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SOLUTIONS OF ELECTROLYTES
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Electrolysis + - + - Cations Anions Cathode (-) reduction Anode (+) oxidation Battery Electrons Ohm’s law transference number 1 coul = 3x10 9 esu Electrolytic cell
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Faraday’s Laws The passage of 96,500 coulombs (= F) of electricity through a conductivity cell produces a chemical change of 1 gram equivalent weight of any substance Charge of electron: 1 gram equivalent 6. 10 23 charges e=96500/ 6. 10 23 =1.6x10 -19 coul/electron =4.8x10 -10 esu/electon
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Theory of Electrolytic Dissociation Arrhenius Theory: Strong electrolytes H 2 O + Na + Cl - = Na + + Cl - + H 2 O H 2 O + HCl = H 3 O + + Cl - Weak electrolytes H 2 O + CH 3 COOH = H 3 O + + CH 3 COO - + - +-+- -+-+ + -- + + -+-+ +-+- + - + - - + + ElectrolytePolar solvent
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Ionization Degree AB = A + + B - = [A + ]/[AB] o = [B - ]/[AB] o [AB] o = [A + ] + [AB] Strong electrolytes: 0.3 Weak electrolytes: < 0.3
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Weak Electrolytes AB = A + + B -
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Colligative Properties i 2 for NaCl and CaSO 4 i 3 for CaCl 2 and K 2 SO 4 i 4 for FeCl 3 and K 3 Fe(CN) 6
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Strong Electrolytes + - + + - + + + + + + + + + - - - - - - - - - + - Ion “atmosphere”
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Debye-Huckel Theory + + + + + + + + + - - - - - - - - - + - Ionic Strength Low concentration of electrolyte High concentration of electrolyte
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Solubility Product What is the solubility of silver chromate (s, M) in aqueous solution containing 0.04M silver nitrate ? Ag 2 CrO 4 = 2Ag + + CrO 4 2- K sp = 2x10 -12 (Reference literature) K sp = [Ag + ] 2 [CrO 4 2- ] = (2s + 0.04) 2. s s = [Ag 2 CrO 4 ] 2 x 10 -12 /1.6 x 10 -3 = 1.25 x 10 -9 M AgCl solid = Ag + + Cl -
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Effect of Ionic Strength on Solubility Calculate solubility of silver chloride (s, M) in 0.1M ammonium sulfate. Ionic strength of 0.1 M (NH 4 ) 2 SO 4 = 0.3; activity coefficient = 0.7. AgCl = Ag + + Cl - K sp = 1.76x10 -10 (Reference literature) s 2 = K sp / 2 s = (1.76 x 10 -10 ) 1/2 /0.7 = 1.9 x 10 -5 M [Ag + ]. [Cl - ] = K sp / 2 AgCl solid = Ag + + Cl - K sp = a Ag +. a Cl - = + [Ag + ]. - [Cl - ]
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