Neutralisation and titration

This page covers one Form 4 content standard: neutralisation and titration. It brings together everything earlier in the chapter, the H⁺ and OH⁻ ions, molarity, and standard solutions, into one experiment and one calculation that appear in almost every SPM Chemistry paper. Master the procedure and the mole-ratio calculation and you secure a reliable block of marks.

Neutralisation: the reaction

Neutralisation is the reaction between an acid and a base (or alkali) to produce a salt and water. The general word equation is:

acid + base → salt + water

For an acid and an alkali in solution, say hydrochloric acid and sodium hydroxide, the full equation is HCl + NaOH → NaCl + H₂O. But if you cancel the ions that appear unchanged on both sides (the “spectator” ions Na⁺ and Cl⁻), what remains is the ionic equation:

H⁺ + OH⁻ → H₂O

This is the heart of every strong acid–strong alkali neutralisation: a hydrogen ion joins a hydroxide ion to make water. That is why the ionic equation is always the same, whatever acid and alkali you start with.

Titration: the technique

Titration is the volumetric method used to find the exact volume of one solution that just neutralises a known volume of another. The standard set-up is:

• A pipette delivers an accurate, fixed volume (e.g. 25.0 cm³) of one solution, often the alkali, into a conical flask.
• A few drops of acid–base indicator (such as phenolphthalein or methyl orange) are added.
• The other solution, often the acid, is run in from a burette, swirling constantly.
• The end point is the moment the indicator just changes colour permanently, showing neutralisation is complete. The burette reading gives the volume used (the titre).

The reading you record is the burette volume at the end point; a repeat titration to get consistent titres is good practice.

The titration calculation

At the end point the acid and alkali have reacted exactly according to the balanced equation. The relationship is:

(Mₐ × Vₐ) / (M_b × V_b) = a / b

where Mₐ, Vₐ are the molarity and volume of the acid, M_b, V_b are those of the base, and a : b is the mole ratio of acid to base in the balanced equation. Rearranged, this lets you find whichever quantity is unknown, usually a concentration.

Worked example

Question. In a titration, 25.0 cm³ of sodium hydroxide solution is exactly neutralised by 20.0 cm³ of 0.10 mol dm⁻³ hydrochloric acid. Calculate the molarity of the sodium hydroxide. The equation is HCl + NaOH → NaCl + H₂O.

Step 1, Moles of acid. Convert the acid volume: 20.0 cm³ = 0.0200 dm³. Moles of HCl = molarity × volume = 0.10 × 0.0200 = 0.00200 mol.

Step 2, Mole ratio. From HCl + NaOH → NaCl + H₂O, the ratio HCl : NaOH is 1 : 1, so moles of NaOH = 0.00200 mol.

Step 3, Molarity of the alkali. Convert its volume: 25.0 cm³ = 0.0250 dm³. Molarity = moles ÷ volume = 0.00200 ÷ 0.0250 = 0.08 mol dm⁻³.

Answer. The molarity of the sodium hydroxide is 0.08 mol dm⁻³.

Practice question

Question. 25.0 cm³ of 0.10 mol dm⁻³ sodium hydroxide is neutralised by 12.5 cm³ of sulfuric acid, H₂SO₄. Calculate the molarity of the sulfuric acid. The equation is 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

Answer. Moles of NaOH = 0.10 × (25.0 ÷ 1000) = 0.00250 mol. From the equation, NaOH : H₂SO₄ = 2 : 1, so moles of H₂SO₄ = 0.00250 ÷ 2 = 0.00125 mol. Molarity of H₂SO₄ = moles ÷ volume = 0.00125 ÷ (12.5 ÷ 1000) = 0.00125 ÷ 0.0125 = 0.1 mol dm⁻³.

Exam tip

The mark that students lose most often is the mole ratio: read it straight from the balanced equation, not from the volumes. A 1 : 1 acid such as HCl with NaOH is not the same as a dibasic acid such as H₂SO₄, which needs 2 mol of NaOH per mole of acid. Always convert cm³ to dm³ before dividing by a volume, and quote the final concentration in mol dm⁻³. For the procedure, remember that the pipette measures the fixed volume into the flask and the burette delivers the variable volume; naming the wrong apparatus costs easy marks.

Working titration questions cleanly

Set every titration calculation out in the same three moves and it becomes routine. First, find the moles of the substance you know completely, the one whose molarity and volume are both given. Second, use the balanced equation’s mole ratio to convert to moles of the unknown substance. Third, divide those moles by the unknown’s volume in dm³ to get its molarity (or multiply by molar mass if a mass or concentration in g dm⁻³ is wanted). Writing “moles known → ratio → moles unknown → answer” as a visible chain means an examiner can follow your reasoning and award method marks even if one figure is mis-keyed.

Watch also for the indicator choice and colour change, which structured questions like to ask alongside the numbers. Phenolphthalein is colourless in acid and pink in alkali; methyl orange is red in acid and yellow in alkali. Being able to state the colour at the end point, and to explain that the end point marks complete neutralisation, turns a pure calculation into a full-mark structured answer.

Where this fits

This is content standard 6.5 of the Acids, Bases and Salts chapter. It draws on molarity (6.3) and standard solutions (6.4) and leads into salt preparation, where neutralisation is one route to a soluble salt. Drill the calculations in the worked examples and test the method with the practice questions. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, our teachers make the mole-ratio step explicit every time, because it is where dibasic-acid titrations most often go wrong.

Quick recap

• Neutralisation: acid + base → salt + water; ionic equation H⁺ + OH⁻ → H₂O.
• Titration: pipette a fixed volume + indicator; run in the other solution from a burette to the end point.
• Calculation: moles known → mole ratio from the balanced equation → moles unknown → molarity.
• Always convert cm³ to dm³; read the ratio from the equation, not the volumes.