Answer the following: What is Kohlrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte? Explain with an example. - Chemistry

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Answer in Brief

Answer the following:

What is Kohlrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte? Explain with an example.

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Solution

1) Kohlrausch law states that “at infinite dilution each ion migrates independent of co-ion and contributes to total molar conductivity of an electrolyte irrespective of the nature of other ions to which it is associated.”

2) Both cation and anion contribute to molar conductivity of the electrolyte at zero concentration and thus ∧0 is the sum of molar conductivity of cation and that of the anion at zero concentration.

Thus, Λ0 `= "n"_+lambda_+^0 + "n"_(_) lambda_(-)^0`

where λ+ and λ_ are molar conductivities of cation and anion, respectively, n+ and n are the number of moles of cation and anion specified in the chemical formula of the electrolyte.

3) Determination of molar conductivity of weak electrolyte at zero concentration:
The theory is particularly useful in calculating ∧0 values of weak electrolytes from those of strong electrolytes.

For example, ∧0 of acetic acid can be calculated by knowing those of HCl, NaCl and CH3COONa as described below:

Λ(HCl) + Λ(CH3COONa) - Λ(NaCl)

`= lambda_("H"^+)^0 + lambda_("Cl"^-)^0 + lambda_("CH"_3"COO"^-)^0 + lambda_("Na"^+)^0 - lambda_("Na"^+)^0 - lambda_("Cl"^-)^0`

`= lambda_("H"^+)^0 + lambda_("CH"_3"COO"_-)^0` = Λ(CH3COONa)

Thus, Λ(CH3COONa) = Λ(HCl) + Λ(CH3COONa) - Λ(NaCl).

Because Λvalues of strong electrolytes, HCl, CH3COONa and NaCl, can be determined by extrapolation method, the Λ0 of acetic acid can be obtained.

Concept: Electrical Conductance of Solution
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Chapter 5: Electrochemistry - Exercises [Page 118]

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Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 5 Electrochemistry
Exercises | Q 4.01 | Page 118

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