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Chemistry: Titration of of a base to solve for the mass of an unknown acid.

  1. Nov 2, 2012 #1
    A student titrated a .4630g of unknown monoprotic acid with .1060M NaOH. The equivalance point volume is 28.70mL.
    a) Calculate the number of moles of NaOH used.
    Im pretty sure I got this one
    I did .1060 M X .02870L to equal 3.04E-3 mol
    b)how many equivalents of unknown acid were titrated? no clue

    c) Determine the equivalent mass of the unknown acid
    Which is the moles times MM


    There is also a table given of acids and there pKa and equivalent masses. If I can figure out what Pka is I should be fine.
     
  2. jcsd
  3. Nov 2, 2012 #2
    If the unknown acid is HA, can you write a balanced equation for the titration?
     
  4. Nov 2, 2012 #3
    would it be

    NaOH + HA= H_2_O+ NaA?
     
  5. Nov 2, 2012 #4
    Yes, so if you have 3.04 mmol of NaOH, how many moles of acid do you have?
     
    Last edited: Nov 2, 2012
  6. Nov 2, 2012 #5
    Well since its one to one it would be the same so 3.04mmol of acid. So would that just be the answer for part b? If so I did not think it could be that simple.
     
  7. Nov 2, 2012 #6
    Part b is more about the recognition that it's 1:1, rather than an actual number.
     
  8. Nov 2, 2012 #7
    Would part c be .4630g/ .00304moles? Which I got to be 161.03
     
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