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Weak acid/strong base titration (Acetic acid/NaOH)

  1. Dec 7, 2008 #1
    I did an acid/base titration with acetic acid and sodium hydroxide to determine the concentration of acetic acid present in vinegar but I'm not quite sure how to go about calculating the concentration of it.

    First things first we dissolved 25ml of vinegar into 250ml of water and filled the burette to its 0 mark with this solution. We then added 25ml of 0.1M NaOH solution to an Erlenmeyer flask and added a few drops of phenolphtalein indicator. Then we titrated and watched as the liquid in the flask went from bright pink to completely clear then recorded the volume in the burette.

    I this formula M2V2/b = M1V1/a for calculating concentration but I'm wondering if I have to calculate the amount of H+ ions from the acetic acid and OH- ions from the NaOH present in the solutions before I can go ahead. I'm not quite sure how this works for a weak acid is there a general rule like "1 dissociated H+ ion for every 2 moles of weak acid while the ratio is 1:1 for a strong acid?"

    One other quick question on the side. I have 25ml of 0.1M NaOH solution so to calculate the moles present I just type 25/1000 x 0.1 into my calculator but this gives me a ridiculously low number which I doubt the teacher would expect us to deal with. Am I doing something wrong here?
     
    Last edited: Dec 7, 2008
  2. jcsd
  3. Dec 7, 2008 #2

    epenguin

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    Homework Helper
    Gold Member

    The calculation is simpler than that.
    Chemists say in their loose fashion :smile: you have neutralised the acetic acid.

    But if you want to be confident you will have to know what a pH against added base curve looks like (so you have to know what pH is), what pK is, and the fact that the pK of phenolphtalein is around 9.
     
  4. Dec 7, 2008 #3

    Borek

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    Staff: Mentor

    No, that's the correct amount. And it is not ridiculously low, it is just 0.25 mmol :wink:
     
  5. Dec 7, 2008 #4
    Thanks. I just did the calculation and my results seem reasonable. The average amount of diluted vinegar it took to neutralize the 25ml 0.1M NaOH solution was 29ml.

    My results were
    Concentration of diluted vinegar = 0.09M
    Concentration of undiluted vinegar = 0.35M

    I'm confused about how to get the percentage of acetic acid in the vinegar if there are 0.35 moles of acetic acid per liter of vinegar. 0.35M is 21 grams but I don't know how to convert mass to volume.
     
    Last edited: Dec 7, 2008
  6. Dec 7, 2008 #5

    Borek

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    Staff: Mentor

    At first you stated 25 mL were diluted to 250 mL. So the general approach is OK, but whether the number is OK - I don't know.

    Watch significant numbers. Were you given 0.1M NaOH, or 0.1000M (or something like that)?
     
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