A white precipitate forms when silver nitrate is added. Which ion is present?

A white precipitate forms when silver nitrate is added. Which ion is present?

• A. Chloride (Cl-)
• B. Sulfate (SO4 2-)
• C. Nitrate (NO3-)
• D. Iodide (I-)

Answer: (A)

When silver ions meet halide ions in solution, they form a precipitate of silver halide. The color of that precipitate tells you which halide is present. A white solid is characteristic of silver chloride, because the reaction A g+ + Cl− → AgCl(s) produces an insoluble white compound. Silver bromide and silver iodide would give pale cream and yellow precipitates, respectively, so they wouldn’t be white. Nitrate doesn’t form an insoluble silver salt under these conditions, so no precipitate appears from nitrate. Silver sulfate is not the typical result in this halide test, and it wouldn’t explain a distinct white precipitate in this context. So the white precipitate points to chloride being present.

When silver ions meet halide ions in solution, they form a precipitate of silver halide. The color of that precipitate tells you which halide is present. A white solid is characteristic of silver chloride, because the reaction A g+ + Cl− → AgCl(s) produces an insoluble white compound. Silver bromide and silver iodide would give pale cream and yellow precipitates, respectively, so they wouldn’t be white. Nitrate doesn’t form an insoluble silver salt under these conditions, so no precipitate appears from nitrate. Silver sulfate is not the typical result in this halide test, and it wouldn’t explain a distinct white precipitate in this context. So the white precipitate points to chloride being present.