The big idea: Concentration tells you how much solute is dissolved in a given volume of solution.
The standard chemistry unit is mol dm⁻³ (moles of solute per cubic decimetre of solution). You will also meet g dm⁻³ (grams per cubic decimetre) and ppm (parts per million) for very dilute solutions.
Square brackets are shorthand for concentration: [NaCl] = 0.10 mol dm⁻³ means 0.10 mol of NaCl in every 1 dm³.
Know your volume units: Almost every mistake here is a volume conversion. The formula needs the volume in dm³, but lab volumes are usually given in cm³.
- 1 dm³ = 1000 cm³ (and 1 dm³ = 1 L) - so divide a volume in cm³ by 1000 to get dm³
Example: 250 cm³ = 250 ÷ 1000 = 0.250 dm³.
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The amount of solute, its concentration and the solution volume are linked by the given equation. Rearrange it for whatever the question asks.
- amount of solute (mol)
- concentration (mol dm⁻³)
- volume of solution (dm³)
Two units, one quick conversion: To convert between the two concentration units, multiply or divide by the molar mass M:
- g dm⁻³ = mol dm⁻³ × M - mol dm⁻³ = g dm⁻³ ÷ M
This is just n = m/M applied to one cubic decimetre of solution.
Worked example — concentration from moles and volume
5.00 g of sodium hydroxide, NaOH, is dissolved and made up to 250 cm³ of solution. Calculate the concentration in mol dm⁻³. (M(NaOH) = 40.00 g mol⁻¹.)
Solution
- First find the amount of solute — formula first:
- Convert the volume to dm³ (this is the step students forget):
- Rearrange the given equation for concentration:
- Substitute and solve — keep the unit:
Final answer
C = 0.500 mol dm⁻³.
Worked example — converting to g dm⁻³
Express the concentration of the 0.500 mol dm⁻³ NaOH solution above in g dm⁻³.
Solution
- Formula first — grams per dm³ is moles per dm³ scaled by molar mass:
- Substitute:
- Work it out:
Final answer
20.0 g dm⁻³.
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Diluting a solution means adding more solvent (usually water). The amount of solute does not change — you just spread it through a larger volume — so the concentration falls. That single idea gives the dilution equation.
- concentration and volume **before** dilution
- concentration and volume **after** dilution
Worked example — dilution
What volume of water must be added to 20.0 cm³ of 0.800 mol dm⁻³ hydrochloric acid to dilute it to 0.100 mol dm⁻³?
Solution
- Formula first — rearrange the dilution equation for the final volume:
- Substitute (volumes can stay in cm³ here, as both sides match):
- That 160 cm³ is the total final volume; subtract the original to get the water added:
Final answer
140 cm³ of water must be added (giving 160 cm³ of solution in total).
Parts per million (ppm): For very dilute solutions chemists use ppm — milligrams of solute per kilogram (or per dm³) of solution:
- 1 ppm = 1 mg dm⁻³ = 1 g per 1 000 000 g of solution.
Dilution still obeys C1V1 = C2V2, so a small sample diluted into a larger flask drops the ppm by the dilution factor (the volume ratio).
Worked example — ppm and dilution factor
A 2.00 cm³ sample of a stock solution is made up to 50.0 cm³ with water. The diluted solution is found to be 4.00 ppm. Calculate the concentration of the original stock solution in ppm.
Solution
- Formula first — rearrange the dilution equation for the stock concentration:
- The dilution factor is V2/V1 = 50.0/2.00 = 25, so concentrating up multiplies by 25:
- Work it out:
Final answer
The stock solution is 100 ppm (it was diluted 25-fold).
Preparing a standard solution: A standard solution has a precisely known concentration. The lab method:
1. Weigh the exact mass of solid on a balance. 2. Dissolve it in a small amount of distilled water in a beaker. 3. Transfer to a volumetric flask (rinse the beaker in too, so no solute is lost). 4. Make up to the calibrated mark with distilled water (final drops from a pipette). 5. Stopper and invert several times to mix thoroughly.
How this is tested: Concentration is the most-tested S1.4 skill (20 questions in the May 2025 corpus), and it dominates the Paper 1B practical/data section.
- Paper 1A (MCQ): a one-step n = CV or dilution calculation. - Paper 1B / Paper 2: 'find the mass to prepare X cm³ of a Y mol dm⁻³ solution', a dilution, a ppm conversion, or a describe of how to make a standard solution.
The near-universal trap: leaving the volume in cm³ instead of converting to dm³.
Banking the marks: (1) Convert the volume to dm³ first. (2) Write n = CV (or m = nM) before the numbers. (3) For dilution, remember V2 is the total volume — subtract to get water added.
IB-style question — preparing a standard solution (a)
A student needs to prepare 500.0 cm³ of a 0.200 mol dm⁻³ solution of anhydrous sodium carbonate, Na2CO3. (a) Calculate the mass of Na2CO3 that must be weighed out. (M = 105.99 g mol⁻¹.)
Solution
- Convert the volume to dm³:
- Formula first — find the amount of solute needed:
- Convert the amount to a mass:
- Work it out:
Final answer
m = 10.6 g of Na2CO3.
IB-style question — preparing a standard solution (b)
(b) Outline how the student would use this weighed solid to prepare the standard solution in a 500.0 cm³ volumetric flask. [3]
How to score the marks
- Mark 1 — dissolve. Dissolve the weighed Na2CO3 in a small volume of distilled water in a beaker (and stir until fully dissolved).
- Mark 2 — transfer fully. Transfer the solution into the 500.0 cm³ volumetric flask, rinsing the beaker into the flask so that no solute is left behind.
- Mark 3 — make up to the mark. Add distilled water up to the calibrated mark (final drops by pipette), then stopper and invert several times to mix.
Final answer
Dissolve in a little distilled water; transfer to the 500 cm³ volumetric flask rinsing the beaker in; make up to the mark with distilled water and invert to mix.