Principles Of Qualitative Analysis

Introduction

The purpose of chemical analysis is to establish the composition of natural or artificially manufactured substances. For the purpose of systematic qualitative analysis cations are classified into five groups on the basis of their behaviour with some reagents. Similarly the anions are also classified as class A or B depending on their behaviour with certain reagents.

Physical Examination of Salt / Mixture

Effect of Heating

Flame Test

Classification of Anions

Methods available for the detection of anions are not as systematic as those described for the detection of cations. Furthermore anions are classified essentially on the basis of the process employed.

Class A

further divided into two sub groups.

i) Gases evolved with dil. HCl/ dil H₂SO₄.

ii) Gases or acid vapours evolved with conc. H₂SO₄

Class A (i): Anions which evolve gases on reaction with dil. HCl/dil. H₂SO₄.

It includes - CO₃^2-, SO₃^2-, S^2-, NO₂^–, CH₃COO^-, S₂O₃^2-

1. Carbonate (CO₃^2- )

i) Dilute HCl : gives effervescence, due to the evolution of carbon dioxide.

CO₃^2- + 2H⁺ $\to$ CO₂$\uparrow$ + H₂O

The gas gives turbidity with lime water and baryta water.

CO₂ + Ca²⁺ + 2OH^- $\to$ CaCO₃ $\downarrow$ + H₂O

CO₂ + Ba²⁺ + 2OH^- $\to$ BaCO₃ $\downarrow$ + H₂O

On prolonged passage of carbon dioxide in lime water, the turbidity slowly disappears due to the formation of soluble hydrogen carbonate.

CaCO₃ $\downarrow$ + CO₂ + H₂O $\to$ Ca(HCO₃)₂

The following tests are performed with the aqueous salt solution.

ii) Barium chloride or Calcium chloride solution: White ppt of barium or calcium carbonate is obtained, which is soluble in mineral acid.

CO₃^2- + Ba²⁺ $\to$ BaCO₃ $\downarrow$

CO₃^2- + Ca²⁺ $\to$ CaCO₃

iii) Silver nitrate solution: White ppt of silver carbonate is obtained.

CO₃^2- + 2Ag⁺ $\to$ Ag₂CO₃$\downarrow$

The ppt so obtained is soluble in nitric acid and in ammonia. The ppt becomes yellow or brown on addition of excess reagent and same may also happen if the mix is boiled, due to the formation of silver oxide

2. Sulphites (SO₃^2-)

i) Dilute HCl or Dilute H₂SO₄ : decomposes with the evolution of sulphur dioxide

SO₃^2- + 2H⁺ $\to$ SO₂ + H₂O

The gas has a suffocating odour of burning sulphur.

The following tests are performed with the aqueous salt solution.

ii) Acidified potassium dichromate solution: Turns filter paper moistened with acidified potassium dichromate solution, green due to the formation of Cr³⁺ions.

3SO₂ + K₂Cr₂O₇ + H₂SO₄ $\to$ K₂SO₄ + Cr₂(SO₄)₃ + H₂O

green

iii) Lime water : On passing the gas through lime water, a milky ppt is formed.

SO₂ + Ca(OH)₂ $\to$ CaSO₃ $\downarrow$ + H₂O

milky

Precipitate dissolves on prolonged passage of the gas, due to the formation of hydrogen sulphite ions.

CaSO₃ $\downarrow$ + SO₂ + H₂O Ca(HSO₃)₂.

iv) Barium chloride or Strontium chloride solution: Gives white ppt. of barium or

strontium sulphite.

$S{O}_3^{2-}+B{a}^{2+}\to BaS{O}_3\downarrow$

$S{O}_3^{2-}+S{r}^{2+}\to SrS{O}_3\downarrow$

3. Sulphide (S^-2)

i) Dil. HCl or Dil. H₂SO₄: A colourless gas smelling of rotten eggs (H₂S) is evolved.

S^2- + 2H⁺ $\to$ H₂S$\uparrow$

ii) The gas turns lead acetate paper black

(CH₃COO)₂Pb + H₂S $\to$ PbS $\downarrow$ + 2CH₃COOH

black

The following tests are performed with the aqueous salt solution.

iii) Gives yellow ppt. with CdCO₃

Na₂S + CdCO₃ $\to$ CdS + Na­₂CO₃

_Yellow

iv) Silver nitrate solution: black ppt. of silver sulphide insoluble in cold but soluble in hot dil nitric acid.

S^2- + 2Ag⁺ $\to$ Ag₂S $\downarrow$

v) Sodium nitroprusside solution : Turns sodium nitroprusside solution purple

Na₂S + Na₂[Fe(CN)₅NO] $\to$ Na₄[Fe(CN)₅NOS]

4. Nitrites (NO₂^-)

i) Dil HCl and Dil. H₂SO₄ : Adding to solid nitrite in cold yields pale blue liquid (due to the presence of free nitrous acid HNO₂ or its anhydride N­₂O₃) & the evolution of brown fumes of nitrogen dioxide, the latter being largely produced by combination of nitric oxide with the oxygen of the air

NO₂^- + H⁺ $\to$ HNO₂

2HNO₂ $\to$ H₂O + N₂O₃

3HNO₂ $\to$ HNO₃ + 2NO$\uparrow$ + H₂O

2NO$\uparrow$ + O₂$\uparrow$ $\to$ 2NO₂$\uparrow$

The following tests are performed with the aqueous salt solution.

ii) Silver nitrate solution : White crystalline ppt. is obtained

NO₂^- + Ag⁺ $\to$ AgNO₂$\downarrow$

iii) Turns acidified KI - starch paper blue

2KI + 2NO₂ $\to$ 2KNO₂ + I₂ $\uparrow$

Starch + I₂ $\to$ Blue colour

iv) Brown ring test: When the nitrite solution is added carefully to a conc. solution of Iron(II) sulphate acidified with dil acetic acid or with dilute sulphuric acid, a brown ring is formed, due to the formation of [FeNO]SO₄ at the junction of the two liquids.

NO₂^- + CH₃COOH $\to$ HNO₂ + CH₃COO^-

3HNO₂ $\to$ H₂O + HNO₃ + 2NO$\uparrow$

FeSO₄ + NO $\to$ [FeNO]SO₄.

5. Acetate (CH₃COO^-)

i) Dilute Sulphuric Acid : Smell of vinegar is observed.

CH₃COO^- + H⁺ $\to$ CH₃COOH $\uparrow$

The following test is performed with the aqueous salt solution.

ii) Iron (III) Chloride Solution: Gives deep - red colouration

CH₃COONa + FeCl₃ $\to$ (CH₃COO)₃Fe + 3NaCl

Brown colour

6. Thiosulphates ($S_2{O}_3^{2-}$)

i) Dil Hydrochloric acid: Gives sulphur & sulphur di oxide

S₂O${}_3^{2-}$+ 2H⁺ $\to$ S $\downarrow$ + SO₂$\uparrow$ + H₂O

ii) Iodine Solution: Gets decolourised due to reduction of iodine to iodide ions.

I₂ + 2S₂O${}_3^{2-}$$\to$ 2I^- + S₄O₆^2-The following tests are performed with the aqueous salt solution.iii)Barium chloride solution: White ppt. ofbarium thiosulphate is formedS₂O${}_3^{2-}$+ Ba²⁺ $\to$ BaS₂O₃ $\downarrow$But no ppt. is obtained with CaCl₂ solution.iv)Silver nitrate solution: Gives white ppt. of silver thiosulphate.S₂O${}_3^{2-}$+ 2Ag⁺ $\to$ Ag₂S₂O₃ $\downarrow$The ppt. is unstable, turning dark on standing, when silver sulphide is formed.Ag₂S₂O₃$\downarrow$ + H₂O $\to$ Ag₂S + H₂SO₄v) Lead acetate or Lead nitrate solution: Gives white ppt.$S_2{O}_3^{2-}$+ Pb²⁺ $\to$ PbS₂O₃ $\downarrow$On boiling it turns black due to the formation of PbS.PbS₂O₃ $\downarrow$ + H₂O $\to$PbS $\downarrow$ + 2H⁺ + SO₄^2-Class A (ii):Gases or acid vapours evolved with conc. sulphuric acidIt includes - Cl^-, Br^-, I^-, NO₃^-.1. Chloride (Cl^-) i) Conc. H₂SO₄ : decomposes with the evolution of HCl.Cl^- + H₂SO₄ $\to$ HCl+ HSO${}_4^{-}$Gas so produced (1) Turns blue litmus paper red2) Gives white fumes of NH₄Cl when a glass rod moistened with ammonia solution is brought near the mouth of test tube.ii) Manganese dioxide and conc. sulphuric acid :When a solid chloride is treated with MnO₂ and conc. H₂SO₄, yellowish green coloured gas is obtained.MnO₂ + 2H₂SO₄ + 2Cl^- $\to$ Mn²⁺ + Cl₂ $\uparrow$+ 2SO₄^2- + 2H₂OThe following tests are performed with the aqueous salt solution.iii)Silver nitrate solution: White, curdy ppt. of AgCl insoluble in water & in dilnitric acid, but soluble in dilute ammonia solution.Cl^- + Ag⁺ $\to$ AgCl $\downarrow$AgCl $\downarrow$ + 2NH₃ $\to$[Ag(NH₃)₂]ClAg(NH₃)₂Cl + 2H⁺ $\to$ AgCl$\downarrow$ + 2NH₄⁺.iv)Lead acetate solution: White ppt. of lead chloride is formed2Cl^- + Pb⁺² $\to$ PbCl₂ $\downarrow$v) Chromyl chloride test: When a salt containing chloride ion is heated with K₂Cr₂O₇ and conc. H₂SO₄ orange red fumes of chromyl chloride (CrO₂Cl₂) are formed.K₂Cr₂O₇ + 4NaCl +6H₂SO₄ $\to$ 2KHSO₄ + 4NaHSO₄ + 2CrO₂Cl₂ $\uparrow$ + 3H₂Oorange – red fumes
Chlorides of mercury, owing to their slight ionization, do not respond to this test and only partial conversion to CrO₂Cl₂ occurs with the chlorides of lead, silver, antimony and tin.When chromyl chloride vapours are passed into sodium hydroxide a yellow solution of sodium chromate is formed which when treated with lead acetate gives yellow ppt. of lead chromate.CrO₂Cl₂ + 2NaOH$\to$ Na₂CrO₄ + 2HClYellow solution
Na₂CrO₄ + (CH₃COO)₂ Pb $\to$ 2CH₃COONa + PbCrO₄ $\downarrow$(yellow ppt.)
2. Bromide (Br^-)i) Conc. H₂SO₄ : Gives reddish brown vapours of bromine accompanying the hydrogen bromide.2KBr + H₂SO₄ $\to$ K₂SO₄ + 2HBr2HBr + H₂SO₄ $\to$ 2H₂O + SO₂ $\uparrow$ + Br₂$\uparrow$ (reddish brown)
ii) Manganese dioxide and conc. sulphuric acid : When a mix of solid bromide, MnO₂ and conc. H₂SO₄ is heated reddish brown vapours of bromine are evolved.2KBr + MnO₂ + 2H₂SO₄ $\to$ Br₂ $\uparrow$ + K₂SO₄ + MnSO₄ + 2H₂OThe following tests are performed with the aqueous salt solution.iii) Silver nitrate solution: Pale yellow ppt. of silver bromide is obtained. This ppt. is sparingly soluble in dil but readily soluble in conc. ammonia solution and insoluble in dil. HNO₃.Br^- + Ag⁺ $\to$ AgBriv) Lead acetate solution: White crystalline ppt. of lead bromide which is soluble in boiling water.2Br^- + Pb⁺² $\to$ PbBr₂ $\downarrow$v)Chlorine water: When chlorine water is added to a mixture of CCl₄ and a solution of bromide liberates free bromine is liberated, which colours the organic layer orange – red.2KBr + Cl₂ (water)$\to$ 2KCl + Br₂Br₂ + CCl₄$\to$ Orange red colourvi) Potassium dichromate & conc. H₂SO₄ : When a mix of solid bromide, K₂Cr₂O₇, and conc. H₂SO₄ is heated and on passing the evolved vapours into water, a yellowish brown solution is obtained.2KBr + K₂Cr₂O₇ + 7H₂SO₄ $\to$ 3Br₂ $\uparrow$ + Cr₂(SO₄)₃ + 4K₂SO₄ + 7H₂O.3. Iodide (I^-) i) Conc. H₂SO₄ : Gives violet vapours of iodine2I^- + 2H₂SO₄$\to$ I₂ + SO₄^2- + 2H₂O + SO₂ $\uparrow$violet vapours
The following tests are performed with the aqueous salt solution.ii) Silver nitrate solution: Yellow ppt. of silver iodide AgI, very slightly soluble in conc. ammonia solution and insoluble in dil nitric acid.I^- + Ag⁺ $\to$ AgIiii) Lead acetate solution: Yellow ppt. of lead iodide soluble in hot water forming a colourless solution & yielding golden yellow plates (spangles) on cooling.2I^- + Pb²⁺ $\to$ PbI₂$\downarrow$iv) Potassium dichromate & conc. sulphuric acid: Iodine is liberated6I^- + Cr₂O₇^-2 + 7H₂SO₄ $\to$ 3I₂ $\uparrow$ + 2Cr³⁺ + 7SO₄^2- + 7H₂O.v) Chlorine water: Iodine is liberated, by the dropwise addition of chlorine water to iodide, and on addition of CHCl₃ violet layer is obtained.2I^- + Cl₂ $\to$ I₂ + 2Cl^-I₂ + chloroform $\to$ violet coloured layer.vi) Copper sulphate solution: Gives brown ppt. consisting of a mixture of copper (I) iodide & iodine and on additon of hypo solution brown ppt changes to white ppt.4I^- + 2Cu²⁺ $\to$ 2CuI + I₂I₂ + 2S₂O₃^2- $\to$ 2I^- + S₆₄O^2-.vii) Mercury (II) chloride solution: Forms scarlet red ppt. of HgI₂2I^- + HgCl₂ $\to$ HgI₂ $\downarrow$ + 2Cl^-.This ppt. dissolves in excess of KI, forming tetraiodo mercurate (II) complex.HgI₂ + 2I^- $\to$ [HgI₄]^2-4. Nitrate (NO₃^- ) Action of heat : The result varies with the metal1. Nitrates of sodium and potassium evolve oxygen (test with glowing splint) & leave solid nitrites (brown fumes with dilute acid)2NaNO₃ $\to$ 2NaNO₂ + O₂$\uparrow$.2.Ammonium nitrate yields dinitrogen oxide & steamNH₄NO₃ $\to$ N₂O $\uparrow$ + 2H₂O.3.Nitrates of the noble metals leave a residue of the metal and a mixture of nitrogen dioxide and oxygen is evolved.2AgNO₃ $\to$ 2Ag + 2NO₂ $\uparrow$ + O₂$\uparrow$.4.Nitrates of other metals, such as those of lead and copper, evolve oxygen and nitrogen dioxide and leave a residue of the oxide.2Pb(NO₃)₂ $\to$ 2PbO + 4NO₂$\uparrow$ + O₂$\uparrow$.i) Conc H₂SO₄ : Gives reddish - brown vapours of nitrogen dioxide4NO₃^- + 2H₂SO₄ $\to$4NO₂$\uparrow$ + 2SO₄^2- + 2H₂O + O₂$\uparrow$The following test is performed with the aqueous salt solution.ii) Brown ring test: When freshly saturated solution of iron (II) sulphate is added to nitrate solution and conc. H₂SO₄ is poured slowly down the side of the test - tube, a brown ring is obtained.2NO₃^- + 4H₂SO₄ + 6Fe²⁺ $\to$ 6Fe³⁺ +2NO + 4SO₄^𠄲 + 4H₂OFe²⁺ + NO$\uparrow$ $\to$ [Fe(NO)]²⁺On shaking and warming the mix, the brown colour disappears, nitric oxide is evolved and a yellow solution of Iron(III) ions remains.Class B
Includes anions that are identified by their reaction in solutions. It is subdivided into two groups:i)Precipitation reactionsii) Oxidation and reduction in solutionClass Bi) Precipitation reaction :SO₄^2-­ii) Oxidation and reduction in solution - CrO₄^{2 -} ,Cr₂O${}_7^{2-}$, MnO₄^-1. Sulphate (SO₄^2-) All sulphates except those of Ba, Pb, Sr are soluble in water. Sulphates of calcium and mercury(II) are slightly soluble.i) Barium chloride solution: White ppt. of barium sulphate BaSO₄ insoluble in warm dil. hydrochloric acid and in dilute nitric acid, but moderately soluble in boiling, conc. hydrochloric acid.SO₄^2-+Ba²⁺ $\to$ BaSO₄ $\downarrow$ii) Mercury (II) nitrate solution: Gives yellow ppt. of basic mercury (II) sulphate.SO₄^2-+3Hg²⁺ +2H₂O $\to$ HgSO₄.2HgO $\downarrow$ +4H⁺2. Chromate CrO₄^{2 -}and Dichromate (Cr₂O${}_7^{2-}$) Metallic chromates gives yellow solution when dissolved in water. In the presence of H⁺ chromates are converted into dichromates (orange-red solution).2CrO${}_4^{2-}$ + 2H⁺ $\rightleftharpoons$ Cr₂O${}_7^{2-}$ + H₂OCr₂O${}_7^{2-}$ + 2OH^- $\rightleftharpoons$ 2CrO${}_4^{2-}$ + H₂OIt may also be expressed as :2CrO${}_4^{2-}$+ 2H⁺ $\rightleftharpoons$ 2HCrO₄^- $\rightleftharpoons$ Cr₂O₇^-2 + H₂Oi) Barium chloride solution: Pale - yellow ppt. of barium chromate soluble in dilute mineralacids but insoluble in water and acetic acid.CrO${}_4^{2-}$ + Ba²⁺$\to$ BaCrO₄ $\downarrow$Dichromate ion also gives the same ppt. but due to the formation of strong acid precipitation is partial.Cr₂O${}_7^{2-}$+ 2Ba²⁺ + H₂O $\rightleftharpoons$ 2 BaCrO₄ $\downarrow$ + 2H⁺If sodium hydroxide or sodium acetate is added, precipitation becomes quantitative.ii) Silver Nitrate Solution: Brownish - red ppt. of silver chromate Ag₂CrO₄ which is soluble in dil. nitric acid & in ammonia solution, but insoluble in acetic acid.CrO₄^2- + 2Ag⁺ $\to$ Ag₂CrO₄ $\downarrow$2 Ag₂CrO₄ + 2H⁺ $\to$ 4 Ag⁺ + Cr₂O₇^2- + H₂OAg₂CrO₄ $\downarrow$ + 4NH₃ $\to$ 2[Ag(NH₃)₂]⁺ + CrO₄^2-Ag₂CrO₄ $\downarrow$ + 2Cl^- $\to$ 2AgCl + CrO₄^2-A reddish brown ppt. of silver dichromate Ag₂Cr₂O₇is formed with a conc. solution of a dichromate.Cr₂O₇^2- + 2Ag⁺ $\to$ Ag₂Cr₂O₇ iii) Lead acetate solution: Yellow ppt. of lead chromate PbCrO₄ insoluble in acetic acid, but soluble in dil nitric acidCrO₄^2- + Pb²⁺ $\to$ PbCrO₄ $\downarrow$.2PbCrO₄ $\downarrow$ + 2H⁺ $\rightleftharpoons$ 2Pb²⁺ + Cr₂O₇^2- + H₂O.iv) H₂O₂ : If an acidic solution of a chromate is treated with H₂O₂ a deep blue solution of chromium penta oxide is obtained.CrO₄^2- + 2H⁺ + 2H₂O₂ $\to$ CrO₅ + 3H₂OCrO₅ is unstable and it decomposes yielding oxygen and a green solution of a Cr⁺³ salt.3. Permanganate MnO ${}_4^{-}$i) Hydrogen peroxide : It decolourises acidified potassium permanganate solution2MnO₄^- + 5H₂O₂ + 6H⁺ $\to$ 5O₂ $\uparrow$ + 2Mn²⁺ + 8H₂O.ii) Iron (II) sulphate, in the presence of sulphuric acid, reduces permanganate to manganese (II). The solution becomes yellowbecause of the formation of iron (III) ionsMnO₄^- + 5Fe²⁺ + 8H⁺ $\to$5Fe³⁺ + Mn²⁺+ 4H₂Oiii) Action of heat: On heating a black residue of potassium manganate K₂MnO₄ and manganese dioxide remains behind. Upon extracting with water and filtering, a green solution of potassium manganate is obtained.2KMnO₄ $\to$ K₂MnO₄ + MnO₂ + O₂$\uparrow$Classification of Cations
For the purpose of systematic qualitative analysis, cations are classified into five groups on the basis of their behaviour with some reagents and classification is based on whether a cation reacts with these reagents by the formation of precipitates or not (solubility difference)Basis of Classification
The division ofdifferent cations in different analytical groups depends on the principle of solubility product. These divisions into groups are altogether different from the groupings in the Periodic Table. To understand fully the application of the principle of solubility product in analytical chemistry we should consider the principles of the Law of Mass Action and Common Ion Effect.Precipitation of Pb⁺⁺, Ag⁺ and $H{g}_2^{++}$ ions as their chlorides in group (I)The metals of the Gr. I are precipitated by adding chloride ions (from dil HCl). Solubility products at laboratory temperature arePbCl₂ = 2.4 x 10^𠄴, AgCl = 1.6 x 10^� andHg ₂Cl₂ = 3.5 x 10^�Let us consider the precipitation of AgCl. From the equilibrium of AgCl in a saturated solution.AgCl $\rightleftharpoons$$\underset{Solution}{\underbrace{A{g}^{+}+C{l}^{-}}}$

$C_{A{g}^{+}}{C}_{C{l}^{-}}=K{s}_{AgCl}$.

By adding dil. HCl, which dissociates into H⁺ and Cl^– ions, the concentration of Cl^– is increased, thus momentarily.$C_{A{g}^{+}}{C}_{C{l}^{-}}$ becomes greater than the solubility product of AgCl $\left(K{s}_{AgCl}\right)$ and hence some Ag⁺ ions and Cl^– ions combine together to form solid AgCl until the product of the concentrations of Ag⁺ and Cl^– ions, i.e., $C_{Ag}^{+}$. $C_{Cl}^{-}$ becomes greater than Ks_AgCl. Ks_AgCl is a constant quantity, the larger is the concentration of Cl^– ions the smaller is the concentration of Ag⁺ ions and hence by adding excess of Cl^– ions it is possible to remove Ag⁺ ions as AgCl. (That is why it is always necessary to add a slight excess of the precipitating reagent). It should be remembered that by increasing the amount of Cl^– ions the value of $C_{A{g}^{+}}$ never becomes equal to zero since the product of the concentrations$C_{A{g}^{+}}{C}_{C{l}^{-}}$ is always equal to Ksp_AgCl but it becomes smaller and smaller and approaches zero.

Similarly by adding dil. HCl i.e., by increasing Cl^– ions concentration, the products of the concentrations of Pb⁺⁺ and Cl^– ions and $H{g}_2^{++}$ and Cl^– ions exceed the respective solubility product of PbCl₂ and Hg₂Cl₂ and hence PbCl₂ and Hg₂Cl₂ are precipitated in group I. But this increased concentration of Cl^– ions cannot exceed the solubility product of the chlorides of the metals of subsequent analytical Groups, and that is why their chlorides remain in solution.

Precipitation of Sulphides of Group (II) and (IIIB)

The sulphides of Gr. II are precipitated by passing H₂S in dil. HCl medium whereas those of Gr. III (B) are precipitated by passing H₂S in NH₄OH medium. Let us select for the sake of convenience, a representative member from two groups, CuS from Gr. II and ZnS from Gr. III (B). Others will follow similarly.

The solubility product of CuS = 1 x10^–14 and that of ZnS = 1 x 10^–23. According to our previous discussions, sulphides of the metals can precipitate only when the product of the concentration of metal ions and sulphide ions i.e.,$C_{M^{++}}$. $C_{S^{--}}$ exceeds the solubility product of the metal sulphide, MS.

Now H₂S in aqueous solution behaves like a weak acid, feebly dissociating into H⁺ ions and S^{– –} ions. H₂S $\rightleftharpoons$ 2H⁺+S^{– –}. This dissociation is still further depressed by the addition of HCl, which being a strong electrolyte dissociates as HCl $\to$ H⁺ + Cl^–.

H⁺ ions, being common to both HCl and H₂S, depresses the dissociation of H₂S i.e., the concentration of sulphide ions is decreased by the addition of common ions H⁺ from the strong electrolyte HCl. But this decreased value of sulphide ion concentrations is sufficient to exceed the low solubility products of the sulphides of Gr.II but insufficient to exceed the comparatively high solubility products of the sulphides of the metals of Gr. III (B). Or in other words with this decreased S^{– –} ions concentrations.

$C_{M^{++}}{C}_{S^{--}}>Ks{p}_{MS}$ of Gr.II

but $C_{M^{++}}{C}_{S^{--}}<$ Ksp_MS of Gr. III (B).

That is why sulphides of Gr.(III) (B) are not precipitated in presence of dil. HCl.

In the presence of NH₄OH, which dissociates into NH₄OH $\rightleftharpoons$ $N{H}_4^{+}$+ (OH)^– the (OH)^– ions combine with H⁺ ions from H₂S giving un-ionised H₂O and since some H⁺ ions are removed in this way from the equilibrium the reaction will proceed in the forward direction, i.e., more H₂S will dissociate giving H⁺ and S^{– –} ions and the concentration of S^{– –} ions will be eventually increased. This increased value of the concentration of sulphide ions is now sufficient to exceed the solubility product of the sulphides of Gr. III (B) and hence the precipitation of the sulphides of Gr. III (B) occurs in the presence of NH₄OH.

Precipitation of Hydroxides of Gr. (III A)

The hydroxides of Gr. III (A) metals are precipitated by adding NH₄Cl and NH₄OH. The latter being a weak base dissociates to a small extent according to

NH₄OH $\rightleftharpoons$ $N{H}_4^{+}+O{H}^{-}$

NH₄Cl being a strong electrolyte dissociates into NH₄Cl $\to$ $N{H}_4^{+}+C{l}^{-}$

The dissociation of NH₄OH is further depressed due to common ion effect (NH₄⁺ ion being common) i.e., some OH^– ions and NH₄⁺ ions recombine together to form undissociated NH₄OH thereby decreasing the (OH)^– concentrations. With this decreased value of the (OH)^– ion concentration the solubility product of the hydroxides of Al. Fe and Cr alone is reached.

i.e., $C_{M^{3+}.}{C^3}_{O{H}^{-}}>Ks{p}_{M{(OH)}_3}$ and under this condition hydroxides of Gr. III (A) are precipitated. But with this low value of $C_{O{H}^{-}}$ the comparatively high value of the solubility product of the hydroxides of the metals of the subsequent groups e.g. Zn, Mn, Ni, Co, Mg etc. are not reached and hence these are not precipitated as their hydroxides in presence of NH₄Cl. If NH₄Cl were not added, then with the undiminished value of $C_{O{H}^{-}}$(as a result of dissociation of NH₄OH) all these metals of Grs. III (A), III (B) and IV will be precipitated as their hydroxides because the solubility products of their hydroxides were reached. This is why NH₄Cl is added to decrease the (OH)^– concentration due to common ion effect so as to reach the solubility products of the hydroxides of Gr. III (A) only but not to reach the solubility product of the hydroxides of Gr. III (B) IV and V.

Precipitation of Carbonates of Gr. (IV)

NH₄OH and (NH₄)₂CO₃ are added to precipitate the carbonates of Gr. IV (Ba, Sr and Ca). By the addition of NH₄Cl and NH₄OH such conditions are maintained so as to render the $C{O}_3^{--}$ ion concentration to be approximately equal to 1.6 x 10^–3 and the lowest limit of concentration of the metal ion necessary for obtaining an appreciable ppt. is 10^–4(M), the ionic product of the metal carbonate will be 1.6 10^–3 x 10^–4 = 1.6 x 10^–7. This value is greater than the solubility product of the carbonates of Ba, Ca and Sr but less than the solubility product of MgCO₃. Hence, the carbonates of Ba, Ca and Sr are precipitated in Gr. IV.

Mg⁺⁺ remains in solution. By the addition of Na₂HPO₄, Mg(NH₄)PO₄ being insoluble is precipitated.

Separation of Cations

Points to Remember

1. Group I radicals (Ag⁺, Pb⁺², Hg₂²⁺) are precipitated as chlorides because the solubility product of these chlorides (AgCl, PbCl₂, Hg₂Cl₂) is less than the solubility products of all other chlorides which remain in solution.

2. Group II radicals are precipitated as sulphides because sulphides of other metals remain in solution because of their low solubility products, HCl acts as a source of H⁺ and thus decreases the conc. of S^2- due to common ion effect. Hence decreased conc. of S^2- is only sufficient to precipitate the Group II radical only.

3. Group III A radicals are precipitated as hydroxides and the NH₄Cl suppresses the ionisation of NH₄OH so that only the group III A radicals are precipitated because of their low solubility product.

Note: i) Excess of NH₄Cl should be added otherwise manganese will be ppt. as MnO₂.H₂O.

ii) (NH₄)₂SO₄ can't be used in place of NH₄Cl because the SO₄^2- will ppt. barium as BaSO₄.

iii) NH₄NO₃ can't be used in place of NH₄Cl because NO₃^- ions will oxidise Mn²⁺ to Mn³⁺ and thus Mn(OH)₃ will be precipitated in III A group.

iv) Only Al(OH)₃ is soluble in excess of NaOH followed by boiling to form sodium meta aluminate while Fe(OH)₃ and Cr(OH)₃ are insoluble.

4. Ammonium hydroxide increases the ionisation of H₂S by removing H⁺ from H₂S as unionised water

H₂S $\rightleftharpoons$ 2H⁺ + S^2-. H⁺ + OH^- $\to$ H₂O

Now the excess of S^2- ions are available and hence the ionic product of Group III B exceed their solubility product and ppt. will be obtained.In case H₂S is passed through a neutral solution, incomplete precipitation will take place due to the formation of HCl which decreases the ionization of H₂S.

MnCl₂ + H₂S $\to$ MnS + 2HCl

Identification of Basic Radicals – Confirmatory Tests

The confirmatory tests are performed with the salt solution.

1. Group I (Pb²⁺, Ag⁺, Hg⁺)

(A) PbCl₂ gives a yellow ppt. with K₂CrO₄. The ppt. is insoluble in acetic acid but soluble in NaOH

PbCl₂ + K₂CrO₄ $\to$ PbCrO₄ $\downarrow$ + 2KCl

Yellow ppt.

PbCrO₄ + 4NaOH $\to$ Na₂[PbO₂] + Na₂CrO₄ + 2H₂O

(B) PbCl₂ + 2KI $\to$ PbI₂ $\downarrow$ + 2KCl

(Yellow)

PbCl₂ + 2KI (excess) $\to$ K₂[PbI₄]

AgCl is soluble in NH₄OH forming a complex while Hg₂Cl₂ forms a black ppt. with NH₄OH.

AgCl + 2NH₄­OH $\to$ Ag(NH₃)₂Cl + 2H₂O

Hg₂Cl₂ + 2NH₄OH $\to$ H₂N $-$Hg $-$ Cl + Hg$\downarrow$ + NH₄Cl + 2H₂O

Amino mercuric Chloride

2. Group II A (Hg²⁺, Cu²⁺, Bi³⁺, Cd²⁺)

i) Hg⁺²ions in solution, on addition of SnCl₂, give a white precipitate turning black.

2Hg⁺² + SnCl₂ $\to$ Sn⁺⁴ + Hg₂Cl₂ $\downarrow$

White

Hg₂Cl₂ + SnCl₂ $\to$ SnCl₄ + 2Hg $\downarrow$

Black

ii) Cu⁺² ions in solution give deep blue colour with excess of NH₄OH

Cu⁺² + 4NH₄OH $\to$ [Cu(NH₃)₄ ]⁺² + 4H₂O

Deep blue in colour

Cu⁺² ions give chocolate precipitate with K₄Fe(CN)₆.

2Cu⁺² + K₄Fe(CN)₆ $\to$ Cu₂[Fe(CN)₆] + 4K⁺

iii) Bi⁺³ ions in solution of HCl on addition of water give white cloudy precipitate.

BiCl₃ + H₂O $\to$ BiOCl $\downarrow$ + 2HCl

White ppt.

When treated with sodium stannite a black ppt. is obtained.

2BiCl₃ + 3Na₂SnO₂ $\to$ 2Bi $\downarrow$ + 3Na₂SnO₃ + 6NaCl + 3H₂O

black

iv) Cd⁺² ions in solution, with NaOH give a white precipitate.

Cd⁺² + 2NaOH $\to$ Cd(OH)₂ $\downarrow$ + 2Na⁺

white

With ammonium hydroxide, Cd²⁺ ions give a white precipitate which dissolves in excess.

Cd²⁺ + 4NH₄OH $\to$ [Cd(NH₃)₄](OH)₂ + 2H₂O

3. Group II B (As³⁺, As⁵⁺, Sb³⁺, Sb⁵⁺, Sn³⁺, Sn⁴⁺)

v) As⁺³ ions in solution give yellow precipitate with ammonium molybdate and HNO₃ on heating.

H₃AsO₄ +12(NH₄)₂MoO₄ +21HNO₃ $\to$ (NH₄)₃ AsMo₁₂O₄₀$\downarrow$+ 21NH₄NO₃ + 12H₂O

Yellow ppt.

vi) Sn²⁺ ions in solution as SnCl₂ give white ppt. with HgCl₂ ,which turns black on standing.

SnCl₂ + 2HgCl₂ $\to$ SnCl₄ + Hg₂Cl₂ $\downarrow$

White

Hg₂Cl₂ + SnCl₂ $\to$SnCl₄ + 2Hg $\downarrow$

Black

vii) Sb⁺³ ions in solution as SbCl₃ , on addition of water give white precipitate.

SbCl₃ + H₂O $\to$ SbOCl$\downarrow$ + 2HCI

White

4. Group III A (Al³⁺, Fe³⁺, Cr³⁺)

i) White precipitate of Al(OH)₃ is soluble in NaOH

Al(OH)₃ + NaOH $\to$ NaAlO₂ + 2H₂O

ii) Precipitate of Cr(OH)₃ is soluble in NaOH + Br₂ water and addition of BaCl₂ to this solution gives yellow precipitate.

Br₂ + H₂O $\to$ 2HBr + (O)

2Cr(OH)₃ + 4NaOH + 3(O) $\to$ 2Na₂CrO₄ + 5H₂O

Na₂CrO₄ + BaCl₂ $\to$ BaCrO₄ $\downarrow$ + 2NaCl

Yellow ppt.

iii) Fe(OH)₃ is insoluble in NaOH

Brown precipitate of Fe(OH)₃ is dissolved in HCl and addition of KCNS to this solution gives blood red colour.

Fe(OH)₃ + 3HCl $\to$ FeCl₃ + 3H₂O

FeCl₃ + 3KCNS $\to$ Fe(CNS)₃ + 3KCl

blood red

Also on addition of K₄Fe(CN)₆ to this solution, a prussian blue colour is obtained.

FeCl₃ + 3K₄Fe(CN)₆ $\to$ Fe₄[Fe(CN)₆]₃ + 12KCl

prussian blue colour

5. Group III B (Ni²⁺, Co²⁺, Mn²⁺, Zn⁺²)

i) Ni⁺² and Co⁺² ions in solution, on addition of KHCO₃ and Br₂ water give apple green colour if Co⁺² is present and black precipitate if Ni⁺² is present.

CoCl₂ + 6KHCO₃ $\to$ K₄[Co(CO₃)₃] + 2KCl + 3CO₂$\uparrow$ + 3H₂O

2K₄[Co(CO₃)₃] + 2KHCO₃ + [O] $\to$ 2K₃[Co(CO₃)₃] + 2K₂CO₃ + H₂O

Apple green colour

NiCl₂ + 2KHCO₃ $\to$ NiCO₃ + 2KCl + H₂O + CO₂$\uparrow$

2NiCO₃ + 4NaOH + [O] $\to$ Ni₂O₃ $\downarrow$ + 2Na₂CO₃ + 2H₂O

Black ppt.

ii) Zn⁺² ions in solution give white precipitate with NaOH, which dissolve in excess of NaOH.

Zn⁺² + 2NaOH $\to$ Zn(OH)₂ $\downarrow$ + 2Na⁺

White

Zn(OH)₂ + 2NaOH $\to$ Na₂ZnO₂ + 2H₂O

Soluble

iii) Mn⁺² ions in solution give pink precipitate with NaOH turning black or brown on heating.

Mn⁺² + 2NaOH $\to$ Mn(OH)₂ + 2Na⁺

Pink

Mn(OH)₂ + [O] $\stackrel{\Delta }{\to }$MnO₂ + H₂O

Brown or black

6. Group IV (Ba²⁺, Sr²⁺, Ca²⁺)

i) Ba⁺² ions in solution give

(A) Yellow precipitate with K₂CrO₄

Ba⁺² + K₂CrO₄ $\to$ BaCrO₄ $\downarrow$ + 2K⁺

Yellow

(B) White precipitate with (NH₄)₂SO₄

Ba⁺² + (NH₄)₂ SO₄ $\to$ BaSO₄$\downarrow$ + $2N{H}_4^{+}$

White

(C) White precipitate with (NH₄)₂ C₂O₄

Ba⁺² + (NH₄)₂C₂O₄ $\to$ BaC₂O₄$\downarrow$ + $2N{H}_4^{+}$

White

ii) Sr⁺² ions give white precipitate with (NH₄)₂SO₄ and (NH₄)₂C₂O₄

Sr⁺² + (NH₄)₂SO₄ $\to$ SrSO₄$\downarrow$ + $2N{H}_4^{+}$

White ppt.

Sr⁺² + (NH₄)₂C₂O₄ $\to$ SrC₂O₄$\downarrow$ + $2N{H}_4^{+}$

White

iii) Ca⁺² ions give white precipitate with (NH₄)₂ C₂O₄ only.

Ca⁺² + (NH₄)₂C₂O₄ $\to$ CaC₂O₄ $\downarrow$ + $2N{H}_4^{+}$

White

7. Group V (NH₄⁺, Na⁺, K⁺, Mg⁺²)

i) All ammonium salts on heating with alkali say NaOH give a colourless gas with a pungent smell (NH₃)

NH₄Cl + NaOH $\to$ NaCl + NH₃ $\uparrow$ + H₂O

(A) Gas evolved gives white fumes with HCl

NH₃ + HCl $\to$ NH₄Cl $\uparrow$

White fumes

(B) Paper soaked in CuSO₄ solution, is turned deep blue by NH₃ due to complex formation

CuSO₄ + 4NH₃ $\to$ [Cu(NH₃)₄]SO₄

deep blue

(C) With Hg₂ (NO₃)₂ , a black colour is obtained

Hg₂(NO₃)₂ + 2NH₃ $\to$ Hg $\downarrow$ + Hg(NH₂)NO₃ $\downarrow$ + NH₄NO₃

black

(D) With Nesslers reagent (alkaline solution of potassium tetraiodomercurate(II) ), a brown ppt. is obtained

ii) Potassium salts give yellow ppt. with sodium cobalt nitrite

Na₃[Co(NO₂)₆] + 3KCl $\to$ K₃[Co(NO₂)₆] + 3NaCl

yellow

iii) Sodium salts give a heavy white ppt. with potassium dihydrogen antimonate

KH₂SbO₄ + NaCl $\to$ NaH₂SbO₄ $\downarrow$ + KCl

White ppt.

iv) Mg²⁺ gives white ppt. of magnesium hydroxide with sodium hydroxide

Mg²⁺ + 2NH₃ + 2H­₂O $\to$ Mg(OH)₂ $\downarrow$ + 2NH₄⁺