Question

The equivalent at dilution of a weak acid such as$HF$ :
(A). Can be determined by extrapolation of measurements on dilute solutions of $HCl$ , $HBr$ .
(B). Can be determined by measurement on very dilute $HF$ solutions .
(C). Can best be determined from measurement on dilute solutions of $NaF$ ,$NaCl$ and $HCl$ .
(D). Is an undefined quantity .

Answer 1

Hint: 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 $NaF$ , $NaCl$ and $HCl$ 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 $HF$ can be dissociated into ions ${H^ + }$ and ${F^ - }$ .
That is ,
$HF \to {H^ + } + {F^ - }$ --------(1)
To calculate the equivalent conductance of $HF$ we will make use of the dilute solutions of $NaF$ , $NaCl$ and $HCl$ respectively .
$NaF$ dissociates into $N{a^ + }$ , ${P^ - }$ .
That is ,$NaF \to N{a^ + } + {F^ - }$ --------(2)
$NaCl$ forms $N{a^ + }$ and $C{l^ - }$ ions. Similarly $HCl$ will form ${H^ + }$ and $C{l^ - }$ ion.
The reaction are :
$NaCl \to N{a^ + } + C{l^ - }$ --------(3)
$HCl \to {H^ + } + C{l^ - }$ ----------(4)
Now comparing (1)with (2,3,4).
We will get it .
$HF \to \left( {NaF - NaCl + HCl} \right)$
We will add ($NaF$ and $HCl$ (2) and (4) then we will subtract it from (3).
We get ,
$HF \to \not{Na^+ } + {F^ - } + {H^ + } + \not{Cl^-} - \not{Na^+} - \not{Cl^-}$
$HF \to {H^ + } + {F^ - }$ which is equal to (1).
So , the equivalent conductance of
$HF = {\text{equivalent conductance of}} \left[ {NaF - NaCl + HCl} \right]{\text{ }}$ .
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.