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Calculating mols carboxylic acid needed to nuetralize NaOH solution

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A 10mL mixture of 2M HCl and carboxylic acid required 43.61 mL of .762 M NaOH to nuetralize it. Calculate the number of mols of carboxylic acid present.


    3. The attempt at a solution
    First figured out the number of mols of Naoh and subtracted the number of mols of HCl:
    .0332molNaOH-.0200molHCl = .0132 mol NaOH that still need nuetralized
    Divided that by total volume so .246M H3O is needed to finish the job.

    With a Ka of 1.75X10^(-5) I get 3458M of acetic acid
    So I think I may be a little off :)
     
  2. jcsd
  3. Oct 23, 2012 #2

    Borek

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    Staff: Mentor

    What do you use Ka for? This is about simple neutralization stoichiometry, no equilibrium calculations needed.
     
  4. Oct 23, 2012 #3
    Hey Borek, okay so the approach is wrong then lol.
    I thought the Ka would affect the H3O delivered by the acetic acid?

    So what do I do?
     
  5. Oct 23, 2012 #4
    Okay, so the carboxylic acid will neutralize the NaOH in a 1:1 ratio?
    If that is the case then it is just .0132 mols, is that right?
     
  6. Oct 23, 2012 #5

    Borek

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    Staff: Mentor

    I have not checked the numbers, but it sounds right and fits what you wrote earlier.
     
  7. Oct 23, 2012 #6
    Good deal, thanks Borek
     
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