Home · A-Level Chemistry 9701 · Atoms, molecules and stoichiometry · Solutions and titrations · Practice
AS · Practice questions · Solutions and titrations

Titrate and calculate.

Six original Cambridge-style questions on solution calculations: concentration, acid-base titrations, a water of crystallisation titration, back titration and practical technique.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

n = c × V, with V in dm3.

01
Calculation
[2 marks]

5.85 g of sodium chloride is dissolved in water and made up to 500 cm3. Calculate the concentration in mol dm−3. (Mr: NaCl 58.5)

  • n = 5.85 ÷ 58.5 = 0.100 mol. ✓
  • c = 0.100 ÷ 0.500 = 0.200 mol dm−3. ✓
02
Calculation
[3 marks]

25.0 cm3 of 0.0500 mol dm−3 sodium carbonate solution is titrated with hydrochloric acid; the mean titre is 22.70 cm3. Na2CO3 + 2HCl → 2NaCl + H2O + CO2. Calculate the concentration of the acid.

  • n(Na2CO3) = 0.0500 × 25.0/1000 = 1.25 × 10−3 mol. ✓
  • n(HCl) = 2 × 1.25 × 10−3 = 2.50 × 10−3 mol. ✓
  • c(HCl) = 2.50 × 10−3 ÷ 0.02270 = 0.110 mol dm−3. ✓
03
Analysis
[4 marks]

1.43 g of hydrated sodium carbonate, Na2CO3·xH2O, is dissolved and made up to 250 cm3. 25.0 cm3 portions need 20.00 cm3 of 0.0500 mol dm−3 HCl. Na2CO3 + 2HCl → 2NaCl + H2O + CO2. Find x. (Mr: Na2CO3 106.0, H2O 18.0)

  • n(HCl) = 0.0500 × 0.02000 = 1.00 × 10−3 mol, so n(Na2CO3) in 25.0 cm3 = 5.00 × 10−4 mol. ✓
  • In 250 cm3: 10 × 5.00 × 10−4 = 5.00 × 10−3 mol. ✓
  • M = 1.43 ÷ 5.00 × 10−3 = 286 g mol−1. ✓
  • 18.0x = 286 − 106.0 = 180, so x = 10. ✓
04
Analysis
[4 marks]

1.50 g of impure calcium carbonate is added to 40.0 cm3 of 1.00 mol dm−3 hydrochloric acid (an excess). The acid remaining needs 15.0 cm3 of 1.00 mol dm−3 sodium hydroxide. Calculate the percentage by mass of calcium carbonate in the sample. (Mr: CaCO3 100.1)

  • HCl added = 1.00 × 0.0400 = 0.0400 mol; HCl left = n(NaOH) = 1.00 × 0.0150 = 0.0150 mol. ✓
  • HCl that reacted with CaCO3 = 0.0400 − 0.0150 = 0.0250 mol. ✓
  • CaCO3 + 2HCl, so n(CaCO3) = 0.0125 mol = 0.0125 × 100.1 = 1.25 g. ✓
  • Percentage = 1.25 ÷ 1.50 × 100 = 83.4%. ✓
05
Calculation
[2 marks]

What volume of 0.100 mol dm−3 sodium hydroxide exactly neutralises 20.0 cm3 of 0.150 mol dm−3 sulfuric acid? H2SO4 + 2NaOH → Na2SO4 + 2H2O.

  • n(H2SO4) = 0.150 × 0.0200 = 3.00 × 10−3 mol, so n(NaOH) = 6.00 × 10−3 mol. ✓
  • V = 6.00 × 10−3 ÷ 0.100 = 0.0600 dm3 = 60.0 cm3. ✓
06
Explain
[3 marks]

Explain why, in a titration, (a) the burette is rinsed with the acid before it is filled, (b) the acid is added dropwise near the end point, and (c) the titration is repeated until concordant titres are obtained.

  • (a) Any water left in the burette would dilute the acid and increase the titre. ✓
  • (b) Adding single drops means the end point is not overshot, so the titre is accurate. ✓
  • (c) Concordant results show the titre is reliable; their mean reduces the effect of random errors. ✓

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.