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Among the following observation, the cor...

Among the following observation, the correct one that differentiates between `SO_(3)^(2-)` and `SO_(4)^(2-)` is

A

Both form precipitates with `BaCl_(2), SO_(3)^(2-)` dissolves in HCl but `SO_(4)^(2-)` does not

B

`SO_(3)^(2-)` forms precipitate with `BaCl_(2), SO_(4)^(2-)` does not

C

`SO_(4)^(2-)` forms precipitate with `BaCl_(2), SO_(4)^(2-)` does not

D

Both forms precipitates with `BaCl_(2), SO_(4)^(2-)` dissolves in HCl but `SO_(3)^(2-)` does not

Text Solution

AI Generated Solution

The correct Answer is:
To differentiate between \( SO_3^{2-} \) (sulfite ion) and \( SO_4^{2-} \) (sulfate ion), we can analyze their behavior in reactions, particularly with barium chloride (\( BaCl_2 \)) and hydrochloric acid (\( HCl \)). ### Step-by-Step Solution: 1. **Reactions with Barium Chloride:** - When \( SO_3^{2-} \) is reacted with \( BaCl_2 \), a white precipitate of barium sulfite (\( BaSO_3 \)) is formed. - When \( SO_4^{2-} \) is reacted with \( BaCl_2 \), a white precipitate of barium sulfate (\( BaSO_4 \)) is formed. - Both precipitates are white, which means this observation alone does not differentiate between the two ions. 2. **Solubility in Hydrochloric Acid:** - When \( SO_3^{2-} \) is added to \( HCl \), it dissolves, indicating that sulfite ions can react with acids to form sulfur dioxide gas (\( SO_2 \)). - Conversely, when \( SO_4^{2-} \) is added to \( HCl \), it does not dissolve, indicating that sulfate ions do not react in the same way with acids. 3. **Conclusion:** - The correct observation that differentiates between \( SO_3^{2-} \) and \( SO_4^{2-} \) is that \( SO_3^{2-} \) dissolves in \( HCl \) while \( SO_4^{2-} \) does not. ### Final Answer: The correct observation that differentiates between \( SO_3^{2-} \) and \( SO_4^{2-} \) is: **"SO3 dissolves in HCl but SO4 does not."**
To differentiate between \( SO_3^{2-} \) (sulfite ion) and \( SO_4^{2-} \) (sulfate ion), we can analyze their behavior in reactions, particularly with barium chloride (\( BaCl_2 \)) and hydrochloric acid (\( HCl \)). ### Step-by-Step Solution: 1. **Reactions with Barium Chloride:** - When \( SO_3^{2-} \) is reacted with \( BaCl_2 \), a white precipitate of barium sulfite (\( BaSO_3 \)) is formed. - When \( SO_4^{2-} \) is reacted with \( BaCl_2 \), a white precipitate of barium sulfate (\( BaSO_4 \)) is formed. - Both precipitates are white, which means this observation alone does not differentiate between the two ions. ...
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