Concentration Calculations Revision Notes | A Level AQA Chemistry

1. Why Concentration Calculations Matter

Picture a hospital pharmacist preparing an intravenous drip. Too concentrated and the solution is toxic; too dilute and it has no effect. Getting the concentration exactly right is not optional — and the maths behind it is exactly what this lesson covers.

Concentration calculations sit at the heart of quantitative chemistry. They appear in titration analysis, preparation of standard solutions, and industrial process control. At A Level, you are expected to move fluently between concentration, volume, and amount of substance — in any direction the question demands.

2. The Core Relationship: Concentration, Amount, and Volume

Think back to the idea of moles from earlier in the course. A mole is a counting unit — it tells you how many particles you have. Concentration tells you how crowded those particles are within a given volume of solution. Always write Mr(NaCl) = 23.0 + 35.5 = 58.5 g mol^-1 before substituting into any concentration calculation involving sodium chloride.

The formula $c = n / V$ is the engine of this topic. Rearranged, it gives you $n = c \times V$ (to find moles) and $V = n / c$ (to find volume). You need all three rearrangements to be automatic — exam questions will ask for any one of the three quantities.

One critical habit: volume must always be in dm³ before you substitute into the formula. Most laboratory measurements are given in cm³, so you will almost always need to convert. Divide cm³ by 1000 to get dm³.

3. Calculating Concentration — Worked Examples

The formula is straightforward, but exam questions are designed to catch you on unit conversions and rearrangements. The worked examples below build from the simplest case to the type of multi-step problem you will see on Paper 1 or Paper 2.

Worked example: Finding Concentration from Moles and Volume

0.250 mol of sodium hydroxide, NaOH, is dissolved in 500 cm³ of solution. Calculate the concentration of the solution in mol dm⁻³.

1. $V = \frac{500}{1000} = 0.500 \text{ dm}^3$ — The formula requires volume in dm³. Dividing 500 cm³ by 1000 gives 0.500 dm³. This step is where most marks are dropped — never skip it.
2. $c = \frac{n}{V}$ — State the formula explicitly. This earns a method mark [M1] in the exam even if arithmetic goes wrong later.
3. $c = \frac{0.250}{0.500}$ — Substitute n = 0.250 mol and V = 0.500 dm³. Both values now carry consistent units.
4. $c = 0.500 \text{ mol dm}^{-3}$ — 0.250 ÷ 0.500 = 0.500. Always include mol dm⁻³ as the unit — a bare number scores no accuracy mark [A1].

Answer: $c = 0.500 \text{ mol dm}^{-3}$

Worked example: Finding Amount of Substance from Concentration and Volume

Calculate the amount of substance, in mol, of hydrochloric acid present in 250 cm³ of a 0.100 mol dm⁻³ solution.

1. $V = \frac{250}{1000} = 0.250 \text{ dm}^3$ — 250 cm³ divided by 1000 gives 0.250 dm³. This conversion is non-negotiable before substitution.
2. $n = c \times V$ — Rearranging c = n/V gives n = c × V. State this rearrangement explicitly for the method mark [M1].
3. $n = 0.100 \times 0.250$ — c = 0.100 mol dm⁻³ and V = 0.250 dm³.
4. $n = 0.100 \times 0.250 = 0.0250 \text{ mol}$ — 0.100 × 0.250 = 0.0250. The answer is 0.0250 mol, which is 2.50 × 10⁻² mol. Either form is acceptable.

Answer: $n = 0.0250 \text{ mol}$

4. Finding Volume from Concentration and Moles

The third rearrangement — finding volume — appears less often but is equally important. It comes up when you need to know what volume of a stock solution to measure out to obtain a specific number of moles.

Worked example: Finding Volume from Concentration and Amount

What volume of a 0.500 mol dm⁻³ sulfuric acid solution contains 0.0750 mol of H₂SO₄? Give your answer in cm³.

1. $V = \frac{n}{c}$ — Starting from c = n/V, multiply both sides by V and divide both sides by c to isolate V.
2. $V = \frac{0.0750}{0.500}$ — n = 0.0750 mol and c = 0.500 mol dm⁻³.
3. $V = 0.0750 \div 0.500 = 0.150 \text{ dm}^3$ — 0.0750 ÷ 0.500 = 0.150. The answer is in dm³ because the formula uses dm³.
4. $V = 0.150 \times 1000 = 150 \text{ cm}^3$ — The question asks for cm³, so multiply by 1000. Always check what unit the question requests for the final answer.

Answer: $V = 150 \text{ cm}^3$

Notice the final step: the question asked for cm³, so the dm³ answer had to be converted back. This is a common source of lost marks — always re-read the question to check the required unit for the answer.

5. Multi-Step Concentration Problems

A Level exam questions rarely stop at one calculation. They chain concentration with molar mass, stoichiometry, or titration data. The strategy is always the same: identify what you know, identify what you need, and work through one step at a time.

The example below links mass, molar mass, and concentration — a combination that appears regularly in AQA Paper 1 and Paper 2 quantitative questions.

Worked example: Multi-Step: Mass to Concentration

5.85 g of sodium chloride, NaCl, is dissolved in water to make 250 cm³ of solution. Calculate the concentration of the solution in mol dm⁻³. (Relative atomic masses: Na = 23.0, Cl = 35.5)

1. $M_r(\text{NaCl}) = 23.0 + 35.5 = 58.5 \text{ g mol}^{-1}$ — Add the relative atomic masses of sodium and chlorine. This gives the mass of one mole of NaCl.
2. $n = \frac{\text{mass}}{M_r} = \frac{5.85}{58.5} = 0.100 \text{ mol}$ — 5.85 ÷ 58.5 = 0.100. This is the amount of substance present in the sample.
3. $V = \frac{250}{1000} = 0.250 \text{ dm}^3$ — Always convert before using the concentration formula.
4. $c = \frac{n}{V} = \frac{0.100}{0.250}$ — Substitute n = 0.100 mol and V = 0.250 dm³ into c = n/V.
5. $c = 0.400 \text{ mol dm}^{-3}$ — 0.100 ÷ 0.250 = 0.400. Include the unit mol dm⁻³ for the accuracy mark [A1].

Answer: $c = 0.400 \text{ mol dm}^{-3}$

6. Dilution and Concentration Changes

When a solution is diluted, the number of moles of solute stays the same — only the volume increases. This means the concentration decreases. You can use $n = c \times V$ to find the moles before dilution, and then recalculate concentration using the new volume.

Worked example: Dilution Calculation

25.0 cm³ of a 2.00 mol dm⁻³ hydrochloric acid solution is diluted to a total volume of 500 cm³. Calculate the concentration of the diluted solution.

1. $V_1 = \frac{25.0}{1000} = 0.0250 \text{ dm}^3$ — Convert 25.0 cm³ to dm³ before calculating moles.
2. $n = c_1 \times V_1 = 2.00 \times 0.0250 = 0.0500 \text{ mol}$ — 2.00 × 0.0250 = 0.0500. The number of moles does not change on dilution.
3. $V_2 = \frac{500}{1000} = 0.500 \text{ dm}^3$ — The total volume after dilution is 500 cm³ = 0.500 dm³.
4. $c_2 = \frac{n}{V_2} = \frac{0.0500}{0.500} = 0.100 \text{ mol dm}^{-3}$ — 0.0500 ÷ 0.500 = 0.100. The concentration has dropped by a factor of 20, which makes sense: the volume increased 20-fold (25 cm³ to 500 cm³).

Answer: $c_2 = 0.100 \text{ mol dm}^{-3}$

A useful sanity check for dilution: the ratio of concentrations should equal the inverse of the ratio of volumes. Here, the volume increased by a factor of 500/25 = 20, so the concentration should decrease by a factor of 20: 2.00 / 20 = 0.100 mol dm⁻³. The answer checks out.

7. Exam Technique and Common Mistakes

Concentration calculations are reliable mark-scorers — if you follow a consistent method. The marks lost in this topic are almost always procedural, not conceptual. The tips below address the patterns that appear most frequently in AQA mark schemes.

Concentration calculations reward methodical working. Every mark in this topic is available to you if you convert units first, state your formula, substitute carefully, and include units throughout. The maths itself is straightforward — the discipline of the method is what separates full marks from partial credit.

Concentration questions become more reliable when you separate the unit conversion from the chemistry. First convert the volume into dm^3. Then calculate moles if necessary from mass and Mr. Only after that should you substitute into c = n / V. Mixing these stages together is what usually creates factors-of-1000 errors.