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pranav adithya Grade: 10
        kohlrausch law applications

6 years ago

Answers : (1)

NIKHIL GARG
38 Points
										

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


 


do approve if answer helps you

6 years ago
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