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Synthesis of metal oxalate complexes

  1. Sep 13, 2004 #1
    I need help trying to come up with chemical reactions for the lab I am doing.
    The target molecule is K3[Mn(C2O4)3]x3H20

    The procedure is Dissovle 11.8g of oxalic acid dihydrate (H2C2O4x2H20) in water. Add potassium permaganate (KMn04). Once addition is complete add K2CO3. Prepare ice-NaCl bath and cool reaction mixture in it.

    I am having trouble trying to figure out what the exact reaction equations are that I have to balance. I am guessing that one is
    MnO4^1- +H2C2O4----> Mn(C2O4)3 ^3- +CO2
    K^+1 +Mn(C2O4)3 ^3- ----> K3[Mn(C2O4)3]

    Can anyone help?
  2. jcsd
  3. Sep 13, 2004 #2


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    Does it say what mass of KMnO4 to add ? That would make things a lot easier.
  4. Sep 13, 2004 #3
    I guess I should write out the entire procedure.

    Dissolve 11.8 g of oxalic acid dihydrate in 75 mL of deionied water in a 250 mL beaker and heat to 70-75. Grind 2.3 g of potassium permanganate to a fine powder using a mortar and pestle. Add this powder in small amounts with viorous stirring to the hot oxalic acid soln. Once the addition is complete, the mixture should be a milky suspension. Add 2.6 g potassium carbonate in small amounts with stirring. Prepare an ice-NaCl bath and cool the rxn mixture in it.

    From this point the rxn mixture should be protected from light as much as possible. Wrap all glass ware that will be used with tin foil. Grind another .6 g of potassium permanganate and add it slowly with stirring to the cold Mn-oxalate soln. Keep the soln in the ice-salt bath and continue stirring for 10 min. Quickly filter the dark red soln using a funnel and a foil-wrapped, ice-cooled filter flask to capture the filtrate. Discard solid. Transfer the filtrate to a foil wrapped and ice cooled beaker and add 95% ethanol to the filtrate slowly with vigorous stirring until the solid is evident. ALlow the mixture to stand for 10 min. Collect the violet crystals by buchner filtration using a fresh piece of filter paper. Was the product with 2 10mL portions of fifty percent cold ethanol, two 10mL portions of cold 95% ethanol and one portion of acetone. Air dry product on watch glass in the dark for at least 15 min. Transfer to vial Record weight.
  5. Sep 14, 2004 #4


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    As I'm a metal-organic chemist, I think I've understood the whole process.

    First, potassium permanganate reacts in an acidic medium (oxalic acid) with another target molecule of oxalic acid to give manganese(II)oxalate and some CO2.

    Potassium carbonate is needed to drive the reaction to the completion.

    There is a much higher amount of oxalate to form a trioxalatomanganate(II) complex; potassium ions are used to provide charge neutralization.

    About the equations, use redox chemistry first:

    Mn7+ + 5e- ---> Mn2+
    5C3+ ---> 5C4+ + 5e-
    2MnO4- + 16H+ + 10e- ---> 2Mn2+ + 8H2O
    5C2O42- ---> 10CO2 + 10e-

    So, the first reaction involving the redox process is as follows:

    2MnO4- + 5C2O42- + 16H+ ---> 2Mn2+ + 10CO2 + 8H2O

    Now that all permanganate is converted into manganese(II)oxalate with the help of excess oxalic acid, now we can think about the complexation:

    MnC2O4 + 2K2C2O4 ---> K4[Mn(C2O4)3]

    However, I don't think that manganese(II) ions are oxidized to manganese(III); but it may happen since oxalate chelates manganese(II) and the reaction may still have permanganate. I may discuss this redox chemistry later.

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