At 298 K, \( \Lambda^{\circ}_{m}(\ce{KCl}) - \Lambda^{\circ}_{m}(\ce{NaCl}) \approx 23.4 \) S cm2 mol−1 and \( \Lambda^{\circ}_{m}(\ce{NaBr}) - \Lambda^{\circ}_{m}(\ce{NaCl}) \approx 1.8 \) S cm2 mol−1. What is the approximate value of \( \Lambda^{\circ}_{m}(\ce{KBr}) - \Lambda^{\circ}_{m}(\ce{NaBr}) \) at 298 K?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 23.4 S cm2 mol−1
By the Kohlrausch law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the contributions of its cation and its anion. So in a difference between a potassium salt and a sodium salt of the same anion, the anion cancels out:
\( \displaystyle \Lambda^{\circ}_{m}(\ce{KBr}) - \Lambda^{\circ}_{m}(\ce{NaBr}) = (\lambda^{\circ}_{\ce{K+}} + \lambda^{\circ}_{\ce{Br-}}) - (\lambda^{\circ}_{\ce{Na+}} + \lambda^{\circ}_{\ce{Br-}}) = \lambda^{\circ}_{\ce{K+}} - \lambda^{\circ}_{\ce{Na+}} \)
\( \displaystyle \Lambda^{\circ}_{m}(\ce{KCl}) - \Lambda^{\circ}_{m}(\ce{NaCl}) = \lambda^{\circ}_{\ce{K+}} - \lambda^{\circ}_{\ce{Na+}} \approx 23.4\ \mathrm{S\ cm^2\ mol^{-1}} \)
Both differences are the same, 23.4 S cm2 mol−1, whatever the anion.
NCERT (2026-27 edition):
2.4.2 Variation of Conductivity and Molar Conductivity with Concentration – “Kohlrausch examined Λ°m values for a number of strong electrolytes and observed certain regularities. He noted that the difference in Λ°m of the electrolytes NaX and KX for any X is nearly constant. For example at 298 K: Λ°m (KCl) – Λ°m (NaCl) = Λ°m (KBr) – Λ°m (NaBr) = Λ°m (KI) – Λ°m (NaI) ≈ 23.4 S cm2 mol–1 and similarly it was found that Λ°m (NaBr) – Λ°m (NaCl) = Λ°m (KBr) – Λ°m (KCl) ≈ 1.8 S cm2 mol–1 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.”