[Rutherford]

On this link: http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt,  below the equation 12.12 is a sentence "Comparing 12.8 and 12.12 it is obvious that both assumptions that allowed to derive equation 12.8 are equivalent to the assumption that quotient [H₂A]/[A^2-] is close to 1." How can they be close to 1, if I had for example 0.001M of a weak amphiprotic substance Ka₁=10^-3 and Ka2=10^-7, [H₂A]  should be much lower than [A^2-]? I need an explanation because that way is shorter and easier to do fast on paper that the one above it.

[Borek]

No idea what kind of explanation you are looking for. First derivation uses well defined assumptions - if they are wrong, ratio is probably far from 1.

Your example (0.001M HA^-, pKa1,2 = 3 & 7) yields a solution with pH 5.15 (instead of expected 5) and ratio

[tex]\frac{[H_2A]}{[A^{2-}]} = \frac{6.9\times10^{-6}}{1.4\times10^{-5}}=0.49[/tex]

Apparently you can't use approximate formula for this case. Hardly surprising, tables below shows that below 0.01M one has to expect problems.

[Rutherford]

I was aiming at the difference between K_b1(10^-11) and K_a2(10^-7). I have the reactions:
HA^- H⁺+A^2- K=10^-7
HA^-+H2O H₂A+OH^- K=10^-11
I don't understand how possibly, whatever concentration I take, [H₂A]≈[A^2-] because H₂A is made tougher than A^2-. This explanation I need.

[Borek]

I don't understand how possibly, whatever concentration I take, [H₂A]≈[A^2-]

Nothing to explain here, as it is not true. It is the other way around - as long as these concentrations are close to each other, you can use formula [itex]pH=\frac 1 2 (pK_{a1}+pK_{a2})[/itex]. When they are not close to each other, the formula fails. As explained on the ChemBuddy page and as I mentioned in the previous post, these concentrations are comparable only as long as starting concentration of HA^- is large enough (and both dissociation constants have reasonable values, neither too high nor too low).

[Rutherford]

(and both dissociation constants have reasonable values, neither too high nor too low).

I just wanted to know this. Therefore, I should find a reserve way to calculate the pH. Thanks.