Testing for ions

The flame test tests for positive ions (metal ions).

Dip the loop in hydrochloric acid and rinse it with distilled water (cleaning).

Dip the wire loop into a sample of the compound.

Put loop in clear blue part of bunsen flame (hottest region).

You can identify the metal ions in the substance by the colour the flame goes.

Lithium (Li+) > crimson

Calcium (Ca2+) > red

Potassium (K+) > lilac

Sodium (Na+) > yellow

Barium (Ba2+) > green

The precipitate test tests for positive ions.

Many metal hydroxides are insoluble and precipitate out of a solution when formed.
Some of these hydroxides have a characteristic colour.

Add a few drops of sodium hydroxide solution to a solution of unknown compound. This should form an insoluble hydoxide.

If you get a coloured insoluble hydoxide you can detect which ions are present.

Calcium ( Ca2+) > white > Ca2+ (aq) + 2OH- (aq) --> Ca(OH)2 (s)

Copper (II) (Cu2+) > blue > Cu2+ (aq) + 2OH- (aq) --> Cu(OH)2 (s)

Iron (II) (Fe2+) > green> Fe2+ (aq) + 2OH- (aq) --> Fe(OH)2 (s)

Iron (III) (Fe3+) > brown > Fe3+ (aq) + 2OH- (aq) --> Fe(OH)3 (s)

Aluminium (Al3+) > white, then redissolves in excess NaOH to form colourless solution > Al3+ (aq) + 2OH- (aq) --> Al(OH)3 (s) THEN Al(OH)3 +OH- (aq) --> Al(OH)4-

Magnesium (Mg2+) > white > Mg2+ (aq) + 2OH- (aq) --> Mg(OH)2 (s)

This test for CO3 2- ions (Carbonate ions).

Carbonates react with dilute acids to form carbon dioxide.

To test for this carbon dioxide pass the gas produce through limewater > if it turns cloudy CO2 is present.

Carbonates are present.

Acid + Carbonate --> Salt + Water + Carbon dioxide

Halide ions.

To test for chloride (Cl-) , bromide (Br-) , iodide (I-) add dilute nitric acid (HNO3).

The nitric acid reacts with, and removes, any ions that may react with the silver and form a precipitate.

Then add silver nitrate solution (AgNO3).

Chloride > white precipitate of silver chloride > Ag+ (aq) + Cl- (aq) --> AgCl (s)

Bromide > cream precipitate of silver bromide > Ag+ (aq) + Br- (aq) --> AgBr (s)

Iodide > yellow precipitate of silver iodide > Ag+ (aq) + I- (aq) --> AgI (s)

Sulfate ion = SO4 2- ion

Add dilute HCl (hydrochloric acid), this gets rid of unwanted ions, such as carbonate ions.

Add barium chloride solution (BaCl2).

White precipitate of barium sulfate means sulfate was present in original compound.

Ba2+ (aq) + So4 2- (aq) --> BaSO4 (s)