Lead – the 82nd element of the table of Mendeleyev – represents itself very dense, but at the same time soft, malleable and fusible metal of dim gray color. Both lead, and alloys on its basis and also its many connections find very broad application in various fields of the industry. This substance was earlier widely used as additive to motor fuel, but was laid off in view of exclusively high toxicity. As one and all derivatives of lead are poisonous, the question of its definition is very relevant.
It is required to you
- - clean test tube;
- - solution of iodide potassium;
- - acetic acid;
- - spirit-lamp or gas burner;
- - ice or container with cold water;
- - sulfuric acid.
Instruction
1. Let's assume, you have a water test. It is necessary to establish whether soluble compounds of lead contain in it. How it can be done? There is very characteristic and highly sensitive reaction which by right it is possible to call one of the most beautiful in chemistry. It is based on ability of lead to enter interaction with iodine, forming slightly soluble PbI2 connection.
2. Pour a little water from this test in a clean test tube from refractory glass, add to it a little solution of iodide potassium – KI, acidify several drops of acetic acid (for the best course of reaction).
3. Stir up test tube contents. If water contained soluble compounds of lead, the yellow deposit of iodide of lead will drop out. It not remarkable in appearance. But if it is good to heat a test tube on a flame of a spirit-lamp or the gas burner (the deposit at the same time has to be dissolved), and then quickly to cool, for example, having placed in ice or container with cold water, then the deposit of PbI2 will drop out again, only now in the form of beautiful golden crystals. This very effective, impressive show therefore such reaction is often used as demonstration experience.
4. How still it is possible to define lead ions in solution? For example, by means of sulfuric acid or its any soluble salt. In interaction to Pb^2 lead ion + there is a reaction of type: K2SO4 + Pb(NO3)2 = PbSO4 + 2KNO3. The formed sulfate of lead drops out in the form of a dense white deposit.
5. But, for example, loss of the same in appearance deposit – characteristic reaction and to barium ion. How it is possible to be sure that it is not sulfate of barium? For this purpose it is necessary to carry out control reaction: to add solution of strong alkali to a deposit then to heat a test tube. If it is lead sulfate, then the deposit will gradually disappear, because of formation of soluble complex salt. Reaction goes according to such scheme: PbSO4 + 4NaOH = Na2 [Pb(OH)4] + Na2SO4. Barium sulfate at the same control test will remain in the form of a deposit.