Moles in Solutions: Concentration and Titrations

When substances are dissolved in water to form solutions, we need a way to describe how much solute is present in a given volume. This is where concentration comes in — and titrations provide a precise experimental method for determining unknown concentrations.

Concentration

Concentration can be expressed in two ways at A-Level:

Mol dm⁻³ (moles per cubic decimetre): This tells you how many moles of solute are dissolved in 1 dm³ (1000 cm³) of solution.

g dm⁻³ (grams per cubic decimetre): This tells you the mass of solute dissolved in 1 dm³ of solution.

The relationship between the two is:

concentration (g dm⁻³) = concentration (mol dm⁻³) × molar mass (g mol⁻¹)

The Key Equation

n = c × V

Where:

n = number of moles (mol)
c = concentration (mol dm⁻³)
V = volume (dm³)

Important: Volume must be in dm³, not cm³. To convert: V (dm³) = V (cm³) / 1000

This is one of the most common sources of error in mole calculations. If you use cm³ directly in the formula, your answer will be 1000 times too large.

Worked Example 1

Calculate the number of moles of NaOH in 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution.

Step 1: Convert volume to dm³. V = 25.0 / 1000 = 0.0250 dm³

Step 2: Use n = c × V. n = 0.100 × 0.0250 = 0.00250 mol (or 2.50 × 10⁻³ mol)

Worked Example 2

What mass of sodium hydroxide is needed to make 500 cm³ of 0.200 mol dm⁻³ solution?

Step 1: n = c × V = 0.200 × 0.500 = 0.100 mol

Step 2: M(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹

Step 3: m = n × M = 0.100 × 40.0 = 4.00 g

Converting Between mol dm⁻³ and g dm⁻³

Concentration (mol dm⁻³)	Molar mass (g mol⁻¹)	Concentration (g dm⁻³)
0.100	40.0 (NaOH)	4.00
0.500	98.0 (H₂SO₄)	49.0
1.00	36.5 (HCl)	36.5
0.0200	158.0 (KMnO₄)	3.16
2.00	58.5 (NaCl)	117.0

To convert: mol dm⁻³ × M = g dm⁻³. To convert back: g dm⁻³ ÷ M = mol dm⁻³.

Dilution Calculations

When you dilute a solution (add more solvent), the number of moles of solute stays the same — only the volume changes. This gives:

c₁V₁ = c₂V₂

Worked Example 3

25.0 cm³ of 2.00 mol dm⁻³ HCl is diluted to 500 cm³. What is the new concentration?

c₁V₁ = c₂V₂ 2.00 × 25.0 = c₂ × 500 c₂ = 50.0 / 500 = 0.100 mol dm⁻³

Worked Example 4

A student needs 250 cm³ of 0.0500 mol dm⁻³ sulfuric acid. They have a 1.00 mol dm⁻³ stock solution. What volume of stock solution should they dilute?

c₁V₁ = c₂V₂ 1.00 × V₁ = 0.0500 × 250 V₁ = 12.5 / 1.00 = 12.5 cm³

They should measure 12.5 cm³ of stock solution and add water to make 250 cm³.

Acid-Base Titrations

A titration is a technique used to determine an unknown concentration by reacting a measured volume of one solution with a known concentration of another. In an acid-base titration:

flowchart TD
    A["Pipette known volume of analyte into conical flask"] --> B["Add indicator"]
    B --> C["Fill burette with standard solution"]
    C --> D["Add solution from burette dropwise"]
    D --> E{"Colour change at endpoint?"}
    E -- No --> D
    E -- Yes --> F["Record burette reading"]
    F --> G["Repeat until concordant results"]
    G --> H["Calculate mean of concordant titres"]
    H --> I["Use n = c × V and mole ratios"]

Common Indicators

Indicator	Acid colour	Alkali colour	Best for
Methyl orange	Red/pink	Yellow	Strong acid + weak base
Phenolphthalein	Colourless	Pink	Weak acid + strong base
Either	—	—	Strong acid + strong base

Note: No indicator works well for weak acid + weak base titrations because the pH change at the equivalence point is too gradual.

Concordant Results

When performing a titration, you repeat the experiment until you obtain at least two titre values that agree to within 0.10 cm³. These are called concordant results. You then take the mean of the concordant titres (not all titres) for your calculation.

Example

Titre	Volume (cm³)	Concordant?
Rough	23.80	No (rough)
1	23.45	Yes
2	23.50	Yes
3	23.40	No

Mean concordant titre = (23.45 + 23.50) / 2 = 23.48 cm³

Note: Titre 3 (23.40) differs by 0.10 from titre 2 — whether to include it depends on your school's interpretation. Strictly, concordant means within 0.10 cm³, so 23.40 and 23.50 differ by 0.10, which is borderline.

Titration Calculations

Worked Example 5 — Basic Titration

25.0 cm³ of sodium hydroxide solution was titrated with 0.100 mol dm⁻³ hydrochloric acid. The mean titre was 24.0 cm³. Calculate the concentration of NaOH.

NaOH + HCl → NaCl + H₂O

Step 1: Moles of HCl = 0.100 × (24.0/1000) = 2.40 × 10⁻³ mol

Step 2: Mole ratio NaOH : HCl = 1 : 1

Step 3: Moles of NaOH = 2.40 × 10⁻³ mol

Step 4: Concentration of NaOH = 2.40 × 10⁻³ / 0.0250 = 0.0960 mol dm⁻³

Worked Example 6 — Diprotic Acid

25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide required 12.5 cm³ of sulfuric acid for neutralisation. Calculate the concentration of H₂SO₄.

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Step 1: Moles of NaOH = 0.100 × 0.0250 = 2.50 × 10⁻³ mol

Step 2: Mole ratio NaOH : H₂SO₄ = 2 : 1, so moles H₂SO₄ = 2.50 × 10⁻³ / 2 = 1.25 × 10⁻³ mol

Step 3: Concentration of H₂SO₄ = 1.25 × 10⁻³ / 0.0125 = 0.100 mol dm⁻³

Worked Example 7 — Multi-Step with Molar Mass

25.0 cm³ of a solution of sodium carbonate required 27.5 cm³ of 0.100 mol dm⁻³ HCl. Calculate the concentration of Na₂CO₃ in g dm⁻³.

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Step 1: Moles of HCl = 0.100 × 0.0275 = 2.75 × 10⁻³ mol

Step 2: Moles of Na₂CO₃ = 2.75 × 10⁻³ / 2 = 1.375 × 10⁻³ mol

Step 3: Concentration of Na₂CO₃ = 1.375 × 10⁻³ / 0.0250 = 0.0550 mol dm⁻³

Step 4: M(Na₂CO₃) = (2 × 23.0) + 12.0 + (3 × 16.0) = 106.0 g mol⁻¹

Step 5: Concentration in g dm⁻³ = 0.0550 × 106.0 = 5.83 g dm⁻³

Common Mistakes in Titration Calculations

Mistake	Why it causes errors	How to avoid it
Forgetting to convert cm³ to dm³	Answer is 1000× too large	Always divide by 1000
Using wrong mole ratio	Wrong answer by a factor of 2, 3, etc.	Write the balanced equation first
Including the rough titre in the mean	Inaccurate mean titre	Only average concordant values
Confusing pipette and burette volumes	Swapping known and unknown	Pipette = analyte, burette = titrant
Not rinsing burette with titrant solution	Dilutes titrant, titre too large	Rinse with the solution it will contain
Rinsing conical flask with analyte	Extra moles of analyte, titre too large	Rinse with distilled water only

Standard Solutions

A standard solution is a solution of precisely known concentration. You can make a standard solution by dissolving a known mass of a primary standard in a volumetric flask and making up to the mark with distilled water.

Requirements for a Primary Standard

High purity
Known, stable formula (not hygroscopic, does not decompose)
High molar mass (minimises weighing errors)
Readily available

Common primary standards include: anhydrous sodium carbonate (Na₂CO₃), potassium hydrogen phthalate (KHP), and oxalic acid dihydrate (H₂C₂O₄·2H₂O).

Worked Example 8 — Triprotic Acid

25.0 cm³ of phosphoric acid (H₃PO₄) required 37.5 cm³ of 0.100 mol dm⁻³ NaOH for complete neutralisation. Calculate the concentration of H₃PO₄.

H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

Step 1: Moles of NaOH = 0.100 × 0.0375 = 3.75 × 10⁻³ mol

Step 2: Mole ratio H₃PO₄ : NaOH = 1 : 3, so moles H₃PO₄ = 3.75 × 10⁻³ / 3 = 1.25 × 10⁻³ mol

Step 3: [H₃PO₄] = 1.25 × 10⁻³ / 0.0250 = 0.0500 mol dm⁻³

The mole ratio is critical here — using 1:1 or 1:2 instead of 1:3 would give the wrong answer.

Titrations are the backbone of quantitative analytical chemistry and appear in nearly every A-Level Chemistry exam paper. The key to mastering them is practice: write the balanced equation, convert volumes, find moles, use the ratio, and calculate the answer.

A-Level Deep Dive: Moles in Solutions: Concentration and Titrations

Spec mapping

Edexcel 9CH0 specification Topic 5 — Formulae, Equations and Amounts of Substance, sub-strand on solutions and titrations, requires students to define concentration in mol dm $^{-3}$ and g dm $^{-3}$ , compute amounts using $n = cV$ , perform single and back-titration calculations, and select appropriate indicators for acid-base reactions (refer to the official specification document for exact wording). The skill is examined heavily on Paper 2 (where titrations underpin Topic 12 acid-base equilibria) and Paper 3 (the synoptic practical paper that includes CP2 and CP13). The data sheet provides standard volumes and unit conversions; concentration formulae must be memorised.

Worked example with full mark scheme

Question (8 marks): A student dissolves 1.500 g of impure sodium carbonate, $\text{Na}_2\text{CO}_3$ , in water and makes the solution up to 250.0 cm $^3$ . A 25.00 cm $^3$ aliquot is titrated with 0.1000 mol dm $^{-3}$ HCl, requiring 26.40 cm $^3$ for complete neutralisation, using methyl orange indicator.

(a) Write the balanced equation for the reaction. (1)

(b) Calculate the percentage purity of the sodium carbonate sample. ( $M_r(\text{Na}_2\text{CO}_3) = 106.0$ .) (6)

(c) Justify the choice of methyl orange (pH range 3.1–4.4) over phenolphthalein (pH range 8.2–10.0) as indicator. (1)

Solution with mark scheme:

(a) $\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$ . B1.

(b) Step 1 — moles of HCl in titre.

$n(\text{HCl}) = 0.02640 \times 0.1000 = 2.640 \times 10^{-3}\,\text{mol}$ .

M1 — $n = cV$ with V in dm $^3$ .

Step 2 — moles of Na $_2$ CO $_3$ in 25.00 cm $^3$ .

Stoichiometry 1 : 2, so $n(\text{Na}_2\text{CO}_3) = 2.640 \times 10^{-3}/2 = 1.320 \times 10^{-3}\,\text{mol}$ .

M1 — using the 1 : 2 mole ratio.

Step 3 — moles of Na $_2$ CO $_3$ in 250.0 cm $^3$ .

$n_{\text{total}} = 1.320 \times 10^{-3} \times (250.0/25.00) = 1.320 \times 10^{-2}\,\text{mol}$ .

M1 — scaling aliquot to total.

Step 4 — mass of Na $_2$ CO $_3$ and percentage purity.

$m(\text{Na}_2\text{CO}_3) = 1.320 \times 10^{-2} \times 106.0 = 1.399\,\text{g}$ .

% purity $= 1.399/1.500 \times 100 = 93.3\%$ .