Mini Chemistry
Part of Mini EducationSwitch subject

Solution Concentration and Dilution

Use n = cV and the dilution rule, converting volume units correctly.

  • GCE A-Level H2 Chemistry 9476-2027

Reading settings

On this page

Most stoichiometry questions use solutions somewhere, such as in titrations, ionic equations or solution preparation. Start by converting V to dm³, then use n = cV. Use c₁V₁ = c₂V₂ only for dilution, when no reaction occurs.

Build on the mole and Avogadro constant, using your course’s Stoichiometry topic navigation to review them when needed. Keep unit conversions and mole ratios together in your working.

What this page is really testing

  • Can you choose the right equation for the scenario (n = cV vs c₁V₁ = c₂V₂)?
  • Can you keep units consistent, especially cm³ to dm³?
  • Can you state clearly when a question is dilution-only versus reaction stoichiometry?

Definitions (Must Know)

A. Amount concentration

The amount concentration is:

c = n/V

where c is in mol dm⁻³ and V is in dm³.

Key Ideas (What Earns Marks)

  • Always convert volume to dm³: 1000 cm³ = 1 dm³.
  • For dilution of the same solute: c₁V₁ = c₂V₂.
  • Moles in solution: n = cV.
Quick Recall (Unit Check)
  • V(dm³) = V(cm³)/1000
  • n = cV (only if V is in dm³)
  • c₁V₁ = c₂V₂ is for dilution only (no reaction)

Dilution: Concentration Falls as Volume Increases (Example)

Dilution: Concentration Falls as Volume Increases (Example). c vs V (same moles) plotted as Concentration against Final volume.

Scroll across the graph to read all labels.

Dilution: Concentration Falls as Volume Increases (Example). c vs V (same moles) plotted as Concentration against Final volume.Dilution: Concentration Falls as Volume Increases (Example). c vs V (same moles) plotted as Concentration against Final volume.
Example: 20.0 cm^3 of 1.50 mol dm^-3 contains 0.0300 mol solute; after dilution, c = n/V so increasing final volume lowers concentration.
Open full-size graph
View figure data
Values for Dilution: Concentration Falls as Volume Increases (Example)
Final volume (cm^3)c vs V (same moles)
201.5
500.6
1000.3
2500.12
5000.06

Detailed Explanations

A. Unit discipline

If a volume is given in cm³, convert:

V(dm³) = V(cm³)/1000

Mini example:

  • 25.0 cm³ = 0.0250 dm³

B. Using n = cV (workflow)

  1. Convert V to dm³.
  2. Calculate moles: n = cV.
  3. If needed, rearrange to find c or V.

C. Dilution equation (why it works + workflow)

Because dilution adds water but does not change moles of solute, n₁ = n₂.

Using n = cV: c₁V₁ = c₂V₂

Use c₁V₁ = c₂V₂ only when:

  • the solute is the same (no reaction), and
  • concentration units match on both sides, and
  • volume units match on both sides.

Mini example:

  • 20.0 cm³ of 1.50 mol dm⁻³ diluted to 250 cm³ gives c₂ = (1.50 × 20.0)/250 = 0.120 mol dm⁻³.

D. Converting between g dm⁻³ and mol dm⁻³

If a concentration is given in g dm⁻³, convert using molar mass:

c(mol dm⁻³) = (mass concentration (g dm⁻³))/(M(g mol⁻¹))

Mini example:

  • 8.00 g dm⁻³ of NaOH is 8.00/40.0 = 0.200 mol dm⁻³.

E. Worked setup method for dilution prep questions

For “how would you prepare…” prompts, this sequence is fast and mark-safe:

  1. Calculate the stock volume required using c₁V₁ = c₂V₂.
  2. Pipette that stock volume into a volumetric flask.
  3. Add distilled water to the calibration mark.
  4. Stopper and invert several times to mix thoroughly.

This avoids the common “beaker + top-up” method that loses accuracy marks.

Worked Examples

Modelled example 1

Calculate Amount in a Solution Aliquot

Core

Problem

How many moles of NaOH are in 25.0 cm³ of 0.200 mol dm⁻³ solution?
Study the worked solution

  1. Convert the volume

    Method

    Divide the volume in cubic centimetres by 1000.

    Reason

    The concentration unit is per cubic decimetre, so V must be expressed in dm ³.

    Working

    V = 25.0/1000 = 0.0250 dm ³

  2. Apply the concentration relationship

    Method

    Use n = cV.

    Reason

    Concentration multiplied by solution volume gives solute amount.

    Working

    n = (0.200)(0.0250) = 0.00500 mol

Guided practice 2

Calculate a Diluted Concentration

About 5 min

Problem

20.0 cm³ of 1.50 mol dm⁻³ HCl is diluted to 250 cm³. Find the new concentration.

Try this before viewing the solution

Unit: mol dm⁻³

Hints

Hint 1: identify what is conserved

Dilution adds water but does not change the amount of HCl.

Hint 2: use matching volume units

Because both volumes are in cm ³, use them directly in c₁V₁ = c₂V₂.

View solution step by step

  1. Use the dilution relationship

    Method

    Set c₁V₁ = c₂V₂.

    Reason

    The solute is unchanged and no reaction occurs.

    Working

    (1.50)(20.0) = c₂(250)

  2. Rearrange

    Method

    Divide by the final volume.

    Reason

    The tenfold-plus volume increase reduces concentration in the same proportion.

    Working

    c₂ = (1.50)(20.0)/250 = 0.120 mol dm⁻³

Common misconception 3

Correct the Cubic-centimetre Error

Find and correct the mistake

Learner attempt

For 25.0 cm³ of 0.200 mol dm⁻³ NaOH, a learner writes n = cV = (0.200)(25.0) = 5.00 mol. Identify the first error and correct the amount.

Diagnose before recalculating

Required volume unit in n = cV

View solution step by step

  1. Correct the volume unit

    Method

    Convert 25.0 cm ³ to 0.0250 dm ³.

    Reason

    Using cubic centimetres directly makes the result too large by a factor of 1000.

    Working

    V = 25.0/1000 = 0.0250 dm ³

  2. Recalculate

    Method

    Multiply concentration by the converted volume.

    Reason

    The units now cancel to moles.

    Working

    n = (0.200)(0.0250) = 0.00500 mol

Examiner practice 4

Prepare a Dilution Accurately

4 marks

Problem

Describe how to prepare 500 cm³ of 0.100 mol dm⁻³ NaCl from 2.00 mol dm⁻³ stock solution. Include the stock volume required. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Calculate stock volume

    1 mark

    Method

    Rearrange c₁V₁ = c₂V₂ for V₁.

    Reason

    The amount of sodium chloride transferred from stock equals the amount in the final solution.

    Working

    V₁ = (0.100)(500)/(2.00) = 25.0 cm³

  2. Transfer accurately

    1 mark

    Method

    Pipette 25.0 cm³ stock into a 500 cm³ volumetric flask.

    Reason

    Volumetric apparatus measures the required aliquot and final volume accurately.

    Working

    Use a volumetric pipette and flask.

  3. Dilute to the mark

    1 mark

    Reason

    The calibration mark defines the final 500 cm³ solution volume.

    Working

    Add distilled water to the calibration mark.

  4. Mix the solution

    1 mark

    Method

    Stopper and invert the flask several times.

    Reason

    Mixing makes the concentration uniform throughout the flask.

    Working

    The prepared solution is 0.100 mol dm⁻³.

Challenge 5

Dilute a Mass-concentration Stock

Minimal support

Problem

A stock solution contains 9.80 g dm⁻³ H₂SO₄. A 25.0 cm³ aliquot is diluted to 250 cm³. Calculate the final concentration in mol dm⁻³. Use M(H₂SO₄) = 98.0 g mol⁻¹.

Try this before viewing the solution

Hints

Hint 1: convert the concentration representation

Divide the mass concentration by molar mass.

Hint 2: identify the dilution factor

The final volume is ten times the aliquot volume.

View solution step by step

  1. Convert to molar concentration

    Method

    Divide grams per cubic decimetre by grams per mole.

    Reason

    The dilution calculation requires the stock concentration on a molar basis.

    Working

    c₁ = 9.80/98.0 = 0.100 mol dm⁻³

  2. Apply the dilution

    Method

    Use c₂ = c₁V₁/V₂.

    Reason

    The same sulfuric acid amount is spread through ten times the volume.

    Working

    c₂ = (0.100)(25.0)/250 = 0.0100 mol dm⁻³

Common Mistakes

  • Plugging V in cm³ directly into n = cV without converting.
  • Using c₁V₁ = c₂V₂ when a reaction happens (it is for dilution only).
  • Mixing mol dm⁻³ with mol L⁻¹ without stating they are equivalent.

Use the topic check in Practise and check below to practise and check your understanding.

Exam Tips

  • Write the conversion line: “25.0 cm³ = 0.0250 dm³”.
  • If the question asks for moles first, do n = cV before any ratios.

A. Phrase-level wording reminders

  • “Using n = cV with volume in dm³, the amount is … mol.”
  • “c₁V₁ = c₂V₂ is valid here because this is dilution of the same solute.”
  • “The limiting reagent is …, so its moles are used for the concentration calculation.”
  • “Answer to 3 s.f. to match given concentration/volume data.”

Mind Stretchers

Connect This To

  • Titration Calculations in your course’s topic navigation for concentration calculations with mole ratios.
  • Reacting Masses and Limiting Reagent in your course’s topic navigation for choosing the correct mole basis in mixed questions.
  • Further H2 practical study: Data Tables, Graphs, and Uncertainty for preparation-method accuracy language in practical writeups.

Practice Route

  1. Solve one n = cV and one c₁V₁ = c₂V₂ question back-to-back and compare method lines.
  2. Complete the topic check in Practise and check below for speed and unit discipline.
  3. Check command-word phrasing in Exam Skills.

Mind stretcher 1Extension

A student prepares 500 cm³ of 0.100 mol dm⁻³ Na₂CO₃ solution. How many grams of Na₂CO₃ are needed? (M(Na₂CO₃) = 106 g mol⁻¹)

Show Hint

For a dilution, moles of the same solute are unchanged. For a reaction, calculate each reactant amount separately.

Show Answer

Mark scheme:

  • V = 500/1000 = 0.500 dm³
  • n = cV = 0.100 × 0.500 = 0.0500 mol
  • Mass = nM = 0.0500 × 106 = 5.30 g

Mind stretcher 2: Concentration after neutralisationExtension

Question. 25.0 cm³ of 0.200 mol dm⁻³ HCl is mixed with 20.0 cm³ of 0.150 mol dm⁻³ NaOH. Find the concentration of excess H⁺, assuming additive volumes.

Show Hint

Find both amounts, subtract using the 1:1 equation, then divide by the total volume.

Show Answer

n(H⁺) = 0.200(0.0250) = 0.00500 mol and n(OH⁻) = 0.150(0.0200) = 0.00300 mol. Excess H⁺ = 0.00200 mol in 0.0450 dm³, so [H⁺] = 0.0444 mol dm⁻³.

Syllabus and review details

Last reviewed: