Chemistry
Give 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]
Practical Chemistry
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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.] + 2NaCl
BaCO3 + 2HCl ⟶ BaCl2 [soluble] + H2O + CO2
When BaCl2 solution is added to sodium sulphate, a white ppt. is formed which is insoluble in dil. HCl.
Na2SO4 + BaCl2 ⟶ BaSO4 ↓ [white ppt.] + 2NaCl
Hence, 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 + KNO3
KNO3 + 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 + Na2ZnO2
Whereas, 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 + Cl2
CuO + 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.] + 2HCl
To 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 + HNO3
HNO3 + AgNO3 ⟶ no white ppt.
Hence, the acids can be distinguished.
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Related Questions
Identify the salts P, Q, R from the following observations:
(i) Salt P has light bluish green colour. On heating, it produces a black coloured residue. Salt P produces brisk effervescence with dil. HCl and the gas evolved turns lime water milky, but no action with acidified potassium dichromate solution.
(ii) Salt Q is white in colour. On strong heating, it produces buff yellow residue and liberates reddish brown gas. Solution of salt Q produces chalky white insoluble ppt. with excess of ammonium hydroxide.
(iii) Salt R is black in colour. On reacting with conc. HCl, it liberates a pungent greenish yellow gas which turns moist starch iodide paper blue black.
The 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.
Identify 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.
Match the 'cations' A to F and the solubility of ppt. G or H with the correct colours from 'X' and 'Y'.
'X' on addition of NaOH in excess Cation Solubility of ppt. in excess 'Y' on addition of NH4OH in excess Cation Solubility of ppt. in excess 1. Reddish brown ppt. A: Ca2+ G: Soluble 6. Dirty green ppt. A: Ca2+ G: Soluble 2. Pale blue ppt. B: Zn2+ H: Insoluble 7. No ppt. formed B: Zn2+ H: Insoluble 3. Gelatinous white ppt. C: Fe2+ 8. Gelatinous white ppt. C: Fe2+ 4. Chalky white ppt. D: Cu2+ 9. Pale blue ppt. D: Cu2+ 5. Milky white ppt. E: Pb2+ 10. Chalky white ppt. E: Pb2+ F: Fe3+ F: Fe3+