4.3.4Concentration in mol/dm3Higher tier
Chemistry only, Higher tier — not assessed in Combined Science: Trilogy.
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
4.3.4Finding an unknown concentrationHigher tier
Chemistry only, Higher tier — not assessed in Combined Science: Trilogy.
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
- This sub-topic is chemistry only and Higher tier only.
- concentration (mol/dm³) = moles of solute ÷ volume of solution in dm³.
- 1 dm³ = 1000 cm³, so divide a volume in cm³ by 1000 first.
- moles = concentration × volume; mass = moles × Mr; and multiplying a mol/dm³ concentration by the Mr gives g/dm³.
- If two solutions react completely and one concentration is known, the other can be calculated.
- The titration route is: moles of the known → mole ratio from the equation → moles of the unknown → divide by its volume.
- Use only concordant titres, agreeing within 0.10 cm³, when calculating the mean.