If one would make no assumtions to try to simplify, except for just assuming OH- were negligible in a solution of weak diprotic acid, H2A, we can create four equations to represent concentrations for : H2A == H + HA- HA- == H + A- Each of those has a K value for equilibrium constant expression. They would be K1 = [H][HA]/[H2A] K2 = [H][A]/[HA] Letting F equal the formality of the H2A then allows the mass balance and charge balance equations: F = [H2A] + [HA] + [A] [H] = [HA] + 2[A] + [OH] but we might usually justly assume hydroxide OH is negligible. An analytical textbook stated that the system of those four equations can be solved for each of the variables. I then tried doing this to obtain an equation having H as the only variable, and I obtain this equation: H^3 + K1H^2 + (K1K2 - FK1)H - 2(FK1K2) = 0 At this point I am not using brackets, but H stands for molarity of hydronium; and the "2" shown in front of the last term is a factor in order to distinguish it from the subscripts on the K values (since I am unable to more conventionally type-set my work here). The point is that the last equation is a cubic equation, not easy to solve. Best to be done, I guess would be use either graphing calculator, or create a computer program to help find the roots for the variable, H. Is that cubic equation practical or useful? Certainly, once a good solution is found for H, the other unknowns will be easy. But does anyone actually work with a cubic equation in a weak-acid equilibrium situation?