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Titration, molecular mass

  1. Oct 15, 2008 #1
    this is from a lab. i titrated NaOH into 2 grams of an unknown acid with NaOH

    the equivalence point was at 38.25ml

    so using 2 grams of an unknown acid and 38.25ml NaOH i have to find molecular mass of the acid.

    i tried adding 2 to 38.25, but i dont think that is right.
     
  2. jcsd
  3. Oct 15, 2008 #2

    symbolipoint

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    You also need the value for the concentration of your NaOH titrant. Volume of titrant in liters multiplied by concentration of titrant in moles per liter gives you the moles of titrant used.
     
  4. Oct 15, 2008 #3
    well the concentration of NaOH was 0.1M. oh and i forgot to mention that i had to make three graphs of the titration.. 2 of them included first and second derivatives. the equivalence point for the first derivative was 37.5ml, and the equivalence point fo the second derivative was 38.125ml. so im not sure which volume i should use to calculate the molecular mass of my unknown, also, once i find the moles of my titrant used... how do i use that to find the molecular mass of an unknown acid? do i just divide 2 by the moles of titrant used?
     
  5. Oct 15, 2008 #4

    symbolipoint

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    So you were performing your titration potentiometricly and you have enough information to find the ratio of moles of hydrogen ions per mass of acid sample. At one time in history, if someone wanted the derivative, he would perform the set of calculations himself; these days, maybe an interactive computer program interfaced with the equipment could calculate and graph the derivatives.
     
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