What is the correct expression for Lambda_{m( | vedclass

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(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for a weak electrolyte like $NH_4OH$ ca

Vedclass · Accepted Answer

$\Lambda_{m(NH_4Cl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0$

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(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for a weak electrolyte like $NH_4OH$ can be calculated using strong electrolytes.$\Lambda_{m(NH_4OH)}^0 = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$To obtain this,we combine the molar conductivities of $NH_4Cl$,$NaOH$,and $NaCl$ as follows:$\Lambda_{m(NH_4Cl)}^0 = \lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0$$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$$\Lambda_{m(NaCl)}^0 = \lambda_{Na^+}^0 + \lambda_{Cl^-}^0$By performing the operation $\Lambda_{m(NH_4Cl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0$,we get:$(\lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0) + (\lambda_{Na^+}^0 + \lambda_{OH^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0) = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0 = \Lambda_{m(NH_4OH)}^0$Therefore,the correct expression is $\Lambda_{m(NH_4Cl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0$.

What is the correct expression for $\Lambda_{m(NH_4OH)}^0$?

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