Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

5.1.4Additional lessons

Concentration Calculations — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 92 of 172

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

Part 1 of 5Expressing Concentration

Square brackets around a formula, such as [HCl], mean the concentration in mol dm⁻³.

Molar concentration
  • Molar concentration c = n ÷ V, where c is in mol dm⁻³, n is amount in moles and V is volume in dm³.
Mass concentration
  • Mass concentration Measured in g dm⁻³, found by dividing the mass of solute by the volume of solution.
Conversion
  • Conversion Concentration in g dm⁻³ equals concentration in mol dm⁻³ multiplied by the molar mass.
Parts per million
  • Parts per million ppm is the mass of solute per million parts by mass of solution, used for very dilute solutions.

Part 2 of 5Dilution

1
Diluting a solution adds solvent, but the number of moles of solute stays unchanged.
2
Because n = cV, the moles before and after dilution are equal: c₁V₁ = c₂V₂.
3
Rearrange to find the unknown concentration or volume, keeping both volumes in the same unit.
4
Prepare the dilution in a volumetric flask by adding solvent up to the graduation mark.
5
Calculate the dilution factor as the final volume divided by the initial volume.

Part 3 of 5Practical Preparation

Practical Preparation
Standard solution
  • Standard solution A solution of accurately known concentration prepared by dissolving a weighed mass in a volumetric flask.
Titration link
  • Titration link Concentration and volume data from titrations give moles of reactants.
Uncertainty
  • Uncertainty Volumetric glassware has set uncertainties that affect the reliability of the concentration.
Units check
  • Units check Always convert cm³ to dm³ by dividing by 1000 before applying c = n ÷ V.

Part 4 of 5Worked Example

Question

  • 5.85 g of sodium chloride (M = 58.44 g mol⁻¹) is dissolved in water to make 250 cm³ of solution.
  • Calculate its concentration, then the concentration after 25.0 cm³ is diluted to 250 cm³.

  1. n = 5.85 ÷ 58.44 = 0.1001 mol; V = 0.250 dm³.
  2. c = 0.1001 ÷ 0.250 = 0.400 mol dm⁻³.
  3. Dilutionc₂ = c₁V₁ ÷ V₂ = 0.400 × 25.0 ÷ 250 = 0.0400 mol dm⁻³.

Answer 0.400 mol dm⁻³, diluted to 0.0400 mol dm⁻³

Part 5 of 5Exam Checklist & Tips

5.1.4 – Key Points for the Exam

  • Concentration The amount of solute per unit volume of solution.
  • Solute The substance dissolved in a solution.
  • Standard solution A solution whose concentration is known accurately.
  • Dilution factor The ratio of final volume to initial volume.

Common Mistakes

  • Using the volume of the solvent instead of the total volume of the solution.
  • Leaving the volume in cm³ in the equation c = n ÷ V.
  • Thinking that dilution changes the number of moles of solute.
  • Mixing up g dm⁻³ and mol dm⁻³ when a question asks for one of them.

Last updated:

\n