1.

(i) Explain the test for sulphate (or) sulphuric acid. (ii) What happens when sulphuric acid reacts with oxalic acid?

Answer»

Solution :(i) Dilute solution of sulphuric acid or aqueous solution of sulphates gives white precipitate with barium chloride solution. It can also be detected using lead ACETATE solution. Here a white precipitate of lead sulphate is OBTAINED.
`BaCl_(2)+H_(2)SO_(4)rarr underset("(White precipitate)")underset("Barium sulphate")(BaSO_(4)darr)+2HCl`
`(CH_(3)COO)_(2)Pb+H_(2)SO_(4)rarr underset("(White precipitate)")underset("Lead sulphate")(PbSO_(4)darr)+2CH_(3)COOH`
(ii) Sulphuric acid reacts with oxalic acid to give `CO and CO_(2)`
`underset("(Oxalic acid)")((COOH)_(2))+H_(2)SO_(4)rarr CO+CO_(2)+H_(2)SO_(4).H_(2)O`


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