5.8CConcentration in mol/dm3Higher tier
Chemistry only, and Higher tier only — not on the Combined Science specification
The concentration of a solution can be measured in mol/dm³.
concentration in mol/dm³ = amount of solute in mol ÷ volume of solution in dm³
This is the same idea as concentration in g/dm³, but counting particles rather than weighing them. It is the more useful measure in a reaction, because a balanced equation is a ratio of moles, not of grams.
The amount in moles of solute, or the mass in grams of solute, in a given volume of solution can be calculated from its concentration in mol/dm³.
You should be able to explain how the concentration of a solution in mol/dm³ is related to the mass of the solute and the volume of the solution.
The chain has three links, and it runs in both directions:
mass in g ⇄ moles (divide or multiply by Mr) ⇄ concentration in mol/dm³ (divide or multiply by the volume in dm³)
So concentration in mol/dm³ increases if more solute is dissolved in the same volume, and decreases if the same solute is dissolved in a greater volume. To go from a concentration to a mass, multiply by the volume to get moles, then by the Mr.
Worked example
Calculate the mass of sodium hydroxide in 250 cm³ of a 0.100 mol/dm³ solution. (Mr of NaOH = 40.)
- Convert the volume to dm³: 250 ÷ 1000 = 0.250 dm³.
- moles = concentration × volume = 0.100 × 0.250 = 0.0250 mol.
- mass = moles × Mr = 0.0250 × 40.
- Check the units at each step: mol/dm³ × dm³ gives mol, and mol × g/mol gives g.
Answer: 1.00 g of sodium hydroxide
- Concentration in mol/dm³
- the amount of solute in moles divided by the volume of solution in dm³; the more useful measure in a reaction, because a balanced equation is a ratio of moles
5.9CTitrations
Chemistry only — not on the Combined Science specification
The volumes of acid and alkali solutions that react with each other can be measured by titration using a suitable indicator.
You should be able to describe how to carry out titrations using strong acids and strong alkalis only — sulfuric, hydrochloric and nitric acids only — to find the reacting volumes accurately.
Required practical 2
Determination of the reacting volumes of a strong acid and a strong alkali by titration
- Use a pipette and pipette filler to transfer 25.0 cm³ of the alkali into a conical flask, and add a few drops of a single indicator such as phenolphthalein.
- Rinse the burette with the acid, fill it, and run some through to remove the air bubble below the tap. Record the initial reading to the nearest 0.05 cm³.
- Run the acid into the flask while swirling, quickly at first, then dropwise as the colour starts to change.
- Stop at the first permanent colour change and record the final reading. Subtract to get the titre.
- Repeat until two or more titres are concordant — within 0.10 cm³ of each other — and take the mean of those only.
Higher tier students should also be able to calculate the chemical quantities in titrations involving concentrations in mol/dm³ and in g/dm³ — the calculation set out under using concentrations of solutions.
Common mistake
Read the burette from the bottom of the meniscus, at eye level, and remember that a burette reads downwards — the volume delivered is the final reading minus the initial one, not the other way round.
- Titration
- a technique for measuring the volumes of acid and alkali solutions that react with each other, using a suitable indicator
5.10CFinding an unknown concentrationHigher tier
Chemistry only, and Higher tier only — not on the Combined Science specification
If the volumes of two solutions that react completely are known, and the concentration of one solution is known, the concentration of the other solution can be calculated.
That is what a titration is for, and it is the standard four-step calculation:
1. Work out the moles of the solution whose concentration you know: moles = concentration × volume in dm³. 2. Use the balanced equation to convert to moles of the other solution. 3. Divide by that solution's volume in dm³ to get its concentration in mol/dm³. 4. If a concentration in g/dm³ is asked for, multiply by the Mr.
Practical skill
Titration: determining the concentration of a strong acid or alkali
- Use a pipette and pipette filler to transfer 25.0 cm³ of the alkali into a conical flask, and add a few drops of a suitable indicator.
- Fill a burette with the acid, run out the air bubble below the tap, and record the initial reading to the nearest 0.05 cm³.
- Add the acid to the flask, swirling constantly, until the indicator changes colour permanently. Record the final burette reading.
- Repeat, adding the acid dropwise near the end point, until at least two titres agree within 0.10 cm³.
- Calculate the mean of the concordant titres only, ignoring the rough first run and any anomaly.
Worked example
25.0 cm³ of sodium hydroxide solution is neutralised by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid. NaOH + HCl → NaCl + H₂O. Calculate the concentration of the sodium hydroxide in mol/dm³ and in g/dm³. (Mr of NaOH = 40.)
- Moles of HCl = concentration × volume = 0.100 × (20.0 ÷ 1000) = 0.00200 mol.
- The equation ratio is 1 : 1, so moles of NaOH = 0.00200 mol.
- Volume of NaOH in dm³ = 25.0 ÷ 1000 = 0.0250 dm³.
- Concentration = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.
- To convert to g/dm³, multiply by the Mr: 0.0800 × 40 = 3.20 g/dm³.
Answer: 0.0800 mol/dm³, which is 3.20 g/dm³.
Common mistake
Do not skip step 2 when the ratio is not 1 : 1. In H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, the moles of sodium hydroxide are twice the moles of sulfuric acid, and forgetting that halves or doubles the final answer.
Common mistake
Average only the concordant titres — those within 0.10 cm³ of each other. Including the rough run drags the mean upwards and makes the calculated concentration too high.
Before you practise
- concentration in mol dm⁻³ = moles ÷ volume in dm³. Divide a volume in cm³ by 1000 first.
- To convert g dm⁻³ to mol dm⁻³, divide by the relative formula mass; multiply to go back.
- Use a pipette for the fixed volume in the flask and a burette for the solution added.
- Add dropwise near the end point, and swirl throughout.
- Repeat until two titres agree within 0.10 cm³, and mean only those concordant results — not the rough one.
- The calculation is moles of the known, then the ratio from the equation, then the unknown.
- Read a burette to the nearest 0.05 cm³, from the bottom of the meniscus at eye level.