This solution was titrated against a solution of 0.110 moldm^-3 NaOH. It was found that 18.6cm^3 of the ethanedioic acid solution was needed to neutralise 25.0cm^3 of the NaOH solution.
Calculate n ???
Moles of NaOH neutralised = = 2.75x10-3 mol
moles of ethanedioic acid required for neutralisation = 1.38x10-3 mol
Moles of ethanedioic acid in 250 cm3 (assuming that the solution volume is about 250 cm3) = = 18.5x10-3 mol
Mr of the hydrate = = 126 gmol-1
Mr of anhydrous ethanedioic acid = 90 gmol-1 Mr of water of crystallisation = 126 - 90 = 36 gmol-1
n = = 2 - this correlates with the fact that ethanedioic acid typically exists as the dihydrate.
Hence, the overall equation is: (COOH)2.2H2O + 2NaOH Na2C2O4 + 4H2O