What are some examples of chemical precipitation

Precipitation reaction

Synonym: precipitation reaction
English: precipitation reaction

1 definition

As Precipitation reactions Chemical reactions in which the reactants are dissolved in the solvent and at least one product that is contained in this solvent is sparingly or even insoluble or the solution becomes supersaturated due to cooling. The product, which has a poorer solubility, precipitates. The precipitate or the precipitated particles are generally referred to as precipitate. This process is marked in reaction equations with a ↓ (arrow pointing downwards) or an (s) for solid (solid) after the empirical formula of the substance.

2 background

The numerical value of the solubility product provides a quantitative statement about the solubility of a compoundL. The ion product of the solution is thus the product of the ion concentrations in the solution - as it is calculated in the expression of the solubility product. If the solution is saturated, the ion product is the same L.. But it is also possible that it is smaller or even larger if the solution is not in equilibrium with the undissolved substance. Thus three cases can be distinguished:

  • The ion product is smaller as L.: The solution is not in a saturated state. So another substance can be dissolved until the value of L. is reached.
  • The ion product is equalL.: In this case the solution is saturated. So it is in equilibrium with the undissolved substance.
  • The ion product is greater as L.: The solution is oversaturated. So there is no equilibrium - precipitation occurs until the value of L. is reached.

With regard to the dissolution behavior, a distinction is made between easily soluble (e.g. alkali and alkaline earth metal halides) and poorly soluble salts (e.g. silver halides, heavy metal sulfates and sulfides).

A typical example of a dynamic, heterogeneous equilibrium is a saturated salt solution that is in contact with the solid body of the salt. Ions constantly go from the sediment into the solution and at the same time ions from the solution are deposited again on the solid.

3 benefits

A salt precipitates when the product of the ion concentration is greater than the solubility product. In analytical chemistry, this process is used specifically for precipitation reactions. Since salts have different solubilities, there are opportunities to selectively precipitate a certain type of ion from a salt solution by adding suitable foreign ions. A precipitate occurs, which can serve as qualitative evidence for an ion type. This can be weighed out and used for the quantitative determination of a type of ion. If you want to precipitate a certain type of ion almost completely, this is possible with an excess of precipitant.

4 examples

  • If 10 mL of a solution of silver nitrate, c [AgNO3] = 0.010 mol / L, mixed with 10 mL of a saline solution, c [NaCl] = 0.00010 mol / L?
L = 1.7 x 10-10 mol2/ L2
If you mix the two solutions, you get a volume of 20 mL. If you consider the concentrations of the ions in the mixture, they are halved by the increase in volume. Ion product:
c (Ag+) · C (Cl-) = 5,0 · 10-3 mol / L x 5.0 x 10-5 minor
= 2,5 · 10-7 mol2/ L2 > L
The result shows that the ion product exceeds the value of L. Thus AgCl precipitates.
  • Falls Mg (OH)2 off, when in a solution of magnesium nitrate, c [Mg (NO3)2] = 0.0010 mol / L, the pH value is set to 9.0?
L [Mg (OH)2] = 8,9 · 10-12 mol3/ L3
The following applies: at pH = 9.0, pOH = 5.0 and c (OH-) = 10-5 minor.
Ion product:
c (Mg2+) · C2(OH-) = 1,0 · 10-3 mol / L · 10-10 mol2/ L2
= 1,0 · 10-13 mol3/ L3
Since the solubility product is not reached, there is also no precipitation.

Do you want Ba2+-Ions are detected, a sodium sulphate solution has to be added. SO42--Ions can be detected by adding a barium chloride solution. In both cases, barium sulfate, which is sparingly soluble, is formed as a colorless precipitate. This barium sulphate precipitate cannot be dissolved in ammonia solution. This can be used to ensure a differentiation from AgCl.

BaCl2 + Well2SO4 → BaSO4↓ + 2 NaCl
Ba2+ + SO42- → BaSO4 L = 10-10 mol2/ L2