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Author: Subject: Paraformaldehyde & formalin question
FriendlyFinger
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[*] posted on 24-9-2005 at 23:32
Paraformaldehyde & formalin question


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?
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Rosco Bodine
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[*] posted on 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 .
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S.C. Wack
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[*] posted on 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.
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[*] posted on 28-9-2005 at 20:49


Watching the paraformaldehyde dissolve in hot water is not the same as depolymerisation is it?
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[*] posted on 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.
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