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.

At 18°C the conductances of H⁺ and CH₃COO⁻ atinfinite dilution are 315 and 35­ cm² eq⁻¹, respectively.

The equivalent conductance of CH₃COOH at infinite dilution is

350

The correct answer is option 1. 350.

The equivalent conductance of CH₃COOH at infinite dilution is given by:

\(\lambda^{\infty} = \lambda^{\infty}_{H^+} + \lambda^{\infty}_{CH_3COO^-} \)

where:

\(\lambda^{\infty} \)is the limiting equivalent conductance

\(\lambda^{\infty}_{H^+} \)is the limiting ionic conductance of H⁺ ion

\(\lambda^{\infty}_{CH_3COO^-} \) is the limiting ionic conductance of CH₃COO^- ion

Plugging in the values given in the question, we get:

\(\lambda^{\infty} = 315 + 35 = 350\text{ }\Omega \text{ }cm^2\text{ eq}^{-1}\)