Sodium Pentobarbital titration

[savagefarmer]

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

 

[sjb-2812]

Seems to be start of a discussion at << link to CF discussion thread deleted by Mod >>

 

[savagefarmer]

Sorry, is there a policy against cross-posting on two entirely different forums?

 

[sjb-2812]

Not per se, but many frequent both; so it may be wise to mention that you've also posted elsewhere

 

[savagefarmer]

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 :)

 

[chemisttree]

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?

 

[savagefarmer]

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⁺ → C₁₁H₁₆N₂O₃^2- + H2O

Does that make sense?

Thanks,
sav

 

[savagefarmer]

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⁺ → C₁₁H₁₆N₂O₃^2- + H2O

Does that make sense?

Thanks,
sav

Sorry that reaction is wrong, it would be:

C₁₁H₁₈N₂O₃ + H₃O⁺ → C₁₁H₁₉N₂O₃⁺ + H2O