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S5 Chemistry Practice Paper: Acids, Bases, Titration and Electrochemistry (12 Questions, 100 Marks, 75 Minutes, Answers Included)
S5 Chemistry: Acids, Bases, Titration, Buffers and Electrochemistry
S5 Chemistry paper: 12 questions, 100 marks, 75 minutes on pH, ionisation, titration, buffers, electrolysis and enthalpy.
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Paper details Level: S5 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Acids, Bases and pH (24 marks)
Each question carries 6 marks; show your working or reasoning.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | At 25 C: (a) calculate the pH of 0.0100 M hydrochloric acid, a strong acid that is fully ionised; (b) calculate the pH of 0.0100 M sodium hydroxide solution; (c) equal volumes of the two solutions are mixed, find the pH and state whether the mixture is acidic, neutral or alkaline. | (a) Hydrochloric acid is fully ionised, so [H+] = 0.0100 M and pH = -log 0.0100 = 2.00. (b) [OH-] = 0.0100 M, pOH = 2.00 and pH = 14.00 – 2.00 = 12.00. (c) Equal volumes of equal concentrations contain equal amounts of H+ and OH-, which neutralise completely to water, so the pH is 7.00 at 25 C and the mixture is neutral. Marking points: full ionisation, pH-pOH conversion, pH after neutralisation. |
| 2 | Ethanoic acid, 0.100 M, has Ka = 1.8 x 10^-5 at 25 C, with Ka much smaller than the concentration. (a) Calculate [H+] from Ka and the concentration. (b) Calculate the pH. (c) Calculate the percentage ionisation and explain why the pH is still below 7. | (a) Ka = [H+]^2/[HA] is approximately [H+]^2/0.100, so [H+] = sqrt(1.8e-5 x 0.100) = 1.34e-03 M. (b) pH = -log 1.34e-03 = 2.87. (c) Percentage ionisation = 1.34e-03/0.100 x 100% = 1.34%, so about 98.7% of the molecules remain unionised; the small amount of H+ released keeps the pH below 7 but far above that of a strong acid of the same concentration, pH 1.00. Marking points: Ka approximation, square root, percentage and comparison. |
| 3 | (a) 25.0 cm3 of 0.200 M hydrochloric acid is diluted to 250.0 cm3; find the new concentration and pH. (b) How much does the pH rise when the acid is diluted ten times? (c) Explain why continued dilution can never raise the pH of a strong acid above 7. | (a) New concentration = 0.200 x 25.0/250.0 = 0.020 M and pH = -log 0.020 = 1.70. (b) The original pH is -log 0.200 = 0.70, so a tenfold dilution raises the pH to 1.70, a rise of 1.00. (c) Dilution lowers [H+] but the self-ionisation of water supplies H+ at about 1e-7 M, so the pH approaches 7 without ever exceeding it. Marking points: dilution formula, pH difference, self-ionisation of water. |
| 4 | 25.0 cm3 of 0.100 M sulphuric acid, a fully ionised diprotic acid, reacts with 0.200 M sodium hydroxide. (a) Find the amount of sulphuric acid. (b) Find the volume of sodium hydroxide needed. (c) If 0.100 M sodium hydroxide were used instead, how would the volume change? Support with a calculation. | (a) n(H2SO4) = 0.100 x 25.0/1000 = 2.50e-03 mol. (b) The diprotic acid supplies 5.00e-03 mol of H+, so the volume of 0.200 M NaOH needed is 5.00e-03/0.200 = 25.0 cm3. (c) With 0.100 M NaOH the concentration is halved and the volume doubled to 50.0 cm3, because the amount of H+ is unchanged and the volume is inversely proportional to the concentration. Marking points: diprotic stoichiometry, volume calculation, inverse relation. |
Section B: Titration, Salts and Buffers (36 marks)
Each question carries 9 marks; note the marks for each part.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 25.0 cm3 of a monoprotic acid HA of unknown concentration is titrated with 0.100 M sodium hydroxide; the end point requires 23.40 cm3. (a) Find the concentration of HA. (b) If the sample came from a 250.0 cm3 volumetric flask, find the total amount of HA in the flask. (c) If the molar mass of HA is 60.0 g/mol, find the mass of HA in the flask. | (a) n(NaOH) = 0.100 x 23.40/1000 = 2.340e-03 mol; the acid is monoprotic so n(HA) is the same and c(HA) = 2.340e-03/0.0250 = 0.0936 M. (b) The flask holds 250.0 cm3, ten times the aliquot, so n(HA) = 0.0234 mol. (c) m = 0.0234 x 60.0 = 1.40 g. Marking points: titration stoichiometry, dilution factor, mass calculation. |
| 6 | 50.0 cm3 of 1.00 M hydrochloric acid and 50.0 cm3 of 1.00 M sodium hydroxide are mixed in a calorimeter; the temperature rises from 24.2 C to 31.0 C. Assume the solution density is 1.00 g/cm3, the specific heat capacity is 4.18 J g-1 K-1 and heat loss to the calorimeter is negligible. (a) Find the heat released in J. (b) Find the amount of water formed. (c) Find the enthalpy change of neutralisation in kJ/mol and explain the meaning of its negative sign. | (a) The total mass of solution is about 100 g, so Q = mcDT = 100 x 4.18 x 6.8 = 2842 J, about 2.84 kJ. (b) There is 0.0500 mol of each reactant, so 0.0500 mol of water forms. (c) DH = -2.84/0.0500 = -56.8 kJ/mol; the negative sign means heat is released, about 56.8 kJ for every mole of water formed. Marking points: Q = mcDT, limiting reagent, enthalpy per mole and sign. |
| 7 | (a) State whether 0.10 M NH4Cl, 0.10 M NaCl and 0.10 M CH3COONa are acidic, neutral or alkaline at 25 C, with one ionic equation for each. (b) Equal volumes of 0.10 M ethanoic acid and 0.10 M sodium ethanoate are mixed; find the pH of the buffer (Ka = 1.8 x 10^-5). (c) Explain why the pH changes very little when a small amount of strong acid is added. | (a) NH4Cl is acidic: NH4+ + H2O <=> NH3 + H3O+; NaCl is neutral because neither ion hydrolyses; CH3COONa is alkaline: CH3COO- + H2O <=> CH3COOH + OH-. (b) After mixing, each solute is 0.050 M, so [HA] = [A-] and [H+] = Ka, giving pH = pKa = 4.74. (c) Added H+ combines with CH3COO- to form a little CH3COOH, so the H+ is consumed; the ratio [A-]/[HA] hardly changes, and by pH = pKa + log([A-]/[HA]) the pH stays almost constant. Marking points: hydrolysis of each salt, buffer pH = pKa, buffer mechanism. |
| 8 | (a) 30.0 cm3 of 0.100 M hydrochloric acid is mixed with 20.0 cm3 of 0.100 M sodium hydroxide; find the pH. (b) Find the pH if 30.0 cm3 of sodium hydroxide is used instead. (c) Hence explain why the pH of a strong acid-strong base titration curve changes abruptly near the equivalence point, and state which type of indicator should be used. | (a) n(H+) = 30.0 x 0.100/1000 = 3.000e-03 mol and n(OH-) = 2.000e-03 mol, so 1.000e-03 mol of H+ remains in 50.0 cm3, giving [H+] = 0.020 M and pH = 1.70. (b) With 30.0 cm3 of NaOH the amounts are equal, neutralisation is complete and the pH is 7.00. (c) Before equivalence a single drop of excess acid leaves a relatively large amount of free H+, and after equivalence a single drop of excess base leaves free OH-, so [H+] changes by several orders of magnitude and the curve is nearly vertical; phenolphthalein or methyl orange, which change over pH 3-10, is suitable. Marking points: excess calculation, equivalence pH, indicator choice. |
Section C: Electrochemistry and Energy Changes (40 marks)
Each question carries 10 marks; give full working or explanations.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A copper(II) sulphate solution is electrolysed with 2.0 A for 30.0 minutes using copper electrodes (Cu = 63.5, F = 96500 C/mol). (a) Find the charge passed and the mass gained at the cathode. (b) Write the half-equations at the cathode and the anode. (c) If inert electrodes were used instead, what would form at the anode? Write its half-equation. | (a) Q = It = 2.0 x 1800 = 3600 C, n(e-) = 3600/96500 = 0.0373 mol and n(Cu) = 0.0187 mol, so the mass gained is 0.0187 x 63.5 = 1.18 g. (b) Cathode: Cu2+ + 2e- -> Cu; anode with copper electrodes: Cu -> Cu2+ + 2e-. (c) With inert electrodes water is oxidised at the anode: 2H2O -> O2 + 4H+ + 4e-, so oxygen is produced. Marking points: Faraday calculation, both half-equations, active versus inert anode. |
| 10 | Dilute sulphuric acid is electrolysed with 1.00 A for 965 s, giving hydrogen at one electrode and oxygen at the other (F = 96500 C/mol, molar volume 24.0 dm3/mol at room temperature). (a) Find the charge passed and the amount of electrons. (b) Find the amount and volume of hydrogen at the cathode. (c) Find the volume of oxygen at the anode and explain why the volume ratio is 2 : 1. | (a) Q = 1.00 x 965 = 965 C and n(e-) = 965/96500 = 0.010 mol. (b) Cathode: 2H+ + 2e- -> H2, so n(H2) = 0.0050 mol and its volume is 0.0050 x 24000 = 120 cm3. (c) Anode: 2H2O -> O2 + 4H+ + 4e-, so n(O2) = 0.0025 mol and its volume is 60 cm3; every 4 mol of electrons gives 2 mol of H2 and 1 mol of O2, so the volume ratio is always 2 : 1. Marking points: charge, electrode stoichiometry, volume conversion. |
| 11 | In a zinc-copper cell, a zinc strip dips into 1 M ZnSO4 and a copper strip into 1 M CuSO4, joined by a salt bridge. E(Zn2+/Zn) = -0.76 V and E(Cu2+/Cu) = +0.34 V. (a) Calculate the cell electromotive force. (b) Identify the anode and cathode, the electron flow in the external circuit and the direction of ion movement in the salt bridge. (c) Write the overall cell reaction and explain how the mass of the zinc strip and the blue colour of the copper(II) sulphate solution change as the cell discharges. | (a) EMF = E(cathode) – E(anode) = (+0.34) – (-0.76) = 1.10 V. (b) Zinc is the anode and is oxidised, copper is the cathode; electrons flow from zinc to copper through the external circuit, while Cl- moves towards the zinc half-cell and K+ towards the copper half-cell to maintain electrical neutrality. (c) Overall reaction: Zn + Cu2+ -> Zn2+ + Cu; the zinc strip dissolves and loses mass while Cu2+ is used up, so the blue colour fades. Marking points: EMF, electrode identification, salt-bridge ion flow, overall reaction and observations. |
| 12 | Burning 2.00 g of methane (CH4, molar mass 16.0 g/mol) releases 111 kJ of heat. (a) Find the enthalpy of combustion of methane in kJ/mol. (b) How much energy is needed to heat 500 g of water from 20 C to 80 C (c = 4.18 J g-1 K-1)? (c) Assuming 100% efficiency, what mass of methane must be burned? | (a) n(CH4) = 2.00/16.0 = 0.1250 mol, so DHc = -111/0.1250 = -888 kJ/mol. (b) Q = mcDT = 500 x 4.18 x 60 = 125400 J = 125.4 kJ. (c) n(CH4) = 125.4/888 = 0.1412 mol and the mass needed is 2.26 g. Marking points: enthalpy of combustion, specific heat capacity formula, energy balance. |
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