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.

Which ion has exceptionally higher ­\(\lambda _{\infty}\) value?

H⁺

The correct answer is option 1. \(H^+\).

The limiting ionic conductivity, \(\lambda _{\infty}\), is the value of the ionic conductivity at infinite dilution. It is a measure of the mobility of an ion in a solution.

The H⁺ ion has an exceptionally higher \(\lambda _{\infty}\) value because it is the smallest ion and has a single positive charge. This makes it very mobile in solution.

The other ions have larger sizes and/or multiple charges, which makes them less mobile in solution.

Here is a table of the limiting ionic conductivities of some common ions:

As you can see, the H⁺ ion has a significantly higher limiting ionic conductivity than the other ions.