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
& 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-1118.5 = 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.
| eq = + |
& 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-1Ans:
(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:
| Kc = |
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 = = = 0.127K =
= = 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-1Ans:
(AgCl) = 0Ag+ + 0Cl- = 61.9 + 76.3 = 138.2 -1 cm2 mol-1Solubility =
= = 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 cm2Sp. 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