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Question: The equivalent at dilution of a weak acid such as\(HF\) : (A). Can be determined by extrapolation ...

The equivalent at dilution of a weak acid such asHFHF :
(A). Can be determined by extrapolation of measurements on dilute solutions of HClHCl , HBrHBr .
(B). Can be determined by measurement on very dilute HFHF solutions .
(C). Can best be determined from measurement on dilute solutions of NaFNaF ,NaClNaCl and HClHCl .
(D). Is an undefined quantity .

Explanation

Solution

We will look at the kohlrausch’s law that defines that the limiting molar conductivity is the sum of the limiting molar conductivity of its constituents .Therefore the given acid can be dissociated into ions and can be compared with equivalent conductance of dilute solutions of NaFNaF , NaClNaCl and HClHCl respectively .

Complete step by step answer:
At infinite dilution , when the dissociation is complete , each ion makes a definite contribution towards equivalent conductance of the electrolyte . The value of the equivalent conductance at infinite dilution is calculated according to the kohlrausch’s law .
Kohlrausch’s law states that the equivalent conductivity of an electrolyte is the sum of the conductances of the anions and cations at infinite dilution .
The weak acid such as HFHF can be dissociated into ions H+{H^ + } and F−{F^ - } .
That is ,
HF→H++F−HF \to {H^ + } + {F^ - } --------(1)
To calculate the equivalent conductance of HFHF we will make use of the dilute solutions of NaFNaF , NaClNaCl and HClHCl respectively .
NaFNaF dissociates into Na+N{a^ + } , P−{P^ - } .
That is ,NaF→Na++F−NaF \to N{a^ + } + {F^ - } --------(2)
NaClNaCl forms Na+N{a^ + } and Cl−C{l^ - } ions. Similarly HClHCl will form H+{H^ + } and Cl−C{l^ - } ion.
The reaction are :
NaCl→Na++Cl−NaCl \to N{a^ + } + C{l^ - } --------(3)
HCl→H++Cl−HCl \to {H^ + } + C{l^ - } ----------(4)
Now comparing (1)with (2,3,4).
We will get it .
HF→(NaF−NaCl+HCl)HF \to \left( {NaF - NaCl + HCl} \right)
We will add (NaFNaF and HClHCl (2) and (4) then we will subtract it from (3).
We get ,
HF↛Na++F−+H++̸Cl−−̸Na+−̸Cl−HF \to \not{Na^+ } + {F^ - } + {H^ + } + \not{Cl^-} - \not{Na^+} - \not{Cl^-}
HF→H++F−HF \to {H^ + } + {F^ - } which is equal to (1).
So , the equivalent conductance of
HF=equivalent conductance of[NaF−NaCl+HCl] HF = {\text{equivalent conductance of}} \left[ {NaF - NaCl + HCl} \right]{\text{ }} .
Hence , option C is correct .

Note:
The conductivity of the ions is constant at infinite dilution and it does not depend on the nature of the co – ions .Each ion has a defined contribution towards equivalent conductance irrespective of its nature.