# Titration calculations

Watch
Announcements
#1
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 ???
0
2 years ago
#2
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.
2
2 years ago
#3

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
Last edited by Kian Stevens; 2 years ago
1
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (581)
34.2%
Yes, I like the idea of applying to uni after I received my grades (PQA) (698)
41.08%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (342)
20.13%
I think there is a better option than the ones suggested (let us know in the thread!) (78)
4.59%