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2gInorganic Chemistry

Acids, bases and salt preparations — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions

10 min read 13 parts Lesson 6 of 23 1 quiz

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Curriculum Equation Key pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: pH + pOH = 14
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Lesson summary

Acids, bases and salt preparations Edexcel O Level Linear Chemistry notes: know the general rules for predicting the solubility of ionic compounds in waterpH + pOH = 14 at 25 °C.Each pH unit is a tenfold change in H⁺.Indicatorslitmus, phenolphthalein, methyl orange.know the gener…

  • Acids, bases and salt preparations Edexcel O Level Linear Chemistry: what to do next
  • Arrhenius acid
  • Arrhenius base
  • Brønsted–Lowry acid
  • Brønsted–Lowry base
  • Strong acid

Section 1 of 13Process & sequence

1
Put acid in a flask with indicator.
2
Add base from a burette.
3
Stop at the colour change (end point).
4
Record the volume.
5
Use M₁V₁ = M₂V₂ (with mole ratio).
  • pH 0
  • Strong acid
  • pH 7 neutral
  • Strong base pH 14

Section 2 of 13Keep going

Acids, bases and salt preparations Edexcel O Level Linear Chemistry: what to do next

These Acids, bases and salt preparations Edexcel O Level Linear Chemistry notes covered the key ideas, worked examples and exam technique for this topic. Check your Acids, bases and salt preparations Edexcel O Level Linear Chemistry understanding with the practice questions below, then reread any Acids, bases and salt preparations Edexcel O Level Linear Chemistry point you were unsure about before moving on.

1
Practice: Acids, bases and salt preparations Edexcel O Level Linear Chemistry practice questions
2
Previous lesson: 2f Acids, alkalis and titrations — Notes
3
Next lesson: 2h Chemical tests — Notes
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Course: Edexcel Linear – OL (chemistry) — all units
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Official syllabus: Pearson Edexcel International GCSEs

Acids, bases and salt preparations Edexcel O Level Linear Chemistry revision notes

Section 3 of 13Preparing copper(II) sulfate (insoluble base method)

The aim is to prepare a pure, dry sample of hydrated copper(II) sulfate crystals. Apparatus: measuring cylinder, 100 cm3 beaker, Bunsen burner, tripod and gauze, glass rod, filter funnel and paper, evaporating basin.

1
Add 50 cm3 of dilute sulfuric acid to a beaker and warm gently with a Bunsen burner.
2
Add copper(II) oxide a spatula at a time with stirring until the base stops dissolving and a black suspension remains (excess).
3
Filter into an evaporating basin to remove the excess oxide.
4
Heat the filtrate gently to evaporate about half of the water, then leave to cool and crystallise.
5
Dry the crystals between filter papers.

The equation is CuO + H2SO4 → CuSO4 + H2O. Safety: warm, do not boil, to avoid spitting; wear eye protection.

Section 4 of 13Titration method for sodium salts

A pipette measures the alkali into a conical flask with a few drops of indicator (phenolphthalein or methyl orange).

1
The acid is added from a burette until the indicator changes colour.
2
The volume of acid is recorded, and the process is repeated without indicator using the same volumes.
3
The neutral solution is evaporated until crystals start to form and then cooled.
4
A diagram should show a burette, a clamp and stand, a conical flask on a white tile.

Section 5 of 13Precipitation of lead(II) sulfate

The pattern soluble salt 1 + soluble salt 2 → insoluble salt + soluble salt 3 applies. Mix 25 cm3 of 0.5 mol/dm3 lead(II) nitrate with 25 cm3 of 0.5 mol/dm3 potassium sulfate, filter, wash with distilled water and dry. The product is the residue.

Section 6 of 13Percentage yield and purity

The theoretical yield is calculated from the balanced equation, and percentage yield = (actual mass ÷ theoretical mass) × 100.

1
Example: 4.0 g of CuO (Mr 80) gives 0.050 mol; CuSO4.5H2O has Mr 250, so theoretical mass = 12.5 g.
2
If 9.0 g is collected, % yield = 9.0 ÷ 12.5 × 100 = 72%.
3
Yields are below 100% because of incomplete reaction, losses during filtration, crystals left in solution and transfers.
4
Purity is checked by melting point or by the crystals having a regular shape.
MethodProductSeparation step
Excess insoluble baseSoluble saltFilter off excess solid
TitrationSoluble salt of Na, K or NH4Evaporate and crystallise
PrecipitationInsoluble saltFilter and wash residue

Section 7 of 13Compare

Acids and bases
TypeExampleIonisationpH (0.1 M)ConductivityReaction with metal
Strong acidHClComplete1HighH₂ gas
Weak acidEthanoic acidPartial~3LowSlow H₂
Strong baseNaOHComplete13High—
Weak baseNH₃Partial~11Low—
NeutralNaCl solution—7High (ions)—

Section 8 of 13Worked calculation: moles to mass of a salt

Moles link every preparation.

1
Example: 2.0 g of magnesium carbonate (Mr 84) reacts with excess hydrochloric acid to make magnesium chloride.
2
Equation: MgCO3 + 2HCl → MgCl2 + H2O + CO2.
3
Step 1: moles MgCO3 = 2.0 ÷ 84 = 0.0238 mol.
4
Step 2: ratio 1:1, so moles MgCl2 = 0.0238 mol.
5
Step 3: Mr of MgCl2 = 95, so mass = 0.0238 × 95 = 2.26 g.
6
If 1.8 g is obtained the yield is 1.8 ÷ 2.26 × 100 = 80%.

To find the volume of acid needed: 0.0238 mol of carbonate needs 0.0476 mol of HCl; with 1.0 mol/dm3 acid this is 47.6 cm3. Always choose a slightly larger volume or an excess of the solid so that the reaction is complete.

Section 9 of 13Worked examples

  1. Calculate the percentage yield when 8.0 g of copper(II) sulfate crystals (theoretical 12.5 g) are obtained. [2]2 marks

  2. Why must the copper(II) oxide be added until it is in excess? [2]2 marks

  3. Give two reasons why the yield is less than 100%. [2]2 marks

  4. Why is the lead(II) sulfate residue washed with water? [1]1 mark

Section 10 of 13Key terms

TermMeaning
Percentage yieldActual mass divided by theoretical mass times 100
Theoretical yieldMaximum mass predicted from the equation
FiltrateSolution passing through filter paper
BuretteApparatus for delivering variable volumes of liquid
PipetteApparatus for delivering a fixed volume accurately
HydratedContaining water of crystallisation

Section 11 of 13Syllabus checklist

1
know the general rules for predicting the solubility of ionic compounds in water
  • - common sodium
  • potassium and ammonium compounds are soluble - all nitrates are soluble - common chlorides are soluble
  • except those of silver and lead(II) - common sulfates are soluble
  • except for those of barium
  • calcium and lead(II) - common carbonates are insoluble
  • except for those of sodium
  • potassium and ammonium - common hydroxides are insoluble except for those of sodium
  • potassium and calcium (calcium hydroxide is slightly soluble)
2
understand acids and bases in terms of proton transfer
3
understand that an acid is a proton donor and a base is a proton acceptor describe the reactions of hydrochloric acid, sulfuric acid and nitric acid with metals, bases and metal carbonates (excluding the reactions between nitric acid and metals) to form salts
4
know that metal oxides, metal hydroxides and ammonia can act as bases, and that alkalis are bases that are soluble in water
5
describe an experiment to prepare a pure, dry sample of a soluble salt, starting from an insoluble reactant
6
practical: prepare a sample of pure, dry hydrated copper(II) sulfate crystals starting from copper(II) oxide

Section 12 of 13Command terms

Command terms
Give / State / Name
Recall a fact; one idea per mark.
Define
Give the exact meaning using the specification keywords.
Describe
Give an account of what happens / the pattern in data, quoting values; no reasons needed.
Explain
Give reasons with linked statements (cause → effect). Each linked step is usually a separate marking point.
Suggest
Use knowledge in a new context; more than one answer may be accepted.
Compare
Similarities and differences; each point must refer to both items.
Calculate
Show working; include units; check significant figures. Method marks are given even if the final answer is wrong.
Discuss
Explore the issue from different angles and link the points.
Evaluate / Assess
Weigh strengths and weaknesses and reach a supported judgement.
Deduce
Draw a conclusion from the information given.
Justify
Support a statement/choice with evidence.

Section 13 of 13Exam technique

1
Evaporating all the water instead of leaving crystals to form.
2
Forgetting to use the hydrated relative formula mass for the theoretical yield.
3
Reading a pipette or burette incorrectly.
4
Using too little base so acid remains in the product.

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