[sodium.dioxid]

Ka = [H+][A-]/[HA]

Why is it necessary to have both [H+] and [A-] on the numerator? It seems like we could have done just as fine using the definition "Ka=[A-]/[HA]" or "Ka=[H+]/[HA]" instead of "Ka=[H+][A-]/[HA]". Am I missing something?

Also, does [H+] include the hydronium ions that existed in water before the acid was added? Or is it exclusively the hydronium ions produced by the acid?

[ramboacid]

Why is it necessary to have both [H+] and [A-] on the numerator? It seems like we could have done just as fine using the definition "Ka=[A-]/[HA]" or "Ka=[H+]/[HA]" instead of "Ka=[H+][A-]/[HA]". Am I missing something?

Well, maybe that plan could plausibly somehow work out for acid-base equilibrium, but for a general reaction how would you make the judgement call to simply ignore a product? Also, that plan would only work if there was no initial concentration of product already in solution. What if you added a weak acid to an already acidic solution? It would dissociate less because of the ambient [H⁺] concentration. In that case, you'd need to factor in the ambient [H⁺] somehow into your equilibrium.

The [H⁺] concentration does include it technically, but consider how small the ambient [H⁺] concentration is in pure water: it's something around 1*10^-7 M. Usually, when you are adding acid or base, they increase or decrease [H⁺] by a way larger amount than the ambient concentration, so the ambient concentration is essentially insignificant. When you are dealing with highly dilute solutions of weak acids where the [H⁺] added by the acid is comparable to the ambient [H⁺] concentration, then you have to take it into effect.

The law for making equilibrium constants and expressions is called the Law of Mass Action. There are a plethora of internet resources describing it if you wish to delve into the nitty gritty parts. 

[Jorriss]

If you want to see reaction equilibria rigorously one must appeal to thermodynamics. Have you taken thermo?