After adding sulfuric acid to copper metal, which copper compound is present in the solution?

• A. CuSO_4 (copper(II) sulfate)
• B. Cu(NO_3)_2 (copper(II) nitrate)
• C. Cu_2SO_4 (copper(I) sulfate)
• D. CuCl_2 (copper(II) chloride)

Answer: (A)

Verified step by step guidance

1. Step 1: Understand the reaction between copper metal and sulfuric acid. Copper is a less reactive metal and does not react with dilute sulfuric acid under normal conditions, but it can react with hot, concentrated sulfuric acid acting as an oxidizing agent.
2. Step 2: Identify the products formed when copper reacts with sulfuric acid. The reaction typically produces copper(II) sulfate (CuSO_4), sulfur dioxide (SO_2), and water (H_2O).
3. Step 3: Recognize that copper forms a +2 oxidation state in this reaction, leading to the formation of copper(II) sulfate, not copper(I) sulfate or other copper salts like nitrates or chlorides.
4. Step 4: Recall that the sulfate ion (SO_4^{2-}) comes directly from sulfuric acid (H_2SO_4), so the copper salt formed will contain the sulfate ion, confirming the compound as CuSO_4.
5. Step 5: Conclude that among the given options, the correct copper compound present in the solution after the reaction is copper(II) sulfate, CuSO_4.