Limiting molar conductivity of H_2O is equal to:

The correct answer is Option (2) → \(\Lambda^o_m HCl + \Lambda^o_m NaOH − \Lambda^o_m NaCl\)

Kohlrausch’s law of independent migration of ions: the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion. For water, a weak electrolyte, it cannot be measured directly, so it is obtained from strong electrolytes:

    \( \Lambda^o_m(H_2O) = \lambda^o_{H^+} + \lambda^o_{OH^-} \)

Adding HCl and NaOH and removing NaCl leaves only these two ions:

    \( \Lambda^o_m HCl + \Lambda^o_m NaOH - \Lambda^o_m NaCl \)

    \( = (\lambda^o_{H^+} + \lambda^o_{Cl^-}) + (\lambda^o_{Na^+} + \lambda^o_{OH^-}) - (\lambda^o_{Na^+} + \lambda^o_{Cl^-}) \)

    \( = \lambda^o_{H^+} + \lambda^o_{OH^-} = \Lambda^o_m(H_2O) \)

Why the other options are incorrect

• Option (1) \(\Lambda^o_m NaCl + \Lambda^o_m HCl − \Lambda^o_m NaOH\) — Incorrect. It gives \( \lambda^o_{H^+} + 2\lambda^o_{Cl^-} - \lambda^o_{OH^-} \); \( \ce{Cl^{-}} \) does not cancel and \( \ce{OH^{-}} \) is subtracted.

• Option (3) \(\Lambda^o_m HCl + \Lambda^o_m NH_4OH − \Lambda^o_m NaCl\) — Incorrect. It gives \( \lambda^o_{H^+} + \lambda^o_{NH_4^+} + \lambda^o_{OH^-} - \lambda^o_{Na^+} \); \( \ce{NH4^{+}} \) and \( \ce{Na^{+}} \) do not cancel.

• Option (4) \(\Lambda^o_m NaCl + \Lambda^o_m HCl − \Lambda^o_m NH_4OH\) — Incorrect. It gives \( \lambda^o_{Na^+} + \lambda^o_{H^+} + 2\lambda^o_{Cl^-} - \lambda^o_{NH_4^+} - \lambda^o_{OH^-} \); \( \ce{OH^{-}} \) is subtracted instead of added.

NCERT (2026-27 edition):
Variation of Conductivity and Molar Conductivity with Concentration – “… Kohlrausch examined Λ°_m values for a number of strong electrolytes and observed certain regularities. … On the basis of the above observations he enunciated Kohlrausch law of independent migration of ions. The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.”