Percent Protonation/Deprotonation

[Ananya Ravikumar 1I]

Is the percent protonation of a base the {(concentration of conjugate acid at equilibrium)/(initial concentration of base)}*100 or the {[OH-] at equilibrium)/(initial concentration of base)}*100?
Similarly, does the percent deprotonation for an acid relate to the equilibrium concentration of its conjugate base or H3O+?

Just making sure since the products of a dissociation reaction may not always be present in the same number of moles
Thanks!

[Shania Hao 3L]

That is a good question, I would also like clarification on this.
In the example given in the textbook for the percent protonation of a base, the concentration of the conjugate acid was used. For the example of the percent deprotonation of a weak acid, the [H3O+] was used. Thus, I believe to calculate the percent protonation/deprotonation, you are considering the part that had [H+] added to it, meaning it would be the [BH+] in a base and [H3O+] in an acid.

[Ananya Ravikumar 1I]

Thank you!

[Payton Zarceno 1E]

Hi! I think Shania is correct, when you're considering a base I believe you need to use [BH+]/[the initial B + new concentration of B].

[Emmy Flores 3F]

Yes, think how percent protonation is the how much a molecule gains a proton while deprotonation is the percent of the molecule that gives away protons.

[Anna_Aldridge_2C]

Hi!
I believe the percent protonation in your first assumption is correct. The protonation will take a proton (H+) from the water and bond with the base, making HB+ and OH-. Because the focus is on the ratio between the initial amount of base and the amount that is protonated, the calculation would be [HB+]/[Binitial]. Similarly, the percent deprotonation is the ratio between the conjugate base (amount that has had protons removed) and the initial amount of acid, and not to equilibrium.