kohlrausch law applications

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 -1}cm^-1. Find its solubility
(Ag⁺) = 61.9^-1cm²mol^-1  &  (Cl^-) = 76.3^-1cm²mol^-1
Ans:  (AgCl) =⁰_Ag+ +⁰_Cl- = 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.    ⁰_n+ = 349.8^-1cm²  &  ⁰_OH- = 198.5^-1cm²


=⁰_H+ +⁰_OH-          = 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 -1}cm^-1


= K x     Molarity i.e.   [H^-1] or [on^-] =
=
= 1.01 x 10^-7mol/L
  K_w= [H^-1] [on^-]         = 1.01 x 10^-7x 1.01 x 10^-7= 1.02 x 10^-14mol/L

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