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titration question help (a level)

Hi, I was wondering how you do question 11 on this paper. I worked out the moles of c2o42- (1.325 x 10^–3) and the moles of h2c2o4 (5.225 x 10^–4) but after that I dont know what to do. the ms says take them away from each other but I have no idea why it says this? Why is this done and what does this calculate? many thanks

https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-74051-QP-JUN17.PDF

Reply 1

charco

Study Forum Helper

Original post by Bertybassett

A white solid is a mixture of sodium ethanedioate (Na2C2O4), ethanedioic acid dihydrate (H2C2O4.2H2O) and an inert solid. A volumetric flask contained 1.90 g of this solid mixture in 250 cm3 of aqueous solution.

In other words, its impure acid and ethandioate salt, both of which can undergo oxidation, but only the acid can be neutralised.

Two different titrations were carried out using this solution.

In the first titration 25.0 cm3 of the solution were added to an excess of sulfuric acid in a conical flask. The flask and contents were heated to 60ºC and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour. The equation for this reaction is

2MnO4−+ 5C2O42−+ 16H+→ 2Mn2++ 8H2O + 10CO2

This is the redox reaction that oxidises all of the ethandioic acid AND the sodium ethandioate.

Total moles of ethandioic acid + ethandioate ions = 5/2 (0.02 x 0.0265) = 1.325 x 10^-3

In the second titration 25.0 cm3 of the solution were titrated with a 0.100 mol dm−3 solution of sodium hydroxide using phenolphthalein as an indicator. The indicator changed colour after the addition of 10.45 cm3 of sodium hydroxide solution.

The equation for this reaction is

H2C2O4+ 2OH−→ C2O42−+ 2H2O

This titration gives you only the moles of ethandioic acid as it is a neutralisation reaction. You should be able to work this part out.

Calculate the percentage by mass of sodium ethanedioate in the white solid.

1st titration gives moles of sodium ethanoate + moles of ethandioic acid
2nd titration gives moles of ethandioic acid only
------------------------------------------------------------------------ subtract 2 from 1
moles of sodium ethandioate