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Homework Help: Find the molecular mass of acid in acid base neutralization

  1. Sep 25, 2010 #1
    1. The problem statement, all variables and given/known data
    1.5000 g of diprotic weak acid H2A was dissolved in 100.00 mL volumetric flask.
    25.00 mL aliqouts of this solution was titrated with a monoprotic strong base NaOH (0.08000 M). The titre volume of NaOH was 40.00 mL. Calculate the molecular weight of H2A.


    2. Relevant equations
    N/A


    3. The attempt at a solution
    1 mole of H2A contains 2 moles of H+
    1 mole of NaOH contains 1 mole of OH-

    2CaVa = CbVb
    2 (Ca)(0.02500 L) = (0.0800 M)(0.04000L)
    Ca = 0.0640 M

    #moles of H2A = CaVa
    = (0.0640 M)(0.10000 L)
    = 0.00640 mol

    molecular mass of H2A = mass/# moles
    = 1.5000 g / 0.00640 mol
    = 234.375 g/mol

    My answer is 234.375 but something seems odd about it. Can anyone tell me if I missed something or did something wrong? Thanks!
     
    Last edited: Sep 26, 2010
  2. jcsd
  3. Sep 26, 2010 #2

    Borek

    User Avatar

    Staff: Mentor

    Question is - is that 2 in the correct place?
     
    Last edited by a moderator: Aug 13, 2013
  4. Sep 26, 2010 #3
    I've always been taught that the exponent in CaVa = CbVb equations is the number of moles of H+ ions or OH- ions which is why I put the 2 infront of the CaVa.

    Another time I attempted this question but I balanced the neutralization first.
    H2A + 2 NaOH --> 2 H2O + Na2A

    CaVa = 2 CbVb
    (Ca)(0.02500 L) = 2(0.0800 M)(0.04000L)
    Ca = 0.256 M

    #moles of H2A = CaVa
    = (0.256 M)(0.10000 L)
    = 0.0256 mol

    molecular mass of H2A = mass/# moles
    = 1.5000 g / 0.0256 mol
    = 58.59 g/mol

    This answer seemed wrong too somehow. One of the questions that came up was that the acid we are neutralizing is a weak acid. 25mL of this weak acid was neutralized by 40mL of strong base. Shouldn't a weak acid take less base to neutralize?
     
    Last edited: Sep 26, 2010
  5. Sep 26, 2010 #4

    Borek

    User Avatar

    Staff: Mentor

    Exponent? I guess you mean stoichiometric coefficient. But you were right the first time, somehow I got it reversed.

    Now, the problem is - 0.00640 moles - is it whole 1.5 g sample?

    Strength of the acid has nothing to do with amount of base that it needs to be neutralized. It is all in stoichiometry.
     
  6. Sep 26, 2010 #5
    :eek: So sorry! Yes, I did mean the coefficient.

    So...I'm using the first attempt...

    2CaVa = CbVb
    2(Ca)(0.02500 L) = (0.0800 M)(0.04000L)
    Ca = 0.0640 M

    #moles of H2A = CaVa
    = (0.0640 M)(0.10000 L)
    = 0.00640 mol

    molecular mass of H2A = mass/# moles
    = 1.5000 g / 0.00640 mol
    = 234.375 g/mol

    So I found the molarity of the H2A that was neutralized. Because a 25.00 mL sample was taken from the 100.00 mL solution, the molarity stays the same. 1.5000 g made the 100.00 mL solution of H2A so I think the 0.00640 mol is the whole 1.5 g sample.
     
  7. Sep 26, 2010 #6

    Borek

    User Avatar

    Staff: Mentor

    have you titrated 100 mL aliquot?
     
  8. Sep 26, 2010 #7
    No, I have not titrated 100mL
     
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