[engelbrekt]

In polyprotic acids, why is the last equivalence point the sharpest on a titration curve?















Let me see if I've understood this now.































Before the first equivalence point there are two acid equilibria and buffer equilibria neutralizing the added NaOH. The capacity to buffer and keep a stable pH diminishes as the acid is neutralized. At the second equivalence point there is no longer any acid to buffer and neutralize NaOH, so the pH shoots up rapidly.































Have I understood this correctly?

[Zor Prime]

Lancenigo di Villorba (TV), Italy































You treated about titrimetry of polyprotic acids.















You stated that :In polyprotic acids, the last equivalence point the sharpest on a titration curve.































You get reason.















Now, I start from a diprotic acid which i will define H2X like a WEAK DIPROTIC ACIDIC SPECIES able to ionize in aqueous solutions like it happen by following equilibria































i) H2X(aq) + H2O(aq) --- HX-(aq) + H3O+(aq)















Ka1 = |HX-| * |H3O+| / |H2X|































ii) HX-(aq) + H2O(aq) --- X--(aq) + H3O+(aq)















Ka2 = |X--| * |H3O+| / |HX-|































So, once you began ACID/BASE TITRIMETRY, you began addiction of chemical equivalent of Strong Base (e.g. usually NaOH) to















Title-Defined aqueous solution of Acidic Species H2X.















So, Acidic Species H2X exists besides its former derivative, e.g. HX- coming from former Chemical Equilibrium Relationship. Since H2X is Acidic Species and HX- is its Salts, it is appearing a Chemical Buffer obeying to HASSELBACH WRITING































pH = pKa1 - log (|H2X| / |HX-|)































Some times later, NaOH added get meaning to Second Chemical Equilibrium : this happen once Chemical Equivalent of Added NaOH moved over half Chemical Equivalent of H2X Acids.















THIS IS LATTER EQUIVALENCE POINT WHICH OBEYS TO ADAPTED HASSELBACH WRITING































pH = pKa2 - log (|HX-| / |X--|)































SINCE NaOH FORMED A SECOND CHEMICAL BUFFER.















This Chemical Buffer has a very low BUFFER CAPACITY BECAUSE EVERY NaOH ADDICTION TO THIS LATTER CONDITION AREN'T ABLE TO NEUTRALIZE ACIDIC SPECIES SO IT AFFECT HEAVILY pH INDEX.































I hope this helps you.