Electro - Chemistry-3
Conductance Behaviour of weak electrolyte:
As weak electrolyte dissociates to a much lesser extent than strong electrolyte. For weak electrolyte, there is a very large increase in conductance with dilution. B/C conc. of electrolyte
& no. of ions in sol.
. So, conductance
.
Illustration: Conductivity of NaCl at 298 K at different conc. is given.
Conc. (m) 0.001 0.01 0.02
10-2 x K (Sm-1 1.237 11.85 23.15
Calculate
for all conc. & find
0 & slope.
Ans: 1 S cm-2 = 100 S m-1
=
- A
................................... (i)
Slope = - A
m =
(S cm2 mol-1)
= 123.7 ............... (ii)
= 118.5 .............. (iii)
123.7 =
- A(0.0316) ...................... (iv)
118.5 =
- A(0.1) ........................... (v)
On solving we get
= 124 S cm2mol-1
118.5 = 124 - A(0.1)
A = 55
Kolilrausch law: Limiting molar conductivity of an electrolyte (molar conductivity at infinite diution) is sum of limiting ionic conductivities of cation & anion each multiplied with no. of ions present in one formula unit of electrolyte.
eg.
for AxBy =
or
Eq. conductivity of an electrolyte at infinite dilution is sum of two values one depending upon cation & other upon anion.
&
Ionic conductivities at infinite dilution.
Applications:
(1) Calculation of Molar Conductivity at infinite dilution (
0) for weak electrolytes:
Illustration: Calculate molar conductance at infinite dilution for CH3COOH.
(HCl) = 425
-1 cm2 mol-1
(NaCl) = 188
-1 cm2 mol-1
(CH3COONa) = 96
-1 cm2 mol-1
Ans:
(CH3COOH) =
(CH3COO-) +
(H+)
Required.
(HCl) =
(H+) +
(Cl) ....................................... (i)
(NaCl) =
(Na+) +
(Cl) .................................... (ii)
(CH3COONa) =
(CH3COO-) +
(Na+) .................. (iii)
By (i) + (iii) - (ii), we get required
(CH3COO-) +
(H+) = 425 + 96 - 188
= 333
-1 cm2 mol-1
(2) Calculation of deg. of dissociation:
in molar conductivity with dilution is due to increase in dissociation of electrolyte.
Deg. of dissociction (
) =
(3) Calculation of dissociation constant:
K can be calculated if
is known.
Illustration: Conductivity of 0.001 M CH3COOH is 4.95 x 10-5 S cm-1. Calculate its dissociation constant. Gijven for acetic acid,
0 is 390.5 S cm2 mol-1.
Ans:
=
=
= 49.5 S cm2 mol-1
=
=
= 0.127
K =
=
= 1.85 x 10-5
(4) Calculation of solubility of sparingly soluble salt:
=
=
Solubility =
Illustration: Conductivity of solurated sol. of Agcl at 288 K is found to be 1.382 x 10-6
-1 cm-1. Find its solubility
(Ag+) = 61.9
-1 cm2 mol-1 &
(Cl-) = 76.3
-1 cm2 mol-1
Ans:
(AgCl) =
0Ag+ +
0Cl- = 61.9 + 76.3 = 138.2
-1 cm2 mol-1
Solubility =
=
= 10-5 mol L-1 = 10-5 x 143.5 g/L
= 1.435 x 10-3 g/L
(5) Calculation of Ionic product of H2O:
Ionic conductance of H+ & on- at infinite dil.
0n+ = 349.8
-1 cm2 &
0OH- = 198.5
-1 cm2
=
0H+ +
0OH-
= 349.8 + 198.5 = 548.7
-1 cm2
Sp. conductance of pure water at 298 K is found to be
K = 5.54 x 10-8
-1 cm-1
= K x
Molarity i.e. [H-1] or [on-] =
=
= 1.01 x 10-7 mol/L
Kw = [H-1] [on-]
= 1.01 x 10-7 x 1.01 x 10-7 = 1.02 x 10-14 mol/L
As weak electrolyte dissociates to a much lesser extent than strong electrolyte. For weak electrolyte, there is a very large increase in conductance with dilution. B/C conc. of electrolyte




Illustration: Conductivity of NaCl at 298 K at different conc. is given.
Conc. (m) 0.001 0.01 0.02
10-2 x K (Sm-1 1.237 11.85 23.15
Calculate


Ans: 1 S cm-2 = 100 S m-1



Slope = - A




123.7 =

118.5 =

On solving we get

118.5 = 124 - A(0.1)


Kolilrausch law: Limiting molar conductivity of an electrolyte (molar conductivity at infinite diution) is sum of limiting ionic conductivities of cation & anion each multiplied with no. of ions present in one formula unit of electrolyte.
eg.


or
Eq. conductivity of an electrolyte at infinite dilution is sum of two values one depending upon cation & other upon anion.
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Applications:
(1) Calculation of Molar Conductivity at infinite dilution (

Illustration: Calculate molar conductance at infinite dilution for CH3COOH.






Ans:













By (i) + (iii) - (ii), we get required


= 333

(2) Calculation of deg. of dissociation:

Deg. of dissociction (


(3) Calculation of dissociation constant:
Kc = ![]() |
K can be calculated if

Illustration: Conductivity of 0.001 M CH3COOH is 4.95 x 10-5 S cm-1. Calculate its dissociation constant. Gijven for acetic acid,

Ans:







K =


(4) Calculation of solubility of sparingly soluble salt:





Illustration: Conductivity of solurated sol. of Agcl at 288 K is found to be 1.382 x 10-6





Ans:




Solubility =


= 10-5 mol L-1 = 10-5 x 143.5 g/L
= 1.435 x 10-3 g/L
(5) Calculation of Ionic product of H2O:
Ionic conductance of H+ & on- at infinite dil.







= 349.8 + 198.5 = 548.7

Sp. conductance of pure water at 298 K is found to be
K = 5.54 x 10-8



Molarity i.e. [H-1] or [on-] =

=


= 1.01 x 10-7 x 1.01 x 10-7 = 1.02 x 10-14 mol/L