
Stanton D. answered 07/04/24
Tutor to Pique Your Sciences Interest
Hi Jordan B.,
you solve this type of problem just as you would solve any other type of equilibrium problem. You have probably been given K values for various ionic association reactions, such as Cu(II) + H2O = Cu(II)OH + H(+) and Cu(II) + CO3(2-) = Cu(II)CO3 , HCO3(-) = H(+) + CO3(2-) and so on. Here, you merely must work your way methodically through each of the possible complex species formations -- Cu(II)HCO3 (+) , Cu(II)CO3 (including the possibility of that precipitating out!), Cu(II)OH(+) , and so on, then go back and view whether tweaking any of the "starting" ICE concentrations is required, and re-solve, etc. This is the iterative method of solving multiple complexation reactions occuring simultaneously. It may seem clunky and roundabout the first time you do this, but you will acquire both calculation speed and estimation finesse as you process more of these types of problems. Don't make the assumption that "somebody else will solve these / an expert system will solve these for me"; it will almost always take you longer to find a program to do this than it will take you to longhand solve yourself.