Kohlrausch's Law | Chemistry Class 12 - NEET PDF Download

Kohlrausch's Law of Independent Migration of Ions: 

Kohlrausch determine ^₀ values of pairs of some strong electrolytes containing same cation say KF and KCl, NaF and NaCl etc., and found that the difference in ^₀  value in each case remains the same

^⁰_m (KCl) - ^⁰_m (KF) = ^⁰_m (NaCl) - ^⁰_m (NaF)

He also determined ^⁰ values of pairs of strong electrolytes containing same anion say KF and NaF, KCl and NaCl etc. and found that the difference in ^₀ values is each case remains the same.

^⁰_m (KF) - ^⁰_m (NaF) = ^⁰_m (KCl) - ^⁰_m (NaCl)

This experimental data led him to formulate the following law called Kohlrausch's law of independent migration of ions.

Variation with Strong and Weak electrolytes

At infinite dilution when dissociation is complete, every ion makes some definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion which with it is associated and that the molar conductance at infinite dilution for any electrolyte is given by the sum of the contribution of the two ions. Thus,

^⁰_m =λ₊^o  + λ_-^o

Where λ⁰  is the contribution of the cation  is the contribution of the anion towards the molar conductance at infinite dilution. These contributions are called molar ionic conductances at infinite dilution. Thus, λ⁰  is the molar ionic conductance of cation and λ_-^o is the molar ionic conductance of anion, at infinite dilution. The above equation is, however, correct only for binary electrolyte like NaCl, MgSO₄ etc.

Application of Kohlrausch's law: 

(1) Determination of ^⁰_m of a weak electrolyte: In order to calculate ^⁰_m of a weak electrolyte say CH₃COOH, we determine experimentally ^⁰_m values of the following three strong electrolytes:

(a) A strong electrolyte containing same cation as in the test electrolyte, say HCl

(b) A strong electrolyte containing same anion as in the test electrolyte, say CH₃COONa

(c) A strong electrolyte containing same anion of (a) and cation of (b) i.e. NaCl.

^⁰_m of CH₃COOH is the given as:

^⁰_m (CH₃COOH) = ^⁰_m (HCl) + ^⁰_m (CH₃COONa) - ^⁰_m (NaCl)

Proof :

^⁰_m (HCl) =  +  ..................(i)

^⁰_m (CH₃COONa) =  +  ..................(ii)

^⁰_m (NaCl) =  +  ..................(iii)

Adding equation (i) and equation (ii) and subtracting (iii) from them:

^⁰_m (HCl) +

(2) Determination of degree of dissociation (α) :

α =

(3) Determination of solubility of sparingly soluble salt:

The specific conductivity of a saturated solution of the est electrolyte (sparingly soluble) made in conductivity water is determined by the method as described above. From this the specific conductivity of conductivity water is deducted.
The molar conductance of the saturated solution is taken to be equal to ^⁰_m as the saturated solution of a sparingly soluble salt is extremely dilute. Hence from equation (4).

^⁰_m =

Where C is the molarity of solution and hence the solubility.

ATLAS