In the systematic separation of cations we perform successive separation of group cations with the use of group reagent. Group reagents react with corresponding group cations and convert them into insoluble salts like chlorides, sulphides, and carbonates. First, we prepare a solution of the given mixture, then we add group reagent of Ist group which converts the cations of Ist group into insoluble chlorides and separate them as precipitate and then we test for IInd group cations in the filtrate of Ist group.
You have learnt in previous post that group reagent of 2nd group is Hydrogen sulphide H2S (gas or saturated aqueous solution). Here you will see how cations get precipitated by common ion effect. Hydrogen sulphide is a weak acid, it dissociates partly. Hydrochloric acid, which we added initially, is a strong acid and dissociates completely.
H2S ⇌ 2H+ + S2-
HCl ⟶ H+ + Cl-
Here H+is a common ion among H2S and HCl. Due to complete dissociation of HCl concentration of H+ increases in the solution which shifts equilibrium of reaction 1 backwards. This results in precipitation of cations in the form of sulphides.
H2S ⟶ 2H+ + S2-
Take the filtrate in a boiling test tube and heat till it is nearly boiling and then pass H2S gas under pressure in excess (30 seconds -1 min). If 2nd group cations are present, you will get coloured precipitate of sulphides.
Black precipitate: Mercury(II) sulphide HgS, lead(II) sulphide PbS, copper(II) sulphide CuS.
Brown precipitate: Bismuth(III) sulphide Bi2S3, tin(II) sulphide SnS.
Yellow precipitate: Cadmium(II) sulphide CdS, arsenic(III) sulphide As2S3, arsenic(V) sulphide As2S5, tin(IV) sulphide SnS2.
Orange precipitate: Antimony(III) sulphide Sb2S3, antimony(V) sulphide Sb2S5.
Filter the precipitate and wash with dilute hydrochloric acid HCl. The precipitate may contain IIA or IIB or both cations. To differentiate them add an excess of (5ml) yellow ammonium polysulphide (NH4)2Sxsolution and heat to 50-60°C for 3-4 minutes with constant stirring. sulphides of sub group IIA (Cu sub group) are insoluble in (NH4)2Sxwhile sulphides of sub group IIB (As sub group) dissolve in it by forming thiosalts. First we will test IIA group in the precipitate and preserve the filtrate for the test of IIB group.
Wash the precipitate with small volume of dilute (1+100ml water) ammonium sulphide (NH4)2Ssolution then with 2% ammonium nitrate NH4NO3solution and reject all washings.
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| Scheme for the separation of IInd Group |
Among Cu sub group IIA, mercury(II) sulphide HgS is the least soluble sulphide or you can say its solubility product is lowest than others. It is insoluble in nitric acid and water.
Separation of Hg(II): Take the precipitate in a boiling test tube or beaker, add 5-10ml 2M nitric acid HNO3, and boil gently for 2-3minutes. Black precipitate of mercury(II) sulphide HgS is obtained. Sulphides of other cations go in to the solution by forming nitrates. Filter and wash the precipitate with a little water. Keep the filtrate to test other cations of IIA.
Confirmatory test for Hg(II): Dissolve the precipitate HgS in aqua regia. Mercury(II) chloride is formed which is soluble.
3HgS + 6HCl + 2HNO3 ⟶ 3HgCl2 + 3S + 2NO + 2H2O
Divide this solution into 3 parts.
Part 1: Add tin(II) chloride SnCl2 solution. white silky precipitate of Mercury(I) chloride is Hg2Cl2 is formed.
2Hg2+ + Sn2+ + 2Cl- ⟶ Hg2Cl2 ↓ + Sn4+
It is an example of oxidation- reduction reaction. Where Hg(II) gets reduced to Hg(I) and Sn(II) gets oxidised to Sn(IV). If more SnCl2 is added, white precipitate turns to black because of Hg(I) gets further reduced to Hg(0) metal..
Hg2Cl2↓ + Sn2+ ⟶ Hg ↓ + Sn4+ + 2Cl-
Part 2: Add sodium hydroxide NaOH solution in small amount, brownish-red precipitate will be obtained and on adding more NaOH, yellow precipitate of mercury(II) oxide HgO will be obtained.
Hg2+ + 2OH- ⟶ HgO ↓ + H2O
This reaction is the characteristic for mercury(II) ions. You can use it to differentiate Hg(II) from Hg(I).
Part 3: Add potassium iodide KI solution slowly, red precipitate of mercury(II) iodide is formed.
Hg2+ + 2I- ⟶ HgI2 ↓
On adding more KI, precipitate will get dissolved by the formation of colourless tetraiodomercurate(II) ion.
HgI2 ↓ + 2I- ⟶ [HgI4]2-
Take the filtrate, it may contain nitrates of other IIA group cations Pb(II), Bi(III), Cu(II), Cd(II). First we will separate lead Pb(II). Test a small portion of filtrate and add dilute sulphuric acid H2SO4 and alcohol. If white precipitate of lead sulphate PbSO4 is obtained (less soluble in presence of alcohol) then take the remaining filtrate and add 1M sulphuric acid H2SO4. Concentrate in fume cupboard until white fumes appear by the decomposition of sulphuric acid. Cool it and add 10ml water, stir and allow to settle. White precipitate of lead sulphate PbSO4 will be obtained. Filter the precipitate and keep the filtrate to test other cations.
Pb2+ + SO42- ⟶ PbSO4 ↓
Confirmatory test for Pb(II): To the precipitate of lead sulphate PbSO4 add 2ml of 6M ammonium acetate CH3COONH4 solution, precipitate will be dissolved by the formation of tetraacetoplumbate(II) ion.
PbSO4 ↓ + 4CH3COO- ⟶ [Pb(CH3COO)4]2- + SO42-
Add few drops of 2M acetic acid and then 0.1M potassium chromate K2CrO4solution, yellow precipitate of lead chromate PbCrO4 will be obtained.
Pb2+ + CrO42- ⟶ PbCrO4 ↓
Filtrate may contain nitrates of sulphates of Bi(III), Cu(II) and Cd(II). Add concentrated ammonia solution in excess. All of them form salts but only the salt of bismuth is insoluble in excess of ammonia. You will get white precipitate of basic salt of bismuth.
Bi3+ + NO3- + 2NH3 + 2H2O ⟶ Bi(OH)2NO3 ↓ + 2NH4+
Copper forms basic copper sulphate salt.
2Cu2+ + SO42- + 2NH3 + 2H2O ⟶ Cu(OH)2.CuSO4 ↓ + 2NH4+
This basic copper sulphate salt is soluble in excess of ammonia and a deep blue colouration is obtained by the formation of tetramminocuprate(II) complex ion.
Cu(OH)2.CuSO4 ↓ + 8NH3 ⟶ 2[Cu(NH3)4]2+ + SO42- + 2OH-
Cadmium forms cadmium(II) hydroxide in ammonia solution.
Cd2+ + 2NH3 + 2H2O ⇌Cd(OH)2 ↓ + 2NH4+
Cd(OH)2 ↓ + 4NH3 ⟶ [Cd(NH3)4]2++ 2OH-
You have seen that only the salt of bismuth is insoluble in excess of reagent. Filter the precipitate and test it for Bi(III) and keep the filtrate to test for remaining cations.
Confirmatory test for Bi(III): Dissolve the precipitate in a little volume of dilute hydrochloric acid HCl and pour into freshly prepared cold sodium tetrahydroxostannate(II) (2ml of 0.25M tin(II) chloride and 2ml of 2M sodium hydroxide). Black precipitate of bismuth metal will be obtained.
Bi3+ + 3OH- ⟶ Bi(OH)3 ↓
Sodium hydroxide present in the reagent first reacts with Bi(III) and then tetrahydroxostannate(II) ion reduces Bi(III) to Bi(0) metal.
2Bi(OH)3 ↓ + 3[Sn(OH)4]2- ⟶ 2Bi ↓ + 3[Sn(OH)6]2-
Take the filtrate it may contain tetramminocuprate(II) and tetramminecadmiate(II). If it is deep blue coloured then the blue colour is because of the presence of Cu(II). Otherwise test it for Cd(II). Divide the filtrate in 2 parts to test Cu(II) and Cd(II) separately.
Confirmatory test for Cu(II): take the first part and acidify it by dilute acetic acid CH3OOH and add potassium hexacyanoferrate(II) K2[Fe(CN)6] solution. Reddish brown precipitate of copper hexacyanoferrate(II) will be obtained.
Cu2+ + [Fe(CN)6]2- ⟶ Cu[Fe(CN)6] ↓
Confirmatory test for Cd(II): Take the second part and add potassium cyanide KCN solution drop by drop until the colour disappears, then add 1 ml more. Pass hydrogen sulphide H2S gas for 30 seconds. Yellow precipitate of CdS will be obtained.
Cd2+ + 2CN- ⟶ Cd(CN)2 ↓
When we add KCN white precipitate of cadmium cyanide is formed which dissolves in excess of reagent by forming tetracyanocadmiate(II) ion.
Cd(CN)2 ↓ + 2CN- ⟶ [Cd(CN)4]2-
This is a colourless complex ion which yields yellow coloured cadmium sulphide CdS precipitate on passing H2S gas.
[Cd(CN)4]2- + H2S ⟶ CdS ↓ + 2H+ + 4CN-
In the next post we will discuss the tests for IIB cations in the filtrate.
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