How is NaOH able to clear a solution of Paraformaldehyde?
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):

$$\ce{(CH2O)_{$n$} + H2O &lt;=&gt; $n$ CH2O + H2O &lt;=&gt;&gt; CH2(OH)2}$$

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.

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