An aqueous solution of a compound (X) when treated with `BaCl_(2)` solution gives a white precipitate in soluble in concentrated HCl. Another sample of (X) gives first white precipitate with NaOH which is soluble in excess of NaOH solution. The solution of (X) does not give the precipitate on passing `H_(2)S` gas. Identify the compound (X) and give necessary reactions.

Explanation

`Al_(2)(SO_(4))_(3)+3BaCl_(2)rarr2AlCl_(3)+underset("white")(3BaSO_(4)darr)`
`Al_(2)(SO_(4))_(3)+6NaOH rarr underset("white")(2Al(OH)_(3))+3Na_(2)SO_(4)`
`Al(OH)_(3)+NaOHrarrunderset("soluble")(NaAlO_(2))+2H_(2)O`
Compound (X) gives white precipitate with `BaCl_(2)` which is insoluble in concentrated HCl hence the anion of (X) must be `SO_94)^(2-)` ion. Because the cation gives white precipitate with NaOH which is soluble in excess of NaOH, therefore cation of compound (X) may be `Zn^(2+), ` or `Al^(3+)` ion. But `Zn^(2+)` cannot be cation of the compound (X).