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S5 Chemistry Practice Paper: Titration, Electrolysis and Rates of Reaction
S5 Chemistry: Titration, Electrolysis and Rates of Reaction
S5 Chemistry paper: titration and indicators, salt hydrolysis pH, electrolysis and Faraday calculations, electroplating, rates and rate equations. 12 questions, 100 marks, 75 minutes.
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Paper details Level: S5 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Acids, Bases and Titration (Q1-Q4)
Section A
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A 25.0 cm3 sample of hydrochloric acid of unknown concentration is titrated with 0.100 M sodium hydroxide; the mean titre is 23.40 cm3. (a) Write the net ionic equation for the reaction. (b) Calculate the molar concentration of the acid (4 significant figures). (c) Find the volume of 0.0500 M sodium hydroxide needed to reach the same end point. (d) Explain why methyl orange is preferred to phenolphthalein here and state the colour change at the end point. | (a) H+(aq) + OH-(aq) -> H2O(l). (b) n(NaOH) = 0.100 x 23.40/1000 = 2.340 x 10-3 mol, so [HCl] = 2.340 x 10-3 / 0.0250 = 0.0936 M. (c) V = 2.340 x 10-3 / 0.0500 x 1000 = 46.8 cm3. (d) The equivalence pH of a strong acid – strong base titration is 7; methyl orange (about 3.1-4.4) changes from yellow to orange-red, which is easy to detect, so it is commonly used. Marking: ionic equation, mole conversion, concentration-volume relation, indicator choice. |
| 2 | At 25 degrees C, Ka of ethanoic acid is 1.8 x 10-5 and Kw is 1.0 x 10-14. (a) Calculate the pH of 0.010 M hydrochloric acid (2 decimal places). (b) Calculate the pH of 0.010 M ethanoic acid and explain the difference (use [H+] = sqrt(Ka x c)). (c) The acid is then diluted tenfold; find the change in pH and state whether it equals 1. | (a) The strong acid is fully ionised, [H+] = 0.010 M, pH = 2.00. (b) Ethanoic acid is only partly ionised: [H+] = sqrt(1.8 x 10-5 x 0.010) = 4.24 x 10-4 M, pH = 3.37; most molecules remain undissociated, so [H+] is far below 0.010 M and the pH is higher. (c) After tenfold dilution c = 0.0010 M, [H+] = 1.34 x 10-4 M, pH = 3.87; the change is 0.50, less than 1, because dilution causes more molecules to ionise and partly restores [H+]. Marking: strong vs weak ionisation, approximation, dilution effect. |
| 3 | 25.0 cm3 of 0.100 M ethanoic acid is titrated with 0.100 M sodium hydroxide. (a) Find the volume of alkali required at the equivalence point. (b) Calculate the pH at the equivalence point (use Kb = Kw / Ka, 2 decimal places). (c) Explain why this pH is greater than 7. (d) How would the equivalence pH differ if hydrochloric acid were used? Explain briefly. | (a) n(CH3COOH) = 0.100 x 0.0250 = 2.50 x 10-3 mol, requiring the same amount of NaOH, so the volume is 25.0 cm3. (b) At the equivalence point the total volume is 50.0 cm3 and [CH3COO-] = 0.0500 M; Kb = Kw/Ka = 5.56e-10, [OH-] = sqrt(Kb x c) = 5.27e-06 M, pOH = 5.28, pH = 8.72. (c) Ethanoate hydrolyses: CH3COO- + H2O <=> CH3COOH + OH-, giving an alkaline solution, so pH > 7. (d) With hydrochloric acid both species are strong electrolytes, leaving only NaCl and water, so the pH is 7, lower than the ethanoic acid case. Marking: equivalence volume, Kb calculation, hydrolysis equation, comparison. |
| 4 | A student analyses an impure sample of sodium hydroxide. 1.25 g of the sample is dissolved and made up to 250.0 cm3. A 25.0 cm3 portion requires 23.50 cm3 of 0.100 M hydrochloric acid. (a) Calculate the amount, in mmol, of sodium hydroxide in the 25.0 cm3 portion. (b) Find the mass of sodium hydroxide in the whole sample. (c) Calculate the percentage purity (3 significant figures). (d) State one source of experimental error and whether it makes the purity too high or too low. | (a) n(HCl) = 0.100 x 23.50/1000 = 2.350 mmol, so the portion contains 2.350 mmol of NaOH. (b) 250.0 cm3 is ten times the portion, n = 23.50 mmol, mass = 23.50 x 10-3 x 40.00 = 0.940 g. (c) Percentage purity = 0.940 / 1.25 x 100% = 75.2%. (d) For example, a burette rinsed with distilled water rather than the alkali dilutes the titre, making the calculated purity too high; or a damp sample lowers the true mass of NaOH, making it too low. Marking: mmol conversion, dilution factor, purity formula, error direction. |
Section B: Electrolysis and Electrochemistry (Q5-Q8)
Section B
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A current of 2.00 A is passed through copper(II) sulphate solution for 60.0 minutes (Faraday constant 96500 C mol-1, Cu = 63.5, molar volume of gas 24.0 dm3 mol-1). (a) Write the reactions at the cathode and at the graphite anode. (b) Calculate the quantity of charge passed. (c) Find the mass of copper deposited (3 significant figures). (d) Find the volume of oxygen produced at the anode in cm3 at room conditions. (e) Explain why the blue colour of the solution fades. | (a) Cathode: Cu2+ + 2e- -> Cu; anode: 2H2O -> O2 + 4H+ + 4e-. (b) Q = It = 2.00 x 3600 = 7200 C. (c) n(e-) = 7200/96500 = 0.0746 mol, n(Cu) = 0.0373 mol, mass = 0.0373 x 63.5 = 2.369 g. (d) n(O2) = 0.0746/4 = 0.0187 mol, volume = 0.0187 x 24000 = 448 cm3. (e) Copper(II) ions are continuously removed at the cathode, so their concentration falls and the blue colour fades. Marking: electrode equations, Q = It, Faraday conversion, molar volume. |
| 6 | A current of 1.00 A is passed through 500 cm3 of 0.100 M sodium chloride solution using platinum electrodes. (a) Write the anode reaction for concentrated and for dilute sodium chloride solution, and explain the difference. (b) Find the time, in hours, needed to oxidise all 0.100 mol of chloride ions (2 decimal places). (c) Describe and explain the change in acidity near the cathode during prolonged electrolysis. | (a) Concentrated NaCl: 2Cl- -> Cl2 + 2e-, because the high chloride concentration favours its discharge; dilute NaCl: 2H2O -> O2 + 4H+ + 4e-, because water is more easily oxidised when chloride is dilute. (b) n(e-) = n(Cl-) = 0.100 mol, Q = 0.100 x 96500 = 9650 C, t = 9650/1.00 = 9650 s = 2.68 hours. (c) At the cathode 2H2O + 2e- -> H2 + 2OH-; the hydroxide produced makes the solution near the cathode alkaline, so phenolphthalein turns pink. Marking: concentration and discharge order, charge-amount relation, cathode alkalinity. |
| 7 | An object is silver-plated in a silver electroplating cell using a current of 1.50 A; 0.500 g of silver is deposited (Ag = 107.9, Faraday constant 96500 C mol-1). (a) Write the cathode reaction. (b) Calculate the charge required in coulombs (nearest integer). (c) Find the time taken in seconds and in minutes. (d) Explain why the anode must be pure silver rather than an inert electrode. (e) If the current efficiency is only 90%, by what percentage does the time increase? | (a) Cathode: Ag+ + e- -> Ag. (b) n(Ag) = 0.500/107.9 = 4.63 x 10-3 mol, Q = 4.63 x 10-3 x 96500 = 447 C. (c) t = 447/1.50 = 298 s, about 4.97 minutes. (d) A pure silver anode dissolves (Ag -> Ag+ + e-) and replenishes the electrolyte, keeping its concentration steady; with an inert anode the silver ions would be used up and plating would stop. (e) At 90% efficiency only nine tenths of the charge plates silver, so the time rises by 1/0.90 – 1, about 11%. Marking: reduction at cathode, Faraday calculation, anode replenishment, efficiency. |
| 8 | Acidified water is electrolysed with a current of 1.00 A for 30.0 minutes (Faraday constant 96500 C mol-1, molar volume 24.0 dm3 mol-1). (a) Write the reactions at both electrodes. (b) Calculate the charge passed and the amount of electrons. (c) Find the volume of hydrogen at the cathode in cm3 (3 significant figures). (d) Find the volume of oxygen at the anode and explain the volume ratio. (e) Explain why a little dilute sulphuric acid is added. | (a) Cathode: 2H2O + 2e- -> H2 + 2OH-; anode: 2H2O -> O2 + 4H+ + 4e-. (b) Q = 1.00 x 1800 = 1800 C, n(e-) = 1800/96500 = 0.0187 mol. (c) n(H2) = 0.0187/2 = 9.33 x 10-3 mol, V = 9.33 x 10-3 x 24000 = 224 cm3. (d) n(O2) = 0.0187/4 = 4.66 x 10-3 mol, V = 112 cm3; 4 mol of electrons give 2 mol H2 and 1 mol O2, so the ratio is always 2 : 1. (e) Pure water conducts very poorly; a little sulphuric acid supplies ions for conduction and is not consumed. Marking: both electrode reactions, electron amount, volume ratio, role of electrolyte. |
Section C: Rate of Reaction (Q9-Q12)
Section C
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Marble chips are added to excess dilute hydrochloric acid and the loss in mass is recorded: 0 s / 0.00 g, 40 s / 0.48 g, 80 s / 0.80 g, 120 s / 0.98 g, 160 s / 1.10 g. (a) Calculate the average rate over 0-40 s and 40-80 s in g s-1. (b) Calculate the average rate over 80-120 s and 120-160 s, and explain why the rate falls. (c) Estimate the instantaneous rate at 60 s and state your method. (d) Suggest two ways to speed up the reaction and explain them with collision theory. | (a) 0-40 s: 0.48/40 = 0.0120 g s-1; 40-80 s: 0.32/40 = 0.0080 g s-1. (b) 80-120 s: 0.18/40 = 0.0045 g s-1; 120-160 s: 0.12/40 = 0.0030 g s-1; the rate falls because the acid is used up, so collisions are less frequent. (c) Using the interval 40-80 s, the instantaneous rate at 60 s is about 0.32/40 = 0.0080 g s-1 (a tangent or the mean of adjacent intervals is also acceptable). (d) Raise the temperature (greater kinetic energy, more frequent and more energetic collisions) and use powdered marble (larger surface area); increasing the acid concentration also works. Marking: average rate formula, reason for fall, estimate, collision theory. |
| 10 | Initial rate data for the reaction A + B -> products ([A] and [B] in mol dm-3, rate in mol dm-3 s-1): experiment 1: 0.10 / 0.10 / 2.0 x 10-3; experiment 2: 0.20 / 0.10 / 8.0 x 10-3; experiment 3: 0.10 / 0.20 / 4.0 x 10-3. (a) Determine the order with respect to A and to B, showing your working. (b) Write the rate equation and calculate the rate constant k with its unit. (c) Find the initial rate when [A] = 0.30 M and [B] = 0.15 M. (d) If the volume of the container is halved, by what factor does the rate change? | (a) Comparing 1 and 2: [B] is constant while [A] doubles and the rate rises from 2.0 x 10-3 to 8.0 x 10-3, a factor of 4, so the order in A is 2; comparing 1 and 3: [A] is constant while [B] doubles and the rate doubles, so the order in B is 1. (b) rate = k[A]2[B]; k = 2.0 x 10-3 / (0.10^2 x 0.10) = 2.0 dm6 mol-2 s-1. (c) rate = 2.0 x 0.30^2 x 0.15 = 0.027 mol dm-3 s-1. (d) Halving the volume doubles both concentrations, so the rate becomes 2^2 x 2 = 8 times larger. Marking: orders from data, rate equation and unit, substitution, concentration-volume relation. |
| 11 | (a) Equal masses of powdered calcium carbonate and of marble chips are added separately to the same acid; the initial rates are 4.8 x 10-3 and 1.2 x 10-3 mol s-1. Find the ratio of the rates and explain it with collision theory. (b) Explain how a catalyst speeds up a reaction and state at least two quantities it does not change. (c) For this reaction the rate roughly doubles for every 10 K rise. If the rate at 300 K is r, find the rate at 330 K in terms of r. (d) Explain why temperature affects the rate far more strongly than concentration. | (a) Ratio = 4.8 x 10-3 / 1.2 x 10-3 = 4.0; the powder has a much larger surface area, so more particles are exposed and the number of effective collisions per second is greater. (b) A catalyst provides an alternative pathway of lower activation energy, so a larger fraction of collisions is successful; it does not change the enthalpy change, the position of equilibrium or the final yield. (c) From 300 K to 330 K is three 10 K steps, so the rate is 2^3 r = 8r. (d) Heating greatly increases the fraction of molecules with energy above the activation energy (the distribution shifts right), whereas concentration only raises the collision frequency proportionally. Marking: surface area and collisions, catalyst role, temperature factor, energy distribution. |
| 12 | The decomposition of hydrogen peroxide is first order with rate constant k = 6.93 x 10-3 s-1. (a) Write the equation with state symbols. (b) Find the half-life (0.1 s). (c) Find the time for 80% decomposition (nearest second). (d) Explain why measuring the gas volume can follow the reaction and state one possible source of error. (e) If manganese(IV) oxide is added, state how k and the half-life change. | (a) 2H2O2(aq) -> 2H2O(l) + O2(g). (b) t1/2 = ln2/k = 0.693/6.93 x 10-3 = 100.0 s. (c) 20% remains, t = -ln(0.20)/k = 1.609/6.93 x 10-3 = 232 s. (d) Each 2 mol of H2O2 gives 1 mol of O2, so the gas volume is proportional to the extent of reaction and can be collected and measured; errors include a leaky apparatus or reading the volume before the gas cools to room temperature. (e) Manganese(IV) oxide is a catalyst: it lowers the activation energy, so k increases and the half-life becomes shorter. Marking: equation with states, first-order formulas, percentage calculation, measurement and error, catalyst effect. |
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