Two solutions are used to investigate the anion in solution C. Solution C is aqueous and colourless. A second solution, solution D, is tested beside it. Sol...

This test identifies a halide anion using the silver nitrate and ammonia sequence, with distilled water as a control. Adding acidified silver nitrate to a halide gives a silver halide precipitate; the colour and its solubility in ammonia then tell you which halide it is: \(\text{Ag}^{+} + \text{Cl}^{-} \rightarrow \text{AgCl}\).

(a) [6] Six results, one mark each:

(The precipitate colour and its ammonia solubility must be consistent: white and soluble in dilute ammonia for chloride; cream and soluble only in concentrated ammonia for bromide; pale yellow and insoluble even in excess for iodide.)

(b) [1] The anion consistent with a white precipitate that dissolves in dilute ammonia is chloride, Cl^- [1].

(c) [3] The ammonia tests separate halides that look similar: a white solid that dissolves in dilute ammonia is a chloride [1]; a cream solid that dissolves only in concentrated ammonia is a bromide [1]; a pale yellow solid that stays even in excess ammonia is an iodide [1]. Maximum 3. The dissolving behaviour therefore confirms the identity beyond the precipitate colour alone.

(d) [2] Distilled water gives no solid [1], so the precipitate seen with solution C must come from the anion in solution C and not from the reagents or the water [1]. This is the purpose of a control.

(e) [2] The dilute nitric acid is added first to remove any carbonate [1], because carbonate would give a white silver carbonate precipitate that could be mistaken for the halide [1].

(f) [2] Left in sunlight the solid darkens [1], turning grey or purple-grey [1], because the silver halide decomposes in light to release silver.

(g) [3] To test for sulfate: add dilute nitric acid to a fresh portion of solution C [1], then add aqueous barium nitrate [1]; a white precipitate shows a sulfate is present [1].

(h) [3] Using the same portion size in both tubes gives the same amount of reagent in each [1], so the two boiling tubes can be compared fairly [1]; a larger portion in one tube would give a heavier precipitate and a false comparison [1].