# JEE Main & Advanced Chemistry Electrochemistry Kohlrausch's Law

Kohlrausch's Law

Category : JEE Main & Advanced

(1) Kohlrausch law states that, “At time infinite dilution, the molar conductivity of an electrolyte can be expressed as the sum of the contributions from its individual ionsi.e., $\Lambda _{m}^{\infty }={{\nu }_{+}}\,\lambda _{+}^{\infty }+{{\nu }_{-}}\lambda _{-}^{\infty }$, where, ${{\nu }_{+}}$ and ${{\nu }_{-}}$ are the number of cations and anions per  formula unit of electrolyte respectively and,$\lambda _{+}^{\infty }$ and $\lambda _{-}^{\infty }$ are the molar conductivities of the cation and anion at infinite dilution respectively. The use of above equation in expressing the molar conductivity of an electrolyte is illustrated as,

The molar conductivity of HCl at infinite dilution can be expressed as,

$\Lambda _{HCl}^{\infty }={{\nu }_{{{H}^{+}}}}\lambda _{{{H}^{+}}}^{\infty }+{{\nu }_{C{{l}^{-}}}}\lambda _{C{{l}^{-}}}^{\infty }$;  For HCl, ${{\nu }_{{{H}^{+}}}}=1$ and ${{\nu }_{C{{l}^{-}}}}=1.$ So, $\Lambda _{HCl}^{\infty }=(1\times \lambda _{{{H}^{+}}}^{\infty })+(1\times \lambda _{C{{l}^{-}}}^{\infty })$;  Hence, $\Lambda _{HCl}^{\infty }=\lambda _{{{H}^{+}}}^{\infty }+\lambda _{C{{l}^{-}}}^{\infty }$

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

(i) Determination of $\Lambda _{m}^{\infty }$ for weak electrolytes : The molar conductivity of a weak electrolyte at infinite dilution $(\Lambda _{m}^{\infty })$ cannot be determined by extrapolation method. However, $\Lambda _{m}^{\infty }$ values for weak electrolytes can be determined by using the Kohlrausch's equation.

$\Lambda _{C{{H}_{3}}COOH}^{\infty }=\Lambda _{C{{H}_{3}}COONa}^{\infty }+\Lambda _{HCl}^{\infty }-\Lambda _{NaCl}^{\infty }$

(ii) Determination of the degree of ionisation of a weak electrolyte : The Kohlrausch's law can be used for determining the degree of ionisation of a weak electrolyte at any concentration. If $\lambda _{m}^{c}$ is the molar conductivity of a weak electrolyte at any concentration C and, $\lambda _{m}^{\infty }$ is the molar conductivity of a electrolyte at infinite dilution. Then, the degree of ionisation is given by, ${{\alpha }_{c}}=\frac{\Lambda _{m}^{c}}{\Lambda _{m}^{\infty }}=\frac{\Lambda _{m}^{c}}{({{\nu }_{+}}\lambda _{+}^{\infty }+{{\nu }_{-}}\lambda _{-}^{\infty })}$

Thus, knowing the value of $\Lambda _{m}^{c}$, and $\Lambda _{m}^{\infty }$ (From the Kohlrausch's equation), the degree of ionisation at any concentration $({{\alpha }_{c}})$ can be determined.

(iii) Determination of the ionisation constant of a weak electrolyte : Weak electrolytes in aqueous solutions ionise to a very small extent. The extent of ionisation is described in terms of the degree of ionisation $(\alpha ).$In solution, the ions are in dynamic equilibrium with the unionised molecules. Such an equilibrium can be described by a constant called ionisation constant. For example, for a weak electrolyte AB, the ionisation 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, $[AB]=C(1-\alpha ),$ $[{{A}^{+}}]=C\alpha$ and $[{{B}^{-}}]=C\alpha$

Then, the ionisation constant of AB is given by, $K=\frac{[{{A}^{+}}][{{B}^{-}}]}{[AB]}=\frac{C\alpha .C\alpha }{C(1-\alpha )}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}$

We know, that at any concentration C, the degree of ionisation $(\alpha )$ is given by, $\alpha =\Lambda _{m}^{c}/\Lambda _{m}^{\infty }$

Then, $K=\frac{C{{(\Lambda _{m}^{c}/\Lambda _{m}^{\infty })}^{2}}}{[1-(\Lambda _{m}^{c}/\Lambda _{m}^{\infty })]}=\frac{C{{(\Lambda _{m}^{c})}^{2}}}{\Lambda _{m}^{\infty }(\Lambda _{m}^{\infty }-\Lambda _{m}^{c})}$; Thus, knowing $\Lambda _{m}^{\infty }$ and $\Lambda _{m}^{c}$ 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 $(\Lambda _{m}^{\infty })$ can be obtained from the relationship,

$\Lambda _{m}^{\infty }={{\nu }_{+}}\lambda _{+}^{\infty }+{{\nu }_{-}}\lambda _{-}^{\infty }$                                                        ........(i)

The conductivity of the saturated solution of the sparingly soluble salt is measured. From this, the conductivity of the salt $({{\kappa }_{salt}})$can be obtained by using the relationship, ${{\kappa }_{\text{salt}}}={{\kappa }_{\text{sol}}}-{{\kappa }_{\text{wate}r}}$, where, ${{\kappa }_{\text{water}}}$ is the conductivity of the water used in the preparation of the saturated solution of the salt.

$\Lambda _{\text{salt}}^{\infty }=\frac{1000\,{{\kappa }_{\text{salt}}}}{{{C}_{m}}}$                                                           ........(ii)

From equation (i) and (ii) ;

${{C}_{m}}=\frac{1000\,{{\kappa }_{\text{salt}}}}{({{\nu }_{+}}\,\lambda _{+}^{\infty }+{{\nu }_{-}}\,\lambda _{-}^{\infty })}$,  ${{C}_{m}}$ is the molar concentration of the sparingly soluble salt in its saturated solution. Thus, ${{C}_{m}}$ 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 ${{C}_{m}}$ with the molar mass of the salt.

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