Posted by u/LoopyBullet
I was reading an article in propofol pain in injection, and I’m having trouble understanding a paragraph on the bottom right portion of the first page, starting with, “It is not clear how…” I don’t understand the author’s rationale from an acid-base standpoint, and was hoping someone could please help me out. I apologize, as I know this is undergraduate stuff. Maybe the way the author’s sentence starts is a clue as to why I don’t understand it (that’s a joke…).
I understand that there are local and/or systemic analgesic effects from lidocaine, but I’ll ignore that for now. It’s just the acid-base stuff that doesn’t make sense to me.
Anyway, this is my understanding:
-Propofol is a weak Brønsted-Lowry acid and releases H+ in solution. This weak acid has a pKa of 11, so when put in a more acidic solution such as physiologic pH of 7.4, the protons don’t dissociate from the propofol molecule, and propofol remains essentially 100% non-ionized (lipid soluble).
-Propofol is a phenol (which is caustic) and irritates the venous endothelium. More propofol in the aqueous phase translates to more venous irritation. As pH of a solution rises closer to (or above) 11 and becomes more basic, more propofol enters the aqueous phase (ionized), irritating the veins.
-Lidocaine is a weak Brønsted-Lowry base, and binds to H+. Putting lidocaine into a solution will INCREASE the pH of the solution.
-This is where I get tripped up – I would think that by putting lidocaine in a mixture with propofol, the lidocaine will raise the pH of the solution by binding to H+. By raising the pH of the solution, the solution’s pH is now closer to propofol’s pKa (11). This causes more propofol to be ionized, and this, be water soluble. However, more water soluble propofol means more propofol available to irritate the venous endothelium, which is counter-intuitive.
-Stated plainly, protons from propofol will be bound to lidocaine molecules. This causes propofol to be ionized and soluble in water, irritating the veins (again, which is counter-intuitive).