Mini Chemistry
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Solution Concentration

Use amount concentration with volume in dm³ in both directions and combine it with a balanced-equation ratio where required.

  • SEC G3 Combined Science Chemistry component 2027
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Learning objectives

  • apply the concept of solution concentration (in mol/dm3 or g/dm3) to process the results of volumetric experiments (e.g. titration) and to solve simple problems. (appropriate guidance will be provided where unfamiliar reactions such as redox are involved. Calculations on % yield and % purity are not required.)

1. Outcome and prerequisites

By the end, you should be able to use amount concentration in both directions and combine concentration data with a balanced-equation ratio. For amount concentration,

c = n/V, n = cV, V = n/c,

where c is in mol dm⁻³, n in mol and V in dm³. Convert cm³ to dm³ before substitution.

Stay within the stated quantity

Amount concentration in mol dm⁻³ is distinct from mass concentration in g dm⁻³. Convert between them only when a molar mass is supplied or can be calculated and the question asks for that conversion.

2. Modelled example

Modelled example 1

Use the balanced ratio between two solutions

Core

Problem

25.0 cm³ of 0.200 mol dm⁻³ H₂SO₄ neutralises 20.0 cm³ NaOH. Calculate the sodium hydroxide concentration.
View solution step by step

  1. Establish the equation

    Method

    Write and balance the neutralisation equation before using solution data.

    Reason

    The two solutions do not necessarily react 1:1.

    Working

    H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

  2. Convert the known solution volume

    Method

    Divide the acid volume in cubic centimetres by 1000.

    Reason

    Concentration is per cubic decimetre.

    Working

    25.0 cm³ = 0.0250 dm³.

  3. Find known moles

    Method

    Multiply known acid concentration by known acid volume.

    Reason

    The relationship n = cV converts solution data to the amount used by equation coefficients.

    Working

    n(H₂SO₄) = cV = 0.200(0.0250) = 5.00 × 10⁻³ mol.

  4. Apply the equation ratio

    Method

    Multiply acid amount by the alkali-to-acid coefficient ratio 2:1.

    Reason

    One mole of acid reacts with two moles of alkali.

    Working

    n(NaOH) = 2(5.00 × 10⁻³) = 1.00 × 10⁻² mol.

  5. Use the target solution volume

    Method

    Convert alkali volume to cubic decimetres and divide alkali amount by that volume.

    Reason

    Concentration belongs to the target solution, so both quantities in c = n/V must describe sodium hydroxide.

    Working

    20.0 cm³ = 0.0200 dm³, so c = 0.0100/0.0200 = 0.500 mol dm⁻³.

3. Guided practice

Guided practice 2

Choose the direction from the requested quantity

About 6 min

Problem

A solution contains 0.0750 mol solute in 250 cm³. Find its concentration. Then find the volume of this solution that contains 0.0300 mol solute.

Show each volume unit

Hints

Hint 1: volume unit
Convert 250 cm³ to 0.250 dm³.
Hint 2: reverse direction
When amount and concentration are known, use V = n/c.
View solution step by step

  1. Concentration

    Method

    Convert volume to cubic decimetres and divide amount by volume.

    Reason

    Amount concentration is amount per cubic decimetre, so c = n/V.

    Working

    250 cm³ = 0.250 dm³; c = 0.0750/0.250 = 0.300 mol dm⁻³.

  2. Required volume

    Method

    Divide the requested amount by the concentration, then convert to cubic centimetres.

    Reason

    Volume is now unknown, so rearrange to V = n/c.

    Working

    V = n/c = 0.0300/0.300 = 0.100 dm³ = 100 cm³.

4. Plausible error contrast

A learner calculates moles in 40.0 cm³ of 0.150 mol dm⁻³ solution as n = 0.150/0.0400 = 3.75 mol. The first invalid step is reversing n = cV. Correctly,

n = (0.150 mol dm⁻³)(0.0400 dm³) = 6.00 × 10⁻³ mol.

The units multiply to mol, and a small dilute sample should not contain several moles.

5. Changed-context transfer

Mind stretcher 1: Select the full pathway without a method cueExtension

30.0 cm³ of 0.120 mol dm⁻³ BaCl₂(aq) reacts completely with Na₂SO₄(aq). Determine the amount of sodium sulfate required and the minimum volume of 0.0900 mol dm⁻³ sodium sulfate solution. Establish the equation yourself.

Show feedback

The equation is BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq). Barium chloride amount is 0.120(0.0300) = 3.60 × 10⁻³ mol. The 1:1 ratio requires the same sodium sulfate amount. Its volume is n/c = 0.00360/0.0900 = 0.0400 dm³ = 40.0 cm³.

6. Independent evidence

25.0 cm³ of sulfuric acid is exactly neutralised by 20.0 cm³ of 0.150 mol dm⁻³ potassium hydroxide solution. Without a named method, construct the balanced equation with states, convert the known solution data to amount, apply the coefficient ratio and calculate the sulfuric acid concentration. Include volume conversions, units and a final magnitude check.

Use the mixed Chemical Calculations practice to choose among all six pathways without a seventh core lesson.