organic chemistry - How is NaOH able to clear a solution of Paraformaldehyde?

While making a solution of $4~\%$ para-formaldehyde for immunohistochemistry (to fix the tissue), we add para-formaldehyde into water and stir. It looks like a chalky supersaturated solution even after stirring with a magnetic stirrer. And as per protocol, you are supposed to add 2 drops of $2~\mathrm{N}\ \ce{NaOH}$ to clear the solution. It surprisingly turned crystal clear!

I was amazed at how just 2 drops could make such a huge difference. Does it react with para-formaldehyde? Does changing pH change solubility? But para-formaldehyde is a polymer, how could it suddenly become soluble? Does adding $\ce{NaOH}$ make it a monomer?

Answer

para-formaldehyde consists of long chains of the following type:

$$\ce{HO-CH2-O-[CH2-O]_{$n$}-CH2-OH}$$

When being dissoluted in water, this needs to be broken down into formaldehyde monomers (and then further into formalin):

$$\ce{(CH2O)_{$n$} + H2O <=> $n$ CH2O + H2O <=>> CH2(OH)2}$$

Acids and bases are both able to speed up the first step. Bases by deprotonating one end and causing a domino effect of bonds being broken and formed with a hydroxide released on the other end. Acids by protonating one end and then the domino occuring in the other direction.

$$\ce{HO-[CH2-O]_{$n$}-CH2-O-CH2-O-CH2-OH ->[\ce{OH-}] \\HO-[CH2-O]_{$n$}-CH2-O-CH2-O-CH2-O- ->\\ HO-[CH2-O]_{$n$}-CH2-O-CH2-O- + H2C=O -> \\HO-[CH2-O]_{$n$}-CH2-O- + 2 H2C=O ->}\\ \dots\\\ce{HO- + ($n$+3) H2C=O}$$