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.
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.
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,
x = 3.6e-5 / 0.30 = 1.2e-4 M
BaCrO4 ® Ba2+ + CrO42-, Ksp = 1.2e-10
y = 1.2e-10 / 0.30 = 4.0e-10 M
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.
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.
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.
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.
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+?
Answer [Fe2+] = 0.079 M
[OH-] = 10(-14+7) = 1.00e-7 (buffer).
[Fe2+] (1.00e-7)2 = Ksp; [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?
Answer [CrO42-] = 2.37e-3 M
Solid BaCrO4 will form first as [CrO42-] increases. The maximum [CrO42-] to precipitate CaCrO4 is estimated as follows.
[CrO42-] = 7.1e-4 / 0.30 = 2.37e-3 M
Estimate [Ba2+] when [CrO42-] = 2.3e-3 M, slightly below the maximum concentration.
Answer [Ca2+] / [Ba2+] = 3e6
[Ba2+] = 1.2e-10/1.2e-3 = 1e-7;
[Ca2+] / [Ba2+] = 0.3 / 1e-7 = ? The ratio of three million is large!