KOHLRAUSCH'S LAW [ ELECTROCHEMISTRY]

KOHLRAUSCH'S LAW  and   APPLICATIONS
  

Kohlrausch's Law

In terms of molar conductivities:-

(a)Statement :-

The limiting molar conductivity of an electrolyte( i.e., molar conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte.

(b)Mathematically:-

Λ⁰ₘ for AₓB_y = xλ⁰ ₊ + yλ⁰ _-
Where,
Λ⁰ₘ = Limiting molar conductivity of electrolyte
AₓB_y = Ionic compound i.e. Chemical formula of an electrolyte
x = Valency of B
y = Valency of A
λ⁰ ₊ = Limiting molar conductivity of cation (A^y+)
λ⁰ _- = Limiting molar conductivity of anion (B^x-)

(c) Examples :-

(i) Λ⁰ₘ for NaCl = λ⁰ _Na+ + λ⁰ _Cl-

(ii) Λ⁰ₘ for BaCl₂ = λ⁰ _Ba²⁺ + 2λ⁰ _Cl^-
(iii) Λ⁰ₘ for Al₂(SO₄)₃ = 2λ⁰ _Al³⁺ + 3λ⁰ _SO4^2-

In terms of equivalent conductivities:-

(a)Statement:-

The equivalent conductivity of an electrolyte at infinite dilution is the sum of two values one depending upon the cation and the other upon the anion.

(b) Mathematically:-

Λ⁰eq = λ⁰ _c + λ⁰ _a
Where,
λ⁰ _c = Limiting ionic conductivities for cation
λ⁰ _a = Limiting ionic conductivities for anion

 Calculation of Λ⁰eq from λ⁰ _m values of ions :-

Λ⁰eq(NaCl) = λ⁰ _m(Na⁺) + λ⁰ _m(Cl^-)
Λ⁰eq(BaCl₂) = 1/2 λ⁰ _m(Ba²⁺) + λ⁰ _m(Cl^-)
Λ⁰eq(AlCl₃) = 1/3 λ⁰ _m(Al³⁺) + λ⁰ _m(Cl^-)
Λ⁰eq(Al₂(SO₄)₃) = 1/3 λ⁰ _m(Al³⁺) + 1/2λ⁰ _m(SO₄^2-)

Relation between molar conductivity (Λₘ) and equivalent conductivity (Λeq)

Λ⁰eq = Λ⁰ₘ /n = Λ⁰ₘ /total charge on anion or cation = Λ⁰ₘ / valency
For example:-
(i) For the salt, sodium potassium oxalate (NaOOC-COOK) , Λ⁰eq = Λ⁰ₘ/2, [ as charge on cations = charge on anions = 2 ]
(ii) For the salt, Potash alum, K₂SO₄.Al₂(SO₄)₃.24H₂O ,
Λ⁰eq = Λ⁰ₘ/8 [ as total charge on cations = total charge on anions = 8 ]

Applications of Kohlrausch's Law

1. Calculation of molar conductivity at infinite dilution for weak electrolyte

The molar conductivity of a weak electrolyte at infinite dilution cannot be determined experimentally due to following two reasons :-
(i) The conductance of weak electrolyte is low.
(ii) The dissociation of weak electrolyte is not complete even at very high dilutions.
The molar conductivity of weak electrolyte at infinite dilution can be calculated using Kohlrausch's law.

EXAMPLE 1. If the molar conductivities at infinite dilution of NaCl, HCl, and CH₃COONa (NaAc) are 126.4, 425.9 and 91.0 S cm² mol^-1 respectively, what will be that of acetic acid (HAc) ?

Solution:-
Λ⁰ₘ for NaCl = λ⁰_Na⁺ + λ⁰_Cl^- = 126.4 Scm² mol^-1 ....(i)
Λ⁰ₘ for HCl = λ⁰_H⁺ + λ⁰_Cl^- = 425.9 S cm² mol^-1 ...(ii)
Λ⁰ₘ for CH₃COONa = λ⁰_CH3COO^- + λ⁰_Na⁺ = 91.0 Scm² mol^-1 ......(iii)
We have to find, Λ⁰ₘ for CH₃COOH = λ⁰_CH3COO^- + λ⁰_H⁺ = ?
Now applying, [(ii) + (iii)] - (i),
 ⟹ [λ⁰_H⁺ + λ⁰_{Cl^{- +}} λ⁰_CH3COO^- + λ⁰_Na⁺ ] - [λ⁰_Na⁺ + λ⁰_Cl^-  ] = [425.9 S cm² mol^-1  + 91.0 Scm² mol^-1] - 126.4 Scm² mol^-1

⟹ λ⁰_CH3COO^- + λ⁰_H⁺ = = 390.5 S cm² mol^-1
⟹ Λ⁰ₘ for CH₃COOH = 390.5 S cm² mol^-1

2. Calculation of the Degree of Dissociation

Degree of dissociation (α ) = Λₘ