Qualitative analysis of Group II(B) cations

  

In the last post we separated group IIB cations from IIA cations by using yellow ammonium polysulphide (NH₄)₂S_x solution, which dissolves the IIB cations. By doing this we obtained them in the form of filtrate. In the filtrate, the cations are in the form of thiosalts (NH₄)₃AsS₄, (NH₄)₂SbS₄, and (NH₄)₂SnS₃ and for further analysis we have to dissociate the salts, remove excess of sulphur and re-precipitate the cations.

How to re-precipitate IIB cations: Add hydrochloric acid HCl to the filtrate until it becomes slightly acidic (test with litmus paper, acid turns blue litmus into red) then warm it and stir for 1-2minutes. It dissociates thiosalts and precipitates excess of sulphur in the form of fine white or yellow precipitate. If a yellow or orange flocculant (colloidal) precipitate is obtained, it may contain As₂S₅, As₂S₃, Sb₂S₅ and sulphur S. Wash the precipitate with a little hydrogen sulphide H₂S water and reject washing.

2AsS₄^3- + 6H⁺ ⟶ As₂S₅ ↓ + 3H₂S ↑

2SbS₄^3- + 6H⁺ ⟶ Sb₂S₅ ↓ + 3H₂S ↑

SnS₃^2- +2H⁺ ⟶ SnS₂ ↓ + H₂S ↑

Separation of Arsenic: Sulphides of arsenic are insoluble in concentrated hydrochloric acid HCl while sulphides of antimony and tin are soluble in it by forming chlorides. Transfer the precipitate in a conical flask and add 5-10ml concentrated hydrochloric acid HCl then place a funnel in the mouth of flask. Boil gently for 5 minutes. Sulphides of antimony and tin will dissolve and you will get arsenic sulphide as residue. Some of the arsenic may have dissolved in it, to re-precipitate it add 2-3ml water and pass hydrogen sulphide H₂S gas for 1 minute. Keep the filtrate to test for antimony and tin.

Tests for arsenic: Arsenic is found in two states As(III) and As(V). In strongly acidic medium As(III) is stable and in strongly alkaline medium As(V) predominates. As(V) exists in solutions as AsO₄^3- ion.

Dissolve the precipitate in 3-4ml warm ammonia NH₃ solution. Add 3-4ml 3% hydrogen peroxide H₂O₂ solution and warm for a few minutes, it oxidizes arsenite As(III) to arsenate As(V). Add 1-2ml of magnesium nitrate reagent (a solution of Mg(NO₃)₂, NH₄Cl and a little NH₃). A whilte precipitate of magnesium ammonium asenate Mg(NH₄)AsO₄.6H₂O is obtained.

            AsO₄^3- + Mg²⁺  + NH₄⁺ ⟶ Mg(NH₄)AsO₄↓

Filter off this white precipitate and add 1ml silver nitrate AgNO₃ solution containing 6-7 drops of 2M acetic acid CH₃COOH. Red precipitate of silver arsenate Ag₃AsO₄ is formed.

            Mg(NH₄)AsO₄ ↓ + 3Ag⁺ ⟶ Ag₃AsO₄ ↓ + Mg²⁺  + NH₄⁺

Scheme for the separation of II^nd group cations

Test for Sb and Sn: Take the filtrate to test for Sb and Sn. Boil to expel H₂S gas then cool it. Divide the solution into three parts. In two parts we will test for Sb and in third part for Sn.

Part 1: add 2M ammonia NH₃ solution to make it just alkaline. Ignore any slight precipitation and add 1-2g solid oxalic acid (COOH)₂, boil and pass H₂S gas for 1 minute into hot solution. Orange precipitate of Sb₂S₃ is obtained.

Part 2: place 1ml solution in a spot plate and add a minute crystal of sodium nitrite NaNO₂ to oxidise Sb(III) to Sb(V) state. Add 1ml of Rhodamine-B reagent (tetraethylrhodamine). A bright red colour of reagent is changed to violet.

Part 3: Partly neutralise the solution with 2M ammonia NH₃ solution. Take 1m solution add 1cm clean iron wire, warm gently to reduce Sn(IV) to Sn(II).

Sn⁴⁺ + Fe ⟶ Fe²⁺  + Sn²⁺

Filter it and pour into a solution of mercury(II) chloride. Tin reduces Hg(II) to Hg(I) state and you will obtain a white precipitate of mercury(I) chloride Hg₂Cl₂.

Sn²⁺  + 2HgCl₂  ⟶ Hg₂Cl₂ ↓ + Sn⁴⁺ + 2Cl^-

If Sn²⁺ ions are in excess, the precipitate turns grey on warming, because of further reduction of Hg(I) ion to Hg metal.

Sn²⁺  + Hg₂Cl₂ ↓  ⟶ Hg↓ + Sn⁴⁺ + 2Cl^-

In the next post we will discuss identifying tests for 3^rd group cations in the filtrate you got after the removal of precipitate of 2^nd group cations. 

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What is oxidation or reduction?

Oxidation and reduction are most common words when you study chemical reactions. What happens in a chemical reaction? Two reactants react together and form the product and by-products. When you view it closely you will find that a chemical reaction is nothing but a kind of business where they exchange electrons instead of money. In any reaction the total number of electrons remains constant, they only move from one species to other so that one species gets profited by gaining electrons and the other suffers a loss by losing electrons. Oxidation and reduction refer to the state of loser and gainer.

Let’s take an example to understand it:

Sn⁴⁺ + Fe ⟶ Fe²⁺  + Sn²⁺

Notice the oxidation number of tin and iron on both sides of the arrow. Oxidation number gives us an idea about an element's willingness to give its electron. Either it likes to donate or accept electrons. In the above reaction:

Sn(IV) ⟶ Sn(II)

Fe(0) ⟶ Fe(II)

Oxidation number of tin was +4 and it becomes +2. It gains 2 electrons. And its oxidation state reduces from 4 to 2. We can say that Sn(IV) is reduced to Sn(II). Now we can define reduction as a process in which element gains electrons and it oxidation number reduces.  

In the case of iron, its oxidation number was (0) and after the reaction it becomes (+2). It loses 2 electrons and its oxidation state increases from 0 to 2. This process is called oxidation. Here iron Fe(0) gets oxidised to Fe(II).

In the above reaction iron Fe(0) gets oxidised by tin Sn(IV), so tin Sn(IV) acts as an oxidising agent. And tin Sn(IV) get reduced by iron Fe(0) so iron Fe(0) acts as a reducing agent.

In the above reaction Sn(IV) gives 2e^- to Fe(0) to get it oxidised to Fe(II) state or you can say Fe(0) gets oxidised by Sn(IV). Here Sn(IV) acts as an oxidising agent. After giving 2e^- to iron tin itself gets reduced to Sn(II) state.

If you see above reaction in a different prospect, Fe(0) accepts electrons from Sn(IV) so that it can get reduced to Sn(II) state, in this way Sn(IV) gets reduced by Fe(0). Now Fe(0) acts as reducing agent which itself gets oxidised to Fe(II) state.

Such reactions are called redox reactions, where oxidation and reduction takes place simultaneously, as happened in the above reaction where Sn(IV) gets reduced and Fe(0) gets oxidised simultaneously. Redox reactions are very useful and important reactions; we will study them in coming posts.

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Qualitative analysis of Group II(A) cations

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 2^nd 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⁺ + S^2-

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⁺ + S^2-

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 2^nd 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)₂S_xsolution 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)₂S_xwhile 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.

Scheme for the separation of IInd 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.

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⁺ + SO₄^2- ⟶ 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- + SO₄^2-

Add few drops of 2M acetic acid and then 0.1M potassium chromate K₂CrO₄solution, yellow precipitate of lead chromate PbCrO₄ will be obtained.

            Pb²⁺ + CrO₄^2- ⟶ 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⁺ + NO^3- + 2NH₃ + 2H₂O ⟶ Bi(OH)₂NO₃ ↓ + 2NH₄⁺

     Copper forms basic copper sulphate salt.   

            2Cu²⁺ + SO₄^2- + 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⁺ + SO₄^2- + 2OH^-

            Cadmium forms cadmium(II) hydroxide in ammonia solution.

            Cd²⁺ + 2NH₃ + 2H₂O ⇌Cd(OH)₂ ↓ + 2NH₄⁺

                    

Cadmium(II) hydroxide dissolves in excess of ammonia due to the formation of tetramminecadmiate(II) ion which is colourless.

                    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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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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