How can you test for the presence of sulfite and carbonate radicals in a sample?

Sulfur dioxide ($\text{SO}_2$) will reduce the orange dichromate solution to a green solution due to the formation of $\text{Cr}^{3+}$ ions. Carbon dioxide ($\text{CO}_2$) does not react with acidified potassium dichromate.

$$\text{Cr}_2\text{O}_7^{2-} (\text{aq}) + 3\text{SO}_2 (\text{g}) + 2\text{H}^{+} (\text{aq}) \longrightarrow 2\text{Cr}^{3+} (\text{aq}) + 3\text{SO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l})$$

The change from orange to green confirms the presence of the sulfite radical ($\text{SO}_3^{2-}$).

If $\text{CO}_2$ is present, the limewater will turn milky due to the formation of insoluble calcium carbonate.

$$\text{Ca(OH)}_2 (\text{aq}) + \text{CO}_2 (\text{g}) \longrightarrow \text{CaCO}_3 (\text{s}) + \text{H}_2\text{O} (\text{l})$$

The milky precipitate confirms the presence of the carbonate radical ($\text{CO}_3^{2-}$).