Respuesta :
Answer:
ll assume that you’re talking about the type of precipitation that occurs when an insoluble salt is formed by the mixture of two salt solutions. A common example is:
AgNO3(aq) + NaCl(aq) - - - - -> AgCl(s) + NaNO3(aq)
Before the reaction, we have two aqueous solutions, one of which contains dissolved Ag+ and NO3- ions, and the other Na+ and Cl- ions. When they’re mixed, there is momentarily a solution with all four ions mixed together. AgCl then immediately precipitates from the solution as a solid, while Na+ and NO3- remain in the solution.
In order to understand why this occurs, we need to understand that solvation is a process that is subject to the usual concepts of equilibrium. Each of the possible salt combinations has a solvation equilibrium. The one for silver chloride is:
AgCl(s) <- - - - -> Ag+(aq) + Cl-(aq)
This reaction has an equilibrium constant, which we write down according to the usual rules, and which is determined by the molar free energy change of reaction:
K_sp = [Ag+][Cl-]/[AgCl]
Since AgCl here represents the pure solid, its activity is unity, therefore
K_sp = [Ag+][Cl-]
Solid silver chloride will be in equilibrium with dissolved Ag+ and Cl- when the ion activity product [Ag+][Cl-] is equal to the equilibrium constant. Therefore,
If [Ag+][Cl-] is less than K_sp, solid AgCl will keep dissolving until [Ag+][Cl-] becomes equal to K_sp (or all the solid AgCl runs out).
If [Ag+][Cl-] is more than K_sp, solid AgCl will precipitate from the solution until [Ag+][Cl-] becomes equal to K_sp. At this point, equilibrium will exist between Ag+, Cl-, and AgCl.
Note that in a chemical system with many different components, all possible reactions must reach equilibrium in order for the entire system to be at equilibrium. Therefore, when [Ag+][Cl-] is more than K_sp for AgCl, precipitation will occur regardless of the other ions in solution. The only way to prevent precipitation from occurring is to have some side reaction that consumes Ag+ or Cl- until their activities fall sufficiently for [Ag+][Cl-] to be less than K_sp for AgCl.
So, while Ag+ and Cl- are happy sitting in solution when those are two separate solutions, once you bring them together, AgCl must form in order for equilibrium to be attained. The solubility equilibrium for NaNO3 must also be considered, but since NaNO3 is highly soluble, it won’t precipitate. Thus, when combining solutions containing different ions, you can predict whether a precipitate forms by identifying the ion combinations that form poorly soluble salts such as AgCl, and determining whether the ion activity product exceeds K_sp for that salt.
However, if you mix aqueous AgNO3 with aqueous NaCl that contains a sufficient quantity of dissolved ammonia, then no precipitate will form. This is because there is a new equilibrium involved:
Ag+(aq) + 2NH3(aq) <- - - - -> [Ag(NH3)2]+
This equilibrium lies far to the right, so almost all the Ag+ will be removed from solution, becoming bound up in the complex diamminesilver ion instead. Because of this, [Ag+][Cl-] might not exceed K_sp when the two solutions are mixed, resulting in no precipitate.
(The astute reader would point out that we also need to check whether {[Ag(NH3)2]+}{Cl-} exceeds the solubility product for [Ag(NH3)2]Cl. The answer is that this salt has a large mismatch in the sizes of its cation and anion, and the N-H bonds in the ammine ligands form strong hydrogen bonds with water molecules, so we can predict that its solvation is very favourable, and it’s unlikely to precipitate.)
