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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
10-2x K (Sm-1
1.237 11.85 23.15
Calculate for all conc. & find0& slope.
Ans: 1 S cm-2= 100 S m-1
 = - A ................................... (i)
Slope = - A
m= (S cm2mol-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 CH
3
COOH.  (HCl) = 425 -1cm2mol-1 (NaCl) = 188-1cm2mol-1
(CH3COONa) = 96
-1cm2mol-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
-1cm2mol-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 ifis known.
Illustration
: Conductivity of 0.001 M CH3COOH is 4.95 x 10-5S cm-1
. Calculate its dissociation constant. Gijven for acetic acid,
0is 390.5 S cm2mol-1.
Ans:== = 49.5 S cm2mol-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 -1cm-1. Find its solubility
 (Ag+) = 61.9-1cm2mol-1 & (Cl-) = 76.3-1cm2mol-1
Ans: (AgCl) = 0Ag+ +0Cl- = 61.9 + 76.3 = 138.2-1cm2mol-1
Solubility == = 10-5mol L-1= 10-5x 143.5 g/L = 1.435 x 10-3g/L
(5) Calculation of Ionic product of H2O
:
Ionic conductance of H+& on-at infinite dil. 0n+ = 349.8-1cm2 & 0OH- = 198.5-1cm2
 =0H+ +0OH- = 349.8 + 198.5 = 548.7-1cm2
Sp. conductance of pure water at 298 K is found to be K = 5.54 x 10-8 -1cm-1
= K x Molarity i.e. [H-1] or [on-] =
=
= 1.01 x 10-7mol/L
 Kw= [H-1] [on-] = 1.01 x 10-7x 1.01 x 10-7= 1.02 x 10-14mol/L
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