5.1.4Additional lessons
Concentration Calculations — IB Chemistry SL Revision Notes & Topical Questions
5 min read 6 parts Lesson 62 of 98
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 6Expressing Concentration
Concentration describes how much solute is dissolved in a given volume of solution.
Molar concentration
- 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 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
- 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
- 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
- Idea Adding solvent changes the concentration but not the amount of solute in the sample.
Equation
- Equation c₁V₁ = c₂V₂, where the volumes must be in the same units on both sides.
Procedure
- Procedure Pipette the required volume of stock solution into a volumetric flask and fill to the mark with distilled water.
Link
- Link A tenfold increase in the volume gives a tenfold decrease in the concentration.
| Quantity | Formula |
|---|---|
| Moles | n = cV |
| Mass concentration | ρ = c × M |
| Dilution | c₁V₁ = c₂V₂ |
Part 4 of 6Preparing and Using Solutions
Preparing and Using Solutions
Standard solution
- Standard solution Prepared by dissolving an accurately weighed mass of solute in a known volume, using a volumetric flask.
Titration
- Titration The volume of one solution which reacts exactly with another is used to calculate unknown concentrations.
Precision
- Precision Volumetric glassware such as pipettes and burettes gives measurements to a higher precision than beakers.
Sense check
- 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.
- n = 5.00 ÷ 40.00 = 0.125 mol and V = 0.250 dm³.
- c = 0.125 ÷ 0.250 = 0.500 mol dm⁻³.
- Mass concentration = 0.500 × 40.00 = 20.0 g dm⁻³.
- 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.
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