Analysis of group 1 cations

Separation and analysis (identification) of group I cations.

Through analysis of cations we are able to separate and identificate the components of an unknown mixture.

First of all, let's get started with a practical flow chart of group 1 cations. Don't worry, we're going to explain step by step; in the end you'll certainly learn how to perform the analysis of group 1 cations!

 

Group 1 cations includes those cations who selectively precipitates as chlorides by addition of diluted hydrochloric acid. These cations are respectevely: Ag⁺,Pb²⁺,Hg₂²⁺ . If we consider the whole periodic table, the only elements whose chlorides are insoluble are those of silver, lead (II) and mercury (I), while chlorides of the other elements are soluble. What does this mean? That even if inside the unknown substance we have a bunch of different inorganic substances, after the addition of HCl all the cations but Ag⁺,Pb²⁺,Hg₂²⁺ will remain in solution. From a practical viewpoint the treatment is carried out with 2N hydrochloric acid (approximately 3mL).

Ok, suppose someone gives you an unknown inorganic substance. How do you find out what's inside?. We take the substance and we put it in a test tube. We add 2N HCl and mix the content of the test tube stirring with a glass rod. To speed up the process, we work in a hot (not boiling) water-bath. At the same time we will dissolve the residue (the unknown substance) and precipitate our insoluble chlorides. Then we cool the test tube under a jet of cold water, rubbing hard the inner walls with the rod. This last step is necessary to precipitate lead chloride, which is not that insoluble. Lead chloride is a crystalline salt whose formation is encouraged by micro holes in the wall (larger surface). Once obtained the white precipitate of chlorides, the solution is centrifuged and decanted to separate the precipitate from the supernatant. The precipitate is "washed" with water added with a few drops of 2N HCl; this will eliminate impurities in the precipitate, without solubilizing again the chlorides (that's why is added with some HCl).

Precipitation mechanism: how to precipitate chlorides of Group I Cations.

The precipitation of these salts takes place because the abundant release of Cl^- ions (from HCl) in solution allows the overcoming of their solubility product. We are obviously talking about the Ksp.

Let's take for example silver chloride, AgCl. The Ksp of this salt is approximately 10^-10. Ksp is the product ot the concentrations of the salt constituent ions, each raised to its stoichiometric coefficient:

[Ag ⁺][Cl ^-] = 10^-10 = K_sp When [Ag ⁺] × [Cl ^-] > 10^-10 we get a precipitate.

We must also ensure a quantitative precipitation. Since we are performing a semi-micro analysis; suppose you have in solution a concentration of [Ag⁺] equal to 10^-2 M. The precipitation of AgCl is quantitative when [Ag⁺] reaches 10^-5 M, a non detectable concentration (for this analysis).

Then, which is the concentration of chloride ions that makes the precipitation quantitative?

  then    it follows that    and so 

a cncentration that without any doubt HCl, a strong acid completely dissociated in water, guarantees. Furthermore, the low concentration of Cl^- necessary justifies the use of diluted (2N) hydrochloric acid instead of the concentrated one.

Why don't we use concentrated HCl? Some theory!

Diluted hydrochloric acid is more than enough to precipitate the cations of group 1 as unsoluble chlorides.

What would happen if we used concentrated hydrochloric acid? We would probably solubilize them. This is because concentradet HCl promote the formation of soluble chlorine complexes of our salts.

As you can see from the scheme, as the concentration of Cl^- raise the soluble complexes formation rate grows. These complexes are soluble because they are electrically charged molecule and therefore efficiently solvated by water molecules,  through thermodynamically advantegeous ion-dipole interactions.

 Separation of Pb ²⁺ from Ag²⁺-Hg₂²⁺

We have seen that among chlorides precipitates of group 1, the most soluble is lead chloride (check the Ksp in the table above). His K_sp (around 10^-5) tells us that it can be dissolved quite easily by dilution. How? Just think to the equilibrium:

Adding water means lowering the concentration of the constituents cations. Considering "le chatelier" principle, the equilibrium will move to the right, and in this context means that the precipitate will solubilize.

Furthermore, the solubility of this species increases with temperature.

Practice

We add 1-2 mL of deionized H₂O and put the test tube in bain-marie, by heating (not boiling) and shaking the solution with the glass rod.

We quickly bring the still hot solution in a centrifuge and separates the supernatant from the precipitate. The treatment is repeated 3-4 times to avoid an undissolved residue of lead chloride on our precipitate of AgCl and Hg₂Cl_2. The first two times you recover the supernatant, which could contain  significative amount of Pb^2+. The two supernatants solutions are then united and here we'll afterwards look for lead.

Then we have to wash the precipitate consisting of AgCl and Hg₂Cl_2. Washing intends once again removing the lead chloride from the precipitate. It is then carried out with hot water (because of the higher solubility of lead chloride in hot water). But how often must the precipitate be washed ?

Enough to be sure that in the washing water is not present Pb²⁺.

To ensure that there is no longer Pb²⁺ is useful an indirect assay. The washing water that we examine is added with some drops of silver nitrate (AgNO₃)_. If there is the Pb²⁺ ion will also be present the chloride ion, indeed, its concentration will be at least double given that the lead-chloride ratio in the chloride lead is 1:2.

So if there is Pb²⁺ and then chloride Cl^- ions in significant amounts, the addition of AgNO₃ will precipitate AgCl, white.

In this case, the precipitate of AgCl and Hg₂Cl₂ must be washed until the assay gives a negative outcome.