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Form 6 Chemistry Practice Paper: Acid-Base Equilibrium, Redox and Electrochemistry
Form 6 Chemistry – 12 questions – 100 marks – 75 minutes
Form 6 Chemistry paper: 12 questions, 100 marks, 75 minutes on pH of strong and weak acids, buffers, titration, electrolysis, oxidation numbers, cells, redox titration and Ksp, with answers.
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Paper details Form 6 Chemistry practice paper with 12 structured questions on acid-base equilibrium, redox and electrochemistry; full answers and two printable PDFs.
Section A: Acid-Base Equilibrium and Titration (Q1-Q5)
Answer all questions; show the expression, substitution and unit for every calculation.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A 0.0100 M hydrochloric acid solution is a strong acid that ionises completely. (a) Find its pH at 25 degrees Celsius. (b) Find the pH of 0.0100 M sodium hydroxide solution (Kw = 1.0 times 10 to the power -14). (c) Find the pH of 0.0100 M barium hydroxide solution and explain why it differs from (b). |
(a) [H+] = 0.0100 M, so pH = -log 0.0100 = 2.00. (b) [OH-] = 0.0100 M, so pOH = 1.70 and pH = 14 – 1.70 = 12.30. (c) Each barium hydroxide unit gives two hydroxide ions, so [OH-] = 0.020 M, pOH = 1.40 and pH = 12.00. At the same molarity it supplies twice as many hydroxide ions, so its pH is higher. Marking notes: pH calculation (3 marks); doubled hydroxide concentration (3 marks); explanation (3 marks). |
| 2 | A weak monoprotic acid HA has Ka = 1.8 times 10 to the power -5. For a 0.100 M solution at 25 degrees Celsius: (a) Find the hydrogen ion concentration by solving the quadratic equation, without assuming a small degree of dissociation. (b) Find the pH of the solution. (c) Find the degree of dissociation of HA as a percentage. |
(a) Let [H+] = x, then Ka = x squared divided by (0.100 – x) = 1.8 times 10 to the power -5, which gives x squared + 1.8e-5 x – 1.8e-6 = 0. The positive root is x = 1.333e-03 M. (b) pH = -log(1.333e-03) = 2.88. (c) Degree of dissociation = x divided by 0.100 times 100 per cent = 1.33 per cent. Marking notes: quadratic set-up (4 marks); pH (3 marks); degree of dissociation (2 marks). |
| 3 | 50.0 cubic centimetres of 0.100 M hydrochloric acid is mixed with 50.0 cubic centimetres of 0.040 M sodium hydroxide solution. Assume the volumes are additive. (a) Find the number of moles of hydrogen ions in excess after mixing. (b) Find the hydrogen ion concentration and the pH of the mixture. (c) If the sodium hydroxide were 0.100 M instead, would the final pH be exactly 7? Explain briefly. |
(a) H+ = 50.0 divided by 1000 times 0.100 = 5.00e-03 mol; OH- = 50.0 divided by 1000 times 0.040 = 2.00e-03 mol. Excess H+ = 5.00e-03 – 2.00e-03 = 3.00e-03 mol. (b) Total volume = 100.0 cubic centimetres = 0.100 cubic decimetres, so [H+] = 3.00e-03 divided by 0.100 = 0.030 M and pH = 1.52. (c) No. Equal moles of a strong acid and a strong base neutralise to water and sodium chloride, giving pH 7, but that only holds when the moles are equal; here the acid is in excess. Marking notes: moles (3 marks); concentration and pH (3 marks); neutral point reasoning (2 marks). |
| 4 | 0.125 M hydrochloric acid is used to titrate 25.0 cubic centimetres of sodium hydroxide solution; 20.0 cubic centimetres of the acid are needed to reach the end point. (a) Find the concentration of the sodium hydroxide solution. (b) Find the number of moles and the mass of sodium hydroxide in the 250.0 cubic centimetre volumetric flask (M(NaOH) = 40.0). (c) State the colour change at the end point when phenolphthalein is used. |
(a) Moles of HCl = 20.0 divided by 1000 times 0.125 = 2.50e-03 mol. Hydrogen ions and hydroxide ions react in a 1 to 1 ratio, so the moles of sodium hydroxide are the same. [NaOH] = 2.50e-03 divided by 0.0250 = 0.100 M. (b) Moles in 250.0 cubic centimetres = 0.100 times 0.250 = 0.025 mol; mass = 0.025 times 40.0 = 1.00 g. (c) Phenolphthalein is pink in alkali and turns colourless at the end point as the solution becomes acidic. Marking notes: concentration (4 marks); moles and mass (3 marks); indicator colour (1 mark). |
Section B: Redox, Electrolysis and Cells (Q6-Q9)
Data: pH = -log[H+], pOH = -log[OH-] and pH + pOH = 14 at 25 degrees Celsius; one faraday is 96500 C.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 0.10 M ethanoic acid (Ka = 1.8 times 10 to the power -5) is mixed with an equal volume of 0.10 M sodium ethanoate to form a buffer. (a) Find the pH of this buffer. (b) 0.010 mol of hydrochloric acid is added to 1.0 cubic decimetre of the buffer, with no volume change. Find the new pH. (c) Compare (a) and (b) and explain how a buffer works. |
(a) Equal volumes of equal concentrations give [CH3COOH] = [CH3COO-] = 0.050 M, a ratio of 1, so pH = pKa + log(1) = 4.74. (b) The added H+ reacts with CH3COO-, giving 0.11 mol of acid and 0.09 mol of salt in 1.0 cubic decimetre. pH = 4.74 + log(0.09 divided by 0.11) = 4.66. (c) The pH falls only from 4.74 to 4.66, a change of about 0.09. A buffer contains a weak acid and its conjugate base, so added H+ is taken up by the conjugate base and added OH- is neutralised by the weak acid, leaving the pH almost unchanged. Marking notes: initial pH (3 marks); pH after adding acid (4 marks); buffer reasoning (2 marks). |
| 6 | A current of 2.0 A passes through copper(II) sulphate solution for 30.0 minutes (96500 C per mole of electrons). (a) Find the quantity of electricity passed. (b) Find the mass of copper deposited at the cathode (M(Cu) = 63.5). (c) How many minutes are needed to deposit 3.175 g of copper? |
(a) Q = I times t = 2.0 times 1800 = 3600 C. (b) Moles of electrons = 3600 divided by 96500 = 0.03731 mol. Since Cu2+ plus 2e- gives Cu, moles of copper = 0.01865 mol and the mass = 0.01865 times 63.5 = 1.18 g. (c) 3.175 g of copper is 0.0500 mol, needing 0.1000 mol of electrons, that is 9650 C, so t = 9650 divided by 2.0 = 4825 s = 80.4 minutes. Marking notes: charge (2 marks); mass (4 marks); time (2 marks). |
| 7 | (a) Find the oxidation number of manganese in MnO4-. (b) Find the oxidation number of chromium in Cr2O7 2- and state the highest common oxidation state of chromium. (c) Find the oxidation numbers of the two nitrogen atoms in NH4NO3 and explain why they differ. |
(a) Let Mn be x: x + 4(-2) = -1, so x = +7. (b) Let Cr be x: 2x + 7(-2) = -2, so 2x = +12 and x = +6. Plus six is the highest common oxidation state of chromium, as in chromates and dichromates. (c) In NH4+ the nitrogen is x + 4(+1) = +1, giving -3; in NO3- it is x + 3(-2) = -1, giving +5. They differ because one nitrogen is bonded to hydrogen while the other is bonded to the more electronegative oxygen, which draws the electrons towards itself. Marking notes: manganese (3 marks); chromium (3 marks); the two nitrogen atoms (2 marks). |
| 8 | A zinc-copper cell is written as Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). Given E standard (Cu2+/Cu) = +0.34 V and E standard (Zn2+/Zn) = -0.76 V: (a) Find the standard cell potential. (b) State which electrode is the anode, whether oxidation or reduction happens there, and write that half-equation. (c) If the copper electrode is replaced by silver (E standard (Ag+/Ag) = +0.80 V), what is the new cell potential? |
(a) E cell = E cathode minus E anode = +0.34 – (-0.76) = 1.10 V. (b) Zinc is the anode and is oxidised: Zn(s) gives Zn2+(aq) plus 2e-. Copper is the cathode and is reduced: Cu2+(aq) plus 2e- gives Cu(s). (c) E cell = +0.80 – (-0.76) = 1.56 V. The electrode with the higher potential is the cathode, so silver is the cathode and zinc remains the anode. Marking notes: cell potential (3 marks); anode and half-equation (4 marks); the silver case (2 marks). |
Section C: Quantitative Analysis and Solubility (Q10-Q12)
Use a molar volume of 24.0 cubic decimetres per mole; give percentages to three significant figures with units.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | In acidic solution permanganate oxidises iron(II): MnO4- + 8H+ + 5Fe2+ gives Mn2+ + 5Fe3+ + 4H2O. (a) Explain why 1 mol of MnO4- can oxidise 5 mol of Fe2+. (b) 25.0 cubic centimetres of an iron(II) solution needs 24.0 cubic centimetres of 0.0200 M KMnO4 to reach the end point. Find the concentration of Fe2+. (c) Find the number of moles of Fe2+ in that 25.0 cubic centimetre portion. |
(a) Manganese falls from +7 to +2, so each MnO4- accepts five electrons, while each Fe2+ releases one electron in going from +2 to +3. Electrons lost must equal electrons gained, so the ratio is 1 to 5. (b) Moles of MnO4- = 0.0240 times 0.0200 = 4.80e-04 mol, so Fe2+ is 5 times that, 2.40e-03 mol. [Fe2+] = 2.40e-03 divided by 0.0250 = 0.0960 M. (c) The number of moles is 2.40e-03 mol, about 2.40 millimoles. Marking notes: electron balance (3 marks); concentration (4 marks); moles (1 mark). |
| 10 | 1.20 g of impure calcium carbonate is dissolved in 50.0 cubic centimetres of 0.500 M hydrochloric acid, which reacts with calcium carbonate in a 2 to 1 ratio. The excess acid is neutralised by 25.0 cubic centimetres of 0.500 M sodium hydroxide. (a) Find the moles of hydrochloric acid that reacted with the calcium carbonate. (b) Find the mass of calcium carbonate in the sample (M(CaCO3) = 100.1). (c) Find the percentage purity of the sample. |
(a) Total acid = 0.0500 times 0.500 = 0.0250 mol; used by the alkali = 0.0250 times 0.500 = 0.0125 mol. Acid that reacted with the carbonate = 0.0125 mol. (b) CaCO3 plus 2HCl gives CaCl2 plus CO2 plus H2O, so CaCO3 is 0.0125 divided by 2 = 0.0063 mol, with a mass of 0.626 g. (c) Purity = 0.626 divided by 1.20 times 100 per cent = 52.1 per cent. Marking notes: acid consumed (4 marks); mass of carbonate (3 marks); purity (1 mark). |
| 11 | The solubility product of silver chloride is Ksp = 1.8 times 10 to the power -10 at 25 degrees Celsius. (a) Find the solubility of silver chloride in pure water in mol per cubic decimetre. (b) Find its solubility in 0.010 M sodium chloride solution. (c) Explain the ratio between (b) and (a) and name the effect responsible. |
(a) Let the solubility be s: Ksp = s squared = 1.8 times 10 to the power -10, so s = 1.342e-05 M. (b) With [Cl-] about 0.010 M, [Ag+] = Ksp divided by 0.010 = 1.80e-08 M, which is the solubility. (c) The solubility is about 745 times smaller. The common ion effect shifts the equilibrium to the left and greatly reduces the solubility. Marking notes: pure water (3 marks); salt solution (3 marks); ratio and name (2 marks). |
| 12 | Dilute sulphuric acid is electrolysed with a current of 5.0 A for 1.0 hour. Oxygen forms at the anode and hydrogen at the cathode (96500 C per mole of electrons; molar volume 24.0 cubic decimetres). (a) Find the moles of electrons passed. (b) Find the volume of hydrogen produced at the cathode in cubic decimetres. (c) Find the volume of oxygen at the anode and explain why the volume ratio is 2 to 1. |
(a) Q = 5.0 times 3600 = 18000 C, so the moles of electrons = 18000 divided by 96500 = 0.18653 mol. (b) Since 2H+ plus 2e- gives H2, the moles of hydrogen = 0.09326 mol and the volume = 0.09326 times 24.0 = 2.24 cubic decimetres. (c) From 4OH- giving O2 plus 2H2O plus 4e-, the moles of oxygen = 0.04663 mol and the volume = 1.12 cubic decimetres. Each mole of oxygen needs four moles of electrons while each mole of hydrogen needs two, so the hydrogen volume is twice the oxygen volume for the same charge. Marking notes: moles of electrons (3 marks); hydrogen volume (3 marks); oxygen volume and explanation (2 marks). |
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