined.