Lidocain pka 7.6

Prolocain pKa 8.4

 

We are told that a decrease in the amount of base facilitates removal of the local anesthetics, resulting in a shorter duration of action. Which has a more rapid onset?

 

The answer was that it was Lidocaine because more exists in its base form at body pH. However, this seems counterintuitive since both pKa's are higher than the pH. I do remember reading a rule somewhere that if pKA>pH, then the group is considered to be in its acidic form when attached to an aa, but I don't know if that rule only applies to aa and what that rule means. Could someone please clarify it?

Are you familiar with the Henderson-Hasselbalch equation?

 

[math] pH=pKa + \log \left ( \frac {[\mathrm{A^{-}}]}{[\mathrm{HA}]} \right ) [/math]

 

So if [math] \mathrm{[A^{-}] \geq [HA] } [/math], then [math] pH \geq pKa [/math]

 

In plain language, if the molar concentration of the conjugate base form of an acid is greater than or equal to the molar concentration of the protonated form then the pH of that acid is greater than it's pKa.

 

The rule you speak of applies in all aqueous solutions [where pH is well defined] if pKa is greater than pH then there will be more of the species in the protonated form.

 

Lidocane's pKa is lower than Prolocain's so Lidocane is more acid. This means that a greater fraction of Lidocane molecules will be in the conjugate base form when compared to the fraction of molecules of Prolocain that are in the conjugate base form in a solution of the same pH.

Edited March 4, 201213 yr by mississippichem