Finding the Value of Kc

0.1mol of ethanol is mixed with 0.1mol of ethanoic acid and allowed to reach equilibrium. The total volume of the system is made up to 20.0cm³ with an inert solvent (not water, because water is a product of the reaction). By titration it was found that 0.033mol of ethanoic acid was present at equilibrium. At the start, 0.033mol of ethanoic acid is left at equilibrium so (0.1 - 0.033)mol must have reacted.(0.1 - 0.033) = 0.067 so 0.067mol of ethanoic acid has reacted with 0.067mol of ethanol to form 0.067mol of water and 0.067mol of ethyl ethanoate since the ratios are 1: 1: 1: 1 at equilibrium. As the the volume is 20cm³ or 0.02dm³ we can work out the concentrations for the equilibrium expression by dividing the moles by 0.02.