Limiting molar conductivity of CH_3COOH (i.e. | vedclass

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(B) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of a weak electrolyte like $CH_3COOH$ can be calcul

Vedclass · Accepted Answer

$\Lambda^o_{m(CH_3COONa)} + \Lambda^o_{m(HCl)} - \Lambda^o_{m(NaCl)}$

Vedclass · Answer

(B) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of a weak electrolyte like $CH_3COOH$ can be calculated using the limiting molar conductivities of strong electrolytes.$\Lambda^o_{m(CH_3COOH)} = \lambda^o_{CH_3COO^-} + \lambda^o_{H^+}$To obtain this,we combine the strong electrolytes $CH_3COONa$,$HCl$,and $NaCl$ as follows:$\Lambda^o_{m(CH_3COONa)} = \lambda^o_{CH_3COO^-} + \lambda^o_{Na^+}$$\Lambda^o_{m(HCl)} = \lambda^o_{H^+} + \lambda^o_{Cl^-}$$\Lambda^o_{m(NaCl)} = \lambda^o_{Na^+} + \lambda^o_{Cl^-}$By performing the operation $\Lambda^o_{m(CH_3COONa)} + \Lambda^o_{m(HCl)} - \Lambda^o_{m(NaCl)}$,we get:$(\lambda^o_{CH_3COO^-} + \lambda^o_{Na^+}) + (\lambda^o_{H^+} + \lambda^o_{Cl^-}) - (\lambda^o_{Na^+} + \lambda^o_{Cl^-}) = \lambda^o_{CH_3COO^-} + \lambda^o_{H^+} = \Lambda^o_{m(CH_3COOH)}$Thus,the correct option is $B$.

Limiting molar conductivity of $CH_3COOH$ (i.e.,$\Lambda^o_{m(CH_3COOH)}$) is equal to:

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