Question:
A saturated solution of \( H_2S \) in 0.1M HCl at 25°C contains \( S^{2-} \) ion concentration of \( 10^{-23} \, \text{mol L}^{-1} \). The solubility product of some sulfides are \( CuS = 10^{-44} \), \( FeS = 10^{-14} \), and \( MnS = 10^{-25} \). If 0.01M solution of these salts in 1M HCl are saturated with \( H_2S \), which of these will be precipitated?
A saturated solution of \( H_2S \) in 0.1M HCl at 25°C contains \( S^{2-} \) ion concentration of \( 10^{-23} \, \text{mol L}^{-1} \). The solubility product of some sulfides are \( CuS = 10^{-44} \), \( FeS = 10^{-14} \), and \( MnS = 10^{-25} \). If 0.01M solution of these salts in 1M HCl are saturated with \( H_2S \), which of these will be precipitated?
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The solubility product can be used to predict whether a salt will precipitate when the concentration of its ions exceeds the product of its solubility constant.
Updated On: Jan 12, 2026
- All
- All except MnS
- All except MnS and FeS
- Only CuS
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The Correct Option is C
Solution and Explanation
The solubility product indicates the solubility of the salts in solution. By comparing the ion product of each sulfide with its solubility product, we determine that CuS will precipitate because its ion product exceeds the solubility product, while MnS and FeS will not.
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