Chemical Analysis PDF

60 932 Chemical Analysis Chapter E3 21 Chemical Analysis (iv) Tests for H2S : It is a colourless gas with a smell of rotten eggs. It turns moist lead acetate paper black. ID Analytical chemistry deals with qualitative and quantitative analysis of substances. (CH 3 COO )2 Pb  H 2 S 2CH 3 COOH  PbS...

60 932 Chemical Analysis Chapter E3 21 Chemical Analysis (iv) Tests for H2S : It is a colourless gas with a smell of rotten eggs. It turns moist lead acetate paper black. ID Analytical chemistry deals with qualitative and quantitative analysis of substances. (CH 3 COO )2 Pb  H 2 S 2CH 3 COOH  PbS Black D YG U Qualitative analysis : A salt consists of two parts known as radicals. The positively charged part of a salt (cation) which has been derived from a base is termed as basic radical and the negatively charged part of salt (anion) which has been derived from an acid is termed as acidic radical. In qualitative inorganic analysis, the given compound is analysed for the basic and acid radicals (i.e., the cations and the anions), that it contains. For example zinc blende is analysed for the Zn 2  and S 2  ions that it contains. Test for Different Gases (1) Colourless gases (v) Tests for SO2 : It is a colourless gas with a suffocating odour of burning sulphur. It turns acidified solution green. K 2 Cr2 O7 3 SO 2  K 2 Cr2 O7  H 2 SO 4 K 2 SO 4  Cr2 (SO 4 )3  H 2 O Green (vi) Tests for NH3 : It is a colourless gas with a characteristic ammonical smell. It gives white fumes of NH 4 Cl with HCl , NH 3  HCl NH 4 Cl. With Nessler’s White fumes reagents, it gives brown ppt.    NH 3 2 K 2 HgI 4 Nessler ' s reagent  KOH NH 2 HgOHgI  7 KI  2 H 2 O Iodine of Millon ' s base ( Brown ppt) CuSO 4  4 NH 3 Cu ( NH 3 )4 SO 4. Ca(OH ) 2  CO 2 CaCO 3   H 2 O water to give NH 4 OH , which being basic, turns red ST U (i) Tests for CO2 : It is colourless and odourless gas. It gives white ppt. with lime water which dissolves on passing excess of CO 2. Lime water White ppt. Excess So lub le (ii) Test for CO : It is colourless and odourless gas. It burns with a blue flame. 2CO  O 2 2CO 2 dissolves in ⇌ NH 4  OH . (vii) Tests for HCl gas : It is colourless gas with a pungent irritating smell. It turns moist blue litmus paper red i.e., it is acidic in nature. It gives white ppt. with AgNO 3 solution. This white ppt. is soluble in HCl  AgNO 3 AgCl  HNO 3 ; NH 4 OH. (iii) Test for O2 : It is colourless and odourless gas. It rekindles a glowing splinter. NH 3 Deep blue litmus blue, NH 3  H 2 O NH 4 OH CaCO 3  CO 2  H 2 O Ca(HCO 3 ) 2 White ppt. It gives deep blue colour with Cu SO 4 solution, White ppt. AgCl  2 NH 4 OH Ag (NH 3 ) 2   2 H 2 O. Soluble (viii) Test for CH 3 COOH vapours : vapours are colourless with a vinegar like smell. These Chemical Analysis (2) Coloured gases Dark brown colourless. It bleaches a moist litmus paper, Blue Cl 2  H 2 O 2 HCl  O ; Colour  O  Colourless. litmus paper colourless. first turns red and then becomes (ii) Tests for Br2 : Brown vapours with a pungent smell. It turns moist starch paper yellow. (but yellow in aq. solution) Pale brown MnCO 3 Light pink Hydrated manganese salts Reddish pink Hydrated cobalt (II) salts Red HgI2 , Pb 3 O4 Yellow CdS , PbI 2 , AgBr , AgI , chromates (iii) Tests for I2 : Violet vapours with a pungent smell. It turns moist starch paper blue. (ii) Dry heating : Substance is heated in a dry test tube. Table : 21.2 E3 (iv) Tests for NO2 : Brown coloured pungent smelling gas. It turns moist starch KI paper blue 2KI  2 NO 2 2KNO 2  I 2 ; I 2  Starch Blue colour. It turns ferrous sulphate solution black, 3 FeSO 4  NO 2  H 2 SO 4 Fe 2 (SO 4 )3  FeSO 4. NO  H 2 O Observation Result (a) A gas or vapour is Compounds with water evolved. of crystallisation Vapour, evolved, with litmus paper. test ID Black brown PbO 2 , Ag 2 O, CdO , Fe 2 O3 , CuCrO 4 , FeCl 3 60 (i) Tests for Cl2 : It is a greenish yellow gas with a pungent smell. In small quantity it appears almost Systematic Procedure for Qualitative Analysis of Inorganic Salts The vapour is alkaline. U It involves the following steps : (1) Preliminary tests (2) Wet tests for acid radicals and (3) Wet tests for basic radicals. D YG (1) Preliminary tests (i) Physical examination : It involves the study of colour, smell, density etc. Black Salt Oxides : MnO 2 , FeO , CuO , Co 3 O4 , Ni2 O3 Sulphides : Readily decomposable salts of strong acids. Oxygen is evolved Nitrates,chlorates certain oxides. Dinitrogen oxide Ammonium nitrate or nitrate mixed with an ammonium salt. Dark-brown or reddish fumes (oxides of nitrogen), acidic in reaction. Nitrates and nitrites of heavy metals. CO 2 is Green Brownish yellow evolved, water becomes turbid. PbS , HgS , Bi2 S 3 (blackish brown) NH 3 is Ammonium salts. evolved lime and Carbonates or hydrogen carbonates. U Orange Ammonium salts. Ag 2 S , CuS , Cu 2 S , FeS , CoS , NiS , ST Blue Ammonium salts, acid salts, and hydroxides. (usually accompanied by change of colour) The vapour is acidic. Table : 21.1 Colour Colour 933 which turns red litmus blue. Hydrated CuSO 4 , anhydrous CoSO 4 SO 2 is evolved, which KO 2 , some dichromate (K2 Cr2 O7 ), Sb 2 S 3 , turns acidified ferricyanides green, decolourises fuschin colour. Nickel salts, hydrated ferrous salts, K 2 Cr2 O7 is evolved, turns Sulphates thiosulphates. Hydrates, sulphides and or potassium permanganate (KMnO 4 ) , some H2S copper (II) salts lead acetate paper black, or cadmium acetate yellow. sulphides in the presence of water. Cl 2 is evolved, yellowish Unstable chlorides e.g., SnS copper chlorides in the 934 Chemical Analysis presence agents. Br2 is Bromides in the presence of oxidising agents. evolved (reddish brown, turns fluorescent paper red). is violet oxidising to Free iodine and certain iodides (b) A sublimate is formed Ammonium and mercury I2 evolved, of vapours condensing black crystals White sublimate As 2 O3 , Sb 2 O3 Grey sublimate Hg Steel grey, garlic odour As Yellow sublimate S , As 2 S 3 HgI2 (Re d) Action of heat on different compounds : Many inorganic salts decomposes on heating, liberating characteristic gases. A few such reactions are tabulated as follows, 60 green gas, bleaches litmus paper, turns KI – starch blue, poisonous. salts. Table : 21.3 (Re d ) 2 Hg  O2 (Silvery deposit ) (Re d )  CuCO 3  CuO  CO 2 (Yellow ) ( Brown )  ZnO 3  ZnO  CO 2 Yellow (hot ) (White ) (Green)  2 PbO 2  2 PbO  O 2  2 Pb 3 O4  6 PbO  O2 E3  2 HgO  White(cold)  CuSO 4.5 H 2 O  CuSO 4  5 H 2 O ( Blue ) (White ) 2 Ag 2 O 4 Ag  O2  2 FeSO 4  Fe 2 O 3  SO 2  SO 3  2 Zn(NO 3 )2  2 ZnO  4 NO 2  O 2 450 C 2 AgNO 3    2 Ag  2 NO 2  O 2 2Cu(NO 3 )2 2CuO  4 NO 2  O2 2 Ag 2 CO 3 4 Ag  2CO 2  O2 2 Pb(NO 3 )2 2 PbO  4 NO 2  O2 ( NH 4 )2 Cr2 O7 N 2  Cr2 O 3  4 H 2 O 2 NaHCO 3 Na 2 CO 3  CO 2  H 2 O NH 4 HCO 3 NH 3  CO 2  H 2 O CaCO 3 CaO  CO 2 2 NaNO 3 2 NaNO 2  O2 MgCO 3 MgO  CO 2 Red hot 2 NH 3   N 2  3 H 2 (White ) ID  CuSO 4  CuO  SO 3 D YG U ( Brown ) (Orange) (Green) 2 Mg(NO 3 )2 2 MgO  4 NO 2  O2 2Ca(NO 3 )2 2CaO  4 NO 2  O2 Red hot Al 2 (SO 4 )3   Al 2 O 3  3 SO 3 2CaSO 4.2 H 2 O 2CaSO 4.H 2 O  2 H 2 O 2 AlCl 3.6 H 2 O Al 2 O3  6 HCl  9 H 2 O 2 BeSO  2 MgSO 4  2 MgO  2 SO 2  O 2  2 ZnSO 4  2 ZnO  2 SO 2  O 2  (COO )2 Sn  SnO  CO 2  CO CaC 2 O4 CaCO 3  CO NH 4 NO 2 N 2  2 H 2 O NH 4 NO 3 N 2 O  2 H 2 O 2 KClO 3 2 KCl  3O2 2 FeCl 3 2 FeCl 2  Cl 2 Li2 CO 3 Li2 O  CO 2 (COO)2 Fe FeO  CO  CO 2 2 KMnO 4 K 2 MnO 4  MnO 2  O2 MgCl 2. 6 H 2 O HgCl 2  Hg NH 4 Cl NH 3  HCl (Plaster of Paris) o ST U 350 C 2 AgNO 3    2 AgNO 2  O 2 Hg(NO 3 )2 Hg  2 NO 2  O2 4   2 BeO  2 SO 2  O 2 1 2 LiNO 3 Li2 O  2 NO 2  O 2 2  2Co ( NO 3 ) 2  2CoO  4 NO 2  O 2  2CuCl 2  Cu 2 Cl 2  Cl 2 4 K 2 Cr2 O7 4 K 2 CrO4 + 2Cr2 O3  3O2  2 Mg( NH 4 )PO 4  Mg 2 P2 O7  H 2 O  2 NH 3  2 Zn( NH 4 )PO 4  Zn 2 P2 O7  H 2 O  2 NH 3  K 4 Fe (CN )6  4 KCN  Fe  2C  N 2  ZnCl 2. 2 H 2 O  Zn(OH )Cl  HCl  H 2 O  2(ZnCl 2.H 2 O)  Zn 2 OCl 2  2 HCl  H 2 O  2[FeCl 3.6 H 2 O]  Fe 2 O 3  9 H 2 O  6 HCl 800 C 2 ZnSO 4    2 ZnO  2 SO 2  O 2 o   Na 2 B4 O7.10 H 2 O  Na 2 B4 O7  2 NaBO 2  B2 O3 (Glassy bead) Chemical Analysis o 935 o Red hot 100 C 160 C  B 2 O 3 H 3 BO 3    HBO 2    H 2 B4 O7  o o o 70 C 100 C 450 C ZnSO 4.7 H 2 O   ZnSO 4.6 H 2 O    ZnSO 4.H 2 O    ZnSO 4  H 2O 5 H 2 O NaPO 3  CoO NaCoPO 4 (Blue); Characteristic flame colour : Certain metals and their salts impart specific colours to Bunsen burner flame. (a) Pb imparts pale greenish colour to the flame. (b) Cu and Cu salts impart blue or green colour to the flame. NaPO3  Cr2O3 NaPO3.Cr2O3 (Green) (v) Charcoal cavity test Table : 21.5 (a) Compound fused in cavity directly Nature and colour of (d) Ba and its salts impart apple green colour to Yellow, brittle bead (g) Na imparts yellow colour to the flame. White, brittle Sb 3  White yellow when hot ZnO White garlic odour As 2 O3 Brown CdO Grey metallic particles attracted by magnet Fe, Ni, CO Maleable beads Ag and Sn (White), Cu (Red flakes) ID K imparts pink-violet (Lilac) colour to the (i) Li imparts crimson-red, Rb imparts violet and Cs imparts violet colours to the flame. U (j) Livid- blue flame is given by As, Sb and Bi. (iv) Borax bead test : The transparent glassy bead (NaBO 2  B2 O3 ) when heated with inorganic salt D YG and the colour produced gives some idea of cation present in it. Table : 21.4 Colour of bead in oxidising flame Colour of bead in reducing flame Pb 2  marks on paper (f) Ca imparts brick red colour to the flame. (h) flame. Bi 3  Yellow, soft bead which (e) Sr imparts crimson red colour to the flame. Cation E3 (c) Borates also impart green colour to the flame. bead the flame. 60 (iii) Flame test Basic radical present (b) Compound mixed with Na 2 CO 3 Crystalline Sustance Decrepitat es Salts, NaCl, KCl ; Red and opaque Cu Dark green in hot and cold Same Cr Deep – blue Deep blue Co Substance infusible, perform test (a) Yellow when hot Green Fe (vi) Cobalt Nitrate test Violet in hot and cold Colourless Mn ST U Greenish when hot, blue in cold. Brown in cold Grey or opaque black or Ni Microcosmic salt bead test : Microcosmic salt, Na(NH 4 )HPO4.4 H 2 O is also used to identify certain cations just like borax. When microcosmic salt is heated in a loop of platinum wire, a colourless transparent bead of sodium metaphosphate is formed. Na (NH 4 )HPO4.4 H 2 O Na(NH 4 )HPO4  4 H 2 O Na(NH 4 )HPO4 NaPO 3  NH 3  H 2 O Now NaPO 3 reacts with metallic oxides to give coloured orthophosphates. NaPO 3  CuO NaCuPO 4 (Blue); Substance deflagrate s Oxidising agents like NO 3 , NO 2 chlorates Table : 21.6 Colour Composition Result Blue residue CoO. Al 2 O3 Al Green residue CoO. ZnO ZnO Pink dirty residue CoO. MgO MgO Blue residue NaCoPO 4 PO 43  in absence of Al. 936 Chemical Analysis (2) Wet tests for acid radicals : Salt or mixture is treated with dil. H 2 SO 4 and also with conc. H 2 SO 4 separately and by observing the types of gases evolved. Confirmatory tests of anions are performed. Table : 21.7 Observations with Dilute H 2 SO 4 Observations Brisk effervescence Acid Radical with evolution of colourless and Confirmatory test CO 32  (carbonate) odourless gas. Gas turns lime water milky but milkyness disappears on passing gas inexcess, Na 2 CO 3  H 2 SO 4  Na 2 SO 4  H 2 O  CO 2 Ca(OH )2  CO 2  CaCO 3  H 2 O lime water milky soluble NO 2 (Nitrite) Brown fumes Add KI and starch 60 CaCO 3  H 2 O  CO 2  Ca(HCO 3 )2 solution blue colour 2 NaNO 3  H 2 SO 4  Na 2 SO 4  2 HNO 2 ; 2 NO  O 2 (air)  2 NO 2 (brown) E3 HNO 2  NO (colourless); 2 KI  H 2 SO 4  2 NHO 2  K 2 SO 4  2 H 2 O  2 NO  I 2 I2  starch   blue colour Smell of rotten eggs S 2  (sulphide) Gas turn lead acetate paper black ID (H 2 S smell) on heating * Sodium carbonate extract (SE ) + sodium nitroprusside – purple colour, Na 2 S  H 2 SO 4   H 2 S  Na 2 SO 4 H 2 S  (CH 3 COO )2 Pb  PbS  2CH 3 COOH U (black) Na 2 S  Na 2 [Fe(CN )5 NO ]  Na 4 [Fe(CN )5 NOS ] D YG sodium nitroprusside Colourless gas with pungent smell of burning sulphur SO 32  (sulphite) (purple) 2 Gas turns acidified K 2 Cr2 O7 solution green [different from CO 3 ] since gas also turns lime water milky  Na 2 SO 3  H 2 SO 4  Na 2 SO 4  H 2 O  SO 2 Cr2 O72  3 SO 2  2 H   2Cr 3   3 SO 42  H 2 O (green) Ca(OH )2  SO 2  CaSO 3 vinegar gives smell of ST White or yellowish white turbidity on warming FeCl 3 blood CH 3 COO  Aq. (acetate) 3CH 3 COONa  FeCl 3  Fe(CH 3 COO )3  3 NaCl U Solution (milky) Solution + neutral neutral red colour (red) S 2 O 32  Aq. Solution + AgNO 3 white ppt. changing to black (viii) on warming (thiosulphate) , Na 2 S 2 O 3  2 AgNO 3   Ag 2 S 2 O 3  2 NaNO 3 white ppt. Ag 2 S 2 O3  H 2 O  Ag 2 S  H 2 SO 4 black ppt. Table : 21.8 Observation with concentrated H 2 SO 4 Observation Colourless pungent gas giving white fumes with Acid Radical Cl  (chloride) Confiramatory Test Add MnO 2 in the same test tube and heat–pale green Cl 2 gas (i) Chemical Analysis aq. NH 4 OH 937 S.E.+ HNO 3  AgNO 3 solution –white ppt. soluble in aq. NH 3 (ii) Chromyl chloride test (iii) Br  (bromide) Reddish brown fumes Add Mn 2 O and heat –yellowish brown Br 2 gas (iv) S.E.+ HNO 3  AgNO 3 solution –pale yellow ppt. partially soluble aq. NH 3 (v) Layer test (vi) 60 S.E.+ HNO 3  AgNO 3 yellow ppt. insoluble in aq. NH 3 (vii) Violet pungent vapours turning starch paper blue. I  (iodide) Brown pungent fumes intensified by the NO 3 (nitrate) Ring test (viii) C 2 O 42  (oxalate) Acidified KMnO 4 solution is decolorised (ix) E3 Layer test (vi) addition of Cu- turnigs. Colourless gases turning lime water milky and burning with blue flame. ID S.E.  CH 3 COOH  CaCl 2 white ppt. decolorising acidified KMnO 4 solution (x)  KHSO 4  H ; Bromide : (iv) KBr  conc. H 2 SO 4  Reactions : KCl  conc. H 2 SO 4  KHSO 4  HCl HCl  NH 3  NH 4 Cl  4 HCl  MnO 2  MnCl 2  Cl 2  2 H 2 O (ii) KCl  AgNO 3  AgCl   KNO 3 white ppt. AgCl  aq. 2 NH 3  [ Ag ( NH 3 )2 ] Cl soluble (iii) Chromyl- chloride test : Chloride  4 HBr  MnO 2  Br2  2 H 2 O  MnBr 2 (v) NaBr  AgNO 3  AgBr   NaNO 3 D YG (white fumes) (i) U Chloride pale yellow ppt. AgBr  aq. 2 NH 3 [ Ag (NH 3 )2 ]Br partially soluble (vi) Layer shake + CHCl 3  Test U vapours of chromyl-chloride (CrO2Cl 2 ). brown Pass these vapours into NaOH , when yellow Na2CrO4 solution is CH 3COOH ST formed. On adding and (CH 3COO)2 Pb, yellow ppt. of lead chromate (PbCrO4 ) is formed.  KCl  conc. H 2 SO 4  KHSO 4  HCl ;  K 2 Cr2 O7  2 H 2 SO 4  2 KHSO 4  2CrO3  H 2 O S. E.  Cl 2 water yellowish orange colour in CHCl 3 layer (CS 2 or CCl 4 can be taken instead of CHCl 3 ); 2 NaBr  Cl 2  2 NaCl  + heat K2Cr2O7 (solid )  conc. H 2 SO 4   reddish : Br2 orange y ellow (soluble in CHCl 3 ) In case of I  , violet colour of I 2 in CHCl 3 layer, 2 NaI  Cl 2   2 NaCl  I 2 (violet) Iodide : (vii)  KI  conc. H 2 SO 4  KHSO 4  HI ; 2 HI  H 2 SO 4   I 2 + 2 H 2 O  SO 2 Nitrate : NaNO 3  H 2 SO 4  NaHSO 4  HNO 3 4 HNO 3  4 NO 2  O 2 2 H 2 O ; brown fumes conc. CrO3  2 HCl  CrO2 Cl 2  2 H 2 O CrO2 Cl 2  4 NaOH  Na 2 CrO4  2 NaCl  2 H 2 O Na 2CrO4  (CH 3 COO )2 Pb  PbCrO 4  2CH 3 COONa yellow ppt. Cu  4 HNO 3   Cu ( NO 3 )2  2 NO 2  2 H 2 O (viii) Ring test : To water extract (all NO 3 are water soluble) add freshly prepared FeSO 4 solution and then conc. H 2 SO 4 carefully by the side of the test- 938 Chemical Analysis tube. A dark brown ring of [Fe (H 2 O)5 NO ] 2  SO 42  at the interface between the two liquids is formed. (ii) Borate : lgnite the mixture containing borate, conc. H 2 SO 4. And ethanol in a china-dish with a burning splinter –green edged flame of ethyl borate. 2 NaNO 3  H 2 SO 4  2 NaHSO 4  2 HNO 3 ; 2 Na 3 BO 3  3 H 2 SO 4 2 H 3 BO 3  3 Na 2 SO 4 ; 2 HNO 3  6 Fe SO 4  3 H 2 SO 4  (conc.)  H 3 BO 3  3C 2 H 5 OH  (C 2 H 5 O)3 B  3 H 2 O 3 Fe 2 (SO 4 )3  2 NO  4 H 2O ethanol Oxalate : Na 2 C 2 O4  H 2 SO 4   Na 2 SO 4  H 2 O  CO  CO 2 CO burns with blue flame and CO 2 turns lime In presence of Cu 2  , perform this test in a test tube since Cu 2  salts are not volatile. (iii) S.E.  HNO 3  ammonium molybdate solution. Heat, yellow crystalline ppt. confirms  Na 3 PO 4  12( NH 4 )2 MoO 4  24 HNO 3  water milky. (ix) 5 C 2 O 42   2 MnO 4  16 H  10 CO 2  2 Mn 2  8 H 2 O ( NH 4 )3 PO 4. 12 MoO 3  21 NH 4 NO 3  3 NaNO 3  12 H 2 O colourless y ellow ppt. CaCl 2  NaC 2O4   CaC 2O4  2 NaCl CaC 2 O 2 white ppt. decolourises acidified KMnO 4. CO 32  until reaction ceases). Add BaCl 2 solution. White in conc. HNO 3 ,  NaF  H 2 SO 4  NaHSO 4  HF  SiO 2  4 HF  SiF4  2 H 2 O U BaCl 2  NaSO 4  BaSO 4  2 NaCl heat and bring a water wetted rod in contact with vapours at the mouth of the test tube. A white deposit on the rod shows the presence to F  ID (i) Sulphate : S.E. add dil. (to decompose insoluble Arsenic also gives this test. Hence presence of phosphate should also be checked after group II. (iv) Fluoride : Sand +salt (F  ) +conc. H 2 SO 4 ; Specific test in solution ppt. E3 (violet) (x) burns with green flame (volatile) 60 [Fe (H 2 O)6 ] SO 4  NO [Fe(H 2 O)5 NO ]2  SO 42   H 2 O 3 SiF4 4 H 2 O   2 H 2 SiF6  H 4 SiO 4 white ppt. white Group I Basic radicals Ag  , Hg 22  (I), Pb 2  Cu 2  , Cd 2  , Pb 2  , 2 3 Hg (II), Bi , As Ppt. as Explanation dil HCl Chloride K SP values of chlorides are low, (AgCl , Hg 2 Cl 2 , PbCl 2 ) hence precipitated. Others have higher K SP values hence not 3 , gas presence HCl of in dil. Sulphides ( CuS , As 2 S 3 etc.) III Al 3  , Cr 3  , Fe 3  precipitated. K SP values of sulphides are low hence precipitated by low [S 2 ] ion. HCl (with common H  ion) decreases ionization of H 2 S ST Sb 3  , Sn 2  Group reagent H2S U II D YG (3) Wet tests for basic radicals : Analysis of Basic Radicals Table : 21.9 which gives low [S 2 ]. Hence II group is precipitated. Others with higher K SP values not precipitated. NH 4 OH in Hydroxide, Al(OH )3 K SP presence of NH 4 Cl etc. low. NH 4 Cl (with common NH 4 ion) values of Al(OH )3 etc. are decreases ionization  of NH 4 OH giving low [OH ]. Hence group III is precipitated. Chemical Analysis Zn 2  , Ni 2  , Mn 2  , Co 2  IV in H2S Sulphides etc.) ammonical ( ZnS values K SP of 939 sulphides of group IV are high hence precipitation takes place in medium higher [S 2 ]. increases Basic ionization increasing [S 2 medium of ] H2S hence precipitation of group IV. Ca 2 , Ba 2 , Sr (NH 4 )2 CO 3  NH 4 Cl Carbonates ( CaCO 3 2 K SP values of carbonate are less 60 V etc.) than that of group VI (Mg 2 ) hence Mg NH 4 OH  Na 2 HPO 4 also (only for Mg included) 0 NH  4 2 White ) ppt. (MgHPO 4 ) – before. – – (Zero) precipitation E3 Mg 2  , (Na  , K  VI 2 Tested independently from original solution. (i) AgCl dissolves in ammonium hydroxide, ID Chemical reactions involved in the tests of basic radicals Group I : When dil. HCl is added to original solution, insoluble chlorides of lead, silver mercurous mercury are precipitated. Pb(NH 3 )2  2 HCl  PbCl 2  2 HNO 3 ; Pb2+ (lead) (i) PbCl 2 is soluble in hot water and on cooling gives a yellow precipitate with potassium chromate solution which is insoluble in acetic acid but soluble in sodium hydroxide. U PbCl 2  K 2 CrO 4  PbCrO 4  2 KCl ; yellow ppt. PbCrO 4  4 NaOH  Na 2 PbO 2  Na 2 CrO 4  2 H 2 O solution ST The of PbCl 2 forms a yellow precipitate with potassium iodide solution. PbCl 2  2 KI  PbI 2  2 KCl Yellow ppt. (iv) White precipitate of lead sulphate is formed with dilute H 2 SO 4. The precipitate is soluble in ammonium acetate, (ii) On adding dilute HNO 3 to the above solution, U D YG Hg( NO 3 ) 2  2 HCl  HgCl 2  2 HNO 3 (iii) Diammine silver (I) chloride white precipitate is again obtained Ag ( NH 3 )2 Cl  2 HNO 3  AgCl  2 NH 4 NO 3 AgNO 3  HCl  AgCl  HNO 3 white crystals are again formed. (ii) The solution of PbCl 2 AgCl  2 NH 4 OH  Ag (NH 3 )2 Cl  2 H 2 O PbCl 2  H 2 SO 4  PbSO 4  2 HCl ; PbSO 4  2CH 3 COONH 4  Pb (CH 3 COO ) 2  ( NH 4 ) 2 SO 4 White ppt. (iii) On adding KI to the complex solution, yellow precipitate is obtained. Ag ( NH 3 ) 2 Cl  KI  AgI  KCl  2 NH 3 Hg 22  (mercurous) (i) Hg 2 Cl 2 turns black with NH 4 OH , Hg2Cl2  2 NH 4 OH Hg  Hg(NH 2 )Cl  NH 4 Cl  2 H 2O    Black (ii) The black residue dissolves in aqua-regia forming mercuric chloride. 3 HCl  HNO 3  NOCl  2 H 2 O  2Cl 2 Hg( NH 2 )Cl  6 Cl  2 HgCl 2  4 HCl  N 2 Hg  2Cl  HgCl 2 (iii) The solution of HgCl 2 forms white or slatecoloured precipitate with stannous chloride. 2 HgCl 2  SnCl 2  Hg 2 Cl 2  SnCl 4 white ppt. Hg 2 Cl 2  SnCl 2  2 Hg  SnCl 4 Grey ppt. Ag+(silver) (iv) The solution of HgCl 2 with copper turning forms a grey deposit. 940 Chemical Analysis HgCl 2  Cu  Hg  CuCl 2 deep blue coloured solution while cadmium forms a colourless soluble complex, Grey ppt. Group II : When hydrogen sulphide is passed in acidified solution, the radicals of second group are precipitated as sulphides. The precipitate is treated with yellow ammonium sulphide. The sulphides of IIB are first oxidised to higher sulphides which then dissolve to form thio-compounds. Bi( NO 3 )3  3 NH 4 OH   Bi(OH )3  3 NH 4 NO 3 White ppt. Cu ( NO 3 )2  4 NH 4 OH [Cu ( NH 3 )4 ] ( NO 3 )2  4 H 2 O; Tetrammin e cupric nitrate (deep blue solution) Cd ( NO 3 )2  4 NH 4 OH  [Cd ( NH 3 )4 ]( NO 3 )2  4 H 2 O Ag 2 S 3  2( NH 4 )2 S 2  2( NH 4 ) 2 S  As 2 S 5 Tetrammin e cadmium nitrate (colourles s solution) SnS  ( NH 4 )2 S 2 ( NH 4 )2 S  SnS 60 Sb 2 S 3  2( NH 4 )2 S 2  2( NH 4 )2 S  Sb 2 S 5 Bi3+ (bismuth) : The precipitate dissolves in 2 dilute HCl, Bi (OH )3  3 HCl  BiCl 3  3 H 2O As 2 S 5  3( NH 4 )S  2( NH 4 )3 AsS 4 Ammonium thioarsena te Sb 2 S 5  3( NH 4 )2 S  2( NH 4 )2 SbS E3 Part I : Addition of excess of water to BiCl 3 solution gives a white precipitate due to hydrolysis. 4 Ammonium thioantimo nate SnS 2  ( NH 4 )2 S  ( NH 4 )2 SnS BiCl 3  H 2 O  BiOCl  2 HCl Bismuth Oxychlorid e (White ppt.) 3 Part II : The solution of BiCl 3 is treated with Ammonium thiostanna te sodium stannite when a black precipitate of metallic  bismuth is formed, 2 BiCl 3  3 Na 2 SnO 2  6 NaOH  ID All the three are soluble. In case, the precipitate does not dissolve in yellow ammonium sulphide, it may be either HgS or PbS or Bi2 S 3 or CuS or CdS. The precipitate is heated with Sod. stannate Cu2+ (copper) : Blue coloured solution is acidified with acetic acid. When potassium ferrocyanide is added a chocolate coloured precipitate is formed, U dilute HNO 3. Except HgS , all other sulphides of IIA are Sod. stannite 3 Na 2 SnO 3  2 Bi  6 NaCl  3 H 2 O soluble. D YG 3 PbS  8 HNO 3 3 Pb(NO3 )2  2 NO  3S  4 H 2O Bi2 S 3  8 HNO3 2Bi(NO3 )3  2 NO  3S  4 H 2O 3CuS  8 HNO 3 3Cu(NO3 )2  2 NO  3S  4 H 2O 3CdS  8 HNO 3 3Cd (NO3 )2  2 NO  3S  4 H 2O Hg2+ (mercuric) HgS is dissolved in aqua-regia, 3 HgS  2 HNO 3  6 HCl 3 HgCl 2  3S  2 NO  4 H 2O The solution is divided into two parts: U Part I : Stannous chloride solution reduces HgCl 2 first into white Hg 2 Cl 2 and then to grey metallic mercury. ST Part II : Copper displaces Hg from HgCl 2 which gets coated on copper turnings as a shining deposit. Pb2+ (lead) In case the sulphide dissolves in dilute HNO 3 , a small part of the solution is taken. Dilute H 2 SO 4 is added. If lead is present, a white precipitate of lead sulphate appears, Pb( NO 3 )2  H 2 SO 4 PbSO 4  2 HNO 3 Cu ( NH 3 )4 ( NO 3 )2  4 CH 3 COOH   Cu(NO 3 )2  4 CH 4 COONH 4 2Cu ( NO 3 )2  K 4 [Fe(CN )6 ]  Cu 2 [Fe(CN )6  4 KNO 3 Chocolate ppt. Cd2+ (cadmium) : H2S is passed colourless solution. The appearance of precipitate confirms the presence of cadmium, through yellow Cd ( NH 3 )4 ( NO 3 )2  H 2 S  CdS  2 NH 4 NO 3  NH 3 Yellow ppt. Group IIB : In case the precipitate dissolves in yellow ammonium sulphide, the tests of the radicals arsenic, antimony and tin are performed. The sulphide is treated with concentrated hydrochloric acid. Antimony and tin sulphide dissolve while arsenic sulphide remains insoluble. As3+ (arsenic) : The insoluble sulphide is treated with concentrated nitric acid which is then heated with ammonium molybdate. Yellow precipitate of ammonium arsenomolybdate is formed. (White ppt.) In absence of lead, the remaining solution is made alkaline by the addition of excess of NH 4 OH. Bismuth forms a white precipitat of Bi(OH )3 , copper forms a As 2 S 5  10 HNO 3  2 H 3 AsO 4  10 NO 2  2 H 2 O  5 S Arsenic acid H 3 AsO 4  12( NH 4 )2 MoO 4  21 HNO 3   Chemical Analysis ( NH 4 )3 AsO 4. 12 MoO 3  21 NH 4 NO 3  12 H 2 O Yellow ppt. Sn2+ or Sn4+ (tin) : Solution of sulphide in concentrated HCl is reduced with iron fillings or granulated zinc. SnS 2  4 HCl   SnCl 4 2 H 2 S 941 is extracted with water or the precipitate is heated with NaOH and bromine water. 2Cr(OH )3  3 KNO  2 Na 2 CO 3   2 Na 2 CrO4  3 KNO 2  2CO 2  3 H 2 O or 2 NaOH  Br2  NaBrO 4  NaBr  H 2 O White ppt. NaBrO  NaBr  [O] SnCl 4  Fe   SnCl 2  FeCl 4 Grey 2Cr(OH )3  4 NaOH  3[O]  2 NaCrO4  5 H 2O 60 HgCl 2 solution is added to above solution which The solution thus obtained contains sodium chromate. The solution is acidified with acetic acid and treated with lead acetate solution. A yellow precipitate appears. gives first a white precipitate that turns to grey. 2 HgCl 2  SnCl 2  HgCl 2  SnCl 4 White ppt. Hg 2 Cl 2  SnCl 2  2 Hg  SnCl 4 E3 Na 2 CrO4  Pb(CH 3 COO )2  PbCrO 4  2CH 3 COONa Grey Yellow ppt. Sb2+ (antimony) : Filtrate of sulphide concentrated HCl is divided into two parts. in Al3+(aluminium) : The gelatinous precipitate dissolves in NaOH , Al (OH )3  NaOH  NaAlO 2  2 H 2 O SbCl 3  H 2 O   SbOCl  2HCl White ppt. The solution is boiled with ammonium chloride when Al(OH )3 is again formed. NaAlO 2  NH 4 Cl  H 2 O   Al (OH )3  NaCl  NH 3 Part II : H 2 S is circulated. Orange precipitate is Orange ppt. D YG Group III : Hydroxides are precipitated on addition of excess of ammonium hydroxide in presence of ammonium chloride. AlCl 3  3 NH 4 OH  Al (OH )3  3 NH 4 Cl Gelatinous ppt. CrCl3  3 NH 4 OH   Cr(OH )3  3 NH 4 Cl Green ppt. FeCl 3  3 NH 4 OH   Fe(OH )3  3 NH 4 Cl Brownish red ppt. U Fe3+ (iron) : The brownish red precipitate dissolves in dilute HCl. The solution is divided into two parts. ST Part I : K4 [Fe(CN )6 ] solution is added which forms deep blue solution or precipitate. the third group, sulphides of fourth group are precipitated. NiS and CoS are black and insoluble in concentrated HCl while MnS (buff coloured), ZnS (colourless) are soluble in conc. HCl. Zn2+ (zinc) : The sulphide dissolves in HCl. ZnS  2 HCl  ZnCl 2  H 2 S When the solution is treated with NaOH , first a white precipitate appears which dissolves in excess of NaOH ZnCl 2  2 NaOH   Zn(OH )2  2 NaCl White ppt. Zn(OH )2  2 NaOH  Na 2 ZnO 2  2 H 2 O (Soluble) On passing H2S , white sulphide Fe(OH )3  3 HCl  FeCl 3  3 H 2 O 4 FeCl 3  3 K4 [Fe(CN )6 ]  Fe 4 [Fe(CN )6 ]3  12 KCl Prussian blue Part II : Addition of potassium thiocyanate solution gives a blood red colouration. FeCl 3  3 KCNS  Fe (CNS )3  3 KCl Blood red colour 3+ Group IV : On passing H 2 S through the filtrate of U formed, 2 SbCl 3  3 H 2 S  Sb 2 S 3  6 HCl Soluble ID Part I : On dilution with excess of water, a white precipitate of antimony oxychloride is obtained. Cr (chromium) : The green precipitate is fused with fusion mixture (Na2CO3  KNO 3 ). The fused product precipitate is of zinc formed Na 2 ZnO 2  H 2 S  ZnS  2 NaOH White ppt. Mn2+ (manganese) dissolves in HCl : Manganese sulphide MnS  2 HCl  MnCl 2  H 2 S On heating the solution with NaOH water, manganese dissolve gets precipitated. MnCl 2  2 NaOH   Mn(OH )2  2 NaCl Mn(OH )2  O   MnO 2 H 2O and Br2 - 942 Chemical Analysis The precipitate is treated with excess of nitric acid and PbO 2 or Pb 3 O4 (red lead). The contents are heated. The formation of permanganic acid imparts pink colour to the supernatant liquid. Br2  H 2 O  2 HBr  O 2 Na 4 Co (CO 3 )3  H 2 O  O  2 Na 3 Co (CO 3 )3  2 NaOH sod. cobalti carbonate (Green colouratio n) 2 MnO 2  4 HNO 3  2 Mn(NO 3 )2 2 H 2O  O2 NiCl 2  2 NaHCO 3   NiCO 3  2 NaCl  H 2O  CO 2 2 Mn( NO 3 )2  5 Pb 3 O 4  26 HNO 3   2 NiCO 3  [O]  Ni2 O3  2CO 2 (Black) 2 HMnO 4  15 Pb (NO 3 )2  12 H 2 O Permanganic acid (pink) The black precipitate is dissolved in aqua- regia. 3 NiS  6 HCl  2 HNO 3   2 NiCl 2  2 NO  3 S  2 H 2 O 3CoS  6 HCl  2 HNO 3   3CoCl 2  2 NO  3 S  4 H 2 O The solution is evaporated to dryness and residue extracted with dilute HCl. It is divided into three parts. Part I : Add NH 4 OH (excess) and dimethyl O  N  C  CH 3 |  2 NH 4 Cl  2 H 2 O N  C  CH 3 D YG OH | CH 3  C  N | CH 3  C  N CaCO 3  2CH 3 COOH (CH 3 COO )2 Ca  CO 2  H 2O Ba2+ (barium) : Barium chromate is insoluble and precipitated by the addition of potassium chromate solution, Ba(CH 3 COO )2  K2CrO4   BaCrO 4  2CH 3 COOK Sr2+ (Strontium) : Strontium insoluble and precipitated by the ammonium sulphate U CH 3  C  NOH NiCl 2  2 |  2 NH 4 OH  CH 3  C  NOH SrCO 3  2CH 3 COOH (CH 3 COO )2 Sr  CO 2  H 2O ID glyoxime. A rosy red precipitate appears, if nickel is present, BaCO 3  2CH 3 COOH (CH 3 COO )2 Ba  CO 2  H 2O E3 Ni2+ (nickel) and Co2+ (cobalt) 60 Group V : Ammonium carbonate precipitates V group radicals in the form of carbonates are soluble in acetic acid. The above test fails in presence of HCl. Ni  O | OH Part II : Add CH 3 COOH in excess and KNO 2. The appearance of yellow precipite confirms the presence of cobalt. U KNO 2  CH 3 COOH  CH 3 COOK  HNO 2 CoCl 2  2 KNO 2  Co (NO 2 )2  2 KCl ST Co (NO 2 )2  2 HNO 2  Co (NO 2 )3  NO  H 2O Sr (CH 3 COO )2  ( NH 4 )2 SO 4  SrSO 4  2CH 3 COONH 4 White ppt. Ca2+ (calcium) : Calcium oxalate is insoluble and precipitated by the addition of ammonium oxalate. Ca(CH 3 COO )2  ( NH 4 )2 C2 O4  CaC 2 O4  2CH 3 COONH 4 White ppt. Group VI : In the filtrate of V group, some quantity of ammonium oxalate is added as to remove Ba, Ca and Sr completely from the solution. The clear solution is concentrated and made alkaline with NH 4 OH. Disodium hydrogen phosphate is now added, a white precipitate is formed. MgCl 2  Na 2 HPO 4  NH 4 OH   Mg( NH 4 )PO 4  2 NaCl  H 2 O Megnesium ammonium phosphate (White ppt.) Co (NO 2 )3  3 KNO 2  K3 [Co (NO 2 )6 ] Part III : Solution containing either nickel or cobalt is treated with NaHCO 3 and bromine water. Appearance of apple green colour is observed, the solution is heated when black precipited is formed, which shows the presence of nickel, CoCl 2  2 NaHCO 3   Co (HCO 3 )2  2 NaCl Co (HCO 3 )2  4 NaHCO 3  Na4 Co (CO3 )3  3 H 2O  3CO 2 sulphate is addition of solution, Zero group NH 4 (ammonium) : The substance (salt or mixture) when heated with evolves ammonia. NaOH solution NH 4 Cl  NaOH   NaCl  NH 3  H 2 O When a rod dipped in HCl is brought on the mouth of the test tube, white fumes of ammonium chloride are formed, NH 3  HCl  NH 4 Cl White fumes Chemical Analysis NH 2 Hg O  7 KI  KCl  3 H 2 O Hg I Iodide of Millon's base (Brown ppt.) Volumetric analysis Mass of substance B Chemical equivalent mass of B or gram equivalent of A  gram equivalent of B or milli-gram equivalent of A  milli-gram equivalent of B The point at which the amounts of the two reactants are just equivalent is known as equivalence point or end point. An auxiliary substance which helps in the usual detection of the completion of the titration or equivalence point or end point is termed as indicator, i.e., substances which undergo some easily detectable changes at the equivalence point are used as indicators. Methods of expressing concentrations of solutions The concentration of a solution can be expressed in various ways. (1) Percent by mass (2) Molarity (3) molality (4) Mole fraction (5) Normality Types of titrations : Titrations can be classified as : (1) Acid base titrations or acidimetry and alkalimetry (2) Oxidation reduction titrations or redox titrations (3) Precipitation titrations (4) Complexometric titrations. (1) Acid-base titrations : When the strength of an acid is determined with the help of a standard solution of base, it is known as acidimetry. Similarly, when the strength of a base (alkali) is determined with the help of a standard solution of an acid, it is known as alkalimetry. Both these titrations involve neutralisation of an acid with an alkali. In these D YG U ID Volumetric analysis is a quantitative analysis. It involves the measurement of the volume of a known solution required to bring about the completion of the reaction with a measured volume of the unknown solution. Titration : The process of addition of the known solution from the burette to the measured volume of solution of the substance to be estimated until the reaction between the two is just complete, is termed as titration. Thus, a titration involves two solutions; (i) Unknown solution : The solution consisting the substance to be estimated is termed unknown solution. The substance is termed titrate. (ii) Standard solution : The solution in which an accurately known amount of the reagent (titrant) has been dissolved in a known volume of the solution is termed standard solution. There are two types of reagents (titrants) : (a) Primary standards : These can be accurately weighed and their solutions are not to be standardised before use. Oxalic acid (H 2 C2 O4.2 H 2 O) , potassium  60 2 K 2 HgI 4  NH 4 Cl  4 KOH  Mass of substance A Chemical equivalent mass of A or E3 To the aqueous solution of ammonium salt when Nessler’s reagents is added, brown coloured precipitate is formed. 943 dichromate (K2 Cr2 O7 ) , silver nitrate ( AgNO 3 ) , copper sulphate (CuSO 4.5 H 2 O) , ferrous ammonium sulphate [FeSO 4 (NH 4 )2 SO 4.6 H 2 O] , sodium thiosulphate U (Na 2 S 2 O3.5 H 2 O) , etc., are the examples of primary ST standards. (b) Secondary standards : The solutions of these reagents are to be standardised before use as these cannot be weighed accurately. The examples are sodium hydroxide (NaOH ) , potassium hydroxide (KOH ) , hydrochloric acid (HCl ) , sulphuric acid (H 2 SO 4 ) , potassium permanganate (KMnO 4 ) , iodine, etc. titrations H  ions of the acid combine with OH  ions of the alkali to form unionised molecules of water. HA  BOH  BA  H 2 O Acid Alkali  Salt  Water  or H  A  B  OH   B   A   H 2 O or H   OH   H 2 O Law of equivalence : It is applied in all volumetric estimations. According to it, the chemical substances react in the ratio of their chemical equivalent masses. The end point in these titrations is determined by the use of organic dyes which are either weak acids or weak bases. These change their colours within a limited range of hydrogen ion concentrations, i.e., pH of the Mass of substance A Chemical equivalent mass of A  Mass of substance B Chemical equivalent mass of B solution. Phenolphthalein is a suitable indicator in the titrations of strong alkalies (free from carbonate) against strong acids or weak acids. Methyl orange is 944 Chemical Analysis 2 KMnO 4  3 H 2 SO 4  K 2 SO 4  2 MnSO 4  3 H 2 O  5[O] KI  I2  KI 3 Potassium tri - iodide This solution is first standardised before use. With the standard solution of I 2. Substances such as 60 sulphite, thiosulphate, arsenite, etc., are estimated. (b) Iodometric titrations : In iodometric titrations, an oxidising agent is allowed to react in neutral medium or in acidic medium, with excess of potassium iodide to liberate free iodine. KI  oxidising agent  I 2 Free iodine is titrated against a standard reducing agent usually with sodium thiosulphate. Halogens, oxyhalogens, dichromates, cupric ion, peroxides, etc., can be estimated by this method. I2  Na 2 S 2 O 3  2 NaI  Na 2 S 4 O6 2CuSO 4  4 KI  Cu 2 I2  2 K 2 SO 4  I2 K 2 Cr2 O7  6 KI  7 H 2 SO 4  ID or MnO 4  8 H   5 e  Mn 2   4 H 2 O the solution of iodine (volatile and less soluble in water), it is dissolved in potassium iodide solution. E3 used as an indicator in the titrations of strong acids against strong and weak alkalies. As no indicator gives correct results in the titrations of weak acids against weak bases, such titrations are performed by some other methods (physical methods). (2) Oxidation reduction titrations : The titrations based on oxidation-reduction reactions are called redox titrations. The chemical reactions proceed with transfer of electrons (simultaneous loss or gain of electrons) among the reacting ions in aqueous solutions. Sometimes these titrations are named after the reagent used, as: (i) Permanganate titrations : These are titrations in which potassium permanganate is used as an oxidising agent in acidic medium. The medium is maintained by the use of dilute sulphuric acid. Potassium permanganate acts as a self-indicator. The potential equation, when potassium permanganate acts as an oxidising agent, is : Before the end point, the solution remains colourless (when KMnO 4 solution is taken in burette) In iodimetric and iodometric titrations, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine. At the end point the blue or violet colour disappears when iodine is completely changed to iodide. U but after the equivalence point only one extra drop of KMnO 4 solution imparts pink colour, i.e., appearance of Cr2 (SO 4 )3  4 K2 SO 4  7 H 2 O  3 I2 D YG pink colour indicates the end point. Potassium permanganate is used for the estimation of ferrous salts, oxalic acid, oxalates, hydrogen peroxide, etc. The solution of potassium permanganate is always first standardised before its use. (ii) Dichromate titrations : These are titrations in which, potassium dichromate is used as an oxidising agent in acidic medium. The medium is maintained acidic by the use of dilute sulphuric acid. The potential equation is K 2 Cr2 O7  4 H 2 SO 4  K 2 SO 4  Cr2 (SO 4 )3  4 H 2 O  3[O] U or Cr2 O 72   14 H   6 e   2Cr 3   7 H 2 O ST The solution of potassium dichromate can be directly used for titrations. It is mainly used for the estimation of ferrous salts and iodides. In the titration of K 2 Cr2 O7 versus ferrous salt either an external indicator (potassium ferricyanide) or an internal indicator (diphenyl amine) can be used. (iii) Iodimetric and iodometric titrations : The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titrations.  I 2  2e  2 I  (reduction) 2 I   I 2  2e  (oxidation) These are divided into two types : (a) Iodimetric titrations : These are the titrations in which free iodine is used. As it is difficult to prepare (3) Precipitation titrations : The titrations which are based on the formation of insoluble precipitates, when the solutions of two reacting substances are brought in contact with each other, are called precipitation titrations. For example, when a solution of silver nitrate is added to a solution of sodium chloride or a solution of ammonium thiocyanate, a white precipitate of silver chloride or silver thiocyanate is formed. AgNO 3  NaCl  AgCl  NaNO 3 AgNO 3  NH 4 CNS  AgCNS  NH 4 NO 3 Such titrations involving silver nitrate are called argentometric titrations. (4) Complexometric titrations : A titration, in which an undissociated complex is formed at the equivalence point, is called complexometric titration. These titrations are superior to precipitation titrations as there is no error due to co-precipitation. Hg 2   2 SCN    Hg(SCN )2  Ag  2CN [ Ag (CN )2 ] EDTA (ethylenediamine tetra-acetic acid) is a useful reagent which forms complexes with metals. In the form of disodium salt, it is used to estimate Ca 2  Chemical Analysis and Mg 2  ions in presence of eriochrome black- T as an indicator. Equivalent masses of acids and bases : Equivalent masses of some acids and bases are given in the following table Table : 21.10  63 g of hydrated oxalic acid Note : Equivalent mass is a variable quantity and depends on the reaction in which the substance takes part. The nature of the reaction should be known before writing the gram equivalent mass of the substance. For example in the reactions. 2 NaCl  2 H 2 SO 4  2 NaHSO 4  2 HCl …..(i) Basicity Mol. Mass Eq. Mass 2 NaCl  H 2 SO 4  Na 2 SO 4  2 HCl HCl 1 36.5 36. 5 =36.5 1 The value of g equivalent mass of H 2 SO 4 in HNO3 1 63 63 =63.0 1 H2SO4 2 98 98 =49.0 2 CH3COOH 1 60 60 =60.0 1 H2C2O4.2H2O 2 126 126 =63.0 2 H3PO4 3 98 98 =32.7 3 H3PO3 2 82 82 =41.0 2 H3PO2 1 66 66 =66.0 1 NaOH KOH NH4OH Acidity Mol. Mass 1 40 40 = 40 1 1 56 56 = 56 1 2 74 74 = 37 2 1 35 35 = 35 1 U Ca(OH)2 60 reaction (i) is 98 g and in reaction (ii) 49 g. E3 (ii) Number of g equivalents Mass of the substance in g  Equivalent mass of the substance M ass of the substance in g Number of g moles  M olecular mass of the substance Volume in litres of the substance at N.T.P. (only for  22.4 gases) Mass in g  1000 Number of milli-equivalent  Equivalent mass M ass in g  1000 Number of milli-moles  M olecular mass No. of moles of the solute w  (iii) Molarity  No. of litres of the solution mV Molarity  molecular mass = strength of the solution (g / L) No. of moles of the solute = Molarity  ID D YG Table : 21.11 Alkali …..(ii) U Acid 945 Eq. Mass ST Calculations of Volumetric analysis The following points should be kept in mind while making calculations of volumetric exercises. (i) 1 g equivalent mass of a substance reacts completely with 1 g equivalent mass of any other substance. 1 g equivalent mass of a substance means equivalent mass of the substance in grams. For example, 1 g equivalent mass of NaOH  40 g of NaOH 1 g equivalent mass of H 2 SO 4  49 g of H 2 SO 4 1 g equivalent mass of KMnO 4 in acidic medium  31.6 g of KMnO 4 1 g equivalent mass of hydrated oxalic acid No. of litres of solution Mass of the solute in g(w)  molarity  No. of litres of solution  mol. mass of solute w No. of g equivalent of the solute  Normality  EV No. of litres of the solution Normality  equivalent mass = strength of the solution (g/L) No. of equivalents of the solute = Normality  No. of litres of solution Mass of the solute in g(w)  Normality  No. of litres of solution  Eq. mass of the solute Molecular mass Normality n Equivalent mass Molarity Normality  n  Molarity (iv) Normality equation : When solutions A and B react completely. N A VA  N B VB Normality of A  volume of A  Normality of B  volume of B Strength A Strength B or  VA   VB Eq. mass A Eq. mass B Wt. of metal hydroxide Eq. wt. of metal hydroxide  wt, of metal oxide Eq. wt. of metal oxide  Eq. wt of metal  Eq. wt of OH Eq. wt. of metal  Eq. wt of O 2  946 Chemical Analysis (v) When the solution is diluted, the following formulae can be applied : N 1 V1  N 2 V2 or M 1 V1  M 2 V2 or S 1 V1  S 2 V2 Before dilution = After dilution (vi) If a number of acids are mixed, the combined normality of the mixture, N x , is given N x Vx  N 1 V1  N 2 V2  N 3 V3 ...... Where Vx is the total volume of the mixture, N 1 60 and V1 are the normality and volume respectively of  Glass rod should never be used in flame test as it ID gives a golden yellow colour due to sodium present in it. An asbestos fibre can be safely used in place of platinum wire. E3 one acid, N 2 and V2 of the second acid and so on.  The transparent bead in borax bead test is made of NaBO2 + B2O3. test and gives better results. U  Filter ash test is an alternative to cobalt nitrate D YG  Both CO2 and SO2 turn lime water milky. Ca(OH )2  CO 2  CaCO 3  H 2 O (Milkiness) Ca(OH )2  SO 2  CaSO 3  H 2 O (Milkiness)  Eq. wt. of KMnO4 is different in acidic, alkaline and neutral medium i.e., 31.6, 158 and 52.67 ST U respectively.