Qualitative analysis of Group I cations

In this post we will learn how to detect I^st group cations in a given mixture. In the last post, you have learnt solution preparation which is the most important step for the successful qualitative analysis.

Before we proceed for the 1^st group, we must check the presence of ammonium ion NH₄⁺ in the given mixture. Although it is 5^th group cation but it is tested first because during successive analysis a number of reagents are added some of which may contain ammonium ions and so considerable amount of ammonium ions may build up in the test solution when we reach up to 5^th group. So, it will be wise to test ammonium ion in the beginning.

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^- ⟶ NH_3(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^- ⟶ PbCl_2(s) ↓

           Hg₂²⁺ + 2Cl^- ⟶ Hg₂Cl_2(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.

            PbCl_2(s) ↓ + 2Cl^-  ⟶ [PbCl₄]^2-

Filter the precipitate and keep filtrate for 2^nd 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 1^st 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⁺ + CrO₄^2- ⟶ 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⁺ + SO₄^2- ⟶ 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- + SO₄^2-

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

Spot tests are just preliminary tests if you get positive test for any cation always do confirmatory test. In the next post we will discuss identifying tests for 2^nd group cations in the filtrate you got after the removal of precipitate of 1^st group cations. 

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