# Kohlrausch's Law

## Kohlrausch's Law

In 1974, Kohlrausch formulated the law of independent migration of ions based on the experimental data of conductivities aof various electrolytes and stated that-*The equivalent conductivity of an electrolyte at infinite dilution (Λ*.

_{o}) is the sum of the ionic conductivities of their cations and anionsΛ

_{o}= λ

_{+}+ λ

_{−}

where- λ

_{+}and λ

_{−}are cationic and anionic conductivities at infinite dilution respectively.

Kohlrausch's Law can also be states in terms of molar conductivities as-

*The limiting molar conductivity of n electrolyte is the sum of individual contributions of limiting molar conductivities of its constituent ions.*

Thus, molar equivalent conductivity of an electrolyte-

μ

_{o}= n

_{+}μ

_{o}

^{+}+ n

_{−}μ

_{o}

^{−}

where, n

_{+}and n

_{−}are the coefficients of positive and negative ions formed during the dissociation of electrolytes and μ

_{o}

^{+}and μ

_{o}

^{−}are limiting molar conductivities of cations and anions respectively.

## Applications of Kohlrausch's Law

Useful incalculating equivalent conductivity at infinite dilution.

Calculation of degree of dissociation of an electrolyte.

Calculation of solubility of sparingly soluble salt,

Calculation of ionic product of water.

### Q. Calculate the limiting molar conductivity of CH_{3}COOH. The molar conductivities of CH_{3}COONa, HCl and NaCl at infinite dilution are 90.1 S.cm^{2}.mol^{-1}, 426.16 S.cm^{2}.mol^{-1} and 126.45 S.cm^{2}.mol^{-1} respectively.

Given-λ

_{CH3COONa}= 90.1 S.cm

^{2}.mol

^{-1}

λ

_{HCl}=426.16 S.cm

^{2}.mol

^{-1}

λ

_{NaCl}=126.45 S.cm

^{2}.mol

^{-1}

According to Kohlrausch law-

λ

_{CH3COOH}= λ

_{CH3COONa}+ λ

_{HCl}– λ

_{NaCl}

or, λ

_{CH3COOH}= 90.1 + 426.16 – 126.45

or, λ

_{CH3COOH}= 390.71 S.cm

^{2}.mol

^{-1}

So, the limiting molar conductivity of CH

_{3}COOH is, 390.71S.cm

^{2}.mol

^{-1}.

### Q. From the given molar conductivities at infinite dilution, calculate λ_{m} for NH_{4}OH.

λ_{m} for Ba(OH)_{2} = 457.6 ohm^{-1} cm^{2}mol^{-1}

λ_{m} for Ba(Cl)_{2} = 240.6 ohm^{-1} cm^{2}mol^{-1}

λ_{m} for NH_{4}Cl = 129.8 ohm^{-1} cm^{2}mol^{-1}

Answer: 238.3 ohm^{-1}cm

^{2}mol

^{-1}

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