as initial remark:
concentration of SO42- (Kb1=8.3 · 10-13 M, Kb2≈0)
K
b2 (SO
42-) would be - 17 , approx.
with respect to your original question:
if we had a polyacidic system , combined with a pH-active cation (like in the hypothetical ammoniumphosphate *
)), we would estimate the pH-outcome by looking for the most relevant species (here: of phosphoric acid) to be considered, and neglect the rest
i.e.: we would judge PO
43- vs. NH
4+ , meaning to compare pKb= 1.67 vs. pKa=9.75
clearly , even with three times the concentration of ammonium around, phosphate will be the winner here, and the solution will be strongly basic
for more detailled calculations , or a situation where the differences pKa / pKb are not that clear or where there are amphotheric species involved (let's say (NH
4)(HCO
3) for example ) good estimations become a problem, and at times you hence can't circumvent charge balance equations based calculations
regards
Ingo
*
)solid tri-ammoniumphosphate is inexistant, as it will decay to diammoniumhydrogenphospohate + ammonia.
only the trihydrate (NH
4)
3(PO4) * 3 H
2O is stable and can be isolated
(if memory serves, this is (NH
4)
3(HPO4)(OH) * 2 H
2O in reality)