# Why do we assume that the successive deprotonation of a polyprotic acid happens when only the previous deprotonation has reached equilibrium?

The steps for finding the concentration of the species at the equilibrium of polyprotic acid are as given in this book in section $$10.16$$ and illustrated on the next page. I have three questions, but closely related:

1. In the second step (see the general step and illustration in the book) in setting up the ICE table the author assumes that the initial concentration of the species is as calculated by the previous deportation. But I have doubt about what does he mean by "initial molarity" here? In the regular ICE table, it means the concentration of species at the point of time from which we want to know how much changes in concentration of different species will be made to reach the equilibrium. My question is at what point of time does this "initial molarity" refer to? Is it referring to the concentration just after the $$1st$$ deprotonation? But we can't assume it since the $$2nd$$ deprotonation doesn't wait for the first to reach equilibrium. It will start even after the formation of the first few products of the reaction.

2. When the second deprotonation happens the concentration of the conjugate base ($$\mathrm{H_2PO^{-}_4}$$) starts to decrease, and hence it must affect the equilibrium of $$1st$$ deprotonation (since the conjugate base is part of the $$1st$$ rxn). But the author is not considering the effects at all. And similarly the third will affect the second and first. This is getting complicated.

3. In the second step since the conjugate base ($$\mathrm{H_2PO^{-}_4}$$) is part of the $$1st$$ equilibrium its concentration must not change, by definition of equilibrium but still in the ICE table the author assumes its reaction to be changing, but on what basis? The conclusion follows that the reaction must stop just after the $$1st$$ deprotonation. But that's absurd.

I think that since in the second deprotonation with a decrease in the conjugate base, the concentration of protons increase too (albeit in very low concentration) and this must be keeping the first in equilibrium. But I am not sure.

// If you think three questions in one post is not a good idea, then comment to me I will post the questions separately. Thank you for reading my long (and maybe dumb and illogical) question. Good day.

• In a phosphoric acid solution, all 4 phosphorus-containing species are present, as shown in the alpha fractions diagram here. At any pH, the 4 species have their appropriate equilibrium concentrations. These are calculated by using the full set of coupled equations for the acid dissociations, mass balance, charge balance and water auto-ionization. Tedious algebra giving expressions best evaluated by computers, etc. Then there are simplifying approximations and reasonable shortcuts.
– Ed V
Nov 16 at 2:54
• We use all the actual expressions and skip the approximations. Here I show the final expression, in a spreadsheet I wrote, for a triprotic acid with optional NaOH added. I used it for carbonic acid, which is simply diprotic. There is no approximation except what you say is in those books you reference, which are probably just freshman chemistry books.
– Ed V
Nov 16 at 3:19
• Try to search for site:stackexchange.com phosphoric acid dissociation // See e.g. chemistry.stackexchange.com/questions/149282/… Nov 16 at 8:43
• After some searching here, I finally found the explicit equation for phosphoric acid plus optional strong base, I.e., NaOH. It is exactly the same as the equation I used, with slightly different notation, in the second spreadsheet screenshot at the link to my answer here. If you follow the top link, there are other relevant links there.
– Ed V
Nov 16 at 22:38
• Thank you for your giving me your time. I can feel you have tried very hard to help me. Thank you again. And thanks to Poutnik as well, now I know another googling technique; I found the answer using his suggested search phrase (it was the first link!). Nov 17 at 1:51