Reactions of s- and p-Block Cations
Na K NH4 Ca Ba Al
Most of salts of alkali metal ions are well soluble in water. The best tests on Na and K presence are based on flame emission: atoms of Na change color of the flame to yellow while K atoms make flame light violet. Ca atoms are making flame red and Ba atoms make it green. Therefore, to determine which of those elements are present in your solution you will need a spectrophotometer. There are several simple tests for Ba and Ca ions which can be performed if you have no access to a flame spectral instrument.

Na Sodium atoms emit light at 589 nm coloring the flame in yellow.

K Potassium atoms emit light at 404 nm coloring the flame in violet. Much stronger emission is at 767 nm , and it can be easily detected by spectrophotometer. We cannnot see this line because it is out of the spectral range of human eyes.

NH4+ There is no good spectral test on ammonia. Nevertheless, it is very easy to detect ammonium in solution. Ammonium is the only ion that forms gas when the strong base is added to a solution:
NH4+ + OH- ® NH3 (­ ) + H2O
Procedure: Place 0.5-1 mL of solution in a test tube. Add an equal ammount of NaOH solution. Place a wet indicator paper above the test tube (just do not touch the tube itsel!). If the indicator changes color to blue, ammonia is present. Its odor is also frequently detected.

Ca. Calcium compounds color flame in a bright red. It can also be detected by flame spectrophotometer.
Chemical test on Ca. Ca2+ ions form an insoluble oxalate when treated with sodium oxalate:
Ca2+ + C2O42- ® Ca C2O4 (white precipitate ¯ )

Ba. Barium ions form an insoluble chromate when they react with K2Cr2O7.

2Ba2+ + Cr2O72- (orange) +2CH3COO- + H2O ®
                 ® 2 BaCrO4 (yellow ¯ ) + 2 CH3COOH

Procedure: To several drops of solution, add an equal amount of potassium dichromate solution and 1-2 drops of sodium acetate solution. Formation of the yellow precipitate shows the presence of barium.

 Al Aluminum. A strong base added to an aqueous solution containing Al(III) precipitates it as gelatinous colorless hydroxide. Al(III), being amphoteric, redissolves in excess base, forming aluminate [Al(OH)4]-:
Al3+ + 3 OH- ®Al(OH)3
Al(OH)3 + OH- ® [Al(OH)4]-
Using this approach, Al can be separated from other metals forming hydroxides: they are simply filtered out from the alkaline solution.
An addition of ammonium chloride re-precipitates aluminum hydroxide:

[Al(OH)4]- + NH4+ ® Al(OH)3 (white gelatinous ¯ ) + NH3 + H2O

No other hydroxides will precipitate at these conditions.

Procedure: To 3-5drops of the unknown solution, add 1 mL of NaOH solution. If any precipitate is formed, filter it out. To filtrate, add solid ammonium chloride (about ¼ by volume) and wait until it will dissolve. A white gelatinous precipitate is formed when Al(III) is present.

Report the cations you have detected in an unknown solution.