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Sodium Pentobarbital titration

  1. Mar 13, 2012 #1
    Hello everyone,

    Today I witnessed the titration of a sodium pentobarbital solution with 0.1M HCl. At first, a lot of white precipitate was formed. When more HCl was added the white precipitate disappeared again completely.

    Now I am trying to figure out what is going on. As I understand it the reaction is:

    C11H17N2O3- + H3O+ -> C11H18N2O3 + H2O

    And the white precipitate is C11H18N2O3. What I don't understand is why the precipitate disappears again when more HCl is added and the pH drops.

    Thanks for any advice!

    -sav
     
  2. jcsd
  3. Mar 13, 2012 #2
    Seems to be start of a discussion at << link to CF discussion thread deleted by Mod >>
     
    Last edited by a moderator: Mar 13, 2012
  4. Mar 13, 2012 #3
    Sorry, is there a policy against cross-posting on two entirely different forums?
     
  5. Mar 13, 2012 #4
    Not per se, but many frequent both; so it may be wise to mention that you've also posted elsewhere
     
  6. Mar 13, 2012 #5
    Ok sorry about that, I didn't realize that there would be a big overlap. I'll leave this up for now and hope for some pointers :)
     
  7. Mar 13, 2012 #6

    chemisttree

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    You will note that this is a di-imide. Imides are generally fairly acidic so the proton attached to the nitrogen is easily replaced with alkali metals like sodium. You have the first step down but what might that second proton do?
     
  8. Mar 14, 2012 #7
    I just read about imides and di-imides, but I'm still clueless about the role they play in this. My guess was that the pentobarbital reacted with another H3O+ ion when a lot of HCl was added and the pH dropped further:

    C11H17N2O3- + H3O+ → C11H16N2O32- + H2O

    Does that make sense?

    Thanks,
    sav
     
  9. Mar 14, 2012 #8
    Sorry that reaction is wrong, it would be:

    C11H18N2O3 + H3O+ → C11H19N2O3+ + H2O
     
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