Engineering Chemistry S Datta

216 ENGINEERING CHEMISTRY
Highlight:
Kohlrausch’s law can be stated in another way: At infinite dilution, when the
dissociation of an electrolyte is considered complete, each ion makes a definite
contribution towards the molar conductance of the electrolyte irrespective of its
association with other ions. Thus, λ m
∞
= r +
. λ +
∞
+ r –
. λ –
∞
where r +
and r –
are numbers of cations and anions and λ &∞
+ and λ &∞ – are molar
conductances of cations and anions respectively. Examples are: λ
∞
m
(KCl) = λ∞
K
∞
λ 3 PO4
+ λ ∞ Cl
– and λ
∞
m
(K 3
PO 4
) = 3λ ∞
K +
+ ′ −
Ionic Conductance of Some ions at 25°C
Na + = 50.11 ohm –1 Br – = 78.4 ohm –1
K + = 53.22 ohm –1 Cl – = 76.34 ohm –1
NH
+
4
= 73.40 ohm –1 NO
–
3
= 71.44 ohm –1
Applications of Kohlrausch’s law
(i) Determination of ionic conductances. Ionic conductance possesses a constant
value at a definite temperature. Its unit is ohm –1 cm –2 and is directly proportional to the
speeds of the ions.
λ c
∝ u or λ c
= ku
λ a
∝ v or λ a
= kv
where k is a constant.
And λ ∞
= λ c
+ λ a
= ku + kv = k(u + v)
∴
λ
c kv
=
λ ku ( + v) u
= t
u+
v +
∞
And λ a
λ
∞
kv v
= = = t
ku ( + v)
u+
v –
= (1 – t +
)
So, the ionic conductances of cations and anions can be determined from their
experimentally determined transport numbers from the above relations.
(ii) Determination of λ ∞
of weak electrolytes: λ ∞
for weak electrolytes cannot be
determined experimentally, because they do not ionise to a sufficient extent in solution,
except at a infinite dilution. But λ ∞
for weak electrolytes can be calculated from
Kohlrausch’s law.
Example: The λ ∞
values for CH 3
COONa, HCl and NaCl are x, y and z, respectively.
Calculate λ ∞
for CH 3
COOH.
We have
(i) λ Na
+ + λ CH 3 COO
− = x
(ii) λ H
+ + λ Cl
− = y
(iii) λ Na
+ + λ Cl
– = z.
So, (λ + λ + λ + λ ) – (λ + + λ
Na + CH COO–
H + Cl − Na Cl – ) = (x + y – z) = ( λ
− + λ + )
CH3COO
H
3
λ ∞
(CH 3
COOH)
(iii) Ionic mobility (Absolute velocity of ions)
We have,
Ah = t I t
+
F . 1000
C