S5 Chemistry Practice Paper: Acids, Bases, Neutralisation and Salts (12 questions, 100 marks, 75 minutes)
- 2026-11-09
12 questions, 100 marks, 75 minutes on acid-base theories, pH and dilution, neutralisation and its heat, titration calculations, salt preparation, acid rain, antacids and vinegar analysis.
Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.
Show all working. Answers without working earn no method marks. Write down a formula before substituting numbers.
Give numerical answers to three significant figures unless stated otherwise, and state the correct units.
Spend about 30 minutes on Section A, then use the remaining time for the titration and salt preparation questions in Section B and the applications in Section C.
Section A Acid-Base Theory and pH Calculations
Section A: Show your working, with units and appropriate significant figures.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (a) State the Arrhenius definitions of an acid and a base, giving one example of each. (b) Use the reaction between NH3 and H2O to explain the meaning of a Bronsted-Lowry conjugate acid-base pair. (c) Compare solutions of a strong acid and a weak acid of the same concentration in terms of pH and conductivity, and explain the difference. | (a) An Arrhenius acid releases hydrogen ions in water, such as hydrochloric acid; an Arrhenius base releases hydroxide ions in water, such as sodium hydroxide (3 marks). (b) NH3 accepts a proton from H2O to form NH4+, and H2O loses a proton to form OH-; NH4+ and NH3 form one conjugate pair and H2O and OH- form another (3 marks). (c) At the same concentration a strong acid ionises almost completely, giving lower pH and higher conductivity, while a weak acid ionises only partly, giving fewer mobile ions, hence higher pH and lower conductivity (3 marks). |
| 2 | (a) Write the definition of pH and state how many times the hydrogen ion concentration changes when pH falls by 1 unit. (b) Calculate the pH of 0.010 M hydrochloric acid. (c) Calculate the pH and the degree of ionisation of 0.010 M ethanoic acid (Ka = 1.8 × 10^-5). | (a) pH equals minus log10 of the hydrogen ion concentration; a fall of 1 pH unit means a tenfold increase in hydrogen ion concentration (3 marks). (b) With [H+] equal to 0.010 M, pH is 2.00 (3 marks). (c) With degree of ionisation a, Ka equals C times a squared, so a equals the square root of (1.8 × 10^-5 divided by 0.010), about 4.24 per cent; [H+] is 4.24 × 10^-4 M and pH is 3.37 (3 marks). |
| 3 | Source 1: a solution at 25 degrees Celsius has a hydrogen ion concentration of 2.5 × 10^-3 M. (a) Calculate its pH to two decimal places. (b) 10.0 cm3 of the solution is diluted with water to 250 cm3; calculate the new pH. (c) Explain why dilution raises the pH. | (a) pH equals minus log10 of 2.5 × 10^-3, which is 2.60 (3 marks). (b) After dilution [H+] equals 2.5 × 10^-3 times 10.0 divided by 250, which is 1.0 × 10^-4 M, so pH is 4.00 (3 marks). (c) Adding water reduces the number of hydrogen ions per unit volume, so the concentration falls and pH rises (2 marks). |
| 4 | (a) State two differences between measuring pH with a pH meter and with an acid-base indicator. (b) Give the colour-change ranges of methyl orange and phenolphthalein. (c) Explain why methyl orange rather than phenolphthalein should be used when titrating aqueous ammonia with hydrochloric acid. | (a) A pH meter gives a direct numerical reading with higher accuracy, whereas an indicator relies on judging colour and is limited by the observer (3 marks). (b) Methyl orange changes from red to yellow between 3.1 and 4.4; phenolphthalein changes from colourless to pink between 8.3 and 10.0 (3 marks). (c) The equivalence point for ammonia and hydrochloric acid is acidic, near pH 5; phenolphthalein changes only in the alkaline region, so the end point lags, while methyl orange changes in the acidic region close to the equivalence point (3 marks). |
Section B Neutralisation, Titration and Salt Preparation
Section B: Show the calculation for titration data and balance all equations.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | (a) Write the ionic equation for the neutralisation of a strong acid by a strong base. (b) Use the hydration of ions to explain why neutralisation is exothermic. (c) Give two uses of neutralisation in daily life or industry. | (a) H+ plus OH- gives H2O (2 marks). (b) The hydration energy released when the ions are hydrated is smaller than the energy released in forming water molecules, so the net process is exothermic (3 marks). (c) Two uses: lime is used to neutralise acidic industrial effluent; aluminium hydroxide or calcium carbonate is used to neutralise excess stomach acid (3 marks). |
| 6 | Source 2: 25.0 cm3 of sodium hydroxide solution needs 22.5 cm3 of 0.100 M hydrochloric acid to reach the end point. (a) Calculate the molar concentration of the sodium hydroxide solution. (b) Calculate the number of moles of sodium hydroxide in the 25.0 cm3 sample. (c) If 0.050 M sulphuric acid were used instead, calculate the volume required. | (a) c(NaOH) equals 22.5 times 0.100 divided by 25.0, which is 0.0900 M (3 marks). (b) n equals 0.0900 times 25.0 divided by 1000, which is 0.00225 mol (2 marks). (c) Each sulphuric acid molecule supplies two protons, so n(H2SO4) equals 0.00225 divided by 2, or 0.001125 mol, and the volume is 0.001125 divided by 0.050, which is 0.0225 dm3 or 22.5 cm3 (3 marks). |
| 7 | (a) Describe two steps in preparing a soluble salt from an acid and a metal oxide. (b) Describe the steps in preparing an insoluble salt by precipitation, using barium sulphate as the example. (c) State two purposes of washing a precipitate. | (a) Add the metal oxide, such as copper(II) oxide, to excess dilute acid and warm gently; filter off the excess oxide, then evaporate the filtrate to concentrate and cool it to crystallise (3 marks). (b) Mix solutions of a soluble barium salt, such as barium chloride, and a soluble sulphate, such as sodium sulphate, to form a white precipitate of barium sulphate, then filter, wash with distilled water and dry (3 marks). (c) Washing removes soluble impurities adsorbed on the precipitate and prevents them from affecting the mass measurement after drying (2 marks). |
| 8 | Source 3: 25.0 cm3 of 0.0900 M sodium hydroxide solution is completely neutralised by excess hydrochloric acid. (a) Calculate the number of moles of sodium chloride formed. (b) Calculate the mass of sodium chloride formed (Mr = 58.5). (c) Explain two reasons why the mass obtained after evaporation is often slightly higher than the theoretical value. | (a) n(NaCl) equals 0.0900 times 25.0 divided by 1000, which is 0.00225 mol (3 marks). (b) m equals 0.00225 times 58.5, which is 0.1316 g, about 0.132 g (3 marks). (c) Two reasons: the solid may not be completely dry, and other soluble impurities in the filtrate crystallise with the product (3 marks). |
Section C Acid-Base Applications in Life and Industry
Section C: Combine chemical principles with the given context and answer in full sentences.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | (a) Explain two main causes of acid rain, naming the gases involved. (b) Write the equation for sulphur dioxide dissolving in water to form sulphurous acid. (c) Suggest two measures to reduce acid rain. | (a) Burning fossil fuels releases sulphur dioxide, and at high temperatures nitrogen and oxygen in the air combine to form nitrogen oxides (3 marks). (b) SO2 plus H2O gives H2SO3 (2 marks). (c) Two measures: use low-sulphur fuel or scrub flue gas with limestone, and fit catalytic converters to vehicles to cut nitrogen oxide emissions (3 marks). |
| 10 | (a) Compare aluminium hydroxide and sodium hydrogencarbonate as antacids, giving one advantage and one disadvantage of each. (b) Write the equation for the reaction between sodium hydrogencarbonate and hydrochloric acid. (c) Explain why long-term heavy use of sodium hydrogencarbonate is not recommended. | (a) Aluminium hydroxide does not produce gas and causes less bloating but acts slowly; sodium hydrogencarbonate acts quickly but the carbon dioxide produced can cause bloating (3 marks). (b) NaHCO3 plus HCl gives NaCl plus H2O plus CO2 (2 marks). (c) Heavy long-term use raises the hydrogencarbonate level in the body, which can cause alkalosis, and the added sodium load can raise blood pressure (3 marks). |
| 11 | Source 4: a brand of vinegar is labelled 5.0 per cent (w/v), that is 5.0 g of ethanoic acid (Mr = 60.0) per 100 cm3. (a) Calculate the molar concentration of ethanoic acid in the vinegar. (b) Calculate the volume of 0.100 M sodium hydroxide needed to neutralise 10.0 cm3 of the vinegar. (c) Explain why the titrant should be added slowly with constant swirling. | (a) c equals (5.0 divided by 60.0) divided by 0.100, which is 0.833 M (3 marks). (b) n(CH3COOH) equals 0.833 times 10.0 divided by 1000, which is 0.00833 mol, so the volume of sodium hydroxide is 0.00833 divided by 0.100, which is 0.0833 dm3 or 83.3 cm3 (3 marks). (c) Adding slowly with swirling avoids local excess that would make the indicator change colour early, so the end point is closer to the equivalence point (2 marks). |
| 12 | (a) Distinguish between the end point and the equivalence point of a titration. (b) Describe the colour change at the end point when phenolphthalein is used to titrate hydrochloric acid with sodium hydroxide. (c) Explain the effect of an air bubble at the jet of the burette on the calculated concentration, and suggest one improvement. | (a) The equivalence point is the theoretical point at which the acid and base have reacted in exact molar proportions; the end point is where the indicator changes colour and titration actually stops (3 marks). (b) The solution turns from colourless to pale pink, and the colour persists for 30 seconds after swirling (3 marks). (c) The bubble occupies part of the burette, so the volume actually delivered is less than the reading, making the calculated concentration of the base too high; the improvement is to expel the bubble and take a new initial reading before titrating (2 marks). |
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