Precipitation Reactions Revision Notes for Grade 10 NSC Matric Physical Sciences

Precipitation Reactions

What are precipitation reactions?

Precipitation reactions occur when ions in solution react with each other to form a new substance that is insoluble in water. This insoluble substance appears as a solid and is called a precipitate.

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

When two aqueous solutions containing different ions are mixed, the ions can sometimes combine to form an insoluble compound. This compound separates from the solution as a visible solid precipitate.

Understanding how precipitation reactions work

Before precipitation reactions can occur, the original compounds must first dissociate (break apart) into their individual ions in solution. For example:

Potassium bromide: $\text{KBr}_{(s)} \rightarrow \text{K}^+_{(aq)} + \text{Br}^-_{(aq)}$
Table salt: $\text{NaCl}_{(s)} \rightarrow \text{Na}^+_{(aq)} + \text{Cl}^-_{(aq)}$
Hydrochloric acid: $\text{HCl}_{(\ell)} + \text{H}_2\text{O}_{(\ell)} \rightarrow \text{H}_3\text{O}^+_{(aq)} + \text{Cl}^-_{(aq)}$
Sodium hydroxide: $\text{NaOH}_{(s)} \rightarrow \text{Na}^+_{(aq)} + \text{OH}^-_{(aq)}$

Once the ions are free in solution, they can react with ions from other compounds to form new combinations. If one of these new combinations is insoluble, a precipitate forms.

Worked example: Investigating precipitation reactions

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Worked Example: Investigating Precipitation Reactions

Aim: To investigate the reactions of ions in solution

Apparatus: 4 test tubes, copper(II) chloride solution, sodium carbonate solution, sodium sulphate solution

Method:

Prepare 2 test tubes with approximately 5 ml of dilute copper(II) chloride solution in each
Prepare 1 test tube with 5 ml sodium carbonate solution
Prepare 1 test tube with 5 ml sodium sulphate solution
Carefully pour the sodium carbonate solution into one of the test tubes containing copper(II) chloride and observe what happens
Carefully pour the sodium sulphate solution into the second test tube containing copper(II) chloride and observe what happens

Results:

A light blue precipitate forms when sodium carbonate reacts with copper(II) chloride
No precipitate forms when sodium sulphate reacts with copper(II) chloride (the solution remains light blue)

Understanding the results using solubility rules

To predict whether a precipitation reaction will occur, we need to understand solubility rules. These rules tell us which compounds are soluble (dissolve in water) and which are insoluble (form precipitates).

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Key solubility rules to remember:

Salt Type	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
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

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

Explaining the experimental results

In Reaction 1 (copper chloride + sodium carbonate), we have these ions in solution:

$\text{Cu}^{2+}$ , $\text{Cl}^-$ , $\text{Na}^+$ , and $\text{CO}_3^{2-}$

According to the solubility rules, carbonates are insoluble except for potassium, sodium and ammonium carbonates. This means $\text{Na}_2\text{CO}_3$ will dissolve, but $\text{CuCO}_3$ will form a precipitate.

The balanced chemical equation is:

2\mathrm{Na}^{+}_{\text{(aq)}} + \mathrm{CO}_{3}^{2-}{}_{\text{(aq)}} + \mathrm{Cu}^{2+}_{\text{(aq)}} + 2\mathrm{Cl}^{-}_{\text{(aq)}} \rightarrow \mathrm{CuCO}_{3\text{(s)}} + 2\mathrm{Na}^{+}_{\text{(aq)}} + 2\mathrm{Cl}^{-}_{\text{(aq)}}

In Reaction 2 (copper chloride + sodium sulphate), we have these ions in solution:

$\text{Cu}^{2+}$ , $\text{Cl}^-$ , $\text{Na}^+$ , and $\text{SO}_4^{2-}$

According to the solubility rules, most chlorides and sulphates are soluble. Since both possible products ( $\text{NaCl}$ and $\text{CuSO}_4$ ) are soluble, no precipitate forms.

The balanced chemical equation is:

2\mathrm{Na}^{+}_{\text{(aq)}} + \mathrm{SO}_{4}^{2-}{}_{\text{(aq)}} + \mathrm{Cu}^{2+}_{\text{(aq)}} + 2\mathrm{Cl}^{-}_{\text{(aq)}} \rightarrow 2\mathrm{Na}^{+}_{\text{(aq)}} + \mathrm{SO}_{4}^{2-}{}_{\text{(aq)}} + \mathrm{Cu}^{2+}_{\text{(aq)}} + 2\mathrm{Cl}^{-}_{\text{(aq)}}

Both reactions are examples of ion exchange reactions.

Tests for anions

Precipitation reactions can be used to identify which anions (negative ions) are present in unknown solutions. Here are the key tests you need to know:

Test for chloride

Method: Add a small amount of silver nitrate solution to the unknown salt solution.

Observation: If a white precipitate forms, the salt contains chloride ions.

Chemical equation:

\mathrm{Cl}^{-}_{\text{(aq)}} + \mathrm{Ag}^{+}_{\text{(aq)}} + \mathrm{NO}_{3}^{-}{}_{\text{(aq)}} \rightarrow \mathrm{AgCl}_{\text{(s)}} + \mathrm{NO}_{3}^{-}{}_{\text{(aq)}}

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Confirmation test: Treat the precipitate with concentrated nitric acid. If the precipitate remains unchanged, it confirms the presence of chloride (since AgCl is insoluble in nitric acid).

$\text{AgCl}_{(s)} + \text{HNO}_{3(\ell)} \rightarrow \text{no reaction (precipitate is unchanged)}$

Test for bromides and iodides

Method: Add silver nitrate solution to the unknown salt solution.

Observation:

Silver bromide precipitate is pale yellow
Silver iodide precipitate is also pale yellow

To distinguish between bromides and iodides, treat the precipitate with chlorine water:

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For bromides:

Chlorine water frees bromine gas and colours the carbon tetrachloride reddish brown
$2\text{Br}^-_{(aq)} + \text{Cl}_{2(aq)} \rightarrow 2\text{Cl}^-_{(aq)} + \text{Br}_{2(g)}$

For iodides:

Chlorine water frees iodine gas and colours the carbon tetrachloride purple
$2\text{I}^-_{(aq)} + \text{Cl}_{2(aq)} \rightarrow 2\text{Cl}^-_{(aq)} + \text{I}_{2(g)}$

Test for sulphate

Method: Add a small amount of barium chloride solution to the unknown salt solution.

Observation: If a white precipitate forms, the salt contains sulphate ions.

Chemical equation:

\mathrm{SO}_{4}^{2-}{}_{\text{(aq)}} + \mathrm{Ba}^{2+}_{\text{(aq)}} + \mathrm{Cl}^{-}_{\text{(aq)}} \rightarrow \mathrm{BaSO}_{4\text{(s)}} + \mathrm{Cl}^{-}_{\text{(aq)}}

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Confirmation test: Treat the precipitate with nitric acid. If it remains unchanged, it confirms the presence of sulphate (since $\text{BaSO}_4$ is insoluble in nitric acid).

$\text{BaSO}_{4(s)} + \text{HNO}_{3(\ell)} \rightarrow \text{no reaction (precipitate is unchanged)}$

Test for carbonate

Method: Treat a sample of the dry salt with a small amount of dilute acid.

Observation: If carbon dioxide gas is produced, the salt is a carbonate.

Chemical equation: $2\text{HCl} + \text{K}_2\text{CO}_{3(aq)} \rightarrow \text{CO}_{2(g)} + 2\text{KCl}_{(aq)} + \text{H}_2\text{O}_{(\ell)}$

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Confirmation test: Pass the gas through limewater (calcium hydroxide solution). If the solution becomes milky, the gas is carbon dioxide.

$\text{Ca}^{2+}_{(aq)} + 2\text{OH}^-_{(aq)} + \text{CO}_{2(g)} \rightarrow \text{CaCO}_{3(s)} + \text{H}_2\text{O}_{(\ell)}$

The insoluble $\text{CaCO}_3$ precipitate makes the limewater appear milky.

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Key Points to Remember:

A precipitate is a solid that forms when ions in solution react to produce an insoluble compound
Use solubility rules to predict whether precipitation reactions will occur - if both possible products are soluble, no precipitate forms
All nitrates and all potassium, sodium and ammonium salts are soluble
Most carbonates are insoluble except those of K⁺, Na⁺, and NH₄⁺
Precipitation reactions can be used as tests to identify specific anions like chloride, sulphate, and carbonate ions

Precipitation Reactions (Grade 10 NSC Matric Physical Sciences): Revision Notes