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K324 / 2027
Chemical Calculations

Lesson 5 of 6

Concentration and titration

Volume tells you moles only when concentration is known.

O-Level 6092 (2026) / SEC G3 K324 (2027)

  • I can use solution concentration.Syllabus

    Syllabus K324, 4.2(g). Use solution concentration

Concentration and titration

Volume tells you moles only when concentration is known.

Molar concentration c = n/V, with V in dm3. Hence n = cV. Mass concentration in g dm-3 = mass/volume; dividing this by molar mass converts to mol dm-3. Always convert cm3 to dm3 by dividing by 1000 before using n = cV.

Worked exampleA titration with a 1:2 ratio

25.0 cm3 of sulfuric acid requires 20.0 cm3 of 0.150 mol dm-3 NaOH. Find the acid concentration.

Try it first. Then show the worked stepsHide the worked steps
  1. H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l).
  2. n(NaOH) = 0.150 x 0.0200 = 0.00300 mol.
  3. n(acid) = 0.00300/2 = 0.00150 mol.
  4. c(acid) = 0.00150/0.0250 = 0.0600 mol dm-3.
Answer

0.0600 mol dm-3. A direct cV = cV shortcut would miss the factor of two.

Try it step by step

Work it out in steps: Moles of the known solution, then Moles from the ratio, then Unknown concentration.

  1. Problem 1. 20.0 cm3 of hydrochloric acid is neutralised by 22.0 cm3 of 0.100 mol dm-3 sodium hydroxide. NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l). Find the concentration of the acid, in mol dm-3 to 3 significant figures.
  2. Problem 2. 25.0 cm3 of sodium hydroxide is neutralised by 12.5 cm3 of 0.0800 mol dm-3 sulfuric acid. H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l). Find the concentration of the sodium hydroxide, in mol dm-3 to 3 significant figures.
  3. Problem 3. 20.0 cm3 of sodium carbonate solution needs 16.0 cm3 of 0.150 mol dm-3 hydrochloric acid. Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g). Find the concentration of the sodium carbonate. Give each answer to 3 significant figures.
Show the answer

Problem 1.

Moles of the known solution: n(NaOH) = c x V = 0.100 x 22.0/1000 = 0.00220 mol. Divide cm3 by 1000 to get dm3 first.

Moles from the ratio: The ratio is 1 NaOH to 1 HCl, so n(HCl) = 0.00220 mol.

Unknown concentration: c(HCl) = n / V = 0.00220 / (20.0/1000) = 0.110 mol dm-3.

Problem 2.

Moles of the known solution: n(H2SO4) = 0.0800 x 12.5/1000 = 0.00100 mol.

Moles from the ratio: The ratio is 1 H2SO4 to 2 NaOH, so n(NaOH) = 2 x 0.00100 = 0.00200 mol.

Unknown concentration: c(NaOH) = 0.00200 / (25.0/1000) = 0.0800 mol dm-3.

Problem 3.

Moles of the known solution: n(HCl) = 0.150 x 16.0/1000 = 0.00240 mol.

Moles from the ratio: The ratio is 1 Na2CO3 to 2 HCl, so n(Na2CO3) = 0.00240 / 2 = 0.00120 mol.

Unknown concentration: c(Na2CO3) = 0.00120 / (20.0/1000) = 0.0600 mol dm-3.

Worked exampleA titration with a 250/25 dilution step

2.65 g of anhydrous sodium carbonate, Na2CO3, is dissolved in water and made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution needs 22.50 cm3 of hydrochloric acid for neutralisation. Calculate the concentration of the acid. (Mr of Na2CO3 = 106)

Try it first. Then show the worked stepsHide the worked steps
  1. Equation: Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g).
  2. Moles of Na2CO3 in the flask = 2.65 ÷ 106 = 0.0250 mol, in 250 cm3.
  3. Only 25.0 cm3 of the 250 cm3 was titrated, one tenth: 0.0250 × 25.0/250 = 0.00250 mol.
  4. Moles of HCl = 2 × 0.00250 = 0.00500 mol.
  5. Concentration of HCl = 0.00500 ÷ (22.50/1000) = 0.222 mol/dm3 (3 s.f.).
Answer

0.222 mol/dm3. The step students miss is the 25.0/250 factor: the titration uses only one tenth of the solution in the flask.

Try it step by step

Work it out in steps: Moles in the flask, then Moles in the portion titrated, then Moles of acid, then Acid concentration.

  1. Problem 1. 1.06 g of anhydrous sodium carbonate is dissolved and made up to 250 cm3. 25.0 cm3 of this solution needs 20.00 cm3 of hydrochloric acid. Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g). Find the acid concentration, in mol dm-3 to 3 significant figures. Use M(Na2CO3) = 106 g mol-1.
  2. Problem 2. 3.18 g of anhydrous sodium carbonate is made up to 250 cm3. 25.0 cm3 of this solution needs 24.00 cm3 of hydrochloric acid. Use the same equation. Find the acid concentration, in mol dm-3 to 3 significant figures. Use M(Na2CO3) = 106 g mol-1.
  3. Problem 3. 2.12 g of anhydrous sodium carbonate is made up to 250 cm3. 25.0 cm3 of this solution needs 20.00 cm3 of hydrochloric acid. Use the same equation. Give each answer to 3 significant figures. Use M(Na2CO3) = 106 g mol-1.
Show the answer

Problem 1.

Moles in the flask: n(Na2CO3) in the flask = 1.06 / 106 = 0.0100 mol, in 250 cm3.

Moles in the portion titrated: Only 25.0 cm3 of the 250 cm3 was titrated: 0.0100 x 25.0/250 = 0.00100 mol.

Moles of acid: The ratio is 1 Na2CO3 to 2 HCl, so n(HCl) = 2 x 0.00100 = 0.00200 mol.

Acid concentration: c(HCl) = 0.00200 / (20.00/1000) = 0.100 mol dm-3.

Problem 2.

Moles in the flask: n(Na2CO3) in the flask = 3.18 / 106 = 0.0300 mol.

Moles in the portion titrated: In the 25.0 cm3 portion: 0.0300 x 25.0/250 = 0.00300 mol.

Moles of acid: n(HCl) = 2 x 0.00300 = 0.00600 mol.

Acid concentration: c(HCl) = 0.00600 / (24.00/1000) = 0.250 mol dm-3.

Problem 3.

Moles in the flask: n(Na2CO3) in the flask = 2.12 / 106 = 0.0200 mol.

Moles in the portion titrated: In the 25.0 cm3 portion: 0.0200 x 25.0/250 = 0.00200 mol.

Moles of acid: n(HCl) = 2 x 0.00200 = 0.00400 mol.

Acid concentration: c(HCl) = 0.00400 / (20.00/1000) = 0.200 mol dm-3.

In a real titration, calculate titre from final minus initial burette reading. Repeat until sufficiently close titres are obtained and average concordant results, excluding the rough run. For example, 19.85 and 19.95 cm3 average to 19.90 cm3. Use the supplied equation for an unfamiliar reaction; the calculation path stays the same.

Check your understanding

A solution contains 8.0 g of NaOH per dm3. What is its concentration in mol dm-3? (M = 40 g mol-1)

  1. 320 mol dm-3, since 8.0 x 40 = 320
  2. 0.20 mol dm-3, since 8.0 / 40 = 0.20
  3. 5.0 mol dm-3, since 40 / 8.0 = 5.0
Show the answer

0.20 mol dm-3, since 8.0 / 40 = 0.20

Divide the mass concentration by the molar mass: (8.0 g dm-3) / (40 g mol-1) = 0.20 mol dm-3. The grams cancel and leave mol dm-3.

Exam-style question

10.0 cm3 of a concentrated sulfuric acid solution is diluted with water to 250 cm3. 25.0 cm3 of the diluted acid is neutralised by 20.00 cm3 of 0.100 mol/dm3 sodium hydroxide.
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Calculate the concentration of the original concentrated acid, in mol/dm3. [4]

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