Question

The sodium salt of unknown action, when treated with $CaC{l}_2$, gives white precipitate only on boiling. The anion is:

• A. $S{O}_4^{2-}$
• B. $HC{O}_3^{-}$
• C. $C{O}_3^{2-}$
• D. $N{O}_3^{-}$

Answer 1

The correct option is A $S{O}_4^{2-}$

The only anion which gives white precipitate with $CaC{l}_2$ is $S{O}_4^{2-}$ since $CaS{O}_4$ is white in colour.

$CaC{l}_2+S{O}_4^{2-}⟶\underset{\left(white\right)}{CaS{O}_4}\downarrow +2C{l}^{-}$