so , i've setup a little lab with few glassware and reagents , and i wanna make ammonia solution (like NH4Cl or NH4OH) to make some qualitative
analysis
so , here are my thoughts about my recent experiment
then i added some 6 M HCl , got some redish brown precipitate (i tought it was some like NH2COOH) , and a strong ammonia smell decantate...so i
filtered it and got this "ammonia liquid" .....but maybe its mixed with some urea that wasn't reacted with OH....
can u help me about key reaction CO(NH2)2 + OH --->NH3 + NH2COOH?? it's the right and unique way??? please some analytical and inorganic chemists
help me out!! thx anywayOzone - 5-2-2007 at 20:59
Welcome to Science Madness.
Please check the Search feature before posting; this can get you to what you want much faster. Since direct NH3 from urea and the reactions involved
have not to my knowledge been addressed, I'll let you off the hook.
Another thread dealing with ammonia generation/concentration is here:
I'll also add that NaOH + amino acid + heat (glutamine, for example) will also yield NH3 (see deamidation). A paper describing the original urea assay
is attached.
Cheers,
O3
Attachment: Urea Assay_01.pdf (599kB) This file has been downloaded 844 times
HenningBasse - 5-2-2007 at 22:11
i realized that there's no NH2COOH precipitate since at low pH , the amine is already protonated NH3+COOH , so its quite soluble in acidic solution
(damn , feel bad for pass out that fault)HenningBasse - 5-2-2007 at 22:13
ehh , yeah , sorry about that Ozone , but really im not a new user here , this is just a new account...but anyway i must entered to some "welcome"
forum....HenningBasse - 5-2-2007 at 22:37
a more theoretical approach
guy - 5-2-2007 at 22:54
Quote:
Originally posted by HenningBasse
a more theoretical approach
It should be NH2COO- + NH3
[Edited on 2/6/2007 by guy]HenningBasse - 5-2-2007 at 23:09
Quote:
Originally posted by guy
Quote:
Originally posted by HenningBasse
a more theoretical approach
It should be NH2COO- + NH3
[Edited on 2/6/2007 by guy]
yeah , that's right , just i did it quickly on chemdraw
.
