“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.

Safe­ty pre­cau­tions

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.

Step-by-step in­struc­tions

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.

Pro­cess­es de­scrip­tion

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.

Try it