A colorless solution contains a mixture of iodide (I⁻) and carbonate (CO₃²⁻) ions. Which sequence of reagents will allow the carbonate ion to be tested without interference from the iodide ion?

Correct answer

D. Add dilute sulfuric acid, then add limewater.

Principle or equation

Carbonate ions react with acids to produce carbon dioxide gas, which turns limewater milky. Iodide ions do not produce a gas with acids. Barium carbonate is insoluble but barium iodide is soluble, so adding barium chloride can precipitate carbonate, but the precipitate dissolves in acid with effervescence. Silver carbonate and silver iodide are both insoluble, so silver nitrate would precipitate both, and dilute nitric acid would dissolve the carbonate but not the iodide, but the initial addition of silver nitrate would not distinguish. The most direct test for carbonate is to add an acid and test the gas with limewater.

Why this answer is correct

Option D: Adding dilute sulfuric acid (or any strong acid) to the mixture will produce carbon dioxide gas from carbonate ions: CO₃²⁻ + 2H⁺ → CO₂(g) + H₂O. The gas is bubbled into limewater (calcium hydroxide solution), which turns milky due to formation of calcium carbonate. Iodide ions do not produce a gas with acid, so they do not interfere. Option A: Adding nitric acid first would liberate CO₂, but then adding silver nitrate would test for iodide (yellow precipitate) and not carbonate. Option B: Barium chloride would precipitate barium carbonate, but also barium sulfate if present; here sulfate is not present, but the precipitate would dissolve in acid with effervescence, but the test is less direct. Option C: Silver nitrate would precipitate both silver carbonate and silver iodide; adding nitric acid would dissolve the carbonate but the iodide remains, but the initial precipitate is ambiguous.

Example

To test for carbonate in a mixture with chloride, add dilute hydrochloric acid and pass the gas through limewater; a milky precipitate indicates carbonate.