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Calculation of Titration problems - Should be Easy?

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data
    1. Calculate the volume (mL) of 0.100 M H3PO4 required to neutralize 30.0 mL of 0.050 M

    2. If 0.2178 g of a diprotic solid acid is neutralized by 44.81 mL of 0.0953 M NaOH, calculate the molar mass of the solid acid.

    2. Relevant equations
    Molar Mass = Mass/Moles
    Concentration = Moles/Volume(mL)
    Neutralization --> Moles of H+ = Moles of OH-

    3. The attempt at a solution
    First Problem
    For this problem, I've already done work to arrive to my answer, I just require someone else's opinion on whether it is right or not. I also felt lazy to type out the work for this one, since this problem is not that important and is easier compared to the second one (which is why I showed work for the second problem and not this one, hopefully you guys understand).

    The volume I derived from the problem was that you needed 10 mL of H3PO4; is this correct?

    Second Problem

    For the second problem, I will show work:

    44.81 mL x 0.0953 M = 4.27 moles of NaOH = 4.27 moles of OH-

    4.27 moles of OH- imply that you need 4.27 moles of H+ for neutralization

    Because the solid acid is diprotic, that means there are two acidic hydogens, which means the chemical formula for the acid is H2X, X being the remaining part of the acid's formula.

    To get 4.27 moles of H+, then you would need (4.27/2), or 2.14 moles of H2X?

    Since there was 0.2178 g of the diprotic solid acid, then the molar mass would be mass/moles, or 0.2178 g/ 2.14 moles = 0.101 g/mol

    At this point, I got confused because the molar mass was so small; can anyone confirm whether or not this is truly the right answer?
     
    Last edited: Nov 8, 2011
  2. jcsd
  3. Nov 8, 2011 #2

    Borek

    User Avatar

    Staff: Mentor

    Of what? Your answer can be right, can be wrong, impossible to tell.

    Your approach is correct in general, but 44.81 mL of this solution doesn't contain 4.27 moles.
     
  4. Nov 8, 2011 #3
    Oh I see, I completely missed the point where molarity is moles/volume(L) and not mL. This gave me, for the first problem, 10 mL still and for the second problem, 102 g/mol.
     
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