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Acid-base reaction

  1. Jun 22, 2009 #1
    Hey guys, I'm having a problem with this question;

    Malonic Acid, C3H4O4, is a diprotic acid. How many grams of it are in a 30mL sample that requires 35mL of 2M NaOH for complete neutralization?

    Since Malonic Acid is diprotic, I think the chemical equation is;
    C3H4O4 + 2NaOH --> Na2C3H2O4 + 2H2O .

    I knw that if the equation is true, the stoichiometric ratio of NaOH : C3H4O4 is 2 : 1 so the number of moles of Malonic Acid is 0.5 x (2 x (35 / 1000)mol = 0.035mol .

    Hence, the mass of Malonic Acid required would be 0.035mol x molar mass. However, how did the 30mL fit in? Thanks.
     
    Last edited: Jun 22, 2009
  2. jcsd
  3. Jun 22, 2009 #2

    symbolipoint

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    Try to think in problem solving steps. First, how many MOLES of the acid are present in the titrated solution? Now use the formula weight of the malonic acid to convert the moles to grams.
     
  4. Jun 22, 2009 #3
    I found out alrdy that the number of moles of Malonic Acid is 0.5 x (2 x (35 / 1000)mol = 0.035mol .

    Yes, the mass of Malonic Acid required would be 0.035mol x molar mass or formula weight. However, I am not sure how I'm supposed to incorporate the 30mL sample part in the solution.

    Thanks for the helping though.
     
  5. Jun 22, 2009 #4

    symbolipoint

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    The volume of solution in which your sample is dissolved is not important. You were interested in the grams (the mass) of the malonic acid present.

    If you were interested in the concentration of malonic acid, then you would have a little more calculation to handle.
     
  6. Jun 25, 2009 #5
    Oh, ok; so I don't hve to include the 30mL part in my calculations, right?
     
  7. Jun 25, 2009 #6

    Borek

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    Yes, you can ignore it. This question is equivalent to:

    Malonic Acid, C3H4O4, is a diprotic acid. How many grams of it are in a sample that requires 35mL of 2M NaOH for complete neutralization?
     
  8. Jul 2, 2009 #7
    Ok then, I get it. Thanks for the help.
     
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