# everything maths & science

You are here: Home Precipitation reactions

## Precipitation reactions

Sometimes, ions in solution may react with each other to form a new substance that is insoluble. This is called a precipitate. The reaction is called a precipitation reaction.

Preparing precipitates

Definition 1: Precipitate

A precipitate is the solid that forms in a solution during a chemical reaction.

## General experiment 1: The reaction of ions in solution

Aim

To investigate the reactions of ions in solutions.

Apparatus

4 test tubes; copper(II) chloride solution; sodium carbonate solution; sodium sulphate solution

Method

1. Prepare 2 test tubes with approximately 5 ml of dilute copper(II) chloride solution in each

2. Prepare 1 test tube with 5 ml sodium carbonate solution

3. Prepare 1 test tube with 5 ml sodium sulphate solution

4. Carefully pour the sodium carbonate solution into one of the test tubes containing copper(II) chloride and observe what happens

5. Carefully pour the sodium sulphate solution into the second test tube containing copper(II) chloride and observe what happens

Results

1. A light blue precipitate forms when sodium carbonate reacts with copper(II) chloride.

2. No precipitate forms when sodium sulphate reacts with copper(II) chloride. The solution is light blue.

It is important to understand what happened in the previous demonstration. We will look at what happens in each reaction, step by step.

For reaction 1 you have the following ions in your solution: $Cu 2+$, $Cl -$, $Na +$ and $CO 3 2-$. A precipitate will form if any combination of cations and anions can become a solid. The following table summarises which combination will form solids (precipitates) in solution.

Table 1: General rules for the solubility of salts
 Salt Solubility Nitrates All are soluble Potassium, sodium and ammonium salts All are soluble Chlorides, bromides and iodides All are soluble except silver, lead(II) and mercury(II) salts (e.g. silver chloride) Sulphates All are soluble except lead(II) sulphate, barium sulphate and calcium sulphate Carbonates All are insoluble except those of potassium, sodium and ammonium Compounds with fluorine Almost all are soluble except those of magnesium, calcium, strontium (II), barium (II) and lead (II) Perchlorates and acetates All are soluble Chlorates All are soluble except potassium chlorate Metal hydroxides and oxides Most are insoluble

## Tip:

Salts of carbonates, phosphates, oxalates, chromates and sulphides are generally insoluble.

If you look under carbonates in the table it states that all carbonates are insoluble except potassium sodium and ammonium. This means that $Na 2 CO 3$ will dissolve in water or remain in solution, but $CuCO 3$ will form a precipitate. The precipitate that was observed in the reaction must therefore be $CuCO 3$. The balanced chemical equation is:

$2Na + (aq)+CO 3 2- (aq)+Cu 2+ (aq)+2Cl - (aq)→CuCO 3 (s)+2Na + (aq)+2Cl - (aq)$(1)

Note that sodium chloride does not precipitate and we write it as ions in the equation. For reaction 2 we have $Cu 2+$, $Cl -$, $Na +$ and $SO 4 2-$ in solution. Most chlorides and sulphates are soluble according to the table. The balanced chemical equation is:

$2Na + (aq)+SO 4 2- (aq)+Cu 2+ (aq)+2Cl - (aq)→2Na + (aq)+SO 4 2- (aq)+Cu 2+ (aq)+2Cl - (aq)$(2)

Both of these reactions are ion exchange reactions.

## Tests for anions

We often want to know which ions are present in solution. If we know which salts precipitate, we can derive tests to identify ions in solution. Given below are a few such tests.

Test for halides

### 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.

$Cl - (aq)+Ag + (aq)+NO 3 - (aq)→AgCl(s)+NO 3 - (aq)$(3)

(AgCl is white precipitate)

$CO 3 2- (aq)+2Ag + (aq)+2NO 3 - (aq)→Ag 2 CO 3 (s)+2NO 3 - (aq)$(4)

($Ag 2 CO 3$ 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.

$AgCl(s)+HNO 3 (ℓ)→$ (no reaction; precipitate is unchanged)

$Ag 2 CO 3 (s)+2HNO 3 (ℓ)→2Ag + (aq)+2NO 3 - (aq)+H 2 O(ℓ)+CO 2 (g)$ (precipitate disappears)

### 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.

$2Br - (aq)+Cl 2 (aq)→2Cl - (aq)+Br 2 (g)$

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

$2I - (aq)+Cl 2 (aq)→2Cl - (aq)+I 2 (g)$

### 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.

$SO 4 2- (aq)+Ba 2+ (aq)+Cl - (aq)→BaSO 4 (s)+Cl - (aq)$ ($BaSO 4$ is a white precipitate)

$CO 3 2- (aq)+Ba 2+ (aq)+Cl - (aq)→BaCO 3 (s)+Cl - (aq)$ ($BaCO 3$ 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).

$BaSO 4 (s)+HNO 3 (ℓ)→$ (no reaction; precipitate is unchanged)

$BaCO 3 (s)+2HNO 3 (ℓ)→Ba 2+ (aq)+2NO 3 - (aq)+H 2 O(ℓ)+CO 2 (g)$ (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.

$2HCl+K 2 CO 3 (aq)→CO 2 (g)+2KCl(aq)+H 2 Oℓ$(5)

If the gas is passed through limewater (an aqueous solution of calcium hydroxide) and the solution becomes milky, the gas is carbon dioxide.

$Ca 2+ (aq)+2OH - (aq)+CO 2 (g)→CaCO 3 (s)+H 2 O(ℓ)$ (It is the insoluble $CaCO 3$ precipitate that makes the limewater go milky)

#### Exercise 1: Precipitation reactions and ions in solution

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. Include the state symbols.

2. What is the name of the insoluble salt that forms?

3. Which of the salts in this reaction are soluble?

a) $AgN{O}_{3}+KCl\to AgCl+KN{O}_{3}$

Alternatively since $KN{O}_{3}$  is soluble we could write:

$AgN{O}_{3}\to AgCl+{K}^{+}+C{l}^{-}$

b) Silver chloride

c) Potassium nitrate.

Barium chloride reacts with sulphuric acid to produce barium sulphate and hydrochloric acid.

1. Write a balanced equation for the reaction that takes place. Include the state symbols.

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.

a) $BaC{l}_{2}+{H}_{2}S{O}_{4}\to BaS{O}_{4}+2HCl$

b) Yes. $BaS{O}_{4}$

c) Barium sulphate is unaffected by nitric acid so if one adds nitric acid the precipitate will remain unchanged.

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. Chlorine water and carbon tetrachloride were added, which resulted in a purple solution. Which one of the following salts was dissolved in the original solution? Write the balanced equation for the reaction that took place between the salt and silver nitrate.

1. $NaI$

2. $KCl$

3. $K 2 CO 3$

4. $Na 2 SO 4$

d.

NaI