How are acid-base equilibria addressed for polyprotic acids such as #H_2SO_4#, and #H_3PO_4#?

2 Answers
Aug 9, 2015

By consideration of #pK_(a2)# for dissociation of the anion.

Explanation:

A case in point is sulfuric acid, #H_2SO_4#. You are no doubt well aware that this species gives rise to 2 equiv #H^+# in solution. Nevertheless, dissociation from the 1st ionization product, #HSO_4^-#, bisulfate anion, is much less favourable than for the first. Even if we knew nothing about chemistry (and the more you learn the more you realize how much you don't know), we realize that it should be harder to remove a positively charged proton from an anionic than from a neutral species. From this link we learn that #pK_(a2)# = 1.99 for #HSO_4^-#; #pK_(a1)# for this acid in water is so high that it is virtually unmeasurable.

We get a similar situation with a formally triprotic acid, #H_3PO_4#, #pK_(a1)# = 2.16; #pK_(a2)# = 7.21; and #pK_(a3)# = 12.32. It has been a long time since I have done an aqueous titration, but I am willing to bet that you couldn't get to a stoichiometric endpoint of #Na_3PO_4#, and would finish with #NaH_2PO_4#.

Aug 10, 2015

Consider the idea that every molecule is in equilibrium with another, but the extent to which that equilibrium remains balanced, or whether or not it's one-sided, depends on the magnitude of difference between the acid and conjugate base or base and conjugate acids' #pK_a#s (or #pK_b#s, if you want to use those).

If you recall that a high #K_a# indicates a strong acid, and #pK_a# is #-log(K_a)#, then #10^(-pK_a)# for a more negative #pK_a# gives a large #K_a# and corresponds to a stronger acid.


In water:

(note that I will call the first conjugate base the second acid, and the second conjugate base the third acid, and so on, provided it is the case that all these acids still have protons)

Case I
When the #pK_(a_1)# (the #pK_a# of the first acid) and the #pK_(a_2)# (the #pK_a# of the second acid) are both lower than the #pKa# of water (which is #~15.7#), the diprotic acid most likely dissociates twice, and once, other times (depending on the molecular orientations; collisions must occur for proton donation, but these are minor details).

Case II
When the #pK_(a_2)# is instead higher than the #pK_a# of water, it's most likely that the diprotic acid only dissociates once.


In other words, the acid with the higher #pK_a# remains in its current form more of the time than the acid with the lower #pK_a# because it's the stronger base or weaker acid that dissociates less than the weaker base or stronger acid.