Which relation is correct for Lambda^0_{m(H_2 | vedclass

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(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the mol

Vedclass · Accepted Answer

$\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}$

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(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.For water $(H_2O)$,the dissociation is $H_2O \rightleftharpoons H^+ + OH^-$.Therefore,$\Lambda^0_{m(H_2O)} = \lambda^0_{H^+} + \lambda^0_{OH^-}$.We can obtain this by combining strong electrolytes:$\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}$$\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$$\Lambda^0_{m(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-}$By calculating $\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}$,we get:$(\lambda^0_{H^+} + \lambda^0_{Cl^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{Cl^-}) = \lambda^0_{H^+} + \lambda^0_{OH^-} = \Lambda^0_{m(H_2O)}$.Thus,option $(B)$ is correct.

Which relation is correct for $\Lambda^0_{m(H_2O)}$?

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