Concentration of Solutions
When a solid (the solute) dissolves in a liquid (the solvent) it forms a solution. The concentration tells you how much solute is packed into a given volume — the more solute in the same volume, the more concentrated the solution.
concentration (g/dm³) = mass of solute (g)volume of solution (dm³)
H Everyone needs to be able to calculate a concentration in g/dm³. Explaining how the concentration depends on the mass of solute and the volume of solution — for example, why dissolving the same mass in a larger volume gives a lower concentration — is a Higher Tier requirement.
Each unit of NaCl dissolves into one Na⁺ ion and one Cl⁻ ion.
Volumes in chemistry are usually measured in cm³, but concentration needs dm³. This conversion is one of the most common places students lose marks:
1 dm³ = 1000 cm³ = 1 litre. To go from cm³ → dm³, divide by 1000; to go from dm³ → cm³, multiply by 1000. Convert before you substitute, and sense-check: have you ended up with a bigger or smaller number than you expected?
Concentration in mol/dm³ T H
Concentration can also be measured in moles per dm³, which is far more useful for calculations because it links directly to the mole ratios in an equation:
concentration (mol/dm³) = amount of solute (mol)volume of solution (dm³)
The concentration–moles relationship is not printed on the exam paper, so you must learn it. It is a good idea to jot it down before you start a calculation so every part is in the right place.
(a) 10 g of sodium hydroxide is dissolved to make 2 dm³ of solution. Find the concentration in g/dm³.
concentration = massvolume = 102 = 5 g/dm³
(b) T H 20 g of sodium hydroxide, NaOH, is dissolved to make 500 cm³ of solution. Find the concentration in mol/dm³. (Mr NaOH = 40.)
- Mass → moles: 20 ÷ 40 = 0.5 mol.
- cm³ → dm³: 500 ÷ 1000 = 0.5 dm³.
- Divide: concentration = 0.5 ÷ 0.5 = 1 mol/dm³.
🧪 Exam-style questions
12 g of sodium chloride is dissolved to make 250 cm³ of solution. Calculate the concentration of the solution in g/dm³.
Show answer
- convert the volume: 250 cm³ ÷ 1000 = 0.25 dm³ 1 mark
- concentration = massvolume = 120.25 = 48 g/dm³ 1 mark
Convert cm³ to dm³ before dividing — the single most common place to drop a mark here.
4.0 g of sodium hydroxide, NaOH, is dissolved to make 250 cm³ of solution. Calculate the concentration of the solution in mol/dm³.Use Mr(NaOH) = 40.
Show answer
- mass → moles: 4.0 ÷ 40 = 0.10 mol 1 mark
- cm³ → dm³: 250 ÷ 1000 = 0.25 dm³ 1 mark
- concentration = moles ÷ volume = 0.10 ÷ 0.25 = 0.4 mol/dm³ 1 mark
Both conversions (mass to moles, and cm³ to dm³) come before the final divide.
Titration Calculations Triple Higher
A titration is a technique for finding the volumes of two solutions that react exactly together. If the concentration of one solution is known and the reacting volumes are measured, the concentration of the other solution can be calculated using the balanced equation — for example finding the concentration of an acid from a known alkali.
In practice, a measured volume of one solution is added to a conical flask with a pipette, together with a few drops of an indicator. The other solution is run in from a burette, swirling all the time, until the indicator just changes colour. This is the end point, where the two solutions have reacted exactly — the volume delivered is the titre. The titration is repeated until the titres agree closely, and the mean titre is used in the calculation.
The apparatus for a titration: acid is run from the burette into a measured volume of alkali plus indicator, drop by drop, until the colour just changes.
Before any calculation you need a reliable titre. The first run is done quickly, just to find the approximate end point — this is the rough (trial) titre and is not used in the mean. The titration is then repeated carefully, adding solution drop by drop near the end point, until the titres are concordant — agreeing to within 0.10 cm³ of each other. Burette readings are taken to the nearest 0.05 cm³, and each titre is the final reading − initial reading:
| Rough | Run 1 | Run 2 | Run 3 | |
|---|---|---|---|---|
| Final reading / cm³ | 20.45 | 40.40 | 20.10 | 20.30 |
| Initial reading / cm³ | 0.00 | 20.45 | 0.05 | 0.30 |
| Titre / cm³ | 20.45 | 19.95 | 20.05 | 20.00 |
Runs 1–3 are concordant (all within 0.10 cm³), so they are used for the mean. The rough titre is too high — it is left out.
Just as in the uncertainty section, the mean is taken only from the concordant results — the rough titre is ignored:
mean titre = 19.95 + 20.05 + 20.003 = 60.003 = 20.00 cm³
This mean titre is the known burette volume that goes into the calculation — it is the 20.0 cm³ of acid used in the worked example below.
- Known solution → moles. moles = concentration × volume in dm³ (convert cm³ to dm³ by dividing by 1000).
- Use the equation ratio. Use the balancing numbers to convert to the moles of the unknown solution.
- Moles → concentration. concentration = moles ÷ volume in dm³ of the unknown solution.
25.0 cm³ of sodium hydroxide solution was exactly neutralised by 20.0 cm³ of dilute hydrochloric acid of concentration 0.100 mol/dm³. Calculate the concentration of the sodium hydroxide in mol/dm³.
Use the same table idea, but now the rows are Ratio, Moles, Concentration and Volume. Write in everything the question gives you, then bridge the two solutions through moles.
HCl + NaOH → NaCl + H₂O
| HClknown | NaOHfind this | |
|---|---|---|
| Ratio | 11 | 11 |
| Moles | 40.002000.100 × 0.0200 | 50.002001 : 1 |
| Conc. | 20.100mol/dm³ · given | 60.08000.00200 ÷ 0.0250 |
| Volume | 320.0 cm³= 0.0200 dm³ | 325.0 cm³= 0.0250 dm³ |
Filling the table in order:
- Ratio — read the big numbers in the equation: HCl : NaOH = 1 : 1.
- Known concentration — write 0.100 mol/dm³ under HCl.
- Volumes — write both in, converting cm³ → dm³ (÷ 1000): HCl = 0.0200 dm³, NaOH = 0.0250 dm³.
- Moles you know — concentration × volume = 0.100 × 0.0200 = 0.00200 mol.
- Moles you want — scale by the ratio (1 : 1): 0.00200 mol of NaOH.
- Answer — concentration = moles ÷ volume = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.
Try a fresh titration the same way. The volumes and the known concentration are already filled in — you fill the rest one box at a time, and each box unlocks once the one before it is correct. Watch the ratio: this acid is diprotic.
25.0 cm³ of sodium hydroxide of concentration 0.200 mol/dm³ was exactly neutralised by 20.0 cm³ of sulfuric acid, H₂SO₄. Calculate the concentration of the sulfuric acid in mol/dm³.
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
| NaOHknown | H₂SO₄find this | |
|---|---|---|
| Ratio | ||
| Moles | ||
| Conc. | 0.200mol/dm³ · given | mol/dm³ |
| Volume | 25.0 cm³= 0.0250 dm³ | 20.0 cm³= 0.0200 dm³ |
🧪 Exam-style questions
In a titration, 25.0 cm³ of sodium carbonate solution of concentration 0.0500 mol/dm³ was exactly neutralised by 25.0 cm³ of dilute hydrochloric acid. Calculate the concentration of the hydrochloric acid in mol/dm³.
Show answer
- moles of Na₂CO₃ = concentration × volume = 0.0500 × (25.0 ÷ 1000) = 0.00125 mol 1 mark
- ratio Na₂CO₃ : HCl = 1 : 2, so moles of HCl = 2 × 0.00125 = 0.00250 mol 1 mark
- concentration of HCl = moles ÷ volume = 0.00250 ÷ (25.0 ÷ 1000) 1 mark
- = 0.00250 ÷ 0.0250 = 0.100 mol/dm³ 1 mark
Convert the 25.0 cm³ volumes to dm³ (÷ 1000) before dividing — a very common place to drop a mark.
In a titration, 20.0 cm³ of sulfuric acid, H₂SO₄, of concentration 0.0500 mol/dm³ exactly neutralised 25.0 cm³ of sodium hydroxide solution. Calculate the concentration of the sodium hydroxide in mol/dm³.
Show answer
- moles of H₂SO₄ = concentration × volume = 0.0500 × (20.0 ÷ 1000) = 0.00100 mol 1 mark
- ratio H₂SO₄ : NaOH = 1 : 2, so moles of NaOH = 2 × 0.00100 = 0.00200 mol 1 mark
- concentration of NaOH = moles ÷ volume = 0.00200 ÷ (25.0 ÷ 1000) 1 mark
- = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³ 1 mark
Sulfuric acid is diprotic, so each H₂SO₄ reacts with two NaOH — the 1 : 2 ratio is the step most often missed.