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Titration of a triprotic acid with strong base NaOH

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data
    A 0.307g sample of an unknown triprotic acid is titrated to its equivilence point using 35.2ml of 0.106 M NaOH. Calculate the molar mass of the acid. Answer is in g/mole.


    2. Relevant equations



    3. The attempt at a solution
    I just dont know where to start with this question.
    35.2 ml * 1L/1000ml * .106M NaOH/1L * 39.9969gNaOH/1mole NaOH = 120.430 g NaOH
    I dont know where to go after this part
     
    Last edited: Oct 22, 2007
  2. jcsd
  3. Oct 22, 2007 #2

    symbolipoint

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    Not "Its" equivalence point; WHICH equivalence point? The first, second, or third? The question is about a triprotic acid.
     
  4. Oct 23, 2007 #3

    chemisttree

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    You don't need to calculate the number of grams of NaOH (which you did incorrectly). You set it up right but probably entered it in your calculator incorrectly. You should have gotten something like 0.149 g NaOH if this was what the question asked, which it wasn't.

    Why don't you calculate the number of moles of NaOH used and find a relationship between that and the number of moles of triprotic acid. Once you have that, you will have the number of moles of acid and the mass. From that you can calculate formula weight.
     
  5. Oct 24, 2007 #4


    It should equal to .149g NaOH
    then I think you just subtract from the total.
     
  6. Oct 24, 2007 #5

    chemisttree

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    no, nope, nein, nyet.
     
  7. Mar 25, 2011 #6
    Re: Titration

    A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 mL of 0.106 M NaOH. Calculate the molar mass of the acid.

    What if the problem is to the Third equivalence point?
     
  8. Mar 26, 2011 #7

    Borek

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