A substance on treatment with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ liberates a colourless gas which produces (i) turbidity with baryta water and (ii) turns acidified dichromate solution green. The reaction indicates the presence of :

Dilute${\mathrm{H}}_2{\mathrm{SO}}_4$ or dilute $\mathrm{HCl}$ are used to identify Group I acidic radicals. Sulphite produces colourless sulphur dioxide gas when it reacts with diluted ${\mathrm{H}}_2{\mathrm{SO}}_4$.
Barium sulfite is formed when this gas is used to treat baryta water $\left(\left[\mathrm{Ba}(\mathrm{OH}{)}_2\right]\right)$, which causes turbidity.

$\mathrm{Ba}(\mathrm{OH}{)}_2+{\mathrm{SO}}_2\to {\mathrm{BaSO}}_3\downarrow +{\mathrm{H}}_2\mathrm{O}$

The solution turns green as chromic sulphate forms when sulphur dioxide is treated with acidified potassium dichromate.

$$\begin{gathered} {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+{\mathrm{H}}_2{\mathrm{SO}}_4+3{\mathrm{SO}}_2\to {\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Green} \end{gathered}$$
The sulphite (${{\mathrm{SO}}^{2-}}_3$) ion is therefore present, as shown by the reaction.