Frans_M_Everaerts_Isotachophoresis_378342.pdf

IONIC MOBILITY AND IONIC EQUIVALENT CONDUCTIVITY
3.3. IONIC MOBIUTY AND IONIC EQUIVALENT CONDUCTIVITY
We speak of an 'equivalent weight' of an electrolyte if, for complete dissociation,
the total amounts of positive and negative charges are eN and -eN respectively, where
N is Avogadro's number and e is the electronic charge. For example, one equivalent
weight of potassium fluoride gives, for complete dissociation, one Avogadro's number
of K’ and of F ions.
The conductance of such an amount of electrolyte is the conductance measured in a
conductance cell with electrodes 1 cm apart and with such cross-sections that the volume
of solution between the electrodes will contain exactly one equivalent of the electrolyte.
This conductance is known as the 'equivalent conductance' and is denoted by A*.
Kohlrausch showed that at a fned temperature the relationship between the equivalent
conductance of an electrolyte and the square root of the concentration is nearly linear,
especially for very low concentrations and strong electrolytes. At infinite dilution, the
equivalent conductances can be interpreted in terms of ionic contributions, whereby the
contribution of an ion is independent of the other ionic species of the electrolyte (the
influence of retardation and relaxation effects can be neglected, as no ionic atmosphere is
present at infinite dilution). At infinite dilution, we can therefore write
A: = Ax’ + Ax- (3.6)
where Ax’ and Ax- are the equivalent ionic conductivities of the anions and cations,
respectively, and A: is the equivalent conductance, all at infinite dilution.
If a voltage V is applied to a cell as mentioned above (see Fig.3.2) a current I flows
through the cell:
I= V/R or I= VA* (3.7)
Assuming that such a cell contains one equivalent of the electrolyte, N/z' positive and
N/z- negative ions are present, where z’ and z- are the valences of the positive and
negative ions, respectively. If the velocities of the ions are represented by v’ and v-,
respectively, the positive ions present in volume B and the negative ions present in
volume C (see Fig.3.2) will have passed the cross-section A in 1 sec. Because the cell is
I cm in length, this means that the volumes B and C will contain v+/l and v-/l parts of
the total amount of the positive and negative ions of the cell, whch is
v’(N/z’) positive ions and v-(N/z-) negative ions (3.9)
The currents corresponding to these flow rates are obtained by multiplying by the ionic
charges ez' and ez- and by this:
r' = ez' v’ (IV/z+) = e Nu+ = Fv'
r = ez- v- (N/z-) = eNv- = Fv-
At infinite dilution, combination of eqns. 3.8 and 3.10 gives
29
(3.10a)
(3. lob)