FriendlyFinger - 24-9-2005 at 23:32
If you want to make 100g 37% formalin from paraformaldehyde, is it as simple as adding 37g (100%) paraformaldehyde and adding water to make it up to
100g and then heating?
I ask because I've seen methylamine synths with paraformaldehyde and NH4CL where a lot more water is used than the corresponding synth with
formalin and NH4CL. Why is that?
Rosco Bodine - 25-9-2005 at 12:02
A small amount of acid catalyzes the polymerization of formalin to paraformaldeyde . Methanol is used as a polymerization inhibitor in formalin ,
particularly to combine with any formic acid traces from manufacture or from oxidation on storage , which would otherwise cause a lowered pH and
reduce the storage life due to polymerization of formalin in the bottle .
Conversely , just as acid catalyzes the polymerization of formalin to paraformaldehyde , strong bases like NaOH in small amount catalyze the
depolymerization of paraformaldehyde in water to formalin , when the mixture is heated .
S.C. Wack - 25-9-2005 at 20:14
The rate of depolymerization in plain water depends of the degree of polymerization, the lower polymers have a mp of 120C IIRC, so one can check what
they have by mp. Some specially prepared lower paraforms are easily soluble, while higher ones might require weeks of stirring in plain water to
dissolve at 15C.
Refluxing for a few hours should depolymerize most paraforms.
FriendlyFinger - 28-9-2005 at 20:49
Watching the paraformaldehyde dissolve in hot water is not the same as depolymerisation is it?
S.C. Wack - 29-9-2005 at 03:03
Although there is little if any monomer in solution ever as such, it isn't polymer either. AFAIK, everything defined as a polymer of formaldehyde
is insoluble.