IB Chemistry SLTopic 5 — Quantifying Chemical ChangePaper 1 & 2Practical skill~13 min read
Calculating Concentration
You cannot weigh a dissolved substance, but you can measure a volume of its solution with astonishing precision. Concentration is what turns that volume into an amount — and a titration into one of the most accurate measurements you will ever make in a school lab.
📚 What you need to know
c = n / V, with concentration in mol dm–3, amount in mol and volume in dm3.
Rearranged: n = c × V. Divide cm3 by 1000 first, every time.
A standard solution has an accurately known concentration, made up in a volumetric flask.
In a titration, a pipette delivers a fixed volume into a conical flask and a burette delivers the variable volume.
Repeat until concordant titres (within 0.10 cm3) are obtained, then average only those.
A back titration reacts the sample with a measured excess, then titrates what is left over.
Concentration
Molar concentration
c = n / V n = c × V
mol dm–3 · mol · dm3
Square brackets are shorthand for this: [HCl] = 0.100 mol dm–3 means exactly that concentration. You will also meet mass concentration in g dm–3, which converts across by dividing by the molar mass.
With concentration added, all four ways of measuring an amount are now in place, and they all meet in the same place.
Whatever a question gives you and whatever it asks for, the path runs through the middle.
WORKED EXAMPLE
5.30 g of anhydrous sodium carbonate is dissolved and made up to 250.0 cm3 in a volumetric flask. Calculate the concentration of the solution.
Step 1 — molar massM(Na₂CO₃) = 2(22.99) + 12.01 + 3(16.00) = 105.99Step 2 — molesn = 5.30 ÷ 105.99 = 0.0500 molStep 3 — volume in dm³250.0 ÷ 1000 = 0.2500 dm³Step 4 — concentrationc = 0.0500 ÷ 0.2500c = 0.200 mol dm⁻³This is how a standard solution is made: weigh accurately, dissolve, then make up to the mark.
Titration
The pipette measures one volume perfectly every time; the burette measures the volume you are actually trying to find.
The technique matters because the calculation is only as good as the titre. The standard procedure:
Use a volumetric pipette to transfer an exact volume (usually 25.0 cm3) into a conical flask.
Add a few drops of indicator — a few drops only, since indicators are themselves weak acids or bases.
Fill the burette with the other solution and record the initial reading to two decimal places.
Run in solution, swirling, until the indicator just changes colour and the colour persists. Do a rough titration first to find the approximate endpoint.
Repeat until you have concordant results, then average only the concordant titres.
Titration
Rough
1
2
3
Final reading / cm3
23.10
22.45
22.40
22.85
Initial reading / cm3
0.00
0.05
0.00
0.50
Titre / cm3
23.10
22.40
22.40
22.35
Here titrations 1 and 2 agree exactly, and titration 3 is within 0.05 cm3 of them, so all three are concordant and average to 22.38 cm3. The rough is always discarded. Note the answer is quoted to two decimal places — an average can never be more precise than the readings it came from.
The endpoint is one drop. Add that drop too many and the titre is wrong by about 0.05 cm3, which is why you swirl constantly and slow to drop-by-drop as the colour starts to linger.
🧩 The titration calculation
Write the balanced equation.
Find moles of the substance you know everything about: n = c × V.
Use the mole ratio to get moles of the unknown.
Divide by its volume in dm3 to get its concentration, or multiply by M to get a mass.
WORKED EXAMPLE
25.0 cm3 of sodium hydroxide solution required 22.4 cm3 of 0.100 mol dm–3 sulfuric acid for neutralisation. Calculate the concentration of the sodium hydroxide.
Step 1 — equation2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂OStep 2 — moles of acid, the one fully knownn = 0.0224 × 0.100 = 2.24 × 10⁻³ molStep 3 — ratio 1 acid : 2 basen(NaOH) = 2 × 2.24 × 10⁻³ = 4.48 × 10⁻³ molStep 4 — concentrationc = 4.48 × 10⁻³ ÷ 0.0250c = 0.179 mol dm⁻³Sulfuric acid is diprotic, so the ratio is 1 : 2. The C₁V₁ = C₂V₂ shortcut would have given 0.0896 and been wrong by a factor of two.
The shortcut C1V1 = C2V2 only works when the ratio is 1 : 1 — both acid and base monoprotic. It is quick and safe for HCl with NaOH, and quietly disastrous for H2SO4. Writing the equation first tells you which case you are in.
Back titration
Some substances cannot be titrated directly: an insoluble solid, a slow reaction, or a sample too impure to weigh meaningfully. The trick is to react it with a measured excess of something, then titrate the excess to find out how much was left — and therefore how much was used.
WORKED EXAMPLE
A 1.50 g indigestion tablet containing calcium carbonate was added to 50.0 cm3 of 0.500 mol dm–3 hydrochloric acid, an excess. The unreacted acid required 21.5 cm3 of 0.400 mol dm–3 sodium hydroxide. Calculate the percentage of CaCO3 in the tablet.
Step 1 — total acid addedn = 0.0500 × 0.500 = 0.02500 molStep 2 — acid left over, from the titrationHCl + NaOH → NaCl + H₂O, a 1 : 1 ratio.n(NaOH) = 0.0215 × 0.400 = 0.00860 mol = n(HCl) excessStep 3 — acid that actually reacted0.02500 − 0.00860 = 0.01640 molStep 4 — moles of carbonateCaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, so divide by 2.0.01640 ÷ 2 = 0.00820 molStep 5 — mass and percentage0.00820 × 100.09 = 0.8207 g(0.8207 ÷ 1.50) × 100 = 54.754.7% calcium carbonateTwo equations, two ratios, one subtraction. The subtraction in step 3 is the whole idea of a back titration.
💡 Exam tip
Divide every cm3 by 1000 before it touches a concentration.
Start from the substance you know both the concentration and volume of.
Average only concordant titres, never the rough, and quote to two decimal places.
For a back titration, label your moles clearly as added, excess and reacted. Mixing them up is the usual failure.
Watch for diprotic acids. H2SO4 supplies two H+, so the ratio is not 1 : 1.
⚠️ Common mix-up
Leaving volumes in cm3, giving a concentration 1000 times too large.
Using C1V1 = C2V2 when the mole ratio is not 1 : 1.
Including the rough titre in the average.
Subtracting the wrong way round in a back titration — it is added minus excess.
Quoting an average titre more precisely than the burette can read.
Up next: Limiting and Excess Reactants — because so far every question has quietly told you which substance to work from, and real ones do not.
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