Qualitative Inorganic Analysis (anions)
A salt solution gives a white precipitate with barium chloride that remains insoluble even after adding dilute hydrochloric acid. Which acid radical is indicated by this observation?
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About This Question
- Subject
- chemistry
- Chapter
- principles related to practical chemistry
- Topic
- qualitative inorganic analysis (anions)
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Sulphate ion
Barium chloride is used to test for the sulphate radical because barium sulphate is a dense white precipitate with a very low solubility product, and it does not dissolve in dilute mineral acids such as hydrochloric or nitric acid. The fact that the white precipitate persists after acidification confirms the sulphate ion. Option Carbonate ion is wrong because barium carbonate, although white, readily dissolves in dilute hydrochloric acid with effervescence of carbon dioxide, so it would not remain. Option Sulphite ion is incorrect because barium sulphite also dissolves in dilute hydrochloric acid, releasing sulphur dioxide gas. Option Nitrate ion is wrong because barium nitrate is soluble and gives no precipitate at all with barium chloride. The acid-insolubility of barium sulphate is the NCERT confirmatory criterion that distinguishes sulphate from acid-soluble anions. In practice the test solution is first acidified with dilute hydrochloric acid before adding barium chloride, precisely so that any carbonate or sulphite is decomposed and only the genuine sulphate survives to form a lasting precipitate, removing ambiguity in the result. Plausibility check: only an extremely insoluble salt of a strong-acid anion survives in acidic medium, and barium sulphate, the salt of strong sulphuric acid, fits this exactly while carbonate and sulphite of weak acids dissolve.
This medium difficulty chemistry question is from the chapter principles related to practical chemistry, covering the topic of qualitative inorganic analysis (anions). It appeared in the 2025 exam.
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