A-Level Chemistry 9701 / Topic 2 / AS

Moles in a measured volume.

Concentration tells you how many moles are in each cubic decimetre of solution. Measure a volume accurately and you know the moles; react it against a solution of known concentration and you can find almost anything else.

Full-screen labStoichiometry lab → Topic 2All atoms, molecules and stoichiometry lessons →

The key idea

Concentration c = n ÷ V, in mol dm⁻³ with V in dm³. So n = c × V, remembering that V in cm³ must be divided by 1000. Concentration in g dm⁻³ = c × M. In a titration: find the moles of the solution of known concentration, use the mole ratio from the equation, then divide by the volume of the other solution. Use the mean of concordant titres, those within 0.10 cm³ of each other.

Fig. 1: an acid-base titration

Section 01

Concentration and dilution.

Every solution calculation starts with n = c × V. The only common slip is forgetting to convert cm³ to dm³.

Moles in a sample: 25.0 cm³ of 0.100 mol dm⁻³ NaOH contains 0.100 × 25.0/1000 = 2.50 × 10⁻³ mol.
Making a solution: 250 cm³ of 0.100 mol dm⁻³ Na₂CO₃ needs 0.0250 mol, which is 0.0250 × 106.0 = 2.65 g.
Mass concentration: 0.200 mol dm⁻³ NaOH is 0.200 × 40.0 = 8.00 g dm⁻³.
Dilution: adding water does not change the moles of solute. 25.0 cm³ of 2.00 mol dm⁻³ acid made up to 250 cm³ becomes 2.00 × 25.0/250 = 0.200 mol dm⁻³.

Stage 1 · Learn

Check what the sim just showed you

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

How many moles of solute are in 20.0 cm³ of a 0.500 mol dm⁻³ solution?

Quick check+10 XP

The usual units of concentration in AS Chemistry are:

Quick check+10 XP

Titres are concordant when they agree to within:

Quick check+10 XP

What is the concentration of 0.200 mol dm⁻³ sodium hydroxide (M_r 40.0) in g dm⁻³?

Section 02

The titration.

Run the titration yourself: add acid from the burette, slow to single drops as the indicator starts to change, and record the titre. Then use it in the calculation the simulator sets up.

1 : 1 example: 25.0 cm³ of NaOH is neutralised by 22.50 cm³ of 0.100 mol dm⁻³ HCl. n(HCl) = 0.100 × 22.50/1000 = 2.25 × 10⁻³ mol = n(NaOH). c(NaOH) = 2.25 × 10⁻³ ÷ 0.0250 = 0.0900 mol dm⁻³.
1 : 2 example: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. 25.0 cm³ of NaOH needs 20.00 cm³ of 0.0500 mol dm⁻³ H₂SO₄: n(H₂SO₄) = 1.00 × 10⁻³ mol, so n(NaOH) = 2.00 × 10⁻³ mol and c = 0.0800 mol dm⁻³.
Concordant titres: for 24.30, 23.85, 23.95 and 23.90 cm³, the first is a rough titre. The mean of the three concordant results is 23.90 cm³.

Section 03

Back titration.

When a solid reacts too slowly or is insoluble, add a known excess of acid, let it react completely, then titrate the acid that is left over.

Setup: 2.00 g of impure calcium carbonate is added to 50.0 cm³ of 1.00 mol dm⁻³ HCl (0.0500 mol, an excess).
Leftover acid: it needs 22.0 cm³ of 1.00 mol dm⁻³ NaOH, so 0.0220 mol of HCl was not used.
Acid used: 0.0500 − 0.0220 = 0.0280 mol. CaCO₃ + 2HCl → ..., so n(CaCO₃) = 0.0140 mol = 1.40 g.
Purity: 1.40 ÷ 2.00 × 100 = 70.1% (using the unrounded 1.401 g).

Examiner trap

Two slips cost the most marks: using a volume in cm³ as if it were in dm³ (answers 1000 times too big), and ignoring the mole ratio when the acid is diprotic, as with H₂SO₄. Also leave out the rough titre and any titre not within 0.10 cm³ of the others when you take the mean.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS Paper 1 and Paper 2 standard.

Finish the four checks above to unlock the exam questions

Exam style+20 XP

25.0 cm³ of sodium hydroxide solution is neutralised by 20.00 cm³ of 0.0500 mol dm⁻³ sulfuric acid. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. What is the concentration of the sodium hydroxide?

Exam style+20 XP

25.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid is diluted with water to 250 cm³. What is the new concentration?

Exam style+20 XP

A student records titres of 24.30, 23.85, 23.95 and 23.90 cm³. Which value should be used in the calculation?

Exam style+20 XP

What mass of anhydrous sodium carbonate (M_r 106.0) is needed to make 250 cm³ of a 0.100 mol dm⁻³ solution?

Skill unlocked

Solutions and titrations, mastered.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics

Go deeper · practice

Six original Cambridge-style questions

Concentration from mass, titrations with 1 : 1 and 1 : 2 ratios, a hydrated salt found by titration, back titration for purity, and good titration technique.