The big idea: A titration is a precise way to find an unknown concentration. You react a measured volume of one solution with a solution of known concentration (a standard solution) until they exactly match — the end point, shown by an indicator colour change.
The maths is just the mole story again: turn each titre into an amount in moles with n = CV, use the balanced equation's mole ratio to cross to the other reagent, then divide by its volume to get the unknown concentration.
Volumetric analysis — the apparatus: Two pieces of glassware do the precise measuring:
- A pipette delivers a fixed, exact volume of the solution being analysed (e.g. 25.0 cm³). - A burette delivers the variable volume of titrant — the titre — read to ±0.05 cm³.
Repeat until two or three titres agree (are concordant); average only those.
| Apparatus | What it measures | Why it is used |
|---|---|---|
| Pipette | a fixed volume (e.g. 25.0 cm³) | delivers one exact, repeatable sample of the analyte |
| Burette | the variable titre added | reads the volume of titrant to ±0.05 cm³ |
| Volumetric flask | an exact total volume | makes up a standard solution of known concentration |
| Indicator | the end point (colour change) | shows when the reacting amounts exactly match |
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The whole calculation hangs on one given equation: n = CV. The only catch is that the volume must be in dm³, so divide a cm³ titre by 1000 first.
- amount of substance in solution (mol)
- concentration (mol dm⁻³)
- volume of solution (dm³) — convert from cm³ by ÷ 1000
The three-step routine: (1) Use n = CV on the reagent whose concentration and volume you know → its amount in mol.
(2) Use the balanced equation's mole ratio to find the amount of the other reagent.
(3) Divide that amount by its volume (in dm³) to get the unknown concentration: C = n/V.
Worked example — finding an acid concentration
25.0 cm³ of hydrochloric acid, HCl, is exactly neutralised by 20.0 cm³ of 0.150 mol dm⁻³ sodium hydroxide, NaOH. Calculate the concentration of the acid.
HCl + NaOH → NaCl + H2O
Solution
- Formula first — start with the reagent you fully know (the NaOH). Convert its volume to dm³:
- Mole ratio is 1 : 1 from the equation, so the amount of HCl is the same:
- C = n/V for the acid — use its own volume in dm³:
- Work it out — keep the unit:
Final answer
C(HCl) = 0.120 mol dm⁻³.
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If the balanced equation has a ratio other than 1 : 1, the only thing that changes is step 2 — scale by the ratio. Watch for diprotic acids (like H2SO4) that need two moles of base per mole of acid.
Worked example — a 1 : 2 ratio
A 25.0 cm³ sample of sulfuric acid, H2SO4, is exactly neutralised by 30.0 cm³ of 0.200 mol dm⁻³ NaOH. Calculate the concentration of the acid.
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Solution
- Formula first — amount of NaOH (volume in dm³):
- Mole ratio is 1 acid : 2 base, so divide by 2:
- C = n/V for the acid:
Final answer
C(H₂SO₄) = 0.120 mol dm⁻³.
What is a back titration?: Sometimes a reaction is too slow, or the sample is an insoluble solid (like a carbonate), so a direct titration is impractical.
Instead you add a known excess of one reagent, let it react, then titrate the leftover excess. The amount that reacted = total added − leftover.
Worked example — a back titration
An impure sample of calcium carbonate, CaCO3, is added to 50.0 cm³ of 1.00 mol dm⁻³ HCl (an excess). The unreacted acid is then titrated and found to be 0.0100 mol. Calculate the amount of HCl that reacted with the carbonate.
Solution
- Formula first — total HCl added, with volume in dm³:
- The titration found the leftover (unreacted) acid:
- Amount reacted = added − leftover:
Final answer
0.0400 mol of HCl reacted with the carbonate.
How this is tested: Titration is the heart of volumetric analysis and runs right through Paper 1B (the practical/data paper) as well as Paper 2.
- Paper 1A: a one-step 'volume to reach equivalence' or 'concentration from the titre' calculation. - Paper 1B / Paper 2: a full multi-step problem — name the apparatus, then calculate an unknown concentration, a reacting mass or volume, or a molar mass of an unknown acid.
The trap that loses marks: forgetting the mole ratio, or leaving a titre in cm³ instead of dm³.
Score it cleanly: (1) Convert every volume to dm³ (÷ 1000) before using n = CV. (2) Apply the balanced equation's mole ratio — never skip it. (3) End with C = n/V (or m = nM), carrying the unit and matching the significant figures.
IB-style question — neutralising a base (a)
A 10.0 g sample of pure potassium hydroxide, KOH, is dissolved and exactly neutralised by sulfuric acid. (a) Calculate the volume of 2.00 mol dm⁻³ H2SO4 required. (M(KOH) = 56.11 g mol⁻¹.)
2KOH + H2SO4 → K2SO4 + 2H2O [3]
How to score the marks
- Mark 1 — amount of base from mass (n = m/M):
- Mark 2 — mole ratio is 2 KOH : 1 H2SO4, so halve it:
- Mark 3 — volume from V = n/C, then convert to cm³:
Final answer
V(H₂SO₄) = 0.0446 dm³ = 44.6 cm³.
IB-style question — molar mass of an unknown acid (b)
(b) A 0.295 g sample of a monoprotic acid, HX, is exactly neutralised by 25.0 cm³ of 0.100 mol dm⁻³ NaOH. Determine the molar mass of HX.
HX + NaOH → NaX + H2O [3]
How to score the marks
- Mark 1 — amount of NaOH (n = CV, volume in dm³):
- Mark 2 — mole ratio is 1 : 1, so the amount of acid is the same:
- Mark 3 — molar mass from M = m/n:
Final answer
M(HX) = 118 g mol⁻¹.