# Dissociation equilibrium for nicotinic acid

Nicotinic acid is an amphoteric compound. Does it mean that in very low $$\mathrm{pH}$$ it will be more soluble?

Would the following be possible?

\begin{align} \ce{H2X+ &<=> H+ + HX}\\ \ce{HX &<=> H+ + X-} \end{align}

• Low pH will drive both reactions to the left so my guess would be yes, since $H_2X^+$ is presumably more soluble than HX. – Night Writer Feb 6 at 10:57
• Thank you. My question arises from the fact that most of the sources state that nicotinic acid is just a monoprotic acid. I would like to know if I really should consider nicotinic acid as positively charged (at the pyridinyl hydrogen) in an aqueous solution of low pH. Would it be very possible according to its structure? or maybe this is just an option, which on reality cannot be observed? – KW1 Feb 7 at 12:33
• I think you can observe that behavior. This is a bit old but may help: chempap.org/file_access.php?file=331a52.pdf – Night Writer Feb 7 at 12:46