“Chemical rainbow” experiment
How to make a series of seven colored solutions
In this interesting experiment, we will examine a series of seven colored solutions, to make a “chemical rainbow”. By mixing different reagents, we will get red, orange, yellow, green, blue, indigo and violet colors.
Wear rubber gloves and protective glasses, to avoid burns to the skin and eyes. Conduct the experiment in a well-ventilated room or under a fume hood.
Warning! Don’t try to repeat this experiment at home. This experiment is supposed to be conducted only by a professional.
Reagents and equipment:
- iron(III) chloride (1 g);
- potassium rhodanide (1 g);
- potassium chromate (1 g);
- potassium dichromate (1 g);
- nickel(II) sulfate (1 g);
- sodium hydroxide (40 ml);
- copper(II) sulfate (3 g);
- 25% ammonia solution (20 ml);
- phenolphthalein (0.5 g);
- distilled water;
- 14 beakers of 150 ml.
Pour 50 ml of solutions into the beakers in succession: iron(III) chloride, potassium chromate, potassium dichromate, copper(II) sulfate, copper(II) sulfate, copper(II) sulfate and phenolphthalein. Then add a solution to each beaker: potassium rhodanide, sulfuric acid, sodium hydroxide, potassium rhodanide, sodium hydroxide, ammonia and sodium hydroxide. In each beaker chemical reactions take place which lead to the solution changing color. The result is a bright rainbow of all the colors.
In the first beaker, iron(III) rhodanide forms, which has a very bright rich red color:
FeCl₃ + 3KSCN = Fe(SCN)₃ (dark red) + 3KCl
In the second beaker, the potassium chromate solution in an acid medium changes to potassium dichromate, which has an orange color:
2K₂CrO₄ + H₂SO₄ = K₂Cr₂O₇ (orange) + K₂SO₄ + H₂O
In the third beaker, a reverse reaction takes place, the orange potassium dichromate in an alkaline medium changes back to potassium chromate, which gives a yellow color to the solution:
K₂Cr₂O₇ + 2NaOH = 2KNaCrO₄ (yellow) + H₂O
In the fourth beaker, a reaction takes place between potassium rhodanide and copper(II) sulfate, in which a complex mixture with a green color is formed:
CuSO₄ + 4KSCN = K₂[Cu(SCN)₄] (green) + K₂SO₄
In the fifth beaker, copper sulfate reacts with sodium hydroxide and a bright blue sediment of copper(II) hydroxide is formed:
CuSO₄ + 2NaOH =Cu(OH)₂ (blue) + 2Na₂SO₄
In the sixth beaker, with the interaction of ammonia solution and copper(II) sulfate, a complex compound with an indigo color is formed:
CuSO₄ + 4NH₃ = [Cu(NH₃)₄]SO₄ (indigo)
In the seventh beaker, the solution turns a violet-pink color, as sodium hydroxide has an alkaline medium, and phenolphthalein is an indicator which turns violet-pink in an alkaline medium.