IGCSE · Practice questions · Concentration and titrations

Measure it, then count it.

Six original Cambridge-style questions on concentration and titrations: units, making a solution, 1 : 1 and 1 : 2 titrations and concordant results. The calculations are Extended content.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Known solution first.

cm³ ÷ 1000 = dm³ before any calculation.
moles = concentration × volume in dm³.
Ratio from the balanced equation; H₂SO₄ reacts 1 : 2 with NaOH.
g/dm³ = mol/dm³ × M_r.

Recall

[2 marks]

State two units in which the concentration of a solution can be given, and say what each one measures.

g/dm³: the mass of solute in grams dissolved in each 1 dm³ of solution. ✓
mol/dm³: the number of moles of solute dissolved in each 1 dm³ of solution. ✓

Calculation

[2 marks]

Extended. Calculate the mass of sodium hydroxide needed to make 250 cm³ of a 0.2 mol/dm³ solution. (A_r: H 1, O 16, Na 23)

Moles = 0.250 dm³ × 0.2 mol/dm³ = 0.05 mol. ✓
M_r(NaOH) = 40, so mass = 0.05 × 40 = 2.0 g. ✓

Calculation

[4 marks]

Extended. 25.0 cm³ of potassium hydroxide solution is neutralised by 22.5 cm³ of 0.100 mol/dm³ hydrochloric acid. KOH + HCl → KCl + H₂O. Calculate the concentration of the potassium hydroxide in mol/dm³ and in g/dm³. (A_r: H 1, O 16, K 39)

Moles of HCl = 0.0225 × 0.100 = 0.00225 mol. ✓
Ratio 1 : 1, so KOH = 0.00225 mol. ✓
Concentration = 0.00225 ÷ 0.0250 = 0.090 mol/dm³. ✓
M_r(KOH) = 56, so 0.090 × 56 = 5.04 g/dm³. ✓

Calculation

[3 marks]

Extended. 20.0 cm³ of 0.050 mol/dm³ sulfuric acid exactly neutralises 25.0 cm³ of sodium hydroxide solution. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the sodium hydroxide.

Moles of H₂SO₄ = 0.0200 × 0.050 = 0.0010 mol. ✓
Ratio 1 : 2, so NaOH = 2 × 0.0010 = 0.0020 mol. ✓
Concentration = 0.0020 ÷ 0.0250 = 0.080 mol/dm³. ✓

Analysis

[2 marks]

A student records these titres: rough 23.40 cm³, then 22.60 cm³, 22.95 cm³ and 22.65 cm³. Choose the titres to use and calculate the mean titre.

Use the concordant titres, those within 0.10 cm³ of each other: 22.60 and 22.65 cm³. Ignore the rough titre and 22.95 cm³. ✓
Mean = (22.60 + 22.65) ÷ 2 = 22.625, which is 22.63 cm³ to 2 decimal places. ✓

Calculation

[3 marks]

Extended. 0.40 g of sodium hydroxide is dissolved in water. Calculate the volume of 0.50 mol/dm³ hydrochloric acid, in cm³, needed to neutralise it. (A_r: H 1, O 16, Na 23)

Moles of NaOH = 0.40 ÷ 40 = 0.010 mol. ✓
Ratio 1 : 1, so 0.010 mol of HCl is needed. ✓
Volume = 0.010 ÷ 0.50 = 0.020 dm³ = 20 cm³. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.