Graseax
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2.33g of a sample of Ethanedioic acid, H2C2O4•nH2O, was dissolved in water and the solution mad early up to 250cm^3
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.

H2C2O4•nH2O+2NaOH—-> Na2C2O4+H2O+nH2O

Calculate n ???
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Maker
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You need to find the molecular mass of ethanoic acid without the water of crystalisation and use that find out how many moles of acid is in 2.33gm.
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Kian Stevens
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I'm guessing 'mad early' means 'made nearly'...


Moles of NaOH neutralised = \frac{25}{1000} * 0.110 = 2.75x10-3 mol
\therefore 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) = \frac{250}{18.6} * 1.38x10^{-3} = 18.5x10-3 mol
\therefore Mr of the hydrate = \frac{2.33}{18.5x10^{-3}} = 126 gmol-1

Mr of anhydrous ethanedioic acid = 90 gmol-1 \therefore Mr of water of crystallisation = 126 - 90 = 36 gmol-1
\therefore n = \frac{36}{18} = 2 - this correlates with the fact that ethanedioic acid typically exists as the dihydrate.


Hence, the overall equation is: (COOH)2.2H2O + 2NaOH \rightarrow Na2C2O4 + 4H2O
Last edited by Kian Stevens; 2 years ago
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