 # kohlrausch law applications

12 years ago

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 if is 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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