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Acid Base Titrations

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Given H2SO4 + 2LiOH --> Li2SO4 + 2H2O, how many mL of a 0.700-M solution of KOH are needed to react with 235 mL of a 0.350-M H2SO4 solution?

    3. The attempt at a solution

    Really, I have none. My problem is that I don't understand what the question is asking. I could probably figure this out but for some reason my brain can't reason out what I'm supposed to be aiming for here? Is it wanting to know how many mL of KOH are needed to neutralize the H2SO4? If so, I don't need the given equation do I? Maybe just a hint would send me in the right direction.... Then again maybe I am just totally lost.

    Thanks in advance for any and all help...
  2. jcsd
  3. Mar 20, 2007 #2


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    At the endpoint, the number of equivalents of base is equal to the number of equivalents of acid.
  4. Mar 20, 2007 #3
    Ok so here's what I've done:

    1) .235L H2O4 solution * .350M = .0823 mol H2SO4
    2).0823 mol KOH/.700M = 117.5mL KOH solution

    Here's where I'm confused, and maybe I've missed a core concept somewhere....but what am I supposed to do with the given reaction equation?
  5. Mar 20, 2007 #4
    Ok, I am starting to think that there is a typo in this question and that it should actually be asking "how many mL of *LiOH* are needed..." not "KOh"... Then it would make sense to me!
  6. Mar 20, 2007 #5


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    Don't forget that molarity (M) is not the same as normality (N). The acid solution concentration needs to be expressed in terms of normality, not molarity.
  7. Mar 20, 2007 #6
    Go to Google, type in "titratiom end points" and look at the top reference. Think that will help you. Good luck.

  8. Mar 21, 2007 #7
    Well I dunno about the normality vs molarity thing, we haven't covered that yet....

    As far as this question goes, I was able to confirm this evening that there is a typo in the question and it should be asking how many mL of LiOH are needed (KOH is a typo). That clears things up for me completely, I was so lost on why the equation had LiOH but the question was asking for KOH! I was afraid I had completely missed something important. I can do it now that I'm not trying to figure out how KOH came into play :)

    Thanks for yalls help tho! Much appreciated :)
  9. Mar 24, 2007 #8


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