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Chemistry

Distinguish between the following pairs of compounds using the reagent given in the bracket.

(i) Manganese dioxide and copper (II) oxide. (using concentrated HCl)

(ii) Ferrous sulphate solution and ferric sulphate solution. (using sodium hydroxide solution)

(iii) Dilute hydrochloric acid and dilute sulphuric acid. (using lead nitrate solution)

Analytical Chemistry

ICSE 2019

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Answer

(i) 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

(ii) When sodium hydroxide is added to the two solns., ferrous sulphate solution gives a dirty green ppt. of Fe(OH)2 whereas, ferric sulphate solution forms a reddish brown ppt. of Fe(OH)3. Hence, the two compounds can be distinguished.

(iii) Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.

H2SO4 (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbSO4 ↓ [white ppt. formed which does not dissolve on warming the mixture].

Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution.

2HCl (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbCl2 ↓ [white ppt. formed which dissolves on warming the mixture].

Hence, dilute hydrochloric acid and dilute sulphuric acid can be distinguished using lead nitrate solution.

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