Respuesta :
According to the equilibrium law, this should move the AgCl equilibrium to the left (i.e., favor the reverse reaction) until the system finds equilibrium once more. This will increase the concentration of Cl-(aq) ions in the solution.
The following equilibrium is established by an AgCl solution that is saturated:
Ag+(aq) + Cl- =>AgCl(s) (aq)
NaCl will entirely separate into ions when added to the solution:
Na+(aq) + Cl=> NaCl(s) (aq)
According to the equilibrium law, this should move the AgCl equilibrium to the left (i.e., favor the reverse reaction) until the system finds equilibrium once more. This will increase the concentration of Cl-(aq) ions in the solution.
What is seen is that as an AgCl precipitate forms, both [Ag+] and [Cl-] will decrease.
Because part of the Ag ions precipitated out, the [Ag+] in the new system will be lower than it was in the original saturated solution, while the [Cl-] ions will be higher (because you added more when you added NaCl). There will have been some precipitation of that excess Cl-, but not all of it. The result will be the same as the Ksp of AgCl if these new ion concentrations are entered into the Ksp equation.
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the question you are looking for is
What would happen to the Ag+ and Cl- concentrations if solid NaCl were dissolved in a saturated solution of AgCl in water?
