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Applications:(1) Calculation of Molar Conductivity at infinite dilution (0) for weak electrolytes:Illustration: Calculate molar conductance at infinite dilution for CH3COOH. (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:
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
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