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    Hi there! I wanted to know the method of deriving the final answer X to check if I did it right.
    Also I have some doubts.
    Before acidification and after acidification of ethanedioic acid what could be the changes?
    And also when something is hydrated does it mean the substance in aqueous state? Or in solid state of crystals with water in it?

    Thanks a ton for any suggestions, solutions in advance!
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    Hi! I admit I'm not 100% sure, but the numbers work out ok, so fingers crossed. First, the acidification is likely just to act as a catalyst, and won't affect the final result. Secondly, hydrated means it is a solid crystal with water in it, i.e. has .xH2O in its formula.
    Now for the maths:
    I balanced the equation the same as you, but with 2 Mn2+ as you have 2Mn on the left hand side. Next, mols(MnO4-) = vol * conc = 0.1 *0.02 = 0.002 moles.
    Ratio of C2O42- to MnO4- is 5:2, so moles of C2O42- in the reacted sample is 0.005. The total moles dissolved from the 6.3g sample is 0.005 * 10 = 0.05 moles as only a tenth of the original solution was used in the reaction.
    Mass / moles = Mr of the H2C2O4.xH2O = 6.3/0.05 = 126. The Mr of H2C2O4 is 90. Therefore the Mr of xH2O is 126-90 = 36. Mr of H20 is 18, so x = 36/18 = 2.
    Hope this helps
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    (Original post by Iluv2edgedsword)
    Also I have some doubts.
    Before acidification and after acidification of ethanedioic acid what could be the changes?
    Conducting the titration under acidic conditions is to ensure that the MnO4- are fully reduced to Mn2+.

    In solutions that are not acidic enough (i.e. neutral, basic), the MnO4- might be reduced to MnO2 or MnO4(2-) instead
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    (Original post by Luxtra)
    Hi! I admit I'm not 100% sure, but the numbers work out ok, so fingers crossed. First, the acidification is likely just to act as a catalyst, and won't affect the final result. Secondly, hydrated means it is a solid crystal with water in it, i.e. has .xH2O in its formula.
    Now for the maths:
    I balanced the equation the same as you, but with 2 Mn2+ as you have 2Mn on the left hand side. Next, mols(MnO4-) = vol * conc = 0.1 *0.02 = 0.002 moles.
    Ratio of C2O42- to MnO4- is 5:2, so moles of C2O42- in the reacted sample is 0.005. The total moles dissolved from the 6.3g sample is 0.005 * 10 = 0.05 moles as only a tenth of the original solution was used in the reaction.
    Mass / moles = Mr of the H2C2O4.xH2O = 6.3/0.05 = 126. The Mr of H2C2O4 is 90. Therefore the Mr of xH2O is 126-90 = 36. Mr of H20 is 18, so x = 36/18 = 2.
    Hope this helps
    Thanks a lot!!!!!! It really really helped! Im sure youre 100% right dw (; Im sorry for the delay in replying tho...
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    (Original post by Metanoia)
    Conducting the titration under acidic conditions is to ensure that the MnO4- are fully reduced to Mn2+.

    In solutions that are not acidic enough (i.e. neutral, basic), the MnO4- might be reduced to MnO2 or MnO4(2-) instead
    Thanks a lot !!!!! Now its more clearer to understand! (: Are they included in redox titrations? I havent reached those lessons yet, so perhaps thats the cause of my 'not understanding part'. Im sorry for the delay in replying..:/
 
 
 
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