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.

  • $\begingroup$ I didn't know that. Do I understand well that bicarbonate is amphoteric? $\endgroup$
    – Hans
    Commented Nov 3, 2020 at 12:59
  • $\begingroup$ I assumed the acid was reacting with the base (calcium or magnesium in the case of the corresponding carbonates), thus releasing carbonate beyond its water solubility such that water expels CO2 $\endgroup$
    – Hans
    Commented Nov 3, 2020 at 13:14
  • 1
    $\begingroup$ Note that $\ce{LiH2PO4}$ solubility is very good. solubility of $\ce{Cd(H2PO4)2}$ and $\ce{CdHPO4}$ is not available. Also $\ce{Cd3(PO4)2}$ solubility in strong acidic environment is much higher than in water. $\endgroup$
    – Poutnik
    Commented Nov 4, 2020 at 10:41
  • $\begingroup$ Hmm. All of the parameters affecting solubility makes it very complicated. I'll try to understand that better. $\endgroup$
    – Hans
    Commented Nov 4, 2020 at 11:19

1 Answer 1


All strong acids dissociate in water: $\ce{HX + H2O -> H3O+ + X-}$.

Metals reacts with hydronium ions, not with a particular acid, e.g. like $\ce{M(s) + 2 H3O+ -> M^2+ + H2(g) + H2O}$. ( With nitric acid, it is more complicated due its reduction to $\ce{NO}$ or $\ce{NO2}$, but it is not important now. )

Metallic cations and acid anions are not paired with a particular counter ion, until they start precipitating. They all form together a "ionic soup".

As a rule of thumb, the guidance is the salt solubility. But in case of mixed anions and/or cations and non stoichiometric cation/anion ratios, more important is solubility product

$$K_\mathrm{sp}=[\ce{M^n+}]^{m} \cdot [\ce{X^m-}]^{n}$$

The salts that would reach their solubility product would start precipitate, but then hard-to-control kinetic factors of crystallization initiation and propagation take part.

The excess of the counter ion decreases solubility of the salt, compared to salt solubility in water. It is similar effect as gravimetric precipitation of very little soluble salts, where excess of the counter ion causes much lower ion residual concentration, compared to being just in water.

Other ions may alternate salt solubility by side reactions ( if applies ) with the considered ions, or by affecting activity coeeficients - see below.

For the case 1, sulphates would have the minor effect of decreasing somewhat the activity conefficients that therefore somewhat increased solubility of the phosphate. But it is general aspect of ions presence.

For the case 2: It depends on if there is formed a double salt like LiCdPO4 or not.

In particular specific cases, there is possibility of forming double salts, where precipitate consists of integrated more than 1 cation or anion. E.g. crystallization from solution containing $\ce{Fe^2+, NH4+, SO4^2-}$ may lead to crystallization of "Mohr salt" $\ce{(NH4)2Fe^{II}(SO4)2 \cdot 6 H2O}$. Similarly for alums with $\ce{K}$ and $\ce{Al/Cr/Fe^{III}}$ as $\ce{M^{I}M^{III}(SO4)2 \cdot 12 H2O}$.

During rapid crystallization there occurs also coprecipitation:

There are three main mechanisms of coprecipitation: inclusion, occlusion, and adsorption. An inclusion occurs when the impurity occupies a lattice site in the crystal structure of the carrier, resulting in a crystallographic defect; this can happen when the ionic radius and charge of the impurity are similar to those of the carrier. An adsorbate is an impurity that is weakly bound (adsorbed) to the surface of the precipitate. An occlusion occurs when an adsorbed impurity gets physically trapped inside the crystal as it grows.

In the case of rather concentrated solutions of well soluble salts, there is complication we should consider the ion activities instead:

$$K_\mathrm{sp}=a_{\ce{M^n+}}^{m} \cdot a_{\ce{X^m-}}^{n}$$

Acticity coefficients for diluted solutions can be calculated by Debye-Huckel theory, but for concentrated solutions it is hard to theoretically predict them. We can use concentrations for rough predictions, but the numbers may not fit well.

  • $\begingroup$ Thanks for this elaborate answer Poutnik! As a conclusion, should one expect most of the metal to precipitate as the least soluble of the possible salts (assuming there is enough of the counter-ion)? Or its solubility to be increased by the presence of counterions for which the corresponding salt has high solubility? $\endgroup$
    – Hans
    Commented Nov 3, 2020 at 12:40
  • $\begingroup$ I'm not sure what the phenomenon of passivation is. Until now I thought it only involved the solubility of a compound (such that upon reacting it would form an insoluble surface layer protecting the metal or alloy from further reaction and dissolution). $\endgroup$
    – Hans
    Commented Nov 3, 2020 at 13:10
  • $\begingroup$ So for case 1, the addition of a counter-ion with which the metal ion forms a salt with very low solubility would not decrease its solubility but very slightly increase the amount of metal ion in solution? For case 2, let's assume there is no double salt formed for the sake of simplicity. (I had initially not thought about the possibility of double salts) $\endgroup$
    – Hans
    Commented Nov 3, 2020 at 14:18
  • 1
    $\begingroup$ Please, delete eventually already obsolete comments. Generally, if you need to clarify an asnwer, use comments. If you have rather followup questions, extend the question and put in a temporary comment about Q update. You can notify the user by @username . $\endgroup$
    – Poutnik
    Commented Nov 3, 2020 at 14:25
  • $\begingroup$ I followed your recommendation about the comments. If you know to provide an answer for case no 2 when the possibility of double salts is left out (for simplicity), it would be really great to have this new tool of knowing what to expect to happen in such a case. $\endgroup$
    – Hans
    Commented Nov 3, 2020 at 17:41

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