ChemistryCore22 min read

Acids, Bases & Salts

The pH scale, neutralisation, and how salts are made

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Practise the full question bank:

01

What acids and bases actually are

An acid releases hydrogen ions, H⁺, when dissolved in water — that single ion is responsible for all acidic behaviour. A base is the opposite: it accepts H⁺, and a base that dissolves in water (releasing OH⁻) is called an alkali.

Strong acids and alkalis (hydrochloric acid, sodium hydroxide) ionise completely in water; weak ones (ethanoic acid, ammonia) only partly ionise. Strong versus weak is about the degree of ionisation, not how concentrated the solution is — a dilute strong acid is still a strong acid.

02

The pH scale

The pH scale runs from 0 to 14 and measures how acidic or alkaline a solution is: below 7 is acidic, 7 is neutral, above 7 is alkaline. It tracks the concentration of H⁺ ions — more H⁺ means lower pH.

It is a logarithmic scale, so each step of 1 is a tenfold change in H⁺ concentration: pH 3 is ten times more acidic than pH 4. You measure pH with an indicator (litmus for a simple acid/alkali test, universal indicator for a full colour range) or a pH meter.

pH < 7 acidic·pH 7 neutral·pH > 7 alkalineeach 1 unit = a ×10 change in H⁺ concentration
03

Neutralisation

Mix an acid with a base and they cancel out — neutralisation — producing a salt and water. Strip away the spectator ions and every neutralisation is really the same reaction:

H⁺ (aq) + OH⁻ (aq) → H₂O (l)the essence of every acid + alkali neutralisation

Neutralisation is everywhere

Indigestion tablets are a base neutralising excess stomach acid; lime is spread on acidic soil to raise its pH; toothpaste neutralises the acid bacteria make on your teeth. Same reaction, three everyday uses.

04

The reactions of acids

Acids follow three reliable patterns, each giving a salt:

Acid reacts withProductsTest / sign
a metalsalt + hydrogenbubbles; a lit splint gives a squeaky pop
a base or alkalisalt + waterneutralisation (no gas)
a carbonatesalt + water + carbon dioxidefizzing; limewater turns milky
05

A worked example

Worked example

Write what forms when (a) magnesium reacts with hydrochloric acid, and (b) calcium carbonate reacts with hydrochloric acid.

  1. (a) Metal + acid → salt + hydrogen. The acid is hydrochloric, so the salt is a chloride: magnesium chloride + hydrogen.Acids give hydrogen with reactive metals; hydrochloric acid always makes a chloride.
  2. Mg + 2HCl → MgCl₂ + H₂.Balanced: two HCl supply the two chlorides and the hydrogen gas.
  3. (b) Carbonate + acid → salt + water + carbon dioxide: calcium chloride + water + CO₂.Carbonates always release CO₂ (the fizzing), plus a salt and water.
  4. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂.The CO₂ turns limewater milky — the standard test.

(a) magnesium chloride + hydrogen (Mg + 2HCl → MgCl₂ + H₂); (b) calcium chloride + water + carbon dioxide (CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂).

06

Indicators and titration

An indicator changes colour depending on pH, letting you tell an acid from an alkali or find the exact end point of a reaction. Litmus is red in acid and blue in alkali; universal indicator gives a full colour range and an approximate pH. For accurate work, a titration adds one solution from a burette to a fixed volume of the other until an indicator just changes colour — methyl orange (red→yellow) and phenolphthalein (colourless→pink) give sharp end points.

From the volume and concentration of the solution you add, you can calculate the concentration of the unknown.

07

Preparing salts

How you make a salt depends on whether it is soluble. For a soluble salt from an insoluble base, metal or carbonate: add the solid in excess to warm acid, filter off the excess, then evaporate and crystallise. For a soluble salt from an alkali, use a titration (no excess to filter) then crystallise the neutral solution.

An insoluble salt is made by precipitation — mixing two soluble solutions so the insoluble salt drops out, which is then filtered, washed and dried.

Salt typeMethodKey steps
Soluble (from base/metal/carbonate)excess solid + acidreact, filter off excess, evaporate, crystallise
Soluble (from alkali)titrationfind the exact volume, remake without indicator, crystallise
Insolubleprecipitationmix two solutions, filter, wash, dry
08

Brønsted–Lowry acids and Ka (A Level)

A more general definition: a Brønsted–Lowry acid is a proton (H⁺) donor and a base is a proton acceptor. When an acid donates a proton it becomes its conjugate base, forming acid–base conjugate pairs. This explains why water can act as both.

For a weak acid, only a fraction of molecules ionise, and the position of that equilibrium is measured by the acid dissociation constant, Ka (or pKa = −log Ka). A larger Ka means a stronger acid. The pH itself comes from the hydrogen-ion concentration.

pH = −log₁₀[H⁺]Ka = [H⁺][A⁻] ÷ [HA][H⁺] in mol/dm³; a bigger Ka (smaller pKa) = stronger acid
09

Another worked example

Worked example

In a titration, 25.0 cm³ of sodium hydroxide is exactly neutralised by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid. NaOH + HCl → NaCl + H₂O. Find the concentration of the sodium hydroxide.

  1. Moles HCl = c × V = 0.100 × (20.0/1000) = 2.0 × 10⁻³ mol.moles = concentration × volume in dm³.
  2. Mole ratio NaOH : HCl = 1 : 1, so moles NaOH = 2.0 × 10⁻³ mol.The equation shows one mole of each reacts.
  3. Concentration NaOH = n ÷ V = 2.0 × 10⁻³ ÷ (25.0/1000).c = n ÷ V with V in dm³.
  4. = 0.080 mol/dm³.2.0 × 10⁻³ ÷ 0.025 = 0.080.

0.080 mol/dm³.

Practice questions

6 questions · 22 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

4 · 9 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State what all acids produce in aqueous solution, and give the pH range of an acid.
Model answer

Acids produce hydrogen ions, H⁺, in aqueous solution; their pH is below 7.

Examiner tip. Acids → H⁺; alkalis → OH⁻. Neutral is pH 7.

SQ2[2 marks]
Explain the difference between a strong acid and a weak acid.
Model answer

A strong acid is fully ionised in water; a weak acid is only partly ionised. It is about degree of ionisation, not concentration.

Examiner tip. Strong vs weak = how completely it ionises; concentrated vs dilute = how much water is present.

SQ3[3 marks]
Write a word and balanced symbol equation for the reaction between hydrochloric acid and sodium hydroxide.
Model answer

Hydrochloric acid + sodium hydroxide → sodium chloride + water. HCl + NaOH → NaCl + H₂O.

Examiner tip. Acid + alkali → salt + water is neutralisation; the salt's name comes from the acid (chloride from HCl).

SQ4[2 marks]
A solution has a hydrogen-ion concentration of 1 × 10⁻³ mol/dm³. Calculate its pH and state whether it is acidic or alkaline.
Model answer

pH = −log₁₀(1 × 10⁻³) = 3. pH 3 is below 7, so the solution is acidic.

Examiner tip. pH = −log[H⁺]; a power of −3 gives pH 3.

Long questions

2 · 13 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[7 marks]
A student prepares dry crystals of copper(II) sulfate from copper(II) oxide and dilute sulfuric acid.
  1. Write the balanced equation for the reaction. [2]
  2. Describe how the student obtains pure dry crystals from the mixture. [4]
  3. Explain why excess copper(II) oxide is used. [1]
Mark scheme
  1. CuO + H₂SO₄ → CuSO₄ + H₂O (correct formulae)[1]
  2. Equation balanced[1]
  3. Warm the acid and add excess copper(II) oxide, then filter off the unreacted solid[1]
  4. Heat the filtrate to evaporate some water / to the point of crystallisation[1]
  5. Leave to cool so crystals form[1]
  6. Filter and dry the crystals (e.g. between filter paper)[1]
  7. Excess ensures all the acid reacts, and the excess solid can simply be filtered off[1]

CuO + H₂SO₄ → CuSO₄ + H₂O; react with excess base, filter, evaporate to crystallisation, cool, filter, dry; excess guarantees all acid used and is easily removed.

Examiner tip. Insoluble base + acid is the classic soluble-salt preparation — the excess solid is removed by filtration.

LQ2[6 marks]
A student is asked to prepare pure, dry crystals of copper(II) sulfate from copper(II) oxide (an insoluble base) and dilute sulfuric acid.
  1. Write the equation for the reaction. [2]
  2. Describe the method to obtain pure dry crystals. [4]
Mark scheme
  1. CuO + H₂SO₄ → CuSO₄ + H₂O (correct formulae)[1]
  2. balanced[1]
  3. Warm the acid and add copper(II) oxide in excess until no more reacts[1]
  4. Filter to remove the excess copper(II) oxide[1]
  5. Heat the filtrate to evaporate water to the point of crystallisation, then leave to cool[1]
  6. Filter off the crystals and dry them (e.g. between filter paper)[1]

CuO + H₂SO₄ → CuSO₄ + H₂O; react with excess base, filter, evaporate to crystallisation, cool, filter, dry.

Examiner tip. Excess insoluble base guarantees all the acid reacts and is easily filtered off.