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Grade: 12th pass
Which is the group reagent used for precipitation of elements of group 1and group 2 in qualitative analysis  ?
one year ago

Answers : (1)

23357 Points
Test for ammonium ion NH4+: take about 1ml of original solution in a boiling test tube and add excess of sodium hydroxide NaOH and boil it gently. Ammonia gas is evolved on warming. Which can be identified by its odour, smell it after removing the test tube from the flame.
           NH4+ + OH- ⟶ NH3(g) ↑ + H2O
If you place red litmus paper at the mouth of test tube, ammonia turns it blue. As it is a Lewis base.
If you bring a glass rod moistened with concentrated hydrochloric acid HCl over the vapours of ammonia, white fumes of ammonium chloride are formed.
           NH4+ + Cl⟶ NH4Cl(g) ↑
With Nessler’s reagent you can perform confirmatory test for ammonium ion. Nessler’s reagent is the alkaline solution of potassium tetraiodomercurate (II) K2HgI4. For this test take a drop of original solution and mix it with a drop of NaOH on a watch glass. Take a drop of this mixture in a separate watch glass and add a drop of Nessler’s reagent. A brown precipitate or yellow or orange-red colouration is produced according to the amount of ammonia of ammonium ions present. This precipitate is a basic mercury(II) amido-iodine.
           NH4+ + 2[HgI4]2- + 4OH-  ⟶ HgO.Hg(NH2)I(s) ↓ + 7I- + 3H2O
Analysis of Group I: After detecting the presence of ammonium ion you can proceed for the group analysis. Take 15-20ml original solution in a boiling test tube or flask add an excess of dilute HCl. White precipitate indicates the presence of chlorides of Pb2+, Hg22+ or Ag+.
            Pb2+ + Cl- ⟶ PbCl2(s) ↓
           Hg22+ + 2Cl- ⟶ Hg2Cl2(s) ↓
            Ag+ + Cl- ⟶ AgCl(s) ↓
But avoid large excess of HCl because lead chloride is soluble in concentrated HCl due to formation of tetrachloroplumbate(II) ion [PbCl4]2- is formed.
            PbCl2(s) ↓ + 2Cl-  ⟶ [PbCl4]2-
Filter the precipitate and keep filtrate for 2nd group analysis. Wash the precipitate with 2ml of 2M HCl, and then wash it 2-3 times with 1ml cold water (because PbCl2 is soluble in hot water) and reject washings.
Strategy to divide cations of group I: Now you have precipitate of 1st group which may contain chlorides of Pb2+, Hg22+ or Ag+ or all. How will you identify them separately? Among them PbCl2 is soluble in hot water but separates again in long needle like crystals on cooling so you can separate it by boiling.
Now transfer the precipitate in boiling test tube and boil with 5-10ml water and filter hot. Thus you can separate PbCl2 from Hg2Cl2 and AgCl. Now you have filtrate and precipitate, in filtrate we will test for Pb2+ and test for Hg+ and Ag+ in precipitate.
Test for Pb(II) ion: Cool the filtrate, long needle like crystals of PbCl2 is obtained if Pb2+ is present in any quantity. For confirmatory tests divide the filtrate in three parts.
Part 1: add 0.1M potassium chromate K2CrO4 solution. Yellow precipitate of lead chromate is obtained which is insoluble in dilute acetic acid.
            Pb2+ + CrO42- ⟶ PbCrO4 ↓
Part 2: Add 0.1M potassium iodide KI solution. Yellow precipitate of lead iodide PbI2 is formed which is soluble in boiling water and deposits golden yellow plates upon cooling.
            Pb2+ + I⟶ PbI2 ↓
            An excess of KI dissolves the precipitate due to formation of tetraplumbate(II) ion.
            PbI2 ↓ + 2I- ⟶ [PbI4]2-
            On diluting with water the precipitate of PbI2 reappears.
Part 3: Add dilute sulphuric acid H2SO4. White precipitate of lead sulphate PbSO4 is obtained.
            Pb2+ + SO42- ⟶ PbSO4 ↓
Hot and concentrated sulphuric acid H2SO4 dissolves the precipitate by the formation of lead hydrogen sulphate Pb(HSO4)2.
            PbSO4 ↓ + H2SO4 ⟶ Pb2+ + 2HSO4-
Lead sulphate PbSO4 is soluble in concentrated solution of ammonium acetate CH3COONH4. On dissolution tetraacetoplumbate(II) is formed.
            PbSO4 ↓ + 4CH3COO- ⟶ [Pb(CH3COO)4]2- + SO42-
How to separate Hg2Cl2 and AgCl: they can be separated by using ammonia solution. On reacting with ammonia, mercury(I) chloride forms insoluble complex while silver chloride forms a soluble complex.
Wash the residue (Hg2Cl2 and AgCl) 3-4 times with boiling water for the complete removal of PbCl2. To ensure that add potassium chromate K2CrO4 solution to the washing, no precipitate indicates absence of Pb2+ ion. Now add 3-4ml hot dilute ammonia NH3 solution to the residue. If Black precipitate appears, it is due to the formation of complex of Hg+ and collect the filtrate which may contains Ag+ ion.
Test for Hg(I) ion : ammonia solution converts the Hg2Cl2 in to a mixture of mercury(II)amidochloride and mercury metal, they both are insoluble and give black precipitate.
            Hg2Cl2 + 2NH3 ⟶ Hg↓ + Hg(NH2)Cl ↓ + NH4+ + Cl-
Mercury(II)amidochloride Hg(NH2)Cl is a white coloured precipitate but finely divided mercury metal makes it shiny black.
Test for Ag(I) ion: Precipitate of AgCl dissolves on adding hot dilute ammonia solution due to formation of diammineargentate complex ion.
            AgCl ↓ + 2NH3  ⇌ [Ag(NH3)2]+ + Cl-
This solution should not be kept for long otherwise a precipitate of silver nitride Ag3N (fulminating silver) is formed which explodes readily even in wet conditions. (Before disposing, add dilute nitric acid HNO3 or hydrochloric acid which neutralises excess ammonia and prevents formation of Ag3N.) For the confirmatory test divide this solution in two parts.
Part 1: Acidify with dilute nitric acid HNO3, white precipitate of AgCl is obtained. When you add acid it neutralises excess ammonia and equilibrium shifts in backward direction and AgCl re-precipitates.
Part 2: Add potassium iodide KI solution, yellow precipitate of silver iodide is obtained.
            Ag+ + I- ⟶ AgI ↓
Now you have learnt systematic separation and identification of group I cations. You can identify them by spot test also.
Spot test for group I cations: Wash the precipitate of group I with cold water. Add ammonia NH3 solution. If precipitate :
Doesn’t change then Pb+ may present
Turns black then Hg22+ may present
Dissolves then Ag+ may present

one year ago
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