We are going now to discover how is it possible to identify and confirmate the presence of lead (Pb2+) in an acqueous solution.
1) Treatment with sodium acetate (CH3COONa) and potassium chromate (K2CrO4)
The aim of this assay is the precipitation of PbCrO4 from a solution that hypothetically contains just Pb2+.
It is an easy test but we should pay some attention considering the amphoteric nature of lead and the pH-dependent equilibrium of the chromate ion;
→ The pH must not be basic otherwise the lead would precipitate as lead hydroxide Pb(OH)2 a white and gelatinous precipitate, indistinguishable from other hydroxides; in addition, an excess of base would solubilize lead as plumbite ion PbO22-.
→ The pH must not be too acidic because chromate CrO42- would turn into dichromate Cr2O72- (and lead dichromate is soluble).
The compromise is an acetate buffer (around pH 5), which prevents the precipitation of lead hydroxide but still ensure a sufficient amount of chromate ion (necessary to precipitate PbCrO4).
From a practical viewpoint we add CH3COONa and subsequently the precipitating reagent, K2CrO4. If lead is present it will precipitate as lead chromate, PbCrO4, yellow.
If we want to make ourselves sure that the precipitate is actually lead chromate we can take advantage of the amphoteric nature of lead. Lead chromate is indeed soluble in NaOH. This wouldn't be true other salts such as barium chromate (BaCrO4) or bismuth oxychloride (BiOCl), both yellow but insoluble in NaOH (cause they're not amphoteric).
2) Formation of lead sulfate (PbSO4)
The solution is added with diluted sulfuric acid (H2SO4). Precipitation of lead sulfate (PbSO4), a white salt, reveals the presence of Pb2+ .
If both adding soda (NaOH) and concentrated acids the white precipitate is solubilized, then the salt must be lead sulfate.
The confirmatory test may be useful, for example, to distinguish between lead sulfate and barium sulfate (BaSO4). The latter also is indeed white and very slightly soluble in water. Fortunately, it is insoluble in NaOH (because Ba2+ isn't amphoteric) and the test thus confirmate the presence of lead in the initial solution.
3) Reaction with potassium iodide (KI)
A super easy way to identify Pb2+ consists in its precipitation as PbI, lead iodide, a slightly soluble yellow precipitate, much less soluble than PbCl2. However, lead iodide could be confused with silver iodide (even if actually the two precipitates are slightly different)
As countercheck we can add an excess of potassium iodide. PbI will solubilize because of the formation of PBI4-. The complex PBI4- colors in yellow the solution and PBI2 can be precipitated again by dilution.
- Cooling of the test tube with cold water
This technique is the exact opposite of what we saw in group 1 analysis when we separated lead (Pb2+) from silver and mercury.
The solubility of PbCl2 decreases lowering the temperature, then we try to have it precipitated cooling the test tube under a stream of cold water. Furthermore the inner walls of the test tube are rubbed in the meanwhile with a glass rod. Doing so we will create microholes in the glass which favor the nucleation of PbCl2 crystals.
With a bit of luck we can observe the formation of needle-like, gleaming, white lead chloride crystals. The formation of the crystals is not immediate, it might also take a few hours.