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Titration finds an unknown concentration by reacting exact volumes of two solutions when one concentration is known. Free notes and 10 questions answered.
Finding Unknown Concentrations Using Reaction Volumes is part of Using concentrations of solutions in mol/dm 3 (chemistry only) (HT only) in GCSE AQA Higher Chemistry.
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Titration finds an unknown concentration by reacting exact volumes of two solutions when one concentration is known. Free notes and 10 questions answered.
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Use this lesson to stay inside Using concentrations of solutions in mol/dm 3 (chemistry only) (HT only) while connecting the detail back to the wider Topic 3: Quantitative chemistry topic.
Read the core explanations from Finding Unknown Concentrations Using Reaction Volumes before testing yourself from memory.
You have two solutions. One has a known concentration, the other is a mystery. When you mix exact volumes of both, they react completely with nothing left over. Here's the question: can you work out the unknown concentration?
This is exactly what chemists do in laboratories every day. The technique is called titration, and it's one of the most powerful tools for finding unknown concentrations.
When two solutions react completely, the moles of each reactant are in the exact ratio shown by the balanced equation. This gives us our route to the unknown concentration.
For example, in the reaction $HCl + NaOH \rightarrow NaCl + H_2O$, the molar ratio is 1:1. One mole of acid reacts with exactly one mole of alkali.
Rearranging this formula gives us $n = c \times V$. This is the key to solving concentration problems.
Every concentration calculation follows the same logical sequence. Master this method and you can tackle any titration problem.
Worked example: Finding Unknown Concentration
25.0 cm³ of 0.100 mol/dm³ HCl exactly neutralises 20.0 cm³ of NaOH solution. Calculate the concentration of the NaOH. [4 marks]
Answer: 0.125 mol/dm³
Notice how we used the molar ratio from the balanced equation. This is the bridge between the known and unknown solutions.
Not all reactions have a 1:1 ratio. Diprotic acids like sulfuric acid can neutralise two moles of alkali for every mole of acid.
Worked example: Handling Non-1:1 Ratios
15.0 cm³ of 0.200 mol/dm³ $H_2SO_4$ neutralises 30.0 cm³ of KOH solution. Find the concentration of KOH. [4 marks]
Answer: 0.200 mol/dm³
The key insight here is that the molar ratio multiplies the moles, not the concentration directly. Always work through moles as the intermediate step.
Students lose marks on concentration calculations for predictable reasons. Here are the main traps and how to avoid them.
Trap 1: Volume units. Many students forget to convert cm³ to dm³. The concentration formula only works with dm³, so always divide by 1000 first.
Trap 2: Molar ratios. Students often assume all reactions are 1:1. Always check the balanced equation and apply the correct ratio to the moles, not the concentrations.
Trap 3: Significant figures. Your final answer should match the least precise measurement in the question. Usually 3 significant figures for GCSE.
Concentration calculations are worth 4-6 marks in exams. The mark scheme rewards method as much as the final answer, so show every step clearly.
Exam tip: convert all volumes to dm3 before using c = n / V, then check the mole ratio from the balanced equation before finding the unknown concentration., always identify the key chemical idea first, then link it to evidence, data, or an equation. A high-scoring GCSE answer names the particles involved, states what changes, and justifies the conclusion using correct scientific language. Where calculations are used, show each step with units and keep significant figures sensible. Finally, check that terminology matches AQA wording so your answer is unambiguous and mark-scheme friendly.
When using reaction volumes to find an unknown concentration, start by writing the balanced equation and the mole ratio. Then convert any measured volume from cm3 to dm3 before using c = n / V. This prevents the common error of making the final concentration 1000 times too large.
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Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
Titration is a technique for finding an unknown concentration by reacting exact volumes of two solutions completely, with nothing left over, when the concentration of one solution is known. The moles that react are in the ratio shown by the balanced equation.
A diprotic acid, such as sulfuric acid, can neutralise two moles of alkali for every mole of acid. This is why H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O has a 1:2 ratio, so the moles of KOH are 2 × the moles of acid.
The equation for HCl reacting with NaOH is HCl + NaOH → NaCl + H₂O. The molar ratio is 1:1, so one mole of acid reacts with exactly one mole of alkali and equal moles of each react.
An unknown concentration is found by converting volumes to dm³, calculating moles of the known solution with n = c × V, using the mole ratio from the balanced equation, then dividing by the unknown volume. If 25.0 cm³ of 0.100 mol/dm³ HCl neutralises 20.0 cm³ of NaOH, c = 0.00250 ÷ 0.0200 = 0.125 mol/dm³.
Sulfuric acid is diprotic, so one mole of acid neutralises two moles of alkali and the ratio is 1:2. In H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O, 0.00300 mol of acid reacts with 2 × 0.00300 = 0.00600 mol of KOH; the ratio multiplies the moles, not the concentration.
The common traps in titration calculations are not converting cm³ to dm³, assuming every reaction is 1:1 instead of checking the balanced equation, and using the wrong number of significant figures. GCSE answers usually use 3 significant figures.
The mole ratio in HCl + NaOH → NaCl + H₂O is 1:1, so one mole of acid reacts with exactly one mole of alkali. When the two solutions react completely, the moles of NaOH equal the moles of HCl.
The balanced equation is H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O. The 1:2 ratio means one mole of sulfuric acid neutralises two moles of KOH, which is crucial in titration calculations.
A final titration answer should match the least precise measurement in the question, which is usually 3 significant figures. For example, 25.0 cm³ of 0.100 mol/dm³ HCl neutralising 20.0 cm³ of NaOH gives a concentration of 0.125 mol/dm³.
The mole ratio from a balanced equation is used with the moles, not the concentrations directly. In a titration, moles of the known solution are found first with n = c × V, multiplied by the ratio, then divided by the unknown volume to give its concentration.