Qualitative Inorganic Analysis (cation Groups)
In the systematic analysis of basic radicals, hydrogen sulphide gas is passed through an acidified solution to precipitate Group II cations. Why is the solution acidified before passing the gas?
Select the correct option:
Solution
To lower the sulphide ion concentration so only the least soluble Group II sulphides precipitate
Hydrogen sulphide is a weak diprotic acid that ionises to give sulphide ions, and this ionisation is governed by an equilibrium sensitive to the hydrogen-ion concentration. Acidifying the solution with dilute hydrochloric acid suppresses the ionisation of H_2S through the common-ion effect, keeping the sulphide ion concentration very low. At this low sulphide concentration the ionic product exceeds the solubility product only for the most insoluble sulphides, namely those of Group II cations such as copper, lead, and cadmium, so only these precipitate while Group IV cations remain in solution. Option B is wrong because raising the sulphide concentration would precipitate Group IV sulphides too, destroying the separation. Option C is incorrect; acidification controls ionisation, not the physical dissolution of the gas. Option D is wrong because hydrogen sulphide is a reducing agent and does not oxidise cations. This selective-precipitation logic is the NCERT and standard analytical basis for cation group separation. For Group IV, the solution is later made alkaline with ammonia, which boosts the sulphide concentration enough to precipitate the more soluble sulphides of zinc, manganese, nickel, and cobalt that survived the acidic stage. Plausibility check: controlling sulphide concentration through pH ensures the solubility-product condition is met selectively, giving a clean Group II versus Group IV split.
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About This Question
- Subject
- chemistry
- Chapter
- principles related to practical chemistry
- Topic
- qualitative inorganic analysis (cation groups)
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
To lower the sulphide ion concentration so only the least soluble Group II sulphides precipitate
Hydrogen sulphide is a weak diprotic acid that ionises to give sulphide ions, and this ionisation is governed by an equilibrium sensitive to the hydrogen-ion concentration. Acidifying the solution with dilute hydrochloric acid suppresses the ionisation of H_2S through the common-ion effect, keeping the sulphide ion concentration very low. At this low sulphide concentration the ionic product exceeds the solubility product only for the most insoluble sulphides, namely those of Group II cations such as copper, lead, and cadmium, so only these precipitate while Group IV cations remain in solution. Option B is wrong because raising the sulphide concentration would precipitate Group IV sulphides too, destroying the separation. Option C is incorrect; acidification controls ionisation, not the physical dissolution of the gas. Option D is wrong because hydrogen sulphide is a reducing agent and does not oxidise cations. This selective-precipitation logic is the NCERT and standard analytical basis for cation group separation. For Group IV, the solution is later made alkaline with ammonia, which boosts the sulphide concentration enough to precipitate the more soluble sulphides of zinc, manganese, nickel, and cobalt that survived the acidic stage. Plausibility check: controlling sulphide concentration through pH ensures the solubility-product condition is met selectively, giving a clean Group II versus Group IV split.
This medium difficulty chemistry question is from the chapter principles related to practical chemistry, covering the topic of qualitative inorganic analysis (cation groups). It appeared in the 2025 exam.
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