qualitative analysis

colour, smell, litmus paper test and test for hydrogen gas

colourless, odourless, litmus paper does not change colour, place a lighted splint near the mouth of the test tube containing the solution, if the gas evolved extinguishes the lighted splint with a ‘pop’ sound, the gas evolved is hydrogen gas

colour, smell, litmus paper test and test for oxygen

colourless, odourless, litmus paper does not change colour, place a glowing splint in the test tube containing the solution, if the gas evolved relights the glowing splint, the gas evolved is oxygen

colour, smell, litmus paper test and test for carbon dioxide gas

colourless, odourless, turns moist blue litmus paper red, bubble the gas evolved from the solution in calcium hydroxide, if the gas evolved forms a white precipitate of calcium carbonate, the gas evolved is carbon dioxide gas

colour, smell and test for chlorine gas

greenish-yellow, swimming pool like smell, place a moist blue litmus paper near the mouth of the test tube containing the solution, if the gas evolved turns the moist blue litmus paper red then bleaches it, the gas evolved is chlorine gas

colour, smell, litmus paper test and test for sulfur dioxide gas

colourless, burning sulfur smell, turns moist blue litmus paper red, bubble gas through a solution of acidified kMnO4 as kMnO4 is an oxidising agent, if gas evolved turns purple kMnO4 colourless, gas evolved is sulfur dioxide gas as it is a reducing agent that turns purple MnO4^- to colourles Mn^ 2+

colour, smell and test for ammonia gas

colourless, urine like smell, place a moist red litmus paper near the mouth of the test tube containing the solution, if the gas evolved turns the moist red litmus paper blue, the gas evolved is ammonia gas

how to identify cation Ca^ 2+

reaction with NaOH: white precipitate is formed and is insoluble in excess NaOH, hence the white precipitate is Ca(OH)2 => Ca^2+ is present, ionic equation: Ca^2+ (aq) + 2OH^- (aq) → Ca(OH)2 (s) reaction with NH3: no precipitate formed, therefore Ca^2+ is present and there is no ionic equation

how to identify cation Zn^2+

reaction with NaOH: white precipitate is formed and is soluble in excess NaOH to form colourless solution, reaction with NH3: white precipitate is formed as is insoluble in excess NH3 to form colourless solution, hence the white precipitate formed is Zn(OH)2 => Zn^2+ is present, ionic equation: Zn^2+ (aq) + 2OH^- (aq) → Zn(OH)2 (s)

how to identify cation Al^3+

reaction with NaOH: white precipitate is formed and is soluble in excess NaOH to form colourless solution, reaction with NH3: white precipitate id formed and is insoluble in excess NH3, hence the white precipitate is Al(OH)3, ionic equation: Al^3+ (aq) + 3OH^- (aq) → Al(OH)3 (s)

how to identify cation Cu^2+

reaction with NaOH: light blue precipitate formed, insoluble in excess NaOH, reaction with NH3: blue precipitate formed, soluble in excess NH3 to form dark blue solution, hence blue precipitate formed is Cu(OH)2 => Cu^2+ is present, ionic equation: Cu^2+ (aq) + 2OH^- (aq) → Cu(OH)2 (s)

how to identify cation Fe^2+

reaction with NaOH: green precipitate formed, insoluble in excess NaOH, green precipitate turns brown on contact with air, reaction with NH3: green precipitate formed, insoluble in excess NH3, green precipitate turns brown on contact with air, hence green precipitate formed is Fe(OH)2 => Fe^2+ is pesent, green Fe(OH)2 oxidises to brown Fe(OH)3, ionic equation: Fe^2+ (aq) + 2OH^- (aq) → Fe(OH)2 (s), Fe(OH)2 (s) → Fe(OH)3 (s)

how to identify cation Fe^3+

reaction with NaOH: reddish-brown precipitate formed, insoluble in excess NaOH, reaction with NH3: reddish-brown precipitate formed, insoluble in excess NH3, hence the reddish-brown precipitate formed is Fe(OH)3 => Fe^3+ is present, ionic equation: Fe^3+ (aq) + 3OH^- (aq) → Fe(OH)3 (s)

how to identify cation NH4+ (NaOH only)

reaction with NaOH: no precipitate formed, on heating, gas evolved turns moist red litmus paper blue, hence the gas evolved is ammonia gas, NH3 => NH4+ is present, ionic equation: NH4+ (aq) + OH^- ( aq) → NH3 (g) + H2O (l)

how to identify anion CO3^2-

add dilute HNO3 to the unknown, bubble the gas evolved through a solution of calcium hyrdoxide, if gas evolved forms a white precipitate of calcium carbonate from calcium hydroxide, the gas evolved is carbon dioxide => CO3^2- is present, cO3^2- + 2HNO3 → 2NO3^- + H2O + CO2, CO2 + Ca(OH)2 → CaCO3 + H2O

how to identify anion NO3^-

add NaOH (aq), Al powder or Al foil to the unknown then warm, test the gas evolved using moist red litmus paper, if the gas evolved turns the moist red litmus paper blue, the gas evolved is ammonia gas, NH3 → NO3^- is present, NO3^- is reduced to NH3

how to identify anion Cl^-

acidify (react away any soluble metal carbonate impurities in the unknown) the unknown with dilute nitric acid and add lead (II) nitrate, Pb(NO3)2 or silver nitrate, AgNO3 solution, a white precipitate will be formed, if Pb(NO3)2 is used, the white precipitate formed is PbCl2 => Cl^- is present, if AgNO3 is used, the white precipitate formed is AgCl => Cl^- is present, Pb^2+ + Cl^- → PbCl2, Ag^+ + Cl^2- → AgCl

how to identify anion I^-

acidify (react away any soluble metal carbonate impurities in the unknown) the unknown with dilute nitric acid and add lead (II) nitrate, Pb(NO3)2 or silver nitrate, AgNO3 solution, a yellow precipitate will be formed, if Pb(NO3)2 is used, the yellow precipitate formed is PbI2 => I^- is present, if AgNO3 is used, the yellow precipitate formed is AgI => I^- is present, Pb^2+ + I^- → PbI2, Ag^+ + I^- → AgI

how to identify anion SO4^2-

acidify (react away any soluble metal carbonate impurities in the unknown) the unknown with dilute nitric acid and add barium chloride, BaCl2 or barium nitrate, Ba(NO3)2 solution, a white precipitate will be formed, the white precipitate formed is barium sulfate => SO4^2- is present, Ba^2+ + SO4^2- → BaSO4