Neutralisation and Preparation of Salts

What you'll learn

Neutralisation and preparation of salts form a substantial portion of the CXC CSEC Chemistry examination, appearing in both Paper 1 (multiple choice) and Paper 2 (structured questions and extended response). This topic connects acid-base theory with practical laboratory techniques, requiring you to predict products, write balanced equations, and describe detailed preparation methods. Mastery of these concepts is essential for scoring marks in the Acids, Bases and Salts section.

Key terms and definitions

Neutralisation — a chemical reaction between an acid and a base that produces a salt and water only, with the release of heat energy (exothermic reaction).

Salt — an ionic compound formed when the hydrogen ion of an acid is replaced by a metal ion or an ammonium ion.

Titration — a laboratory technique used to determine the exact volume of one solution needed to react completely with a known volume of another solution, using an indicator to show the endpoint.

Precipitation — the formation of an insoluble solid (precipitate) when two aqueous solutions are mixed, used to prepare insoluble salts.

Soluble salt — a salt that dissolves readily in water to form a clear solution; examples include all sodium, potassium, and ammonium salts, all nitrates, and most chlorides and sulfates.

Insoluble salt — a salt that does not dissolve appreciably in water; examples include most carbonates, most sulfides, and silver chloride.

Crystallisation — the process of forming solid crystals from a saturated solution by evaporation or cooling, used to obtain pure dry salt crystals.

Excess reagent — when more than the required amount of a reactant is added to ensure the other reactant is completely used up, with the excess removed by filtration.

Core concepts

Understanding neutralisation reactions

Neutralisation occurs when hydrogen ions (H⁺) from an acid combine with hydroxide ions (OH⁻) from a base to form water:

H⁺(aq) + OH⁻(aq) → H₂O(l)

The general equations for neutralisation reactions are:

• Acid + Base → Salt + Water
• Acid + Metal oxide → Salt + Water
• Acid + Metal hydroxide → Salt + Water
• Acid + Metal carbonate → Salt + Water + Carbon dioxide

For CXC CSEC examinations, you must write both word equations and balanced symbol equations. The salt formed always takes its first name from the metal (or ammonium) and its second name from the acid:

• Hydrochloric acid produces chlorides
• Sulfuric acid produces sulfates
• Nitric acid produces nitrates
• Ethanoic acid produces ethanoates

Example balanced equations:

• 2HCl(aq) + CuO(s) → CuCl₂(aq) + H₂O(l)
• H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
• 2HNO₃(aq) + CaCO₃(s) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g)

Classification of salts by solubility

Before preparing any salt, you must determine whether it is soluble or insoluble, as this dictates the preparation method.

Soluble salts:

• All sodium, potassium, and ammonium salts
• All nitrates
• Most chlorides (except silver chloride, lead chloride)
• Most sulfates (except barium sulfate, calcium sulfate, lead sulfate)

Insoluble salts:

• Most carbonates (except sodium, potassium, ammonium carbonates)
• Most hydroxides (except sodium, potassium, calcium hydroxides — slightly soluble)
• Most sulfides (except sodium, potassium, ammonium sulfides)
• Silver halides (AgCl, AgBr, AgI)
• Barium sulfate
• Lead chloride, lead sulfate, lead iodide

Method 1: Preparing soluble salts using an acid and an insoluble base

This method is used when preparing a soluble salt from an acid and an insoluble base (metal, metal oxide, metal hydroxide, or metal carbonate). The Caribbean alumina industry in Jamaica, for instance, produces aluminium oxide which could theoretically be neutralised with acids to produce aluminium salts.

Procedure for preparing copper(II) sulfate crystals:

1. Add excess copper(II) oxide to warm dilute sulfuric acid in a beaker, stirring continuously with a glass rod
2. Continue adding the black copper(II) oxide powder until no more dissolves and excess solid remains (solution turns blue)
3. Heat gently to speed up the reaction, but do not boil
4. Filter the mixture to remove excess unreacted copper(II) oxide
5. Collect the filtrate (blue copper(II) sulfate solution) in an evaporating dish
6. Heat the solution gently to evaporate water until crystals begin to form at the surface (point of crystallisation)
7. Leave to cool and crystallise — blue crystals form
8. Filter to collect crystals, wash with cold distilled water, and dry between filter papers

Chemical equation: CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)

Why use excess base? Using excess ensures all acid is neutralised. The excess can be removed by filtration because it is insoluble.

Method 2: Preparing soluble salts by titration

This method is essential when both the acid and base are soluble (e.g., preparing sodium chloride from hydrochloric acid and sodium hydroxide solution). Since both reactants are aqueous, you cannot use excess and filter. Instead, use titration to find exact volumes.

Procedure for preparing sodium chloride:

1. Pipette exactly 25.0 cm³ of sodium hydroxide solution into a conical flask
2. Add 2-3 drops of a suitable indicator (methyl orange or phenolphthalein)
3. Fill a burette with dilute hydrochloric acid and record the initial reading
4. Add acid slowly to the alkali, swirling the flask constantly
5. Stop adding acid when the indicator changes colour permanently (endpoint reached)
6. Record the final burette reading and calculate the volume of acid used
7. Repeat the titration without indicator, using the exact volumes determined
8. Evaporate the resulting solution to crystallisation point
9. Cool, filter, wash, and dry the crystals

Chemical equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

The repetition without indicator is crucial — indicators are organic compounds that would contaminate the salt crystals.

Method 3: Preparing insoluble salts by precipitation

Precipitation is the only method for preparing insoluble salts. Mix two soluble salt solutions containing the required ions, and the insoluble salt forms immediately as a precipitate.

Procedure for preparing barium sulfate:

1. Add barium chloride solution to a beaker
2. Add dilute sulfuric acid (or sodium sulfate solution)
3. A white precipitate of barium sulfate forms immediately
4. Filter to collect the precipitate
5. Wash thoroughly with distilled water to remove soluble impurities
6. Dry in a warm oven or between filter papers

Chemical equation: BaCl₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2HCl(aq)

Ionic equation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Caribbean context: Barium sulfate is used as "barium meal" in medical X-rays. Silver chloride, another insoluble salt, is light-sensitive and used in photographic film and paper still employed in some regional laboratories.

Selecting the appropriate preparation method — exam decision tree

For exam questions asking "describe how you would prepare [specific salt]":

Step 1: Determine if the salt is soluble or insoluble (use solubility rules)

Step 2 (if soluble): Identify suitable starting materials

• If you can use an acid + insoluble base (metal, metal oxide, metal hydroxide, or carbonate) → Use excess base method
• If both reactants are soluble (acid + alkali) → Use titration method

Step 3 (if insoluble): Use precipitation method

• Identify two soluble compounds that contain the required ions
• Mix solutions, filter, wash, and dry the precipitate

Writing ionic equations for neutralisation

Full ionic equations show all ions present:

H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l)

The net ionic equation removes spectator ions (ions that don't change):

H⁺(aq) + OH⁻(aq) → H₂O(l)

This shows that all neutralisation reactions between strong acids and strong bases are fundamentally the same — hydrogen ions neutralising hydroxide ions.

Worked examples

Example 1: Choosing a preparation method (Paper 2 style, 6 marks)

A student wishes to prepare a sample of zinc sulfate crystals, ZnSO₄·7H₂O.

(a) Name a suitable acid and a suitable base for this preparation. (2 marks)

(b) Explain why excess base should be used. (2 marks)

(c) Describe how the excess base is removed. (2 marks)

Model answer:

(a) Acid: Sulfuric acid / dilute H₂SO₄ (1 mark) Base: Zinc / zinc oxide / zinc hydroxide / zinc carbonate (1 mark)

(b) Excess base ensures that all the acid is used up / completely neutralised (1 mark), preventing acidic impurities in the final crystals (1 mark).

(c) The excess base is removed by filtration (1 mark), since it is insoluble and remains as solid residue / does not dissolve (1 mark).

Example 2: Precipitation reaction (Paper 2, 7 marks)

Lead(II) iodide is an insoluble yellow salt that can be prepared by precipitation.

(a) Name two suitable soluble compounds that could be mixed to prepare lead(II) iodide. (2 marks)

(b) Write a balanced chemical equation for the formation of lead(II) iodide. (2 marks)

(c) Describe how the lead(II) iodide would be separated from the reaction mixture and purified. (3 marks)

Model answer:

(a) Lead(II) nitrate / Pb(NO₃)₂ (1 mark) and potassium iodide / sodium iodide / KI / NaI (1 mark)

(b) Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) (2 marks — 1 for correct formula, 1 for balanced equation)

(c)

• Filter the mixture to collect the yellow precipitate (1 mark)
• Wash the residue with distilled water to remove soluble impurities / KNO₃ (1 mark)
• Dry the precipitate in a warm oven / between filter papers (1 mark)

Example 3: Titration calculation (Paper 2, 5 marks)

In a titration, 25.0 cm³ of sodium hydroxide solution of unknown concentration required 22.5 cm³ of 0.100 mol/dm³ hydrochloric acid for complete neutralisation.

(a) Write a balanced equation for the reaction. (2 marks)

(b) Calculate the concentration of the sodium hydroxide solution in mol/dm³. (3 marks)

Model answer:

(a) HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) (2 marks)

(b)

• Moles of HCl = concentration × volume = 0.100 × (22.5/1000) = 0.00225 mol (1 mark)
• From equation: mole ratio HCl : NaOH = 1:1 (1 mark)
• Therefore moles NaOH = 0.00225 mol
• Concentration NaOH = moles/volume = 0.00225 / (25.0/1000) = 0.090 mol/dm³ (1 mark)

Common mistakes and how to avoid them

Mistake: Using the titration method for preparing salts when an insoluble base is available. Correction: If you can use an acid and insoluble base (metal, oxide, hydroxide, carbonate), always choose this method — it's simpler and doesn't require precise volume measurements. Reserve titration for when both reactants are soluble.

Mistake: Failing to wash precipitates with distilled water after filtration. Correction: Always wash insoluble salts to remove soluble impurities that remain on the crystal surface. State "wash with distilled water" explicitly in exam answers for the mark.

Mistake: Not removing the indicator before crystallisation in titration preparations. Correction: After finding the endpoint volume in a trial titration, repeat the experiment without indicator using the exact volumes determined. This prevents indicator contamination of your salt.

Mistake: Writing incomplete or unbalanced equations for carbonate neutralisation. Correction: Neutralisation of carbonates produces three products: salt + water + carbon dioxide. Include CO₂(g) and balance the equation carefully. Example: 2HCl + CaCO₃ → CaCl₂ + H₂O + CO₂ (not just salt and water).

Mistake: Confusing filtrate and residue in separation descriptions. Correction: The filtrate is the liquid that passes through the filter paper (contains dissolved salt). The residue is the solid that remains on the filter paper (excess base or precipitate). Use these terms precisely in exam answers.

Mistake: Claiming that barium sulfate or silver chloride are soluble. Correction: Memorise the exceptions to solubility rules. While most chlorides are soluble, silver chloride and lead chloride are not. While most sulfates are soluble, barium sulfate, calcium sulfate, and lead sulfate are not.

Exam technique for Neutralisation and Preparation of Salts

Recognising command words and mark allocation:

• "Name" requires only the substance or method (1 mark each) — no explanation needed
• "Describe" requires a sequence of steps with apparatus and observations — typically 4-6 marks for full preparation
• "Explain why" requires reasons and understanding — expect 2-3 marks, give cause and effect
• "State what you would see/observe" requires colour changes, precipitate formation, effervescence — be specific about colours (e.g., "blue solution" not just "coloured")

Structuring preparation descriptions: Write in clear numbered steps. Include: reactants used, apparatus (beaker, filter funnel), process (heating, filtering), and final purification (crystallisation, washing, drying). For a 6-mark description, aim for 6 distinct points.

Balancing equations efficiently: Count the acid ions first (chloride, sulfate, nitrate), then the metal ions, then hydrogen and oxygen. Remember that sulfuric acid and carbonic acid need coefficient 2 for complete balance.

Using solubility rules decisively: When asked to choose a preparation method, the first decision is solubility. State this explicitly: "Zinc sulfate is soluble, therefore..." or "Barium sulfate is insoluble, therefore precipitation is used."

Quick revision summary

Neutralisation reactions produce salts and water when acids react with bases, metal oxides, hydroxides, or carbonates. Soluble salts are prepared either by reacting acids with excess insoluble bases (filtering off excess) or by titration when both reactants are soluble. Insoluble salts are prepared only by precipitation — mixing two soluble solutions to form an insoluble product. Always determine solubility first using the rules (all nitrates, sodium, potassium, and ammonium salts are soluble; most carbonates and hydroxides are insoluble). Master the step-by-step procedures, write balanced equations including state symbols, and use precise terminology (filtrate, residue, crystallisation) for maximum marks.