O Level Chemistry Qualitative Analysis


Test for anions


  1. Carbonate CO32-
Test with dilute acid Test result
Add dilute acid to the unknown solution. Observe for effervescence. Test for carbon dioxide if effervescence occurs. Effervescence (fizzling or bubbling) occurs.
White ppt (precipitate) appears when gas is bubbled through limewater => Gas produced is carbon dioxide.
Ionic equation: CO32- + H+ --> H2O + CO2 (neutralization)
Note: For the test for carbon dioxide, scroll down to refer to Tests for gases.
FYI: Adding barium nitrate Ba(NO3)2(aq) solution to a carbonate solution will give a white ppt.
Ba2+(aq) + CO32-(aq) --> BaCO3(s) (precipitation)
FYI: Adding silver nitrate AgNO3(aq) solution to a carbonate solution gives a white ppt.
2Ag+(aq) + CO32-(aq) --> Ag2CO3(s) (precipitation)
  1. Chloride Cl-
Test with AgNO3(aq) Test result
Acidify the unknown solution with dilute nitric acid to get rid of any carbonate impurity in the solution. Then add aqueous silver nitrate AgNO3(aq). Observe for any ppt. White ppt appears.


Ppt darkens on standing.
Ionic equation: Ag+(aq) + Cl-(aq) --> AgCl(s) (precipitation)
Comment: Halides (e.g. NaCl and CuCl2) are ionic compounds. Ionic compounds are soluble in water. Silver chloride AgCl, however, is insoluble (or sparingly soluble). Only 0.0019 g of silver chloride can dissolve in 1 dm3 of water. This is because AgCl has more covalent character than ionic character.
Why does AgCl darken? On exposure to light, silver ions are reduced to silver atoms: This is used in photography.
Ag+ + e- --> Ag (reduction)
FYI: Adding lead nitrate Pb(NO3)2(aq) to a chloride solution gives a white ppt.
Pb2+(aq) + 2Cl-(aq) --> PbCl2(s) (precipitation)
Note: Nitric acid is an irritant and silver chloride can stain skin and clothes.


  1. Iodide I-
Test with AgNO3(aq) Test result
Acidify the unknown solution with dilute nitric acid to get rid of any carbonate impurity in the solution. Then add aqueous silver nitrate. Observe for any ppt. Yellow ppt appears.


Ppt darkens on standing.
Ionic equation: Ag+(aq) + I-(aq) --> AgI(s) (precipitation)
FYI: Adding lead nitrate Pb(NO3)2(aq) to an iodide solution gives a yellow ppt.
Pb2+(aq) + 2I-(aq) --> PbI2(s) (precipitation)
  1. Nitrate NO3-
Because all nitrates are soluble, there is no precipitation test for nitrate.
Test with NaOH and Al foil Test result
Add aqueous NaOH(aq) and then Al foil (or Devarda's alloy which contains aluminium, copper). Warm carefully. Test any gas using moist red litmus paper. Characteristic smell of ammonia. Gas turns moist red litmus paper blue.
Ionic equation: 3NO3-(aq) + Al(s) + 5OH-(aq) + 18H2O --> 3NH3(g) + 8[Al(OH)4]-(aq) (redox)
Comment: Aluminium is a reducing agent in the reaction. The oxidation state of aluminium increases from 0 in Al(s) to +3 in [Al(OH)4]-. The oxidation state of nitrogen decreases from +5 in NO3- to -3 in NH3.


  1. Sulfate SO42-
Test with BaNO3(aq) Test result
Acidify the known solution with dilute nitric acid to get rid of any carbonate impurity and then add barium nitrate Ba(NO3)2(aq). Observe for any ppt formed. White ppt appears.
Ionic equation: SO42-(aq) + Ba2+(aq) --> BaSO4(s) (precipitation)
FYI: Adding lead nitrate Pb(NO3)2(aq) to an unknown solution containing SO42-(aq) can also produce a white ppt. Pb2+(aq) + SO42-(aq) --> PbSO4(s) (precipitation)
FYI: Adding silver nitrate AgNO3(aq) to a sulfate solution may give a white ppt.
Ag+(aq) + SO42-(aq) --> Ag2SO4(s) (precipitation)
Test for cations


When the pH of a salt solution is increased by adding an alkali (NaOH(aq) (a strong alkali) or NH3(aq) (a weak alkali)), an insoluble compound is formed as a ppt. These ppt are metal hydroxides M(OH)x(s), especially the hydroxides of transition metals. Because the hydroxides of many transition metals have distinct color, they can be used to tell the identity of metal ions present in unknown solutions.


FYI: Hydroxides of Group I metals (e.g. NaOH, KOH) are soluble. They are colorless alkaline solutions. Hydroxides of Group II metals (e.g. Ca(OH)2, Mg(OH)2) are slightly soluble.


Some of these insoluble hydroxides (e.g. Al(OH)3, Pb(OH)2, Zn(OH)2) are amphoteric, meaning they can react with either acids or alkalis. That is why excess NaOH(aq) is added to see if they dissolve.


Amphoteric hydroxides (e.g. Al(OH)3, Pb(OH)2, Zn(OH)2) redissolve in excess NaOH(aq)... 


Aqueous ammonia also provides OH-(aq) to increase the pH of unknown solution. Ammonia, being a weak alkali, partially dissociates to produce OH-(aq): H2O(l) + NH3(aq) <==> NH4+(aq) + OH-(aq). Aqueous ammonia also provides ammonia molecules. In excess NH3(aq), the ammonia molecules bond to the metal ions in the insoluble hydroxide to form charged complex ions. This makes some of the insoluble hydroxides to dissolve in excess aqueous ammonia.


Excess NH3 molecules bond with some metal ions (Cu2+, Zn2+) in insoluble hydroxides to form soluble complexes... 


Test: Divide the unknown solution into different test tubes. Add either alkali to each portion of the unknown solution. Observe for ppt. Add excess and observe if precipitate dissolves.


Note: Sodium hydroxide is an irritant and ammonia can irritate the eyes and lungs.


  1. Aluminium Al3+(aq) (Group III element)
NaOH(aq) NH3(aq)
Test result White ppt appears.
Ppt dissolves in excess NaOH(aq). 		White ppt appears.
Ppt does not dissolve in excess NH3(aq).
Ionic equation(s) Al3+(aq) + 3OH-(aq) --> Al(OH)3(s)


The ppt Al(OH)3(s) is amphoteric. In excess NaOH, Al(OH)3 reacts to form a complex.
Al(OH)3(s) + OH-(aq) --> Al(OH)4-(aq). 		Al3+(aq) + 3NH3(aq)+ 3H2O(aq) --> Al(OH)3(s) + 3NH4+(aq)



The ppt Al(OH)3(s) is amphoteric but does not dissolve in excess NH3(aq) which is a weak alkali. There is not enough OH-(aq) to react with Al(OH)3(s). 


  1. Ammonium NH4+(aq) (a polyatomic cation)
Test result with NaOH(aq): No ppt produced. On warming, a gas with a characteristic smell is produced. The gas turns moist red litmus paper blue.
Ionic equation: NH4+(aq) + OH-(aq) <==> H2O(l) + NH3(aq)
Note: For the test on NH3(g), refer to Tests for gases.


  1. Calcium Ca2+(aq) (Group II element)
NaOH(aq) NH3(aq)
Test result White ppt is formed.
Ppt is insoluble in excess alkali. 		No ppt is observed.
Ionic equation(s) Ca2+(aq) + 2OH-(aq) --> Ca(OH)2(s) --
Comment The hydroxides of Group II elements are slightly soluble. For example, only 1 g of Ca(OH)2 can dissolve in 1 dm3 of water. NH3(aq) partially dissociates into NH4+(aq) and OH-(aq). Any Ca(OH)2 formed will be in very small amount. Ca(OH)2 is slightly soluble. A small amount of it would be able to dissolve in water.
Note: The calcium hydroxide Ca(OH)2(aq) solution is our familiar limewater.
FYI: Adding ammonium oxalate (NH4)2C2O4(aq) to a calcium solution gives a white ppt.
Ca2+(aq) + C2O42-(aq) --> CaC2O4(s)


  1. Copper Cu2+(aq) (transition metal)
NaOH(aq) NH3(aq)
Test result Light blue / turquoise ppt appears.
Ppt does not dissolve in excess. 		Light blue ppt appears.
Ppt dissolves in excess to give a dark blue solution.
Ionic equation(s) Cu2+(aq) + OH-(aq) --> Cu(OH)2(s) Cu2+(aq) + OH-(aq) --> Cu(OH)2(s)


Cu(OH)2(s) + 4NH3(aq) --> [Cu(NH3)4]2+(aq) +2OH-(aq)
Comment

The blue ppt dissolves as the copper ion forms a complex ion with ammonia molecules.


  1. Iron(II) Fe2+(aq) (transition metal)
NaOH(aq) NH3(aq)
Test result Green ppt appears. Precipitate is insoluble in excess NaOH(aq).


Ppt darkens on standing. 		Green ppt appears. Precipitate is insoluble in excess NH3(aq).


Ppt, especially that is near the surface, turns reddish-brown on standing.
Ionic equation(s) Fe2+(aq) + 2OH-(aq) --> Fe(OH)2(s) Fe2+(aq) + 2OH-(aq) --> Fe(OH)2(s)


Ammonia molecules does not bond with the metal ion to form complex. 


Fe(OH)2(s) is then oxidized to reddish-brown Fe(OH)3(s) by atmospheric oxygen.


  1. Iron(III) Fe3+(aq) (transition metal)
NaOH(aq) NH3(aq)
Test result Red-brown ppt appears.
Ppt is insoluble in excess NaOH(aq). 		Red-brown ppt appears.
Ppt is insoluble in excess NH3(aq).
Ionic equation(s) Fe3+(aq) + 3OH-(aq) --> Fe(OH)3(s) Fe3+(aq) + 3OH-(aq) --> Fe(OH)3(s)


Ammonia molecules does not bond with the metal ion to form a soluble complex.
FYI: Another test for Fe3+(aq) is by adding a thiocyanide solution SCN-. A blood-red complex will be formed.


  1. Lead(II) Pb2+(aq) (transition metal)
NaOH(aq) NH3(aq)
Test result White ppt appears.
Ppt dissolves in excess giving a colorless solution. 		White ppt appears.
Ppt does not dissolve in excess.
Ionic equation(s) Pb2+(aq) + 2OH-(aq) --> Pb(OH)2(s)
Pb(OH)2(s) is amphoteric. In excess NaOH(aq), Pb(OH)2(s) dissolves to form a soluble complex ion.


Pb(OH)2(s) + 2OH-(aq) --> [Pb(OH)4]2-(aq) 		Pb2+(aq) + 2OH-(aq) --> Pb(OH)2(s)
The ppt Pb(OH)2(s) does not dissolve in NH3(aq) which is a weaker alkali than NaOH(aq). There is not enough OH-(aq) in NH3(aq) to react with Pb(OH)2(s).


Ammonia molecules does not bond with the metal ion to form complex.
Note: Notice that Pb2+(aq) gives the same result with NaOH(aq) and with NH3(aq) as Al3+(aq). To distinguish between Pb2+(aq) and Al3+(aq), we use a reagent such as HCl(aq) containing Cl-(aq). PbCl2(s) will be precipitated out, while AlCl3(aq) is soluble.
FYI: Adding a carbonate solution or sulfate solution to a Pb2+(aq) solution also give a white ppt.
Pb2+(aq) + CO32-(aq) --> PbCO3(s)
Pb2+(aq) + SO42-(aq) --> PbSO4(s)


  1. Zinc Zn2+(aq) (transition metal)
NaOH(aq) NH3(aq)
Test result White ppt appears.
Ppt dissolves in excess to give a colorless solution. 		White ppt appears.
Ppt dissolves in excess to give a colorless solution.
Ionic equation(s) Zn2+(aq) + 2OH-(aq) --> Zn(OH)2(s)
Zn(OH)2(s) is amphoteric. In excess NaOH(aq), it forms a soluble complex ion.


Zn(OH)2(s) + 2OH-(aq) --> [Zn(OH)4]2-(aq) 		Zn2+(aq) + 2OH-(aq) --> Zn(OH)2(s)
Zn(OH)2(s) dissolves in excess NH3(aq) to form a soluble, colorless complex.
Zn(OH)2(s) + 4NH3(aq) --> Zn(NH3)4]2+(aq) + 2OH-(aq)
Note: Most of the transition metals such as iron and copper forms colored solutions. If you observe a colorless solution, you can deduce that the solution may not contain a transition metal.




Test for Gases


  1. Ammonia NH3(g)
Test Test result
Introduce damp red litmus paper. Red litmus paper turns blue.


Reason: NH3(g) is an alkaline gas, dissolving in water to form a weak alkali.
NH3(g) + H2O(l) <==> NH4OH(aq)
Note: Aqueous ammonia is preferably written as NH3(aq), not NH4OH.


  1. Carbon dioxide CO2(g)
Test Test result
Pass gas through or shake gas with limewater. White ppt appears.
The ppt dissolves when CO2(g) is being blown through continuously.
Equations: CO2(g) + Ca(OH)2(aq) --> CaCO3(s) + H2O(l) (neutralization)
CaCO3(s) + CO2(g) + H2O(l) <==> Ca(HCO3)2(aq)
Explanation: Carbon dioxide, an acidic gas, undergoes an acid-base reaction with limewater (calcium hydroxide) to form calcium carbonate which is the white ppt and water. In excess carbon dioxide, a soluble bicarbonate is formed.


  1. Chlorine Cl2(g)
Test Test result
Introduce damp litmus paper. Litmus paper is bleached.
Equation: Cl2(g) + H2O(l) <==> HClO(aq) + HCl(aq) (disproportionation)
Explanation: The reaction is called disproportionation because the chlorine Cl2 (oxidation state = 0) is both oxidized to HClO (oxidation state = +1) and reduced to HCl (oxidation state = -1).
The HCl(aq) formed turns blue litmus red, while the hypochlorous acid HClO(aq) attacks the molecule of the litmus and bleaches litmus.
Comment: This reaction has been used in water treatment plants to kill bacteria in water.


  1. Hydrogen H2(g)
Test Test result
Introduce a lighted splint. The flame extinguishes with a “pop” sound.
Chemical equation: H2(g) + O2(g) --> H2O(l) (combustion)
Explanation: Hydrogen is a flammable gas and readily burns in air to give water vapor.
Comment: This reaction is used to launch rockets into outer space.


  1. Oxygen O2(g)
Test Test result
Introduce a glowing splint. The glow rekindles into a flame.
Reason: Oxygen supports combustion.


  1. Sulfur dioxide SO2(g)
Test Test result
Pass gas through acidified potassium dichromate(VI). Orange potassium dichromate(VI) turns green.
Ionic equation: Cr2O72- + 2H+ + 3SO2 --> 2Cr3+ + H2O + 3SO42- (redox)
Explanation: SO2(g) is a strong reducing agent while Cr2O72-(aq) is a strong oxidizing agent. Orange Cr2O72-(aq) is reduced to green Cr3+(aq).
Constructing the equation: From SO2 --> SO42-
Add 2H2O on the LHS to balance the oxygen atoms: SO2 + 2H2O --> SO42-
Add 4H+ on the RHS to balance the hydrogen atoms: SO2 + 2H2O --> SO42- + 4H+
There are -2 + 4(+1) = +2 charges on the RHS.
Add 2e on the RHS to balance the +2: SO2 + 2H2O --> SO42- + 4H+ + 2e –(1)
From Cr2O72- --> 2Cr3+
Add 7H2O on the RHS to balance the oxygen atoms: Cr2O72- --> 2Cr3+ + 7H2O
Add 14H+ on the LHS to balance the hydrogen atoms: Cr2O72- + 14H+ --> 2Cr3+ + 7H2O
There are -2 + 14(+1) = +12 on the RHS and 2(+3) = +6 on the LHS.
Add 6e on the LHS to balance the changes: Cr2O72- + 14H+ + 6e --> 2Cr3+ + 7H2O –(2)
To eliminate the electrons, we multiply (1) x 3 before we add to (2):
3SO2 + 6H2O --> 3SO42- + 12H+ + 6e –(3)
(2) + (3): Cr2O72- + 14H+ + 6e + 3SO2 + 6H2O --> 2Cr3+ + 7H2O + 3SO42- + 12H+ + 6e


1 comment:

  1. qwertyuiop5 April 2021 at 07:07

    ooo! tysm for this! :) <3 - student taking Os this yr

    ReplyDelete

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