A sample of air turns lime water milky and also turns acidified potassium dichromate green in aqueous solution has low `pH`. This is due to the presence of pollutants

Options

A. `CO_(2)`
B. `SO_(2)`
C. Both (a) and (b)
D. None of these

Answer

B

Explanation

`SO_(2)` turns lime water milky.
`SO_(2)` in aquenous solution gives `H_(2)SO_(3)`
`H_(2)SO_(3) hArrH^(+)+HSO_(3)^(-)`
`CO_(2)` also turns lime water milky.
`CO_(2)` in aqueous solution gives `H_(2)SO_(3)`
`H_(2)CO_(3)hArrH^(+)+HCO_(3)^(-)`
`H_(2)SO_(3)` is stronger acid than `H_(2)CO_(3)`. Thus, ionization of `H_(2)SO_(3)` is larger than `H_(2)CO_(3)`.
Thus, `pH` of `H_(2)SO_(3)` is lower than that of `H_(2)CO_(3)`
`SO_(2)` also turns acidified `K_(2)Cr_(2)O_(7)` solution green .
`3SO_(4)+Cr_(2)O_(7)^(2-)+2H^(+) rarr 3SO_(4)^(2-)+2underset(Green)(Cr^(3+))+H_(2)O`.
Thus, air sample has `SO_(2)` as pollutants.