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Phosphoric acid mixture titration

  1. Nov 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A mixture of phosphoric acid and NaH2PO4 are titrated against 0.151 M NaOH. The first equivalence point is at 0.00345 L and the second equiv point is at 0.01288 M.
    The volume of the total mixture was 60 ml.
    Calculate the concentration of both the phosphoric acid mixtures (H3PO4 and NaH2PO4)

    2. Relevant equations

    H3PO4 + NaOH => NaH2PO4 + H2O

    3. The attempt at a solution
    Iv made the titration graphs and figured out the endpoints (mentioned above).
    Here's an attempt at the solution...

    moles H3PO4 = Conc of NaOH x volume of NaOH at 1st eq.point.
    = 0.151 x 0.00345 = 0.000521 mol

    moles H2PO4 = conc of NaOH x vol of NaOH at 2nd eq. point
    = 0.00143 mol

    Im still unsure and pretty confused... Any help please? Thanks!!
  2. jcsd
  3. Nov 15, 2010 #2


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    Homework Helper
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    Gold Member

    The second equivalence point included the titration for monohydrogen phosphate from the formal amount PLUS the monohydrogen phosphate from the formal amount of original phosphoric acid from the first equivalence point.
  4. Nov 15, 2010 #3
    ok.. and is the calculation for the first end point correct?
    and if I know that the second endpoint gives the amount for HPO4- from both the sources, how do i actually go about the calculation..?

  5. Nov 15, 2010 #4


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    Staff: Mentor

    Yes, your calculation if phosphoric acid is right.

    Come on, you know amount of one of the components and you know the sum, how can you calculate amount of the other component?

    You have a 5 pound bag of apples and oranges. You know there is 2 pounds of oranges inside, what is weight of apples?
  6. Nov 15, 2010 #5
    oh.. really? Im still not sure because that seems too easy!

    so.. the volume (and moles) of NaOH added at the first quivalence point will give the moles of H3PO4, and the volume (and moles) of the NaOH added at the 2nd equivalence point will give the amount of H3PO4 + H2PO4-.

    So i can subtract the mol of H3PO4 from the moles found at the 2nd endpoint, in order to get the moles of H2PO4-. Is that correct?

    Thanks a lot people!!
  7. Nov 15, 2010 #6


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    Staff: Mentor

    Yes, it is REALLY that simple.
  8. Nov 15, 2010 #7
    omg thanku guys so much!!! :D
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