1. Rossita Radzak : SASER
SALTS
Preparation of soluble salt
1. Acid + reactive metal(Zn / Mg)  salt + H2 / 2H+ + Mg  Mg2+ + H2
2. Acid + base ( metal oxide)  salt + water
3. Acid + alkali  salt + water / H+ + OH-  H2O ( NaOH, KOH, NH4OH)
4. Acid + carbonate metal  salt + CO2 + H2O / 2H+ + CO32+  CO2 + H2O
Procedure:
1. pour ( 25 – 100cm3) acid ( 0.5 – 2.0 mol dm-3) into a beaker
2. warm /heat slowly
3. add solid (metal / base/ carbonate ) a little until excess / no more dissolve
4. stir
5. filter the mixture into evaporating dish
6. heat (slowly) the filtrate until 1/3 from original volume / saturated solution formed
7. cool down the saturated solution (until crystallized )
8. filter (to separate the crystals)
9. dry / transfer onto filter paper / dry between sheets of filter paper
Observation
Chemical equation
Preparation of insoluble salt – precipitation reaction / double decomposition reaction
Pb2+ + SO42-  PbSO4
Procedure :
1. pour ( 25 – 50cm3) of soluble salt (Pb(NO3)2 into a beaker
2. add ( 25 – 50cm3) of soluble salt (Na2SO4)
3. stir
4. filter the mixture
5. rinse residue / solid / precipitate
6. dry between sheets of filter paper
Observation
Chemical equation
Ionic equation
Action of heat on salt
Carbonate  oxide metal (base) + CO2 except Na, K and NH4+
Example: CuCO3  CuO + CO2
Nitrate  oxide metal + nitrogen oxide + oxygen except Na, K,
Examples:
(2NaNO3  2NaNO2 + O2 )
2Mg(NO3)2  2MgO + 4NO2 + O2
(Brown gas)
Ammonium chloride  ammonia gas + hydrogen chloride gas, (NH4Cl  NH3 + HCl )
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2. Rossita Radzak : SASER
Confirmatory test
1. State the material / chemical / reagent
2. procedure
3. observation
4. conclusion
Example: You are given a bottle of ammonium chloride solution. Describe chemical test to
verify the cation and anion.
(a) test for cation (NH4+)
1. pour 2 cm3 the solutions into a test tube
2. add 1 cm3 copper (II) sulphate solution / Nessler reagent into the test tube
3. blue precipitate soluble in excess to form dark blue solution / brown precipitate
4. Ammonium ions (NH4+) present
(b) test for anion (Cl-)
1. pour 2 cm3 the solution into a test tube
2. add 1 cm3 of dilute nitric acid and silver nitrate solution
3. white precipitate formed
4. confirm the presence of chloride ions
Example: You are given lead (II) nitrate and aluminium nitrate solution. Describe chemical test
to verify the cation and anion.
(c) test for cation
1. pour 2 cm3 the solutions into different test tubes
2. add 1 cm3 potassium iodide solution into the test tubes
3. yellow precipitate formed
4. lead (II) ion present.
(d) test for anion
1. pour 2 cm3 of lead (II) nitrate solution into a test tube
2. add 1 cm3 of dilute sulphuric acid
3. add 1 cm3 of iron (II) sulphate solution
4. shake the mixure
5. tilt the test tube, add concentrated sulphuric acid carefully // drop by drop down
the side of the test tube
6. the brown ring formed
7. nitrate ion, NO3- present.
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3. Rossita Radzak : SASER
Paper 3
Aim : To construct the ionic equation for the formation of lead (II) chromate(VI)
Problem Statement:
How does an ionic equation for the formation of lead (II) chromate (VI) can be constructed?
Hypothesis :
As the volume potassium chromate (VI) used is increases, the height of the yellow precipitate
increases until it achieves a maximum height.
Variables:
Manipulated : Volumes of 0.5 mol dm-3 potassium chromate (VI) solution.
Responding : Height of yellow precipitate.
Fixed : Size of test tubes, volume and concentration of lead (II) nitrate solution,
concentration of potassium chromate (VI) solution.
Apparatus : Test tubes of the same size, test tube rack, burette, retort stand with
clamp, ruler, glass rod, dropper.
Material : 0.5 mol dm-3 potassium chromate (VI) solution, 0.5 mol dm-3 lead (II)
nitrate solution.
Procedure :
1. Seven test tubes of the same size were labelled from number 1 to 7. They were placed in
a test tube rack.
2. A burette was filled 0.5 mol dm-3 lead (II) nitrate solution, 5.00 cm3 of the lead (II)
nitrate solution was run into each the seven tubes.
3. Another burette was filled with 0.5 mol dm-3 potassium chromate (VI) solution.
4. Potassium chromate (VI) solution from the burette was added into each of the seven test
tubes according to the volumes specified in the table.
5. The mixture in each test tube was stirred with a clean glass rod.
6. The test tubes were left aside for about an hour.
7. The height of the precipitate in each test tube was measured. The colour of the solution
above the precipitate in each test tube was observed and recorded.
Results:
Test tube 1 2 3 4 5 6 7
Volume of 0.5 mol 5.00 5.00 5.00 5.00 5.00 5.00 5.00
dm-3
Pb(NO3)2 /cm3
Volume of 0.5 mol 1.00 2.00 3.00 4.00 5.00 6.00 7.00
dm-3
K2Cr O4 /cm3
Height of precipitate 0.60 1.20 1.80 2.40 3.00 3.00 3.00
(cm)
Colour of solution colourless colourless colourless colourless colourless yellow yellow
above the precipitate
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4. Rossita Radzak : SASER
Paper 2
Discussion
The volume of 0.5 mol dm-3 potassium chromate (VI), solution required to exactly react with
5.00 cm3 of 0.5 mol dm-3 lead (II) nitrate solution is 5.00 cm3.
Calculation:
Number of moles lead (II) ions = MV
= 0.5 x 5.00/1000 = 0.0025 mol.
Number of moles chromate (VI) ions = MV
= 0.5 x 5.00/1000 = 0.0025 mol.
Simplest mole ratio of lead (II) ions : chromate (VI) ions
0.0025 : 0.0025
1 : 1
Discussions:
1. A yellow precipitate of lead (II) chromate (VI) is formed in each of the seven test tubes.
2. The height of the precipitate increases gradually from test tubes 1 to 5 because more and more
lead (II) chromate (VI) is formed due to the increasing amount of potassium chromate (VI)
added to the test tubes.
3. The colour of solution above the precipitate in test tubes 1 to 4 are colourless due to the excess
lead (II) nitrate.
4. The colour of solution above the precipitate in test tubes 6 to 7 is yellow due to the excess
potassium chromate (VI).
5. Ionic equation : Pb2+ + Cr2O72-  Pb Cr2O7
Conclusion:
As / when the volume of potassium chromate (VI) solution used increases, the height of the
precipitate increases until it achieves a maximum height.
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