Separating mixtures by precipitation

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Intros
Lessons
  1. How can we separate mixtures of compounds?
  2. Using qualitative analysis.
  3. Using precipitation to identify unknown ions.
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Examples
Lessons
  1. Use tables of solubility to suggest ways to separate ions from aqueous mixtures.
    Below is a table of solubilities, showing which combinations of aqueous ions will result in a precipitate when mixed.

    A series of experiments were performed with solutions containing different unknown combinations of the four cations in the first column.

    Cl-/Br-/I-

    SO42-

    S2-

    OH-

    PO43-

    Ba2+ (aq)

    -

    Ppt

    -

    Ppt

    Ppt

    Pb2+ (aq)

    Ppt

    Ppt

    Ppt

    Ppt

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    Ag+ (aq)

    Ppt

    Ppt

    Ppt

    Ppt

    Ppt

    Ni2+ (aq)

    -

    -

    Ppt

    Ppt

    Ppt

    Sr2+ (aq)

    -

    Ppt

    -

    -

    Ppt

    Mg2+ (aq)

    -

    -

    -

    Ppt

    Ppt

    1. The anions (across the first row) are added in the form of soluble salts that dissolve to give the aqueous anion we need.
      Explain why sodium salts are a good source of the anions in this experiment.
    2. A solution contains Pb2+(aq), Sr2+(aq) and Ni2+ (aq). Which anions and in which order should be added to separate these cations?
  2. Use tables of solubility to predict precipitates formed from experiments.
    Below is a table of solubilities, showing which combinations of aqueous ions will result in a precipitate when mixed.

    A series of experiments were performed with solutions containing different unknown combinations of the four cations in the first column.

    Cl-/Br-/I-

    SO42-

    S2-

    OH-

    PO43-

    Ba2+ (aq)

    -

    Ppt

    -

    Ppt

    Ppt

    Pb2+ (aq)

    Ppt

    Ppt

    Ppt

    Ppt

    Ppt

    Ag+ (aq)

    Ppt

    Ppt

    Ppt

    Ppt

    Ppt

    Ni2+ (aq)

    -

    -

    Ppt

    Ppt

    Ppt

    Sr2+ (aq)

    -

    Ppt

    -

    -

    Ppt

    Mg2+ (aq)

    -

    -

    -

    Ppt

    Ppt

    1. A solution with two of the above cations present was mixed with the following anions, with the results below:
          • Chloride (Cl-) ions were added. No precipitate was observed.
          • Sulfate ions (SO42-) were added. No precipitate was observed.

      Which two cations must be present in the solution?
    2. A solution with two of the above cations present was mixed with the following anions, with the results below:
          • Chloride (Cl-) ions were added. No precipitate was observed.
          • Sulfide (S2-) ions were added. A precipitate formed in this mixture and was filtered off.
          • Hydroxide (OH-) ions were added. No precipitate was observed.

      What is the precipitate formed with the sulfide ion?

      What is the identity of the remaining cation?
Topic Notes
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In this lesson, we will learn:

  • To use ion solubility and precipitation as a form of qualitative analysis.
  • Using precipitation to identify unknown ions.

Notes:

  • The fact that one compound might be soluble (for example NaCl), while a similar compound with the same anion but different cation (for example AgCl) might have low solubility is very useful to chemists.
    • When trying to identify an unknown compound, reacting it with a known compound and observing a precipitate forming could tell us if a particular ion is present or not, because the ion we are looking for may be known to be insoluble.
      • For example, if we suspect silver ions are present, reacting the sample with a soluble halide (Cl-, Br- or I-) compound could be useful because the product would be a silver halide – AgCl, AgBr and AgI are all insoluble.
    • The same method could be used to rule out possible compounds, observing no precipitate will rule out a lot of possible identities of the compound.
    • Analyzing chemicals in this way is known as qualitative analysis – the results of your investigation produce a binary, yes/no result (is the ion present or not?).
  • Identifying some ions will require multiple precipitation reactions to be done:
    • For example, in an experiment where we need to precipitate Ba2+ ions, using a solubility table we can see that barium sulfate, BaSO4, is insoluble and will form a precipitate but this will also precipitate other ions, for example lead ions, Pb2+:

      Cl-/Br-/I-

      SO42-

      S2-

      OH-

      CO32-

      Ba2+

      -

      Ppt

      -

      Ppt

      Ppt

      Pb2+

      Ppt

      Ppt

      Ppt

      Ppt

      Ppt


      The problem is that there is no anion test to make a barium precipitate exclusively. If SO42- ions alone were added, both ions will form precipitates and there will be no certainty which ion specifically caused it.
    • One way to separate and identify is to add another anion (a slight excess) which causes only one species to precipitate. Do this to successively eliminate possible candidates. Taking our example of possible Ba2+ or Pb2+:
      • If halide (Cl-, Br-, I-) or sulfide (S2-) ions were added to the mixture, the result would be Pb2+ ions precipitating, while the Ba2+ would remain in solution. Filter this precipitate off and discard it.
      • The resultant solution should have no Pb2+ remaining. Now, the addition of SO42- ions will still produce BaSO4 precipitate, however with the Pb2+ already removed we can be certain that it is Ba2+present in the solution.