Question

When $BaC{l}_2$ is added to an aqueous salt solution, a white precipitate is obtained. The anion among $C{O}_3^{2-},S{O}_3^{2-}$ and $S{O}_4^{2-}$ that was present in the solution can be:

• A. $C{O}_3^{2-}$ but not any of the other two
• B. $S{O}_3^{2-}$ but not any of the other two
• C. $S{O}_4^{2-}$ but not any of the other two
• D. None of these

Answer 1

The correct option is D None of these

$B{a}^{2+}$ ion will form white ppt of either $BaC{O}_3$ or $BaS{O}_3$ or $BaS{O}_4$.
$B{a}^{2+}+C{O}_3^{2-}\to BaC{O}_3\downarrow$
$B{a}^{2+}+S{O}_3^{2-}\to BaS{O}_3\downarrow$
$B{a}^{2+}+S{O}_4^{2-}\to BaS{O}_4\downarrow$

Hence, no ion will be present in the solution.