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.

   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-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:

   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
        =      = = 0.127
       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



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