Kohlrausch law- Applications

Kohlrausch law-

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Applications of Kohlrausch law-

1)  Calculation of degree of ionisation and ionisation constant-

Degree of ionisation ‘∝’ =  Λ_{m /  Λ⁰m}

Λ_m   = molar conductance at any conc.

Λ⁰_m = molar conductance at infinite dilution

∝ =  Λ_{eq /  Λ⁰eq}

ionisation constant= C ∝2 /(1 -∝)

2) Calculation of solubility product of sparingly soluble salt-

Λ_{eq = K x 1000 /C}

C = conc. of solution or molarity

Since solution is very dilute so,

Λ_{eq =   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 =  Λ⁰_CH3COO^–    +    Λ⁰_H⁺————eq1

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

Λ⁺_CH3COONa =  Λ⁰_CH3COO^–    +    Λ⁰_Na⁺————eq2

Λ⁺_HCl =  Λ⁰_H⁺ + Λ⁰_Cl^–        ————eq3

Λ⁺_NaCl =  Λ⁰_Cl^–    +    Λ⁰_Na⁺————eq4

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

Λ⁺_CH3COONa  + Λ⁺_HCl   –  Λ⁺_NaCl        =  Λ⁰_CH3COO^–    + Λ⁰_Na⁺ +   Λ⁰_H⁺ + Λ⁰_Cl–   –  Λ⁰_Cl^–    –    Λ⁰_Na⁺

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 ⁺ + Λ⁰_OH^– ———-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 =  Λ⁰_OH^–    +    Λ⁰_NH4⁺————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  =  Λ⁰_Cl^–    +    Λ⁰_NH4⁺————eq2

Λ⁺_NaOH =  Λ⁰_Na⁺ + Λ⁰_OH^–        ————eq3

Λ⁺_NaCl =  Λ⁰_Cl^–    +    Λ⁰_Na⁺————eq4

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

Λ⁺_NH4Cl  + Λ⁺_NaOH   –  Λ⁺_NaCl        =  Λ⁰_Cl^–    + Λ⁰_NH4⁺ +   Λ⁰_Na⁺ + Λ⁰_OH–   –  Λ⁰_Cl^–    –    Λ⁰_Na⁺

Λ⁺_NH4Cl  + Λ⁺_NaOH   –  Λ⁺_{NaCl  =  Λ⁰ OH^–  +  Λ⁰NH4⁺}