Preparation of salts

1. Overview

The preparation of salts is a fundamental laboratory process in chemistry. It involves selecting the correct method based on the solubility of the reactants and the desired salt product. Understanding these techniques is crucial for controlling chemical reactions and purifying substances for industrial and medicinal use.

Key Definitions

• Salt: A compound formed when the hydrogen ions ($H^+$) of an acid are replaced by metal ions or ammonium ($NH_4^+$) ions.
• Hydrated substance: A substance that is chemically combined with water molecules within its crystal structure.
• Anhydrous substance: A substance containing no water, often formed by heating a hydrated salt to drive off the water of crystallisation.
• Precipitate: An insoluble solid that emerges from a liquid solution during a chemical reaction.
• Filtrate: The liquid that passes through a filter.
• Residue: The solid left behind on the filter paper after filtration.

Core Content

Solubility Rules for Salts

To choose the correct preparation method, you must know if the salt is soluble:

• Sodium, potassium, and ammonium salts: All are soluble.
• Nitrates: All are soluble.
• Chlorides: All are soluble except lead(II) chloride ($PbCl_2$) and silver chloride ($AgCl$).
• Sulfates: All are soluble except barium sulfate ($BaSO_4$), calcium sulfate ($CaSO_4$), and lead(II) sulfate ($PbSO_4$).
• Carbonates: All are insoluble except sodium, potassium, and ammonium carbonates.
• Hydroxides: All are insoluble except sodium, potassium, ammonium, and calcium hydroxides (calcium hydroxide is partially soluble).

Preparation of Soluble Salts

There are two main methods for making soluble salts:

Method A: Reaction of an acid with an insoluble reactant (Excess Metal, Base, or Carbonate) Used when the starting material (metal, base, or carbonate) does not dissolve in water.

1. Add excess insoluble solid to the acid to ensure all acid is neutralised.
2. Heat gently to speed up the reaction (for bases and metals).
3. Filter the mixture to remove the unreacted excess solid.
4. Heat the resulting filtrate in an evaporating dish to the "point of crystallisation" (saturated solution).
5. Allow to cool so crystals form; then filter, wash with a little distilled water, and dry.

• Example (Excess Metal): $Magnesium + Sulfuric\ acid \rightarrow Magnesium\ sulfate + Hydrogen$ $Mg(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2(g)$
• Example (Excess Insoluble Base): $Copper(II)\ oxide + Hydrochloric\ acid \rightarrow Copper(II)\ chloride + Water$ $CuO(s) + 2HCl(aq) \rightarrow CuCl_2(aq) + H_2O(l)$
• Example (Excess Insoluble Carbonate): $Calcium\ carbonate + Nitric\ acid \rightarrow Calcium\ nitrate + Water + Carbon\ dioxide$ $CaCO_3(s) + 2HNO_3(aq) \rightarrow Ca(NO_3)_2(aq) + H_2O(l) + CO_2(g)$

Method B: Reaction of an acid with an alkali (Titration) Used when both reactants are soluble (e.g., sodium hydroxide and hydrochloric acid).

1. Use a pipette to add a fixed volume of alkali to a conical flask.
2. Add an indicator (e.g., methyl orange or phenolphthalein).
3. Add acid from a burette until the indicator changes colour (the end-point).
4. Note the volume of acid used, then repeat the process without indicator using the same volumes.
5. Evaporate and crystallise the resulting solution as described in Method A.

• Example (Titration): $Sodium\ hydroxide + Hydrochloric\ acid \rightarrow Sodium\ chloride + Water$ $NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$

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Extended Content (Extended Only)

Preparation of Insoluble Salts (Precipitation)

Insoluble salts are prepared by mixing two solutions of soluble salts.

1. Mix the two soluble salt solutions.
2. A precipitate (the insoluble salt) forms immediately.
3. Filter the mixture to collect the precipitate (residue).
4. Wash the residue with distilled water to remove traces of soluble impurities.
5. Dry the precipitate in a warm oven or by patting with filter paper.

• Example (Precipitation): To make silver chloride ($AgCl$): $Silver\ nitrate(aq) + Sodium\ chloride(aq) \rightarrow Silver\ chloride(s) + Sodium\ nitrate(aq)$ $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$

Water of Crystallisation

Many salts form crystals that contain water molecules as part of their fixed chemical structure. This is called water of crystallisation.

• Definition: The water molecules present in hydrated crystals.
• Hydrated Copper(II) sulfate: $CuSO_4 \cdot 5H_2O$ (Blue crystals).
• Hydrated Cobalt(II) chloride: $CoCl_2 \cdot 6H_2O$ (Pink crystals).

When these are heated, the water is lost, leaving the anhydrous salt (Anhydrous $CuSO_4$ is white; anhydrous $CoCl_2$ is blue).

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Key Equations

Reaction Type	General Equation
Acid + Metal	$Acid + Metal \rightarrow Salt + Hydrogen$
Acid + Base	$Acid + Base \rightarrow Salt + Water$
Acid + Carbonate	$Acid + Carbonate \rightarrow Salt + Water + Carbon\ Dioxide$
Precipitation	$Soluble\ salt\ A + Soluble\ salt\ B \rightarrow Insoluble\ salt + Soluble\ salt\ C$

Specific Balanced Symbols Equations to Know:

1. $Zn(s) + H_2SO_4(aq) \rightarrow ZnSO_4(aq) + H_2(g)$
2. $NaOH(aq) + HNO_3(aq) \rightarrow NaNO_3(aq) + H_2O(l)$
3. $BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)$

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Common Mistakes to Avoid

• ❌ Wrong: Forgetting to use "excess" solid when making a soluble salt from an insoluble base.
• ✓ Right: Always state "add excess solid" to ensure all the acid is used up, so the salt isn't contaminated with acid.
• ❌ Wrong: Washing a soluble salt with lots of water after filtration.
• ✓ Right: Only wash insoluble precipitates. If you wash a soluble salt crystal with too much water, it will dissolve and be lost.
• ❌ Wrong: Heating a solution to dryness to get crystals.
• ✓ Right: Heat to the "point of crystallisation" (saturation) and let it cool slowly to get large, well-defined crystals.

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Exam Tips

• Command Word "Describe": When asked to describe a preparation, list the practical steps in order (Add, Stir, Filter, Evaporate, Dry).
• Identifying Methods: If the salt is insoluble, use Precipitation. If it is soluble and contains Na, K, or $NH_4$, use Titration. For all other soluble salts, use the "Excess Insoluble Base/Metal" method.
• State Symbols: In precipitation questions, the salt you are making must have the state symbol (s), while the reactants are (aq).
• Real-world Context: Barium sulfate is often used as a "Barium meal" in X-rays because it is insoluble and won't be absorbed into the blood, making it safe despite barium's toxicity.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0620 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student prepares a sample of zinc sulfate crystals using the reaction between dilute sulfuric acid and excess zinc oxide.

(a) State why the student uses excess zinc oxide. [1]

(b) Describe how the student would obtain a pure, dry sample of zinc sulfate crystals from the resulting mixture. [4]

Worked Solution:

(a)

1. To ensure all the sulfuric acid reacts. Using excess zinc oxide guarantees that the sulfuric acid is the limiting reactant and is completely used up.

How to earn full marks:

• Mention the key idea of 'all sulfuric acid reacts', or 'zinc oxide is in excess'.

(b)

1. Filter the mixture. To remove the excess zinc oxide, which is insoluble.

2. Heat the filtrate to evaporate some of the water. To concentrate the zinc sulfate solution and promote crystallisation. Do not heat to dryness.

3. Leave the solution to cool. To allow crystals to form.

4. Filter the crystals and dry them between filter paper. To separate the crystals from the remaining solution and remove any surface water.

How to earn full marks:

• Filtration to remove excess zinc oxide.
• Evaporation of water (not to dryness!).
• Cooling to crystallise.
• Drying the crystals.
• Each step must be in the correct order.

Common Pitfall: Students often forget to specify that the solution should not be heated to dryness during evaporation. Also, the order of steps is crucial for obtaining a pure sample; ensure you understand why each step is performed in the given sequence.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

(a) Define the term "hydrated salt". [2]

(b) Cobalt(II) chloride exists as both an anhydrous salt (CoCl₂) and a hydrated salt (CoCl₂.6H₂O). Describe a chemical test to show the presence of water in hydrated cobalt(II) chloride. Include the expected observation. [4]

Worked Solution:

(a)

1. A hydrated salt is a salt that contains water molecules chemically combined within its crystal structure. Definition of hydrated salt

2. The water molecules are present in a fixed ratio. Specifies the fixed ratio aspect of water of crystallisation.

How to earn full marks:

• Mention that a hydrated salt contains water molecules chemically combined.
• State that the water molecules are in a fixed ratio.

(b)

1. Add the hydrated cobalt(II) chloride to anhydrous copper(II) sulfate. Anhydrous copper(II) sulfate is white and turns blue in the presence of water.

2. Observation: The anhydrous copper(II) sulfate will turn blue. The color change confirms the presence of water.

3. Alternatively, heat the hydrated cobalt(II) chloride strongly. Heating releases water.

4. Observation: A colour change of solid from pink to blue is seen, and water droplets are seen at the top of the test tube. The colour change and visible water confirms the presence of water.

How to earn full marks:

• Correctly state the addition of substance and the expected observation.
• Alternatively, state the heating and the correct observations.

Common Pitfall: When describing the test for water, be sure to include both the reagent used (anhydrous copper(II) sulfate or heating) and the expected observation (color change or water droplets). Forgetting either part will result in a loss of marks.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A student wants to prepare a pure sample of lead(II) chloride, PbCl₂, which is an insoluble salt.

(a) State the general solubility rules that explain why lead(II) chloride is insoluble, while other chlorides are generally soluble. [2]

(b) Describe how the student would prepare a pure, dry sample of lead(II) chloride starting from solutions of lead(II) nitrate and sodium chloride. Include relevant steps for separation and purification. [6]

Worked Solution:

(a)

1. Chlorides are generally soluble. States the general rule for chlorides

2. Except for lead and silver chlorides. Identifies the exceptions to the rule

How to earn full marks:

• State the general rule: "Chlorides are soluble".
• State the exceptions: "Except lead and silver".

(b)

1. Mix solutions of lead(II) nitrate and sodium chloride. Mixing the solutions will result in a precipitation reaction.

2. A precipitate of lead(II) chloride will form. Identifies the formation of the precipitate of lead(II) chloride.

3. Filter the mixture to collect the lead(II) chloride precipitate. Filtration separates the solid from the solution.

4. Wash the precipitate with distilled water. Washing removes any soluble impurities from the precipitate.

5. Dry the precipitate in a warm oven or between filter paper. Drying removes the water from the precipitate, resulting in a pure, dry sample.

6. Write a balanced chemical equation for the reaction. $Pb(NO_3)_2(aq) + 2NaCl(aq) \rightarrow PbCl_2(s) + 2NaNO_3(aq)$ Equation shows the reaction with correct formula and state symbols.

How to earn full marks:

• Mix the correct solutions.
• Formation of precipitate.
• Filtration to collect the precipitate.
• Washing to remove impurities.
• Drying to obtain a dry sample.
• Balanced chemical equation, including state symbols.

Common Pitfall: Many students forget to include state symbols in the balanced chemical equation. Also, remember that washing the precipitate is crucial for removing soluble impurities, leading to a purer final product.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A student performs a titration experiment to determine the concentration of a hydrochloric acid solution. They react the hydrochloric acid with a standard solution of sodium hydroxide.

(a) State the name of a suitable indicator for this titration. State the colour change observed at the end point. [2]

(b) In one experiment, 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide solution is neutralised by 20.0 cm³ of the hydrochloric acid. Calculate the concentration of the hydrochloric acid in mol/dm³. [5]

(c) Suggest two reasons why the experimental value obtained by the student might be slightly different from the actual value. [2]

Worked Solution:

(a)

1. Methyl orange or phenolphthalein. Acceptable indicators for a strong acid/strong base titration.

2. Methyl orange: red to yellow; Phenolphthalein: pink to colourless. Correct colour change at the end point for the stated indicator.

How to earn full marks:

• State a correct indicator (methyl orange or phenolphthalein).
• Give the correct colour change at the end point for the stated indicator.

(b)

1. Calculate moles of NaOH. $moles \ NaOH = concentration \times volume = 0.100 \ mol/dm^3 \times \frac{25.0}{1000} \ dm^3 = 0.00250 \ mol$ Correctly converts volume to dm³ and calculates moles of NaOH.

2. Write the balanced equation. $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$ Correct balanced equation for the reaction.

3. Determine moles of HCl. From the equation, 1 mole of HCl reacts with 1 mole of NaOH. Therefore, moles of HCl = moles of NaOH = 0.00250 mol.

4. Calculate concentration of HCl. $concentration \ of \ HCl = \frac{moles}{volume} = \frac{0.00250 \ mol}{\frac{20.0}{1000} \ dm^3} = 0.125 \ mol/dm^3$ Correctly calculates the concentration of HCl.

How to earn full marks:

• Calculates moles of NaOH correctly.
• Writes the balanced equation correctly.
• Determines the moles of HCl correctly.
• Calculates the concentration of HCl correctly.
• Includes the correct unit (mol/dm³).

(c)

1. Incomplete mixing of the solutions. This could lead to local variations in concentration.

2. Error in reading the burette/pipette. Parallax error or misreading the meniscus.

3. Some solution was lost during transfer. Solution spilled or stuck to the sides of the glassware.

4. Indicator not added correctly, or overshot the endpoint. Adding too much indicator can affect the endpoint volume.

How to earn full marks:

• Gives two valid reasons why the experimental value might differ from the actual value.
• The reasons must be related to practical errors in the titration process.

Common Pitfall: When calculating concentrations in titrations, always double-check that you've converted volumes from cm³ to dm³ correctly. Also, remember to relate the moles of acid and base using the stoichiometry of the balanced chemical equation. The final answer is $\boxed{0.125 \ mol/dm^3}$.