Molar conductivity : Molar conductivity
Slide 2 : Molar conductivity The molar conductance is defined as the conductance of all the ions produced by ionization of 1 g mole of an electrolyte when present in V mL of solution. It is denoted by.
Molar conductance ?m = k ×V
Where V is the volume in mL containing 1 g mole of the electrolyte. If c is the concentration of the solution in g mole per litre, then
?m = k × 1000/c
It units are ohm-1 cm2 mol-1.
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Slide 4 : Cell constant G* = Conductivity X Resistance
=100 O x 1.29 S/m=129 m-1=1.29cm-1
Conductivity of 0.02molL-1 KCl= 129 m-1/520O = 0.248Sm-1
= 0.248x10-2Scm-2
Molar conductivity ?m = k × 1000/c
=0.248 x1000/0.02
=0.248x 0.248x10-2Scm-2x1000cm-3L-1x/0.02molL-1
= 124Scm2mol
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Slide 24 : The degree of ionization is given by
a = ?cm /?8m = ?cm / ( v+ ?8+ + v- ?8- )
Thus, knowing the value of ?cm, and ?8m (From the Kohlrausch's equation), the degree of ionization at any concentration (ac) can be determined.
Slide 25 : ( K ) = C(?cm / ?8m )2 / [ 1 - ( ?cm / ?8m )] = C(?cm)2 / ?8m - ?cm ) We know ?8m and ?cm at any concentration, the ionisation constant (K) of the electrolyte can be determined
Slide 26 : the molar conductivity of a sparingly soluble salt at infinite dilution
?8m = V+?8+ + V-?8-
?8salt = 1000 ksalt / Cm
Cm = 1000 ksalt / ( V+?8+ + V-?8- ),
Cm is the molar concentration of the sparingly soluble salt in its saturated solution.
Thus,Cm is equal to the solubility of the sparingly soluble salt in the mole per litre units. The solubility of the salt in gram per litre units can be obtained by multiplying Cm with the molar mass of the salt. Determination of the solubility of a sparingly soluble salt
Slide 27 : The molar conductivity of a weak electrolyte at infinite dilution (?8m) cannot be determined by extrapolation method. However, ?8m values for weak electrolytes can be determined by using the Kohlrausch's equation.
?8CH3 COOH = ?8CH3COONa + ?8HCI - ?8NaCI Determination of ?8m for weak electrolytes