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Heterogeneous Equilibrium Review

Review skills for Quiz

Selected Precipitation

The strategy for chemical analysis has been discussed in Separation by Precipitation. More examples are given here.

Separate, identify, and quantitatively determine the compounds or ions present in a sample are essential in the study of chemistry. Separate and determine amounts of different types of ions are important industrial and scientific applications. For example, if a solution contains 0.010 M of both Cl- and Cr2O42-, you may want to find a cation that reacts with them forming compounds of different solubility. From a table of Ksp values, you find the following:

[Ag+] [Cl-] = 1.8e-10
[Ag+]2 [Cr2O42-] = 1.9e-12

Does this difference in Ksp values afford a separation for Cl- and CrO42-?

To answer this question, we calculate the lowest concentrations of Ag+ at which AgCl and Ag2CrO4 start to precipitate when you increase [Ag+] in the solution.

Lets calculate [Ag+] when AgCl starts to form a precipitate:

[Ag+] = 1.8E-10 / [Cl-]
      = 1.8E-10 / 0.010
      = 1.8E-8
If [Ag+] > = 1.8E-8 M, then AgCl precipitates.

Now, lets calculate [Ag+] when Ag2CrO4 begins to precipitate:

[Ag+]2 = (1.9E-12) / [Cr2O42-]
      = (1.9E-12) / 0.010
      = 1.9EE-10
[Ag+] = 1.4e-5 M
If [Ag+] > 1.4e-5, then Ag2CrO4 will precipitate

AgCl will precipitate at a lower Ag+ concentration than Ag2CrO4 will. As you increase the concentration of Ag+, a series of event will happen as the diagram below shows.

[Ag+] 1.8e-8 1.8e-6 1.4e-5 1.4e-3 0.014
[Cl-] 1.0e-2 1.0e-4 1.3e-5 1.3e-7 1.3e-8
[Cr2O42-] 0.01 0.01 0.01 9.7e-7 9.7e-9
Reaction Forming AgCl solid Forming Ag2CrO4 solid
plus little AgCl solid
What is the [Cl-] when Ag2CrO4 begins to precipitate?

  [Cl-] = 1.8E-10 / 1.4E-5
        = 1.3E-5 M
When Ag2CrO4 starts to precipitate, [CrO42-] = 0.01, thus the precipitate at this point has the molar ratio of:
  Ag2CrO4      0.010       200
  -------  =  -------  =  ------
   AgCl       0.00005       1
and note that the Cl concentration has been drastically reduced when Ag2CrO4 begins to precipitate.

Separating Metal Ions as Sulfide

Sulfide Ksp
MnS 7e-16
FeS 1e-19
ZnS 4.5e-24
CdS 1.4e-28
HgS 3e-53
Many metal ions form insoluble compounds with the sulfide ion, S2-. Some examples are given in a table on the right.

In this group, MnS is the most soluble, whereas HgS is the least soluble. These metal ions can be separated by careful control of sulfide ion concentration, [S2-].

Hydrogen sulfide is a diprotic acid. Its two-stage ionizations are shown in the following equilibrium equations:

H2S = H+ + HS- Ka1 = 1.1E-7
HS- = H+ + S2- Ka2 = 1E-14
H2S = 2 H+ + S2- Koverall = 1E-21
[S2-] = 1e-21 / 10-pH*2
Thus, by careful control of pH the concentration of sulfide ion can be adjusted to precipitate a group of metal ions. This was the main technique of qualitative analysis before 1960s. Today, many spectroscopic methods have been introduced and minute amount of metal ions can be detected using for example atomic absorption (AA) spectroscopy.

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