Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

5.1.4Additional lessons

Concentration Calculations — IB Chemistry SL Revision Notes & Topical Questions

5 min read 6 parts Lesson 62 of 98

📚 IB Chemistry SL Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
0%read

Part 1 of 6Expressing Concentration

5.1.4 Concentration Calculations

Concentration describes how much solute is dissolved in a given volume of solution.

Molar concentration
  • Concentration c = n ÷ V, where n is the amount in moles and V is the volume in dm³, with the unit mol dm⁻³.
Mass concentration
  • Mass of solute divided by volume of solution in dm³ gives a value in g dm⁻³, which equals c × M.
Very low concentrations
  • Parts per million (ppm) gives the mass of solute per million parts of solution and is used for trace amounts.
Volume units
  • Volumes in cm³ must be divided by 1000 to give dm³ before they are used with mol dm⁻³.

Part 2 of 6Using Concentration

1
Convert any volume in cm³ to dm³ by dividing by 1000.
2
Calculate the amount of solute using n = cV.
3
Use the balanced equation to link the amount of solute to the amount of another reactant or product.
4
Convert the result to the requested quantity, such as mass, volume or concentration.
5
Check that the units and significant figures are consistent.

Part 3 of 6Dilution

Dilution
Idea
  • Adding solvent changes the concentration but not the amount of solute in the sample.
Equation
  • c₁V₁ = c₂V₂, where the volumes must be in the same units on both sides.
Procedure
  • Pipette the required volume of stock solution into a volumetric flask and fill to the mark with distilled water.
Link
  • A tenfold increase in the volume gives a tenfold decrease in the concentration.
QuantityFormula
Molesn = cV
Mass concentrationρ = c × M
Dilutionc₁V₁ = c₂V₂

Part 4 of 6Preparing and Using Solutions

Preparing and Using Solutions
Standard solution
  • Prepared by dissolving an accurately weighed mass of solute in a known volume, using a volumetric flask.
Titration
  • The volume of one solution which reacts exactly with another is used to calculate unknown concentrations.
Precision
  • Volumetric glassware such as pipettes and burettes gives measurements to a higher precision than beakers.
Sense check
  • A concentration calculation should give a sensible result in the range of typical laboratory solutions.

Part 5 of 6Worked Example

Question

  • 5.00 g of NaOH (Mr = 40.00) is dissolved to make 250 cm³ of solution.
  • Calculate its concentration in mol dm⁻³ and g dm⁻³, and the concentration after 25.0 cm³ is diluted to 100 cm³.

  1. n = 5.00 ÷ 40.00 = 0.125 mol and V = 0.250 dm³.
  2. c = 0.125 ÷ 0.250 = 0.500 mol dm⁻³.
  3. Mass concentration = 0.500 × 40.00 = 20.0 g dm⁻³.
  4. Dilutionc₂ = 0.500 × 25.0 ÷ 100 = 0.125 mol dm⁻³.

Answer 0.500 mol dm⁻³, 20.0 g dm⁻³ and after dilution 0.125 mol dm⁻³

Part 6 of 6Exam Checklist & Tips

5.1.4 – Key Points for the Exam

  • Concentration The amount of solute per unit volume of solution.
  • Dilution Reducing the concentration of a solution by adding solvent.
  • Standard solution A solution of accurately known concentration.
  • ppm Parts per million, used for very small concentrations.

Common Mistakes

  • Using the volume in cm³ without converting to dm³.
  • Using the volume of solvent instead of the volume of the final solution.
  • Confusing g dm⁻³ with mol dm⁻³.
  • Using c₁V₁ = c₂V₂ with volumes in different units on each side.

Last updated:

\n