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 A_xB_y =
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) = ⁰_Ag+ + ⁰_Cl- = 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.
    ⁰_n+ = 349.8 ^-1 cm²   &   ⁰_OH- = 198.5 ^-1 cm²
     = ⁰_H+ + ⁰_OH-
          = 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
  K_w = [H^-1] [on^-]
         = 1.01 x 10^-7 x 1.01 x 10^-7 = 1.02 x 10^-14 mol/L