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KMnO4 Titration Calculation help Watch

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    My question is:

    1.44g of sodium ethanedioate was dissolved in water and the volume made up to 200cm3. 20.00cm3 of this solution, after acidification with dilute sulphuric acid, reduced 17.90cm3 of a solution of potassium manganate(VII).

    a) Calculate the conc. of the sodium ethanedioate solution in mold dm-3

    1.44/134 = 0.0107.../0.2 = 0.054 mol dm-3

    I don't know how to do b) please explain

    b) Calculate the conc. of the potassium manganate(VII) solution, in mol dm-3 and g dm-3.
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    Do you have a balanced equation for the reaction that took place?

    Or the half equation for the ester being oxidized? (I know the MnO4 ion needs 5 electrons to reduce)
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    (Original post by dannyboy383)
    Do you have a balanced equation for the reaction that took place?

    Or the half equation for the ester being oxidized? (I know the MnO4 ion needs 5 electrons to reduce)
    2MnO4- + 5C2O42- + 16H+ = 2Mn2+ + 10CO2 + 8H2O
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    Okay, so you know 5 moles of sodium ethanedioate reacts with 2 moles of potassium permanganate. You know how much ethanedioate you used (1.44g in 200cm^3, you used 20cm^3 so 0.144g) so you can work out that there's going to be 2/5 of that many moles of permanganate, which you can also get in grams, and divide these by the volume of permanganate you used.
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    sweeeeeeeeeeeet.

    So, two moles of manganate react with 5 moles of ester.

    5 moles of ester : 2 moles manganate
    1 mole ester : 2/5 mole manganate


    We assume that part (a) is correct (assumpition is the mother of all confusion)

    Concentration of ester: 0.054 mol dm-3

    FOR ESTER:

    1000 cm3 contains 0.054 mol
    1000 cm3 : 0.054 mol
    20.00 cm3 : (0.054 X 20.00/1000) mol
    (we simply multiply both sides of ratio by 20/1000)

    (0.054 X 20.00/1000) mol ester reacts


    now if 1 mol ester reacts, then 2/5 manganate mol reacts (see above)

    so if (0.054 X 20.00/1000) mol ester reacts
    Then (2/5 X 0.054 X 20.00/1000) mol manganate reacts (but 17.90 cm3 reacts.)

    FOR MANGANATE:

    17.90cm3 contains (2/5 X 0.054 X 20.00/1000) mol manganate
    17.90cm3 : (2/5 X 0.054 X 20.00/1000) mol manganate
    1000cm3 : 1000/17.90 X (2/5 X 0.054 X 20.00/1000) mol manganate
    (we simply multiply both sides of the ratio by 1000/17.90)

    BUT 1000cm3 = 1 dm3

    So concentration of manganate IS:
    1000/17.90 X (2/5 X 0.054 X 20.00/1000) mol dm-3

    This is how I work out these type of questions. It may be a bit long and tedious, but I find I understand everything this way. If you don't understand, I will try to explain. Other's may have a much better way at working out the problem. PM me anytime.
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    Thanks Danny! You explained it really well.
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    By the same token

    1 mol of KMnO4 contains 197.13g
    1 mol : 197.13 g
    (multiply both sides of ratio by same factor)
    1000/17.90 X (2/5 X 0.054 X 20.00/1000) mol : 1000/17.90 X (2/5 X 0.054 X 20.00/1000) X 197.13 g

    (1000/17.90 X (2/5 X 0.054 X 20.00/1000) mol was the number of moles in 1 dm3 of manganate)

    The blue is the answer.
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    No probs!
 
 
 
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