Removal of interfering radicals before IIIrd group analysis

What are the interfering radicals? How do they interfere in systematic separation of cationic radicals? Why is it necessary to remove them before III^rd gr analysis? Why don't they interfere in I^st or II^nd group analysis? Interfering radicals are oxalate, tartrate, fluoride, borate and phosphate and they are anionic radicals. They form complex with III^rd gr group reagent ammonium chloride and ammonium hydroxide. This leads to incomplete precipitation of III^rd group cations and causes immature precipitation of IV^th and V^th group cations in alkaline medium. Let’s try to understand it.

Oxalate, tartrate, fluoride, borate, silicate and phosphate of the metals are soluble in acidic medium. 

If you remember, for 1^stand 2^nd analysis medium remain acidic (dilute HCl) that’s why they do not interfere then. But for 3^rd group analysis the medium becomes alkaline by group reagents ammonium chloride and ammonium sulphide. Here interfering radicals come into action and disturb the solubility product of cations which causes their premature or incomplete precipitation.

In acidic medium these salts produce their corresponding acids like oxalic acid, phosphoric acid, hydrofluoric acid, boric acid and tartaric acid. For example, barium oxalate reacts with HCl and produces oxalic acid.

            BaC₂O₄ + 2HCl ⟶ BaCl₂ + H₂C₂O₄

These interfering acids are weak acids so they do not dissociate completely and remain in solution in their unionised form. Equilibrium is developed between dissociated and un-dissociated acid.

            H₂C₂O₄ ⇌ 2H⁺ + C₂O₄^2-

Hydrochloric acid is a strong acid and is ionised completely.

            HCl ⟶ H⁺ + Cl^-

Hydrogen ions acts as common ion among them and higher concentration of H⁺ suppresses the ionization of interfering acid. Therefore, ionic product of C₂O₄^2- and Ba²⁺ doesn’t exceed the solubility product of barium oxalate which is why Ba²⁺ remains in the solution as barium oxalate. That’s how interfering radicals do not interfere as long as the medium remains acidic enough. But when we make the medium alkaline by adding 3^rd group reagent ammonium hydroxide NH₄OH, OH^- ions combine with H⁺and neutralise them. This decreases the concentration of H⁺ ions which shifts the equilibrium of dissociation of interfering acid forward and increases the concentration of C₂O₄^2- . Thus the ionic product of C₂O₄^2- and Ba²⁺ exceeds the solubility product of barium oxalate and Ba²⁺ gets precipitated in the 3^rd group, which actually belongs to the 4^th group.

One or more interfering radicals can be present in the solution. They have to be removed in the following order: first we remove oxalate and tartrate, then borate and fluoride, then silicate and in the last phosphate.

Scheme for the Removal of Interfering Radicals

Procedure for the removal of oxalate and tartrate:  Oxalate and tartrate of metals are soluble in acid and they decompose on heating. Take the filtrate of 2^nd group and boil off H₂S gas from it. Add 4-5ml concentrated nitric acid HNO₃ and heat it till it is almost dry. Repeat this treatment for 2-3 times.

            (COO)₂^2- + H⁺ ⟶  (COOH)₂

                (COOH)₂  ⟶ HCOOH + CO₂↑

            HCOOH ⟶ CO↑ + H₂O↑

Tartrate and tartaric acid decomposes in a complex manner; charring takes place on heating and a smell of burnt sugar develops. Extract the with dilute HCl and filter. Use this filtrate for analysis of 3^rd group or use for removal of other interfering radicals.

Procedure for the removal of borate and fluoride: Take the filtrate and evaporate it to dryness. Add concentrated HCl and again evaporate to dryness.

            F^- + H⁺ ⟶ HF

            CaF₂ + 2HCl ⟶ CaCl2 + 2HF

  

On heating with HCl fluoride forms hydrofluoric acid and Borate forms orthoboric acid which evaporate on heating.

               BO₃^3- + 3H⁺ ⟶ H₃BO₃

            Na₃BO₃ + 3HCl ⟶ 3NaCl + H₃BO₃

Extract the residue with dilute HCl and filter. Use this filtrate for analysis of 3^rd group or use for removal of other interfering radicals.

If fluoride is absent and borate is present then residue use a mixture of 5ml ethyl alcohol and 10ml conc. HCl and evaporate to dryness.

BO₃^3- + 3H⁺ ⟶ H₃BO₃

H₃BO₃ + 3C₂H₅OH ⟶ (C₂H₅O)₃B↑ + H₂O

Procedure for the removal of silicate: Evaporate the filtrate of 2^nd group or residue obtained from removal of interfering radicals with concentrated HCl to dryness. Repeat this treatment for 3-4 times.

            SiO₃^2- + 2H⁺ ⟶ H₂SiO₃ ↓

               H₂SiO₃ ↓ ⟶ SiO₂ ↓  + H₂O

On heating with HCl silicate converts to metasilicic acid (H₂SiO₃) which is converted into white insoluble powder silica (SiO₂) on repetitive heating with concentrated HCl.

Test for phosphate HPO₄^2-:  test 0.5ml of the filtrate with 1ml ammonium molybdate reagent and a few drops of concentrated HNO₃, and warm gently, yellow precipitate indicates the presence of phosphate. Its composition is not known exactly.

Procedure for the removal of phosphate: Ferric chloride is generally used for the removal of phosphate. Fe(III) combines with phosphate and removes all phosphate as insoluble FePO₄. Fe(III) is also a member of 3^rd group so first we have to test its presence in the filtrate of 2^nd group then we can proceed for the removal of phosphate.

            HPO₄^2-  + Fe³⁺ ⟶ FePO₄ ↓ + H⁺

Test for Fe: To the filtrate of 2^nd group add ammonium chloride NH₄Cl and a slight excess of ammonia NH₃ solution. If precipitate appears, it indicates the presence of 3^rd group. It may contain hydroxides Fe(OH)₃, Cr(OH)₃, Al(OH)₃, MnO₂.xH₂O, traces of CaF₂ and phosphates of Mg and IIIA, IIIB and IV group metals. Dissolve the precipitate in minimum volume of 2M HCl. Take 0.5ml solution and add potassium hexacyanoferrate (II) K₄[Fe(CN)₆] solution. If iron is present, you will get prussian blue coloured precipitate of iron(III) hexacyanoferrate.

4Fe³⁺ + 3[Fe(CN)₆]^4- ⟶ Fe₄[Fe(CN)₆]₃

Removal of phosphate: To the main solution add 2M ammonia NH₃ solution drop wise, with stirring, until a faint permanent precipitate is just obtained. Then add 2-3ml 9M acetic acid and 5ml 6M ammonium acetate solution. Discard any precipitate if obtained at this stage. If the solution is red or brownish red, sufficient iron Fe(III) is present here to combine with phosphate. If the solution is not red or brownish red in colour then add ferric chloride FeCl₃ solution drop wise with stirring, until the solution gets a deep brownish red coloured. Dilute the solution to about 150ml with hot water, boil gently for 1-2min, filter hot and wash the residue with a little boiling water. Residue may contain phosphate of Fe, Al and Cr. Keep the filtrate for test of IIIB group. Rinse the residue in porcelain dish with 10ml cold water, add 1-1.5g sodium peroxoborate and boil gently until the evolution of O₂ ceases (2-3min). Filter and wash with hot water. Reject the residue to remove phosphate in the form of FePO₄. Keep the filtrate and test for IIIA group.

To test the presence of interfering radicals you need to prepare sodium carbonate extract and then test them separately. Scheme for the test of anionic radicals is not as systematic as cationic radicals. We will study them in coming posts. In the next post we will discuss the analysis of IIIA group cations. 

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Thermodynamics

As the name suggests thermodynamics is the study of transfer of energy/heat. Why do we need to study it? What’s the role it plays in chemistry? We all know energy is involved whenever a reaction takes place. So it is as important as money for any business. When someone starts a business, he calculates the possible profit and loss, and he starts it only after ensuring that it is profitable. Similarly a reaction happens only when it is profitable. In terms of chemistry what does it mean and how can we calculate it?

Internal energy

The profit in a reaction can be calculated in terms of its internal energy. Internal energy is the sum of all types of energies of the molecules which is very difficult to calculate. But there is an easy way to calculate the difference in internal energy of a system using thermodynamics.

If we want to calculate internal energy of a reaction, we have to focus on the reaction and where it is taking place. If the reaction is taking place in a reaction vessel such as a beaker, flask or container then we consider the vessel as a system. And we consider everything other than the vessel as surrounding.

Types of System

In order to calculate internal energy we have to observe properties of a system. First it is necessary to understand the system. System is the place where the reaction takes place and any place in this universe other than system is the surrounding. System and the surrounding are separated by a boundary. You can understand it by taking an example of your room. Suppose you are playing with a ball in your room. Playing with ball is a reaction and it is taking place in your room so your room is the system and the walls of the room are the boundary. All the other places in the house are surrounding. You and ball may or may not go out of the room; it depends on the design of your room. If you keep all of the windows and doors open, you or the ball can go outside but if you close all the windows and doors, no one can go out of the room, only the sound of the ball hitting the walls can be heard from outside. If your room is closed and sound proof too, then neither sound nor the ball or you can go outside. Similarly a system can be of three types.

• Open system where energy and matter can be exchanged with the surrounding.
• Closed system where energy can be exchanged with the surrounding but there is no exchange of matter.
• Isolated system is the one which doesn’t allow any exchange of matter or energy.

Types of Thermodynamic System

State Variables

How can we define the state of the system? We can define it by measurable quantities like pressure, temperature, volume and amount. If you have to define the state of the room in above example, you will define it by its present pressure, temperature, volume, you and ball. How this room gets that temperature or pressure doesn’t affect the state of room. Such variables are called as state variables; those values depend on the state of system not on the path how it is reached. Internal energy “U” is also a state variable.

Possibility of any reaction happening depends on the benefit of internal energy it gets. If a reaction gets benefit of internal energy then it happens. That means to check the feasibility of any reaction we have to calculate the benefit of U or ΔU. Let’s see how internal energy of the system changes. If we add or remove any kind of energy to/from the system we can increase/ decrease the internal energy of the system. Energy deference can be created as a result of some work just like when you workout you lose energy or when someone gives you massage your body gets relaxed and you feel energetic again. If you add some heat to the system it also increases the internal energy of the system and vice versa. Removal or addition of matter also affects the internal energy of the system. Now we will discuss different ways to create difference in internal energy of the system.

Internal Energy of a Closed System

Let’s take hot water in a beaker and cover it with a lid. It’s an example of closed system. Now define the state of the system at time t1 by pressure, volume and temperature. Note the state of the system after time t2. You will find only difference in temperature of initial state and final state of the system. So the change in internal energy of the system can be given by the temperature difference. The energy which is a result of temperature difference is called heat q.

   

         ΔU = T_final - T_initial

            ΔU = q

In above example T_final is lesser than T_initial because heat is transferred from the system to the surrounding and you will get q negative. Similarly when you give heat to the system, q will get a positive sign.

Internal Energy of a Adiabatic System

The other way to bring change in the internal energy of the system is work. Let’s see how you can change the U by the work done on the system or by the system. To observe the energy change by doing work we have to keep other factors constant, that means there should be no heat transfer between system and surrounding. To make this sure we can take adiabatic system. Take some water in a flask with thermally insulated walls. Make some arrangement to add paddle and thermometer to it. Note the initial temperature of the water.  Now stir up water for some time with the help of paddles and note the temperature at the final state. This change in temperature will give us change in internal energy.

ΔU = T_final - T_initial

Here you will find T_final is greater than T_initial so the difference in internal energy will be positive. Here mechanical work is done on the system which adds some more energy to the system and you get positive change in U. Difference in internal energy is equal to the work done in an adiabatic system.

ΔU = w

When work is done on the system it will get positive sign and when it is done by the system it will get the negative sign.

First law of thermodynamics

We have studied two types of systems one is closed system which allows heat transfer and the other is adiabatic system which doesn’t allow heat transfer. In the first system, change in internal energy is carried out by heat and in the latter it is done by doing work on the system. What happens when we put both in order to change the internal energy?

First Law of Thermodynamics

In that case internal energy difference is calculated by the equation:

 ΔU = q + w

This equation is the mathematical expression of “First law of thermodynamics” it states that “The energy of an isolated system is constant.”

Internal energy is a state function; its value depends only on initial and final state. It will be independent of the way the change is carried out. In the above example you did mechanical work on the system but if you replace paddle with an emersion rod and do electrical work on the system you will get the same results. In the next post we will study its applications.

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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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