Given limiting molar conductivities (in S cm^2 mol^-1): HCl = 426.2, NaCl = 126.5, CH3COONa = 91.0. The limiting molar conductivity of CH3COOH is

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Given limiting molar conductivities (in $S c m^{2} mo l^{- 1}$ ): $HCl = 426.2$ , $NaCl = 126.5$ , $CH X_{3} COONa = 91.0$ . The limiting molar conductivity of $CH X_{3} COOH$ is

By Kohlrausch's law,

Λ^{\circ} (CH X_{3} COOH) = λ^{\circ} (H X^{+}) + λ^{\circ} (CH X_{3} COO X^{-}) = Λ^{\circ} (HCl) + Λ^{\circ} (CH X_{3} COONa) - Λ^{\circ} (NaCl)

= 426.2 + 91.0 - 126.5 = 390.7 S c m^{2} mo l^{- 1}

. NaCl is subtracted because

Na X^{+}

and

Cl X^{-}

must be removed.