I googled a bit about what $\ce{AgNO3 + AlCl3}$ will yield and found out the following:

$\ce{3AgNO3 + AlCl3 -> Al(NO3)3 + 3AgCl}$

Why is that? I know that $\ce{Ag}$ is higher up in the reactivity series than $\ce{Al}$ but that does not make sense to me in this problem. Is it because $\ce{Cl}$ is just one when $\ce{NO3}$ is multiple?


3 Answers 3


Le Châtelier's principle only states that a system previously at equilibrium will want to stay at equilibrium - that is, if we perturb it, it will try to go back to equilibrium. This system is initially not at equilibrium, and therefore we don't need his principle.

I think the reaction proceeds because the formation of AgCl(s) is enthalpically favored. Molecular cohesion (enthalpy) is greater in AgCl than in AgNO3, thus promoting its formation. How do I know that? Well, it seems AgCl is less ionic than AgNO3 since it is insoluble in water.

This reaction is a type of metathesis reaction. It can be called double displacement reaction or salt metathesis reaction, depending on who you're talking with.


The other answers make an error, I think. Both suggest that the solubility product of AgCl is of importance in determining why the reaction takes place. My opinion is that the argument, as presented, is circular, in the following manner:

The solubility product of AgCl in water is very low, therefore we observe formation of a precipitate.


AgCl is insoluble in water, therefore its solubility product is low.

It's easy to see that the fact AgCl is insoluble in water implies that it has a low Ksp in water. If someone can predict and quantify the solubility product of AgCl in water from first principles, then that's another story, but the explanations offered only rely on observation (experiment) to quantify the Ksp.

My argument goes down the ladder. If you can answer why AgCl is insoluble in water, then you can definitely predict that Ag+ and Cl- ions from different sources (i.e. in different solutions) put into contact will spontaneously form a precipitate.

Now you maybe thinking this argument is also circular, but it's not.

We have access to experiment, which clearly informs us that AgCl is insoluble in water. We can confidently assume (but we might be wrong!) that the AgCl bond is less ionic than the AgNO3 bond, since it practically does not dissociate/has a low Ksp in water. A bond with a greater covalent character is more stable, since the electrons are shared and not transfered.

Perhaps the crystal structure of AgCl makes it especially stable to solubilization from H2O. I just think it's obvious that the enthalpy of formation of AgCl is favorable, since it has to counter an unfavorable entropy loss in the formation of an ordered crystal lattice.

As you can see, I cannot provide you with a thorough answer, it would be extremely time consuming (for me) and I'd have to conduct many many calculations. I really hope someone can clarify the why. Maybe you can do your own research from the answers provided here and come back with a more "elaborate" answer.

  • 1
    $\begingroup$ I talked to my teacher today and she said it exactly as @CHM did. AgCl<sub>(s)</sub> is favored due to it is has a more covalent fashion than NO<sub>3</sub> $\endgroup$
    – Mattias
    May 15, 2012 at 19:24

Ag is higher up in the Reactivity serie than Al but that do not make sense to me in this problem

  • indeed this is no redox problem, oxidation states do not change.

  • instead it is a phase transition equilibrium $$\ce{Ag+ + Cl- <=> AgCl v}$$

As you marked this question as homework:

  • Edit the question and put the oxidation numbers to confirm that it is not a redox reaction.

  • Look up the solubility constant of AgCl. Together with the total initial concentration of $\ce{Ag+}$ and $\ce{Cl-}$, you can then calculate the equilibrium concentrations.

  • You'll see that the precipitation moves the reaction towards equilibrium (unless the concentrations in the aqeous solution are not extremely low).

Here are two experiments that you can do to check that it is really a dynamic equilibrium (though much on the side of the precipitate) $$\ce{Ag+_(aq) + Cl(aq)- <=>> AgCl_(s)}$$

  1. If you disturb the equilibrium by removing either $\ce{Ag+}$ or $\ce{Cl-}$, the AgCl will dissolve again.
    Here are some ideas:

    • Complex (mask) the $\ce{Ag+}$, e.g. by $\ce{NH3}$ or $\ce{CN-}$
      have you ever done this? If not, you should really try. It is really fast.
    • reduce the $\ce{Ag+}$
    • oxidize the Cl-, so that $\ce{Cl2 ^}$ is removed out of the system.
      (you could also remove the AgCl, but that isn't practical, as you'd never be sure whether you observe formation of more precipitate or whether you didn't remove all of the AgCl)
  2. Crystal growth in saturated solutions. You may have to wait quite a while for this one, though...

F'x pointed us to a paper Paradigms and Paradoxes: The Solubility of AgCl in Water. As it is behind a paywall, I'll try to summarize a few points out of it.

First of all, it's actually kind of an overdue homework question:

It is well established in the pedagogical literature that AgCl is insoluble in water while NaCl and KCl are soluble [...] What is usually left unsaid, however, is why AgCl is so much less soluble than NaCl or KCl.

Here are some interesting points for the question:

  • Liebman concludes from thermodynamic data (and contrary to HSAB expectations) that $\ce{Ag+}$ is better solvated than $\ce{Na+}$ or $\ce{K+}$

  • Lattice energy for AgCl ($\ce{Ag_{(g)}^+ + Cl_{(g)}^- -> AgCl_{(s)}}$) is more negative than for NaCl and KCl.

  • The paper estimates that the enthalpy of $\ce{AgCl_{(g)} -> AgCl_{(s)}}$ is comparable to that for the alkali chlorides.

  • Liebman concludes from that, that the AgCl crystal is nothing special but is rather similar to NaCl and KCl

  • So the point that is different is that the formation of diatomic AgCl is much more favorable than formation of diatomic NaCl or KCl.

My personal conclusions: It's a really tough question. I'll stick to the experiment with measurable properties such as equilibrium constants. ;-)

That is, insolubility of AgCl is no surprise to me: after all, I'm used to the fact for years. And given the insolubility of AgBr and AgI, I'd guesstimate unsoluble even though AgF is soluble (after all $\ce{F-}$ is usually different from the other 3). But I find the particular situation far away from any textbook kind of reasoning that should introduce general chemical concepts to students. Unless the message is supposed to be that reality is complicated... (which OTOH allows for the beauty of reality and makes it interesting).


To analyse such situations, you first must realize that nearly everything is present as ions. For example, if you took the reaction $\ce{NaCl +KBr->NaBr +KCl}$, the "direction" it goes in has no real meaning since $\ce{Na+}$, $\ce{Cl-}$, $\ce{K+}$, $\ce{Br-}$ are present as ions. There is no actual reaction going on here, just a mixing of ions.

In this case, we have something slightly different. $\ce{AgCl}$ precipitates here:

$$\ce{3AgNO3_{(aq)} + AlCl3_{(aq)} -> Al(NO3)3_{(aq)} + 3AgCl v}$$

Similar to the $\ce{NaCl}$ case above, $\ce{NO3-}$ and $\ce{Al^{3+}}$ are present as ions on both sides--so nothing to worry about there. We can safely remove them from the equation. This reduces the reaction to:

$$\ce{3Ag+ + 3Cl- -> 3AgCl v}$$

Now, we have reduced the question to "why does $\ce{AgCl}$ precipitate?". The easiest way to decide is by looking at thermodynamic stability--solid silver chloride is more stable. We can experimentally prove this, or use standard enthalpies.

Reactions proceed in the direction of more stability, so the above reaction goes forward.

Of course, I sort of "assumed" that $\ce{AgCl}$ precipitates in the beginning--this was so that I wouldn't have to complicate stuff. Nevertheless, we can, in a similar way, experimentally prove that the others are found in ionized form and generally do not precipitate. There are some nice arguments why $\ce{AgCl}$ is less soluble than $\ce{AgNO3}$ in this post.


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