Find the moles of the known solution with n = MV (volume in dm3), apply the acid-to-base mole ratio from the equation, then find the unknown concentration or volume.
This page walks through titration calculation for SPM Chemistry, step by step: the formula you need, the units to watch, and a worked example.
When you use this
Titration is the core Form 4 experiment for finding an unknown concentration. In SPM Chemistry you are given a burette and pipette reading and asked to work back to a concentration or a volume, using the balanced neutralisation equation.
The formula and units
For any solution, n = M x V, where M is the molar concentration and V is the volume in dm3. The link between acid and alkali is the mole ratio from the equation:
n(acid) / n(base) = coefficient of acid / coefficient of base
Units. Concentration M in mol dm−3, volume V in dm3 (so divide a cm3 reading by 1000), amount n in mol. A common short form for a fully worked titration is (Ma x Va) / (Mb x Vb) = a / b, where a and b are the acid and base coefficients.
Worked example 1 (easy)
NaOH + HCl → NaCl + H2O. 25.0 cm3 of 0.10 mol dm−3 sodium hydroxide is neutralised by 20.0 cm3 of hydrochloric acid. Find the concentration of the acid.
- n(NaOH) = 0.10 x (25.0 / 1000) = 0.0025 mol.
- Ratio NaOH : HCl = 1 : 1, so n(HCl) = 0.0025 mol.
- M(HCl) = n / V = 0.0025 / (20.0 / 1000) = 0.125 mol dm−3.
Worked example 2 (medium)
H2SO4 + 2NaOH → Na2SO4 + 2H2O. 25.0 cm3 of 0.20 mol dm−3 sodium hydroxide is neutralised by 15.0 cm3 of sulfuric acid. Find the concentration of the acid.
- n(NaOH) = 0.20 x (25.0 / 1000) = 0.0050 mol.
- Ratio NaOH : H2SO4 = 2 : 1, so n(H2SO4) = 0.0050 / 2 = 0.0025 mol.
- M(H2SO4) = 0.0025 / (15.0 / 1000) = 0.167 mol dm−3.
Worked example 3 (SPM level)
Find the volume this time. What volume of 0.25 mol dm−3 sodium hydroxide neutralises 25.0 cm3 of 0.10 mol dm−3 sulfuric acid? (H2SO4 + 2NaOH → Na2SO4 + 2H2O)
- n(H2SO4) = 0.10 x (25.0 / 1000) = 0.0025 mol.
- Ratio H2SO4 : NaOH = 1 : 2, so n(NaOH) = 0.0025 x 2 = 0.0050 mol.
- V(NaOH) = n / M = 0.0050 / 0.25 = 0.020 dm3 = 20.0 cm3.
The end point and the average titre
In the experiment an indicator marks the end point, where acid and alkali have exactly reacted. You repeat the titration and average only the concordant titres (readings within 0.10 cm3 of each other), then use that average volume in the calculation. Understanding this is why the numbers in the question look the way they do, and it is often examined as a short explanation alongside the arithmetic.
Reading a titration table
Many questions give a table of rough and accurate titrations. Ignore the rough trial, then pick the readings that agree to within 0.10 cm3, add them and divide to get the average titre. That single averaged volume is the number you put into n = MV. Getting the average right before any calculation begins is often worth as much as the arithmetic that follows, so treat choosing the concordant readings as the real first step of the sum.
Common traps
- Volume left in cm3. Divide every cm3 reading by 1000 before using n = MV, or work in consistent units throughout.
- Ignoring the ratio for a diprotic acid. H2SO4 reacts with twice as much NaOH; the 2 : 1 ratio is where marks are lost.
- Using a rough titre. Only the average of concordant titres belongs in the calculation.
- Confusing M and n. Concentration is per dm3; moles is the total in the volume you used.
In our online one-to-one lessons we practise reading the equation for the ratio first, then slotting the burette figures in, so a titration question becomes routine rather than a puzzle each time it appears.
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