Chemical Separation by Precipitation

Skills to develop

Calculate ion concentrations to maintain a heterogeneous equilibrium.

Calculate pH required to precipitate a metal hydroxide.

Design experiments to separate metal ions in a solution of mixtures of metals

Chemical Separation by Precipitation

A mixture of metal ions in a solution can be separated by precipitation with anions such as Cl^-, Br^-, SO₄^2-, CO₃^2-, S^2-, Cr₂O₄^2-, PO₄^2-, 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.

All nitrates are soluble. The singly charge large NO₃^- ions form salts with high solubilities. So do ClO₄^-., ClO₃^-, NO₂^-, HCOO^-, and CH₃COO^-.
All chlorides, bromides, and iodides are soluble except those of Ag⁺, Hg₂²⁺, and Pb²⁺. CaF₂, BaF₂, and PbF2 are also insoluble.
All sulfates are soluble, except those of Ba²⁺, Sr²⁺, and Pb²⁺. The doubly charged sulfates are usually less soluble than halides, and nitrates.
Most singly charge cations K⁺, Na⁺, NH₄⁺ form soluble salts. However, K₃Co(NO₂)₆ and (NH₄)₃Co(NO₂)₆ are insoluble.

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 K_sp 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 K_sps of some salts.

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

Example 1

The K_sp for strontium chromate is 3.6E-5 and the K_sp 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 K_sp for barium chromate is smaller, we know that BaCrO₄ will form a precipitate first as [CrO₄^2-] increases so that Q_sp for BaCrO₄ also increase from zero to K_sp of BaCrO₄, at which point, BaCrO₄ precipitates. As [CrO₄^2-] increases, [Ba²⁺] decreases. Further increase of [CrO₄^2-] till Q_sp for SrCrO₄ increases to K_sp 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 K_sp we have,

SrCrO₄ ® Sr²⁺ + CrO₄^2-, K_sp = 3.6e-5
0.30 x

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

BaCrO₄ ® Ba²⁺ + CrO₄^2-, K_sp = 1.2e-10
0.30 y

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

The K_sp'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 [CrO₄^2-] is very difficult.

Example 2

The K_sp for strontium chromate is 3.6E-5 and the K_sp 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, [CrO₄^2-] = 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 K_sp 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 [CrO₄^2-] = 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 K_sp's for salts of silver and lead are required. We list the K_sp'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 K_sp Salt K_sp
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 K_sp's AgCl and PbCl₂ are very different, chloride, Cl^-, apppears a good choice of negative ions for their separation.

Salt	K_sp	Salt	K_sp
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

The literature also indicates that PbCl₂ is rather soluble in warm water, and by heating the solution to 350 K (80^oC), 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⁺?

Confidence Building Questions

Iron(II) hydroxide is only sparingly soluble in water at 25^oC; its K_sp = 7.9E-16. Calculate the solubility of iron(II) hydroxide in a buffer solution with pH = 7.00.
Answer [Fe²⁺] = 0.079 M
Consider...
[OH^-] = 10^(-14+7) = 1.00e-7 (buffer).
[Fe²⁺] (1.00e-7)² = K_sp; [Fe²⁺] = ?
This Fe²⁺ concentration is low, it is not very soluble in a neutral solution (pH = 7).
What is [Fe²⁺] in a solution whose pH = 6.00?
A solution contains 0.60 M Ba²⁺ and 0.30 M Ca²⁺; K_sp values for BaCrO₄ and CaCrO₄ are 1.2e-10 and 7.1e-4 respectively. What value of [CrO₄^2-] will result in a maximum separation of these two ions?
Answer [CrO₄^2-] = 2.37e-3 M
Consider...
Solid BaCrO₄ will form first as [CrO₄^2-] increases. The maximum [CrO₄^2-] to precipitate CaCrO₄ is estimated as follows.
[CrO₄^2-] = 7.1e-4 / 0.30 = 2.37e-3 M
Estimate [Ba²⁺] when [CrO₄^2-] = 2.3e-3 M, slightly below the maximum concentration.
A solution contains 0.60 M Ba²⁺ and 0.30 M Ca²⁺; K_sp values for BaCrO₄ and CaCrO₄ are 1.2e-10 and 7.1e-4 respectively. Calculate the [Ca²⁺] / [Ba²⁺] ratio in the solution when [CrO₄^2-] is maintained at 1.2e-3 M.
Answer [Ca²⁺] / [Ba²⁺] = 3e6
Consider...
[Ba²⁺] = 1.2e-10/1.2e-3 = 1e-7;
[Ca²⁺] / [Ba²⁺] = 0.3 / 1e-7 = ? The ratio of three million is large!