# Displacement of salts due to higher solubility

I was wondering why mixtures of salts in the same solution precipitate excactly in order of their solubilities. Imagine a solution with NaCl, NaBr, NaI in equal moles. If AgNO₃ solution is now added in excess, first silver iodide, then bromide, then chloride will precipitate out of solution in that order due to their different solubilities (Ksp=1.8x10⁻¹⁰, 5x10⁻¹³, 8.3x10⁻¹⁷ respectively).

First of all, what is the thermodynamic background of this phenomenon? It's obvious that for example the iodide is less soluble than let's say the silver chloride, but under the given conditions, all of them are basically insoluble, so I don't really get why the silver iodide selectively precipitates first. Secondly, would it matter if the sodium salts wouldn't be there in equal moles?

And on another note; would it be possible to dissolve silver chloride just by adding another, more insoluble salt? For example:

$$\ce{AgCl(s) + NaBr(aq) -> AgBr(s) + NaCl(aq)}$$

A saturated solution of AgCl with already precipitated, solid AgCl is treated with sodium bromide (NaBr) salt . Given the explanation above, would it be possible that the silver(I) ions form a precipitate with Br- ions giving NaBr, and let the AgCl precipitate dissolve and form NaCl? Thanks in advance.

would it be possible to dissolve silver chloride just by adding another, more insoluble salt? For example: $$\ce{AgCl(s) + NaBr(aq) -> AgBr(s) + NaCl(aq)}$$

It is possible to get the chloride ion back into solution by adding NaBr. The effect is to lower the silver ion concentration so drastically that chloride goes back into solution. This kind of process is used in analytical chemistry to get ions of insoluble salts into solution. Here is a related example:

$$\ce{Mg(OH)2 (s) + 2HCl(aq) -> Mg^2+ + 2H2O + 2Cl-(aq)}$$

In this case, you lower the hydroxide concentration by neutralizing it with hydrochloric acid.

Imagine a solution with NaCl, NaBr, NaI in equal moles. If AgNO₃ solution is now added in excess, first silver iodide, then bromide, then chloride will precipitate out of solution in that order due to their different solubilities (Ksp=1.8x10⁻¹⁰, 5x10⁻¹³, 8.3x10⁻¹⁷ respectively).

If you add the silver nitrate dropwise and wait a bit inbetween, you will observe solid silver iodide after the first drop. You should not make any statement about what forms first (question of kinetics) but rather what is present at equilibrium. Once the iodide concentration is 360-fold lower than the bromide concentration, you will also form a silver bromide precipitate when adding more silver ions. When most of the bromide is precipitated, silver chloride precipitate will also be observed.

So in that sense, they form stable precipitate in that order, with barely noticable overlap.