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Chemical Separation by Precipitation

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Chemical Separation by Precipitation

A mixture of metal ions in a solution can be separated by precipitation with anions such as Cl-, Br-, SO42-, CO32-, S2-, Cr2O42-, PO42-, OH- etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated form others by precipitation.

We can also separate the anions by precipitating them with appropriate metal ions.

There is no definite dividing lines between insoluble salts, sparingly soluble, and soluble salts, but concentrations of their saturated solutions are small, medium, and large. Solubility products are usually listed for insoluble and sparingly soluble salts, but they are not given for soluble salts. Solubility products for soluble salts are very large.

What type of salts are usually soluble, sparingly soluble and insoluble? The following are some general guidelines, but these are not precise laws.

These are handy rules for us to have if we deal with salts often. On the other hand, solubility is an important physical property of a substance, and these properties are listed in handbooks.

Chemical Separation of Metal Ions

Formation of crystals from a saturated solution is a heterogeneous equilibrium phenomenon, and it can be applied to separate various chemicals or ions in a solution.

When solubilities of two metal salts are very different, they can be separated by precipitation. The Ksp values for various salts are valuable information, and some data are given in the Handbook of this website. On the Handbook Menu, click the item Salts Ksp will give the Ksps of some salts.

In the first two examples, we show how barium and strontium can be separated as chromate.

Example 1

The Ksp for strontium chromate is 3.6E-5 and the Ksp for barium chromate is 1.2E-10. What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.30 M solution of barium and strontium ions without precipitating the other?

Solution
Since the Ksp for barium chromate is smaller, we know that BaCrO4 will form a precipitate first as [CrO42-] increases so that Qsp for BaCrO4 also increase from zero to Ksp of BaCrO4, at which point, BaCrO4 precipitates. As [CrO42-] increases, [Ba2+] decreases. Further increase of [CrO42-] till Qsp for SrCrO4 increases to Ksp of SrCrO4, it then precipitates.

Let us write the equilibrium equations and data down to help us think. Further, let x be the concentration of chromate to precipitate Sr2+, and y be that to precipitate Ba2+. According to the definition of Ksp we have,

SrCrO4 ® Sr2+ + CrO42-,     Ksp = 3.6e-5
                0.30       x

x = 3.6e-5 / 0.30 = 1.2e-4 M

BaCrO4 ® Ba2+ + CrO42-,     Ksp = 1.2e-10
                0.30       y

y = 1.2e-10 / 0.30 = 4.0e-10 M

The Ksp's for the two salts indicate BaCrO4 to be much less soluble, and it will precipitate before any SrCrO4 precipitates. If chromate concentration is maintained a little less than 1.2e-4 M, Sr2+ ions will remain in the solution.

Discussion

In reality, to control the increase of [CrO42-] is very difficult.

Example 2

The Ksp for strontium chromate is 3.6E-5 and the Ksp for barium chromate is 1.2E-10. Potassium chromate is added a small amount at a time to first precipitate BaCrO4. Calculate [Ba2+] when the first trace of SrCrO4 precipitate starts to form in a solution that contains 0.30 M each of Ba2+ and Sr2+ ions.

Solution
From the solution given in Example 1, [CrO42-] = 3.6e-4 M when SrCrO4 starts to form. At this concentration, the [Ba2+] is estimated as follows.

[Ba2+] 3.6e-4 = 1.2e-10 (The Ksp of BaCrO4). Thus, [Ba2+] = 3.33e-7 M

very small indeed, compared to 0.30; In the fresh precipitate of SrCrO4, the mole ratio of SrCrO4 to BaCrO4 is 0.30 / 3.33e-7 = 9.0e5. In other words, the amount of Ba2+ ion in the solid is only 1e-6 (1 ppm) of all metal ions, providing that all the solid were removed when [CrO42-] = 3.6e-4 M.

Discussion

The calculation shown here indicates that the separation of Sr and Ba is pretty good. In practice, an impurity level of 1 ppm is a very small value.

Example 3

What reagent should you use to separate silver and lead ions that are present in a solution? What data or informatin will be required for this task?

Solution
The Ksp's for salts of silver and lead are required. We list the Ksp's for chlorides and sulfates in a table here. These value are found a handbook and in the Handbook Menu of our website as Salts Ksp.

Salt Ksp Salt Ksp
AgCl 1.8e-10 Ag2SO4 1.4e-5
Hg2Cl2 1.3e-18 BaSO4 1.1e-10
PbCl2 1.7e-5 CaSO4 2.4e-5
PbSO4 6.3e-7
SrSO4 3.2e-7
Because the Ksp's AgCl and PbCl2 are very different, chloride, Cl-, apppears a good choice of negative ions for their separation.

The literature also indicates that PbCl2 is rather soluble in warm water, and by heating the solution to 350 K (80oC), you can keep Pb2+ ions in solution and precipitate AgCl as a solid. The solubility of AgCl is very small even at high temperatures. This separation experiment is performed in the laboratory course CHEM123L.

Discussion

Find more detailed information about the solubility of lead chloride as a function of temperature.

Can sulfate be used to separate silver and lead ions? Which one will form a precipitate first as the sulfate ion concentration increases? What is the [Pb2+] when Ag2SO4 begin to precipitate in a solution that contains 0.10 M Ag+?

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