Preparation of salts Revision Notes

When preparing soluble salts from an acid and an alkali, titration is used to find the exact volume of alkali needed to neutralise a fixed volume of acid. This ensures the salt formed is pure and not contaminated by excess acid or alkali.

The procedure involves adding a measured volume of alkali to a conical flask with a few drops of a suitable indicator (e.g., methyl orange or phenolphthalein). Acid is added from a burette until the indicator changes colour, signalling neutralisation.

The neutralisation reaction forms a soluble salt and water. For example, reacting hydrochloric acid with sodium hydroxide: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) (balanced). After titration, the salt solution is evaporated gently to crystallise pure sodium chloride, which is then dried.

• Use a pipette to measure alkali accurately.
• Add acid slowly from burette until indicator changes colour.
• Repeat titration to obtain concordant results (within 0.1 cm³).
• Mix the exact volumes without indicator for salt preparation.
• Evaporate solution to crystallise salt, then dry crystals.

Some metals react with acids to produce soluble salts and hydrogen gas. To prepare a salt, add excess metal to the acid to ensure complete reaction.

The reaction continues until all acid is used up; excess metal remains unreacted and is removed by filtration.

Example: Zinc reacts with sulfuric acid to form zinc sulfate and hydrogen gas: Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g). The solution is then evaporated to crystallise zinc sulfate.

• Add metal in excess to acid in a beaker.
• Stir and allow reaction to complete (no more bubbles).
• Filter to remove unreacted metal.
• Evaporate filtrate to crystallise salt.
• Dry crystals between filter paper.

Insoluble bases like copper(II) oxide react with acids to form soluble salts and water.

Add excess base to acid and stir until no more base dissolves, indicating complete reaction.

Filter off excess base, then evaporate the filtrate to crystallise the salt. For example: CuO(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l).

• Add excess insoluble base to acid in a beaker.
• Stir mixture until base no longer dissolves.
• Filter to remove excess base.
• Evaporate filtrate to crystallise salt.
• Dry crystals carefully.

Insoluble carbonates react with acids to form soluble salts, carbon dioxide gas, and water.

Add excess carbonate to acid; bubbles of CO₂ gas are observed.

Continue until bubbling stops, indicating acid is fully reacted. Filter off excess carbonate, then evaporate filtrate to crystallise salt. Example: CuCO₃(s) + 2HCl(aq) → CuCl₂(aq) + CO₂(g) + H₂O(l).

• Add excess insoluble carbonate to acid.
• Observe effervescence of CO₂ gas.
• Stir until no more bubbles form.
• Filter to remove excess carbonate.
• Evaporate filtrate to crystallise salt.
• Dry crystals thoroughly.

After reaction and filtration, the salt solution contains dissolved salt and water.

Heat the solution gently to evaporate water until crystals begin to form (saturated solution).

Allow solution to cool slowly to form pure crystals of the salt (crystallisation). Avoid evaporating to dryness to prevent decomposition or impurities concentrating in crystals. Dry crystals between filter paper or in a warm place.

• Heat solution gently to concentrate.
• Cool to crystallise salt.
• Filter crystals from solution.
• Dry crystals carefully.
• Store in a dry container.

Apparatus: burette, pipette, conical flask, beaker, evaporating basin, tripod, gauze, Bunsen burner, filter funnel, filter paper, white tile, indicator (e.g., methyl orange).

Method: 1. Rinse burette with acid, pipette with alkali. 2. Fill burette with acid, record initial volume. 3. Pipette fixed volume of alkali into conical flask, add indicator. 4. Add acid from burette to alkali, swirling until indicator changes colour (end point). 5. Record final burette reading; calculate volume used. 6. Repeat titration until concordant results obtained. 7. Mix exact volumes without indicator, heat to evaporate water and crystallise salt. 8. Filter crystals and dry.

Variables: independent – volume of acid added; dependent – indicator colour change; control – concentration of acid and alkali, temperature, volume of alkali pipetted. Repeat titrations improve accuracy. Safety includes handling acids carefully, using heat sources cautiously, and wearing eye protection.

• Use accurate volumetric apparatus for measurement.
• Repeat titrations for reliability.
• Heat solution gently to avoid splattering.
• Filter carefully to separate solids.
• Dry crystals in a warm, dry place.
• Wear safety goggles and gloves.

Solubility refers to the ability of a substance to dissolve in a solvent, commonly water in chemistry. Salts vary widely in their solubility, which affects their behaviour in aqueous solutions, including precipitation reactions and salt preparation methods. Understanding general solubility rules helps predict whether a salt will dissolve or form a precipitate when mixed with water or other solutions.

• Solubility is measured qualitatively as soluble or insoluble, or quantitatively in grams per 100 cm³ of water.
• Soluble salts dissolve to give clear solutions; insoluble salts form precipitates.
• Solubility depends on the ions present in the salt and their interactions with water molecules.

Salts containing sodium (Na⁺), potassium (K⁺), or ammonium (NH₄⁺) ions are generally soluble in water. This is due to the strong hydration of these ions, which overcomes the lattice energy of the salt, allowing it to dissolve readily.

• Examples include sodium chloride (NaCl), potassium nitrate (KNO₃), and ammonium sulfate ((NH₄)₂SO₄).
• These salts dissolve to form colourless solutions.
• This rule applies regardless of the accompanying anion.

Nitrate salts (NO₃⁻) are all soluble in water without exception. This universal solubility is important in qualitative analysis and salt preparation.

Chloride salts (Cl⁻) are mostly soluble, but notable exceptions include lead(II) chloride (PbCl₂) and silver chloride (AgCl), which are insoluble or sparingly soluble in cold water.

• Lead(II) chloride is sparingly soluble in cold water but more soluble in hot water.
• Silver chloride forms a white precipitate when silver ions react with chloride ions.
• Balanced equation for silver chloride precipitation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s) (white precipitate)

Most sulfate salts (SO₄²⁻) are soluble in water, but exceptions include barium sulfate (BaSO₄), calcium sulfate (CaSO₄), and lead(II) sulfate (PbSO₄), which are insoluble or only slightly soluble.

These exceptions are important in qualitative analysis and industrial processes.

• Barium sulfate forms a white precipitate, used in tests for sulfate ions.
• Calcium sulfate is sparingly soluble, often forming a slight precipitate.
• Lead(II) sulfate is insoluble, forming a white precipitate.

Carbonates (CO₃²⁻) are generally insoluble in water, except for those of sodium, potassium, and ammonium, which are soluble.

Hydroxides (OH⁻) are mostly insoluble, but sodium, potassium, and ammonium hydroxides are soluble. Calcium hydroxide is only slightly soluble, producing a weakly alkaline solution known as limewater.

• Sodium carbonate (Na₂CO₃) dissolves readily in water.
• Calcium carbonate (CaCO₃) is insoluble and forms a white precipitate.
• Calcium hydroxide (Ca(OH)₂) is sparingly soluble; its solution turns red litmus paper blue.
• Balanced equation for calcium hydroxide dissolving: Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq)

A hydrated substance is a compound that contains water molecules chemically bonded within its crystalline structure. These water molecules are known as water of crystallisation. They are an integral part of the compound's formula and affect its physical and chemical properties.

The water molecules in a hydrated substance are not merely trapped physically; they are incorporated into the crystal lattice through coordinate bonds or hydrogen bonding, which stabilises the structure.

Hydrated substances are often represented by a chemical formula that includes the number of water molecules, for example, copper(II) sulfate pentahydrate is written as CuSO₄·5H₂O, indicating five water molecules per formula unit.

• Water molecules are chemically bonded in hydrated substances.
• Water of crystallisation is part of the compound's formula.
• Example: CuSO₄·5H₂O (copper(II) sulfate pentahydrate).

An anhydrous substance is a compound that contains no water molecules within its structure. It is the form of the compound after all water of crystallisation has been removed, usually by heating.

Anhydrous substances often have different physical properties compared to their hydrated forms, such as colour, solubility, and density.

For example, anhydrous copper(II) sulfate is written as CuSO₄ and is white or greyish, whereas the hydrated form CuSO₄·5H₂O is blue.

• Anhydrous substances contain no water molecules.
• They result from removal of water of crystallisation.
• Example: CuSO₄ (anhydrous copper(II) sulfate) is white or greyish.

Water of crystallisation refers to the fixed number of water molecules chemically bonded within a crystalline solid.

This water affects the compound's colour, mass, and sometimes reactivity. For example, hydrated copper(II) sulfate is blue due to the water molecules, while the anhydrous form is white or grey.

When heated, hydrated substances lose their water of crystallisation, often accompanied by a colour change and mass loss. This process is reversible if water is reabsorbed from the atmosphere or solution.

• Water of crystallisation is chemically bonded water in crystals.
• It influences colour and mass of the compound.
• Heating removes water, causing colour change and mass loss.

The removal of water from a hydrated substance can be represented by a chemical equation showing the loss of water molecules.

For example, heating copper(II) sulfate pentahydrate causes it to lose water and become anhydrous copper(II) sulfate:

CuSO₄·5H₂O (s) → CuSO₄ (s) + 5H₂O (g) (on heating)

• Heating drives off water of crystallisation as water vapour.
• The compound changes from hydrated to anhydrous form.
• Example equation: CuSO₄·5H₂O (s) → CuSO₄ (s) + 5H₂O (g)

Hydrated substances can be identified by their characteristic colour and mass, which differ from their anhydrous counterparts.

In practical chemistry, heating a hydrated salt gently until it stops losing mass can produce the anhydrous form. The mass difference corresponds to the water lost.

Careful heating is required to avoid decomposition of the salt itself. The process is used to determine the amount of water of crystallisation experimentally.

• Colour change indicates hydration state.
• Mass loss on heating corresponds to water lost.
• Heating must be controlled to avoid salt decomposition.

Precipitation is a process where two soluble salts in aqueous solution react to form an insoluble salt that separates out as a solid, called a precipitate. This occurs because the product salt has very low solubility in water and thus cannot remain dissolved.

When two solutions containing ions are mixed, if a product salt is insoluble, it forms a solid that appears as a cloudy suspension or solid particles in the mixture. This solid can be collected by filtration.

For example, mixing aqueous solutions of silver nitrate (AgNO₃) and sodium chloride (NaCl) produces silver chloride (AgCl), an insoluble salt, as a white precipitate: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

• Precipitation occurs when an insoluble salt forms from two soluble salts in solution.
• The insoluble salt appears as a solid precipitate.
• The reaction is a double displacement (metathesis) reaction.

To prepare a specific insoluble salt, select two soluble salts whose ions combine to form the desired insoluble salt.

Solubility rules help predict which salts are insoluble. For example, most chlorides are soluble except silver chloride and lead chloride; most sulfates are soluble except barium sulfate, calcium sulfate, and lead sulfate.

For example, to prepare barium sulfate (BaSO₄), mix barium chloride (BaCl₂) solution with sodium sulfate (Na₂SO₄) solution. Both are soluble, but BaSO₄ is insoluble and precipitates out.

• Use solubility rules to identify insoluble salts.
• Mix solutions containing ions that form the insoluble salt.
• Example: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

Precipitation reactions can be represented by balanced symbol equations and ionic equations.

Example: Preparation of silver chloride precipitate

Symbol equation: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) (balanced as written) Ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s) Spectator ions (Na⁺ and NO₃⁻) remain in solution and do not participate in precipitate formation.

• Write full balanced symbol equations including state symbols.
• Write ionic equations showing only ions involved in precipitate formation.
• Identify and exclude spectator ions in ionic equations.

Apparatus needed: two clean beakers, stirring rod, filter funnel, filter paper, conical flask, wash bottle with distilled water, drying oven or warm place.

Method: 1. Measure equal volumes (e.g., 50 cm³) of two soluble salt solutions using a measuring cylinder. 2. Pour one solution into a beaker, then add the other solution slowly while stirring continuously. 3. A precipitate forms immediately if an insoluble salt is produced. 4. Stir the mixture for a few minutes to complete the reaction. 5. Set up filtration apparatus with filter paper in a funnel over a conical flask. 6. Pour the mixture through the filter paper to separate the precipitate from the solution. 7. Wash the precipitate on the filter paper with small amounts of distilled water to remove soluble impurities. 8. Transfer the wet precipitate to a watch glass and dry in a warm oven or air dry.

Variables: Independent variable: type or concentration of solutions. Dependent variable: amount or mass of precipitate formed. Control variables: volume of solutions, temperature, stirring time. Repeat the experiment to ensure reproducibility and calculate average mass of precipitate for accuracy. Safety: Handle chemicals with care, wear safety goggles and gloves, avoid ingestion or inhalation of powders and solutions.

• Measure equal volumes of two soluble salt solutions.
• Mix solutions and stir to form precipitate.
• Filter mixture to collect precipitate.
• Wash precipitate with distilled water to remove impurities.
• Dry precipitate before weighing or further use.
• Control variables and repeat for accuracy.
• Wear appropriate safety equipment.

When the two solutions are mixed, a cloudy suspension or solid particles appear, indicating precipitate formation.

The filtrate (liquid passed through filter paper) is clear, showing removal of the solid.

The precipitate is usually a fine solid that can be dried to a powder or crystals depending on the salt and drying conditions. Example: Silver chloride precipitate is a white solid that darkens on exposure to light due to decomposition to silver and chlorine gas (not part of preparation but important to note). Mass of dried precipitate can be measured to calculate yield or purity if required.

• Cloudy suspension or solid forms on mixing solutions.
• Filtrate is clear after filtration.
• Precipitate is a solid that can be dried and weighed.
• Observations confirm formation of insoluble salt.

Water of crystallisation is the term used to describe water molecules that are chemically bound within the crystal lattice of a hydrated salt. These water molecules are not merely trapped physically but are an integral part of the crystal structure, often influencing its shape and stability.

The water molecules are present in fixed stoichiometric amounts and are represented in the chemical formula of the compound, for example, CuSO₄·5H₂O indicates five water molecules per formula unit of copper(II) sulfate.

• Water of crystallisation = water molecules chemically incorporated in a crystal lattice.
• Present in fixed ratios, shown in the formula after a dot (·).

Two common examples of hydrated salts are copper(II) sulfate pentahydrate and cobalt(II) chloride hexahydrate. Their formulas explicitly show the number of water molecules associated with each formula unit:

Copper(II) sulfate pentahydrate: CuSO₄·5H₂O (five water molecules)

Cobalt(II) chloride hexahydrate: CoCl₂·6H₂O (six water molecules)

• CuSO₄·5H₂O is blue due to water of crystallisation.
• CoCl₂·6H₂O is pink; loss of water changes colour.

Water of crystallisation helps maintain the three-dimensional arrangement of ions in the crystal lattice. It affects the physical properties such as colour, shape, and solubility of the salt.

For example, copper(II) sulfate pentahydrate is blue because of the water molecules; when the water is removed, the anhydrous form is white or greyish.

The water molecules can also influence the melting point and stability of the compound.

• Water molecules stabilise crystal structure.
• Affects colour and physical appearance.
• Influences melting point and solubility.

When hydrated crystals are heated gently, the water of crystallisation is driven off as water vapour. This process is called dehydration.

For example, heating CuSO₄·5H₂O causes it to lose water and change from blue to white anhydrous CuSO₄.

The reaction can be represented as: CuSO₄·5H₂O (s) → CuSO₄ (s) + 5H₂O (g)

• Heating removes water of crystallisation (dehydration).
• Colour changes indicate loss of water (e.g., blue to white for CuSO₄).
• Water vapour is released during heating.

Copper(II) sulfate pentahydrate has the formula CuSO₄·5H₂O. This means each formula unit contains one copper(II) sulfate unit and five water molecules.

If 100 g of CuSO₄·5H₂O is heated and loses all its water of crystallisation, the mass of anhydrous CuSO₄ remaining can be calculated.

Molar masses: CuSO₄ = 159.6 g/mol, H₂O = 18.0 g/mol, so 5H₂O = 90.0 g/mol. Total molar mass of CuSO₄·5H₂O = 159.6 + 90.0 = 249.6 g/mol.

• Calculate mass of anhydrous CuSO₄ in 100 g hydrated salt:
• (159.6 / 249.6) × 100 g = 63.9 g CuSO₄
• Mass of water lost = 100 g - 63.9 g = 36.1 g