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Titration Questions

  1. Aug 12, 2011 #1
    Hi. I'm currently an intern in a biology lab and I have a few questions about titration.

    In the lab we are performing an experiment where the goal is to find the concentration of levulinic acid (H8C5O3). To do this, we decided to titrate the acid with NaOH.

    Using a KHP standard, we performed the appropriate tests and found that the normality of our NaOH is 1.3570799 N. We then tested it with a pH probe and found that it has a pH of 13.86.

    First question; is normality = molarity?​

    We then took 25mL of the levulinic acid and titrated it with our NaOH solution. We used phenolphthalein as our indicator, and the whole solution turned pink after the addition of 114.9mL of NaOH solution.

    Second question; I know that levulinic acid has a carboxylic acid group. Does this mean that there is only one, reactive, "acidic" hydrogen that reacts with the OH- in the titration mixture?​

    Now I am at home and trying to find the concentration of the levulinic acid myself and I'm just unwary/unsure about how to go and find the answer. Here's what I have so far:

    Using a neutralization reaction chemical equation for

    H8C5O3 (aq) + OH- (aq) <--> H2O (l) + H7C5O3-(aq)

    I found that we have 0.15592848051 moles of NaOH (volume of NaOH * normality). If at equilibrium all the moles of reactants fully react with no excess, I assumed then that at the end of the reaction, there are no moles of H8C5O3 and OH- at equilibrium; instead, only negligible moles of H2O and 0.15592848051 created moles of H7C5O3- (aq).

    From the moles of H7C5O3-, I found the concentration of H7C5O3-. It would be 1.114570982916 M (moles of the anion / total volume).

    If I were to use this concentration to find the concentration of levulinic acid, then I would:

    H2O (l) + H7C5O3-(aq) <--> H8C5O3 (aq) + OH- (aq)

    From this step, I would use an ICE chart. The initial for H7C5O3- would be the 1.11457 M. Then, using the Kb of levulinic acid,

    (pKa of levulinic acid = 4.78, so then pKb = 14 - 4.78 = 9.22. Kb therefore = 10-9.22)

    I would have

    Kb = x2 / (1.11457 - x)

    Third question: does this all sort out to be okay???

    Last, final, kicker question. If I have a pH probe, can I just pH probe the levulinic stock solution we have and then find the number of moles of H+ ions? We know the volume of the stock solution. I know the formula of levulinic acid. If I knew the pH, I could find the [H+], multiply by the volume and get moles of H+. Could the moles of H+ then be used in a 1/8 ratio to find the moles of levulinic acid in stock??

    Thank you.
    Last edited: Aug 12, 2011
  2. jcsd
  3. Aug 12, 2011 #2
    Holy ****.

    I just realized I'm going insane.

    At the equivalence point, the moles of OH that reacted with the levulinic acid should be equal, no?

    So the moles of OH = 0.1559284805 moles.

    Then that means that 0.1559284805 moles of levulinic acid reacted as well.

    How many mLs did I have of levulinic acid? 25.

    So the [Levulinic acid] = 0.1559284805 moles / 0.025L = 6.23713922 M.

    Hope I'm right. :D
  4. Aug 13, 2011 #3
    The definition of equivalence point -

    When a stoichiometrically equivalent amount of titrant has been added to the analyte. So, for example, hydroxide ion and levulinic acid.

    Your numbers look OK from a very quick scan (have to run soon), although are there really that many significant figures? :)

    (Also, I would think that there'd be an established protocol for doing titrations in your lab - something else to put on my "make sure I have this worked out" list when I have my own lab.....)

    FYI, normality is not the same as molarity, at least in theory. There needs to be a correction via an equivalence factor. You can look this up to see a more detailed explanation, but for something like HCl or NaOH, normality is equivalent to molarity since for every one mole of HCl or NaOH in one liter of solution, one mole of each ion is produced (H+, Cl-, Na+, and OH-). For something where one mole of the molecule would give rise to one mole of anions and two moles of cations (or something) is where that correction comes in.

    P.S. - I had an extra moment to reread, as I thought I had seen something potentially worrisome, but it didn't quite click on that first read. Endpoints aren't always equivalence points. Ideally, they should be really close, of course, in a well-planned experiment. And in a range where your pH indicator will be helpful. And 115 mL of a strong base for 25 mL of a presumably weak acid? I know nothing about the particular chemistry of levulinic acid, so maybe that's appropriate, but my chemical intuition makes me wonder.
    Last edited: Aug 13, 2011
  5. Aug 13, 2011 #4


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    Staff: Mentor

    Mike already addressed most of the important things, but just in case: levulinic acid seems to be monoprotic, so calculation of the titration results is particularly simple. No need for ICE table (actually it is never necessary for titration results). See


    and browse the site for answers to your other questions.

    Note: if you had to use 115 mL of titrant, your sample size was badly selected. See discussion here: http://www.titrations.info/titrant-and-sample-volume
  6. Aug 13, 2011 #5
    Thanks for all the responses. I will perhaps try again with a stronger base in order to not refill the buret over and over (I knew that it would arise to errors).

    Once again, thanks so much!

    @Mike: I kept all the significant figures for figures' sake. I was taught that I should keep all of them until the "answer" step (when you declare an answer)... but once again, that's just my experience.

    Thanks once again, really grateful. :D
  7. Aug 13, 2011 #6


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    Staff: Mentor

    You should keep them for calculations, but not report them when you list intermediate result.
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