Tough AQA A Level question

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b.r.2018
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#1
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#1
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.

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

Can someone try this question and see if they get the same answer, as I have no access to the MS?
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charco
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#2
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#2
(Original post by b.r.2018)
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.

My attempt: https://drive.google.com/open?id=1qL...B_-DAwjLX2Xe9w

Can someone try this question and see if they get the same answer, as I have no access to the MS?

Calculating total ethanedioate ions

mol manganate(VII) = 0.02 x 0.0265 = 5.3 x 10-4
mol ethanedioate in 25ml = 5/2 x 5.3 x 10-4 = 1.325 x 10-3
Hence mol ethanedioate in 250 ml = 1.325 x 10-2

This includes both ethanedioate ions and ethanedioic acid

Calculating ethanedioic acid

mol NaOH = 0.1 x 0.01045 = 0.001045 mol
mol acid in 25 ml = 0.001045/2 = 5.225 x 10-4
Hence mol acid in 250 ml = 5.225 x 10-3


mol sodium ethanedioate = total moles ethanedioate – mol acid

= 1.325 x 10-2 – 5.225 x 10-3 = 8.025 x 10-3

Mr of sodium ethanedioate = 134

Mass sodium ethanedioate = 1.075g


Total mass = 1.90g

Mass sodium ethanedioate = 1.075g

Percentage by mass sodium ethanedioate = 100 x 1.075/1.90 = 56.6%
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Wolfram Alpha
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#3
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(Original post by charco)
Calculating total ethanedioate ions

mol manganate(VII) = 0.02 x 0.0265 = 5.3 x 10-4
mol ethanedioate in 25ml = 5/2 x 5.3 x 10-4 = 1.325 x 10-3
Hence mol ethanedioate in 250 ml = 1.325 x 10-2

This includes both ethanedioate ions and ethanedioic acid

Calculating ethanedioic acid

mol NaOH = 0.1 x 0.01045 = 0.001045 mol
mol acid in 25 ml = 0.001045/2 = 5.225 x 10-4
Hence mol acid in 250 ml = 5.225 x 10-3


mol sodium ethanedioate = total moles ethanedioate – mol acid

= 1.325 x 10-2 – 5.225 x 10-3 = 8.025 x 10-3

Mr of sodium ethanedioate = 134

Mass sodium ethanedioate = 1.075g


Total mass = 1.90g

Mass sodium ethanedioate = 1.075g

Percentage by mass sodium ethanedioate = 100 x 1.075/1.90 = 56.6%
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

I understand everything up to that point...
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charco
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#4
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(Original post by Wolfram Alpha)
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

I understand everything up to that point...
In the redox reaction the manganate(VII) reacts with both the ethanedioate ions and the ethanedioic acid

But in the acid-base reaction the sodium hydroxide reacts with only the ethanedioic acid.
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Wolfram Alpha
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#5
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(Original post by charco)
In the redox reaction the manganate(VII) reacts with both the ethanedioate ions and the ethanedioic acid

But in the acid-base reaction the sodium hydroxide reacts with only the ethanedioic acid.
Sorry, I don't understand - would you not add them together instead?
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charco
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#6
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(Original post by Wolfram Alpha)
Sorry, I don't understand - would you not add them together instead?
You want the value of the ethanedioate ions only

Manganate(VII) ==> mol ethanedioate + ethanedioic acid
Sodium hydroxide ==> mol ethanedioic acid

Hence Manganate(VII) - Sodium hydroxide ==> mol ethanedioate
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Wolfram Alpha
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#7
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(Original post by charco)
You want the value of the ethanedioate ions only

Manganate(VII) ==> mol ethanedioate + ethanedioic acid
Sodium hydroxide ==> mol ethanedioic acid

Hence Manganate(VII) - Sodium hydroxide ==> mol ethanedioate
Ohhhh I get you, thanks a lot!
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jazz_xox_
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#8
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(Original post by charco)
Calculating total ethanedioate ions

mol manganate(VII) = 0.02 x 0.0265 = 5.3 x 10-4
mol ethanedioate in 25ml = 5/2 x 5.3 x 10-4 = 1.325 x 10-3
Hence mol ethanedioate in 250 ml = 1.325 x 10-2

This includes both ethanedioate ions and ethanedioic acid

Calculating ethanedioic acid

mol NaOH = 0.1 x 0.01045 = 0.001045 mol
mol acid in 25 ml = 0.001045/2 = 5.225 x 10-4
Hence mol acid in 250 ml = 5.225 x 10-3


mol sodium ethanedioate = total moles ethanedioate – mol acid

= 1.325 x 10-2 – 5.225 x 10-3 = 8.025 x 10-3

Mr of sodium ethanedioate = 134

Mass sodium ethanedioate = 1.075g


Total mass = 1.90g

Mass sodium ethanedioate = 1.075g

Percentage by mass sodium ethanedioate = 100 x 1.075/1.90 = 56.6%
Hi! I know this post was a while ago but I just have a few questions because I didn't understand this question at all when I first did it (understand it a bit more from your post though )

From the first titration, can you conclude that the C2O4 2- ions come from BOTH the dicarboxylic acid, and the sodium ethanedioate?
Is this because the dicarboxylic acid is a weak acid so will partially dissociate into ethanedioate ions? (and the Na2C2O4 will dissolve in solution)

And for the second titration, it appears only the acid is reacting, so does this mean the Na2C2O4 just doesn't take place in any reactions here?

Thanks so much
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charco
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#9
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#9
(Original post by jazz_xox_)
Hi! I know this post was a while ago but I just have a few questions because I didn't understand this question at all when I first did it (understand it a bit more from your post though )

From the first titration, can you conclude that the C2O4 2- ions come from BOTH the dicarboxylic acid, and the sodium ethanedioate?
Is this because the dicarboxylic acid is a weak acid so will partially dissociate into ethanedioate ions? (and the Na2C2O4 will dissolve in solution)

And for the second titration, it appears only the acid is reacting, so does this mean the Na2C2O4 just doesn't take place in any reactions here?

Thanks so much
More or less, but the first part is less ot do with the ions and more to do with the fact that both ethandioic acid (and ethandioate ions) have carbon in an oxidation state that can be oxidised by manganate(VII) ions. So they BOTH react with manganate(VII).

In the second part of the reaction only the acid can perform an acid base reaction.
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DoctorFeuer
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#10
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#10
How do you know Manganate (VII) reacts with ethanedioic acid as well as the ethanedioate ions(which is shown in the equation)?
(Original post by charco)
You want the value of the ethanedioate ions only

Manganate(VII) ==> mol ethanedioate + ethanedioic acid
Sodium hydroxide ==> mol ethanedioic acid

Hence Manganate(VII) - Sodium hydroxide ==> mol ethanedioate
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Cheea076
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#11
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Could you please explain what to fo after we calculate mr of sodium ethandioate Sorry ik its very later but i came across this x
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Ayahshxh
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#12
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I know this thread is 2 years old but you did not put brackets when taking away moles of acid in excess and moles left. This therefore gives you a different value than the true value. Other than that the method is spot on.
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greenie emsaph
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#13
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#13
as mol times mr will give the mass of sodium ethandioate
(Original post by Cheea076)
Could you please explain what to fo after we calculate mr of sodium ethandioate Sorry ik its very later but i came across this x
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