RP4 - Identification of cations and anions Flashcards
How do you test for group 2 metal cations?
1.Place 10 drops of a group 2 compound in test tube (0.1 mol dm-3).
2.Add 10 drops of 0.6 mol dm-3 NaOH to test tube. Record any observations.
3.Continue to add NaOH so it is in excess. Record any
observations.
4.Repeat with other group 2 compounds.
How would you distinguish between calcium bromide and strontium chloride?
- Place 10 drops of BaCl2 (0.1 mol dm-3) in a test tube.
- Add 10 drops of sulfuric acid (1 mol dm-3). Record observations.
- Continue to add H2SO4 until in excess. Record observations.
- Repeat for other the group 2 compounds
How do you test for ammonium (NH4+) ions?
- Place 10 drops of NH4Cl into a test tube.
- Add 10 drops of NaOH solution. Shake.
- Warm the solution in the test tube.
- Test the gas released with damp red litmus paper.
- If it goes blue, ammonium ions are present.
How do you test for group 7/halide ions?
1.To the compound being tested, add nitric acid and silver nitrate. Record observations. 2.To samples of this solution, add dilute and then concentrated ammonia.
How do you test for hydroxide (OH-) ions?
- Test a 1 cm depth of solution in a test tube with red litmus paper or universal indicator paper.
- Record observations.
- Sodium hydroxide will turn damp red litmus paper blue.
How do you test for carbonate (CO32-) ions?
1.Put 2 cm3 of Ca(OH)2 into a test tube.
2.Add 3 cm3 Na2CO3
(0.5 mol dm–3) in another test tube
then add an equal volume of dilute HCl (1.0 mol dm–3).
3.Immediately put in delivery tube with open end into the
Ca(OH)2 test tube.
4.If Ca(OH)2 goes cloudy, carbonate ions were present in the other test tube.
How do you test for sulfate (SO42-) ions?
1.Add HCl and BaCl2 to the suspected sulfate solution. 2.If sulfate ions are present, a white precipitate of BaSO4 will form.
What is the order of testing ions?
Carbonate, halide and sulfate
Carbonate → Sulfate → Halide
This prevents false positive results i.e. Unexpected
insoluble precipitates such as Ag2SO4,Ag2CO3 and BaCO3 could form.