Shree Jayendrapuri Arts & Science College, Bharuch Inorganic Qualitative Analysis Presented by:Dr.Nitinkumar B. Patel Associate Professor,Dept. of ChemistryID: web : 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). Key Concepts Solubility Product, K sp refers to an ionic compound dissolving to form ions For the reaction: MA(s) M + (aq) + A - (aq) The equilibrium constant, K is: equilibrium constant, K Since the concentration of the solid is a constant, it is effectively incorporated into the equilibrium constant: K sp = [M + ][A - ] [M + ][A - ] is referred to as the ion product At equilibrium, the ion product = solubility product (K sp ) 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 - ] > K sp [M+] [A-] ________ [MA] K = ConditionNature of SolutionPossibility of Precipitation ion product < K sp unsaturated solutionno precipitation occurs ion product = K sp saturated solution saturated solution at equilibrium ion product > K sp supersaturated solutionprecipitation occurs Example : Calculating the solubility of an ionic compound (MA). Calculate how much silver bromide will dissolve in 1 L of water given K sp = 5.0 x at 25 o C Write the equation for dissolving AgBr in water: AgBr(s) Ag + (aq) + Br - (aq) Write the equilibrium expression: K sp = [Ag + ][Br - ] = 5.0 x 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 = [x] 2 Solve for x: x = 5.0 x = 7 x mol L -1 So, [Ag + ] = [Br - ] = 7 x mol L -1 The solubility of AgBr at 25 o C is 7 x mol L -1 7 x moles AgBr will dissolve in 1L of water at 25 o C. Example : Calculating the solubility of an ionic compound (MA 2 ). Calculate how much strontium fluoride will dissolve in 1 L of water given K sp = 2.5 x at 25 o C. Write the equation for dissolving SrF 2 in water: SrF 2 (s) Sr 2+ (aq) + 2F - (aq) Write the equilibrium expression: K sp = [Sr 2+ ][F - ] 2 = 2.5 x Determine the relative concentrations of each ion: at equilibrium [Sr 2+ ] = x and [F - ] = 2x (from the balanced chemical equation) Substitute these vales into the equilibrium expression: 2.5 x = [x][2x] 2 = 4x 3 Solve for x: 2.5 x 4 = x 3 = 6.25 x x = 3 6.25 x = 8.5 x mol L -1 The solubility of SrF 2 at 25 o C is 8.5 x mol L x moles of SrF 2 will dissolve in 1L of water at 25 o C. Example : Deciding whether a precipitate will form Will a precipitate form if 25.0 mL of 1.4 x mol L -1 NaI and 35.0 mL of 7.9 x mol L -1 AgNO 3 are mixed? (K sp for AgI at 25 o C is 8.5 x ) 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 - ] : c 1 = [I - ] before mixing = 1.4 x mol L -1 (assuming full dissociaiton of NaI) V 1 = initial volume = 25.0 mL = 25.0 x L V 2 = final volume after mixing = = 60.0 mL = 60.0 x L (assuming volumes are additive) c 2 = [I - ]after mixing=c 1 x V 1 V 2 = (1.4 x x 25.0 x )(60.0 x )=5.8x mol L -1 [Ag + ]: c 1 = [Ag + ] before mixing = 7.9 x mol L -1 (assuming full dissociaiton of AgNO 3 ) V 1 = initial volume = 35.0 mL = 35.0 x L V 2 = final volume after mixing = = 60.0 mL = 60.0 x L (assuming volumes are additive) c 2 = [Ag + ] after mixing = c 1 x V 1 V 2 = (7.9 x x 35.0 x ) (60.0 x ) = 4.6 x Calculate the ion product: ion product = [Ag + ][I - ] = 5.8 x x 4.6 x = 2.7 x Decide whether a precipitate forms: If ion product > K sp a precipitate will form If ion product K sp (8.5 x ) so a precipitate will form Ksp equation for different stoichometric inorganic substances : AgCl Ksp = [Ag + ] [Cl - ] = (S) x (S)=S 2 SrF 2 Ksp = [Sr +2 ] [F -2 ] = (S) x(2S) 2 =4S 3 Na 2 SO 4 2Na + +SO 4 -2 Ksp = (2S) 2 x (S) = 4S 3 Ksp = [Na + ] [SO 4 -2 ] = (2S) 2 x (S) = 4S 3 Ag 2 CrO 4 Ksp = [Ag + ] 2 [CrO 4 -2 ] = (2S) 2 x (S) =4S 3 Cr(OH) 3 Ksp = [S]x[3S] 3 = 27S 4 where, S = (Ksp /27) 1/4 Molar solubility Bi 2 S 3 Ksp = [2S] 2 x [3S] 3 = 4S 2 x 27S 3 = 108 S 5 Hg 2 I 2 Hg I -, Ksp = [S]x[2S] 2 = 4S 3 Ca 3 (PO 4 ) 2 Ksp = [Ca +2 ] 3 x[PO 4 -3 ] 2 = [3S] 3 x[2S] 2 = 108 S 5 NiS Ksp = [Ni +2 ][S -2 ] = (S) x (S)=S 2 CaCO 3 Ksp = [Ca +2 ] [CO 3 -2 ] = (S) x (S)=S 2 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). NH 4 +, 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).CO 3 -2,CrO 4 -2,PO 4 -3,SiO 4 -2 (silicates) are insoluble. Exception: NH 4 +,Na +,K + soluble,MgCrO 4 (6}.S -2 : All sulphides are insoluble. Exceptions: Na +,K +,NH 4 +,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 H 2 S water in the qualitative analysis of second group ions. The solubility of S -2 ions decreases. H 2 S(aq) + H 2 O(l) H 3 O + (aq)+S -2 (aq) HCl(aq) +H 2 O(l) H 3 O + (aq)+Cl - (aq) and so, the metal sulphides whose solubility product are less will only be precipitated in the second group. NH 4 Cl is added to a solution before adding NH 4 OH in the qualitative analysis of III-A group ions. The solubility of OH - ions decreases. NH 4 Cl(aq) NH 4 + ((aq) +Cl - (aq) NH 4 OH(aq) NH 4 + (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 ) Brick red flame(Ca +2 ) Pink colour flame(K + ) Crimson red flame(Sr +2 ) Golden yellow flame (Na + ) Borax bead test: Na 2 B 4 O 7 2NaBO 2 + B 2 O 3 CuO + B 2 O 3 Cu(BO 2 ) 2 (Copper(II) meta borate) CuO+NaBO 2 NaCuBO 3 (ortho borate) In the reducing flame ortho borate probably occurs in the presence of carbon. Two reaction may take place. 1.The colored Cu(II) is reduced to colorless Cu(I) meta borate 2 Cu(BO 2 ) 2 + 2NaBO 2 + C 2CuBO 2 + Na 2 B 4 O 7 +CO 2.The Copper(II) borate is reduced to metallic copper, so that bead appear red and opaque. 2 Cu(BO 2 ) 2 + 4NaBO 2 + 2C 2Cu+ 2Na 2 B 4 O 7 +2CO Borax bead test Lime Water turns milky Speedy effervesence of CO 2 gas (CrO 4 -2 ) Yellow color changes to orange 2CrO H + Cr 2 O H 2 O CO H+CO 2 +H 2 O (Limewater) Ca(OH) 2 +CO 2 CaCO 3 + H 2 O (Baryta water )Ba(OH) 2 +CO 2 BaCO 3 + H 2 O CaCO 3 +excess CO 2 +H 2 O Ca(HCO 3 ) 2 (soluble) Mix +dil.HCl -Effervesence of SO 2 gas which turns acidic K 2 Cr 2 O 7 Paper green -White ppts of PbCl 2 obtained, O.S is prepared in dil.HNO 3 Sodium bisulphite + K 2 Cr 2 O 7 Sol n + Con. HCl Green Coloration Without heating reddish brown gas evolved Starch iodide paper turns blue black Test for SO 3 -2 Test for NO 2 -, NO 3 - 2NO HCl 2NO 2 + H 2 +2Cl - 2NO 2 + 2KI I 2 + 2KNO 2 I 2 + starch Dark blue color Mixture + conc. HCl Evolution of H 2 S gas Shining Black paper Colorless gas evolved which turns PbAc 2 paper shining black Test for sulphide ( S -2 ) PbAC 2 + H 2 S PbS + CH 3 COOH shining black ppts Mixture + MnO 2 powder +conc.H 2 So 4 Heat it Colorless gas which gives white fumes with NH 4 OH ( on glass rod ) Brown gas for Br - Violet gas For I - Cl - + H 2 SO 4 HSO HCl NH 3 + HCl NH 4 Cl (White Fumes) Mno 2 + 4HCl MnCl 2 + Cl 2 +2H 2 O Br - +H 2 SO 4 HBr +HSO 4 - MnO 2 + 4HBr MnBr 2 + Br 2 + 2H 2 O I - + H 2 SO 4 HSO HI H 2 SO 4 + 2HI I 2 + SO 3 + 2H 2 O 2I - + MnO 2 + H 2 SO 4 I 2 + Mn +2 +2H 2 O + 2SO 4 -2 Brown Gas Brown gas which turns starch iodide paper black Brown gas which turns moist starch paper Yellow 2NO Cu + 4H + Cu NO 2 + 2H 2 O 2NO Cu + 4H + Cu NO 2 + 2H 2 2NO 2 + 2KI I 2 + 2KNO 2 I 2 + starch Dark blue color Mixture + Cu-Foil + conc.H 2 So 4 evolution of brown gas NO 3 - (aq)+3Fe +2 +4H + (aq) 3Fe +3 +2H 2 O+NO(g) Fe +2 (aq) +NO(g) Fe(NO)] +2 (aq) Test for borate (BO 3 -3 ) ion Mixture + conc.H 2 SO 4 + C 2 H 5 OH burns with green flame of ethyl borate Boil & ignite Mixture + conc.HNO 3 (heat it) + (NH 4 ) 2 MoO 4 canary Yellow ppts Ba 3 (PO 4 ) 2 + 6HNO 3 2H 3 PO 4 + 3Ba(NO 3 ) 2 H 3 PO 4 (aq) + 12(NH 4 ) 2 MoO 4 (aq)+21HNO 3 (l) (NH4) 3 [PO 4 Mo 12 O 36 ](s)+21NH 4 NO 3 +12H 2 O Ammonium phospho molybladate Ksp Value at 25 C 1 st Group PbCl 2 : 1.70 x AgCl : 1.7 x Hg 2 Cl 2 : 1.43 x nd Group: 2 nd A HgS :- 2 x PbS : 3 x PbCrO 4 : 3 x Bi 2 S 3 : 1.6 x CuS : 8 x CdS : 1 x rd Group 3 rd A Fe(OH) 2 : 5.0 x Fe(OH) 3 : 6.0 x Cr(OH) 3 : 3 x Al(OH) 3 : 1.8 x th Group : BaCO 3 : 2.58 x BaCrO 4 : 1.17 x BaSO 4 : 1.08 x BaSO 3 : 5 x SrCO 3 : 5.60 x CaCO 3 : 3.36 x th Group Mg 3 (PO 4 ) 2 : 1.04 x MgNH 4 PO 4 : 3 x rd B CoS : 5 x NiS : 4 x MnS : 3 x ZnS : 2 x FeS : 5 x nd B As 2 S 3 : 4.4 x Sb 2 S 3 : 1.7 x SnS : 1 x PrecipitatesMolecular WeightKsp at 25 0 CSolubility Mol/LitSolubility Gm/Lit PbCl x PbS X X X Bi 2 S x x x CuS X X X CdS x x x AS2S3AS2S x x x Sb 2 S x x x Fe[OH] X X X Fe[OH] X X X Al[OH] X X X 10-7 PrecipitatesMolecular WeightKsp at 25 0 CSolubility Mol/LitSolubility Gm/Lit Cr[OH] x x x CoS x x x NiS x x x MnS x x x ZnS x x x BaCO x x SrCO x x x CaCO X X X Mg[NH 4 ]PO X X X 10 -8 Preparation of water extract(W.E.) : 1.0 gm mixture + 1T.T. distillied water Filtrate is W.E. Boil & Filter Water insoluble carbonates, phoshphates and sulphides do not give test with W.E. perform 5 th -B group test for NH 4 +, Na +, K + with W.E. W.E. + AgNO 3 Insoluble Yellow precipitate in dil.HNO 3 Cl -, Br -, I - Present Yellow precipitate soluble in dil.HNO 3 Po 4 -3 Present Chocolate colored precipitate soluble in dil.HNO 3 & NH 4 OH Cro 4 -2 Present White precipitate soluble in dil.HNO 3, with Co 2 effervesence CO 3 -2 Present W.E. + Ba(NO 3 ) 2 White precipitate Insoluble in dil.HNO 3 So 4 -2 is Present White precipitate soluble in dil.HNO 3 So 4 -2, Po 4 -3, Bo 3 -3 Present Yellow precipitate soluble in dil.HNO 3 Cro 4 -2 Present White precipitate soluble in dil.HNO 3 with Co 2 effervesence CO 3 -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 H 2 S gas oxidizing sulphide ion of H 2 S 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 SO 3 +H 2 + 2Cl -, S HCl S + H 2 +2Cl - 2NO HCl 2NO 2 + H 2 + 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 1 st Group AgCl,HgCl 2,PbCl 2 (White ppts) Explanation for Lead coming in group 1 as well as group 2. Lead is precipitated in group 1 as PbCl 2. 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 H 2 S gas, it give a black ppts of PbS. PbCl 2 + H 2 S PbS + 2HCl Explanation for Addition of HCl before proceeding to test Group 2 radicals : HCl is added to suppress the ionization of H 2 S. Thus, the lesser concentration of S 2- 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 2 nd group (pH