IITPrep

Ques. Paper code : CINO1
solutions code : SCINO1

Chemistry - Qualitative Analysis

SOLUTIONS


  1. The given reactions point out that the gas X is SO2. The given reactions of SO2 are as follows:
    1. K2Cr2O7 + 3H2SO4 ® K2SO4 + Cr2(SO4)3 + 3H2O +5O
      H2O + SO2 + O ® H2SO4
    2. H2O2 ® H2O + O
      H2O + SO2 + O ® H2SO4
      BaCl2 + H2SO4 ® BaSO4 + 2HCl
    3. 2H2S + SO2 ® 3S¯ + 2H2O
  2. The given facts of this question are :-
    Mixture of 2 salts ® (heat with NaOH) Gas ® (alk K2HgI4) Reddish brown ppt ® (K2Cr2O7 conc H2SO4) Red vap of A
    Aq soln of mixture ® (BaCl2) White ppt¯ ® (K3[Fe(CN)6]) Deep blue coloration of B
    The given reaction sequences leads to the following conclusions:
    1. Heating of mix with NaOH to give NH3 gas (indicated by reddish brown ppt with alk soln of K2HgI4) indicates the presence of NH4+ ion in the mixture.
    2. Heating the mixture with K2Cr2O7 and conc. H2SO4 to give red vapours of chromyl chloride indicates the presence of Cl- ion in the mix.
    3. Reaction of aq soln of the mix with BaCl2 soln to give white ppt of BaSO4 sparingly soluble in conc HCl indicates the presence of SO42- ions in the mixture.
    4. Reaction of aq soln of the mix with K3[Fe(CN6)] soln to give deep blue color indicates the presence of Fe2+ ions in the mixture.

    Hence the mix contains the following 4 ions: Fe2+, NH4+, SO42-, Cl-
    Equations for the formation of A and B:
    4NaCl + K2Cr2O7 + 3H2SO4 ®(heat) K2SO4 + 2Na2SO4 + 2CrO2Cl2 (A) + 3H2O
    3Fe2+ + 2K3[Fe(CN)6]) ® Fe3[Fe(CN)6]2 (B) + 6K+
  3. Enlist the facts for yourself. The conclusions that can be drawn are:
    1. Reactions of aq soln of the compd E with BaCl2 to give white ppt insoluble in HNO3 indicates that the salt E contains SO42- ion
    2. Reaction of compd G with K4[Fe(CN)6] in presence of acetic acid to give choc ppt H indicates that G must contain Cu2+ and hence H must be cupric ferrocyanide
    3. Since compd G is derived from the reaction of gas B and compd E, E must contain Cu2+ ion. Ppt of Cu2+ ion soluble in dil HNO3 should be CuS; hence F must be CuS and thus B is H2S
    4. According to pt (i) E contains SO42- hence E must be CuSO4
    5. Gas B which has now been identified as H2S is obtained by decomposition of a black colored compd A with dil H2SO4. A must be sulphide of Cu, Pb, Hg, Fe, Co, Ni, etc.
    Thus the various compds A to H can be identified as follows:
    FeS (A) + H2SO4 ® FeSO4 + H2S (B)
    H2S + HNO3 (C) ® 2NO2 + 2H2O + S (D)
    CuSO4 (E) + H2S ® (HCl dil) CuS (F) + H2SO4
    3CuS + 8HNO3 ® 3Cu(NO3)2 + 2NO + 3S + 4H2O
    Cu(NO3)2 + H2SO4 ® CuSO4 + 2HNO3
    Cu(SO4) + 4NH4OH ® [Cu(NH3)4]SO4 (G)+ 4H2O
    [Cu(NH3)4]SO4 + 4CH3COOH ® CuSO4 +4CH3COONH4
    2CuSO4+K4[Fe(CN)6] ® Cu2[Fe(CN)6] (H) + 2K2SO4
    CuSO4+BaCl2 ® BaSO4 + CuCl2
  4. Test (i) indicates that the mix contains Cl- ion which is liberated as Cl2 when heated with MnO2 and conc H2SO4.Test (ii) indicates the presence of NH4+ ion in the mixture which gives ammonia when heated with NaOH soln. Since NH3 is basic in nature, it turns red litmus blue. Presence of NH4+ in the mix is further confirmed by test (iv) according to which the gas NH3 gives brown ppt with Nesslers reagent. Test (iii) indicates Fe2+ in the mix which gives blue ppt with pot. ferricyanide. Red colouration with ammonium Thiocyanate indicates that the mix also contains Fe3+ ions which are believed to be formed by the oxidation of Fe2+ ions by air. Thus the mix contains FeCl2 and NH4Cl. Ionic reactions can be easily written.
  5. Since the compd A on stg heating gives 2 oxides of S which might be SO2, SO3, it must be a sulphate. The reacn of E with thiocyanate indicates Fe3+. Thus the compd A must be FeSO4.7H2O; B-Fe2O3; C-SO2; D-SO3; E-FeCl3; F-S; G-FeCl2; H-Fe(CNS)3
  6. Reactions of B and C indicate A must be CrCl3. Reactions are as follows:
    CrCl3+ NaOH + H2O2 ® Na2CrO4 (B) + NaCl + H2O
    Na2CrO4 + H2SO4 ® Na2Cr2O7 (C) + Na2SO4 + H2O
    Na2Cr2O7 + KCl + H2SO4 ® CrO2Cl2 (D) + NaHSO4 + H2O + KHSO4
    Na2Cr2O7 + NH4Cl ® (NH4)2Cr2O7 (E) + NaCl
    (NH4)2Cr2O7 ® (heat) N2 (F) + 4H2O + Cr2O3 (G)
    Mg + N2 ® Mg3N2 (H)
    Mg3N2 + H2O ® MgO + 2NH3 (I)
  7. The given compd is heated no gas comes out but white powder on heating and agan white on cooling. When the white solid is heated with a soln of cobalt nitrate a green coloured product called as Rinman’s Green is formed. It is ZnO.
    The white powder on heating gives a gas which turns lime water milky and the residue gives the above tests. It is ZnCO3.
    The white powder on heating gives a vapour which can be tested by anhydrous CuSO4 which becomes blue and the residue gives the tests. It is Zn(OH)2
    The white substance on heating with dil H2SO4 gives H2S which turns lead acetate paper black. The compd is ZnS.
    Since tetracyano complex is more stable than tetramine complex hence if a soln of NaCN is added ammonia is liberated which can be tested by heating the mixture and testing the gas with HCl which gives white fumes.
    Zn(NH3)4Cl2 + 4NaCN + H2O ® Na2[Zn(CN)4] + 2NaCl + NH4OH
    NH4OH ® NH3 + H2O
    NH3 + HCl ® NH4Cl
  8. Quite straight forward A-ZnCO3; B-CO2; C-ZnO; D-ZnS; E-Zn(OH)2
  9. Gas D reacts with Ca forming a compd which on hydrolysis gives ammonia indicating that D must be N2. Residues E and G give yellow flame on burning indiacting Na salts. Hence B and A must also be Na salts.
    The colourless solid B gives reddish brown fumes with dil acids the reddish brown fumes are probably of NO2. Hence compd B must be nitrate.(Recall NO3- are not attacked by dil acids). Hence A must be NO3- and B must be NO2-
    NaNO3 (A) ® NaNO2 (B) + O2 (C)
    NaNO2 + H2SO4 ® Na2SO4 + 2HNO2
    HNO2 ® HNO3 + H2O + NO
    NO + O2 ® NO2
    NaNO2 + NH4Cl ® NaCl (E)+N2 (D)+H2O
    NaNO3 + (NH4)2SO4 ® Na2SO4 (G) + NH3 (F) + HNO3
    Mg + O2 ® MgO
    MgO + H2O ® Mg(OH)2
    Ca + N2 ® Ca3N2
    Ca3N2 + H2O ® Ca(OH)2 + NH3
  10. Gas A could be H2S or SO2 since it turns acidicfied dichromate green. But H2S is ruled out because the mix does not turn lead acetate paper black. Hence A is SO2. Since B turns lime water milky, it is CO2. C is O2 which is absorbed by pyragallol.
    K2Cr2O7 + H2SO4 + SO2 ® K2SO4 + Cr2(SO4)3 + H2O
    Ca(OH)2 + CO2 ® CaCO3 + H2O