_bonnie
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Report Thread starter 11 years ago
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The following equation :

5Fe^2^+ + MnO_4^- + 8H^+ \Rightarrow Mn^2+ + 5Fe^3^+ + 4H_2O

Now, according to my book it says
The colour change is from purple to pale pink, but it is usually so faint that the solution appears colourless.


I was doing a past paper, and the question came up asking for the colourchange.
It said "State the colour change observed at the end"
So i thought it was purple to pale pink / colourless.
But the markscheme said "Colourless / Pale green to purple / pink.
The equation was written exactly the same way round as how the book had stated it.

Just wondered if anyone could clear it up for me?
Thanks
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Wenzel
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#2
Report 11 years ago
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Mmm, in my notes it says it goes colourless and then permantly pink at the end point.
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charco
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The equation as shown goes from dark purple to pale pìnk/colourless

The mark scheme is wrong if you have stated the question correctly
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Suspect
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#4
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If its a titration MnO4 is being added to Fe2 solution
so every time you add the MnO4 ion it reacts and goes colourless like you said.
When you reach the end point it won't react anymore because all the Fe2 ions have reacted and been oxidised to Fe3.
The colour at the endpoint is pink because of the unreacted MnO4 added just after the end-point.

Hope thats right and makes sense
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_bonnie
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Ohh, i get it now. I think...
Because you are adding the MnO4, the solution in the flask is green (fe2+). At the end it will be pink (Mn2+)
However, it you were to add the Fe2+ to the MnO4, the solution in the flask would be purple (MnO4), and the end will be pink again (Mn2+).

Thankkkks
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charco
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OK - If you are told which solution is in the conical flask and which is in the burette fair enough.

Adding KMnO4 from a burette into a solution in a conical flask ...

The flask will not change colour until an excess of KMnO4 is present, in which case it will go from pale green to deep purple.
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