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(C) The equivalent conductance at infinite dilution for a weak electrolyte like $HF$ cannot be determined by direct extrapolation of its own conductan

Vedclass · Accepted Answer

Can best be determined from measurements on dilute solutions of $NaF$,$NaCl$ and $HCl$

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(C) The equivalent conductance at infinite dilution for a weak electrolyte like $HF$ cannot be determined by direct extrapolation of its own conductance measurements because the degree of dissociation does not approach $1$ at finite concentrations.According to Kohlrausch's Law of independent migration of ions,the molar conductivity at infinite dilution can be determined using strong electrolytes.Specifically,$\Lambda^{\circ}_{m}(HF) = \Lambda^{\circ}_{m}(NaF) + \Lambda^{\circ}_{m}(HCl) - \Lambda^{\circ}_{m}(NaCl)$.Therefore,the correct approach is to use measurements on dilute solutions of $NaF$,$NaCl$,and $HCl$.

The equivalent conductance at infinite dilution of a weak acid such as $HF$:

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