assume you had a solution that contained a mixture of a silver ions and stronium ions. come up with three different reaction plans that would allow you to precipitat either the silver ion or strontium ion, but the other ion would remain in solution

Question

assume you had a solution that contained a mixture of a silver ions and stronium ions. come up with three different reaction plans that would allow you to precipitat either the silver ion or strontium ion, but the other ion would remain in solution

Answer 1

This question is designed to help you use the solubility tables. Here is a simplified table of solubilities.

http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html

As an example, AgCl is not soluble in water but SrCl2 is. SrCrO4 is soluble in water but Ag2CrO4 is not. Now you think of the third one.

Answer 2

To selectively precipitate either the silver ions or the strontium ions while leaving the other ion in solution, you can utilize different precipitation reactions. Here are three reaction plans for achieving this:

Reaction Plan 1: Precipitate Silver Ions

1. Add a few drops of dilute hydrochloric acid (HCl) to the mixture solution. Silver ions (Ag+) will react with chloride ions (Cl-) to form insoluble silver chloride (AgCl) precipitate.
Ag+ (aq) + Cl- (aq) → AgCl (s)

2. Filter the solution, separating the precipitated silver chloride (AgCl) from the remaining solution. The silver chloride is retained on the filter paper, while the strontium ions remain in the filtrate.

Reaction Plan 2: Precipitate Strontium Ions

1. Add a few drops of dilute sulfuric acid (H2SO4) to the mixture solution. Strontium ions (Sr2+) will react with sulfate ions (SO42-) to form insoluble strontium sulfate (SrSO4) precipitate.
Sr2+ (aq) + SO42- (aq) → SrSO4 (s)

2. Filter the solution, separating the precipitated strontium sulfate (SrSO4) from the remaining solution. The strontium sulfate is retained on the filter paper, while the silver ions remain in the filtrate.

Reaction Plan 3: Precipitate Silver Ions

1. Add a few drops of a sodium chloride (NaCl) solution to the mixture solution. Silver ions (Ag+) will react with chloride ions (Cl-) to form insoluble silver chloride (AgCl) precipitate.
Ag+ (aq) + Cl- (aq) → AgCl (s)

2. After the formation of silver chloride precipitate, add excess ammonia solution (NH3). Silver chloride dissolves in excess ammonia due to the formation of a soluble complex, called diamminesilver(I) complex.
AgCl (s) + 2NH3 (aq) → [Ag(NH3)2]+ (aq) + Cl- (aq)

3. Filter the solution to separate the retained silver chloride (AgCl) precipitate. The strontium ions remain in the filtrate.

In all three reaction plans, by selectively precipitating one ion, the other ion remains in the solution. The choice between the reaction plans can depend on the specific laboratory setup, available chemicals, or preferences.