Kohlrausch law is useful in calculating ­\(\Lambda ^0\) for any electrolyte from the ­\(\lambda ^0\) of individual ions. The molar conductivities of H⁺ and OH⁻ ions are very high because these ions are passed from one molecule to another and released at the electrodes without travelling. Equivalent conductance of weak electrolytes can be calculated from the conductances of completely dissociated strong electrolytes e.g.,

\(\Lambda^0_{CH_3COOH} = \Lambda^0_{CH_3COONa} + \Lambda^0_{HCl} - \Lambda^0{NaCl}\)

Or,\(\Lambda^0_{CH_3COOH} = \Lambda^0_{Na^+} + \Lambda^0_{CH_3COO^-} + \Lambda^0_{H^+} + \Lambda^0_{Cl^-} - \Lambda^0_{Na^+} - \Lambda^0_{Cl^-}\)

Or,\(\Lambda^0_{CH_3COOH} = \Lambda^0_{CH_3COO^-} + \Lambda^0_{H^+} \)

Solubility of sparingly soluble salts can be calculated from the specific conductance of its saturated solution and from the equivalent conductivity at infinite dilution obtained from ­ ­

\[\Lambda ^0_e = \frac{1000 K_{salt}}{C}\]

\[\text{Absolute ionic mobility =} \frac{\text{Ionic conductance}}{96500} \]

\[\text{Absolute ionic mobility =} \frac{\Lambda _0}{96500} \]

Using ionic conductance measurements, the ionic product of water can be determined as 1 × 10⁻¹⁴ at 25°C.

The speed of migration of Ag⁺ ion and NO₃⁻ ion are 0.00057cm sec⁻¹ and 0.00063cm sec^-1 at infinite dilution. The equivalent conductance of AgNO₃ at infinite dilution is

115.8

The correct answer is option 4. 115.8.

We have,

\(\lambda ^{\infty}_{cation} = u^{\infty}_{cation} × Faraday\)

or, \(\lambda ^{\infty}_{cation} = 0.00057 × 96500\)

or, \(\lambda^{\infty}_{Ag^+} = 55.005 S\text{ }cm^2\text{ }mol^{-1}\)

Also,

\(\lambda ^{\infty}_{anion} = u^{\infty}_{anion} × Faraday\)

or, \(\lambda ^{\infty}_{anion} = 0.00063 × 96500\)

or, \(\lambda^{\infty}_{NO_3^-} = 60.795 S\text{ }cm^2\text{ }mol^{-1}\)

Now from Kohlrausch law,

\(\Lambda ^{\infty}_{AgNO_3} = \lambda^{\infty}_{Ag^+} + \lambda^{\infty}_{NO_3^-} = 55.005 + 60.795\)

\(\Lambda ^{\infty}_{AgNO_3} = 115.8 \text{ }S\text{ }cm^2\text{ }mol^{-1}\)