I read on another equilibrium question that if two equilibria occur at the same time, then it is not accurate with actual stoichiometry to add the two equations and multiply the equilibrium constants $K_1$ and $K_2$ in order to get an equation with stoichiometric consistency. In , multiple ICE tables are used to get around the problem. This makes the most sense to me.
However, when dealing with solubility of compounds with complex formation, many textbooks (Atkins) and websites (libretexts Oxtoby) show that you just add the two equations together and calculate solubility from there. From the libretexts site, Example 16.5.2, Equation 16.5.15 - 17 does the same operation as what I thought was forbidden/not accurate above. Here, the website says $\ce{{AgCl} + Cl^- <=> [AgCl_2^{-}]}$, which I do not think is correct. edit: I understand that this equilibrium is correct in the way that $\frac{\ce{[AgCl_{2}^{-}]}}{\ce{[Cl^-]}}$ is constant, but using the new equilibrium to solve based on stoichiometry (when the website says that
$K=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]}=\dfrac{x}{1.0 - x}\approx1.9\times10^{-5}=x$
can be used to solve for $x = \ce{[AgCl_{2}^{-}]}$) is off to me.
Which one is right? Can you actually add two simultaneous equilibria together to simplify the problem with easier stoichiometry? Or is this some sort of approximation that works because of the specific conditions that most solubility problems have? I feel like I have a big misunderstanding here.
EDIT: I figured it out. For solubility-complex reactions, the above is an approximation based on the relative magnitudes of the equilibrium constants. Because complex formation is so favorable the majority of the time, we can assume that almost every mol of solid dissolving into solution will be consumed to form complex ions. I didn't see this anywhere I looked so I figure I might as well post it here.
