Question:

Consider the sulphides \( HgS, PbS, CuS, Sb_2S_3, As_2S_3 \) and \( CdS \). Number of these sulphides soluble in \( 50% \ HNO_3 \) is _____________ 
 

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\( HgS \) has the lowest solubility product among common Group II sulfides, making it the unique one that resists nitric acid.
Updated On: Jan 2, 2026
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Correct Answer: 4

Solution and Explanation

Step 1: Understanding the Concept:
Metal sulfides are classified in qualitative inorganic analysis by their solubility in different acids. Nitric acid (\( HNO_3 \)) acts as an oxidizing agent, converting sulfide ions (\( S^{2-} \)) into elemental sulfur (\( S \)), which helps dissolve the precipitate.
Step 2: Detailed Explanation:
1. \( HgS \): Mercuric sulfide is extremely insoluble and does not dissolve in boiling \( 50% \ HNO_3 \). It requires aqua regia (a mixture of \( HCl \) and \( HNO_3 \)) for dissolution.
2. \( PbS, CuS, CdS \): These Group IIA sulfides are soluble in hot dilute or \( 50% \ HNO_3 \) with the evolution of \( NO \) gas and formation of elemental sulfur.
3. \( Sb_2S_3, As_2S_3 \): These Group IIB sulfides are also soluble in concentrated or \( 50% \ HNO_3 \), though they are often separated using yellow ammonium sulfide first in standard schemes.
Counting the soluble ones: \( PbS, CuS, Sb_2S_3, As_2S_3 \), and \( CdS \).
Step 3: Final Answer:
The number of sulfides soluble in \( 50% \ HNO_3 \) is 5.
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