Titration Calculations: Concentration

Introduction

A titration gives you two volumes: the fixed volume pipetted into the flask and the mean titre from the burette. On their own, these are just numbers. Combined with a balanced equation and one known concentration, they reveal the concentration of the other solution. This is how laboratories check the strength of acids, cleaning products and medicines. The calculation always follows the same three-stage pattern, and once you have mastered it you can tackle almost any titration question.

Core explanation

Concentration and moles. Concentration in mol/dm³ tells you how many moles of solute are dissolved in each cubic decimetre (1 dm³ = 1000 cm³) of solution. The key relationship is:

moles = concentration (mol/dm³) × volume (dm³)

Because titration volumes are measured in cm³, divide by 1000 to convert to dm³. For example, 25.0 cm³ = 0.0250 dm³.

The three stages.

1. Moles of the known solution. Use the solution whose concentration you know (the standard solution) and its volume: n = c × V. 2. Mole ratio. Use the balanced equation to find the moles of the other reactant. For HCl + NaOH → NaCl + H₂O the ratio is 1 : 1. For H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, one mole of sulfuric acid reacts with two moles of sodium hydroxide, so the moles of acid are half the moles of alkali. 3. Unknown concentration. Divide the moles of the second reactant by its volume in dm³: c = n ÷ V.

Converting to g/dm³. Sometimes the answer is required as a mass concentration. Multiply the concentration in mol/dm³ by the relative formula mass (Mr): concentration (g/dm³) = concentration (mol/dm³) × Mr. For NaOH, Mr = 23 + 16 + 1 = 40.

Using the right volume. Always use the mean of the concordant titres, never the rough titre. Keep all the figures in the calculator between steps, and round only the final answer, usually to three significant figures to match the precision of the data.

Checking the answer. A quick sense check helps. If equal volumes react in a 1 : 1 ratio, the concentrations must be equal. If a smaller titre was needed, the burette solution must be more concentrated than the flask solution.

Formulae

n = c × V, where n is moles, c is concentration in mol/dm³ and V is volume in dm³. Volume in dm³ = volume in cm³ ÷ 1000. Mass concentration (g/dm³) = c (mol/dm³) × Mr.

Step-by-step reasoning

For any titration calculation:

1. Write the balanced equation. 2. Convert both volumes to dm³. 3. Find moles of the known substance with n = c × V. 4. Apply the mole ratio. 5. Find the unknown concentration with c = n ÷ V.

Visual explanation

Draw three boxes in a row joined by arrows: "moles of known" → (× mole ratio) → "moles of unknown" → (÷ volume) → "concentration of unknown". Write c × V above the first box. The titration simulation displays these values as you complete each run.

Real-world analogy

It is like working out how strong a jug of squash is. If you know how much concentrate a full bottle holds and how many glasses it takes to use one bottle, you can work out how much concentrate is in each glass. Titration uses a known "bottle" of moles to measure an unknown one.

Real-world example

Pharmaceutical laboratories titrate samples of indigestion remedies with standard hydrochloric acid to confirm the amount of base in each tablet matches the label. Environmental chemists titrate river samples to measure acidity or alkalinity.

Why?

Why is the mole ratio essential? Titration finds when reactants have reacted exactly, which means in the ratio given by the equation, not necessarily one to one. Sulfuric acid releases two H⁺ per formula, so each mole needs two moles of NaOH.

Common misconception

"Moles = concentration × volume in cm³." Using cm³ without dividing by 1000 gives answers a thousand times too large. Concentration in mol/dm³ must be multiplied by a volume in dm³.

Worked example

Question: 25.0 cm³ of sodium hydroxide solution is neutralised by a mean titre of 20.00 cm³ of 0.100 mol/dm³ sulfuric acid. Find the concentration of the sodium hydroxide.

Reasoning: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Moles of H₂SO₄ = 0.100 × 0.02000 = 0.00200 mol. Mole ratio 1 : 2, so moles of NaOH = 0.00400 mol. Concentration = 0.00400 ÷ 0.0250 = 0.160 mol/dm³.

Answer: 0.160 mol/dm³ (0.160 × 40 = 6.40 g/dm³).

Quick check

1. How many moles of HCl are in 25.0 cm³ of 0.200 mol/dm³ hydrochloric acid? Answer: 0.200 × 0.0250 = 0.00500 mol.

Exam focus

Show every step with units: equation, moles known, ratio, moles unknown, concentration. Marks are often awarded for each stage, so a slip in one step does not lose everything. Watch for 1 : 2 ratios with sulfuric acid and for questions asking for g/dm³.

Advanced insight

In back titration, an excess of acid of known amount is added to an insoluble base such as a limestone sample, and the leftover acid is titrated with alkali. Subtracting the leftover from the total gives the acid that reacted, and hence the amount of base. This is used when a substance does not dissolve or reacts too slowly for direct titration.

Summary

Titration calculations use n = c × V with volumes in dm³. Find the moles of the known solution, use the mole ratio from the balanced equation to find the moles of the unknown, then divide by its volume to get the concentration. Multiply by Mr to convert mol/dm³ to g/dm³. Use the mean concordant titre.

Practice questions

1. 25.0 cm³ of hydrochloric acid is neutralised by 22.50 cm³ of 0.100 mol/dm³ sodium hydroxide. Find the concentration of the acid. Answer: Moles NaOH = 0.100 × 0.02250 = 0.00225 mol; ratio 1 : 1, so moles HCl = 0.00225 mol; concentration = 0.00225 ÷ 0.0250 = 0.0900 mol/dm³. 2. Convert 0.250 mol/dm³ sodium hydroxide into g/dm³. Answer: 0.250 × 40 = 10.0 g/dm³. 3. 20.0 cm³ of 0.150 mol/dm³ potassium hydroxide reacts with 24.00 cm³ of nitric acid. Find the acid concentration. Answer: Moles KOH = 0.150 × 0.0200 = 0.00300 mol; ratio 1 : 1; concentration HNO₃ = 0.00300 ÷ 0.02400 = 0.125 mol/dm³. 4. 25.0 cm³ of sulfuric acid needs 30.00 cm³ of 0.200 mol/dm³ sodium hydroxide. Find the acid concentration. Answer: Moles NaOH = 0.00600 mol; moles H₂SO₄ = 0.00300 mol; concentration = 0.00300 ÷ 0.0250 = 0.120 mol/dm³.