IB Chemistry HLTopic 5 — How Much? Quantifying Chemical ChangePaper 1 & 2Practical skill~12 min read
Calculating Concentration
You cannot weigh out 0.00250 mol of hydrochloric acid. What you can do is measure a volume of solution very precisely, and that is what titrations are for. The chemistry is simple; the marks are won and lost on volumes, units and mole ratios.
📚 What you need to know
c = n / V — concentration in mol dm−3, moles, volume in dm3.
Always convert cm3 to dm3 by dividing by 1000 before using this equation.
Concentration in g dm−3 = concentration in mol dm−3 × M.
A titration uses a solution of known concentration (the standard solution) to find an unknown one.
Volume is measured with a pipette into the flask and a burette for the solution you add.
Concordant titres agree within 0.10 cm3; average only the concordant ones.
A back titration adds a known excess, then titrates what is left to find what reacted.
The equation, and the unit that catches everyone
Concentrationc = n ÷ V — n = c × V — V = n ÷ c
One equation, three rearrangements, and one trap: V must be in dm3. Burettes and pipettes are marked in cm3, so nearly every question hands you the wrong unit on purpose. 25.0 cm3 is 0.0250 dm3.
Do the division by 1000 the moment you write the number down, not later. Write “V = 25.0 cm3 = 0.0250 dm3” as one line and the problem disappears for the rest of the question.
The apparatus, and why each piece is used
The pipette measures one fixed volume very accurately into the flask. The burette measures however much you needed to add — that reading is your titre.
🧩 How a titration actually runs
Pipette a known volume (usually 20.0 or 25.0 cm3) of one solution into a conical flask.
Add a few drops of indicator. A few drops only — indicator is itself a weak acid or base.
Fill the burette with the other solution and take the starting reading.
Add quickly at first, then drop by drop, swirling, until the colour just changes and stays changed.
Record the final reading and subtract to get the titre.
Repeat until two titres are concordant, then average those.
Four steps that solve any titration calculation
Step 3 is the one that separates the grades. A 2 : 1 acid like H2SO4 needs half as many moles as the base, and that halving is easy to forget.
The c1V1 = c2V2 shortcut. When the acid and base react 1 : 1 (one H+ against one OH−), you can use this directly — and because the volume units cancel, cm3 is fine. It fails the moment the ratio is not 1 : 1, so check the equation before you reach for it.
Worked examples
WORKED EXAMPLE
25.0 cm3 of 0.100 mol dm−3 NaOH was exactly neutralised by 22.4 cm3 of sulfuric acid. Calculate the concentration of the acid.
Step 1: balanced equation2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂OStep 2: moles of the solution you know
25.0 cm3 = 0.0250 dm3 n(NaOH) = 0.0250 × 0.100 = 2.50 × 10⁻³ molStep 3: the ratio is 2 NaOH to 1 acidn(H₂SO₄) = 2.50 × 10⁻³ ÷ 2 = 1.25 × 10⁻³ molStep 4: divide by the acid’s volume in dm³c = 1.25 × 10⁻³ ÷ 0.0224 = 0.0558c(H₂SO₄) = 0.0558 mol dm⁻³Miss the halving and you get 0.112, exactly double. Examiners set 2 : 1 acids for precisely this reason.
WORKED EXAMPLE
What mass of sodium hydroxide is needed to make 250 cm3 of a 0.200 mol dm−3 solution?
Step 1: volume in dm³250 ÷ 1000 = 0.250 dm³Step 2: moles neededn = c × V = 0.200 × 0.250 = 0.0500 molStep 3: moles to mass
M(NaOH) = 22.99 + 16.00 + 1.01 = 40.00 m = 0.0500 × 40.00 = 2.00 g2.00 g, dissolved and made up to 250 cm³“Made up to” matters: you dissolve the solid in a little water, then top up to the mark. You do not add 250 cm³ of water to it.
WORKED EXAMPLE
A 0.500 g sample of marble was added to 50.0 cm3 of 0.250 mol dm−3 HCl, an excess. The unreacted acid needed 40.0 cm3 of 0.100 mol dm−3 NaOH. Find the percentage of CaCO3 in the marble.
Step 1: total acid addedn = 0.0500 × 0.250 = 0.01250 molStep 2: acid left over, from the titration (1 : 1 with NaOH)n = 0.0400 × 0.100 = 0.00400 molStep 3: acid that reacted with the marble0.01250 − 0.00400 = 0.00850 molStep 4: CaCO₃ reacts 1 : 2 with HCln(CaCO₃) = 0.00850 ÷ 2 = 0.00425 molStep 5: mass, then percentagem = 0.00425 × 100.09 = 0.4254 g 0.4254 ÷ 0.500 × 100 = 85.185.1% calcium carbonateThe logic of a back titration: total minus leftover equals reacted. Write those three words down and the steps write themselves.
Units at a glance
Quantity
Symbol
Unit
Equation
Amount
n
mol
n = c × V
Concentration
c
mol dm−3
c = n ÷ V
Volume
V
dm3
V = n ÷ c
Mass concentration
—
g dm−3
c (mol dm−3) × M
💡 Exam tip
Convert every volume to dm3 as you write it down. One line, done, and no factor of 1000 errors.
Write the balanced equation even when the question looks simple. It is where the ratio comes from and it is usually a mark.
Average only concordant titres, and say why you rejected any rough or outlying value.
Round the average titre to two decimal places, because that is the precision of a burette reading.
State units in the final answer. “0.0558” on its own may not score.
For back titrations, label the three amounts: total, excess, reacted. It keeps you from subtracting the wrong pair.
⚠ Common mix-up
Using cm3 in c = n/V. Your answer comes out 1000 times too big.
Skipping the mole ratio because both volumes look similar. Check the equation every time.
Using c1V1 = c2V2 on a 2 : 1 reaction. It only works for a 1 : 1 ratio.
Dividing by the wrong volume at the end. Use the volume of the solution whose concentration you want.
Including a rough titre in the average. The first run is nearly always overshot.
Adding too much indicator. It reacts too, and it blurs the endpoint.
Every calculation so far has quietly assumed one reactant runs out and the other is in excess. Time to look at that assumption properly, because it decides how much product you can possibly make. Up next: Limiting and Excess Reactants.
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