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state Kohlrausche's law? what are its applications?
Kohlrausch’s law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If a salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations. 

Applications of Kohlrausch's law: Some typical applications of the Kohlrausch's law are described below,

(i) Determination of ??m for weak electrolytes: The molar conductivity of a weak electrolyte at infinite dilution (??m) cannot be determined by extrapolation method. However, ??m values for weak electrolytes can be determined by using the Kohlrausch's equation.

??CH3 COOH = ??CH3COONa + ??CHI - ??NaCI

(ii) Determination of the degree of ionization of a weak electrolyte: The Kohlrausch's law can be used for determining the degree of ionization of a weak electrolyte at any concentration. If ?cm is the molar conductivity of a weak electrolyte at any concentration C and, ?cm is the molar conductivity of a electrolyte at infinite dilution. Then, the degree of ionization is given by, ac = ?cm = ??m = ?cm/(v+ ??+ + v- ??-)

Thus, knowing the value of ?cm , and ??m (From the Kohlrausch's equation), the degree of ionization at any concentration (ac) can be determined.

(iii) Determination of the ionization constant of a weak electrolyte : Weak electrolytes in aqueous solutions ionize to a very small extent. The extent of ionization is described in terms of the degree of ionization ( a ). In solution, the ions are in dynamic equilibrium with the unionized molecules. Such an equilibrium can be described by a constant called ionization constant. For example, for a weak electrolyte AB, the ionization equilibrium is, AB ? A+ + B-; If C is the initial concentration of the electrolyte AB in solution, then the equilibrium concentrations of various species in the solution are, [AB} = C (1 - A), [A+] = Ca and [B-] = Ca

Then, the ionisation constant of AB is given by, K = [A+] [B-]/[AB] = Ca.Ca/C (1 - A) = Ca2/( 1 - a )

We know, that at any concentration C, the degree of ionisation ( a ) is given by, a = ?cm / ??m

Then, (K) = C(?cm/??m)2/[1 - (?cm/??m)] = C(?cm)2/??m - ?cm); Thus, knowing ??m and ?cm at any concentration, the ionisation constant (K) of the electrolyte can be determined.

(iv) Determination of the solubility of a sparingly soluble salt : The solubility of a sparingly soluble salt in a solvent is quite low. Even a saturated solution of such a salt is so dilute that it can be assumed to be at infinite dilution. Then, the molar conductivity of a sparingly soluble salt at infinite dilution (??m) can be obtained from the relationship,

??m = V+??+ + V-??- ........(i)

The conductivity of the saturated solution of the sparingly soluble salt is measured. From this, the conductivity of the salt (Ksalt) can be obtained by using the relationship, Ksalt = Ksol - Kwater, where, Kwater is the conductivity of the water used in the preparation of the saturated solution of the salt.

??salt = 1000 ksalt/Cm ........(ii)

From equation (i) and (ii) ;

Cm = 1000 ksalt/(V+??+ + V-??-), Cm is the molar concentration of the sparingly soluble salt in its saturated solution. Thus,Cm is equal to the solubility of the sparingly soluble salt in the mole per litre units. The solubility of the salt in gram per litre units can be obtained by multiplying Cm with the molar mass of the salt.


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