[1z3]

Question:

A 300mg sample of UO(C2O4).6H20 (Mr=450) was dissolved in hot dilute sulphuric acid and titrated with 0.5 mol dm^-3 potassium manganate (VII) solution. The end point was obtained after the addition of 10.70cm^3 of the KMnO4 solution. In this titration the KMnO4 oxidises both the C2O4 2- ions and the uranyl(IV) ions.

Use these data to calculate the change in oxidation state of the uranium and hence deduce the final ixidation state of uranium.

Equations given: MnO4 + 8H+ + 5e- ---> Mn2+ + 4H20
                       C2O4 2- ---> 2CO2 + 2e-

My attempt:

The question is worth 6 marks so there's probably around 6 steps to do. As a result I thought that I couldn't just say that when UO 2+ gives out 3 electrons when oxidised (and then the oxidation state would be +7).

I calculated the no. of moles of MnO4 to be 5.35*10^-3 (n=cv --> 0.5*(10.7/1000))
So the number of electrons it takes is 5.35*10^-3 *5 = 0.02675

I then calculated the number of moles of UO 2+ to be 6.667*10^-4 (from no. of moles = m/Mr where the mass is 300x10^-3 (to be in grams) times the ratio of Mr of anhydrous form to hydrated form). As a result the number of moles of C2O4 2- is also 6.667*10^-4 as when 1 mole of UO(C2O4) dissolves it forms 1 mole of UO 2+ and 1 mole of C2O4 2- aqueous ions.

I then tried to calculate the number of electrons given out per each mole of C2O4 2- and UO 2+ to be 20.06... (I did no of moles of electrons taken in by MnO4 divided by the total number of moles of C2O4 2- plus UO 2+).

I feel like I either am doing strange steps that I don't have to do or I have to manipulate this value of 20.06 to answer the question. However I am stuck.

Help would be very much appreciated.

[Vidya]

You have number of moles of oxalate ion so calculate the number of electrons given by oxalate ion.Subtract this number from the total number of electrons used in the reduction of the MnO4- ion.Now this is number of electrons which must be provided by the UO2+ ions .You its number of moles ..so find the number of electrons given by each mole and use it to calculate the oxidation number. 

[Borek]

I then calculated the number of moles of UO 2+ to be 6.667*10^-4 (from no. of moles = m/Mr where the mass is 300x10^-3 (to be in grams) times the ratio of Mr of anhydrous form to hydrated form).

No idea what you mean by the "ratio of anhydrous to hydrated form". You are just given molar mass, use it without any other tricks.

The problem is the molar mass as given is wrong, it should be 463 g/mol.

And then Vidya's hints are spot on.

[AWK]

Molar mass of  UO(C2O4)·6H2O is 450.13899.
Error is in H20 instead of H2O.
0.5 M KMnO₄ is not used for titration, rather 0.05.

[Borek]

Ah, thanks.

Note to self: never copy/paste formulas to calculate molar mass.

[1z3]

Okay so I calculated the number of moles given out by UO2+ to be 0.0254166666 (by subtracting number of moles of electrons given out by C2O4 2- from the total number of electrons taken in by MnO4). I then divide this by the number of moles of UO2+ to find number of moles of electrons given out per mole of UO2+ to be 38. This number is way too high.

Another commented saying the solution of MnO4 for titration should be 0.05 which would mean 2 moles of electrons are given out per mole of UO2+ so the oxidation state would go from +4 to +6. However the question that I am reading from my paper does say 0.5 mol dm^-3.


What I meant by ratio of molar masses was to work out the number of moles of UO(C2O4) in the 300mg sample of UO(C2O4).6H2O which is (342/450)*300mg = 228mg.

[AWK]

Result for 0.05 M KMnO₄ is reliable but for your starting data (0.5) is a pure nonsense.
This should be a printing error in yout textbook.

[Borek]

Okay so I calculated the number of moles given out by UO2+ to be 0.0254166666 (by subtracting number of moles of electrons given out by C2O4 2- from the total number of electrons taken in by MnO4). I then divide this by the number of moles of UO2+ to find number of moles of electrons given out per mole of UO2+ to be 38. This number is way too high.

Another commented saying the solution of MnO4 for titration should be 0.05 which would mean 2 moles of electrons are given out per mole of UO2+ so the oxidation state would go from +4 to +6. However the question that I am reading from my paper does say 0.5 mol dm^-3.

Must be a mistake in the paper, I am with AWK on this one. Note, that the result you got doesn't make sense, but if you will try 0.05 M you will get a very nice and logical result.

What I meant by ratio of molar masses was to work out the number of moles of UO(C2O4) in the 300mg sample of UO(C2O4).6H2O which is (342/450)*300mg = 228mg.

This is not necessary - number of moles of UO(C₂O₄) in UO(C₂O₄)·6H₂ is exactly the same as the number of moles of UO(C₂O₄)·6H₂O.

[1z3]

Does this mean my calculation for the number of moles of UO 2+ and C2O4 2- is wrong?

EDIT: Don't worry you get the same number of moles either way. I just did an extra step.

Assuming that the proper concentration should be 0.05mol dm^-3, should the new oxidation state of uranium be +6?

[AWK]

5 UO11C2H12 + 4 KMnO4 + 11 H2SO4 = 5 UO2SO4 + 4 MnSO4 + 2 K2SO4 + 41 H2O + 10 CO2
Note: UO11C2H12 = UO(C2O4)·6H2O

Your problem (with concentration 0.5 M) comes from
General Certificate of Education
June 2008
Advanced Extension Award
CHEMISTRY

I did not found no answer sheet or comments for this paper.