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VCE Chemistry Question Thread
lzxnl:
--- Quote from: Penguuu on December 28, 2013, 11:45:39 am ---Please help! :) (this thread kind of died, not sure where to post haha)
A solution of potassium permanganate can be standardised using pure iron wire. in a particular experiment 0.317 g of iron wire was dissolved such that Fe2+ ions were formed, and the resulting solution was made up to 250.0 mL.
20ml aliquots of this solution were then taken, with 11.72ml of the permanganate solution being required.
calculate the molarity of the permanganate solution.
--- End quote ---
--- Quote from: DJALogical on December 28, 2013, 12:21:49 pm ---So a Volumetric analysis question:
The reaction between the potassium permanganate and the iron wire can be written omitting spectator ions in the ionic form as follows (Due to the potassium permanganate being composed of K+ and MnO4-)
MnO4- + Fe2+ ---> Fe(MnO4)2
If we balance this equation we put a 2 in front of the MnO4 and get a 2:1 mole ratio:
2MnO4- + Fe2+ ---> Fe(MnO4)2
Keep this in mind for later.
Looking at the question lets convert everything to moles first as it is easier!
n(Fe2+)= 0.317/55.8 = 0.005681mol total in 250.0 mL of solution.
Thus in a 20mL aliquot, we use a down-scaling factor to find out the amount of Fe2+ in 20 mL
ie. n(Fe2+) in a 20mL aliquot = 20/250 x 0.005681 = 0.0004545 mol
Now remember the mole ratio worked out earlier, we can see that we need 2 times more MnO4- than Fe2+ for the reaction to go to completion (the 2:1 ratio).
Thus we can conclude:
n(MnO4-) required = 2 x n(Fe2+)
= 2 x 0.0004545 = 0.0009090 mol
Now to find the concentration of the permanganate solution. We know that 11.72mL of the solution reacted to completion with the Fe2+ solution. We also know the mol which reacted from our working so far. So we are left with an arbitrary calculation:
C(permanganate solution) = 0.0009090/0.01172 = 0.07756M
The concentration of the permaganate solution is 0.07756M rounded to 4 sf.
Can anyone else confirm my answer by checking the caluculations?
(This should be right, disclaimer I haven't done volumetric analysis for ages so I may have errors but hopefully not ;))
--- End quote ---
Firstly...I don't quite get your first step, DJALogical. The reaction between iron(II) and potassium permanganate is not a precipitation reaction but a redox reaction. Permanganate is a pretty powerful oxidant, enough to oxidise iron(II) to iron(III).
MnO4-(aq) + 8H+(aq) + 5 Fe2+(aq) => 5 Fe3+(aq) + Mn2+(aq) +4H2O(l)
I can just write this equation out because I've dealt with this equation too many times in trial exams, and because you can just replace any electrons in the permanganate reduction equation with Fe(II) ions :D
So...0.317 g iron wire eh? Molar mass of Fe = 55.8 g/mol => n(Fe2+) = 0.317/55.8 mol = 5.681 mmol
As you make a 250 mL solution of this, and then take 20 mL of it, you really just have 20mL/250 mL*5.681 mmol = 0.45448 mmol
Mol ratio suggests that n(Fe2+)=5n(MnO4-) => n(MnO4-)=9.0896*10-5 mol
Concentration = n/V = 9.0896*10-5 mol / 0.01172 L = 7.76*10-3 M
Note that as the mass of the wire is to 3 s.f., the answer is only to 3 s.f.
Interesting. My answer differs from yours by a factor of ten exactly.
DJA:
--- Quote from: lzxnl on December 28, 2013, 03:46:31 pm ---Interesting. My answer differs from yours by a factor of ten exactly.
--- End quote ---
:) lol this is what happens when I haven't done redox titrations since like midyear chem olympiad-totally missed on on the possibility of it being a redox reaction.
Yep since the mole ratio if you realise it is redox is 5:1 whereas if you incorrectly assume precip, you get a mole ratio of 1:2.
Otherwise theoretically the steps following the redox reaction are the same.
Thanks for the pick up!! I'm modifying my post so I dont screw over the next poor sod who comes along.
DJA:
lzxnl just for my own understanding could you write out the two separate ionic equations for the permaganate and Iron (II) so I can see what they look like and balance out? I've gotten it written down on paper I just want to check if its right with you. :)
Its been ages since I did redox.
lzxnl:
--- Quote from: DJALogical on December 28, 2013, 03:59:53 pm ---lzxnl just for my own understanding could you write out the two separate ionic equations for the permaganate and Iron (II) so I can see what they look like and balance out? I've gotten it written down on paper I just want to check if its right with you. :)
Its been ages since I did redox.
--- End quote ---
MnO4-(aq) + 8H+(aq) + 5e- => Mn2+(aq) +4H2O(l)
5Fe2+(aq) => 5Fe3+(aq) + 5e-
DJA:
--- Quote from: lzxnl on December 28, 2013, 04:03:41 pm ---MnO4-(aq) + 8H+(aq) + 5e- => Mn2+(aq) +4H2O(l)
5Fe2+(aq) => 5Fe3+(aq) + 5e-
--- End quote ---
Cheers!
Also I wanted to ask, is there any 'atmospheric chemistry' in Chemistry 3/4? Because I hated that stuff in year 11 and am sincerely hoping I don't have to do it again.
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