Ion identification Flashcards
(1) A test was carried out to find which anion is present in substance X.
Dilute hydrochloric acid was added to a sample of substance X.
There was effervescence and the gas given off turned limewater milky.
The anion present in substance X is
carbonate ion CO3
2-
(2) Describe how sodium hydroxide solution can be used to show that
ammonium ions are present in substance X.
warm / heat / boil (1) {gas/ammonia} turns (damp red/pink) litmus blue / (damp red/pink) litmus turns blue when held above (the mixture)(1)
(3) Aluminium ions, Al3+, react with hydroxide ions in solution to give a white precipitate of aluminium hydroxide
Write the ionic equation for this reaction.
Al3+ + 3OH- → Al(OH)3
(6) A technician found some colourless crystals of a substance left, unlabelled, in a
beaker in a laboratory.
She knew the substance was one of potassium sulfate, potassium iodide, sodium
sulfate or sodium iodide.
Explain how, using chemical tests, the technician could find out if the substance
left in the beaker was potassium sulfate, potassium iodide, sodium sulfate or
sodium iodide.
You may include equations in your answer.
test for cation
• flame test
• if the flame is yellow/not lilac, sodium ions are present
• if the flame is lilac/not yellow, potassium ions are
present
test for iodide ions
• make a solution of the crystals in water
• add dilute nitric acid
• add silver nitrate solution
• if there is a yellow precipitate, iodide ions are present
• if there is no precipitate, sulfate ions are present
• Ag+ + I¯ → AgI
OR
• make a solution of the crystals in water
• add chlorine water
• then cyclohexane
• if the cyclohexane/top layer turns purple, iodide ions
were present
• if there is no colour change, sulfate ions are present
• Cl2 + 2I-
→ 2Cl-
+ I2
test for sulfate ions
• make a solution of the crystals in water
• add dilute {hydrochloric/nitric} acid
• add barium {chloride/nitrate} solution
• if there is a white precipitate, sulfate ions are present
• if there is no precipitate, iodide ions are present
• Ba2+ + SO4
2- → BaSO4
(2) Solid A is potassium iodide.
A small amount of solid A is dissolved in water to form a solution.
Describe the test to show that the solution of A contains iodide ions.
add dil nitric acid then silver
nitrate solution (1)
yellow precipitate/solid (1)
(3) Solid B is ammonium chloride.
Describe the test to show that solid B contains ammonium ions.
add to sodium hydroxide (solution) and warm (1) test gases with (moist ) (red) litmus paper (1) (litmus paper) turns blue (1)
(1) When sodium hydroxide solution is added to the solution of C, a red-brown
precipitate is formed.
This test shows that the ion present in solid C is
iron(III), Fe3+
(3) Sodium hydroxide solution can be used to test for aluminium ions and for calcium
ions in solution.
Describe the results of these tests for aluminium ions and for calcium ions,
explaining how the results distinguish between the two ions.
white {precipitate/solid} with calcium (ions) (1) white {precipitate/solid} with aluminium (ions) (1) (precipitate/solid) dissolves in excess for aluminium ions / (precipitate/solid) remains in excess for calcium ions (1)
(1) A solution of one salt was made and some dilute nitric acid was added.
Drops of silver nitrate solution were added.
A yellow precipitate formed.
This test shows the anion in the salt is
iodide, I-
(2) The technician wanted to find out which bottle contained the sodium salt and
which bottle contained the potassium salt.
Explain how the technician should do this.
flame test / description of flame test mentioning in flame (1) sodium gives a yellow flame (1) potassium gives a {lilac/purple/violet} flame (1)
(2) When sodium hydroxide solution is mixed with a solution containing copper ions, Cu2+,
copper hydroxide, Cu(OH)2
, is formed.
Describe what you would see when these solutions are mixed.
blue (1)
precipitate / solid (1)
(3)When sodium hydroxide solution is mixed with a solution containing copper ions, Cu2+,
copper hydroxide, Cu(OH)2
, is formed.
Write the ionic equation for this reaction.
Cu2+ + 2OH- → Cu(OH)2 (3)