In the test for distinguishing copper sulfate from copper carbonate, which test indicates the presence of sulfate ions after acidifying with hydrochloric acid and adding barium chloride?

• A. White precipitate forms with sulfate, none with carbonate.
• B. Green precipitate forms with sulfate.
• C. Gas is evolved with carbonate.
• D. No reaction occurs.

Answer: (A)

When testing for sulfate ions, you rely on barium ions forming a white, insoluble precipitate with sulfate. The acidified step with hydrochloric acid removes carbonate interference because any carbonate is converted to carbon dioxide and leaves the solution. If sulfate ions are present, adding barium chloride yields BaSO4 as a white solid, signaling sulfate. Copper sulfate contains sulfate, so this white precipitate appears after acidification and BaCl2 is added. Copper carbonate, once acidified, loses carbonate as CO2, so there’s no sulfate left to react with barium chloride, and no precipitate forms. The gas you might see with carbonate comes from the carbonate reacting with the acid, not from the BaCl2 test. So the test that indicates sulfate ions is the appearance of a white precipitate with barium chloride after acidification; its absence indicates carbonate.