Question

Separation of chlorides by adding HCl

Answer 1

If a dilute solution of HCl is added to a solution containing ALL of the common metal ions, a white precipitate will be formed. This precipitate will contain AgCl(s), Hg₂Cl₂(s), and PbCl₂(s). If this solution is centrifuged to collect the solid material at the bottom of a test tube, the supernatant liquid can be removed (decanted). The remaining solid can be washed with distilled water so that the ONLY metal ions that remain are Ag⁺, Hg₂²⁺, and Pb²⁺. The reactions that occur are:

    Ag⁺(aq) + Cl^-(aq) -> AgCl(s)

    Hg₂²⁺(aq) + 2 Cl^-(aq) -> Hg₂Cl₂(s)

    Pb²⁺(aq) + 2 Cl^-(aq) -> PbCl₂(s)

The reactions are conducted in acidic solution, to avoid the possibility that these and other metal ions might be precipitated as hydroxides or oxides. This reaction of this group of ions with chloride ions allows them to be separated from all other metal ions.  Silver, mercurous, and plumbous ions have been designated as the Group I Cations.

    If addition of dilute HCl to a solution of metal cations results in a precipitate, this is conclusive evidence that at least one - and possibly all - of the Group I Cations were are present in the original solution. Additional tests are necessary to determine which of these three cations are present in the precipitate.

    While these three chlorides are relatively insoluble in cold water, lead chloride is found to be quite soluble in hot water. Therefore, cold water may be added to the precipitate containing any or all of the three chlorides without any of them re-dissolving. However, heating this mixture will cause lead chloride to dissolve, if any is present. While still hot, the tube can be centrifuged and the supernatant solution can be decanted into another test tube. If any lead ion is present, addition of cold HCl solution to the liquid will cause white lead chloride to precipitate.

    Silver ion has a special property which distinguishes it from the other two cations of Group I. When mixed with ammonia in basic solution, complex ions are formed (these complexes are not formed in acidic solution in which the ammonia exists as ammonium ion):

    Ag⁺(aq) + NH₃(aq) = Ag(NH₃)⁺(aq)

    Ag(NH₃)⁺(aq) + NH₃(aq) = Ag(NH₃)₂⁺(aq)

The formation of these complex ions lowers the concentration of free silver ion, allowing precipitated silver chloride to dissolve:

    AgCl(s) + x NH₃(aq) -> Ag(NH₃)_x⁺(aq) + Cl^-(aq) ;   x = 1 to 2 .

Thus, when the Group I cations are precipitated with HCl, ammonia may be added to the precipitate until litmus indicate that the solution is basic. If the precipitate contains silver ion, it will dissolve and may be removed from the others by centrifuging and decanting the supernatant solution. Addition of HCl to this solution until litmus indicates that the solution is acidic will re-precipitate silver ion, if present.

    Mercurous chloride may be distinguished from the other two Group I chlorides by its ability to resist re-dissolving by either heating or the addition of ammonia.

    Once the Group I cations have been separated by heating and the addition of ammonia, their presence(s) may be confirmed by their precipitates with chromate ion:

    Ag⁺(aq) + CrO₄^2-(aq) -> Ag₂CrO₄(s)     Red precipitate

    Hg₂²⁺(aq) + CrO₄^2-(aq) -> Hg₂CrO₄(s)   Yellow-brown precipitate

    Pb²⁺(aq) + CrO₄^2-(aq) -> PbCrO₄(s)       Yellow precipitate
  

Suggested Experiment:

Observations with solutions containing known cations:

  1. Place a drop or two of a sample of Ag⁺ ion in one test tube, a sample of Hg₂²⁺ in a second tube, and a sample of Pb²⁺ in a third tube.

  2. Add 2 drops of dilute HCl (3 M) to each tube and observe the precipitates.

  3. Centrifuge the tubes. Decant and discard the supernatant liquid from each tube.

  4. Wash the precipitates by adding 2 drops of distilled water to each tube. Centrifuge, decant and discard the supernatant liquids.

  5. Add 2 drops of distilled water to each tube, then heat the tubes in a boiling water bath.

  6. Note that the solution containing Pb²⁺ clears. The tube may be centifuged and no solid will appear. If the solution is cooled by adding a drop of distilled water (or simply by standing for a while) the white precipitate will reappear.

  7. Add two drops of ammonia (NH₃) solution to each tube, and test the acidity of each solution by dipping a clean stirring rod into the liquid and touching it to litmus paper. If the solutions are not basic, add more ammonia until they turn red litmus blue. Note any change in the precipitates.

  8. Note that the solution containing Ag⁺clears. Again, the tube may be centifuged and no solid will appear.

  9. Add a drop of dilute HCl (3 M) to the clear solution, and test with litmus. If the solution is not acidic, add more HCl dropwise until the solution turns blue litmus red. Note that the white AgCl precipitate reappears as soon as the solution becomes acidic.

10. Add a drop or two of sodium chromate solution (Na₂CrO₄) to each tube. Note the colors of the precipitates.