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In qualitative inorganic analysis of metal ions, the ion which precipitates as sulfide in the presence of H2S in warm dilute HCl is
- a)Cr3+
- b)Al3+
- c)Co2+
- d)Bi3+
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In qualitative inorganic analysis of metal ions, the ion which precipi...
- Group II → Bi3+
Group III A → Al3+, Cr3+
Group III B → Co2+ - Second group cations are precipitated as sulphides because of their low solubility products whereas sulphides of other metals remain in solution because of their high solubility products.
- HCl acts as a source of H+ which decrease the concentration of S2− ion that exceeds only the solubility products of the metal sulphide of the IInd group.
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In qualitative inorganic analysis of metal ions, the ion which precipi...
In inorganic qualitative analysis the second group radical cu Will be precipitated and third group Al precipitated..may I correct
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In qualitative inorganic analysis of metal ions, the ion which precipi...
In qualitative inorganic analysis of metal ions, the ion which precipitates as sulfide in the presence of H2S in warm dilute HCl is Bi3+
Explanation:
1. Introduction:
In qualitative inorganic analysis, the identification and separation of different metal ions is based on their characteristic reactions with specific reagents. The use of H2S gas in warm dilute HCl is a common method for the precipitation of metal sulfides. Each metal ion has a specific solubility product constant (Ksp) for the formation of its sulfide, which determines its precipitation behavior.
2. Precipitation of Metal Sulfides:
When H2S gas is passed through warm dilute HCl, it reacts with metal ions to form insoluble metal sulfides. The solubility of metal sulfides varies depending on the metal ion, pH, temperature, and concentration of reagents. Some metal sulfides are readily precipitated, while others may require specific conditions for their precipitation.
3. The Precipitation of Bi3+:
Among the given options (Cr3+, Al3+, Co2+, Bi3+), the ion that precipitates as sulfide in the presence of H2S in warm dilute HCl is Bi3+. Bismuth (Bi) forms an insoluble sulfide, Bi2S3, in acidic conditions. The solubility product constant (Ksp) for the formation of Bi2S3 is relatively low, resulting in the precipitation of bismuth sulfide even at low concentrations of Bi3+ ions.
4. Other Options:
- Cr3+ (chromium) does not precipitate as a sulfide in the given conditions. It forms a stable complex with H2S, resulting in the formation of a green precipitate of Cr2S3.
- Al3+ (aluminum) does not precipitate as a sulfide in warm dilute HCl. It requires more alkaline conditions for the precipitation of aluminum sulfide (Al2S3).
- Co2+ (cobalt) does not precipitate as a sulfide in the given conditions. It forms a stable complex with H2S, resulting in the formation of a pink precipitate of CoS.
5. Conclusion:
Based on the solubility behavior of metal sulfides, the ion that precipitates as sulfide in the presence of H2S in warm dilute HCl is Bi3+. This precipitation reaction can be used as a specific test for the presence of bismuth ions in qualitative inorganic analysis.
Explanation:
1. Introduction:
In qualitative inorganic analysis, the identification and separation of different metal ions is based on their characteristic reactions with specific reagents. The use of H2S gas in warm dilute HCl is a common method for the precipitation of metal sulfides. Each metal ion has a specific solubility product constant (Ksp) for the formation of its sulfide, which determines its precipitation behavior.
2. Precipitation of Metal Sulfides:
When H2S gas is passed through warm dilute HCl, it reacts with metal ions to form insoluble metal sulfides. The solubility of metal sulfides varies depending on the metal ion, pH, temperature, and concentration of reagents. Some metal sulfides are readily precipitated, while others may require specific conditions for their precipitation.
3. The Precipitation of Bi3+:
Among the given options (Cr3+, Al3+, Co2+, Bi3+), the ion that precipitates as sulfide in the presence of H2S in warm dilute HCl is Bi3+. Bismuth (Bi) forms an insoluble sulfide, Bi2S3, in acidic conditions. The solubility product constant (Ksp) for the formation of Bi2S3 is relatively low, resulting in the precipitation of bismuth sulfide even at low concentrations of Bi3+ ions.
4. Other Options:
- Cr3+ (chromium) does not precipitate as a sulfide in the given conditions. It forms a stable complex with H2S, resulting in the formation of a green precipitate of Cr2S3.
- Al3+ (aluminum) does not precipitate as a sulfide in warm dilute HCl. It requires more alkaline conditions for the precipitation of aluminum sulfide (Al2S3).
- Co2+ (cobalt) does not precipitate as a sulfide in the given conditions. It forms a stable complex with H2S, resulting in the formation of a pink precipitate of CoS.
5. Conclusion:
Based on the solubility behavior of metal sulfides, the ion that precipitates as sulfide in the presence of H2S in warm dilute HCl is Bi3+. This precipitation reaction can be used as a specific test for the presence of bismuth ions in qualitative inorganic analysis.
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In qualitative inorganic analysis of metal ions, the ion which precipitates as sulfide in the presence of H2S in warm dilute HCl isa)Cr3+b)Al3+c)Co2+d)Bi3+Correct answer is option 'D'. Can you explain this answer?
Question Description
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Solutions for In qualitative inorganic analysis of metal ions, the ion which precipitates as sulfide in the presence of H2S in warm dilute HCl isa)Cr3+b)Al3+c)Co2+d)Bi3+Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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