GCSE Chemistry Titration questions by topic

Complete the concentration calculations.

1. (a)Convert 250 cm³ to dm³.[1]
2. (b)12 g of solute is dissolved in 200 cm³ of solution. Calculate the concentration in g/dm³.[2]
3. (c)State the unit of concentration used in (b).[1]

A cook dissolves 50 g of sugar in water to make 400 cm³ of syrup.

1. (a)Calculate the concentration in g/dm³.[2]
2. (b)State the mass of sugar in 100 cm³ of this syrup.[1]

Put the titration steps in order and answer the questions.

1. (a)Number these steps 1–4: run in acid dropwise near the colour change · fill the burette with acid · record concordant titres and average · add 25.0 cm³ of alkali with a pipette.[4]
2. (b)Name the piece of apparatus that measures the acid volume used.[1]

A titration table shows: rough 23.90 cm³; run 1: 24.20 cm³; run 2: 24.10 cm³; run 3: 24.20 cm³.

1. (a)Identify which runs are concordant.[1]
2. (b)Calculate the mean titre from the concordant runs.[2]

Why are titrations repeated, and why is the first run treated differently?

1. (a)State the purpose of the first, quick titration.[2]
2. (b)Explain why repeating improves the result.[2]

Complete the table of solution concentrations.

1. (a)Calculate the concentration in row 1.[1]
2. (b)Calculate the concentration in row 2.[1]
3. (c)Calculate the missing mass in row 3.[1]
4. (d)State the relationship between cm³ and dm³ used in these calculations.[1]
5. (e)A student divides by 250 instead of 0.25 in row 1 and writes 0.04 g/dm³. Explain the error.[1]

A student prepares ammonium chloride by titrating ammonia solution against hydrochloric acid. The steps of the method are jumbled.

1. (a)Put these steps in order: P — evaporate the neutral solution to crystallisation point; Q — titrate with indicator to find the reacting volume; R — cool the solution so crystals form; S — repeat the titration with the same volumes but no indicator.[2]
2. (b)Name the salt formed and write the two reactants’ names as a word equation skeleton.[2]
3. (c)Explain why the first titration (with indicator) is not used to make the crystals.[2]

A student makes 250 cm³ of a sodium chloride solution containing 14.6 g of NaCl.

1. (a)Calculate the concentration in g/dm³.[2]
2. (b)Calculate the mass of NaCl needed to make 750 cm³ of a solution of the same concentration.[2]
3. (c)State what happens to the concentration if the same 14.6 g is made up to 500 cm³ instead.[1]

Explain these titration technique choices.

1. (a)Why is the burette rinsed with acid, but the conical flask only with distilled water?[3]
2. (b)Why must the pipette and burette readings be taken at eye level to the bottom of the meniscus?[2]
3. (c)Why are only 2–3 drops of indicator added?[1]

A 250 cm³ solution contains 8.0 g of sodium hydroxide (Na 23, O 16, H 1).

1. (a)Calculate the concentration in g/dm³.[2]
2. (b)Calculate the Mr of NaOH and the concentration in mol/dm³.[3]
3. (c)State how many moles are in 25.0 cm³ of this solution.[1]

A titration finds that 22.0 cm³ of hydrochloric acid neutralises 25.0 cm³ of 0.16 mol/dm³ NaOH.

1. (a)Calculate the moles of NaOH used.[2]
2. (b)Using NaOH + HCl → NaCl + H₂O, calculate the moles and concentration of the acid.[3]
3. (c)State the concentration in mol/dm³ to 2 significant figures.[1]

25.0 cm³ of potassium hydroxide solution is neutralised by 30.0 cm³ of nitric acid in a titration. The alkali is 0.12 mol/dm³ and the reacting ratio is 1 : 1.

1. (a)Calculate the moles of KOH pipetted.[2]
2. (b)State the moles of HNO₃ that reacted.[1]
3. (c)Calculate the concentration of the nitric acid in mol/dm³.[3]

A solution of sulfuric acid has a concentration of 9.8 g/dm³ (H₂SO₄: H 1, S 32, O 16).

1. (a)Calculate the Mr of sulfuric acid and hence its concentration in mol/dm³.[2]
2. (b)Calculate the moles of H₂SO₄ in 25.0 cm³ of the solution.[1]
3. (c)The 25.0 cm³ sample needs 20.0 cm³ of sodium hydroxide solution for complete neutralisation (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O). Calculate the concentration of the alkali in mol/dm³.[3]

25.0 cm³ of sulfuric acid is titrated with 0.10 mol/dm³ sodium hydroxide. The concordant mean titre is 30.0 cm³. (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.)

1. (a)Calculate the moles of NaOH delivered.[2]
2. (b)Calculate the moles and concentration of the sulfuric acid.[3]
3. (c)Express the acid concentration in g/dm³ (H 1, S 32, O 16).[2]

A student titrates vinegar (ethanoic acid) against 0.50 mol/dm³ NaOH. 25.0 cm³ of vinegar needs 20.0 cm³ of the alkali. (CH₃COOH + NaOH → CH₃COONa + H₂O; Mr of ethanoic acid = 60.)

1. (a)Calculate the concentration of the vinegar in mol/dm³.[3]
2. (b)Calculate its concentration in g/dm³.[2]
3. (c)The bottle claims “5% acetic acid” — about 50 g/dm³. Comment on the measured value.[2]

A concentrated 2.0 mol/dm³ acid stock must be diluted to exactly 0.10 mol/dm³, making 500 cm³.

1. (a)Calculate the moles of acid needed in the final solution.[2]
2. (b)Calculate the volume of stock solution required.[2]
3. (c)Describe how to make the dilution safely and accurately.[2]

A student runs five titrations. The rough run is discarded. The remaining titres are 24.60 cm³, 24.50 cm³, 25.90 cm³, 24.55 cm³ and 24.50 cm³.

1. (a)Identify the anomalous titre and justify the choice.[2]
2. (b)Calculate the mean of the remaining titres and the range of the concordant set.[3]
3. (c)Explain why the rough run is excluded and comment on the precision of the result.[3]

The concentration of sulfuric acid is found by titration. 25.0 cm³ of the acid is neutralised by a mean titre of 20.0 cm³ of 0.10 mol/dm³ sodium hydroxide: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

1. (a)Calculate the moles of sodium hydroxide in the mean titre.[2]
2. (b)Calculate the moles of sulfuric acid in the flask.[2]
3. (c)Calculate the concentration of the sulfuric acid in mol/dm³ and in g/dm³ (H 1, S 32, O 16).[2]
4. (d)The student’s rough titre was 19.90 cm³ and they want to use it “because it is close to the mean”. Evaluate this argument.[2]