Mini Chemistry
Part of Mini EducationSwitch subject

Solution concentration and titration calculations

Use mass and molar concentration, conserve solute amount during dilution, and calculate an unknown concentration from a titration and its balanced equation.

  • SEC G3 Pure Chemistry 2027

Reading settings

On this page

A solution’s concentration compares its solute amount or mass with the total solution volume. More solute in the same volume means a higher concentration; the same solute spread through a larger volume means a lower concentration.

Be comfortable with moles and molar mass. Gas molar volume, such as 24 dm³ mol⁻¹ at RTP, applies to gases and must not be used to find the amount in an aqueous solution.

Choose mass concentration or molar concentration

QuantityMeaningEquationUsual unit here
Mass concentration, cₘMass of solute per unit solution volumecₘ = m/Vg dm⁻³
Molar concentration, cAmount of solute per unit solution volumec = n/Vmol dm⁻³

Here m is the solute mass in g, n is its amount in mol, and V is the final solution volume in dm³. This is not necessarily the volume of water added. For example, “dissolve to make 250 cm³ of solution” means the solute and water together have that final volume.

Use 1000 cm³ = 1 dm³: divide a volume in cm³ by 1000 before using concentration in mol dm⁻³ or g dm⁻³.

Rearrange the molar concentration equation to suit the unknown:

n = cV V = n/c

For example, 0.0200 mol of glucose is dissolved to make 100.0 cm³ of solution. The volume is 100.0/1000 = 0.1000 dm³, so its molar concentration is:

c = 0.0200/0.1000 = 0.200 mol dm⁻³

If the solute molar mass is M in g mol⁻¹, then n = m/M. This connects the two concentration measures:

c = cₘ/M cₘ = cM

For an ionic solute, the stated concentration usually refers to its formula amount. For example, 0.100 mol dm⁻³ of dissolved NaCl supplies 0.100 mol dm⁻³ of each of Na⁺ and Cl⁻. Counting all ions together would be a different quantity.

Dilution conserves the solute amount

If you add water without losing solute or causing a reaction, the solute amount stays the same while the total solution volume increases. Using n = cV before and after gives:

c₁V₁ = c₂V₂

The final volume V₂ is the total volume after dilution, not just the added water. Both volumes must use the same unit in this shortcut. To calculate an actual amount using n = cV, use dm³ with mol dm⁻³.

Dilution keeps solute amount unchangedBefore dilution, 0.040 mole of solute in 50.0 cubic centimetres gives 0.80 mole per cubic decimetre. After dilution to a total of 200 cubic centimetres, the same 0.040 mole gives 0.20 mole per cubic decimetre. Each panel shows six identical solute markers; the solution volume increases fourfold.Before adding water50.0 cm³ of solutionn = 0.040 mol in bothc = 0.80 mol dm⁻³c = n/VAfter dilution200 cm³ total solutionn = 0.040 mol in bothc = 0.20 mol dm⁻³c = n/V
Dilution spreads the same amount of a non-ionising solute through four times the volume, so concentration becomes one-quarter as large. Each marker represents the same number of solute molecules; solvent molecules are omitted. The cylinders have the same cross-section, so four times the liquid height represents four times the volume. This is a counting model, not a laboratory procedure or a drawing of molecular sizes.
Dilution is different from a reaction

During dilution, the same solute is spread through more solution. In a titration, one substance reacts with another: use the balanced equation to connect their amounts. Do not assume their amounts or cV values are equal unless the reacting ratio is 1:1.

Use the balanced equation in titration calculations

A titre is the volume delivered from the burette to reach the endpoint. Use the measured titre and known concentration to find one reactant’s amount, then use the equation’s mole ratio to find the other’s amount. Divide by that reactant’s own solution volume to calculate its concentration.

For titration calculations, assume the chosen endpoint corresponds to the stated stoichiometric reaction. For the measurement method, see Titration technique and concordant results.

Worked examples

Guided practice 1

Concentration from mass and final solution volume

About 7 min

Problem

5.60 g of potassium hydroxide, KOH, is dissolved to make 250.0 cm³ of solution. Calculate its molar concentration.

Use Aᵣ(K) = 39, Aᵣ(O) = 16 and Aᵣ(H) = 1.

Complete the mass–mole–concentration chain

Hints

Hint 1: find moles first

Use M(KOH) = 39 + 16 + 1, then n = m/M.

Hint 2: make volume compatible

Convert 250.0 cm³ to 0.2500 dm³ before using c = n/V.

View solution step by step

  1. Find molar mass

    Method

    Add the K, O and H relative masses.

    Reason

    Mass must be converted into moles before concentration can be calculated.

    Working

    M(KOH) = 39 + 16 + 1 = 56 g mol⁻¹.

  2. Find amount of solute

    Method

    Divide mass by molar mass.

    Reason

    n = m/M.

    Working

    n = 5.60/56 = 0.100 mol.

  3. Convert solution volume

    Method

    Divide 250.0 cm³ by 1000.

    Reason

    Concentration in mol dm⁻³ requires volume in dm³.

    Working

    V = 250.0/1000 = 0.2500 dm³.

  4. Calculate concentration

    Method

    Divide moles by solution volume.

    Reason

    c = n/V.

    Working

    c = 0.100/0.250 = 0.400 mol dm⁻³.

Common misconception 2

Check a dilution claim

Find and correct the mistake

Learner response

50.0 cm³ of 0.80 mol dm⁻³ HCl(aq) is diluted to 200 cm³. A student says the concentration becomes four times larger because the volume is four times larger. Explain the error and calculate the new concentration.

Connect volume change to concentration

What remains constant?

View solution step by step

  1. Identify the conserved quantity

    Method

    Keep the moles of HCl unchanged.

    Reason

    Dilution adds solvent, not solute.

    Working

    n(HCl) = c₁V₁ = 0.80 × 0.0500 = 0.040 mol.

  2. Use compatible volume units

    Method

    Convert the final volume to dm³.

    Reason

    Concentration in mol dm⁻³ requires volume in dm³.

    Working

    V₂ = 200/1000 = 0.200 dm³.

  3. Correct the direction

    Method

    Divide the initial concentration by four.

    Reason

    Spreading unchanged moles through four times the volume makes concentration four times smaller.

    Working

    c₂ = n/V₂ = 0.040/0.200 = 0.20 mol dm⁻³.

Examiner practice 3

Find an unknown concentration from a 1:1 titration

5 marks

Examination question

25.0 cm³ of NaOH(aq) is neutralised by 0.100 mol dm⁻³ HCl(aq). The average titre is 20.0 cm³. Find the concentration of NaOH(aq). Reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). [5 marks]

Show conversions, mole ratio and concentration

View solution step by step

  1. Convert acid volume

    1 mark

    Method

    Convert the titre to dm³.

    Reason

    n = cV requires volume in dm³.

    Working

    V(HCl) = 20.0/1000 = 0.0200 dm³.

  2. Find acid moles

    1 mark

    Method

    Multiply acid concentration by acid volume.

    Reason

    n = cV.

    Working

    n(HCl) = 0.100(0.0200) = 0.00200 mol.

  3. Apply mole ratio

    1 mark

    Method

    Transfer the amount through the 1:1 equation.

    Reason

    One mole of H⁺ reacts with one mole of OH⁻.

    Working

    n(NaOH) = 0.00200 mol.

  4. Convert alkali volume

    1 mark

    Method

    Convert 25.0 cm³ to dm³.

    Reason

    The final concentration uses volume in dm³.

    Working

    V(NaOH) = 0.0250 dm³.

  5. Calculate alkali concentration

    1 mark

    Working

    c(NaOH) = 0.00200/0.0250 = 0.0800 mol dm⁻³.

Challenge 4

Convert mass concentration to molar concentration

Minimal support

Representation transfer

A solution contains 5.85 g of sodium chloride per dm³. Find its concentration in mol dm⁻³. Given Aᵣ(Na) = 23, Aᵣ(Cl) = 35.5.

Convert the quantity per dm3

Hints

Hint 1: interpret per dm3

Treat the stated quantity as a 1 dm³ sample containing 5.85 g NaCl.

Hint 2: convert grams to moles

Divide the mass in that 1 dm³ by M(NaCl).

View solution step by step

  1. Calculate molar mass

    Method

    Add the Na and Cl relative masses.

    Reason

    Molar mass converts the mass quantity into moles.

    Working

    M(NaCl) = 23 + 35.5 = 58.5 g mol⁻¹.

  2. Convert the per-volume quantity

    Method

    Divide grams per dm³ by grams per mole.

    Reason

    The gram units cancel, leaving moles per dm³.

    Working

    c = 5.85/58.5 = 0.100 mol dm⁻³.

Try it yourself

Mind stretcher 1: Check the volume unitExtension

Question: A student calculates c = 0.10/250 = 4.0 × 10⁻⁴ mol dm⁻³ for a 250 cm³ solution containing 0.10 mol. Explain what is wrong and give the correct answer.

Show answer

They used 250 cm³ as if it were 250 dm³.

Correct volume: 250 cm³ = 0.250 dm³.

c = 0.10/0.250 = 0.40 mol dm⁻³

Mind stretcher 2: A titration with a different mole ratioExtension

25.0 cm³ of sodium hydroxide solution is neutralised by 15.0 cm³ of 0.100 mol dm⁻³ sulfuric acid:

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

Find the sodium hydroxide concentration. Explain why equating the acid and alkali amounts would give the wrong result.

Show answer

The acid amount is 0.100(15.0/1000) = 0.00150 mol. The equation requires two moles of NaOH per mole of sulfuric acid, so n(NaOH) = 2(0.00150) = 0.00300 mol.

c(NaOH) = 0.00300/(25.0/1000) = 0.120 mol dm⁻³

Equating the reactant amounts would ignore the coefficient 2 and give a concentration half as large.

Mind stretcher 3: Use the final volume, not the water volumeExtension

20.0 cm³ of 0.600 mol dm⁻³ solution is diluted with water to a final volume of 100.0 cm³. Find its new concentration. Would “add 100.0 cm³ of water” describe the same instruction?

Show answer

The solute amount is 0.600(20.0/1000) = 0.0120 mol. The final solution volume is 0.1000 dm³, so c = 0.0120/0.1000 = 0.120 mol dm⁻³.

Adding 100.0 cm³ of water is a different instruction: it does not set the final solution volume to 100.0 cm³. In laboratory preparation, dilute to the specified final volume rather than assuming separately measured solution and water volumes add exactly.

Practise and check

Topic check

See what you know across this topic, then go back to anything you got wrong.

Take the topic check

Syllabus and review details

Last reviewed: