Hmmm...solution of copper sulfide? copper sulfide has apretty low solubility product (Ksp(CuS) = 1.3×10
-3 6)
Sulfide is partially oxidized to sulfate with dil. nitric acid.
H
2S is oxidized to S
8 with oxygen:
8 H
2S + 4 O
2 ----> S
8 + 8 H
2O
Why is this so when both have an oxidation number of 2 to begin with?
I don't know if I understood the question, but I think that it depends on the strength of the oxidation agent.