Please help me clarify the title if you come up with a better way to put it (you certainly will).
My question is the following:
If you make a mix of equal parts of various acids (Bicarbonate, Hydrochloric, Nitric, Sulfuric, etc.) and throw some metal in there that will ionize and wait for a while until you have an equilibrium, what will that resulting equilibrium look like vs. the known solubilities of the expected salts? Where and in what form will the metal ions most likely end up? Will the metal's solubility tend towards an average of the expected solubilities? Or tend towards the solubility of the least soluble metal salt? Or of the most soluble metal salt? Or exceed the solubility of the most soluble salt?
This is assuming that no acid is / would have been limiting in reacting with the metal in the first place.
My assumption is that most metal ions would end up precipitating as the least soluble salt but I'm very much unsure of myself.
Upon Poutnik's advice, here are two examples I provided in the comments:
Case No 1: I drop 1 mole Cadmium metal in a solution with 5 moles of sulfuric acid and a great excess of water beyond the solubility limit of the product (I assume it would dissolve entirely), I then add 5 moles of phosphoric acid. Would that (a) precipitate most of the nickel (almost as if only phosphoric acid was present), or will (b) the solubility be somewhere between that of the sulfate and phosphate salts, or (c) most of the nickel remain in solution as if only the sulfate was present?
Case No 2: I drop 1 mole of lithium and 1 mole of cadmium metal into 2 moles of sulfuric acid with enough water for both to be entirely dissolved. I then add one mole of phosphoric acid, leave the solution in a sealed container with heavy stirring and come back 1000 years later. I then filter out and analyze the precipitate: Will it be all cadmium / mainly cadmium / an even mix of both cadmium and lithium phosphate salt? (knowing both cadmium and lithium form insoluble phosphate salts, but the solubility of the lithium salt is several orders of magnitude above that of the cadmium salt). Let's assume there is no double salt formed for the sake of simplicity.