[Brian Lin]

I know that for a diprotic acid with fairly close ka values result in a titration curve of only one break, because "protons will be titrated almost simultaneously". I know their half way points will be so close to each other, further making the curve just one curve, instead of two.

What I don't get is that how can two acids be deprotonated at the same time if "the best base reacts with the best acid since acidity is relative" (quoted directly from Klein's Organic chemistry book). Therefore, shouldn't the base react with the most acidic proton first before reacting with the 2nd proton that is weaker?

I'm not sure what concept I have down wrong, so please help.

[Borek]

What I don't get is that how can two acids be deprotonated at the same time if "the best base reacts with the best acid since acidity is relative" (quoted directly from Klein's Organic chemistry book). Therefore, shouldn't the base react with the most acidic proton first before reacting with the 2nd proton that is weaker?

Sadly, the book is misleading as quoted. Yes, the strongest bases/acids reacts first, but it doesn't mean other bases/acids don't react at the same time, just to a lesser amount.

Imagine a diprotic acid H₂A with dissociation constants K_a1 and K_a2.

[tex]K_{a1} = \frac {[H^+][HA^-]}{[H_2A]}[/tex]

[tex]K_{a2} = \frac {[H^+][A^{2-}]}{[HA^-]}[/tex]

Simple rearrangement of these formulas gives:

[tex]\frac {[HA^-]}{[H_2A]} = \frac {K_{a1}}{[H^+]}[/tex]

[tex]\frac {[A^{2-}]}{[HA^-]} = \frac {K_{a2}}{[H^+]}[/tex]

Assume pK_a1 = 4 and pK_a2 = 5 (so a diprotic weak acid, with close dissociation constants). Assume pH = 5 (so [H⁺] = 10^-5). You can plug both dissociations constants and concentration of H⁺ into the right had sides and see, that all three forms of acid (H₂A, HA^-, A^2-) are present, and what their relative amounts are.

Apparently while the strongest acid (H₂A) reacted first, the weaker one (HA^-) reacted as well.