O-Level Chemistry · Secondary 3 and 4
Salts and Their Preparation
A salt is formed when the hydrogen ions of an acid are replaced by metal ions or ammonium ions. In O-Level Chemistry the focus is practical: deciding whether a salt is soluble, choosing the right method to make it, and describing that method step by step so the product is pure and dry. Questions on salt preparation often appear as structured planning questions, and they draw on acid reactions, solubility rules and mole calculations together.
What the syllabus expects
- Use the solubility rules to decide whether a salt is soluble or insoluble in water
- Describe the preparation of a soluble salt by reacting an acid with an excess of an insoluble metal, base or carbonate
- Describe the preparation of sodium, potassium and ammonium salts by titration
- Describe the preparation of an insoluble salt by precipitation
- Describe filtration, crystallisation, washing and drying to obtain a pure salt
- Write balanced equations, including ionic equations for precipitation reactions
Solubility rules
Choosing a preparation method starts with knowing whether the salt and the reactants dissolve in water. All sodium, potassium and ammonium salts are soluble. All nitrates are soluble. Chlorides are soluble except silver chloride and lead(II) chloride. Sulfates are soluble except barium sulfate, lead(II) sulfate and calcium sulfate, which is only slightly soluble. Carbonates are insoluble except sodium, potassium and ammonium carbonates. Most metal oxides and hydroxides are insoluble, but sodium and potassium hydroxides are soluble and calcium hydroxide is slightly soluble. Learn these rules as a set, because you need them to pick a method and to predict precipitates in qualitative analysis.
Method 1: acid with an excess of an insoluble solid
Use this method for a soluble salt that is not a sodium, potassium or ammonium salt, such as copper(II) sulfate, magnesium chloride or zinc sulfate. Warm a measured volume of dilute acid and add the insoluble solid, which can be a metal, a metal oxide or a metal carbonate, a little at a time with stirring until no more dissolves. The excess solid shows that all the acid has reacted. Filter to remove the excess solid and keep the filtrate. Heat the filtrate to evaporate some water until it is saturated, which you can check by dipping in a glass rod and seeing crystals form as it cools. Leave it to cool so that crystals form, filter them off, wash with a little cold distilled water and dry between sheets of filter paper. Do not evaporate to dryness, because hydrated crystals would lose their water of crystallisation and any soluble impurities would end up in the solid. Copper is too unreactive to react with dilute acid, so use copper(II) oxide or copper(II) carbonate to make copper salts.
Method 2: titration for sodium, potassium and ammonium salts
Sodium, potassium and ammonium salts are made from an acid and an alkali (or an aqueous carbonate), and both reactants are soluble. There is no excess solid to filter off, so you must add exactly the right amount. Put a measured volume of alkali in a conical flask with a few drops of indicator and add acid from a burette until the indicator just changes colour. Record the volume of acid used. Repeat the titration with the same volumes but without the indicator, so that the salt is not contaminated. Then evaporate the solution to saturation and crystallise as in method 1. For example, potassium nitrate is made from potassium hydroxide and nitric acid: KOH + HNO₃ gives KNO₃ + H₂O.
Method 3: precipitation for insoluble salts
An insoluble salt, such as barium sulfate, silver chloride or lead(II) iodide, is made by mixing two solutions, one containing the metal ion and one containing the non-metal ion. Both starting compounds must be soluble, so nitrates and sodium salts are common choices. For barium sulfate, mix aqueous barium chloride with aqueous sodium sulfate: BaCl₂ + Na₂SO₄ gives BaSO₄ + 2NaCl. The ionic equation shows only the ions that form the solid: Ba²⁺ + SO₄²⁻ gives BaSO₄. Filter the mixture and keep the residue this time, because the salt is the solid. Wash the residue with distilled water to remove soluble impurities such as sodium chloride, then dry it in a warm oven or between filter papers.
Worked example
Describe how to prepare pure, dry crystals of zinc sulfate starting from zinc oxide and 50.0 cm³ of 1.00 mol/dm³ sulfuric acid. Show that 5.0 g of zinc oxide is enough to be in excess. ZnO + H₂SO₄ gives ZnSO₄ + H₂O. (Ar: Zn = 65, O = 16)
- Choose the method: zinc sulfate is soluble and zinc oxide is an insoluble base, so react the acid with excess zinc oxide.
- Check the excess: moles of H₂SO₄ = 1.00 × 50.0/1000 = 0.0500 mol, and the ratio is 1 : 1, so 0.0500 mol of ZnO is needed.
- Mr of ZnO = 65 + 16 = 81, so 0.0500 × 81 = 4.05 g reacts, and 5.0 g is therefore in excess.
- Warm the acid, add the zinc oxide a little at a time with stirring until some solid remains undissolved.
- Filter off the excess zinc oxide and collect the filtrate, which is aqueous zinc sulfate.
- Heat the filtrate until it is saturated, then leave it to cool so that crystals form.
- Filter off the crystals, wash them with a little cold distilled water and dry them between filter papers.
Answer: React the warm acid with excess zinc oxide (4.05 g is needed, so 5.0 g is an excess), filter, heat the filtrate to saturation, cool to crystallise, then filter, wash with a little cold distilled water and dry.
Common mistakes
- Using method 1 with a soluble reactant such as sodium carbonate. The excess would dissolve and could not be filtered off, so sodium salts need titration.
- Evaporating the solution to dryness. Heat only until saturated, then cool, so the crystals keep their water of crystallisation and impurities stay in solution.
- Trying to make copper(II) sulfate from copper metal and dilute sulfuric acid. Copper does not react with dilute acids, so use copper(II) oxide or carbonate.
- Keeping the filtrate in a precipitation method. The insoluble salt is the residue left in the filter paper.
- Washing crystals of a soluble salt with lots of water. Use a small amount of cold distilled water so the crystals do not dissolve.
- Choosing an insoluble starting compound for precipitation, such as calcium carbonate. Both reactants must be in solution so their ions can meet.
Exam tips
- Start every preparation answer by stating whether the salt is soluble, and whether it is a sodium, potassium or ammonium salt. That one sentence decides the method and earns credit.
- Give a reason for each key step, such as 'add excess so that all the acid is used up' or 'wash to remove soluble impurities'. Planning questions reward the reasons as well as the steps.
- In precipitation questions, write the ionic equation with only the two ions that form the solid and add the state symbol (s) to the salt if asked for state symbols.
Practice questions
1. Easy: State whether each salt is soluble or insoluble in water: lead(II) sulfate, potassium carbonate, silver nitrate, calcium carbonate.
Show answer
Lead(II) sulfate is insoluble (an exception among sulfates). Potassium carbonate is soluble (all potassium salts are). Silver nitrate is soluble (all nitrates are). Calcium carbonate is insoluble (only sodium, potassium and ammonium carbonates dissolve).
2. Medium: Suggest two reagents and a method to prepare a sample of silver chloride, and write the ionic equation.
Show answer
Silver chloride is insoluble, so use precipitation. Mix aqueous silver nitrate with aqueous sodium chloride, filter, wash the white residue with distilled water and dry it. Ionic equation: Ag⁺ + Cl⁻ gives AgCl.
3. Harder: A student makes potassium nitrate from 25.0 cm³ of 0.160 mol/dm³ nitric acid and 0.200 mol/dm³ potassium hydroxide. Explain why titration is needed and calculate the volume of potassium hydroxide required.
Show answer
Titration is needed because both reactants are soluble, so any excess could not be removed by filtration. KOH + HNO₃ gives KNO₃ + H₂O, a 1 : 1 ratio. Moles of HNO₃ = 0.160 × 25.0/1000 = 0.00400 mol, so 0.00400 mol of KOH is needed. Volume = 0.00400 ÷ 0.200 = 0.0200 dm³ = 20.0 cm³.
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