Question

Define Kohlrausch’s law?

Answer 1

Hint: According to Kohlrausch’s law “when the process of dissociation is stopped at its infinite dilution, each and every ion present in the solution makes its individual contribution to the molar conductivity of electrolyte irrespective of other ions present in the solution”.

Complete Step By Step Answer:
Kohlrausch’s law relates the ions and molar conductivity of solution. Therefore, it is also known as Kohlrausch’s law of independent migration of ions. This law helps to calculate the molar conductivity of solution by adding the individual contribution of the ions from the electrolytes.
Molar conductivity of anion present in electrolyte is expressed as $ \left( {{\lambda ^ \circ }_ - } \right) $ and Molar conductivity of cation present in electrolyte is expressed $ \left( {{\lambda ^ \circ }_ + } \right) $ . Then molar conductivity of solution at infinite dilution will become-
 $ {\lambda ^ \circ } = \left( {{\lambda ^ \circ }_ - } \right) + \left( {{\lambda ^ \circ }_ + } \right) $
If there are more than one similar ions present in a compound then a term $ \left( v \right) $ is used to denote the number of particular ions. So, the formula modified and become-
 $ {\lambda ^ \circ } = {v_ - }\left( {{\lambda ^ \circ }_ - } \right) + {v_ + }\left( {{\lambda ^ \circ }_ + } \right) $
Where $ \left( {{v_ - }} \right) $ represent the number of anions in a compound and $ \left( {{v_ + }} \right) $ represent the number of cations in a compound.
For example, $ NaCl $ is made up of $ \left( {N{a^ + }} \right) $ and $ \left( {C{l^ - }} \right) $ . Since there are only one cation and one anion the value $ \left( v \right) $ of becomes $ 1 $ . So, the molar conductivity of $ NaCl $ electrolyte solution will become-
 $ {\lambda ^ \circ }_{\left( {NaCl} \right)} = {\lambda ^ \circ }\left( {N{a^ + }} \right) + {\lambda ^ \circ }\left( {C{l^ - }} \right) $
In the case of compound like $ \left[ {A{l_2}{{\left( {S{O_4}} \right)}_3}} \right] $ , it is made up of $ \left( {A{l^{ + 3}}} \right) $ and $ \left( {S{O_4}^{ - 2}} \right) $ ions. But as we see from the molecular formula there are two atoms of $ \left( {A{l^{ + 3}}} \right) $ and three atoms of $ \left( {S{O_4}^{ - 2}} \right) $ . So the value of $ \left( {{v_ + }} \right) $ is $ 2 $ and the value of $ \left( {{v_ - }} \right) $ is $ 3 $ . The molar conductivity of $ \left[ {A{l_2}{{\left( {S{O_4}} \right)}_3}} \right] $ electrolyte solution will become-
 $ {\lambda ^ \circ }_{\left( {NaCl} \right)} = 2 {\lambda ^ \circ }\left( {A{l^{ + 3}}} \right) + 3{\lambda ^ \circ }\left( {S{O_4}^{ - 2}} \right) $ .

Note:
Kohlrausch’s law is used to calculate the degree of dissociation of weak electrolyte because it depends on its molar conductivity. Kohlrausch’s law is also used to calculate the solubility aspect of sparingly soluble salts like silver chloride.