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S4 Chemistry Practice Paper: Acids, Bases, Salts and Electrolysis (12 questions, 100 marks, 75 minutes)
S4 Chemistry: Acids, Bases, Salts and Electrolysis
S4 Chemistry paper: titration and pH, salt preparation, electrolysis, water of crystallisation and Faraday calculations. 12 questions, 100 marks, 75 minutes, with answers.
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Paper details Level: S4 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A (Q1–Q4): Acids, Bases and Salt Preparation (28 marks)
Section A
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. A beaker contains 250.0 cm3 of sodium hydroxide solution of concentration 0.200 mol/dm3 (relative molecular mass 40.0). (a) Calculate the number of moles and the mass of sodium hydroxide in the solution. (b) The whole solution is diluted with water to 500.0 cm3. Calculate the new concentration. (c) What volume of the diluted solution contains 0.0200 mol of sodium hydroxide? | (a) n = cV = 0.200 x 0.250 = 0.0500 mol; m = nM = 0.0500 x 40.0 = 2.00 g. (3 marks) (b) The amount is unchanged on dilution, so c = 0.0500 / 0.500 = 0.100 mol/dm3. (2 marks) (c) V = n / c = 0.0200 / 0.100 = 0.200 dm3 = 200.0 cm3. (2 marks) Marking note: part (a) 1 mark for moles, 1 for mass, 1 for units. |
| 2 | 2. Hydrochloric acid of concentration 0.120 mol/dm3 is used to titrate 25.0 cm3 of sodium hydroxide solution of unknown concentration. Three titres gave 24.10 cm3, 23.95 cm3 and 24.00 cm3 of acid. (a) State which titres are concordant and calculate the mean titre. (b) Calculate the amount of hydrochloric acid used and the concentration of the alkali. (c) If the original alkali solution had a volume of 250.0 cm3, calculate the mass of sodium hydroxide it contained. | (a) 23.95 and 24.00 differ by 0.05 cm3 and are concordant; 24.10 is discordant and is discarded. Mean = (23.95 + 24.00) / 2 = 23.975 cm3. (2 marks) (b) n(HCl) = 0.120 x 0.023975 = 0.0028770 mol; from NaOH + HCl -> NaCl + H2O the ratio is 1:1, so c(NaOH) = 0.0028770 / 0.025 = 0.115 mol/dm3. (3 marks) (c) m = 0.115 x 0.250 x 40.0 = 1.15 g. (2 marks) Marking note: part (b) 1 mark for the equation or ratio, 1 for substitution, 1 for the answer. |
| 3 | 3. An acidic solution X has a hydrogen ion concentration of 4.0e-04 mol/dm3. (a) Calculate the pH of solution X. (b) X is diluted 100 times with water. Calculate the new pH and explain the change. (c) Which indicator should be used for the titration of ethanoic acid with sodium hydroxide? State the colour change at the end point and explain your choice. | (a) pH = -log(4.0e-04) = 3.40. (2 marks) (b) After a 100-fold dilution the hydrogen ion concentration becomes 4.0e-06 mol/dm3, so pH = 3.40 + 2 = 5.40; dilution lowers the hydrogen ion concentration and raises the pH. (2 marks) (c) Phenolphthalein: ethanoic acid and sodium hydroxide give sodium ethanoate, which is weakly alkaline (pH about 8-9) and falls inside the phenolphthalein range (pH 8.3-10.0); the end point is colourless to pink. (3 marks) Marking note: 1 mark for the name, 1 for the colour change, 1 for the reason. |
| 4 | 4. Three salts are prepared by different routes. (a) Preparation of copper(II) sulphate from copper(II) oxide: write the equation and outline the three main steps from the starting materials to the dry solid. (b) Preparation of silver chloride from silver nitrate solution and sodium chloride solution: name the type of reaction and the method used to separate the solid. (c) Explain why copper(II) sulphate cannot be prepared by titrating an acid with a base until the indicator changes colour. | (a) CuO + H2SO4 -> CuSO4 + H2O. Add excess copper(II) oxide to warm dilute sulphuric acid with stirring, filter off the unreacted oxide, evaporate the filtrate until crystals appear, then cool, filter and dry the crystals. (3 marks) (b) Precipitation (double decomposition): AgNO3 + NaCl -> AgCl(s) + NaNO3; filter, wash with distilled water and dry. (2 marks) (c) Titration only works for an acid with a soluble base or carbonate in one phase, where an indicator shows the end point. Copper(II) oxide is an insoluble solid, so the reaction happens at a solid surface, no indicator can signal the end point and excess oxide is hard to avoid; adding excess solid and filtering is used instead. (2 marks) |
Section B (Q5–Q8): Electrolysis and Chemical Calculations (32 marks)
Section B
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 5. Copper(II) sulphate solution is electrolysed using copper electrodes at a current of 0.50 A for 40 minutes. (a) Write the half-equation at the anode and at the cathode. (b) Describe the observations at both electrodes and in the solution. (c) Calculate the mass of copper deposited at the cathode (Cu = 63.5, F = 96500 C/mol). (d) How would the product at the anode change if graphite electrodes were used? | (a) Anode: Cu -> Cu2+ + 2e-; cathode: Cu2+ + 2e- -> Cu. (2 marks) (b) The anode dissolves and loses mass; a reddish-brown deposit forms on the cathode, which gains mass; the blue colour of the solution stays about the same because the copper ions dissolving at the anode equal those removed at the cathode. (2 marks) (c) Q = It = 0.50 x 2400 = 1200 C; n(e-) = 1200 / 96500 = 0.0124 mol; n(Cu) = 0.0124 / 2 = 0.0062 mol; m = 0.0062 x 63.5 = 0.395 g, about 0.40 g. (3 marks) (d) With graphite the anode does not dissolve; hydroxide ions are discharged instead and oxygen is produced (4OH- -> O2 + 2H2O + 4e-), while copper ions continue to be removed, so the blue colour fades. (1 mark) Marking note: part (c) 1 mark for charge, 1 for moles of electrons, 1 for the mass. |
| 6 | 6. Concentrated sodium chloride solution (brine) is electrolysed using graphite electrodes. (a) State the product and the half-equation at each electrode. (b) Describe a test for the product at the anode and the expected observation. (c) Explain why the solution left behind is alkaline. (d) If very dilute sodium chloride solution were used, what would be produced at the anode? Explain. | (a) Anode: chlorine, 2Cl- -> Cl2 + 2e-; cathode: hydrogen, 2H2O + 2e- -> H2 + 2OH-. (3 marks) (b) Hold damp starch-iodide paper near the anode: it turns blue-black; damp blue litmus paper is first bleached after turning red, which also confirms chlorine. (2 marks) (c) Hydroxide ions are produced at the cathode while chloride ions are removed at the anode, so the remaining sodium and hydroxide ions form sodium hydroxide and the solution becomes alkaline. (1 mark) (d) Oxygen is produced at the anode: in a very dilute solution the chloride concentration is too low, so hydroxide ions are discharged more easily, 4OH- -> O2 + 2H2O + 4e-. (2 marks) Marking note: part (a) 1 mark per half-equation, 1 mark for correct product names. |
| 7 | 7. 5.00 g of hydrated copper(II) sulphate (CuSO4.5H2O) is heated in an evaporating dish to constant mass (Cu = 63.5, S = 32.1, O = 16.0, H = 1.0). (a) Calculate the relative molecular mass of CuSO4.5H2O and the percentage by mass of water. (b) Calculate the theoretical mass of anhydrous copper(II) sulphate left and the mass of water lost. (c) Describe the colour change on heating and how to show that the gas given off is water vapour. | (a) M(CuSO4.5H2O) = 63.5 + 32.1 + 4 x 16.0 + 5 x 18.0 = 249.6; percentage of water = 90.0 / 249.6 x 100% = 36.1%. (3 marks) (b) M(anhydrous) = 159.6; m = 5.00 x 159.6 / 249.6 = 3.20 g; water lost = 5.00 – 3.20 = 1.80 g. (2 marks) (c) The blue crystals turn into a white powder; after cooling, colourless droplets condense on the dish, and white anhydrous copper(II) sulphate turns blue when touched with them, showing that the gas is water vapour. (3 marks) Marking note: part (a) 1 mark for the molar mass, 2 for the percentage. |
| 8 | 8. A student determined the concentration of a sodium hydroxide solution by titration and obtained a mass of 1.21 g, while the correct value calculated from the data is 1.15 g. (a) Calculate the percentage error of this experiment. (b) The student rinsed the burette with distilled water only, not with the acid. Would the result be too high or too low? Explain. (c) State two other faulty techniques and the direction of their effect on the result. | (a) Percentage error = (1.21 – 1.15) / 1.15 x 100% = 5.2%. (2 marks) (b) Too high. Water left in the burette dilutes the acid, so its real concentration is below the stated value and more volume is needed to reach the end point; the calculated amount and mass of sodium hydroxide are therefore too large. (3 marks) (c) Any two, for example: an air bubble in the jet before titration escapes during it, so the reading is too large and the result too high; reading the meniscus with the eye above the level gives a smaller reading and a result that is too low; adding acid past the end point without stopping makes the result too high. (3 marks, 1.5 each) Marking note: no credit for naming a mistake without its direction of effect. |
Section C (Q9–Q12): Integrated Applications and Experimental Design (40 marks)
Section C
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | 9. 25.0 cm3 of sodium carbonate solution of concentration 0.100 mol/dm3 is titrated with hydrochloric acid of concentration 0.150 mol/dm3 until reaction is complete. (a) Write the full equation for the reaction. (b) Calculate the amount of sodium carbonate, the amount of acid required and the volume of acid. (c) Calculate the volume of carbon dioxide released at room temperature and pressure (molar volume 24.0 dm3/mol). (d) Should methyl orange or phenolphthalein be used? State the colour change at the end point. | (a) Na2CO3 + 2HCl -> 2NaCl + H2O + CO2. (2 marks) (b) n(Na2CO3) = 0.100 x 25.0/1000 = 0.0025 mol; the ratio is 1:2, so n(HCl) = 0.0050 mol; V(HCl) = 0.0050 / 0.150 = 0.033 dm3 = 33.3 cm3. (4 marks) (c) V(CO2) = 0.0025 x 24.0 = 0.060 dm3 = 60.0 cm3. (2 marks) (d) Methyl orange: at the end point the solution is acidic (pH about 4) because of dissolved carbon dioxide and traces of acid, so methyl orange changes from yellow to orange-red; phenolphthalein would change at pH 8.3, long before the reaction is complete. (2 marks) Marking note: part (b) 2 marks for the ratio, 1 for the moles of acid, 1 for the volume. |
| 10 | 10. Silver is used as the anode and an iron plate as the cathode for electroplating at a current of 2.0 A for 30 minutes. (a) State which metal ion the electrolyte must contain and write the half-equation at the cathode. (b) Calculate the mass of silver deposited (Ag = 108, F = 96500 C/mol). (c) Give two reasons why the actual mass deposited is less than the theoretical value. | (a) The electrolyte must contain silver ions, for example silver nitrate solution; cathode: Ag+ + e- -> Ag. (3 marks) (b) Q = 2.0 x 1800 = 3600 C; n(e-) = 3600 / 96500 = 0.0373 mol; n(Ag) = 0.0373 mol; m = 0.0373 x 108 = 4.03 g. (4 marks) (c) Any two: some energy drives side reactions such as hydrogen evolution at the cathode; the silver layer is loose and some metal falls off; current efficiency is below 100% because of resistance heating. (3 marks) Marking note: part (b) 1 mark for charge, 2 for moles of electrons and the 1:1 ratio, 1 for the mass. |
| 11 | 11. Sodium hydroxide solution of concentration 0.100 mol/dm3 is added dropwise to 25.0 cm3 of hydrochloric acid of the same concentration. The pH after adding 0, 5, 10, 15, 20, 22, 24, 25, 26 and 30 cm3 is 1.2, 1.4, 1.7, 2.0, 2.4, 3.0, 4.6, 7.0, 10.4 and 12.0 respectively. (a) Describe the shape of the titration curve and identify its steepest part. (b) Find the volume of alkali at the equivalence point and explain why the pH there is 7. (c) Calculate the pH jump between 24 cm3 and 26 cm3. (d) Which indicator should be used? State the colour change at the end point. | (a) The curve rises slowly at first (pH 1.2 to about 2.4), shoots up near the equivalence point with an almost vertical section near 25.0 cm3, then flattens towards the pH of a strong alkali. (2 marks) (b) Equivalence volume = 25.0 cm3; the acid and alkali have then reacted exactly, giving sodium chloride and water, and neither ion hydrolyses, so the pH is 7. (2 marks) (c) pH jump = 10.4 – 4.6 = 5.8. (2 marks) (d) Phenolphthalein: its range pH 8.3-10.0 lies inside the jump, and the end point is colourless to pink. (2 marks) Remaining 2 marks: showing that at 25.0 cm3 the acid and alkali contain equal amounts (0.0025 mol each), so c1V1 = c2V2 can be used as a check. Marking note: the steep section must be mentioned for full marks in (a). |
| 12 | 12. A student determines the purity of a sample of washing soda (mainly sodium carbonate, relative formula mass 106.0). A 1.00 g sample is dissolved and made up to 250.0 cm3; 25.0 cm3 of this solution needs 18.00 cm3 of 0.100 mol/dm3 hydrochloric acid. (a) Explain why the sample is made up to 250.0 cm3 and only 25.0 cm3 is titrated, rather than titrating the whole sample at once. (b) Calculate the amount of sodium carbonate in 25.0 cm3 of the sample solution. (c) Calculate the mass of sodium carbonate in the sample and its percentage purity. (d) Suggest one improvement for accuracy and one safety precaution. | (a) Titrating the whole sample would need a very large volume of acid and the end point would be hard to see; making the solution up to a fixed volume and titrating a fixed portion allows repeat titrations and a mean value, and the result can be scaled up by proportion using c1V1 = c2V2. (2 marks) (b) n(HCl) = 0.100 x 18.00/1000 = 0.0018 mol; from Na2CO3 + 2HCl -> 2NaCl + H2O + CO2, n(Na2CO3) = 0.0018 / 2 = 0.0009 mol. (3 marks) (c) In 250.0 cm3, n = 0.0009 x 10 = 0.0090 mol; m = 0.0090 x 106.0 = 0.954 g; percentage purity = 0.954 / 1.00 x 100% = 95.4%. (3 marks) (d) Improvement: measure the 25.0 cm3 with a pipette and repeat the titration, taking the mean of concordant titres; safety: wear safety spectacles, since hydrochloric acid is corrosive and splashes must be washed off with plenty of water. (2 marks) Marking note: part (c) 1 mark for the dilution factor, 1 for the molar mass, 1 for the percentage. |
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