Differences in solubility are widely used to selectively remove one species from a solution containing several kinds of ions.

Example 9

The solubility products of AgCl and Ag₂CrO₄ are 1.8E–10 and 2.0E–12, respectively. Suppose that a dilute solution of AgNO₃ is added dropwise to a solution containing 0.001M Cl^– and 0.01M CrO₄^2–.

Which solid, AgCl or Ag₂CrO₄, will precipitate first? What fraction of the first anion will have been removed when the second just begins to precipitate? Neglect any volume changes.

Solution: The silver ion concentrations required to precipitate the two salts are found by substituting into the appropriate solubility product expressions:

to precipitate AgCl: [Ag⁺] = 1.8E-10 / .001 = 1.8E-7 M

to precipitate Ag₂CrO₄: [Ag⁺] = (2.0E-12 / .01)^½ = 1.4E–5 M

The first solid to form as the concentration of Ag⁺ increases will be AgCl. Eventually the Ag⁺ concentration reaches 1.4E-5 M and Ag₂CrO₄ begins to precipitate. At this point the concentration of chloride ion in the solution will be 1.3E-5 M which is about 13% of the amount originally present.

 The preceding example is the basis of the Mohr titration of chloride by Ag⁺, commonly done to determine the salinity of water samples. The equivalence point of this precipitation titration occurs when no more AgCl is formed, but there is no way of observing this directly in the presence of the white AgCl which is suspended in the container. Before the titration is begun, a small amount of K₂CrO₄ is added to the solution. Ag₂CrO₄ is red-orange in color, so its formation, which signals the approximate end of AgCl precipitation, can be detected visually.