Qualitative Analysis (Cations) Exercise -1

Qualitative Analysis (Cations) Marked questions are recommended for Revision. PART - I : SUBJECTIVE QUESTIONS Section (A) : Ist Group A-1. Name of one chloride which is soluble in hot water as well as in excess of HCl. A-2. Why do lead salts turn black on keeping for a long time in the laboratory ? A-3. A metal salt forms a yellow precipitate (P) with potassium iodide solution and black precipitate (Q) with acidified H2S. Yellow precipitate (P) dissolve in excess of concentrated solution of KI and ppt.reappears on dilution. Then metal salt could be and also write the chemical equation. A-4. When calomel reacts with ammonia solution, a black precipitate is formed. Write the chemical equation and also name the reaction nature. A-5. What products are formed when precipitate formed by the reaction of Hg22+ ions and excess of sodium hydroxide solution is boiled ? A-6. A metal salt forms a green precipitate with KI solutiion and white ppt with KCl. This green precipitate under goes disproportionation reaction in excess of reagent which results in a soluble compound and black precipitate then write the chemical equations and identify the cation in the original sample. A-7. Why bright yellow precipitate of AgI is soluble in KCN and Na 2S2O3 solution. A-8. Solid (A) is used in photographic film, Identify (A) to (E) ? Section (B) : II A Group B-1. Why Na2S cannot be used in place of H2S (in presence of HCl) as a reagent for IInd group cations ? B-2. Does sodium hydroxide solution can be used to differentiate Hg(I) from Hg(II) ? B-3. What happens when white precipitate of Bi(OH)3 is boiled ? B-4. Identify A to F. (F) K 4 [Fe(CN)6 ] BaCl2 (A) Chocolate ⎯⎯⎯⎯⎯ ⎯ ⎯⎯⎯⎯ → Blue solution brown Hypo (D) ⎯ → Brown ppt. ⎯⎯⎯ Solution (B) + (C) White ppt. (E) White ppt. Qualitative Analysis (Cations) Section (C) : IIB Group C-1. An original solution is prepared in conc. HCl when diluted a white ppt. is formed. What does it indicate ? C-2. Why HNO3 can not be used in place of HCl as a reagent (H2S in presence of HCl) for IInd group cations? C-3. A chloride of an element (X) forms yellow precipitate with H 2S in acidic medium. This yellow precipitate is soluble in Conc. HNO3 due to formation of compound (Y). Compound (Y) gives ammonium molybdate test then find the metal salt (X). C-4. Identify (A) based on the following facts : (i) (A) reduces HgCl2 solution to white ppt. changing to grey. (ii) (A) turns FeCl3 yellow coloured solution to green. (iii) (A) gives white ppt. with NaOH soluble in excess of NaOH. (iv) (A) gives yellow dirty ppt. on passing H2S gas, soluble in yellow ammonium sulphide (YAS). (v) (A) gives chromyl chloride test. Section (D) : IIIrd Group D-1. Why is the original solution boiled with conc. HNO 3 in group  ? D-2. Do Fe(III) salts and Fe(II) salts both give red colouration with dimethylglyoxime in ammonical solution. If not then which iron salt gives red colouration with dimethylglyoxime ? D-3. Which colour precipitate is formed by Fe(II) salt with potassium ferrocyanide, (i) in complete absence of air and (ii) under ordinary atmospheric conditions ? Section (E) : IVth Group E-1. Partial precipitation of Mn2+ as Mn(OH)2 occurs with ammonia solution but the precipitate is soluble in ammonium salts. Explain ? E-2. Why Zn(II) salt is not precipitated as Zn(OH)2 by ammonia solution in the presence of excess of ammonium chloride ? E-3. In the absence of copper ion, zinc ion form a white precipitate with the ammonium tetrathiocyanatomercurate(II) How the presence of Cu2+ ion affect this test ? E-4. Identify (A) to (D).  ⎯⎯→ + (On cooling (C) turns white) (B) (C) (A) (Brown gas) (Yellow residue) Colourless water soluble compound (D) (White ppt.) Section (F) : Vth, VIth and Zero Group F-1. What is the formula of iodide of Millon's base ? F-2. What happens when ammonia gas is passed into a solution of sodium cobaltinitrite ? F-3. What will happen if the precipitation of Vth group cation by ammonium carbonate is replaced by Na2CO3? F-4. What happens when ammonium sulphate solution is added to a solution containing both Sr 2+ and Ca2+ ions? F-5. Which colour precipitate is obtained when a solution of Ca2+ ions reacts with potassium ferrocyanide. Qualitative Analysis (Cations) PART - II : ONLY ONE OPTION CORRECT TYPE Section (A) : Ist Group A-1. A metal nitrate reacts with KI solution to give yellow precipitate which on addition of excess of more concentrated solution (6 M) of KI dissolves forming a solution. The cation of metal nitrate is : (A) Hg22+ (B) Ag+ (C) Pb2+ (D) Cu2+ A-2. Three separate samples of a solution of a single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dilute NaCl solution and one formed a black precipitate with H2S. The salt could be : (A) AgNO3 (B) Pb(NO3)2 (C) Hg(NO3)2 (D) Mn(NO3)2 A-3. Consider the following observation : 2– CrO4  → water soluble ⎯⎯⎯ Mn+ + HCl (dilute) ⎯⎯→ white precipitate ⎯⎯ → yellow precipitate. n+ The metal ion M will be : (A) Hg2+ (B) Ag+ (C) Pb2+ (D) Sn2+ A-4. When calomel reacts with NH4OH solution the compound formed is : (A) Hg2Cl2 (B) Hg(NH2)Cl (C) Hg(NH3)2Cl A-5. Consider the following equilibrium : AgCl  + 2NH3 [Ag(NH3)2]+ + Cl– White ppt of AgCl appears on adding (A) NH3 (B) aq. NaBr A-6. AgCl with NH3 forms a complex : (A) [Ag(NH3)2]Cl (B) AgNO3 (D) HgCl2.NH3 (C) aq. HNO3 (D) aq. NH4I (C) [Ag(NH2)2]Cl (D) Ag mirror Section (B) : IIA Group B-1. Sometimes yellow turbidity appears while passing H2S gas even in slightly acidic medium in the absence of  group radicals. This is because : (A) sulphur is present in the mixture as impurity. (B) IV group radicals are precipitated as sulphides. (C) of the oxidation of H2S gas by some acid radicals. (D)  group radicals are precipitated as hydroxides. B-2. H2S in the presence of HCl precipitates  group but not V group because : (A) HCl activates H2S (B) HCl increases concentration of Cl– (C) HCl decreases concentration of S2– (D) HCl lowers the solubility of H2S in solution B-3. When small amount of SnCl2 is added to a solution of Hg2+ ions, a silky white precipitate is obtained. The silky white precipitate is due to the formation of : (A) Hg2Cl2 (B) SnCl4 (C) Sn (D) Hg B-4. When excess of dilute NH4OH is added to an aqueous solution of copper sulphate an intense blue colour is developed. This is due to the formation of : (A) [Cu(NH3)6]2+ (B) Cu(OH)2 (C) [Cu(NH3)4]2+ (D) (NH4)2SO4 B-5. A black sulphide is formed by the action of H2S on : (A) cupric chloride (B) cadmium chloride (C) zinc chloride (D) ferric chloride. B-6. Which one of the following salts will produce clear and transparent original solution in 2M HCl ? (A) Ag2CO3 (B) Pb(CO3) (C) Hg2CO3 (D) CuCO3 B-7. When bismuth chloride is poured into a large volume of water the white precipitate produced is of : (A) BiO.OH (B) Bi2 O3 (C) BiOCl (D) Bi(OH)3 B-8. In which of the following pairs the precipitates are red and black coloured respectively and both precipitates are soluble in excess KI solution ? (A) HgI2, Hg2I2 (B) HgI2, BiI3 (C) Cu2I2, AgI (D) CdI2, PbI2 Qualitative Analysis (Cations) B-9. A metal chloride original solution (i.e. O.S) on mixing with K2CrO4 solution gives a yellow precipitate soluble in aqueous sodium hydroxide. The metal may be : (A) mercury (B) iron (C) silver (D) lead Section (C) : IIB Group C-1. Which of the following metal sulphide is soluble in YAS (yellow ammonium sulphide) : (A) HgS (B) PbS (C) Bi2S3 (D) Sb2S3 C-2. When white crystalline precipitate of magnesium ammonium arsenate is treated with acidified silver nitrate solution, precipitate of silver arsenate is formed. The colour of precipitate is : (A) Yellow (B) Brownish red (C) White (D) Brownish black Section (D) : IIIrd Group D-1. When NH4Cl is added to a solution of NH4OH : (A) the dissociation of NH4OH increases. (B) the concentration of OH– increases. (C) the concentrations of both OH– an NH4+ increase. (D) the concentration of OH– ion decreases. D-2. To avoid the precipitation of Hydroxides of Zn2+, Mn2+ and Ni2+ along with those of Fe3+, Al3+ and Cr3+ the third group solution should be : (A) Concentrated HNO3 is added (B) Treated with excess of NH4Cl (C) Concentrated H2SO4 is added (D) Treated with excess of NH4OH D-3. An inorganic lewis acid (X) fumes in moist air, and intensity of fumes increases when and rod dipped in NH4OH is brought near to it. An acidic solution of (X) on addition of NH 4Cl and NH4OH gives a precipitate which dissolves in NaOH solution. A solution of (X) does not give precipitate with H 2S Hence, compound (X) is (A) FeCl3 (B) AlCl3 (C) CrCl3 (D) ZnCl2 D-4. An original solution of an inorganic salt in dilute HCl gives a brown colouration with potassium hexacyanidoferrate (III) and reddish brown colouration with sodium acetate solution. The cation of the salt is : (A) Ni2+ (B) Fe3+ (C) Cu2+ (D) none D-5. Fe(OH)3 and Cr(OH)3 precipitates can be completely separated by : (A) Aq. NH3 (B) HCl (C) NaOH/H2O2 (D) H2SO4 Section (E) : IVth Group E-1. A coloured solution of a salt gives following reactions. (i) It gives white precipitate with sodium hydroxide which becomes brown on exposure to air. (ii) It gives white precipitate with ammonia solution which is soluble in ammonium salts. (A) Mn2+ (B) Zn2+ (C) Al3+ (D) Ni2+ E-2. A metal salt form precipitate with H2S in presence of (NH4OH + NH4Cl) and this precipitate is soluble in acetic acid then metal sulphide could be : (A) ZnS (B) CoS (C) MnS (D) NiS E-3. To increase significantly the concentration of free Zn2+ ion in a solution of the complex ion [Zn(NH3)4]2+ Zn2+ (aq) + 4NH 3 (aq) [Zn(NH3)4]2+ (aq) add to the solution some : (A) H2O (B) HCl (aq) (C) NH3(aq) (D) NH4Cl (aq) E-4. An aqueous solution of colourless metal sulphate M, gives a white precipitate with NH 4OH. This was soluble in excess of NH4OH. On passing H2S gas through this solution a white precipitate is formed. The metal M in the salt is : (A) Ca (B) Ba (C) Al (D) Zn E-5. A metal salt solution when treated with dimethyl glyoxime and NH 4OH gives a rose red complex. The metal is : (A) Ni (B) Zn (C) Co (D) Mn. Qualitative Analysis (Cations) E-6. The ion that can not be precipited by H 2S in presence of dil. HCl. (A) Pb2+ (B) Bi3+ (C) Cu2+ (D) Ni2+ Section (F) : Vth, VIth and Zero Group F-1. Aqueous Solution of BaBr2 gives yellow precipitate with : (A) K2CrO4 (B) AgNO3 (C) (CH3COO)2Pb (D) (A) and (B) both F-2. The addition of K2CO3 (aq) to the following solution is expected to produce a precipitate in every case but that one which does not produce precipitate is : (A) BaCl2(aq) (B) CaBr2(aq) (C) Na2SO4(aq) (D) Pb(NO3)2 (aq) F-3. An aqueous solution of salt gives white precipitate with AgNO 3 solution as well as with dilute H2SO4. It may be (A) Pb(NO3)2 (B) Ba(NO3)2 (C) BaCl2 (D) CuCl2 F-4. Mg is not precipitated in V group because : (A) MgCO3 is soluble in water. (C) MgCO3 is soluble in NH4OH. (B) Ksp of MgCO3 is high. (D) None. F-5. A metal salt solution form a yellow ppt with potassium chromate in acetic acid, a white ppt with dilute sulphuric acid but gives no ppt with sodium chloride or iodide it is : (A) Lead carbonate (B) Basic lead carbonate (C) Barium nitrate (D) Strontium nitrate F-6. A white crystalline substance dissolves in water. On passing H 2S in this solution, a black precipitate is obtained. The black precipitate dissolves completely in hot HNO 3. On adding a few drops of concentrated H2SO4, a white precipitate is obtained. This precipitate is that of : (A) BaSO4 (B) SrSO4 (C) PbSO4 (D) CdSO4 F-7. The yellow precipitate formed by passing ammonia into Nessler’s reagent in due to the formation of (A) Hgl42– (B) NH2O–Hg–Hgl (C) NH2–Hg–O–Hg–I (D) NH3–Hg–I PART - III : MATCH THE COLUMN 1. Match the precipitates listed in column-I with their suitable solvents listed in column-II. Column I Column II (precipitate) (solvent) (A) (p) Concentrated HCl. AgCl (white) (B) (q) Dilute ammonia solution (excess). CuS (black) (C) (r) Potassium cyanide solution. Zn(OH)2 (white) (D) (s) Hot 50% nitric acid. BaCO3 (white) (t) Sodium hydroxide solution. 2. Match the basic radicals listed in column-I with the properties listed in column-II. Column-I Column-II (Basic radical) (Properties) (A) Mn2+ (p) Forms coloured metaborate in oxidising flame in borax bead test. Forms white precipitate with sodium hydroxide but on exposure to air (B) Cr3+ (q) turns rapidly brown. With both potassium cyanide (not in excess) and ammonia solution (C) Al3+ (r) separately forms reddish brown precipitate. With excess of sodium hydroxide forms soluble complex but on adding (D) Fe3+ (s) acid to soluble complex, a precipitate is obtained which redissolves on adding excess of acid. Qualitative Analysis (Cations) EXERCISE - 1 PART - I A-1. PbCl2 is soluble in hot water and it also dissolves in concentrated HCl and KCl solutions forming soluble complex. ; PbCl2 + 2HCl ⎯→ H2PbCl4 (soluble complex). A-2. Pb2+ + H2S (atmospheric) ⎯→ PbS  (black) + 2H+. A-3. PbCl2 + 2KI ⎯→ PbI2 (yellow) + 2KCl ; PbCl2 + H2S ⎯→ PbS (Black) PbI2 + KI(Conc.) K2(PbI4) (aq) yellow precipitate of Pbl2 dissolves in excess KI(conc.) and give PbI42– , ppt reappears on dilution. A-4. Hg2Cl2 + 2 NH4OH ⎯→ HgNH2Cl  (white) + Hg  (black) + NH4Cl + 2 H2O 1 4 4 4 4 4 4 44 2 4 4 4 4 4 4 4 43 black Disproportionation reaction. A-5. Hg22+ + 2OH– ⎯→ Hg2O  (black) + H2O Hg2O  ⎯→ HgO  + Hg  (grey) ; Disproportionation reaction. A-6. Hg22+ + 2 – ⎯⎯ → Hg2  2  green A-7. AgI + 2CN– ⎯→ [Ag(CN)2]– + I–, AgI + 2S2O32– ⎯→ [Ag(S2O3)2]3– + I– for formation of soluble complex A-8. (A) is AgBr (present in photographic plate) AgBr + 2Na2S2O3 ⎯→ Na3[Ag(S2O3)2] + NaBr soluble (B) NH3 HCl+HNO3 [Ag(NH3 )2 Cl] ⎯⎯ ⎯ AgCl ⎯⎯⎯⎯⎯ ⎯ Ag2S (D) (C) (E) B-1. With Na2S, along with 2nd group cations, some cations of higher groups i.e., III rd and IVth groups may be precipitated because the higher concentration of S 2– ions is obtained according to following reaction. Na2S 2 Na+ + S2– nd To precipitate II group cations lower concentration of S2– ions is required (because of low Ksp of IInd group sulphides) and this can be obtained easily by H2S in presence of dilute HCl. H2S 2 H+ + S2– ; HCl H+ + Cl– Due to common ion effect the ionisation of H 2S is suppressed and the concentration of S2– ions obtained is just sufficient to precipitate the cations of II nd group. B-2. Yes, B-3. It turns yellowish–white owing to the formation of BiO.OH . Bi(OH)3 ⎯→ BiO.OH  + H2O B-4. (A) : CuSO4 (B) : BaSO4 (E) (Cu2I2 + I– + S4O62–) (F) Cu2 [Fe(CN)6] Hg2+ + 2OH– ⎯→ HgO + H2O yellow Hg22+ + 2OH– ⎯→ Hg2O + H2O black (C) : HCl (D) : (Cu2I2 + I3–) Qualitative Analysis (Cations) C-1. Presence of Sb3+ or Bi3+, their chloride hydrolyse to oxychlorides in presence of excess of water. BiCl3 + H2O ⎯→ BiOCl + 2HCl SbCl3 + H2O ⎯→ SbOCl + 2HCl C-2. HNO3 is a powerful oxidising solution. Causing confusion with CdS, AS 2S3 (both yellow ppt). Colloidal solution can not be filtered causing unnecessary trouble. C-3. → As2S3  (yellow) + 6HCl AsCl3 (x) + 3H2S ⎯⎯⎯ 3As2S3 + 28HNO3+4H2O ⎯→ 6H3AsO4(y) + 18H + 9SO2– 4 + 28 NO H3 AsO4(y) + 12 NH4MoO4+21 HNO3 ⎯→ (NH4) AsO4.12MoO3(yellow) +21 NH4NO3+12H2O C-4. (A) is SnCl2 D-1. Fe2+ salt are oxidised to Fe3+ salt by boiling with conc. HNO3, before adding NH4Cl and NH4OH, otherwise Fe2+ would not be precipitated in group . D-2. Only Fe(II) salts give soluble red iron(II) dimethylglyoxime in ammonical solution. D-3. (i) White precipitate, K2Fe[Fe(CN)6] and (ii) Pale blue precipitate. E-1. Mn2+ + 2 NH3 + 2 H2O Mn(OH)2  (white) + NH4+ . In presence of ammonium salts, the reaction proceeds in backward direction. E-2. Zn2+ + 2 NH3 + 2 H2O Zn(OH)2 + 2 NH4+ – It is due to the lowering of OH ion concentration because of common ion effect of NH 4+ to such a value that the Ksp of Zn(OH)2 is not attained. E-3. → H+ → Cu2+ + [Hg(SCN)4]2– ⎯→ Cu[Hg(SCN)4] * In the presence of the copper ions, the copper complex Co-precipitated with that of zinc complex and colour become violet E-4. (A) : Zn (NO3)2 (B) : NO2 F-1. Hg2ONH2I or F-2. A yellow precipitate of ammonium cobaltinitrite is formed. 3 NH4+ + Na3[Co(NO2)6] ⎯→ (NH4)3 [Co(NO2)6]  (yellow) + 3 Na+ F-3. Mg2+ ions will also be precipitated. F-4. Sr2+ + NH4+ + SO42– ⎯→ SrSO4  (white) + NH4+ Ca2+ + 2 NH4+ + 2 SO42– ⎯→ (NH4)2[Ca(SO4)2] (soluble complex). F-5. White precipitate is obtained. Ca2+ + 2K+ + [Fe(CN)6]4– ⎯→ K2Ca[Fe(CN)6]  or (C) : ZnO (D) : ZnS . HgO PART - II A-1. (C) A-2. (B) A-3. (C) A-4. (B) A-5. (C) A-6. (A) B-1. (C) B-2. (C) B-3. (A) B-4. (C) B-5. (A) B-6. (D) B-7. (C) B-8. (B) B-9. (D) C-1. (D) C-2. (B) D-1. (D) D-2. (B) D-3. (B) D-4. (B) D-5. (C) E-1. (A) E-2. (C) E-3. (B) E-4. (D) E-5. (A) E-6. (D) F-1. (D) F-2. (C) F-3. (C) F-4. (B) F-5. (C) F-6. (C) F-7. (C) PART - III 1. (A - p,q,r ; B - r,s ; C - p,q,r,s,t ; D - p,s) 2. (A - p,q ; B - p,s ; C - s ; D - p,r)