Posted by gss on Sunday, March 13, 2011 at 5:48am.
Ethanol + acetic acid ==> ethyl acetate + H2O
Ba(OH)2 + 2CH3COOH ==> Ba(CH3COO)2 + 2H2O
moles Ba(OH)2 = 0.02880*0.1040 = 0.002995.
moles CH3COOH remaining after the rxn (in the aliquot taken for analysis) = 2*0.002995 = 0.00599 moles.
Since that was exactly 1/100 of the equilibrium mixture, mol CH3COOH after the rxn = 0.599. I don't have enough room to write the equation unless we shorten how we write the chemicals. Therefore CH3COOH becomes HAc and ethanol becomes EtOH.
............HAc + EtOH ==> EtAc + H2O
initial..... 1.0...0.50.....0......0
change.......-x......-x......+x....+x
final........0.599....0.099..401...401
So if we start with 1.0 mol HAc and we have 0.599 at equilibrium, that means x, the amount reacted must be 0.401 and you can complete the ICE table as I've done above.
Then substitute equilibrium values into the Kc expression for the reaction and calculate the Keq.
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