# Can precipitation reactions attain equilibrium?

While studying about the characteristics of Chemical Equilibrium, I came across a point in my book which read

A chemical equilibrium can be established only if none of the products is allowed to escape out or separate out as a solid (precipitate)

I understand that if one of the products is a gas and escapes out the reaction won't be able to proceed in the backward direction since that gas is no more a part of the system.

But what I'm not able to get is why it's written that precipitation reactions can't attain equilibrium.

Is there no precipitation reaction which can attain equilibrium? Or is it that the precipitate can't react back chemically with the solution to produce the reactants in reverse?

My teacher told me about about a reaction $$\ce{Ca(HCO3)2(aq) + heat -> CaCO3 +CO2 +H2O}$$

and on cooling the solution and bubbling $$CO_2(g)$$ through it the $$CaCO_3(ppt.)$$ reacts back to form calcium bi-carbonate.

I'm still a bit confused as the reverse reaction doesn't happen spontaneously. Moreover if the reaction is carried out in a closed vessel and constant temperature (after heating), will the reaction reach equilibrium?

# edit

in the answer recently provided, the solid-solution equilibrium between $$\ce{AgCl}$$ and $$\ce{Ag+}$$ $$\ce{Cl-}$$ ions is discussed. But I want to know if in a reaction like $$\ce{AgNO3(aq) + NaCl(aq)\to AgCl(s,ppt) + NaNO3(aq)}$$ the precipitate $$\ce{AgCl(s,ppt)}$$ can react with $$\ce{NaNO3(aq)}$$ in the solution to give the reactants? Can it reach chemical equilibrium?

I've seen this and this but couldn't understand.

Any of your help is of great value to me.

• They simply confused you. Solubility of ionic salts or in general dissolution of solid is not an equilibrium covered by their assertion. They simply means that the precipitation of a solid can shift the equilibrium of a reaction occurring in solution . The same would be for dissolution if you would stop your analysis to individually solvated particles, eg solvated molecule of sugar. – Alchimista Jul 26 '19 at 9:35
• @Alchimista Isn't solid-solution equilibrium a thing?...like dis-solution and precipitation?... Moreover I want to know if there are such reactions where the precipitated product can react back chemically with other products in the solution to produce the reactants under the same conditions? Thank you! – user8718165 Jul 26 '19 at 9:43
• – Alchimista Jul 26 '19 at 9:43
• I said related. You can find what are you looking for, there. – Alchimista Jul 26 '19 at 9:45
• You are right about the fact that there is precipitation equilibrium. Perhaps I weren't clear but it is why I said they confused you. The link you gave as well the much recent question that I have mentioned cover your original doubt. The editing you did meantime seems relates but another question . Yes that would reach equilibrium, if the vessel confines the chemicals in it. – Alchimista Jul 26 '19 at 9:58

The statement is inaccurate. A better statement would be something like:

A chemical equilibrium can not be established if one of the products is continuously removed (i.e. its concentration always decreases).

If one of the products is a gas and can mix with earth's entire atmosphere, that would make its concentration (or partial pressure) drop to almost zero. In order to establish an equilibrium, however, all concentrations have to be non-zero. Otherwise, the equilibrium constant will not be equal to the reaction quotient, and there will continue to be a net reaction.

In case of the precipitation, the solute is not continuously removed. At some point, the solution will no longer be supersaturated, so then the concentration of solute will be constant.

$$\ce{Ca(HCO3)2(aq) + heat -> CaCO3 +CO2 +H2O}$$

and on cooling the solution and bubbling CO2(g) through it the CaCO3(ppt.) reacts back to form calcium bi-carbonate.

Once you change the temperature, you don't expect the reaction to stay at equilibrium. So this does not support either the original statement or the opposite statement.

But I want to know if precipitation reactions can attain equilibrium. Why does the book state otherwise?

Yes, it can. Here is an example:

$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$$

The equilibrium constant expression for this reaction is:

$$K = \frac{1}{[\ce{Ag+(aq)}][\ce{Cl-(aq)}]}$$

(I could have written the chemical equation in the other direction, and then the equilibrium constant would have been the solubility product.) So the product is a solid, and its concentration does not change. The silver and chloride ion concentrations will drop until the equilibrium has been reached.

Here is another set of reactions:

$$\ce{CoCl4^2-(aq) + 6 H2O(l) <=> Co(H2O)6^2+(aq) + 4 Cl-(aq)}\tag{1}$$

$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}\tag{2}$$

Here, the equilibrium of the first reaction will be far on the right side because the second reaction removes a lot (but not all) chloride ions from the solution if the silver ions are in excess of the chloride ions. Still, the system will reach equilibrium.

Why does the book state otherwise?

I don't know. Maybe they consider an equilibrium where one of the species is at extremely low concentration not an equilibrium. Or maybe it is a typo.

It is great to have a critical mindset when reading books and hearing people talk about chemistry. Even if all the statements are true, it helps you learn.

edit

I want to know if in a reaction like $$\ce{AgNO3(aq) + NaCl(aq)\to AgCl(s,ppt) + NaNO3(aq)}$$ the precipitate $$\ce{AgCl(s,ppt)}$$ can react with $$\ce{NaNO3(aq)}$$ in the solution to give the reactants?

Yes. I would rather write it like this, though:

$$\ce{Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) <=> AgCl(s) + Na+(aq) + NO3-(aq)}$$

and then leave out the spectator ions:

$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$$

The following "reaction" might serve as illustration why leaving the spectator ions in the equation is not a good idea in preparation for writing the equilibrium constant expression:

$$\ce{KCl(aq) + NaNO3(aq) <=> KNO3(aq) + NaCl(aq)}$$

For this "reaction", the concentration of species is undefined. Writing the equilibrium constant would be strange because how can I tell apart $$\ce{KCl(aq) + NaNO3(aq)}$$ from $$\ce{KNO3(aq) + NaCl(aq)}$$ if they fully ionize in solution? If I show the ions separately, it becomes clear that the "reaction" is not a reaction:

$$\ce{K+(aq) + Na+(aq) + NO3-(aq) + Cl-(aq) <=> \mathrm{the\ same???}}$$

I've seen this and this but couldn't understand.

Those two Q&A's are quite confusing, I agree.

• Thanks....I upvoted you. But I want to know if precipitation reactions can attain equilibrium. Why does the book state otherwise...please tell me...i get the escaped gas point...the ppt. point is troubling me... thanks – user8718165 Jul 26 '19 at 14:03
• @user8718165 I added your question in the comment to my answer. – Karsten Theis Jul 26 '19 at 15:55
• @user8718165 I updated my answer to respond to your updated question. Next time, maybe ask a new question. – Karsten Theis Jul 27 '19 at 2:32
• Thank you so much. Now I fully understand it. I upvoted and accepted your answer. Hope I didn't cause you much trouble. – user8718165 Jul 27 '19 at 2:49