Labelling four unlabelled sodium salt solutions
Four unlabelled bottles of colourless solutions are known to be sodium chloride, sodium sulfate, sodium carbonate and sodium iodide, but their labels are missing. A student flame-tests all four and sees a persistent yellow flame from each, which she finds unhelpful. She then adds dilute nitric acid to samples of each solution: only S fizzes, and the gas turns limewater milky. With silver nitrate solution, P gives a pale yellow precipitate that is almost insoluble in concentrated ammonia solution, Q gives a white precipitate that dissolves readily in dilute ammonia, R gives no precipitate, and S gives a white precipitate that dissolves with effervescence when dilute nitric acid is added. Barium chloride solution gives a white precipitate with R that does not dissolve in dilute hydrochloric acid; with S it gives a white precipitate that does dissolve, releasing carbon dioxide. The student must label the four bottles and write a short justification for the laboratory technician, who will check the labels before the bottles are returned to the stock cupboard.
1. Assign a name to each of P, Q, R and S, justifying each assignment from the observations. Explain why the flame test gave no useful information.
Read the worked answer
S = sodium carbonate. Only S reacts with dilute nitric acid to give a gas that turns limewater milky, and the gas is carbon dioxide: CO₃²⁻ + 2H⁺ → CO₂ + H₂O. The white precipitate S gives with silver nitrate is Ag₂CO₃, which dissolves in acid because the carbonate is protonated and the CO₂ escapes.
P = sodium iodide. Silver iodide is the only pale yellow silver halide, and it is essentially insoluble in concentrated ammonia, unlike AgCl.
Q = sodium chloride. Silver chloride is white and dissolves readily in dilute ammonia because the diamminesilver(I) complex forms: AgCl(s) + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻. P (pale yellow, insoluble) and Q (white, soluble) separate the two halides.
R = sodium sulfate. It gives no precipitate with silver nitrate under these conditions, but with barium chloride it gives BaSO₄, which is white and insoluble in dilute acid.
The flame test failed because all four compounds are sodium salts: the intense yellow emission from excited sodium atoms at about 589 nm masks any other colour, so the test cannot distinguish them. A test that relies on the cation is useless when the cation is the same in every sample.
2. Write the ionic equations for the four confirmatory reactions the student relies on, and state the observation that accompanies each.
Read the worked answer
Carbonate with acid (S): CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l). Observed: steady effervescence; the gas turns limewater milky (Ca(OH)₂ + CO₂ → CaCO₃ + H₂O).
Iodide with silver nitrate (P): Ag⁺(aq) + I⁻(aq) → AgI(s). Observed: pale yellow precipitate, essentially insoluble in concentrated ammonia.
Chloride with silver nitrate (Q): Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Observed: white precipitate, curdy, dissolving readily in dilute ammonia.
Sulfate with barium chloride (R): Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Observed: dense white precipitate that survives boiling with dilute hydrochloric acid.
The carbonate/silver reaction that explains S can also be written: 2Ag⁺(aq) + CO₃²⁻(aq) → Ag₂CO₃(s), followed by Ag₂CO₃(s) + 2H⁺(aq) → 2Ag⁺(aq) + CO₂(g) + H₂O(l), which is why the precipitate appears at first and then dissolves with fizzing when acid is added.
3. Explain why the student acidified with nitric acid rather than hydrochloric acid, and why the barium chloride test must be done on an acidified sample. What error would she make in each case?
Read the worked answer
Nitric acid is used because it contributes neither chloride nor sulfate ions. If she added hydrochloric acid, she would introduce Cl⁻ into every sample; adding silver nitrate would then give a white AgCl precipitate in all four solutions, and she would wrongly conclude that every bottle contained a chloride. Sulfuric acid would be equally bad in the other direction, adding SO₄²⁻ and producing a false barium sulfate precipitate.
The barium test must be acidified because several other anions also give white barium salts: BaCO₃ from carbonate, BaSO₃ from sulfite, and barium phosphate or hydroxide in some solutions. All of these are decomposed by dilute acid, releasing CO₂, SO₂ or simply dissolving, whereas BaSO₄ is genuinely insoluble in dilute acid. Acidifying first, or adding acid after the precipitate forms and seeing which precipitate survives, means the only precipitate left is BaSO₄. The student's own data show this working: S gives a white precipitate with barium chloride that dissolves in acid (BaCO₃, consistent with S being the carbonate), while R's precipitate survives, confirming sulfate.
One caution for the write-up: silver sulfate is slightly soluble, so a concentrated sulfate solution (R) can give a faint silver precipitate; this is why the conclusion for R should rest on the barium test rather than on the absence of a silver precipitate.