Question

what can one deduce from the reaction of chloride, bromide and iodide ions with the addition of silver nitrate solution in water? a if a cream - colored solid appears, bromide is present. b if a white solid appears, iodide is present. c if a yellow solid appears, chloride is present. d if no solid appears, multiple ions are present.

Explanation:

To solve this, we analyze the reactions of halide ions with silver nitrate:

• Chloride (\(Cl^-\)) reacts with \(AgNO_3\) to form \(AgCl\), a white precipitate.
• Bromide (\(Br^-\)) reacts to form \(AgBr\), a cream - colored precipitate.
• Iodide (\(I^-\)) reacts to form \(AgI\), a yellow precipitate.

Now let's analyze each option:

• Option A: Bromide forms \(AgBr\) (cream - colored solid), so if a cream - colored solid appears, bromide is present. This is correct.
• Option B: Iodide forms a yellow solid, not a white solid. White solid is for chloride. So this is incorrect.
• Option C: Chloride forms a white solid, iodide forms a yellow solid. So this is incorrect.
• Option D: If no solid appears, it means none of the halide ions (chloride, bromide, iodide) are present (or in very low concentration), not that multiple ions are present. So this is incorrect.

Answer:

A. If a cream - colored solid appears, bromide is present.