Second Deprotonation? (6E.3)

[Chesca Legaspi 2E]

The textbook problem, 6E.3, asks us to calculate the pH of a couple of diprotic acids, ignoring second deprotonations only when approximation is justified. How do we know when approximation is justified and the second deprotonation can be ignored? The textbook seems to imply you can ignore the second deprotonation whenever Ka2 < Ka1, but in the chart, Ka2 is always less than Ka1. Does this mean that you can always ignore the second deprotonation because it will always be smaller than the first, or is the chart given just an exception?

[Chem_Mod]

Ka2 will always be less than Ka1, so that is not the rule to follow. Generally if Ka2 is 3 magnitudes or more lower than Ka1, it can be ignored.
One can also check using the 5% rule.

[Natalie Do 3F]

The book usually defines Ka2<<Ka1 to be a difference of 1000 or 1x10^3. It is similar to how subtracting x is arbitrary in calculating the equilibrium constant. The 5% rule applies to both cases.

[Kelly Yun 2I]

Ka2 is always less than Ka1, because once it is deprotonated, it will have -1 charge from its original state and will not try to lose more protons.

[vanessanguyen3I]

I agree with all the posts above. Just make sure you know that the second deprotonation of H2SO4 is an exception!

[Karl Yost 1L]

It is difficult to remove a positively charged proton from a negatively charged species (in this case, a di-/triprotic acid that has already been protonated once). Therefore, the acid that has already been protonated once will dissociate to a lesser extent in the second deprotonation, as reflected in the significantly smaller Ka2 values (H2SO4 is an exception!).

[Ellison Gonzales 1H]

My understanding is to keep an eye out for sulfuric acid and when you come a cross it, know it requires a 2nd deprotonation