Kohlrausch law-

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Applications of Kohlrausch law-

1)  Calculation of degree of ionisation and ionisation constant-

Degree of ionisation ‘∝’ =  Λm /  Λ0m

Λm   = molar conductance at any conc.

Λ0m = molar conductance at infinite dilution

∝ =  Λeq /  Λ0eq

ionisation constant= C ∝2 /(1 -∝)

2) Calculation of solubility product of sparingly soluble salt-

Λeq = K x 1000 /C

C = conc. of solution or molarity

Since solution is very dilute so,

Λeq =   Λ0eq

Λ0eq = K x 1000 / C

We can calculate solubility of sparingly soluble salt by this formula.

C = solubility of salt

Λ0eq can be calculated by Kohlrausch law,

Λ+eq =  Λ0+    +    Λ0

3) Calculation of equivalent or molar conductance of weak electrolyte at infinite dilution-

Weak electrolytes are not completely ionises at any dilution therefore their equivalent or molar conductance can not be determine experimentely . So their molar or equivalent conductance at infinite dilution can be calculated by Kohlrausch law.

Ex 1) Molar conductivity of CH3COOH  can be calculated by Kohlrausch law,

Λ+CH3COOH =  Λ0CH3COO    +    Λ0H+————eq1

The above eq. can be obtained by  knowing the molar conductivity at infinite dilution for strong electrolyte like sodium acetate-

Λ+CH3COONa =  Λ0CH3COO    +    Λ0Na+————eq2

Λ+HCl =  Λ0H+ + Λ0Cl        ————eq3

Λ+NaCl =  Λ0Cl    +    Λ0Na+————eq4

Adding eq (2) and (3) and subtracting eq(4)

Λ+CH3COONa  + Λ+HCl   –  Λ+NaCl        =  Λ0CH3COO    + Λ0Na+ +   Λ0H+ + Λ0Cl–   –  Λ0Cl    –    Λ0Na+

Λ+CH3COONa  + Λ+HCl   –  Λ+NaCl  =  Λ0CH3COO  +  Λ0H+

Ex 2) Molar conductivity of NH4OH can also be calculated by Kohlrausch law-

Λ+NH4OH =  Λ0 NH4 + + Λ0OH ———-eq 1

The above eq. can be obtained by the knowing the molar conductivities at infinite dilution for strong electrolyte NH4Cl, NaOH , NaCl.

Λ+NH4OH =  Λ0OH    +    Λ0NH4+————eq1

The above eq. can be obtained by the knowing of molar conductivity at infinite dilution for strong electrolyte like NH4Cl,NaOH,NaCl.

Λ+NH4Cl  =  Λ0Cl    +    Λ0NH4+————eq2

Λ+NaOH =  Λ0Na+ + Λ0OH        ————eq3

Λ+NaCl =  Λ0Cl    +    Λ0Na+————eq4

Adding eq (2) and (3) and subtracting eq(4)

Λ+NH4Cl  + Λ+NaOH   –  Λ+NaCl        =  Λ0Cl    + Λ0NH4+ +   Λ0Na+ + Λ0OH–   –  Λ0Cl    –    Λ0Na+

Λ+NH4Cl  + Λ+NaOH   –  Λ+NaCl  =  Λ0 OH  +  Λ0NH4+