Distinguish between the following pairs of compounds using the reagent given in the bracket , Dilute hydrochloric acid & dilute sulphuric acid [using lead nitrate solution]

Explanation

Lead nitrate reacts with dilute HCI to form the insoluble salt lead chloride, which appears as the white precipitate. The insoluble lead chloride reacts with excess `Cl^(-)` ions (of HCl) to form a soluble complex, the tetrachloroplumbate(II) ion,
`Pb(NO_(3))_(2)+2HCl to [PbCl_(4)]^(2-)+2H^(+)(aq)`
Soluble
Lead nitrate solution reacts with `H_(2)SO_(4)` to give lead sulphate precipitate, which does not dissolve further in sulphuric acid solution.
`Pb(NO_(3))_(2)+H_(2)SO_(4) to PbSO_(4)darr+HNO_(3)`