Unit 4 assignment 2
Task 1: Anion and Cation Analysis
1.1
Anion and Cation book completed and attached.
1.2 The chemical basis of the anion and cation tests
a ) Flame colours are produced via the movement of electrons in the metal ions within a
compound. When heat is applied, the electrons jump to a high energy state. This is due to the
added heat energy transferring onto the electrons as kinetic energy, providing them with more
energy. At this point, as the electrons have a high energy state - and eventually they fall back
down to where they were originally, however this occurs at different times. The exact sizes of
the possible jumps in energy terms differ from one metal ion to another. Electrons get excited by
different amounts into different energy levels. When they fall back down and return to their
ground state, they emit their energy as visible light and this is how different colours are formed
as a consequence. Colours are dependent on the element as different ions produce different
patterns of spectral lines. Metal ions emit a specific color when heated, which identifies them in
the sample. The colour of the light is dependent on the location of the electrons and the
attraction of the outer shell electrons to the nucleus. For example, a flame turns lilac to
represent the presence of potassium. Sources of human error are contamination which will
affect the flames colour.
This test involves heating the nichrome wire, then putting the loop into hydrochloric acid. This
cleans the wire avoiding contamination. Next, the wire is dipped into the compound and held
directly in the flame. The compound present will then form a colour for identification of the metal
ion present. Colours are very bright and look like this:
,Unit 4 assignment 2
B and C) Cation chemical equations
● Cu2+ (s) + 2NaOH (aq) = Cu(OH)2 +2Na
= copper + sodium hydroxide = copper hydroxide + sodium
When copper reacts with sodium hydroxide, a pale blue precipitate will form due to the
copper hydroxide and sodium formation (reactants of the reaction). If excess sodium
hydroxide is added, the precipitate does not dissolve and there is no further reaction.
The blue allows us to identify the cation present.
● Fe3+ (s) + NaOH (aq) = Fe(OH)3 + Na
= iron + sodium hydroxide = iron hydroxide + sodium
When iron reacts with sodium hydroxide, a reddish/brown precipitate is formed as a
consequence of iron hydroxide and sodium being produced. If excess sodium hydroxide
is added to this, the precipitate does not dissolve and there is no further reaction. The
reddish/brown precipitate allows us to identify the cation present.
● Zn2+ (s) + NaOH (aq) = Zn(OH)3 + Na
= zinc + sodium hydroxide = zinc hydroxide + sodium
When zinc reacts with sodium hydroxide, a white precipitate is formed as a result of zinc
hydroxide and sodium being produced. If excess sodium hydroxide was added to this
the precipitate will dissolve and a further reaction has occurred. The white ppt allows us
to infer what action is involved.
.
● Cu2+ (s) + NH4OH (aq) = Cu(OH)2 +NH4
= copper + ammonium hydroxide = copper hydroxide + ammonia
When copper reacts with ammonium hydroxide, a pale blue precipitate is formed due to
reactants of ammonia and copper hydroxide being produced. If excess ammonium
hydroxide is added, the precipitate will dissolve to produce a pale blue solution, showing
that there has been a further reaction. The blue precipitate formed identifies the cation
copper.
● Fe3+ (s) + NH4OH (aq) = Fe(OH)3 +NH4
= iron + ammonium hydroxide = iron hydroxide + ammonia
Iron and ammonium hydroxide should not be tested in this laboratory because the fumes
produced are dangerous. This should be done by a specialist in a scientific laboratory
with special PPE and equipment.
● Zn2+ (s) + NH4OH (aq) = Zn(OH)2 + NH4
= zinc + ammonium hydroxide = zinc hydroxide + ammonia
When zinc reacts with ammonium hydroxide, it forms a white precipitate alongside the
products of ammonia and zinc hydroxide. If excess ammonium hydroxide is added, the
precipitate will dissolve and a further reaction has taken place. The white precipitate
helps identify the zinc cation.
, Unit 4 assignment 2
D and E) Anion chemical equations
● Tests for Cl- chloride
1. If the chloride is soluble, then dilute nitric acid will be added to silver nitrate
solution. The nitric acid acidifies the nitrate to prevent the precipitation of other
non-halide silver salts.
Ag+ (aq) + Cl- (aq) = AgCl (s)
This reaction will produce a white precipitate of silver chloride, soluble in dilute
ammonia
(Any soluble silver salt added to any soluble chloride produces a white silver
chloride precipitate which darkens in the light)
2. If the chloride is insoluble, then add concentrated sulfuric acid and wam up if
needed to then test the gas for HCL.
Then … Ag+(aq) + Br- (aq) = AgCl (s)
This reaction will produce fumes of hydrogen chloride which turns blue litmus
paper red and produces a white precipitate alongside a silver nitrate solution.
● Tests for SO42-
1. This is for a solution of suspected sulphate. Add dilute hydrochloric acid and a
few drops of barium chloride solution.
Ba2+ (aq) + SO42- (aq) = BaSO4 (s)
This reaction will produce a white precipitate of barium sulphate
(Any soluble barium salt mixed with any soluble sulphate forms a white dense
barium sulphate precipitate)
2. Adding lead (11) to nitrate solution.
Pb2+ (aq) + SO42- (aq) = PbSO4 (s)
This reaction will produce a white precipitate of lead (11) sulphate.
Neither white precipitates from either reaction are soluble in excess hydrochloric
acid.
● Tests for CO32-
1. Add any dilute strong acid into the suspected solid carbonate - if colourless gas is
given off, the then test with limewater
Any carbonate + acid = salt = water + carbon dioxide + ( white ppt with
limewater)
E.g CO32- (s) + 2H+ (aq) = H20 (l) + CO2 (g)
Task 1: Anion and Cation Analysis
1.1
Anion and Cation book completed and attached.
1.2 The chemical basis of the anion and cation tests
a ) Flame colours are produced via the movement of electrons in the metal ions within a
compound. When heat is applied, the electrons jump to a high energy state. This is due to the
added heat energy transferring onto the electrons as kinetic energy, providing them with more
energy. At this point, as the electrons have a high energy state - and eventually they fall back
down to where they were originally, however this occurs at different times. The exact sizes of
the possible jumps in energy terms differ from one metal ion to another. Electrons get excited by
different amounts into different energy levels. When they fall back down and return to their
ground state, they emit their energy as visible light and this is how different colours are formed
as a consequence. Colours are dependent on the element as different ions produce different
patterns of spectral lines. Metal ions emit a specific color when heated, which identifies them in
the sample. The colour of the light is dependent on the location of the electrons and the
attraction of the outer shell electrons to the nucleus. For example, a flame turns lilac to
represent the presence of potassium. Sources of human error are contamination which will
affect the flames colour.
This test involves heating the nichrome wire, then putting the loop into hydrochloric acid. This
cleans the wire avoiding contamination. Next, the wire is dipped into the compound and held
directly in the flame. The compound present will then form a colour for identification of the metal
ion present. Colours are very bright and look like this:
,Unit 4 assignment 2
B and C) Cation chemical equations
● Cu2+ (s) + 2NaOH (aq) = Cu(OH)2 +2Na
= copper + sodium hydroxide = copper hydroxide + sodium
When copper reacts with sodium hydroxide, a pale blue precipitate will form due to the
copper hydroxide and sodium formation (reactants of the reaction). If excess sodium
hydroxide is added, the precipitate does not dissolve and there is no further reaction.
The blue allows us to identify the cation present.
● Fe3+ (s) + NaOH (aq) = Fe(OH)3 + Na
= iron + sodium hydroxide = iron hydroxide + sodium
When iron reacts with sodium hydroxide, a reddish/brown precipitate is formed as a
consequence of iron hydroxide and sodium being produced. If excess sodium hydroxide
is added to this, the precipitate does not dissolve and there is no further reaction. The
reddish/brown precipitate allows us to identify the cation present.
● Zn2+ (s) + NaOH (aq) = Zn(OH)3 + Na
= zinc + sodium hydroxide = zinc hydroxide + sodium
When zinc reacts with sodium hydroxide, a white precipitate is formed as a result of zinc
hydroxide and sodium being produced. If excess sodium hydroxide was added to this
the precipitate will dissolve and a further reaction has occurred. The white ppt allows us
to infer what action is involved.
.
● Cu2+ (s) + NH4OH (aq) = Cu(OH)2 +NH4
= copper + ammonium hydroxide = copper hydroxide + ammonia
When copper reacts with ammonium hydroxide, a pale blue precipitate is formed due to
reactants of ammonia and copper hydroxide being produced. If excess ammonium
hydroxide is added, the precipitate will dissolve to produce a pale blue solution, showing
that there has been a further reaction. The blue precipitate formed identifies the cation
copper.
● Fe3+ (s) + NH4OH (aq) = Fe(OH)3 +NH4
= iron + ammonium hydroxide = iron hydroxide + ammonia
Iron and ammonium hydroxide should not be tested in this laboratory because the fumes
produced are dangerous. This should be done by a specialist in a scientific laboratory
with special PPE and equipment.
● Zn2+ (s) + NH4OH (aq) = Zn(OH)2 + NH4
= zinc + ammonium hydroxide = zinc hydroxide + ammonia
When zinc reacts with ammonium hydroxide, it forms a white precipitate alongside the
products of ammonia and zinc hydroxide. If excess ammonium hydroxide is added, the
precipitate will dissolve and a further reaction has taken place. The white precipitate
helps identify the zinc cation.
, Unit 4 assignment 2
D and E) Anion chemical equations
● Tests for Cl- chloride
1. If the chloride is soluble, then dilute nitric acid will be added to silver nitrate
solution. The nitric acid acidifies the nitrate to prevent the precipitation of other
non-halide silver salts.
Ag+ (aq) + Cl- (aq) = AgCl (s)
This reaction will produce a white precipitate of silver chloride, soluble in dilute
ammonia
(Any soluble silver salt added to any soluble chloride produces a white silver
chloride precipitate which darkens in the light)
2. If the chloride is insoluble, then add concentrated sulfuric acid and wam up if
needed to then test the gas for HCL.
Then … Ag+(aq) + Br- (aq) = AgCl (s)
This reaction will produce fumes of hydrogen chloride which turns blue litmus
paper red and produces a white precipitate alongside a silver nitrate solution.
● Tests for SO42-
1. This is for a solution of suspected sulphate. Add dilute hydrochloric acid and a
few drops of barium chloride solution.
Ba2+ (aq) + SO42- (aq) = BaSO4 (s)
This reaction will produce a white precipitate of barium sulphate
(Any soluble barium salt mixed with any soluble sulphate forms a white dense
barium sulphate precipitate)
2. Adding lead (11) to nitrate solution.
Pb2+ (aq) + SO42- (aq) = PbSO4 (s)
This reaction will produce a white precipitate of lead (11) sulphate.
Neither white precipitates from either reaction are soluble in excess hydrochloric
acid.
● Tests for CO32-
1. Add any dilute strong acid into the suspected solid carbonate - if colourless gas is
given off, the then test with limewater
Any carbonate + acid = salt = water + carbon dioxide + ( white ppt with
limewater)
E.g CO32- (s) + 2H+ (aq) = H20 (l) + CO2 (g)
