Making Salts Decision Tree

Salt preparation is one of the most heavily examined practical topics in 0620. The challenge is not remembering the procedures but knowing which procedure to use for which salt. This decision tree makes that choice systematic.

Step 1: Is the salt soluble or insoluble?

Use the solubility rules:

Soluble	Exceptions (insoluble)
All sodium, potassium, ammonium salts	None
All nitrates	None
Most chlorides	Lead(II) chloride, silver chloride
Most sulfates	Barium sulfate, calcium sulfate, lead(II) sulfate

Insoluble	Exceptions (soluble)
Most carbonates	Sodium, potassium, ammonium carbonates
Most hydroxides	Sodium, potassium, (calcium slightly)

If the salt is insoluble, use precipitation (Method C below).

If the salt is soluble, go to Step 2.

Step 2: Is the reactant you will add to the acid soluble or insoluble?

For soluble salts, you react an acid with a metal, base, or carbonate. The question is whether the second reactant dissolves in water:

• Insoluble reactant (most metal oxides, insoluble carbonates, reactive metals): use the excess solid method (Method A)
• Soluble reactant (NaOH, KOH, Na2CO3, NH3): use titration (Method B)

Method A: Excess solid + acid (crystallisation)

Use for: Soluble salts where one reactant is an insoluble solid (metal oxide, metal hydroxide, metal carbonate, or reactive metal).

Examples: Making CuSO4 from CuO + H2SO4. Making ZnCl2 from ZnO + HCl. Making MgSO4 from Mg + H2SO4.

Procedure

1. Warm the acid gently in a beaker
2. Add the insoluble solid (e.g. CuO) in small portions, stirring after each addition
3. Keep adding until the solid is in excess (some solid remains undissolved). This ensures all the acid has reacted
4. Filter to remove the excess solid. The filtrate is the pure salt solution
5. Evaporate the filtrate gently (heat on a water bath or evaporate until crystallisation point)
6. Leave to crystallise by cooling
7. Filter to collect crystals, wash with distilled water, dry

Why add excess solid? You cannot see when the acid is used up (the solution may be coloured or not). By adding excess, you guarantee all the acid reacts. The excess unreacted solid is simply filtered off.

Why not boil dry? Boiling would produce a powder, not crystals. Gentle evaporation followed by cooling produces well-formed crystals. Some hydrated salts decompose if overheated (e.g. CuSO4.5H2O loses water).

Method B: Titration

Use for: Soluble salts where both reactants are soluble (typically alkali + acid, or soluble carbonate + acid).

Examples: Making NaCl from NaOH + HCl. Making K2SO4 from KOH + H2SO4. Making Na2SO4 from Na2CO3 + H2SO4.

Why is titration needed?

Both reactants are dissolved in solution and both solutions may be colourless. You cannot add one in excess because you would contaminate the product. You need to find the exact volume of each solution that reacts completely, with nothing left over.

Procedure

1. Use a burette and pipette to find the exact volume of acid that neutralises a known volume of alkali, using an indicator
2. Note the exact volumes
3. Repeat the experiment using those exact volumes but without the indicator (indicator would contaminate the salt crystals)
4. Evaporate and crystallise the solution to obtain pure salt crystals

The critical detail: the final batch is made without indicator. Students often forget this step in exams.

Method C: Precipitation

Use for: Insoluble salts.

Examples: Making BaSO4, PbCl2, AgCl, PbI2, CaCO3.

Procedure

1. Mix two soluble solutions that contain the ions needed for the insoluble salt
2. The insoluble salt forms immediately as a precipitate
3. Filter to collect the precipitate
4. Wash with distilled water to remove soluble impurities
5. Dry in a warm oven or between filter papers

Example: Making lead(II) iodide (PbI2)

Mix lead(II) nitrate solution (Pb(NO3)2) with potassium iodide solution (KI):

Pb(NO3)2(aq) + 2KI(aq) -> PbI2(s) + 2KNO3(aq)

A bright yellow precipitate of PbI2 forms. Filter, wash, and dry.

Choosing the two solutions

You need two soluble compounds: one containing the cation, one containing the anion of the target salt. Nitrate salts are always soluble, so they make convenient starting reagents.

Insoluble salt target	Solution 1	Solution 2
BaSO4	BaCl2(aq)	Na2SO4(aq)
AgCl	AgNO3(aq)	NaCl(aq)
PbI2	Pb(NO3)2(aq)	KI(aq)
CaCO3	CaCl2(aq)	Na2CO3(aq)

Decision tree summary

1. Is the salt soluble?
   • No -> Precipitation (Method C)
   • Yes -> go to 2
2. Is the base/carbonate/metal soluble?
   • No (e.g. CuO, ZnCO3, Mg metal) -> Excess solid + acid, filter, crystallise (Method A)
   • Yes (e.g. NaOH, KOH, Na2CO3) -> Titration (Method B)

Common exam mistakes

1. Choosing the wrong method: CuSO4 cannot be made by titration because CuO is insoluble. NaCl cannot be made by the excess solid method because NaOH is soluble (you could not add excess and filter).
2. Forgetting to filter in Method A: The excess solid must be removed.
3. Using indicator in the final batch for titration: The indicator must be omitted in the second run, or the salt will be contaminated.
4. Boiling to dryness: This gives a powder, not crystals, and may decompose the salt.
5. Not washing the precipitate: Soluble impurities remain unless washed with distilled water.

Worked exam questions

Describe how you would prepare a pure, dry sample of copper(II) sulfate crystals starting from copper(II) oxide and dilute sulfuric acid. [6 marks]

• Add excess copper(II) oxide to warm dilute sulfuric acid and stir [1]
• The black CuO reacts and the solution turns blue [1]
• Filter to remove excess (unreacted) copper(II) oxide [1]
• Heat the blue filtrate gently / evaporate on a water bath to reduce volume / until crystallisation point [1]
• Allow to cool so crystals form [1]
• Filter to collect crystals, wash with distilled water, and dry / pat dry with filter paper [1]

Explain why sodium chloride cannot be made by adding excess sodium hydroxide to hydrochloric acid and filtering. [2 marks]

• Sodium hydroxide is soluble, so it cannot be removed by filtering [1]
• Excess NaOH would remain dissolved in the solution and contaminate the NaCl product [1]

Describe how to prepare a sample of barium sulfate from barium chloride solution and sodium sulfate solution. [4 marks]

• Mix barium chloride solution with sodium sulfate solution [1]
• A white precipitate of barium sulfate forms: BaCl2 + Na2SO4 -> BaSO4 + 2NaCl [1]
• Filter to collect the precipitate [1]
• Wash with distilled water and dry [1]