Preparation of Salts Exam Questions

• Add excess copper oxide to warm dilute sulfuric acid and stir [1]
• The copper oxide reacts with the acid: CuO + H2SO4 → CuSO4 + H2O [1]
• Filter the mixture to remove the excess (unreacted) copper oxide [1]
• Heat the blue filtrate gently to evaporate some water, then leave to cool and crystallise [1]
• Filter the crystals and pat dry with filter paper / dry in a warm oven [1]

Examiner note: Excess copper oxide is used to ensure all the acid has reacted — unreacted acid would contaminate the crystals. The excess solid is removed by filtration. A common error is boiling the solution to dryness — this produces a powder, not crystals. Gentle heating followed by cooling produces proper crystals.

(a) Acid + excess metal (or excess metal oxide/carbonate): zinc + hydrochloric acid [1]

(b) Titration (acid + alkali): sodium hydroxide + hydrochloric acid [1]

(c) Precipitation: mix lead nitrate solution with sodium sulfate solution [1]

Examiner note: The method depends on the salt’s solubility and the nature of the metal. For soluble salts of reactive metals that form soluble hydroxides (Na, K), use titration. For soluble salts of less reactive metals, use acid + excess solid. For insoluble salts, use precipitation.

• Mix solutions of lead(II) nitrate and sodium sulfate (both soluble) [1]
• Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq) [1]
• Filter the mixture to collect the insoluble precipitate of lead sulfate [1]
• Wash the precipitate with distilled water (to remove soluble impurities) and dry in a warm oven or between sheets of filter paper [1]

Examiner note: For insoluble salts, you must use two soluble reagents that contain the required ions. Lead sulfate is insoluble, so mixing Pb2+(aq) with SO42-(aq) produces a precipitate. The washing step is essential to remove contaminating Na+ and NO3- ions.

• Sodium hydroxide is soluble, so you cannot filter off excess to obtain a pure product [1]
• In titration, you measure the exact volumes needed for complete neutralisation [1]
• This ensures no excess acid or alkali remains, giving a pure salt solution [1]

Examiner note: The “excess reactant” method only works when the excess reagent is an insoluble solid that can be filtered off. Since NaOH is soluble, excess NaOH would remain dissolved and contaminate the salt. Titration finds the exact ratio.

(a) Zn + H2SO4 → ZnSO4 + H2 [1]

(b) Excess zinc ensures all the sulfuric acid has reacted, so no acid contaminates the crystals. The excess zinc can be removed by filtration because it is an insoluble solid [1].

(c) Heat the filtrate gently to evaporate about half the water [1]. Allow the concentrated solution to cool slowly — crystals form as the solution becomes saturated. Filter and dry the crystals [1].

Examiner note: Gentle evaporation until crystals start to form at the edge of the solution is the signal to stop heating. Cooling slowly produces larger, better-formed crystals. Rapid boiling gives small, impure crystals.

(a) Soluble [1] — all nitrates are soluble; all potassium salts are soluble

(b) Insoluble [1] — barium sulfate is an exception to sulfate solubility

(c) Insoluble [1] — silver chloride is an exception to chloride solubility

Examiner note: Solubility rules to memorise: all nitrates are soluble; all sodium/potassium/ammonium salts are soluble; most chlorides are soluble EXCEPT AgCl and PbCl2; most sulfates are soluble EXCEPT BaSO4, PbSO4, and CaSO4; most carbonates and hydroxides are insoluble EXCEPT Na/K/NH4.

(a) Na2CO3 + 2HCl → 2NaCl + H2O + CO2 [1]

(b) When no more bubbles (CO2) are produced after adding the acid [1]

(c) The production of gas acts as a visible indicator — when bubbling stops, neutralisation is complete and all the carbonate has reacted [1]

Examiner note: For acid + carbonate reactions, the effervescence provides a natural endpoint indicator. However, for acid + alkali titrations (no gas produced), an indicator such as methyl orange or phenolphthalein is needed to detect the endpoint.

(a) Full: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) [1] Ionic: Ba2+(aq) + SO42-(aq) → BaSO4(s) [1]

(b) Full: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) [1] Ionic: Ag+(aq) + Cl-(aq) → AgCl(s) [1]

Examiner note: The ionic equation removes spectator ions (ions that appear unchanged on both sides). Only the ions that form the precipitate are shown. These ionic equations are also used as the basis for tests for sulfate and chloride ions.