Question

Among the following observations, the correct one that differentiates between $S{O}_3^{2-}$ and $S{O}_4^{2-}$ is:

• A. Both form precipitate with $Ba{Cl}_2,S{O}_3^{2-}$ dissolves in $HCl$ but $S{O}_4^{2-}$ does not
• B. $S{O}_3^{2-}$ forms precipitate with $Ba{Cl}_2,S{O}_4^{2-}$ does not
• C. $S{O}_4^{2-}$ forms precipitate with $Ba{Cl}_2,S{O}_3^{2-}$ does not
• D. Both form precipitate with $Ba{Cl}_2,S{O}_4^{2-}$ dissolves in $HCl$ but $S{O}_3^{2-}$ does not

Answer 1

The correct option is A Both form precipitate with $Ba{Cl}_2,S{O}_3^{2-}$ dissolves in $HCl$ but $S{O}_4^{2-}$ does not

$BaC{l}_2+S{O}_4^{2-}\to BaS{O}_4\downarrow +2C{l}^{-}$

$BaC{l}_2+S{O}_3^{2-}\to BaS{O}_3\downarrow +2C{l}^{-}$

$S{O}_3^{2-}$ and $S{O}_4^{2-}$ when treated with $Ba{Cl}_2$, give white precipitate of $BaS{O}_4$ respectively.
Out of these two, $S{O}_3^{2-}$ is soluble in $HCl$ but $S{O}_4^{2-}$ does not.