Question

1. a student was performing the test for chloride in his unknown. when he added the five drops of hno₃, he observed bubbles forming (effervescence) in his unknown. was chloride present in his unknown? how do you know? 2. a student added barium chloride to her unknown and immediately witnessed the formation of a precipitate. she concluded that this white precipitate indicated the presence of chloride ion in her unknown. is she correct? how do you know? 3. a student has two unlabeled bottles of solution. one contains hpo₄²⁻ and the other so₄²⁻. how can this student identify which is which?

Explanation:

Step1: Analyze first - test for chloride

Chloride ions do not react with nitric acid to produce bubbles. Bubbles with nitric acid usually indicate the presence of carbonate or bicarbonate ions reacting to form carbon - dioxide gas. So chloride was not indicated by the bubbling.

Step2: Analyze second - test for chloride

When barium chloride is added, a white precipitate does not necessarily mean chloride is present. Barium can form white precipitates with sulfate ($SO_{4}^{2 - }$), carbonate ($CO_{3}^{2 - }$) and phosphate ($PO_{4}^{3 - }$) ions etc. A proper test for chloride is adding silver nitrate in acidic medium.

Step3: Identify $HPO_{4}^{2 - }$ and $SO_{4}^{2 - }$

Add barium chloride solution. Barium sulfate ($BaSO_{4}$) forms a very insoluble white precipitate in acidic medium, while barium hydrogen - phosphate ($BaHPO_{4}$) is more soluble in acidic solutions. So if a white precipitate forms immediately in acidic medium, it is the $SO_{4}^{2 - }$ solution. If no or little precipitate forms in acidic medium, it is the $HPO_{4}^{2 - }$ solution.

Answer:

1. No, chloride was not indicated. Chloride does not react with $HNO_{3}$ to produce bubbles.
2. No, she is not correct. Barium forms precipitates with other anions too, not just chloride.
3. Add barium chloride in acidic medium. The solution that forms a white precipitate is $SO_{4}^{2 - }$, and the one that doesn't or forms little precipitate is $HPO_{4}^{2 - }$.