8.3 Precipitation reactions

[link] shows some of the general rules about the solubility of different salts based on a number of investigations:

Testing for common anions in solution

It is also possible to carry out tests to determine which ions are present in a solution.

Test for a chloride

Prepare a solution of the unknown salt using distilled water and add a small amount of silver nitrate solution. If a white precipitate forms, the salt is either a chloride or a carbonate.

Cl ^$-$ + Ag ⁺ + NO ₃ ^$-$ $\to$ AgCl + NO ₃ ^$-$ (AgCl is white precipitate)

CO ₃ ^{2 $-$} + 2Ag ⁺ + 2NO ₃ ^$-$ $\to$ Ag ₂ CO ₃ + 2NO ₃ ^$-$ (Ag ₂ CO ₃ is white precipitate)

The next step is to treat the precipitate with a small amount of concentrated nitric acid . If the precipitate remains unchanged, then the salt is a chloride. If carbon dioxide is formed, and the precipitate disappears, the salt is a carbonate.

AgCl + HNO ₃ $\to$ (no reaction; precipitate is unchanged)

${\mathrm{Ag}}_2{\mathrm{CO}}_3+2{\mathrm{HNO}}_3\to 2{\mathrm{AgNO}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$ (precipitate disappears)

Test for a sulphate

Add a small amount of barium chloride solution to a solution of the test salt. If a white precipitate forms, the salt is either a sulphate or a carbonate.

${\mathrm{SO}}_4^{2-}+{\mathrm{Ba}}^{2+}+{\mathrm{Cl}}^{-}\to {\mathrm{BaSO}}_4+{\mathrm{Cl}}^{-}$ (BaSO ${}_4$ is a white precipitate)

${\mathrm{CO}}_3^{2-}+{\mathrm{Ba}}^{2+}+{\mathrm{Cl}}^{-}\to {\mathrm{BaCO}}_3+{\mathrm{Cl}}^{-}$ (BaCO ${}_3$ is a white precipitate)

If the precipitate is treated with nitric acid, it is possible to distinguish whether the salt is a sulphate or a carbonate (as in the test for a chloride).

${\mathrm{BaSO}}_4+{\mathrm{HNO}}_3\to$ (no reaction; precipitate is unchanged)

${\mathrm{BaCO}}_3+2{\mathrm{HNO}}_3\to \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$ (precipitate disappears)

Test for a carbonate

If a sample of the dry salt is treated with a small amount of acid, the production of carbon dioxide is a positive test for a carbonate.

Acid + CO ₃ ^{2 $-$} $\to$ CO ₂

If the gas is passed through limewater and the solution becomes milky, the gas is carbon dioxide.

Ca(OH) ₂ + CO ₂ $\to$ CaCO ₃ + H ₂ O (It is the insoluble CaCO ₃ precipitate that makes the limewater go milky)

Test for bromides and iodides

As was the case with the chlorides, the bromides and iodides also form precipitates when they are reacted with silver nitrate. Silver chloride is a white precipitate, but the silver bromide and silver iodide precipitates are both pale yellow. To determine whether the precipitate is a bromide or an iodide, we use chlorine water and carbon tetrachloride (CCl ${}_4$ ).

Chlorine water frees bromine gas from the bromide and colours the carbon tetrachloride a reddish brown.

Chlorine water frees iodine gas from an iodide and colours the carbon tetrachloride purple.

Precipitation reactions and ions in solution

1. Silver nitrate (AgNO ${}_3$ ) reacts with potassium chloride (KCl) and a white precipitate is formed.
   1. Write a balanced equation for the reaction that takes place.
   2. What is the name of the insoluble salt that forms?
   3. Which of the salts in this reaction are soluble?
2. Barium chloride reacts with sulphuric acid to produce barium sulphate and hydrochloric acid.
   1. Write a balanced equation for the reaction that takes place.
   2. Does a precipitate form during the reaction?
   3. Describe a test that could be used to test for the presence of barium sulphate in the products.
3. A test tube contains a clear, colourless salt solution. A few drops of silver nitrate solution are added to the solution and a pale yellow precipitate forms. Which one of the following salts was dissolved in the original solution?
   1. NaI
   2. KCl
   3. K ${}_2$ CO ${}_3$
   4. Na ${}_2$ SO ${}_4$
   (IEB Paper 2, 2005)