Limiting Molar Conductance
Limiting
Molar Conductance
From Kohrausch’s equation (eqn. 9) and
Onsager’s equation (eqn. 10) when C=0, m =m .. Thus extrapolation to
zero concentration (infinite dilution) givesm .. the limiting molar
conductance at infinite dilution. As figure 2. shows, this is practicable with
strong electrolytes. In the case of weak electrolytes, there is strong
dependence of m on concentration at low concentrations and hence
extrapolations do not give correct m, values. For weak electrolytes
an indirect method is used to obtain m values.
Table Molar Conductances for various Electrolytes at Infinite Dilution at 250C
Electrolyte
Pairs
|
m/-1cm2mol-1
|
Difference/-1cm2mol-1
|
KCI
NaCl
KNO3
NaNO3
KCI
KNO3
BaCl2
Ba(NO3)2
|
149.46
126.45
144.96
121.55
149.86
144.96
139.94
135.04
|
23.41
23.41
4.90
4.90
|
For pairs of salts that either have a
common anion or a common cation (Table 3) the difference between the m
values are constant. For those that have a common anion, the differences in m
values can be attributed to cations and similarly for those with a common
cation, the difference in m values is due to the anions.
These observations led Kohlrausch to
postulate Kohlrausch’s law of independent migration of ions. “Each ion is
assumed to make its own contribution to the molar conductance irrespective of
the nature of the other ion with which it is associated”. Thus in general;
m
=v++ + v--
(11)
where + and - are the ionic conductances of the cation and
anion, respectively at infinite dilution and v+ and v-
are the numbers of moles of cations and anions needed to form one mole of the
salt (e.g. v+ = v- =1 for NaCI, but v+ 1, v-
= 2 for MgCl2).
Table Limiting Molar Conductances of Ions at 25°C
Cation
|
+/-1cm2mol-1
|
Anion
|
-/-1cm2mol-1
|
H+
Na+
K+
Ag+
Mg2+
|
349.8
50.11
73.52
61.92
53.06
|
OH-
Cl-
NO3-
SO42-
CH3COO-
|
197.6
76.34
71.44
80.0
40.9
|
Using the values in table 4, the
limiting molar conductances of various electrolytes may be determined, e.g.
m .(KCI)=k+Cl
= 73.52 + 76.34
=149.86-1cm2mol-1
and
m (KCI)-m
(NaCl) = K+ + Cl- - Na+
- Cl-
=
K+ - Na+
=
73.52 – 50.00
=23.42
-1cm2mol-1
This difference, 23.42 -1cm2mol-1
will be the same whatever the nature of the anion. .
unabQer_i-Portland Brenda Anderson https://wakelet.com/wake/a_dkhLn2sj881lxsI8lam
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