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Offline zimrock

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volumetric
« on: February 01, 2008, 11:30:24 AM »
potassium oxalate is titrated against KMnO4 solution to give ferric ion , carbon dioxide and Mn2+ ion. 100 ml of .1M KMnO4 solution is used. The resulting carbon dioxide is treated with V ml of NaOH (.1M) solution. what is V?

according to the answer, the number of equvalents of potassium permanganate is not equal to number of equvalents of carbon dioxide liberated. how?

Offline Borek

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Re: volumetric
« Reply #1 on: February 01, 2008, 11:34:33 AM »
potassium oxalate is titrated against KMnO4 solution to give ferric ion

Huh?

Write all balanced reaction equations.
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Offline zimrock

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Re: volumetric
« Reply #2 on: February 01, 2008, 11:37:47 AM »
FeC2O4  +  KMnO4 +  H+  ---> Fe3+  + CO2   + Mn2+.

solve the problem

Offline zimrock

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Re: volumetric
« Reply #3 on: February 01, 2008, 11:40:26 AM »
and,,
CO2 + NaOH ---> Na2CO3

Offline Borek

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Re: volumetric
« Reply #4 on: February 01, 2008, 11:47:19 AM »
So it is ferric oxalate, not potassium.

Start balancing reactions.
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Offline zimrock

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Re: volumetric
« Reply #5 on: February 01, 2008, 11:52:55 AM »
oops. sorry about that typo.

why is it necessary to balance the reactions?

isnt it right to say, the equvalents of KMnO4= eq of CO2 = eq of NaOH ?


Offline Borek

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Re: volumetric
« Reply #6 on: February 01, 2008, 01:43:45 PM »
why is it necessary to balance the reactions?

isnt it right to say, the equvalents of KMnO4= eq of CO2 = eq of NaOH ?

You have to know what equivalents are and thet's defined by the reaction equation. And balanced one. Depending on the reaction 1 mole of MnO4- can be 1, 3 or 5 equivalents (at least).
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Offline zimrock

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Re: volumetric
« Reply #7 on: February 02, 2008, 08:08:27 AM »
that doesnt exactly help me solving the problem.

Please post a solution

Offline Borek

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Re: volumetric
« Reply #8 on: February 02, 2008, 09:10:11 AM »
that doesnt exactly help me solving the problem.

As far as I can tell you have not tried yet.

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Please post a solution

Please read forum rules. And please balance the reaction equations, as thats the first step to solution.
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Offline zimrock

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Re: volumetric
« Reply #9 on: February 02, 2008, 10:59:14 AM »
5C2O4 (2-) + 16H+ + MnO4-  ---> 10CO2 + 8H2O + 2 Mn2+

i dont really see why this would help

Offline Borek

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Re: volumetric
« Reply #10 on: February 02, 2008, 12:58:22 PM »
Almost correct, but I suppose it is just a typo (lacking 2).

Now, looking at this reaction equation - what is ratio of permanganate to oxalate? How many moles of permanganate per mole of oxalate? What is permanganate equivalent?
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Offline zimrock

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Re: volumetric
« Reply #11 on: February 04, 2008, 07:25:34 AM »
two moles of permanganate for every 5 moles of oxalate.

one sec.
i think i've got my definition of equivalents wrong.
Isnt there 5 moles in one equavalent of permanganate and 2 moles in one equivalent of oxalate by just seeing the change in oxidation state?

Offline Borek

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Re: volumetric
« Reply #12 on: February 04, 2008, 08:07:53 AM »
i think i've got my definition of equivalents wrong.

That's I supposed from teh very beginning :)

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Isnt there 5 moles in one equavalent of permanganate and 2 moles in one equivalent of oxalate by just seeing the change in oxidation state?

Quite the opposite - in this case there are 5 equivalents in 1 mole of permanganate (thus 1 equivalent is 0.2 mole). But you are right about the oxidation state part.

Problem with equivalents is that they can't be calculated once and for ever, as the substance can behave differently in different reactions. So 1 mole of permanganate is 5 equivalents for redox reactions as long as we are talking about oxidation is acidic solution.
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Offline zimrock

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Re: volumetric
« Reply #13 on: February 04, 2008, 09:46:51 AM »
ok got that.

but what is the purpose of balancing this equation and what is the next step in solving this problem

Offline Borek

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Re: volumetric
« Reply #14 on: February 04, 2008, 09:56:53 AM »
Now it is simple stoichiometry - and stoichiometry always require balanced equations. You know amount of permanganate used, that plus the reaction equation is enough to calculate amount of CO2 produced - that goes into the second reaction and gives amount of NaOH necessary for neutralization.
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