Chemistry
Give a chemical test to distinguish between the following pairs of compounds:
(i) Sodium chloride solution and sodium nitrate solution.
(ii) Hydrogen chloride gas and hydrogen sulphide gas.
(iii) Ethene gas and ethyne gas.
(iv) Calcium nitrate solution and zinc nitrate solution.
(v) Potassium carbonate and potassium sulphite.
Analytical Chemistry
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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.] + NaNO3
NaNO3 + 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 ammoniacal silver nitrate is added to the two solutions ethyne forms a white ppt. of silver acetylide, whereas, no change appears in ethene.
(iv) When NaOH is added to the given soln., Zinc nitrate, reacts to form a gelatinous white ppt. which dissolves in excess of NaOH whereas, Calcium nitrate forms a milky white ppt. which is insoluble in excess of NaOH. Hence, the two can be distinguished
(v) When dil. sulphuric acid is added to potassium 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 potassium 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.
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