Question 1

Question 1The following materials are provided – solutions of cobalt chloride, ammonia, potassium permanganate, lime water, starch-iodide, sodium hydroxide, lead acetate, potassium iodide. Also provided are litmus and filter papers, glowing splinters and glass rods. Using the above how would you distinguish between :(a) a neutral, acidic and a basic gas(b) oxygen and hydrogen gas(c) carbon dioxide and sulphur dioxide gas(d) chlorine and hydrogen chloride gas(e) hydrogen sulphide and nitrogen dioxide gas(f) ammonia and carbon dioxide gas(g) zinc carbonate and potassium nitrate(h) hydrated copper sulphate and anhydrous copper sulphate(i) ammonium sulphate and sodium sulphate.

Accepted Answer

(a) Litmus paper test is done to distinguish between neutral, acidic and basic gas. Neutral gas does not effect litmus paper. Acidic gas will turn blue litmus paper red and basic gas will turn red litmus blue.(b) A burning wooden splinter is extinguished in hydrogen whereas oxygen gas rekindles a glowing wooden splinter. Hence, the two gases can be distinguished using a wooden splinter.(c) There is no effect of carbon dioxide (CO2) gas on potassium permanganate soln. whereas sulphur dioxide (SO2) turns acidified potassium permanganate from pink to clear colourless.(d) Chlorine turns moist blue litmus red and then bleaches it.Cl2 + H2O ⟶ HOCl + HClHOCl ⟶ HCl + [O]Colouring matter + [O] ⟶ Colourless or bleached product.Whereas, HCl gas only turns moist blue litmus paper red and does not bleach it.(e) Nitrogen dioxide (NO2) liberates iodine [violet vapours] with potassium iodide KI soln., and turns potassium iodide paper brown.2KI + 2NO2 ⟶ 2KNO2 + I2Hydrogen sulphide (H2S) gas turns potassium permanganate (KMnO4) from pink to colourless.2KMnO4 + 3H2SO4 + 5H2S ⟶ K2SO4 + 2MnSO4 + 8H2O + 5S(f) In the presence of ammonia moist red litmus turns blue and carbon dioxide turns moist blue litmus faint red.(g) If we heat the given salts, zinc carbonate turns yellow, evolving carbon dioxide gas which turns lime water milky and has no effect on potassium permanganate solution.ZnCO3→ΔZnO+CO2[Zinc[Zinc oxide][Carboncarbonate][yellow - hot]dioxide][white][white - cold]\begin{matrix} 	ext{ZnCO}_3 &\xrightarrow\Delta & 	ext{ZnO} & + & 	ext{CO}_2 \ 	ext{[Zinc} & & 	ext{[Zinc oxide]} & & 	ext{[Carbon} \ 	ext{carbonate]} & & 	ext{[yellow - hot]} & & 	ext{dioxide]} \ 	ext{[white]} & & 	ext{[white - cold]} \end{matrix}ZnCO3​[Zinccarbonate][white]​Δ​​ZnO[Zinc oxide][yellow - hot][white - cold]​+​CO2​[Carbondioxide]​Whereas, potassium nitrate emits oxygen gas on heating. Oxygen gas rekindles a glowing wooden splinter.2KNO3 ⟶ 2KNO2 + O2Hence, the two salts can be distinguished.(h) Hydrated copper sulphate is bright blue in colour, whereas anhydrous copper sulphate appears as a white powder. Hence, the two can be distinguished by their colour and appearance.(i) When ammonium sulphate is heated with NaOH, ammonia gas will be produced which turns red litmus blue whereas, sodium sulphate will not react with sodium hydroxide.(NH4)2SO4 + 2NaOH ⟶ Na2SO4 + 2H2O + 2NH3Na2SO4 + NaOH ⟶ no reaction.

Question 2

Question 2Give a chemical test to distinguish between the following:(i) Sodium carbonate and sodium sulphate(ii) Potassium chloride and potassium nitrate(iii) Copper carbonate and copper sulphite(iv) Lead chloride and lead sulphide(v) Iron (II) sulphate and iron (III) sulphate(vi) Calcium sulphate and zinc sulphate(vii) Lead nitrate and zinc nitrate(viii) Copper sulphate and calcium sulphate(ix) Manganese dioxide and copper (II) oxide(x) dil. HCl, dil. HNO3, dil. H2SO4.[explain the procedure for the preparation of the solutions for the above tests wherever required]

Accepted Answer

(i) When BaCl2 solution is added to sodium carbonate, a white ppt. is formed which is soluble in dil. HCl.Na2CO3 + BaCl2 ⟶ BaCO3 ↓ [white ppt.] + 2NaClBaCO3 + 2HCl ⟶ BaCl2 [soluble] + H2O + CO2When BaCl2 solution is added to sodium sulphate, a white ppt. is formed which is insoluble in dil. HCl.Na2SO4 + BaCl2 ⟶ BaSO4 ↓ [white ppt.] + 2NaClHence, the two compound can be distinguished.(ii) Add silver nitrate soln. to the given solns., potassium chloride reacts to form a white ppt. which is soluble in NH4OH and insoluble in dil. HNO3. The other soln. is potassium nitrate.KCl + AgNO3 ⟶ AgCl + KNO3KNO3 + AgNO3 ⟶ no white ppt.(iii) Add dil. H2SO4 to both the solns. and heat. When a colourless, odourless gas is evolved which has no effect on acidified KMnO4 or K2Cr2O7 solns., then the gas is carbon dioxide and the compound is copper carbonate.When a colourless gas with a suffocating odour is evolved which turns pink acidified KMnO4 to colourless then the gas is sulphur dioxide and the compound is copper sulphite.(iv) Add dil. H2SO4 to both the solns. and heat. When a colourless gas with a smell of rotten eggs is evolved which turns pink acidified KMnO4 colourless, then the gas is hydrogen sulphide and the compound is lead sulphide.Now, add conc. H2SO4 to both the solns. and heat. When a gas with a pungent smell is evolved, which gives dense white fumes when a rod dipped in ammonia soln. is brought near it then the gas is HCl and compound is lead chloride.(v) When sodium hydroxide is added to the two solns., Iron (II) sulphate solution gives a dirty green ppt. of Fe(OH)2 whereas, Iron (III) sulphate solution forms a reddish brown ppt. of Fe(OH)3. Hence, the two compounds can be distinguished.(vi) When NaOH is added to the given soln., zinc sulphate reacts to form a gelatinous white ppt. which dissolves in excess of NaOH.ZnSO4 + 2NaOH ⟶ Na2SO4 + Zn(OH)2 ↓Zn(OH)2 + 2NaOH [excess] ⟶ 2H2O + Na2ZnO2Whereas, calcium sulphate forms a milky white ppt. which is insoluble in excess of NaOH. Hence, the two can be distinguished.CaSO4 + 2NaOH ⟶ Ca(OH)2 ↓ + Na2SO4(vii) When NaOH is added to each of the compounds, lead nitrate forms a chalky white precipitate of lead hydroxide [Pb(OH)2]Pb(NO3)2 + 2NaOH ⟶ 2NaNO3 + Pb(OH)2 ↓Whereas a gelatinous white precipitate of zinc hydroxide [Zn(OH)2] is formed in case of zinc nitrate.Zn(NO3)2 + 2NaOH ⟶ 2NaNO3 + Zn(OH)2 ↓(viii) When NaOH is added to each of the compounds, copper sulphate forms a pale blue precipitate of copper [II] hydroxide [Cu(OH)2]CuSO4 + 2NaOH ⟶ Na2SO4 + Cu(OH)2 ↓Whereas a milky white precipitate of calcium hydroxide [Ca(OH)2] is formed in case of calcium sulphate.CaSO4 + 2NaOH ⟶ Na2SO4 + Ca(OH)2 ↓(ix) When each of the compound is heated with conc. hydrochloric acid, greenish yellow (chlorine) gas is evolved in case of manganese dioxide and filtrate is brownish in colour. Whereas, no chlorine gas is evolved in case of copper (II) oxide and filtrate is bluish in colour.MnO2 + 4HCl ⟶ MnCl2 + 2H2O + Cl2CuO + 2HCl ⟶ CuCl2 + H2O(x) When BaCl2 is added to the three acids, dil sulphuric acid reacts with BaCl2 to give a white ppt. of BaSO4 but with dil HCl and dil HNO3 no white ppt. is produced.BaCl2 + H2SO4 ⟶ BaSO4 ↓ [white ppt.] + 2HClTo distinguish between dil HCl and dil HNO3, we add Silver Nitrate (AgNO3) solution to the two acids. Dil. HCl reacts with AgNO3 to give a curdy white ppt. of Silver chloride (AgCl) but with dil HNO3, no white ppt. is produced.HCl + AgNO3 ⟶ AgCl + HNO3HNO3 + AgNO3 ⟶ no white ppt.Hence, the acids can be distinguished.

Question 3

Question 3Identify the cation [positive ion] and anion [negative ion] in - A, B and C. Also identify P, Q, R, S, T, U, V, W.(a) Substance 'A' is water soluble and gives a curdy white precipitate 'P' with silver nitrate solution. 'P' is soluble in ammonium hydroxide but insoluble in dil. HNO3. Substance 'A' reacts with ammonium hydroxide solution to give a white precipitate 'Q' soluble in excess of NH4OH.(b) A solution of substance 'B' is added to barium chloride solution. A white ppt. 'R' is formed, insoluble in dil. HCl or HNO3. A dirty green ppt. 'S' is formed on addition of ammonium hydroxide to a solution of 'B' and the precipitate is insoluble in excess of ammonium hydroxide.(c) Substance 'C' is a coloured, crystalline salt which on heating decomposes leaving a black residue 'T'. On addition of copper turnings and conc. H2SO4 to 'C' a coloured acidic gas 'U' is evolved on heating. A solution of 'C' is added to NaOH soln. until in excess. A pale blue ppt. 'P' is obtained insoluble in excess of NaOH. A solution of 'C' is then added to NH4OH soln. in excess to give an inky blue solution 'V'. A solution of 'C' is warmed and hydrogen sulphide gas is passed through it. A black ppt. 'W' appears.

Accepted Answer

(a) A is ZnCl2Cation and anion in ZnCl2 : Zn2+ and Cl-ZnCl2 + 2AgNO3 ⟶ 2AgCl ↓ [curdy white ppt.] + Zn(NO3)2P is AgClThe precipitate AgCl is soluble in ammonium hydroxide but insoluble in dil. HNO3.ZnCl2 + 2NH4OH ⟶ 2NH4Cl + Zn(OH)2 ↓ [white ppt.]2NH4Cl + Zn(OH)2 + 2NH4OH [excess] ⟶ 4H2O + [Zn(NH3)2]Cl2Q is Zn(OH)2(b) B is ferrous sulphate (FeSO4)Cation and anion in FeSO4 : Fe2+ and SO42-FeSO4 + BaCl2 ⟶ BaSO4 ↓ [white ppt.] + 2NaClR is BaSO4FeSO4 + 2NH4OH ⟶ (NH4)2SO4 + Fe(OH)2 ↓S is Fe(OH)2(c) C is copper nitrate Cu(NO3)2Cation and anion in Cu(NO3)2 : Cu2+ and NO31-Black residue (T) : CuO - copper oxide2Cu(NO3)2[Copper nitrate (blue)]⟶Δ2CuO[copper oxide - black]+4NO2[Nitrogen dioxide]+O2[Oxygen]\underset{	ext{[Copper nitrate (blue)]}}{2	ext{Cu(NO}_3)_2} \overset{\Delta}{\longrightarrow} \underset{	ext{[copper oxide - black]}}{2	ext{CuO}} + \underset{	ext{[Nitrogen dioxide]}}{4	ext{NO}_2} + \underset{	ext{[Oxygen]}}{	ext{O}_2}[Copper nitrate (blue)]2Cu(NO3​)2​​⟶Δ​[copper oxide - black]2CuO​+[Nitrogen dioxide]4NO2​​+[Oxygen]O2​​U is Nitrogen dioxide (NO2)P is Copper [II] hydroxide Cu(OH)2Cu(NO3)2 + 2NaOH ⟶ 2NaNO3 + Cu(OH)2 ↓V is [Cu(NH3)4]SO4Cu(NO3)2 + 2NH4OH ⟶ Cu(OH)2 ↓ + 2NH4NO32Cu(OH)2 + 2NH4NO3 + 2NH4OH ⟶ 2Cu(NH3)4NO3 + 4H2OW is CuSCu(NO3)2 + H2S ⟶ CuS + 2HNO3

Question 4

Question 1(2007)Salts A, B, C, D and E undergo reactions (i) to (v) respectively. Identify the anion present in each salt.(i) When AgNO3 solution is added to a soln. of A, a white precipitate, insoluble in dilute nitric acid, is formed.(ii) Addition of dil. HCl to B produces a gas which turns lead acetate paper black.(iii) When a freshly prepared solution of FeSO4 is added to a soln. of C and conc. H2SO4 is gently poured from the side of the test-tube, a brown ring is formed.(iv) When dil. H2SO4 is added to D a gas is produced which turns acidified K2Cr2O7 soln. from orange to green.(v) Addition of dil. HCl to E produced an effervescence. The gas produced turns limewater milky but does not effect acidified K2Cr2O7 soln.

Accepted Answer

(i) Chloride (Cl1-)(ii) Sulphide (S2-)(iii) Nitrate (NO31-)(iv) Sulphite (SO32-)(v) Carbonate (CO32-)

Question 5

Question 2(2007)How will the addition of barium chloride soln. help to distinguish between dil. HCl and dil. H2SO4.

Accepted Answer

Barium chloride reacts with dil H2SO4 to form a white ppt. of barium sulphate.BaCl2 + H2SO4 [dil.] ⟶ 2HCl + BaSO4 ↓ [white ppt.]Whereas, dil. HCl does not form a white ppt. Hence, the two can be distinguished.BaCl2 + HCl [dil.] ⟶ No white ppt.

Question 6

Question 1(2008)The salt which in solution gives a pale green ppt. with NaOH soln. and a white ppt. with barium chloride solution is :Iron (III) sulphateIron (II) sulphateIron (II) chlorideIron (III) chloride

Accepted Answer

Iron (II) sulphatePale green ppt. of Fe(OH)2 and white ppt. of barium sulphate are formed.

Question 7

Question 1(2009)CO2 and SO2 gas can be distinguished using :moist blue litmus paperlime wateracidified K2Cr2O7 papernone of the above.

Accepted Answer

acidified K2Cr2O7 (potassium dichromate) paperReasonThere is no effect of CO2 gas on potassium dichromate whereas SO2 turns acidified potassium dichromate from orange to clear green.K2Cr2O7 + H2SO4 + 3SO2 ⟶ K2SO4 + Cr2(SO4)3 + H2O

Question 8

Question 2(2009)Identity the substance 'R' based on the information given below : The pale green solid 'R' turns reddish brown on heating. It's aqueous solution gives a white precipitate with barium chloride solution. The precipitate is insoluble in mineral acids.

Accepted Answer

The substance 'R' is Ferrous sulphate (FeSO4.7H2O).ReasonWhen heated strongly, the hydrous pale green ferrous sulphate (FeSO4.7H2O), loses it's water of crystallization and decomposes to form brown ferric oxide Fe2O3 along with sulphur dioxide (SO2) and sulphur trioxide (SO3).FeSO4.7H2O →Δ\xrightarrow{\Delta}Δ​ FeSO4 + 7H2O2FeSO4 →Δ\xrightarrow{\Delta}Δ​ Fe2O3 + SO2 + SO3Ferrous sulphate on reaction with barium chloride soln., forms a white ppt. of barium sulphate (BaSO4) which is insoluble in mineral acids.

Question 9

Question 3(2009)Give one test each to distinguish between the following pairs of chemical solutions:(i) ZnSO4 and ZnCl2(ii) FeCl2 and FeCl3(iii) Calcium nitrate soln. and Calcium chloride soln.

Accepted Answer

(i) When BaCl2 soln. is added to ZnSO4, a white ppt. of BaSO4 is formed, whereas, no ppt. is formed in case of ZnCl2. Hence, the two solns. can be distinguished.BaCl2 + ZnSO4 ⟶ BaSO4 ↓ [white ppt.] + ZnCl2BaCl2 + ZnCl2 ⟶ No white ppt.(ii) When NaOH soln. is added to the given solns., FeCl2 reacts to form a dirty green ppt. of Fe(OH)2 whereas, FeCl3 reacts to form a reddish brown ppt. of Fe(OH)3.(iii) Add silver nitrate soln. to the given solns., calcium chloride reacts to form a white ppt. which is soluble in NH4OH and insoluble in dil. HNO3. The other soln. is calcium nitrate.CaCl2 + 2AgNO3 ⟶ 2AgCl ↓ [white ppt.] + Ca(NO3)2Ca(NO3)2 + AgNO3 ⟶ no white ppt.

Question 10

Question 1(2010)Select the correct answer from A, B, C, D and E –(A) : Nitroso Iron (II) sulphate(B) : Iron (III) chloride(C) : Chromium sulphate(D) : Lead (II) chloride(E) : Sodium chloride.The compound which is responsible for the green colour formed when SO2 is bubbled through acidified potassium dichromate solution.

Accepted Answer

Chromium sulphateK2Cr2O7 + H2SO4 + 3SO2 ⟶ K2SO4 + Cr2(SO4)3 + H2O

Question 11

Question 2(2010)State your observation – (i) A piece of moist blue litmus paper (ii) paper soaked in potassium permanganate solution - is introduced in each case into a jar of sulphur dioxide.

Accepted Answer

(i) Moist blue litmus turns red.(ii) Sulphur dioxide turns potassium permanganate from pink to clear colourless.2KMnO4 + 2H2O + 5SO2 ⟶ K2SO4 + 2MnSO4 + 2H2SO4

Question 12

Question 3(2010)Write the equation for the reaction of magnesium sulphate solution with barium chloride solution.

Accepted Answer

MgSO4 + BaCl2 ⟶ BaSO4 ↓ [white ppt.] + MgCl2

Question 13

Question 1 (2011)Choose from the list of substances – Acetylene gas, aqua fortis, coke, brass, barium chloride, bronze, platinum.An aqueous salt solution used for testing sulphate radical.

Accepted Answer

Barium chloride.

Question 14

Question 1(2012)Name the gas which turns acidified potassium dichromate clear green.

Accepted Answer

Sulphur dioxide gas (SO2)K2Cr2O7 + H2SO4 + 3SO2 ⟶ K2SO4 + Cr2(SO4)3 + H2O

Question 15

Question 2(2012)Identify the anion present in the following compounds :(i) Compound X on heating with copper turnings and concentrated sulphuric acid liberates a reddish brown gas.(ii) When a solution of compound Y is treated with silver nitrate solution a white precipitate is obtained which is soluble in excess of ammonium hydroxide solution.(iii) Compound Z which on reacting with dilute sulphuric acid liberates a gas which turns lime water milky, but the gas has no effect on acidified potassium dichromate solution.(iv) Compound L on reacting with Barium chloride solution gives a white precipitate insoluble in dilute hydrochloric acid or dilute nitric acid.

Accepted Answer

(i) Nitrate ion, NO3-(ii) Chloride ion, Cl-(iii) Carbonate ion, CO32-(iv) Sulphate ion, SO42-

Question 16

Question 3(2012)State one chemical test between each of the following pairs :(i) Sodium carbonate and Sodium sulphite(ii) Ferrous nitrate and Lead nitrate(iii) Manganese dioxide and Copper (II) oxide

Accepted Answer

(i) When dil. sulphuric acid is added to sodium carbonate and heated, colourless, odourless gas is evolved which turns lime water milky and has no effect on KMnO4 or K2Cr2O7 solutions.When dil. sulphuric acid is added to sodium sulphite and heated, colourless gas with suffocating odour is evolved which turns lime water milky. It turns acidified K2Cr2O7 from orange to clear green and pink coloured KMnO4 to clear colourless.Hence, the two compounds can be distinguished.(ii) When NaOH is added to each of the compounds, a dirty green precipitate of Iron [II] hydroxide [Fe(OH)2] is formed in case of Ferrous nitrate whereas a chalky white precipitate of lead hydroxide [Pb(OH)2] is formed in case of lead nitrate. Iron [II] Hydroxide [Fe(OH)2] is insoluble in excess of NaOH, whereas Lead Hydroxide [Pb(OH)2] is soluble in excess of NaOH. Hence, the two compounds can be distinguished.(iii) When each of the compound is heated with conc. hydrochloric acid, greenish yellow (chlorine) gas is evolved in case of manganese dioxide and filtrate is brownish in colour whereas, no chlorine gas is evolved in case of copper (II) oxide and filtrate is bluish in colour.MnO2 + 4HCl ⟶ MnCl2 + 2H2O + Cl2CuO + 2HCl ⟶ CuCl2 + H2O

Question 17

Question 4(2012)State one observation : A zinc granule is added to copper sulphate solution.

Accepted Answer

Zinc displaces copper from copper sulphate and forms zinc sulphate soln. because zinc is more reactive than copper. Hence, the blue colour of copper sulphate solution changes to colourless.

Question 18

Question 5(2012)Give balanced equation for the reaction :Silver nitrate solution and Sodium chloride solution.

Accepted Answer

AgNO3 + NaCl ⟶ AgCl ↓ [white ppt.] + NaNO3

Question 19

Question 1(2013)Give a chemical test to distinguish between :(i) NaCl soln. and NaNO3 soln.(ii) HCl gas and H2S gas.(iii) Calcium nitrate soln. and zinc nitrate soln.(iv) Carbon dioxide gas and sulphur dioxide gas.

Accepted Answer

(i) Add silver nitrate soln. to the given solns., sodium chloride reacts to form a white ppt. which is soluble in NH4OH and insoluble in dil. HNO3. The other soln. is sodium nitrate.NaCl + AgNO3 ⟶ AgCl ↓ [white ppt.] + NaNO3NaNO3 + AgNO3 ⟶ no white ppt.(ii) Hydrogen sulphide gas turns moist lead acetate paper silvery black or black whereas, no change is observed in case of HCl gas.Pb(CH3COO)2 [colourless] + H2S ⟶ PbS [black] + 2CH3COOH(iii) When NaOH is added to the given soln., Zn(NO3)2 reacts to form a gelatinous white ppt. which dissolves in excess of NaOH whereas, Ca(NO3)2 forms a milky white ppt. which is insoluble in excess of NaOH. Hence, the two can be distinguished.(iv) Carbon dioxide gas has no effect on acidified potassium permanganate (KMnO4) and acidified potassium dichromate (K2Cr2O7) solution whereas sulphur dioxide gas turns acidified potassium permanganate from pink to clear colourless and acidified potassium dichromate from orange to clear green.2KMnO4 + 2H2O + 5SO2 ⟶ K2SO4 + 2MnSO4 + 2H2SO4K2Cr2O7 + H2SO4 + 3SO2 ⟶ K2SO4 + Cr2(SO4)3 + H2O

Question 20

Question 2(2013)FromA : COB : CO2C : NO2D : SO3State which will not produce an acid on reaction with water.

Accepted Answer

COReason — On reaction with water, CO2 produces carbonic acid, NO2 forms nitric acid and SO3 forms sulphuric acid whereas CO forms carbon dioxide gas.CO + H2O ⟶ CO2 + H2

Practical Chemistry

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