Solutions and Concentrations

Concentration measurement underpins every piece of volumetric analysis a chemist performs, from a school titration to industrial process control and clinical assays. This lesson develops the three core concentration equations — c = n/V (the molar concentration definition), c₁V₁ = c₂V₂ (the dilution equation, derived from conservation of moles), and the molar-mass route between mol dm⁻³ and g dm⁻³ — and embeds them in the practical context of preparing standard solutions, evaluating uncertainty, and combining concentrations with stoichiometric mole ratios. The lesson is the conceptual prerequisite for lesson 8 (titration calculations and AQA Required Practical 1) and feeds downstream into Kc and Kp (equilibrium), the rate equation (kinetics), pH and buffer calculations (acids and bases), and the standard electrochemical-cell convention (electrochemistry).

Spec mapping (AQA 7405): The primary anchor for this lesson is §3.1.2.4 (the use of concentrations of solutions in mol dm⁻³ and g dm⁻³). The closely related §3.1.2.5 (balanced equations and associated calculations) was developed in lesson 6 and is reapplied here whenever a volumetric calculation involves a mole ratio. Titration calculations — the primary application of §3.1.2.4 in the laboratory and on the exam paper — together with AQA Required Practical 1 (RP1, preparation of a standard solution from a solid and titration of a strong acid against a strong base) form the deepening content of lesson 8. The empirical-formula and balanced-equation skills from §3.1.2.3 and §3.1.2.5 (lessons 5 and 6) are assumed as prerequisites. Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives: Definitions of solute, solvent, solution, concentration, and standard solution are AO1 recall items. Numerical calculations using n = cV, the unit conversion between mol dm⁻³ and g dm⁻³, and the dilution equation c₁V₁ = c₂V₂ are AO2 staples — they appear on every Paper 1 and Paper 3 paper. Multi-step problems combining serial dilution with a stoichiometric mole ratio, or evaluating the percentage uncertainty in a prepared concentration from balance and volumetric-flask tolerances, test AO3 synthesis and reasoning, and are typical Grade A/A* discriminators.

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Key Definitions

• Solute: the substance being dissolved (e.g. NaOH(s), H₂SO₄(l), CO₂(g) — solutes can be solid, liquid, or gas).
• Solvent: the substance doing the dissolving. In A-Level Chemistry the solvent is usually water (giving an aqueous solution, state symbol (aq)), but ethanol, propanone, and other organic solvents also occur — particularly in organic-chemistry workups.
• Solution: a homogeneous (single-phase) mixture of solute and solvent. "Homogeneous" means that, at any sampling point in the solution, the composition is the same.
• Concentration: the amount of solute dissolved per unit volume of solution. At A-Level this is quoted in either mol dm⁻³ (preferred by chemists, sometimes written as M and read "molar") or g dm⁻³ (preferred for regulatory or food-and-drug contexts).
• Standard solution: a solution of accurately known concentration, prepared either by direct weighing of a primary standard or by titration against a primary standard. Standard solutions are the calibrating reference for every quantitative volumetric measurement.

It is worth flagging two adjacent measures used outside the A-Level spec but often encountered in further reading. Molality (mol kg⁻¹) is the moles of solute per kilogram of solvent (not solution); because mass — unlike volume — does not change with temperature, molality is preferred in high-precision physical-chemistry work. Mass fraction (or weight percent) reports the mass of solute as a fraction of the total mass of solution; pharmacists and metallurgists routinely quote concentrations this way. Neither is examined at A-Level, but a Grade A* student should know they exist and why they are sometimes more useful than molarity.

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Concentration in mol dm⁻³

The most useful concentration measure in chemistry is moles of solute per cubic decimetre of solution, with the formal definition:

c = n / V

where c has units mol dm⁻³, n has units mol, and V has units dm³. The equation rearranges to n = c × V (the mole equation most often used at A-Level, alongside n = m/M and pV = nRT) and V = n / c.

Unit consistency — the single most common source of marks lost

A dm³ is a cubic decimetre, equal to a litre and to 1000 cm³. To convert between cm³ and dm³, divide by 1000 (cm³ → dm³) or multiply by 1000 (dm³ → cm³). To convert between dm³ and m³, divide by 1000 (dm³ → m³). The volume in the concentration equation must be in dm³ — not cm³, not m³. Writing the unit-conversion step explicitly is the simplest way to secure that mark on the paper.

Key Point: Volume must be in dm³ for c = n/V to give mol dm⁻³. To convert cm³ to dm³, divide by 1000. Write this conversion as a separate line — examiners frequently award a mark just for the conversion.

Worked Example 1: Finding concentration in mol dm⁻³

0.400 g of sodium hydroxide (NaOH) is dissolved in water to make 250 cm³ of solution. Calculate the concentration in mol dm⁻³.

M(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹

n(NaOH) = m / M = 0.400 / 40.0 = 0.0100 mol

V = 250 / 1000 = 0.250 dm³

c = n / V = 0.0100 / 0.250 = 0.0400 mol dm⁻³

Note three things: the three-step structure (moles, volume conversion, concentration), the three-significant-figure precision in the final answer (matching the input data), and the explicit cm³ → dm³ conversion. Each of these steps would attract a mark in a typical exam scheme.

Worked Example 2: Finding moles from concentration and volume

Calculate the number of moles of H₂SO₄ in 25.0 cm³ of 0.100 mol dm⁻³ sulfuric acid.

V = 25.0 / 1000 = 0.0250 dm³

n = c × V = 0.100 × 0.0250 = 2.50 × 10⁻³ mol (= 0.00250 mol)

This is the canonical pre-calculation for a titration: take an aliquot of known volume of an acid of known concentration, then react it with an alkali whose concentration is to be determined.

Worked Example 3: Finding the volume of a solution

What volume of 0.200 mol dm⁻³ HCl contains 0.0500 mol of HCl?

V = n / c = 0.0500 / 0.200 = 0.250 dm³ = 250 cm³

A note on activity — qualitative only

In dilute aqueous solution the concentration c (mol dm⁻³) is, to a very good approximation, the chemically active concentration — the quantity that appears in the equilibrium constant Kc and the rate equation. In concentrated solution this breaks down: ion-ion interactions (think Na⁺ and Cl⁻ in seawater) cause the effective concentration, called the activity, to differ from the formal concentration. The activity coefficient γ (a number between 0 and 1 for non-ideal ionic solutions) relates them: activity = γ × c / c°, where c° = 1 mol dm⁻³ is the standard concentration. A-Level treats γ = 1 throughout — equilibrium and rate constants are always written in terms of concentrations, not activities — but a Grade A* student should know the approximation exists. This idea will resurface in §3.1.6 (equilibrium constant Kc), §3.1.9 (kinetics), and §3.1.12 (acids, bases, and Kw) — all of which use mol dm⁻³ as the unit of choice.

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Concentration in g dm⁻³

Concentration can equivalently be expressed as grams of solute per cubic decimetre of solution, with units g dm⁻³ (or g/L — the two are numerically identical). The defining equation is:

concentration (g dm⁻³) = mass of solute (g) / volume of solution (dm³)

Converting between mol dm⁻³ and g dm⁻³

Multiply by the molar mass to go from mol dm⁻³ to g dm⁻³, and divide by the molar mass to go in reverse:

c (g dm⁻³) = c (mol dm⁻³) × M (g mol⁻¹)

c (mol dm⁻³) = c (g dm⁻³) / M (g mol⁻¹)

The molar-mass conversion is dimensionally consistent (mol dm⁻³ × g mol⁻¹ = g dm⁻³), and proceeds via exactly one multiplication or division.

Why two units? Audience and convention

Chemists prefer mol dm⁻³ because mole ratios in balanced equations let a concentration in mol dm⁻³ feed directly into a stoichiometric calculation. A solution at 0.100 mol dm⁻³ HCl contains exactly 0.100 mol of HCl per dm³ — no further arithmetic needed before the mole ratio comes into play.

Food-and-drug regulators, by contrast, usually quote g dm⁻³ (or, more often, g/L, mg/100 mL, or simply % w/v). The reason is practical: a regulator is interested in how much of a substance a person consumes, and mass is more intuitive than moles. Physiological saline — the 0.9% w/v sodium chloride solution used as an intravenous fluid — has a concentration of approximately 9.0 g dm⁻³ NaCl, equivalent to (9.0 / 58.5) ≈ 0.154 mol dm⁻³. The mass figure 9.0 g dm⁻³ is the one quoted on the label; the molar figure 0.154 mol dm⁻³ is what a clinical pharmacologist would compute when reasoning about osmotic balance or sodium-channel pharmacology. Both are correct; the choice of units depends on the audience.

Worked Example 4: Converting from mol dm⁻³ to g dm⁻³

A solution of sodium carbonate (Na₂CO₃) has a concentration of 0.150 mol dm⁻³. Express this in g dm⁻³.

M(Na₂CO₃) = 2(23.0) + 12.0 + 3(16.0) = 46.0 + 12.0 + 48.0 = 106.0 g mol⁻¹

c = 0.150 × 106.0 = 15.9 g dm⁻³

Worked Example 5: Converting from g dm⁻³ to mol dm⁻³

A glucose (C₆H₁₂O₆) solution has a concentration of 36.0 g dm⁻³. Calculate the concentration in mol dm⁻³.

M(C₆H₁₂O₆) = 6(12.0) + 12(1.0) + 6(16.0) = 72.0 + 12.0 + 96.0 = 180.0 g mol⁻¹

c = 36.0 / 180.0 = 0.200 mol dm⁻³

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Dilution

When a solution is diluted, the number of moles of solute stays constant — what changes is the volume of solvent (and therefore the concentration). Since n = cV, conservation of moles before and after dilution gives:

n(before) = n(after)

c₁V₁ = c₂V₂

where the subscripts 1 and 2 denote initial and final states. This is the dilution equation. Note two features:

• The equation is unit-agnostic for V: as long as V₁ and V₂ are in the same units (both cm³ or both dm³), the answer comes out correctly in the units of c. This is because the volume appears on both sides of the equation and the units cancel.
• The equation is unit-consistent: c and V on each side must combine to give moles in the same unit on both sides. Mixing mol dm⁻³ on one side with mol cm⁻³ on the other will give a wrong answer by a factor of 1000.

Worked Example 6: Single-step dilution

50.0 cm³ of 2.00 mol dm⁻³ HCl is diluted to a total volume of 500 cm³. Calculate the new concentration.

c₁V₁ = c₂V₂

2.00 × 50.0 = c₂ × 500

c₂ = (2.00 × 50.0) / 500 = 0.200 mol dm⁻³

The dilution factor here is 10× (50 cm³ diluted to 500 cm³ is a tenfold increase in volume), and the concentration falls from 2.00 to 0.200 — a tenfold reduction, exactly as expected.

Worked Example 7: Volume needed to prepare a diluted solution

What volume of 0.500 mol dm⁻³ NaOH is needed to prepare 250 cm³ of 0.100 mol dm⁻³ NaOH?

c₁V₁ = c₂V₂

0.500 × V₁ = 0.100 × 250

V₁ = (0.100 × 250) / 0.500 = 50.0 cm³

So pipette 50.0 cm³ of the 0.500 mol dm⁻³ stock into a 250 cm³ volumetric flask, then make up to the mark with distilled water.

Worked Example 8: Serial dilution

A 1.00 mol dm⁻³ stock of CuSO₄ is diluted by taking 10.0 cm³ and making up to 100 cm³. A 10.0 cm³ aliquot of that diluted solution is then taken and made up to 100 cm³ again. What is the final concentration?

First dilution: c₁V₁ = c₂V₂ → 1.00 × 10.0 = c₂ × 100 → c₂ = 0.100 mol dm⁻³.

Second dilution: c₂V₂ = c₃V₃ → 0.100 × 10.0 = c₃ × 100 → c₃ = 0.0100 mol dm⁻³.

The overall dilution factor is 10 × 10 = 100×, matching the 100-fold reduction in concentration. Serial dilution is the standard technique for preparing very dilute solutions accurately — pipetting a tiny 10 μL aliquot directly into a litre flask amplifies pipetting error catastrophically, whereas two careful 10× dilutions reduce the cumulative percentage uncertainty considerably.

Exam Tip: When using c₁V₁ = c₂V₂, you do not need to convert to dm³, provided both volumes are in the same units. The equation is unit-agnostic for V because volume appears on both sides.

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Standard Solutions

A standard solution is a solution of accurately known concentration. The full preparation method below is one of the techniques anchored by AQA Required Practical 1 and is the basis of every titration in the A-Level syllabus.

Primary and secondary standards

A primary standard is a solute that can be weighed accurately to provide a directly traceable known concentration. Required properties:

• High purity (≥ 99.95%) so the weighed mass corresponds exactly to the chemical formula.
• Stable in air — does not absorb moisture, CO₂, or oxygen, and does not decompose.
• High molar mass so that the percentage uncertainty in weighing is small (weighing 5.000 g of a high-M solid carries a smaller percentage error than weighing 0.050 g of a low-M solid on the same balance).
• Soluble in the chosen solvent at the concentration required.
• Non-hygroscopic (does not absorb water from the atmosphere — hygroscopic solids gain mass between being weighed and being dissolved, distorting concentration).

Typical primary standards used in A-Level Chemistry are anhydrous sodium carbonate (Na₂CO₃, M = 106.0), potassium hydrogen phthalate (KHP, KC₈H₅O₄, M = 204.2), and oxalic acid dihydrate ((COOH)₂·2H₂O, M = 126.1). Note that the last is hydrated — but it is a stable, well-defined hydrate, and so still functions as a primary standard provided you use the molar mass of the hydrate.

A secondary standard is a solute whose concentration is determined indirectly, typically by titration against a primary standard. Sodium hydroxide is the canonical example: NaOH absorbs CO₂ and water from air, so it cannot be weighed accurately to give a standard solution directly. Instead, a roughly-concentrated NaOH solution is prepared, then titrated against a primary standard such as KHP or oxalic acid to determine its exact concentration. The titrated value is then used as the "standard" in subsequent work.

Method for preparing a standard solution from a primary standard

1. Calculate the required mass. Use n = c × V and m = n × M to find the mass of solute needed for the chosen final volume (typically 250 cm³ for a school standard solution).
2. Weigh by difference. Place a weighing boat (or weighing bottle) on the balance, tare (zero) it, then add slightly more than the calculated mass. Record the mass. Tip the solid into a beaker (or use the weighing bottle and wash through). Re-weigh the empty boat — the difference is the actual mass transferred. This eliminates the error of solid clinging to the boat.
3. Dissolve in a small volume of distilled water (around 50–100 cm³, less than the final volume) in a beaker. Stir with a glass rod until fully dissolved. For sparingly soluble solids, gentle warming may help — cool back to room temperature before the next step.
4. Transfer quantitatively to a volumetric flask of the correct size (250 cm³, 500 cm³, or 1 dm³). "Quantitatively" means making sure all the solute reaches the flask — rinse the beaker and stirring rod with distilled water and add the rinsings to the flask. Repeat at least twice.
5. Make up to the mark. Add distilled water until the volume approaches the calibration mark on the neck of the flask. Use a dropping pipette for the final few cm³ so that the bottom of the meniscus sits exactly on the graduation mark, at eye level.
6. Stopper and invert the flask at least 10 times to mix thoroughly. Inadequate mixing is a common cause of inconsistent titration results — denser solution will settle in the lower part of the flask.

Worked Example 9: Mass of solute for a standard solution (anhydrous)

Calculate the mass of anhydrous sodium carbonate (Na₂CO₃) needed to prepare 250 cm³ of a 0.100 mol dm⁻³ solution.

n = c × V = 0.100 × 0.250 = 0.0250 mol

m = n × M = 0.0250 × 106.0 = 2.65 g

Worked Example 10: Mass of solute for a standard solution (hydrated)

Calculate the mass of hydrated sodium carbonate (Na₂CO₃·10H₂O) needed to prepare 500 cm³ of a 0.0500 mol dm⁻³ solution of anhydrous Na₂CO₃ equivalent.

n(Na₂CO₃) = c × V = 0.0500 × 0.500 = 0.0250 mol

Each mole of Na₂CO₃·10H₂O contains 1 mole of Na₂CO₃, so n(hydrate) = 0.0250 mol.

M(Na₂CO₃·10H₂O) = 106.0 + 10 × 18.0 = 106.0 + 180.0 = 286.0 g mol⁻¹

m = 0.0250 × 286.0 = 7.15 g

Common Misconception: When preparing a standard solution from a hydrated salt, weigh out the mass of the hydrated compound (using its full molar mass including water of crystallisation), but quote the concentration in terms of the anhydrous solute. The reason: each formula unit of Na₂CO₃·10H₂O delivers exactly one Na₂CO₃ unit to solution — the waters of crystallisation are released into the bulk solvent and are not "extra" solute.

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Volumetric Calculations with Stoichiometry

The full power of c = n/V emerges when it is combined with the mole-ratio reasoning from §3.1.2.5 (lesson 6).

Worked Example 11: 1:1 acid-base reaction

25.0 cm³ of 0.100 mol dm⁻³ NaOH reacts exactly with hydrochloric acid. What volume of 0.0500 mol dm⁻³ HCl is needed?

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

n(NaOH) = c × V = 0.100 × 25.0 / 1000 = 2.50 × 10⁻³ mol

Mole ratio NaOH : HCl = 1 : 1, so n(HCl) = 2.50 × 10⁻³ mol.

V(HCl) = n / c = 2.50 × 10⁻³ / 0.0500 = 0.0500 dm³ = 50.0 cm³

A useful sanity check: the HCl is half as concentrated as the NaOH, so twice the volume is needed to provide the same number of moles — exactly what the calculation gives.

Worked Example 12: Diprotic acid neutralisation

25.0 cm³ of 0.100 mol dm⁻³ Na₂CO₃ reacts with excess HCl. Calculate the volume of CO₂ produced at RTP.

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

n(Na₂CO₃) = c × V = 0.100 × 0.0250 = 2.50 × 10⁻³ mol

Mole ratio Na₂CO₃ : CO₂ = 1 : 1, so n(CO₂) = 2.50 × 10⁻³ mol.

V(CO₂) at RTP = n × 24.0 = 2.50 × 10⁻³ × 24.0 = 0.0600 dm³ (= 60.0 cm³)

Note that the 1 : 2 ratio of Na₂CO₃ : HCl in the equation would matter if the question asked for the volume of HCl required — but here CO₂ is the target and the mole ratio is 1 : 1 with the carbonate.

Worked Example 13: Diprotic acid titration

A 25.0 cm³ sample of 0.150 mol dm⁻³ H₂SO₄ is titrated with NaOH. What volume of 0.200 mol dm⁻³ NaOH is needed to reach the equivalence point?

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

n(H₂SO₄) = 0.150 × 25.0 / 1000 = 3.75 × 10⁻³ mol

Mole ratio H₂SO₄ : NaOH = 1 : 2, so n(NaOH) = 2 × 3.75 × 10⁻³ = 7.50 × 10⁻³ mol.

V(NaOH) = 7.50 × 10⁻³ / 0.200 = 0.0375 dm³ = 37.5 cm³

H₂SO₄ is diprotic, so each mole of acid neutralises two moles of NaOH. Missing the factor of two is a classic Grade C-to-Grade B discriminator.

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Uncertainty in Concentration

Every measurement in solution preparation carries an instrumental uncertainty. The two dominant contributors at A-Level are the balance (typical resolution ± 0.005 g for a 2-decimal-place balance, or ± 0.0005 g for an analytical 4-d.p. balance) and the volumetric flask (typical tolerance ± 0.30 cm³ for a Class B 250 cm³ flask). These contributions are at the heart of Required Practical 1 and lesson 8.

Calculating percentage uncertainty

The percentage uncertainty in a single measurement is:

% uncertainty = (absolute uncertainty / measured value) × 100%

When two measurements are combined (e.g. a mass and a volume), the percentage uncertainties add for the simple A-Level treatment (technically they add in quadrature for independent errors, i.e. √(u₁² + u₂²), but the linear sum is standard for A-Level and is conservative — it slightly over-estimates the combined uncertainty).

Worked Example 14: Percentage uncertainty in a prepared concentration

A student prepares 250 cm³ of approximately 0.100 mol dm⁻³ NaOH by weighing out 1.0000 g of solid NaOH on a 4-decimal-place balance (uncertainty ± 0.0005 g) and dissolving in a 250 cm³ volumetric flask (uncertainty ± 0.30 cm³). Calculate the percentage uncertainty in the prepared concentration.

Step 1 — mass uncertainty. The balance reading uses two weighings (tare and final, or weighing-by-difference), so the absolute uncertainty in the mass is 2 × 0.0005 = 0.0010 g.

% uncertainty in mass = (0.0010 / 1.0000) × 100% = 0.10%

Step 2 — volume uncertainty.

% uncertainty in volume = (0.30 / 250) × 100% = 0.12%

Step 3 — combined percentage uncertainty.

% uncertainty in c = % uncertainty in n + % uncertainty in V

= % uncertainty in mass + % uncertainty in volume (the molar mass is a known constant — no measurement uncertainty)

= 0.10 + 0.12 = 0.22%