The Student Room Group

A-Level Chemistry Question

This question is about compounds containing ethanedioate ions. 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.

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 60 oC 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
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

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

Give your answer to the appropriate number of significant figures. Show your working.

Solution:
https://drive.google.com/open?id=1ei...0rOE383l1dCPfl
https://drive.google.com/open?id=1AS...FEzeN0NohDOJ1M

Can someone explain why do you do:

mol sodium ethanedioate = total moles ethanedioate mol acid

= 1.325 x 10-2 5.225 x 10-3 = 8.025 x 10-3
In the first titration, the KMnO4 will react with ethanedioate ions. Hence total number of ethanedioate ions from both sodium ethanedioate and ethanedioic acid is found:
n(C2O42-) = 1.33 x 10^-2 mol

In the second titration, the NaOH will react with the ethanedioic acid.
2NaOH + H2C2O4 -> Na2C2O4 + 2H2O
n(H2C2O4) = 5.23 x 10^-3 mol

1.33 x 10^-2 mol = ethanedioate from Na2C2O4 + ethanedioate from H2C2O4
5.23 x 10^-3 mol = ethanedioate from H2C2O4
hence ethanedioate from Na2C2O4 = 1.325 x 10-2 5.225 x 10-3 = 8.025 x 10-3 mol
hence number of moles of Na2C2O4 = 8.025 x 10^-3 mol
Ah! So just to confirm, in the first reaction MnO4 will react with both the acid and the dehydrate but in the second only with the acid, also why do we not account for the unreacted sodium ethandioate in the second titration.
Original post by BobbJo
In the first titration, the KMnO4 will react with ethanedioate ions. Hence total number of ethanedioate ions from both sodium ethanedioate and ethanedioic acid is found:
n(C2O42-) = 1.33 x 10^-2 mol

In the second titration, the NaOH will react with the ethanedioic acid.
2NaOH + H2C2O4 -> Na2C2O4 + 2H2O
n(H2C2O4) = 5.23 x 10^-3 mol

1.33 x 10^-2 mol = ethanedioate from Na2C2O4 + ethanedioate from H2C2O4
5.23 x 10^-3 mol = ethanedioate from H2C2O4
hence ethanedioate from Na2C2O4 = 1.325 x 10-2 5.225 x 10-3 = 8.025 x 10-3 mol
hence number of moles of Na2C2O4 = 8.025 x 10^-3 mol
Also how do you infer that the MnO4 reacts with both the acid and dehydrate, the equation or prior knowledge? does the .2h20(dehydrate) have anything to do with it?
Original post by DoctorFeuer
Ah! So just to confirm, in the first reaction MnO4 will react with both the acid and the dehydrate but in the second only with the acid, also why do we not account for the unreacted sodium ethandioate in the second titration.
Original post by DoctorFeuer
Ah! So just to confirm, in the first reaction MnO4 will react with both the acid and the dehydrate but in the second only with the acid, also why do we not account for the unreacted sodium ethandioate in the second titration.

[dihydrate not dehydrate]

Yes, in the first titration the KMnO4 reacts with ethanedioate from both the acid and the sodium salt

In the second titration, NaOH will not react with sodium ethanedioate. NaOH is a base. It will react with acids. Only acid present is ethanedioic acid. Hence NaOH reacts with ethanedioic acid and the number of moles of ethanedioic acid is found
Original post by DoctorFeuer
Also how do you infer that the MnO4 reacts with both the acid and dehydrate, the equation or prior knowledge? does the .2h20(dehydrate) have anything to do with it?

The equation. MnO4- reacts with ethanedioate ions as shown in the equation.
2H2O does not affect the reaction. It only affects the molar mass of the (hydrated) acid.
Thanks so much. Also Ik its dihydrate, I think autocorrect changed it to dehydrate lol
Original post by BobbJo
[dihydrate not dehydrate]

Yes, in the first titration the KMnO4 reacts with ethanedioate from both the acid and the sodium salt

In the second titration, NaOH will not react with sodium ethanedioate. NaOH is a base. It will react with acids. Only acid present is ethanedioic acid. Hence NaOH reacts with ethanedioic acid and the number of moles of ethanedioic acid is found

The equation. MnO4- reacts with ethanedioate ions as shown in the equation.
2H2O does not affect the reaction. It only affects the molar mass of the (hydrated) acid.
(edited 5 years ago)

Quick Reply