Titration and Solubility Exam Questions
Practice IGCSE Chemistry exam questions on titration and solubility. Covers titration procedures, titration calculations, solubility rules, solubility curves, and interpreting titration data with mark schemes and examiner notes.
Published by IGCSEChemistry.com.my
Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
These questions cover titration and solubility. Show all working in calculations.
Question 1 (4 marks, Core)
Describe the procedure for carrying out a titration between sodium hydroxide solution and hydrochloric acid to find the exact volume of acid needed to neutralise the alkali.
Mark scheme
- Measure a known volume of sodium hydroxide solution using a pipette into a conical flask [1]
- Add a few drops of a suitable indicator (e.g. methyl orange or phenolphthalein) [1]
- Fill a burette with hydrochloric acid and record the starting volume [1]
- Add the acid from the burette to the flask, swirling, until the indicator changes colour (endpoint). Record the final burette reading and calculate the volume of acid used [1]
Examiner note: Use a pipette (not a measuring cylinder) for accuracy. Swirl the flask while adding acid. The titre is the volume of acid added (final reading - initial reading). Repeat to get concordant results (within 0.10 cm3 of each other).
Question 2 (4 marks, Supplement)
25.0 cm3 of 0.10 mol/dm3 sodium hydroxide solution is exactly neutralised by 20.0 cm3 of sulfuric acid.
NaOH + H2SO4 → Na2SO4 + H2O (unbalanced)
(a) Balance the equation. (1)
(b) Calculate the concentration of the sulfuric acid. (3)
Mark scheme
(a) 2NaOH + H2SO4 → Na2SO4 + 2H2O [1]
(b) Moles of NaOH = (25.0/1000) x 0.10 = 0.0025 mol [1] Mole ratio NaOH : H2SO4 = 2 : 1, so moles of H2SO4 = 0.0025/2 = 0.00125 mol [1] Concentration of H2SO4 = 0.00125 / (20.0/1000) = 0.00125/0.02 = 0.0625 mol/dm3 [1]
Examiner note: The mole ratio from the balanced equation is critical — NaOH : H2SO4 is 2:1, not 1:1. Many candidates forget to divide by 2 and get double the correct answer. Always balance the equation first.
Question 3 (3 marks, Core)
Using the solubility curve below (described in words), answer the following:
At 20 degC, the solubility of potassium nitrate is 32 g per 100 g water. At 60 degC, the solubility is 110 g per 100 g water. At 80 degC, the solubility is 170 g per 100 g water.
(a) How much potassium nitrate dissolves in 100 g of water at 60 degC? (1)
(b) If a saturated solution at 80 degC is cooled to 20 degC, how much solid crystallises out from 100 g of water? (1)
(c) Define the term saturated solution. (1)
Mark scheme
(a) 110 g [1]
(b) 170 - 32 = 138 g [1]
(c) A solution in which no more solute can dissolve at that temperature / a solution that contains the maximum amount of dissolved solute at a given temperature [1]
Examiner note: When a saturated solution is cooled, the excess solute that can no longer remain dissolved crystallises out. The mass that crystallises = solubility at high temperature - solubility at low temperature.
Question 4 (4 marks, Supplement)
A student titrates 25.0 cm3 of potassium hydroxide solution with 0.20 mol/dm3 hydrochloric acid. The results are:
| Titration | Initial reading (cm3) | Final reading (cm3) | Titre (cm3) |
|---|---|---|---|
| Rough | 0.00 | 23.50 | 23.50 |
| 1 | 0.50 | 22.90 | 22.40 |
| 2 | 1.20 | 23.55 | 22.35 |
| 3 | 0.00 | 22.40 | 22.40 |
(a) Identify the concordant results and calculate the mean titre. (2)
(b) Calculate the concentration of the KOH solution. (2)
Mark scheme
(a) Concordant results: titrations 1, 2, and 3 (22.40, 22.35, 22.40) — all within 0.10 cm3 [1] Mean titre = (22.40 + 22.35 + 22.40) / 3 = 22.38 cm3 (accept 22.40) [1]
(b) Moles of HCl = (22.38/1000) x 0.20 = 0.004476 mol KOH + HCl → KCl + H2O, ratio 1:1 Moles of KOH = 0.004476 mol [1] Concentration = 0.004476 / (25.0/1000) = 0.179 mol/dm3 (accept 0.18) [1]
Examiner note: The rough titration is always excluded from the average. Only use concordant results (within 0.10 cm3 of each other). Calculate the titre for each run as final - initial reading, not from the final reading alone.
Question 5 (3 marks, Core)
Explain why the following salts are insoluble in water:
(a) Barium sulfate (1)
(b) Lead(II) chloride (1)
And state a salt that IS soluble despite containing the sulfate ion. (1)
Mark scheme
(a) Barium sulfate is insoluble because it is an exception to the general solubility of sulfates [1]
(b) Lead(II) chloride is insoluble because it is an exception to the general solubility of chlorides [1]
(c) Any soluble sulfate, e.g. sodium sulfate, copper sulfate, magnesium sulfate, zinc sulfate [1]
Examiner note: Solubility rules must be memorised. Most sulfates are soluble EXCEPT BaSO4, PbSO4, (and CaSO4 is slightly soluble). Most chlorides are soluble EXCEPT AgCl and PbCl2.
Question 6 (3 marks, Supplement)
Explain why solubility generally increases with temperature for solid solutes but decreases with temperature for gaseous solutes.
Mark scheme
- For solids: higher temperature provides more energy to break apart the solute lattice / to overcome intermolecular forces, allowing more solute to dissolve [1]
- For gases: higher temperature gives gas molecules more kinetic energy, allowing them to escape from the solution surface more easily [1]
- Gas molecules move too fast at higher temperatures to remain dissolved / the kinetic energy exceeds the attractive forces holding them in solution [1]
Examiner note: This explains why hot water holds less dissolved oxygen than cold water (relevant to aquatic ecosystems) and why heating a fizzy drink causes it to go flat (CO2 escapes).
Question 7 (4 marks, Supplement)
20.0 cm3 of 0.050 mol/dm3 barium chloride solution is added to 30.0 cm3 of sodium sulfate solution. A white precipitate forms.
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
(a) Calculate the moles of BaCl2 used. (1)
(b) What is the minimum concentration of Na2SO4 needed to precipitate all the barium ions? (3)
Mark scheme
(a) Moles of BaCl2 = (20.0/1000) x 0.050 = 0.001 mol [1]
(b) Mole ratio BaCl2 : Na2SO4 = 1:1, so moles of Na2SO4 needed = 0.001 mol [1] Volume of Na2SO4 = 30.0/1000 = 0.030 dm3 [1] Concentration = 0.001/0.030 = 0.033 mol/dm3 [1]
Examiner note: This combines precipitation with concentration calculations. The 1:1 ratio means equal moles of each reagent are needed. If less Na2SO4 is used, not all Ba2+ will be precipitated.
What to revise if you scored below 5
If titration calculations were difficult, memorise the four steps: moles of known → mole ratio → moles of unknown → concentration. If solubility rules were forgotten, use the mnemonic and practise predicting solubility for 10 different salts. Revisit the titration and solubility notes.
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