Inorganic Qualitative Analysis

Inorganic Qualitative Analysis
Shree Jayendrapuri Arts & Science College, Bharuch Inorganic Qualitative Analysis Presented by:Dr.Nitinkumar B. Patel Associate Professor ,Dept. of Chemistry ID: web :

What is an inorganic qualitative analysis
What is an inorganic qualitative analysis ? Inorganic qualitative analysis is a systematic methods of analysis to confirm the presence of cation(s) and anion(s) present in an inorganic substance or mixture of substances. Soluble substance: The solubility of a substance is the amount of that substance that will dissolved in a given amount of solvent. Solubility greater than 0.1M. Slightly soluble substance having solubility between M Sparingly Soluble substance: The salts which has solubility less than 0.01M are said to be sparingly soluble substance. e.g. AgCl

Solubility Product Ksp:
The product of the concentrations of ions in mole/liter present in a saturated solution at definite temperature is called solubility product Ksp. The value of Ksp changes with temperature. The solubility product is the equilibrium constant representing the maximum amount of solid that can be dissolved in aqueous solution. The higher the Ksp value, the greater the solubility. Ionic Product: On mixing aqueous solutions of two different substances, the product of the concentrations of positive and the negative ions present in the mixed solution is called ionic product (Ip).

K = Key Concepts [M+] [A-] ________ [MA]
Solubility Product, Ksp refers to an ionic compound dissolving to form ions For the reaction: MA(s) M+(aq) + A-(aq) The equilibrium constant, K is: [M+] [A-] ________ [MA] K = Since the concentration of the solid is a constant, it is effectively incorporated into the equilibrium constant: Ksp = [M+][A-] [M+][A-] is referred to as the ion product At equilibrium, the ion product = solubility product (Ksp) and the solution is said to be saturated. A precipitate will form when two solutions are mixed if the ion product is greater than the solubility product, i.e, [M+][A-] > Ksp

Possibility of Precipitation
Condition Nature of Solution Possibility of Precipitation ion product < Ksp unsaturated solution no precipitation occurs ion product = Ksp saturated solution saturated solution at equilibrium ion product > Ksp supersaturated solution precipitation occurs Example : Calculating the solubility of an ionic compound (MA). Calculate how much silver bromide will dissolve in 1 L of water given Ksp = 5.0 x 10-13 at 25oC Write the equation for dissolving AgBr in water: AgBr(s) Ag+(aq) + Br-(aq) Write the equilibrium expression: Ksp = [Ag+][Br-] = 5.0 x 10-13

Determine the relative concentrations of each ion:
at equilibrium [Ag+] = [Br-] (from the balanced chemical equation) So, [Ag+] = [Br-] = x Substitute these vales into the equilibrium expression: 5.0 x 10-5 = [x]2 Solve for x: x = √5.0 x 10-5 = 7 x 10-7 mol L-1 So, [Ag+] = [Br-] = 7 x 10-7 mol L-1 The solubility of AgBr at 25oC is 7 x 10-7 mol L-1 7 x 10-7 moles AgBr will dissolve in 1L of water at 25oC.

Example : Calculating the solubility of an ionic compound (MA2).
Calculate how much strontium fluoride will dissolve in 1 L of water given Ksp = 2.5 x 10-9 at 25oC. Write the equation for dissolving SrF2 in water: SrF2(s) Sr2+(aq) + 2F-(aq) Write the equilibrium expression: Ksp = [Sr2+][F-]2 = 2.5 x 10-9 Determine the relative concentrations of each ion: at equilibrium [Sr2+] = x and [F-] = 2x (from the balanced chemical equation) Substitute these vales into the equilibrium expression: 2.5 x 10-9 = [x][2x]2 = 4x3 Solve for x: 2.5 x 10-9 ÷ 4 = x3 = 6.25 x 10-10 x = 3√6.25 x 10-10 = 8.5 x 10-4 mol L-1 The solubility of SrF2 at 25oC is 8.5 x 10-4 mol L-1 x 10-4 moles of SrF2 will dissolve in 1L of water at 25oC.

Example : Deciding whether a precipitate will form
Will a precipitate form if 25.0 mL of 1.4 x 10-9 mol L-1 NaI and 35.0 mL of 7.9 x 10-7 mol L-1 AgNO3 are mixed? (Ksp for AgI at 25oC is 8.5 x 10-17) Write the net ionic equation for the dissolving of the precipitate: AgI Ag+ + I- Write the expression for the ion product: ion product = [Ag+][I-] Calculate the concentration each reactant after mixing: [I-] : c1 = [I-] before mixing = 1.4 x 10-9 mol L-1 (assuming full dissociaiton of NaI) V1 = initial volume = 25.0 mL = 25.0 x 10-3 L V2 = final volume after mixing = = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive) c2 = [I-]after mixing=c1 x V1 ÷ V2= (1.4 x 10-9 x 25.0 x 10-3)÷(60.0 x 10-3)=5.8x10-10 mol L-1

[Ag+]: c1 = [Ag+] before mixing = 7
[Ag+]: c1 = [Ag+] before mixing = 7.9 x 10-7 mol L-1 (assuming full dissociaiton of AgNO3) V1 = initial volume = 35.0 mL = 35.0 x 10-3 L V2 = final volume after mixing = = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive) c2 = [Ag+] after mixing = c1 x V1 ÷ V2 = (7.9 x 10-7 x 35.0 x 10-3) ÷ (60.0 x 10-3) = 4.6 x 10-7 Calculate the ion product: ion product = [Ag+][I-] = 5.8 x 10-10 x 4.6 x 10-7 = 2.7 x 10-16 Decide whether a precipitate forms: If ion product > Ksp a precipitate will form If ion product < Ksp a precipitate will not form In this case, 2.7 x 10-16 > Ksp (8.5 x 10-17) so a precipitate will form

Ksp equation for different stoichometric inorganic substances :
AgCl Ksp = [Ag+] [Cl-] = (S) x (S)=S2 SrF Ksp = [Sr+2] [F-2] = (S) x(2S)2 =4S3 Na2SO Na+ +SO Ksp = (2S)2 x (S) = 4S3 Ksp = [Na+] [SO4-2] = (2S)2 x (S) = 4S3 Ag2CrO Ksp = [Ag+] 2[CrO4-2] = (2S)2 x (S) =4S3 Cr(OH) Ksp = [S]x[3S]3 = 27S4 where, S = (Ksp /27)1/ Molar solubility Bi2S Ksp = [2S]2x [3S]3 = 4S2x 27S3 = 108 S5 Hg2I Hg I- , Ksp = [S]x[2S]2 = 4S3 Ca3(PO4) Ksp = [Ca+2]3x[PO4-3]2= [3S]3x[2S]2 = 108 S5 NiS Ksp = [Ni+2][S-2] = (S) x (S)=S2 CaCO Ksp = [Ca+2] [CO3-2] = (S) x (S)=S2 Solubility mol/lit = gm/lit X 1/molar mass(g/mol) = g/l x mol/g = mol/lit g/l = mol/l x molarmass(g/mol)/l = mol/l x g/mol = g/l

Solubility of inorganic compounds dissolved in distilled water
(1) NH4+, Na+, K+ all salts are soluble Exceptions: some transition metal compounds (2). Br -, Cl-, I- most are soluble. Exceptions: salts contains Ag+,Pb+2 and Hg+1 (3). Nitrates: All nitrates are soluble. (4). Sulphates: All sulphates are soluble. Exceptions: sulphates of Ba+2,Pb+2,Ag+,Hg+,Sr+2 are insoluble. (5). CO3-2,CrO4-2,PO4-3,SiO4-2 (silicates) are insoluble. Exception: NH4+,Na+,K+ soluble,MgCrO4 (6}. S-2: All sulphides are insoluble. Exceptions: Na+,K+,NH4+,Mg+2,Ca+2,Ba+2 are soluble. Al+3,Cr+3 sulphides are hydrolysed and precipitated as hydroxides.

Common Ion Effect: The concentration of the other uncommon ion decreases when an aqueous solutions of strong electrolyte is added to a solution of weak electrolyte or sparingly soluble salt decrease. i.e. the solubility of sparingly soluble salt decreases. This phenomenon is called common ion effect. The dil.HCl is added to a solution before adding H2S water in the qualitative analysis of second group ions. The solubility of S-2 ions decreases. H2S(aq) + H2O(l) H3O+(aq)+S-2(aq) HCl(aq) +H2O(l) H3O+ (aq)+Cl-(aq) and so, the metal sulphides whose solubility product are less will only be precipitated in the second group. NH4Cl is added to a solution before adding NH4OH in the qualitative analysis of III-A group ions. The solubility of OH- ions decreases. NH4Cl(aq) NH4+((aq) +Cl-(aq) NH4OH(aq) NH4+ (aq)+OH-(aq) The solubility of OH- ions decrease and so, the metal hydroxide whose solubility products are less will only be precipitated in the III-A group

Flame test Apple green flame(Ba+2) Pink colour flame(K+)
Brick red flame(Ca+2) Apple green flame(Ba+2) Golden yellow flame (Na+) Crimson red flame(Sr+2)

Borax bead test: Na2B4O7 2NaBO2 + B2O3
CuO + B2O Cu(BO2)2 (Copper(II) meta borate) CuO+NaBO NaCuBO3 (ortho borate) In the reducing flame ortho borate probably occurs in the presence of carbon. Two reaction may take place. The colored Cu(II) is reduced to colorless Cu(I) meta borate 2 Cu(BO2) 2 + 2NaBO2+ C CuBO2 + Na2B4O7 +CO The Copper(II) borate is reduced to metallic copper, so that bead appear red and opaque. 2 Cu(BO2) 2+ 4NaBO2+ 2C Cu+ 2Na2B4O7 +2CO

Borax bead test

Dry Test For Negative Radicals
Mixture + dil.HCl Speedy effervesence of CO2 gas Lime Water turns milky (CrO4-2) Yellow color changes to orange 2CrO H Cr2O7-2 + H2O CO H+ CO2+H2O (Limewater) Ca(OH) 2+CO CaCO3 + H2O (Baryta water )Ba(OH) 2+CO BaCO3 + H2O CaCO3+excess CO2+H2O Ca(HCO3) 2(soluble)

Mix +dil.HCl Test for NO2- , NO3- Test for SO3-2 Starch iodide paper
turns blue black Without heating reddish brown gas evolved -Effervesence of SO2 gas which turns acidic K2Cr2O7 Paper green -White ppts of PbCl2 obtained, O.S is prepared in dil.HNO3 Sodium bisulphite + K2Cr2O7 Soln + Con. HCl Green Coloration 2NO HCl NO2 + H2 +2Cl- 2NO2 + 2KI I2 + 2KNO2 I2 + starch Dark blue color

Mixture + conc. HCl Evolution of H2S gas
Test for sulphide ( S-2 ) Shining Black paper PbAC2 + H2S PbS + CH3COOH shining black ppts Colorless gas evolved which turns PbAc2 paper shining black

Mixture + MnO2 powder +conc.H2So4 Heat it
Colorless gas which gives white fumes with NH4OH ( on glass rod ) Brown gas for Br- Violet gas For I- Br- +H2SO HBr +HSO4- MnO2 + 4HBr MnBr2 + Br2 + 2H2O I- + H2SO HSO4- + HI H2SO4 + 2HI I2 + SO3 + 2H2O 2I- + MnO2 + H2SO I2 + Mn H2O + 2SO4-2 Cl- + H2SO HSO4- + HCl NH3 + HCl NH4Cl (White Fumes) Mno2 + 4HCl MnCl2 + Cl2 +2H2O

Mixture + Cu-Foil + conc.H2So4 evolution of brown gas
Brown gas which turns starch iodide paper black Brown gas which turns moist starch paper Yellow 2NO3- + Cu + 4H Cu NO2 + 2H2O 2NO2- + Cu + 4H Cu NO2 + 2H2 2NO2 + 2KI I2 + 2KNO2 I2 + starch Dark blue color NO3- (aq)+3Fe+2+4H+(aq) Fe+3+2H2O+NO(g) Fe+2 (aq) +NO(g) Fe(NO)] +2(aq)

Test for borate (BO3-3) ion
Mixture + conc.H2SO4+ C2H5OH burns with green flame of ethyl borate Boil & ignite

Mixture + conc.HNO3 (heat it) + (NH4)2MoO4 canary Yellow ppts
Ba3 (PO4)2 + 6HNO H3PO4 + 3Ba(NO3) 2 H3PO4 (aq) (NH4) 2MoO4 (aq)+21HNO3 (l) (NH4)3[PO4Mo12O36](s)+21NH4NO3+12H2O Ammonium phospho molybladate

Ksp Value at 25C 1st Group PbCl2 : 1.70 x 10-5 AgCl : 1.7 x 10-10
Hg2Cl2 : 1.43 x 10-18 4th Group : BaCO3 : 2.58 x 10-9 BaCrO4 : 1.17 x 10-10 BaSO4 : 1.08 x 10-10 BaSO3 : 5 x 10-10 SrCO3 : 5.60 x 10-10 CaCO3 : 3.36 x 10-9 2nd Group: 2nd A HgS :- 2 x 10-53 PbS : 3 x 10-28 PbCrO4 : 3 x 10-13 Bi2S3 : 1.6 x 10-72 CuS : 8 x 10-37 CdS : 1 x 10-27 3rd Group 3rd A Fe(OH)2 : 5.0 x 10-18 Fe(OH)3 : 6.0 x 10-38 Cr(OH)3 : 3 x 10-21 Al(OH)3 : 1.8 x 10-33 2nd B As2S3 : 4.4 x 10-27 Sb2S3 : 1.7 x 10-93 SnS : 1 x 10-25 3rd B CoS : 5 x 10-22 NiS : 4 x 10-20 MnS : 3 x 10-11 ZnS : 2 x 10-25 FeS : 5 x 10-18 5th Group Mg3(PO4)2 : 1.04 x 10-24 MgNH4PO4 : 3 x 10-13

Precipitates Molecular Weight Ksp at 250 C Solubility Mol/Lit Solubility Gm/Lit PbCl2 1.70 x 10-5 5.0004 PbS 239.26 3.0 X 10-28 X 10-14 X 10-12 Bi2S3 514.02 1.60 x 10-72 x 10-15 x 10-13 CuS 95.60 8.0 X 10-37 X 10-19 X 10-17 CdS 144.26 1.0 x 10-27 x 10-14 x 10-12 AS2S3 4.4 x 10-27 x 10-6 x 10-4 Sb2S3 339.56 1.7 x 10-93 x 10-19 x 10-17 Fe[OH]2 89.861 5.0 X 10-18 X 10-6 X 10-4 Fe[OH]3 6.0 X 10-38 X 10-10 X 10-8 Al[OH]3 78.003 1.8 X 10-33 X 10-9 X 10-7

Precipitates Molecular Weight Ksp at 250 C Solubility Mol/Lit Solubility Gm/Lit Cr[OH]3 3.0 x 10-21 x 10-6 x 10-4 CoS 90.993 5.0 x 10-22 x 10-11 2.0346x 10-9 NiS 90.77 4.0 x 10-20 2.0 x 10-10 x 10-8 MnS 86.998 3.0 x 10-11 x 10-6 x 10-4 ZnS 97.44 2.0 x 10-25 x 10-13 x 10-11 BaCO3 2.58 x 10-9 x 10-5 SrCO3 5.6 x 10-10 x 10-5 x 10-3 CaCO3 3.36 X 10-9 X 10-5 X 10-3 Mg[NH4]PO4 3.0 X 10-13 X 10-5 X 10-8

Preparation of water extract(W.E.) :
Boil & Filter 1.0 gm mixture + 1T.T. distillied water Filtrate is W.E. Water insoluble carbonates, phoshphates and sulphides do not give test with W.E. perform 5th -B group test for NH4+ , Na+ , K+ with W.E. W.E. + AgNO3 Insoluble Yellow precipitate in dil.HNO3 Cl- , Br- , I- Present White precipitate soluble in dil.HNO3 , with Co2 effervesence CO3-2 Present Yellow precipitate soluble in dil.HNO3 Po4-3 Present Chocolate colored precipitate soluble in dil.HNO3 & NH4 OH Cro4-2 Present

soluble in dil.HNO3 with Co2 effervesence
W.E. + Ba(NO3)2 White precipitate Insoluble in dil.HNO3 So 4-2 is Present White precipitate soluble in dil.HNO3 So4-2 , Po4-3 , Bo3-3 Present Yellow precipitate soluble in dil.HNO3 Cro4-2 Present White precipitate soluble in dil.HNO3 with Co2 effervesence CO3-2 Present

Preparation of original solution(O.S):
-Why oxidizing negative radicals is removed while preparing O.S ? If nitrite , nitrate ions is present in O.S. while passing H2S gas oxidizing sulphide ion of H2S to free sulphur. Off-white precipitate of sulphur is obtained, indicating second group is absent. Second group radicals gives black, yellow and orange colour precipitates. SO HCl SO3 +H2 + 2Cl- , S HCl S + H2 +2Cl- 2NO2- +2HCl NO2 + H2 + 2Cl- If sulphide ion is not removed from O.S. then during precipitation of (III)-A group,(III)-B group cations get precipitated. If borate ion is not removed from O.S. then further group cations get precipitated in (III)-A group. If chromate ion is not removed from O.S. then cation like Ba+2 get precipitated in (III)-A group. Hence chromate ion is reduced to Cr+3 by evaporated to dryness with conc. HCl. If sulphite ion is not removed from O.S. then (IV)th group cations precipitated as insoluble sulphite in (III)-A group

Separation Of Positive Radicals in to Group
O.S. + dil. HCl White Residue of 1st Group AgCl,HgCl2,PbCl2 (White ppts) Heavy Residue of PbCl2

Explanation for Lead coming in group 1 as well as group 2.
Lead is precipitated in group 1 as PbCl2. Since it is partly soluble in dil.HCl, a part of it goes into the filtrate which is to be tested for group 2 radicals. Thus, on passing H2S gas, it give a black ppts of PbS. PbCl2 + H2S PbS + 2HCl Explanation for Addition of HCl before proceeding to test Group 2 radicals : HCl is added to suppress the ionization of H2S. Thus, the lesser concentration of S2- ions is available which is just sufficient to precipitate group 2 radicals as their sulphides because their solubility products are low, on the other hand , the solubility products of 3-B group radicals are very high and therefore they are not precipitated at all.

Analysis of 2nd group (pH <0.5)
2nd A Group HgS :- black colour PbS :- black colour Bi2S3 :- black brown colour CuS:- black colour CdS :- yellow colour 2nd B Group As2S3,As2S5 :- yellow colour Sb2S3,Sb2S5 :- orange colour SnS2 :- yellow colour SnS :- brown colour HGS BAKI

O.S./Fil.+ dil.HCl warm & pass H2S gas colored precipitate of 2nd group
Filtrate of 2nd group use for 3-A group analysis, boil off H2S Residue of 2nd group

II-group ppts ppts. of II group+15ml yellow ammonium sulphide + dil
II-group ppts ppts. of II group+15ml yellow ammonium sulphide + dil. NaOH Heat 60C & filter Residue may contain II-A group Filtrate Fil.+ dil.HCl colored ppts II-B group present Separation of II-A Group Ppts of II-A group+10 ml 50% HNO3 boil & filter Blue soln.+ excess of dil.H2SO4 +1 gm iron fillings Warm& filter Colorless filtrate+NH4OH(till alkaline) Pass H2S gas Yellow ppts. of Cd+2 (CdS) Filtrate+dil.H2SO4 White ppts of Pb+2 (PbSO4) Black residue -Hg+2 is confirmed, HgS Filtrate + NH4OH(excess) White ppts -Bi+3 (Bi(OH)3) Filtrate may contain Cu+2 & Cd+2. If solution (filtrate) is blue colour -Cu+2present

Separation of II-B group
ppts of II-B group+10ml conc. HCl boil till H2S is evolved completely, filter it Filtrate: Boil off H2S gas+ NH4OH (till alkaline)+ oxalic acid boil & pass H2S gas filter it Residue yellow ppts -As+3(As2S3) orange ppts. -Sb+3 (Sb2S3) Filtrate: Boil of H2S gas +Zn dust + dil.HCl (keep it for four mins.) Fil. + HgCl white ppts (turn grey in excess HgCl2) -Sn+2 is confirmed.

2nd A Group Analysis Hg +2 Pb+2 Bi+3 Cd+2 Cu+2
2nd A group radicals possible mixture Pbs + Bi2S Pbs + Bi2S3 + Cus Pbs + Cds Pbs + Bi2S3 + Cds Bi2S3 + Cus Bi2S3 + CuS + CdS Bi2S3 + Cds PbS + CuS + CdS Pbs + Cus Cus + Cds Mixture of two cations precipitates Mixture of three cations precipitates

2nd A group Precipitate Hgs Pbs Bi2s3 Cus Cds Turmeric yellow
BLACK BROWN Turmeric yellow Black colour Shining Black Black Brown Black colour

Pbs Bi2S3 Pbs+Bi2s3

PbS CdS PbS + CdS

Bi2S3 CuS Bi2S3 + CuS

Bi2S3 CdS Bi2S3 + CdS

PbS CuS PbS + CuS

CuS CdS Cus + CdS

Pbs + Bi2S3 + CuS Bi2S3 CuS PbS

PbS Bi2S3 CdS PbS + Bi2S 3 + CdS

Bi2S3 PbS CdS Bi2S3 + PbS + CdS

CuS CdS CuS + CdS+ PbS PbS

Comparison of Cus & PbS PbS CuS Shining Black Pure Black

O.S./Fil. + dil.HCl +dist.water warm it pass H2S gas
Colored ppts precipitates of 2nd group Radicals

Precipitation of 2nd A group
O.S./Fil. + dil. HCl + pass H2S gas Residue of ( Hgs + PbS + Bi2S3 + CuS + CdS) Residue of 2nd –A group Residue Filter it Filtrate of 2nd –A group

Filtrate of 2nd A group add dil.HCl
Yellow , orange precipitates 2nd B group present AS2S3 + Sb2S3 precipitates

ppts. of 2nd A group + conc. HNO3
Analysis of 2nd –A group ppts. of 2nd A group + conc. HNO3 Residue Boil Filter It Filtrate Insoluble residue of HgS Dissolve all residue in conc. HNO3 Filtrate:- PbS Bi2S3 CuS CdS Soluble Residue of Hgs

C.T. For Pb+2 Above Fil. + K2CrO4 Yellow Ppts PbCrO4 Fil + dil.H2SO4
WHITE RESIDUE OF PbSO4 Filtrate of 2nd –A group

C.T for Bi+3 :- Residue dissolve in dil HCl Filtrate Add KI soln
Fil + NH4OH Brown Ppts. Residue of Bi(OH)3

Blue Fil. +dil. H2SO4 + Iron(Fe) Fillings Warm & Filter Colorless Soln
colorless Soln + NH4OH pass H2S gas ,Yellow ppts of CdS Blue filtrate of Cu+2 & Cd+2 soluble as Cu(NH3) 4SO4 & Cd(NH3) 4SO4

Iron fillings

Cu+2 (aq) +[Fe(CN)6]-4(aq) Cu2[Fe(CN)6](s)
Filtrate [Cu+2] + dil.potassium ferrocyanide K4Fe(CN) Reddish brown ppts. Cu+2 (aq) +[Fe(CN)6]-4(aq) Cu2[Fe(CN)6](s) CuSO4.5H2O Cu(OH)2 [Cu(NH3) 4] +2 Cu NH4OH [Cu(NH3) 4] H2O Tetramine copper(II) ion(blue colour solution)

2nd B group radicals Orange Dark Yellow As2S3 Sb2S3 SnS2 Sns
( Light Yellow) (Brown)

As2S3 Sb2S3 As2S3 + Sb2S3

Analysis of 2nd B group Boil & FIltrate Fil Residue of 2nd B group
(As2S3 + Sb2S3) Residue of As2S3 + Sb2S3 Dissolve in con. HCl Removale of H2S gas

Fil+NH4OH(till alkaline) + Oxalic Acid
Residue + conc.HNO3(Boil) + (NH4)2MoO Yellow Ppts of AsO3-3 & AsO4-3 Insoluble Yellow Residue of (As2S3) & Filtrate contain soluble (SbCl3) Fil+NH4OH(till alkaline) + Oxalic Acid +d.w.(Boil it) + pass H2S gas Sb2S3 orange ppts. . Filtrate contain Soluble (SbCl3)

Analysis of 3rd –A group 3 Fe+2 + 2[Fe(CN)6]-3 Fe3[Fe(CN)6]2 dark blue
Filtrate of 2rd group Filtrate of (II) group(Fe+3 )+ KCNS Blood Red Colouration Filtrate of (II) group(Fe+3 )+ K4Fe(CN) Blue Ppts 3 Fe+2 + 2[Fe(CN)6] Fe3[Fe(CN)6]2 dark blue Iron(II) hexacyno ferrate(III) 4Fe+3 + 3[Fe(CN)6] Fe4[Fe(CN)6]3 prussion blue Iron(III) hexacyno ferrate(II)

Filtrate[Fe+2]+ K4Fe(CN) 6 whitish blue ppts
2Fe+2 (aq)+ [Fe(CN) 6]-4 (aq) Fe2 [Fe(CN)6](s) Filtrate [Fe+2] + dil.potassium ferricyanide K3Fe(CN) Dark blue ppts Fe+2(aq)+ 2[Fe(CN)6] Fe3[Fe(CN)6]2(s) Filtrate [Fe+3]+ K4Fe(CN) Dark blue ppts 4Fe+3(aq)+ 3[Fe(CN)6] Fe4[Fe(CN)6](s) (ferric ferro cynide) Filtrate[Fe+3] + potassium thiocynate (KCNS) Blood red colour Fe+3 (aq)+SCN [Fe(SCN)] +2(aq)

Precipitation of Fe+3 not Fe+2 in 3rd group of hydroxide :
If Fe+2 ions are present in the electrolyte mixture under analysis, they must be oxidized to Fe+3 ions(with HNO3) prior to the precipitation of the group 3rd hydroxide. This is because Ksp for Fe(OH)2 is 5 x compare to Ksp for Fe(OH)3 is 6 x The [OH-] produced on addition of NH4OH and NH4Cl is there for insufficient to reach that required by the solubility product for Fe(OH)2. Fe+2 ions are not there for precipitated in group 3-A. but appear in group 3-B as FeS, for which Ksp is 5 x comparable with the solubility products of the group 3-B sulphides . FeS will then interfere with the further analysis of these sulphides.

Role of Addition of Nitric Acid before the prepetition of Group 3 :
3FesO4 + 4HNO Fe2(SO4)3 + NO + 2H2O The oxidation of ferrous ion to ferric ion is essential due to the following reasons : Fe+2 is not completely precipitated as Fe(OH)2 in presence of NH4Cl by addition of NH4OH whereas Fe+3 is completely precipitated as Fe(OH)3 under the same conditions. Further Fe(OH)2 is oxidized slowly to ferric hydroxide. Instead of getting a precipitate of a define colour, the precipitate is differently colored. Moreover, Fe(OH)2 is green in colour like Cr(OH)3 whereas Fe(OH)3 helps to distinguish it from Cr(OH)3.

Role of NH4Cl in 3-A group:
Ammonium hydroxide is a weak base and ionizes only to a small extent. NH4OH NH4+ + OH- In presence of highly ionized ammonium chloride which gives a large amount of NH4+ ions, the ionization of NH4OH is further suppressed due to common ion effect, so that concentration of OH- ions falls considerably low. Under such conditions, the solubility products of the hydroxide of 3rd –A group cations, are reached while those of 3-B , 4 and 5-A group cations, which possess comparatively high value of solubility products, are prevented from precipitating in 3rd –A group.

Why insoluble phosphate is removed before analysis of 3rd-A group ?
If insoluble phosphate is present in a mixture, O.S is prepared in dil. HCl. O.S. is acidic in nature, while precipitation of 3rd –A group is carried out with NH4Cl and NH4OH, acidic O.S. is first neutralize and then made alkaline. Under such conditions phosphate of 3rd-A ,3rd-B ,4 and 5th-A group cations get precipitated in 3rd-A group.

PHOSHPHATE removal scheme
O.S/Fil. after 2nd gr.(Boil of H2S completely) drops conc. HNO3 + 1 gm NH4Cl(s) + ½ T.T Zirconium nitrate Reagent (heat & Filter) Poshphate Free Filtrate Add Zr(NO3)2 Remove for Zr3(PO4)2 Residue for Zr3(PO4)2

Analysis of 3rd (A) Group pH : 9.0
Al(OH)3 :- white Fe(OH)3 :- red – brown Cr(OH)3 :- gray-green Mn(OH)2 :- white-brown

Analysis of III A group Mn+2,Fe+3,Cr+3,Al+3 ppts of III A group+ 2g Na2O2 (or NaOH + H2O2)in evaporating dish and boil & filter Precipitate: If reddish brown-Fe+3 is present and if black Fe+3 or Mn+2may be present. Divided in to two parts. Filtrate: If colourless, Al+3 may be present and if yellow Cr+3 and Al+3 may also be present . Divide filtrate into two parts. (i).Dissolve the ppts. in dil.HCl and prepare solution. Soln.+K4[Fe(CN)6] dark blue ppts. or colouration. -Fe+3 is confirmed (ii)Soln.+ KCNS blood red colourisation (iii)Soln.+ NaOH Brown ppts. (ii) ppts+conc.HNO3 + Pb3 O4 or PbO2 boil & cool purple colour -Mn+2 is confirmed. Dissolve ppts in dil.HNO3 and use for test. Soln.+K4 [Fe(CN) 6] pink ppts. Soln. + NaOH pink ppts turns black when exposed to air. Soln.+ sodium bismuthate shake purple or pinkish colour Filtrate+NH4Cl & boil the soln for few minute white gelatinous ppts. -Al+3 is confirmed Filtrate + dil.HCl till acidic & prepare Soln. (i) Soln.+Na2HPO4 white ppts. Soluble in HCl (ii) Soln. + NH4OH white gelatinous ppts. Fil+ CH3 COOH +lead acetate yellow ppts. Fil+dil H2SO4+ Ether +H2O2 (few drops) Blue colour in upper layer Fil+CH3 COOH+BaCl yellow ppts.

Fe(OH)3 Cr(OH)3 Al(OH)3 Fe(OH)3 + Cr(OH)3 + Al(OH)3 Red Brown
Grey Green White Fe(OH) Cr(OH) Al(OH)3 Fe(OH)3 + Cr(OH)3 + Al(OH)3

Fe(OH)3 + Cr(OH)3 Fe(OH)3 + Al(OH)3 Cr(OH)3 + Al(OH)3

Ppts of 3rd A group + H2O2 + NaOH Boil & Filter
Residue of Fe(OH)3 Reddish Brown ppts Residue + dil.HCl clear Soln New slide phosphate upper Solution of Fe+3 + KCNS red Blood coloration Solution of Fe+3 + K4Fe(CN) Blue Ppts

Filtrate of Cr+3 & Al+3 If filtrate is Yellow Cr+3 present.
Filtrate is colorless Al+3 present. C.T For Cr+3 :- Fil + HAc + PbAc Yellow Ppts

Filtrate of Cr+3 & Al+3 If filtrate is Yellow Cr+3 present.
Fil + dil.HCl + NH4(OH) (Boil) Filtrate of Cr+3 & Al+3 If filtrate is Yellow Cr+3 present. Filtrate is Colorless Al+3 present. Gelatinous White Residue of Al(OH)3

3rd B Group Analysis O.S./Fil. + NH4Cl + NH4OH (excess) Pass H2S(g)
Co+2 , Ni+2 , Mn+2 , Zn+2 O.S./Fil. + NH4Cl + NH4OH (excess) Boil it Pass H2S(g) CoS Black NiS Black MnS Pink ZnS White

Analysis of III B group Co+2,Ni+2,Mn+2,Zn+2 ppts of IIIB group + dil
Analysis of III B group Co+2,Ni+2,Mn+2,Zn+2 ppts of IIIB group + dil.HCl warm & filter Black ppts: Dissolve ppts. In aquaregia and evaporate to dryness. If residue has Blue or Bluish green colour Co+2is present .Dissolve residue in water & divide two parts. Filtrate: Boil off H2S + NaOH in excess boil & Filter Soln.+NH4Cl+NH4OH till basic + K3[Fe(CN)6] Red colour or ppts. Co+2 is confirmed Soln.+ glacial CH3COOH +KNO yellow ppts (ii) Soln.+NaOH Blue ppts. (iii) Borax bead test Blue bead Soln.+ NH4OH till basic +D.M.G Brilliant red ppts. -Ni+2 is confirmed (i)Soln.+NaOH+Br2 water black ppts. (ii) Borax bead test: reddish brown bead. Precipitate: ppts. +conc. HNO3+PbO2 boil &cool Purple colour -Mn+2 is confirmed (ii) ppts+HNO3+sodium bismuthate shake pink colour in clear liquid (iii) ppts + Br2 water+ NaOH soln. Boil black ppts. Filtrate: Fil+CH3COOH+H2S(gas) dirty white ppts. -Zn+2 is confirmed (ii)Fil+ dil.HCl +K4 [Fe(CN) 6] white ppts-Zn+2 is confirmed

CoS + NiS + MnS + ZnS CoS + NiS NiS + MnS MnS + ZnS CoS +MnS

Comparision of MnS & (MnS + ZnS) Precipitates
NiS + ZnS CoS + ZnS Comparision of MnS & (MnS + ZnS) Precipitates Nis and zns

MnS + ZnS + NiS CoS + NiS +MnS CoS + NiS + ZnS CoS + MnS + ZnS

Ppts of 3rd B group + dil.HCl
Aquaregia solution = 3:1 (HCl +HNO3) Boil & evaporate to dryness add dilute HCl 3 ml. Warm & filter Dissolved in aquaregia soln residue Black Ppts of CoS and NiS C.T for Co+2: Soln + NaOH Blue Ppts filtrate For Mn+2 & Zn+2 colourless Soln +NH4Cl(s)+NH4OH+K3Fe(CN) Red Colour or Ppts

Filtrate of Mn+2and Zn+2 boil off H2S and add dil. NaOH,filter it.
C.T for Ni+2 Filtrate of Mn+2and Zn+2 boil off H2S and add dil. NaOH,filter it. Soln + NH4OH(till basic) + D.M.G Brilliant red ppts Brown ppts. of Mn+2 C.T for Mn+2 fi;ltrate Fil. + dil. HCl +K4[Fe(CN)6] White ppts. { Zn2[Fe(CN)6]} Brown ppts.(Mn+2)+ conc.HNO3 + PbO purple coloration Mn+2 (aq)+5PbO2 (s)+H2O(l) HMnO4 (aq) + 5Pb+2 (aq)+3O-2 (aq)

Analysis of 3-B group pH-9.0
ZnS :- white CoS :- black NiS :- black MnS :- pink

Separation of group 5-A from 4
group 4 radicals are precipitated as insoluble carbonates in ammonical solution containing ammonium chloride. The carbonates of Ba+2 , Sr+2 and Ca+2 are precipitated while MgCO3 does not. NH4Cl NH4+ + Cl- (NH4)2CO NH4+ + CO3-2 Due to common ion effect , ammonium chloride suppresses the ionization of ammonium carbonates. The low concentration of CO3-2 ions is sufficient to exceed the solubility products of the carbonates of calcium, strontium and barium, while magnesium remains in solution, because the solubility products of magnesium carbonates is comparatively high.

Separation of IV th group Ba+2, Sr+2,Ca+2 Dissolve ppts of IV group in dilute CH3COOH First check Ba+2 as follows: Soln of ppts in one test tube + K2CrO4 drop by drop. If yellow ppts obtained add more K2CrO4 to whole bluk warm & filter (appeareance of yellow ppts.shows presence of Ba+2) Yellow ppts: Dissolve ppts. In dil. HCl and prepare soln. Soln.+H2SO white ppts.-Ba+2 is confirmed Soln.+CaSO white ppts (at once) Soln.+ flame test green flame Filtrate: Fil.+NH4 OH+(NH4)2SO4 (solid)excess warm and Filter White ppts: -Sr+2 is confirm Make paste of white ppts + Conc. HCl & apply this paste to platinum wire & do flame test : crimson red flame O.S + CaSO white ppts (after scratching) Filtrate: (i) Fil+ NH4OH+(NH4)2C2O4 white ppts. – Ca+2is confirmed (ii) Fil+NH4OH+K4 [Fe(CN)6] yellow ppts (iii) Fil+ Conc. H2SO white ppts (iv) Fil—Flame test-Brick red flame

Ppts. of 4th group radicals
Fil + NH4Cl(s) + NH4OH + (NH4)2CO white ppts 4th group radicals BaCO3 SrCO3 CaCO3 Dissolve the ppts. in Hot diluted acetic acid &add K2CrO4 Soln Yellow ppts. of BaCrO4 Ksp value : 1.17 x 10-10

(White ppts of calcuim oxalate) White Residue SrSO4
Filtrate of Sr+2 and Ca+2 Add NH4OH +(NH4)2SO4(S) Warm & Filter Filter yellow ppts Colourless pic Fil + NH4OH + (NH4)2C2O4 White ppts of calcuim oxalate (White ppts of calcuim oxalate) White Residue SrSO4

Why is Mg+2 not precipitated in group 4 ?
Magnesium is not precipitated as MgCO3 in 4 group because the concentration of CO3-2 ions is such that the solubility products of the MgCO3 is not reached, while Ca(II), Sr(II) and Ba(II) can be easily precipitated with such a low concentration of CO32- ions .

Analysis of group V-A Mg+2 O. S. /Fil
Analysis of group V-A Mg+2 O.S./Fil. + NH4Cl +NH4OH +Na2HPO White precipitates White ppts of V-A group indicates presence of Mg+2 Dissolved ppts of V-A group in dil. HCl and test as follows. (I) solution +NaOH+I2 (till yellow colour) red ppts or colour (II) solution + NaOH white ppts (III) solution +Na2CO white ppts (IV) solution + 8-hydroxy quinoline +NaOH yellow ppts

Residue of Mg+2 as MgNH4PO4
Filtrate [Mg+2] +NaOH+I2 (NaOI) sodium hypoiodide Reddish brown ppts Mg+2(aq)+2IO-(aq) Mg(IO) 2 (s) Soln +NaOH + I2 Soln Reddish brown ppts or colour Residue of Mg+2 as MgNH4PO4 Water pic Disslove the residue of Mg+2 in dil.HCl

Analysis of V-B group : Na+,K+,NH4+
Note: Test of NH4+ should be performed before group I in the beginning as follows: W.E +NaOH heat the solution, then add Nessler’s Reagent Brown ppt. -NH4+ is present. -Filtrate of Group-V-A: Filtrate may contain Na+ and K+ Evaporate this filtrate to dryness in the presence of a few drops of conc. HNO3 continue heating till white fumes of ammonia ceases. No residue Na+ and K+ absent -The residue shows the presence of Na+ and /or K+. keep the portion of the residue from flame test while dissolve the remaining residue in the small quantity of distilled water and perform the wet test for Na+ and K+ by dividing in two parts. Part-I for Na+ Confirm To a portion add magnesium uranyl acetate solution Yellow crystalline ppts. To another portion add potasium pyroantimonate K2H2Sb2o7 and starch the side of the tube with glass rod white ppts. or milkyness (iii) Flame test: Perform the flame test with the residue (kept reserved) A golden yellow flame observed through blue glass Part-II for K+ (i)To a portion add sodium cobaltinitrite sol yellow ppts. (ii) To another portion add picric acid solution Yellow ppts. (iii) To the third portion add saturated sol. of tartaric acid and a little alcohol White ppts.. (iv) Perform the flame test with residue violet flame Pink through blue glass -K+ present

5th B Group Analysis Nesslers Reagent :
NH4+ K+ Na+ Test For NH4+ :- W.E + NaOH Smell of NH3 gas & turns turmeric paper Red With dil NaOH turmeric paper turns Red Nesslers Reagent : HgCl2 + KI Orange Ppts Dissolve it in KI solution then add dil. NaOH NH3 gas, turns moist turmeric paper red

W. E. + Nesslers Reagent Brown Ppts
HgI2 + 2NH Hg NH4I K2HgI HgI KI NH2 I

NH3 free W.E + Na[Co(NO2)3] Yellow ppts
NH3 free W.E. + Picric Acid Yellow Ppts. (Shake Well) C.T. for K+ W.E. + dil. NaOH boil till complete removal of NH3 gas C.T. for Na+ W.E. + K2H2Sb2O White Ppts

W.E (NH4+) + Picric acid yellow ppts
NH4+ (aq)+ C6H2 (NO3) 2OH(aq) C6H2 (NO2) 3ONH4 (s) +H+ (aq) W.E (K+) + Picric acid yellow ppts K+ (aq)+ C6H2 (NO3) 3OH(aq) C6H2 (NO2) 3OK(s) +H+ (aq) 3K+ +[Co(NO2)6]-3(aq) K3 [Co(NO2) 6](s) yellow ppts. Na+ + K2H2Sb2O white ppts. 2Na+ (aq) + K2H2Sb2O7(aq) Na2H2Sb2O7 (s) + 2K+ (aq)

self Self assembled H2S gas generation plant with six nozzles

Guided by Dr . Nitin B. Patel
Worked By Solanki yashpal (research student) Solanki sandip Solanki praful Rana Jaimin Patanvadiya Hitesh Tailor krunal Akash Thakor Sheth EZAZ VALI

Acknowledgement We are very thankful to Principal, Shree Jayendrapuri Arts & Science college, Bharuch, head of the chemistry department ,all teaching and nonteaching staff members and students to carried out work in the laboratory and for computer facilities.

Thank you for patience

Inorganic Qualitative Analysis

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