A titration calculation often goes wrong before the final division. A volume stays in cm³ when the concentration is in mol/dm³, or a 1:2 reaction is treated as though the reactants combine 1:1.
A clear route is: write the balanced equation, calculate the known amount in moles, use the reaction ratio, then find the unknown concentration. Keep the chemical reasoning visible rather than relying on a memorised concentration-times-volume shortcut.
This is an original calculation exercise, not instructions to conduct a titration outside a supervised laboratory. Check your full course requirements on the Cambridge Chemistry 9701 page.
Begin with the equation
An acid solution contains sulfuric acid. A 25.0 cm³ sample is neutralised completely by 20.0 cm³ of 0.100 mol/dm³ sodium hydroxide solution. Find the sulfuric acid concentration, assuming the stated complete reaction:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
One mole of sulfuric acid reacts with two moles of sodium hydroxide. Writing this ratio before the arithmetic helps prevent an unnoticed factor-of-two error.
Convert the known volume
Use n = cV, where n is amount in moles, c is concentration in mol/dm³ and V is volume in dm³. Since 1000 cm³ = 1 dm³, 20.0 cm³ = 0.0200 dm³.
The amount of sodium hydroxide is therefore 0.100 × 0.0200 = 0.00200 mol. If you had multiplied by 20.0 without converting, your answer would be a thousand times too large.
You can use other consistent unit systems, but the units must match throughout. Writing the conversion explicitly is a useful safeguard when learning the method.
Use the ratio in the correct direction
There are two moles of sodium hydroxide for every one mole of sulfuric acid. The acid amount in the sample is therefore 0.00200 ÷ 2 = 0.00100 mol.
Ask yourself whether the acid amount should be larger or smaller than the sodium hydroxide amount. The equation says fewer acid particles are required, so halving is sensible. Multiplying by two would contradict the balanced equation.
Divide by the volume of the unknown solution
The acid sample volume is 25.0 cm³, or 0.0250 dm³. Its concentration is c = n/V = 0.00100/0.0250 = 0.0400 mol/dm³.
Use the acid sample volume at this step, not the sodium hydroxide volume. Each concentration describes the amount per volume of its own solution.
Check backwards: 0.0400 × 0.0250 gives 0.00100 mol acid. Double that to find 0.00200 mol sodium hydroxide. At 0.100 mol/dm³, the required sodium hydroxide volume is 0.0200 dm³, matching the question.
When repeated titres are supplied
Select and average readings according to the question's instructions and the method you have been taught. Do not automatically average a rough trial with all other readings. Keep unrounded values through intermediate calculations, then present the final precision appropriately for the supplied data.
If the original solution was diluted to a known volume before an aliquot was taken, use that dilution factor to recover the original concentration. For example, making 10.0 cm³ up to 100.0 cm³ gives a tenfold dilution. Adding distilled water to the titration flask after measuring the aliquot is different: it changes the volume but not the moles of acid already transferred, so it does not change the amount of titrant needed for complete reaction.
Try a 1:1 reaction
A 20.0 cm³ sample of hydrochloric acid is neutralised by 16.0 cm³ of 0.150 mol/dm³ sodium hydroxide. Use HCl + NaOH → NaCl + H₂O to find the acid concentration.
The sodium hydroxide amount is 0.150 × 0.0160 = 0.00240 mol. The ratio is 1:1, so the acid amount is also 0.00240 mol. Divide by 0.0200 dm³ to obtain 0.120 mol/dm³.
Continue with StudyForA's Chemistry 9701 lessons. Before checking your next numerical answer, check its units and the direction of its mole ratio.