1. Propanoic acid is a weak acid; explain the term weak. incompletely / partially / slightly dissociated OR dissociation varies with concentration allow poor proton donor OR conjugate base strong (1) 2. (i) Give an equation for the dissociation of propanoic acid and hence an expression for its dissociation constant, k2. CH3CH2COOH CH3CH2COO- + H+ CH3CH2COO- + H3O+ (1) OR CH3CH2COOH + H2O Ka = [CH3CH2COO − ][H+ ] [CH3CH2COOH] Mark consequentially for second mark if acid is incorrect but MUST have H+ or H3O+. (ii) At 25 °C ka for propanoic acid is 1.30 × 10-5 mol dm-3. Find the pH of a solution of propanoic acid of concentration 0.0100 mol dm-3. State any assumptions you make. (iii) w. pH down means [H+] up (1) dissociation endothermic (1) equilibrium shifts in endothermic direction when temperature rises (1) mark consequentially ch Increasing the temperature of the propanoic acid solution causes the pH to decrease. what does this tell you about the enthalpy of dissociation? Justify your answer. em ac ti [H+] = K a c = 1.30 × 10 −5 × 0.01 = 3.6 × 10-4 (1) pH = –lg[H+] (1) = -lg(3.6 ×10-4 = 3.4(4) any rounding should be consistent (1) assumes dissociation of acid is small / concentration of acid does not change (1) if made no assumptions ie solved quadratic and correct answer (4) penalise arithmetic errors ONCE ONLY ve .co if H2O in equation: Ka = [CH3CH2COO − ][H3O + ] [CH3CH2COOH] m (9) (Total 2 marks) 1 ww 1 3. (i) (ii) pH = –lg[H+] (1) or lg [ H + ] Define Kw , the ionic product of water. Kw = [H+] [OH–] (1) (Accept [H3O+] instead of [H+]) NT Exampro 4. (i) What is the principal property of a buffer solution? maintains pH nearly constant (1) if small amounts of acid or base added (1) accept resists change to pH (ii) The dissociation constant for ethanoic acid is 1.80 × 10–5 mol dm–3. Calculate the pH of a buffer solution which has a concentration of 0.0150 mol dm–3 with respect to ethanoic acid and 0.0550 mol dm–3 with respect to sodium ethanoate. [ H + ][CH 3 CO2 ] [ H + ][salt ] (1) = Ka = 1.8 × 10–5 = [CH 3 CO2 H ] [acid ] [H+] = 1.8 × 10–5 × 0.015/0.055 = 4.91 × 10–6 (mol dm–3) − .co Mark consequentially if salt and acid wrong way up Incorrect if used 0.015 twice but would get final mark consequentially (Total 5 marks) Write a formula of the potassium salts of this acid. KH2PO4 K2HPO4 K3PO4 (1) do not allow if ions only em ac ti ve (Total 1 marks) (Total 3 marks) 5. Phosphoric(V) acid, H3 PO4 , is a tribasic acid. 6. For the reaction CaCO3 (s) CaO(s) +CO2 (g) the equilibrium total pressure at 1200 K is 4 atm. Determine Kp. 7. (a) State Raoult’s Law for an ideal binary liquid mixture. p(A) = po(A) × (A) (1) definition of symbols (1) (or words to that effect) (2) NT Exampro ww w. Kp = p(CO2) (1) 4 (1) atm (1) so Kp = 4 atm Do not allow square brackets ch m pH = –lg 4.91 × 10–6 = 5.3 (1) 2 (b) Liquid oxygen and liquid nitrogen form an ideal liquid mixture. At 70 K the vapour pressures are 38.3 kPa for N2 and 6.40 kPa for O2. Find the composition of the vapour in equilibrium with a liquid mixture which at equilibrium is an equimolar mixture of the two elements. p(N 2 ) = 38.3 x 0.5 = 19.15 kPa p(O 2 ) = 6.4 x 0.5 = 3.2 kPa x(N2 ) = 19.15/22.35 = 0.857 ∴ x(O 2 ) = 0.143 (by any method) (1) Ptot = 22.35kPa (1) OR recognition of ratio 6:1 (1) OR ∴ 6/7 and 1/7 of final composition (1) (1) Answer must be quoted to 2 or more significant figures but penalise 6 or more significant figures once only on the paper (c) diagrams to show wider difference between b.p. (1) if b.p's close ∴ v.p's each component close ∴ effective separation more difficult ∴more stages in the fractionating column necessary (or words to that effect) (2) (3) (d) vapour pressure ww x(A) = 1 w. ch The vapour pressure / composition diagram at constant temperature for a mixture of two liquids A and B which shows a strong deviation from Raoult’s Law is shown below. (x(A) = mole fraction of A. em ac ti mole fraction NT Exampro ve .co x(A) = 0 Show by means of sketches of boiling point / composition diagrams and a brief comment why ideal mixtures of liquids of similar boiling temperature are more difficult to separate by fractional distillation than those with boiling temperatures more widely separated. m (3) 3 (i) Sketch on the axes below the boiling point / composition diagram for this system, showing the curves for both liquid and vapour. θ b (A) < θ b (B) (1) azeotrope in more or less correct % position (1) azeotrope is minimum (1) liquid/vapour lines correct and labelled (1) (labels need only show vap. and liq. once) θb θb x(A)=1 composition x(A)=0 (ii) .co distillate is azeotrope (1) residue is A (1) suitable discussion (1) referred to diagram (1) (eg use of tie lines) (8) (Total 16 marks) 8. State Raoult’s Law for an ideal binary liquid mixture. p(A) = po(A)x(A) (1) definition of symbols (1) (or words to that effect) ve m consider a mixture for which the mole fraction of A is initially 0.75. Use your diagram to explain what happens when this mixture is fractionally distilled, making clear the nature of the distillate and of the residue. 9. Liquid oxygen and liquid nitrogen form an ideal liquid mixture. At 70 K the vapour pressures are 38.3 kPa for N2 and 6.40 kPa for O2. Find the composition of the vapour in equilibrium with a liquid mixture which at equilibrium is an equimolar mixture of the two elements. p(N 2 ) = 38.3 x 0.5 = 19.15 kPa p(O 2 ) = 6.4 x 0.5 = 3.2 kPa (1) Ptot = 22.35kPa (1) OR recognition of ratio 6:1 (1) x(N2 ) = 19.15/22.35 = 0.857 ch ∴ x(O 2 ) = 0.143 (by any method) w. Answer must be quoted to 2 or more significant figures but penalise 6 or more significant figures once only on the paper (Total 3 marks) NT Exampro ww em ac ti (1) (Total 2 marks) OR ∴ 6/7 and 1/7 of final composition (1) 4 10. Show by means of sketches of boiling point / composition diagrams and a brief comment why ideal mixtures of liquids of similar boiling temperature are more difficult to separate by fractional distillation than those with boiling temperatures more widely separated. m x(A) = 0 diagrams to show wider difference between b.p. (1) if b.p's close ∴ v.p's each component close ∴effective separation more difficult ∴more stages in the fractionating column necessary (or words to that effect) (2) (Total 3 marks) 11. vapour pressure x(A) = 1 (i) Sketch on the axes below the boiling point / composition diagram for this system, showing the curves for both liquid and vapour. θ b (A) < θ b (B) (1) azeotrope in more or less correct % position (1) azeotrope is minimum (1) liquid/vapour lines correct and labelled (1) (labels need only show vap. and liq. once) x(A)=1 ww θb w. composition x(A)=0 ch (ii) Consider a mixture for which the mole fraction of A is initially 0.75. Use your diagram to explain what happens when this mixture is fractionally distilled, making clear the nature of the distillate and of the residue. distillate is azeotrope (1) residue is A (1) suitable discussion (1) referred to diagram (1) (eg use of tie lines) (Total 8 marks) NT Exampro em ac ti mole fraction θb ve The vapour pressure / composition diagram at constant temperature for a mixture of two liquids A and B which shows a strong deviation from Raoult’s Law is shown below. (x(A) = mole fraction of A. .co 5 12. (a) Ammonia is manufactured by direct synthesis in the Harber Process: N2(g) + 3H2(g) = 2NH3(G) (i) ∆H = 92Kj mol–1 Write an expression for the equilibrium constant, Kc, for this reaction and give its units. Kc [ NH 3 ]2 = [ N 2 ][ H 2 ]3 (1) must be square brackets mol–2 dm6 (1) units marked consequentially on above Kc ( or Kp if used) the unit mark can be obtained if the correct units are mentioned in (ii) (ii) Kc = (146) 2 . (0.27)(0.81) 3 (1) = 14.86 or 14.9 or 15 (1) penalise more than 4 sig figs (lose (1)) em ac ti ve .co (4) (6) (Total 10 marks) Calculate Kc at this temperature. (b) Predict and explain the effect of an increase in temperature on: (i) the value of Kc: Kc decreases (1) since reaction exothermic / reverse reaction endothermic (1) (increase in temperature) moves equilibrium to left / causes endothermic reaction to occur (1) (ii) the rate of forward reaction. 13. When ammonium salts are dissolved in water, the following reaction occurs. NH4 + (aq) + H2O(1) (a) NH3(aq) + H3O+ (aq) Identify the acid/ base conjugate pairs in this reaction by writing appropriate symbols under each of the species in the equation above. A1 B2 B1 A2 (1) must show related pairs NT Exampro ww w. increases (1) energies / speed of molecules increased (1) more have energy > EAct (1) or sufficient energy to react instead of EAct reference to EAct or sufficient energy to react essential. ch m When 3 mol of hydrogen and 1 mol of nitrogen were allowed to reach equilibrium in a vessel of 1 dm3 capacity at 500°C and 1000 atm pressure, the equilibrium mixture contained 0.27 mol of N2, 0.81 mol of H2 and 1.46 mol of NH3. 6 (b) Write an expression for the dissociation constant, K2, for NH 4 + (aq) Ka = [ NH 3 ][ H 3 0 + ] [ NH 4 ] + allow H+, water must not be included (c) Calculate the pH of a solution of ammonium chloride 0.100 mol dm–3 at 298 K, the Ka value for NH4+ being 5.62 × 10–10 mol dm–3 at this temperature. 5.62 × 10–10 = [Total H30+ marks]2 /0.1 (1) √(5.62 × 10–11) [Total H3O+ marks] = = 7.497 × 10–6 (1) ∴ pH = 5.1 (1) if H+ concentration not correct but correct pH calculated from it (1) correct answer only (1) m 14. temperature /ºC b.p. of benzene 0 (Total 5 marks) Benzene and methylbenzene may be separated by fractional distillation. Sketch the general form of the boiling point/composition diagram for such a mixture and use it to explain the basis on which fractional distillation rests. em ac ti CH3CO2– + H3O+ CH3CO2H2 + Cl– I II ve ww w. ch mole fraction of methylbenzene 1 vapour line (1) liquid line (1) tie line (1) vapour richer in more volatile component (1) must be a verbal statement repeat tie lines (1) ∴ distillate is benzene (½) and residue is methylbenzene (½) must be a verbal statement (6) (Total 6 marks) 15. (a) When ethanoic acid is dissolved in water, the following equilibrium is established: CH3CO2H + H2O CH3CO2H +HCI When hydrogen chloride dissolves in ethanoic acid, the equilibrium established is: Comment on the role of the ethanoic acid in: NT Exampro .co b.p. of methylbenzene 7 (i) (ii) I II (acid) –proton/H+ donor (1) if just acid (½). (1) (base) – proton/H+ acceptor/remover (1) if just base (½) (1) (b) What is the relationship between the species CH3CO2H2+ and CH3CO2H? conjugate (acid/base) pair (1) (1) (c) The value of Ka for ethanoic acid at 298 K is 1.74 × 10–5 mol dm–3 and for methanoic acid, HCO2H, it is 1.60 × 10–4 mol dm–3 at the same temperature. (i) Write an expression for Ka for XH3CO2H. + − Ka m [ H ][ CH 3CO2 ] (1) if [H2O] included (0) accept [H3O+] [CH 3CO2 H ] (ii) Hence calculate the pH of a 0.100 mol dm–3 solution of CH3CO2H at 298 K. 1.74 × 10–5 = pH = 2.88 / 2.9 / 2.90(1) ve [ H + ]2 or [H+] = 1.32 × 10–3 (1) . 01 .co (1) (2) (d) The pH of a 0.050 mol dm–3 solution of HCO2H is 2.55. (i) state which of the two acids is the stronger; methanoic (1) Using this, together with the data in (c) and your answer to (c)(i): em ac ti (1) (ii) comment on the relative pH values of the two acids. pH less for methanoic (½) despite it being more dilute (½) strength depends on Ka / degree of dissociation (1) pH depends on concentration as well as strength / pH is measure of concentration of hydrogen ions sensible comment on relative electron release effect of the methyl group (1) w. ch (3) correct start pH 2 – 3 (1) line must be rising for this mark to be awarded correct place for vertical line at 40 cm3 (1) correct range for vertical portion (or slightly off vertical) ie start at 6/7 end at 10/11 (1) (3) NT Exampro ww (e) (i) Sketch with reasonable accuracy on the axes below, how the pH changes during the titration of 20.0 cm3 of a 0.100 mol dm–3 solution of methanoic acid with 0.050 mol dm–3 sodium hydroxide solution. 8 (ii) Select using the data below a suitable indicator for this titration. Give a brief reason for your choice based on the curve drawn in (e)(i). Indicator Bromocresol green Bromothymol blue Phenol red pH Range 3.5 – 5.4 6.0 – 7.6 6.8 – 8.4 Phenol red or bromothymol blue (1) pH range for colour change in correct pH range for vertical portion of graph (1) note: the choice of indicator is consequential on the straight vertical portion of the graph drawn (2) (Total 15 marks) 16. This question concerns redox behaviour and the following data will be found useful. .co Bi3+ + 3H2O – (a) In potassium manganate(VII) titrations, the solutions are acidified with dilute sulphuric acid. (i) Using the data above, explain why dilute hydrochloric acid is not used for this purpose. CI– oxidised (by manganate(VII) / CI– ions form CI2 (1) quantitative relevance of redox potentials (1) but calculations not required hence volume of manganate(VII) unreliable / too large (1) (3) colour change would be colourless to pink/purple (1) which is very easy to see / very sharp end-point (1) or inverse argument ww w. not clear instead of colourless ch (ii) Why is potassium manganate(VII) usually placed in the burette, despite the difficulties it presents in reading the burette? em ac ti ve Fe3+ (aq) + e– Fe2+ (aq ½Cl2(g) + e– Cl– (aq) mNo4– (aq) + 8H+ (aq) + 5E– Mn2+ (aq) + 4H2O(I) + m E /V +0.77 +1.36 +1.52 (2) (b) A test for Mn2+ ions in solution is to react them with sodium bismuthate(V), NaBiO3, in the presence of nitric acid. A purple colour will develop owing to the formation of MnO4– ions in the solution. The ionic half equation for the reduction of BiO3– ion is: BiO3– + 6H+ + 2e– (i) Use this half equation and the data above to write an ionic equation for the oxidation of the Mn2+ ion. 5BiO3– + 14H+ + 2Mn2+ → 5Bi3+ + 2MnO4 multiplication 5 to 2 (1) correct overall equation (1) (2) + 7H2O (ii) Suggest qualitatively, how the E MnO4–| Mn2+ value for BiO3– | Bi3+ compares with that for NT Exampro 9 E for bismuthate must be more positive/greater/bigger/higher than that for manganate(VII) (1) (1) (c) An alloy contains iron and manganese only. One arming with dilute nitric acid 2.30 g of this alloy gave a solution containing iron(III) ions and manganese (II) ions. Treatment of this solution with excess sodium bismuthate(V) completely oxidised all the Mn2+ ions present to MnO4– ions. The excess bismuthate(V) ions were then completely destroyed and the solution made up to 250cm3 with distilled water and thoroughly shaken. Titration of 25.0cm3 portions of this solution required 25.0cm3 of standard 0.100 mol dm–3 iron (II) sulphate solution. (i) Write the equation for the reaction occurring during the titration. MnO4 + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O multiplication 5 to 1 (1) correct overall equation (1) – .co ve 2.5 × 10 −3 (1) = 5 × 10–4 5 m (2) (ii) Calculate the percentage of manganese present in the alloy. moles Fe2+ = 0.1 × 25 (1) = 2.5 × 10–3 1000 moles manganate(VII) = moles Mn2+ = moles Mn2+ in 250 cm3 = 10 × (0.5 × 10–3) (1) = 5 × 10–3 mass Mn in alloy = 55 × (5 × 10–3) (1) = 0.275g % Mn = (55 × 5 × 10–3) × = 12% marks are for the processes but if the final answer is incorrect ie ≠ 12% to 2 – 4 (inc.) significant figures then penalise (–1) if answer is correct but not given to 2–4 (inc.) significant figures then penalise (–1) correct answer with some working (5) correct answer and no working (1) if equation is incorrect in (c)(i) but the processes are correct for that equation then (max 4) (5) (Total 15 marks) NT Exampro ww w. ch em ac ti 100 (1) 2.3 10 17. Hydrogen and iodine react together to give an equilibrium: H2(g) + I2(g) (a) 2HI(g) Write an expression for Kp for this equilibrium, giving consideration to its units. correct expression for Kp (1) no square brackets must state “no units” or atm 2 (1) (2) atm 2 (b) When 0.50 mol of I2 and 0.50 mol of H2 were mixed in a closed container at 723 K and 2 atm pressure, 0.11 mol of I2 were found to be present when equilibrium was established. (i) Calculate the partial pressures of I2 and 0.50 mol of H2 were mixed in a closed container at 723 K and 2 atm pressure, 0.11 mol of I2 were found to be present when equilibrium was established. moles HI = 0.78 (1) calculation of the mole fraction(s) (1) calculation of the partial pressure(s) (1) .co m (3) (ii) Hence calculate the value of Kp at 723 K. Kp = (1.56) 2 (0.22) 2 (consequential on values in (b)(i) and expression in (a)) (1) = 50.28 or 50.3 or 50 (1) this answer only em ac ti ve (2) (c) In an experiment to establish the equilibrium concentrations in (b), the reaction was allowed to reach equilibrium at 723 K and then quenched by addition to a known, large volume of water. The concentration of iodine in this solution was then determined by titration with standard sodium thiosulphate solution. (i) Write an equation for the reaction between sodium thiosulphate and iodine. species (1) balance (1) conditional on correct species ionic equation acceptable (2) (ii) What indicator would you use? Give the colour change at the end point. starch (1) blue/black to colourless (1) not white or clear instead of colourless (2) NT Exampro ww w. ch 2Na2S2O3 + I2 → Na2S4O6 + 2NaI 11 (iii) In this titration and in titrations involving potassium manganate(VII), a colour change occurs during reaction. Why is an indicator usually added in iodine/thiosulphate titrations but not in titrations involving potassium manganate(VII)? ease of discernibility compared (1) actual colour change for iodine / thiosulphate ie yellow → colourless (1) (2) (Total 13 marks) 18. (a) .co (b) (i) CrO42– and Cr2O72– both contain Cr(+6) (1) statement that no change implies not a redox reaction (1) 2nd mark is not conditional on the 1st if show a change in oxidation state for H or O then (max 1) m Cr2O72– + 3H2O2 + 8H+ → 2Cr3+ + 3O2 + 7H2O (2) allow doubled if equation is incorrect but some evidence of the multiplication by 3 then (1) only if equation correct but electrons are left in then penalise (–1) 2 2 1 (iii) acid converts chromate to dichromate (1) em ac ti which oxidises H2O2 / reacts with H2O2 / reduced by H2O2 (1) Cr3+ formed / reduces yield (1) (iv) ve (ii) no more bubbles/fizzing/effervescence or glowing splint does not relight (1) 3 CrO42– + 4H2O + 3e– → Cr3+ + 8OH– or Cr3+ + 8OH– → CrO42– + 4H2O + 3e– (1) H2O2 + 2e– → 2OH– or 2OH– → H2O2 + 2e– (1) equations must be completely correct to score the mark allow multiples 2 [10] 19. (a) (i) KP = pPCl5 ch (1) pPCl3 × pCl 2 square brackets (0) 1 (ii) no. of moles PCl3 = 0.39 (1) no. of moles PCl5 = 0.61 (1 - 0.39) (1) total number of moles = 1.39 (1 + 0.39) (1) x 2 (1) ie in each case ’process’ marked allow if percentages used if just answers with no working then (max 2) (ie (1) for PCl5 and (1) for PCl3) correct answers: PCl5 PCl3 - ww w. 1 − 0.39 × 2 = 0.878 (atm) 1 + 0.39 0.39 × 2 = 0.561 (atm) 1 + 0.39 4 (iii) pCl2 = p PCl3 = 0.561 NT Exampro 12 KP = (0.561) 2 = 0.359 atm (accept 0.35 or 0.36) 0.878 answer (1) mark consequentially on answer to (ii) units (1) (b) (i) (ii) HCl / hydrogen chloride / hydrochloric acid gas (1) AlCl3 + 3H2O Al(OH)3 + 3HCl (1) or balanced partial hydrolysis but must start with AlCl3 and end with HCl water bonds to Al via co-ordinate bond or lone pair on H2O attacks Al (1) OWTTE comment about Al being electron deficient / AlCl3 molecule is electron deficient / Al has accessible or empty orbitals (1) can be shown by diagram 2 1 1 (iii) m 3+ 2 1 (c) (i) (ii) ionic / not covalent (1) 3+ H O H em ac ti • • OH 2 [Al(H2O)6]3+ + H2O [Al(H2O)5OH]2+ + H3O+ / [Al(H2O)6]3+ [Al(H2O)5OH]2+ + H+ (1) comment on polarising power of Al3+ / comment on charge density of Al3+ (1) it must be clear that it is the charge that is important or if shown diagrammatically ve Al .co OH – + Al HOH 2 2 [14] if the 3+ charge is missing on the Al then (1) only 20. (a) (i) KP = pPCl3 × pCl 2 pPCl5 ch (1) square brackets (0) (1) (ii) no. of moles PCl3 = 0.39 (1) no. of moles PCl5 = 0.61 (1 - 0.39) (1) total number of moles = 1.39 (1 + 0.39) (1) x 2 (1) ie in each case ’process’ marked allow if percentages used if just answers with no working then (max 2) (ie (1) for PCl5 and (1) for PCl3) NT Exampro ww w. 13 correct answers: PCl5 PCl3 1 − 0.39 × 2 = 0.878 (atm) 1 + 0.39 0.39 × 2 = 0.561 (atm) 1 + 0.39 (4) (iii) pCl2 = p PCl3 = 0.561 KP = (0.561) 2 = 0.359 atm (accept 0.35 or 0.36) 0.878 answer (1) mark consequentially on answer to (ii) units (1) .co m (2) (Total 7 marks) em ac ti ve 21. (a) PbO2(s) + Pb(s) + 4H+(aq) + 2SO42–(aq) 2PbSO4(s) + 2H2O(l) or PbO2(s) + Pb(s) + 2H+(aq) + 2HSO4–(aq) 2PbSO4(s) + 2H2O(l) or PbO2(s) + Pb(s) + 2H2SO4(aq) 2PbSO4(s) + 2H2O(l) species (1) balanced (1) (i) (ii) presence of (dissolved) oxygen/air (1) not just H+ / acid 2 1 (b) (iii) (iv) ww E Mg more –ve than E Fe / Mg is more reactive than Fe (1) Mg reacts preferentially (1) aluminium similarly but has protective oxide layer (1) Parts (f) (ii) and (iii) marked to a (max 4)(f)(ii) + (iii) w. surface irregularities due to working / crystal structure of iron changes under stress (1) cause corrosion at boundaries of crystals / small E variation for Fe sets up cell action (1) ch iron dissolves or reacts at same place (one point) / cathodic region / pit holds water (1) or any sensible chemical point 1 4 [8] 22. (a) (b) Kp= (i) p CO 2 p CO Cr2O72–(aq) + 14H4(aq) + 6e → 2Cr3+ (aq) + 7H2O(I) (1) State symbols not necessary 1 1 NT Exampro 14 (ii) (c) 6Fe2+(aq)+Cr2O72–(aq) + 14H+ → 6Fe3+(aq) + 2Cr34(aq) + 7H2O(I) (2) Mark consequentially from d(i). 2 27.40 × 0.0220 = 6.03 × 10–4 1000 Use of titration data to get moles of Cr2O72– (1) mols of Cr2O72–(aq) = Use of stoichiometry above (1) mols of Fe2+ (≡ Mols of Fe) = 6.03 × 10–4 × 6 = 3.62 × 10–3 moles of mass (1) mass of Fe = 3.62 × 10–3 × 56 = 0.203 g mass to % (1) 0.203 × 100 %Fe = % =99.5 % 0.204 m ve .co (1) 4 [8] 23. (a) HSO4– + H3O+ (1) H2SO4 + H2O can be expressed in acid 1 base 2 base 1 acid 2 (1) words need identification of pairs by number or letter, not A B A B alone sulphuric acid acts as proton donor or water acts as proton acceptor i.e. base (1) 3 (ii) weak (acid) or incompletely dissociated (1) needs to ionise during reaction which is an endothermic process or some explanation of this in candidates own words (1) 2 solution resists change in pH (NOT constant pH ) or candidates’ own words expressing same idea (1) for small addition of acid or base (1) 2 (c) (i) (ii) (d) (i) ww Used to introduce hydrophilic or ionic or polar(1) sulphonate or sulphonic acid group (1) or introduces sulphonate group (1) into non–polar or hydrophobic molecules (1) w. accept balanced non–ionic equations. ch +H+ : SO42− + H+ → HSO4− +OH − : HSO4− + OH − → SO42− + H2O em ac ti (b) (i) both strong acids or both completely dissociated (1) the reaction in both is H+ + OH− → H2O (1); must have equation for second mark 2 2 2 (ii) Do not form scum (1) (with hard water) and reason e.g. form soluble calcium or magnesium salts or reference to soap forming insoluble salts (1) or less detergent is needed (1) [Max 2] references to lather or environment or clothes cleaner do not score [15] 24. (a) (i) selects appropriate iron redox potentials / uses data (1) suggests that Fe(II) is more likely than Fe(III) since –0.44 is more negative than –0.04 or some equivalent explanation (1) If comparisons are made with the chlorine potential then max 2 NT Exampro 15 (ii) React metal with chlorine (1) dissolve in water (1) or react metal with hydrochloric acid oxidise with chlorine (1) or react with chlorine water (2) (1) Justification, e.g. chlorine will convert Fe to Fe3+ referred to Eo values (1) A route Fe → Fe2+ → Fe(OH)2 → Fe(OH)3 → FeCl3 is acceptable Justification must not be in terms of hydrochloric acid. Other oxidising agents may not be used. 3 (b) Fe → Fe2+ + 2e (1) ½ O2 + H2O + 2e– → 2OH– (1) or ½ O2 + H2O + Fe → Fe(OH)2 or Fe2+ + 2OH– (2) followed by: Fe(OH)2 + OH– → Fe(OH)3 + e – (1) or in words: iron oxidised to Fe2+ by oxygen and water (1) iron(II) hydroxide or Fe2+ and 2OH– formed (1) further oxidation gives iron(III) hydroxide or Fe2O3.xH2O or Fe(OH)3 ve .co m (1) 3 [8] 25. (a) (i) HCl: pH = 1.13 ∴ [H+] = 0.074 mol dm–3 ∴ [HCl] = 0.074 mol dm–3 [0.074 to 0.07413] HOCl: pH = 4.23 ∴ [H+] = 5.89 × 10–5 mol dm–3 (1) Ka = [H+] [OCl–] (1) [HOCl] em ac ti Must be H3O+ + SO42– ignore state symbols + 1 (ii) [H+] = [OCl–] (1) or implied later in calculation [HOCl] = [H+]2/ Ka = 0.0932 mol dm–3 (1) (b) (i) (ii) [H+] = 0.10 / 0.1047 / 0.105 (1) → or H2SO4 → H+ + HSO4–(1) +O + HSO4– ignore state symbols H2SO4 + H2O → H3 HSO4– ch 4 H2SO4 + H2O (iii) (c) (i) (ii) second ionisation suppressed by the first ionisation (1) ww w. H+ + SO42– (1) 4 1 Kc = [Cl 2 ] 2 × [H 2 O] 2 [HCl]4 × [O 2 ] 4HCl + O2 2Cl2 equilibrium mols 0.20 [ ] eq ÷ 10 (1) 0.020 2H2O 0.050 (1) 0.30 and 0.30 (1) 0.0050 0.030 0.030 4 2 16 (d) (i) Kc = [0.030]2 x [0.030]2 = 1010 or 1012 or 1013 or 1012.5 [0.020]4 x [0.005] (mol–1 dm3) (1) As reaction (left to right) is exothermic (1) Decrease in temperature drives equilibrium to from left to right (1) NT Exampro (ii) (iii) As more (gas) molecules on the left (1), equilibrium is driven from left to right (1) 2 A catalyst has no effect (1) As it only alters the rate of the reaction not the position of equilibrium / it alters the rate of the forward and reverse reactions equally (1) 2 [20] 26. (a) (i) CH3COOH + H2O If no water present (0) CH3COO– + H3O+ (1) 1 (ii) allow [H+] instead of [H3O+] (iii) m ve H H H C C C H H H H or CH 2 CH 3 [H 3 O + ][CH 3 COO – ] Ka = (1) [CH 3 COOH] 1 [H+]2 = Ka [acid] / some recognition that [H+] = [anion] (1) [H+] = 1.8 × 10–5 × 2.00 (1) pH = –log10 (6 × 10–3) = 2.2(2) (1) or pH = ½ pKa – ½ log10 [acid] (1) = ½ × 4.74 - ½ × 0.3 (1) = 2.2(2) (1) .co 3 2 1 1 2 1 3 17 C O (1) C H (b) (i) CH 3 (ii) (iii) (c) (i) (ii) (iii) (iv) propyl ethanoate (1) catalyst (1) 1cm3 scores 1 mark 33 or (35 – ans on (i)) (1) 33 cm3 = 3.3 × 10–2 mol NaOH = 3.3 × 10–2 mol remaining ethanoic acid (1) 1 3.3 × 10–2 mol propanol (1) 6.7 × 10–2 mol ester (1) 6.7 × 10–2 mol water (1) NT Exampro ww w. 0.001 × 1000 (1) × 2 (1) = 2 cm3 ch em ac ti C C O H (2) O H O O C O CH 2 (v) Keq = 0.67 × 0.67 = 4.1(1) scores (2 marks) 0.33 × 0.33 or 6.7 × 10 –2 × 6.7 × 10 –2 = 4.12 (1) 3.3 × 10 – 2 × 3.3 × 10 – 2 + Some reference to volume or 100 cm3 used e.g volumes cancel (1) (d) (i) (ii) 1st (1) and 1st (1) 6 × 10–4 = k × 1 × 0.1 (1) k = 6 × 10–3 (1) mol–1 dm3 sec–1(1) Consequential on (i) 2 2 3 27. (a) (i) (ii) pH = –lg[H+] or lg 1/[H+] or –lg([H+]/mol dm–³ .co m (1) (2) (2) 3 An acid that is partially ionised (1) e.g. HCOOH + H2O → HCOO– + H3O+ /accept H+ (1) (ii) [23] em ac ti [OH–] = 0.747 (mol dm–³) (1) then either [H+] = 1.0 × 10–14/0.747 (1) = 1.34 × 10–14 pH = 13.87 or pOH = 0.127 (1) pH = 14 – pOH pH = 13.87 (1) Ka = ve (b) (i) [H+] = 0.152 (mol dm–³) (1) pH = 0.82 (accept 0.818) (1) (3) w. ch (iii) [H + ][HCOO – ] (1) [HCOOH] [H+] =[HCOO–] (1) [H+] = √(Ka .[HCOOH]) = √(2.706 × 10–5) (1) = 5.20 × 10–³ pH = 2.28 (1) (4) 9 (c) (i) (ii) [H + ] . [CH 3 COO – ] [H+ ] . [salt] = = (1) Ka = 1.8 × [CH 3 COOH] [acid] 0.105 [H+] = 1.8 × 10–5 × = 5.526 × 10–6(mol dm–3) (1) 0.342 pH = –lg[H+] = 5.26 (1) ww 10–5 maintains pH nearly constant or resists change in pH (1) when small amounts of acid or base /H+ or OH– added (1) (2) (3) 5 [17] NT Exampro 18 28. (a) (i) (ii) Kp = p(NO2)²/p(N2O4) (1) units are atm or any SI unit of pressure (1) Kp = p(CO2) (1) units are atm or any SI unit of pressure (1) moles at start change but x = 0.81 equilibrium moles mole fraction partial pressure /atm N2O4 1 –x 1 –0.81 = 0.19 0.19 /1.81 = 0.105 0.105 × 1.2 = 0.126 NO2 0 + 2x (2) (2) 4 (b) (i) ∴total moles = 1 + x 1.62 ∴total moles = 1.81 1.62 /1.81 = 0.895 0.895 × 1.2 = 1.07 .co marking points For correct equilibrium moles (1) for consequential total moles (1) for consequential mole fractions (1) for partial pressure N2O4(1) for partial pressure NO2 (1) (ii) (c) (i) (ii) (iii) Kp = (1.07)² /0.126 = 9.1 (atm)(accept 9.2) Kc = m (5) (1) (1) (1) 6 (3) 5 [15] [SO 3 ]² initial [SO 2 ]² initial [O 2 ]initial Catalyst has no effect on the position of equilibrium [SO 3 ] 2 [SO 2 ] 2 [O 2 ] = (0.20/2) 2 (1) = 50 (0.40/2) 2 .(0.010/2) this is 〈〈 the value of Kc (1.7 × 106) or this is not = Kc (1) so reaction will move to the right to increase value of [SO3] /or to get value up to 1.7 × 106 (1) 29. (a) only partially dissociated / ionised / not fully dissociated (1) into H+ ions / H3+O / proton donor (1) Ka = (i) (ii) (iii) [H 3 O + ][A – ] (1) [HA] ch em ac ti ve 2 1 1 1 2 (b) (c) 9.0 to 9.4 (1) 9.0 to 9.4 (1) or same answer as (c)(ii) pKa = 5.6 (1) Ka = 2.5 × 10–6 (1) consequential (a solution that) resists change in pH / retains an almost constant pH (1) on addition of small quantities of acid or alkali (1) 5.2 to 5.8 (1) 5.5 or 5.6 (1) or answer from (c) (iii) based on misreading scale of graph, eg. 4.8 (d) (i) ww w. 2 (ii) 2 NT Exampro 19 (e) Phenolphthalein (1) indictor changes colour between pH 7 and 10 this is vertical part of graph (1) methyl orange would change in acid / give pH between pH4 and pH6 (1) n. b. must be +ve statement about methyl orange exothermic reaction / heat (energy) released during reaction (1) HCl is strong acid fully ionised (1) this is weak acid so some energy used for dissociation (1) (i) (ii) pH = −log(10) [H+] or in words [H + ] 2 (1) 1 [H+] = √1.8 × 10−5 = 4.24 × 10−3 (1) pH = −log (4.24 × 10−3) = 2.37/2.4 (1) 2 to 4 sig. figs. 3 (f) 3 1 (g) 1.8 × 10−5 = m ve .co → 2SO2 0 1.5 + 0 0.75 O2 (1) 3 [21] 30. (a) (i) Kp = PSO2 × PO2 2 PSO3 2 (1) [ ] no mark ( ) OK (ii) Mols at start mols at equ 1 Mark by process 1 mark for working out mole fraction 1 mark for × 10 1 mark for correct substitution in Kp and answer 1 mark for unit i.e. PSO2 = n.b. could show mole fraction for all 3 and then × 10 later to give partial pressure. Kp = (5.46)2 × (2.73) / (1.83)2 = 24.5 (1) atm (1) (b) (i) (ii) No effect (1) No effect (1) 5 1 1 [8] NT Exampro ww w. 1.5 × 10 = 5.46 2.75 0.7 × 10 = 2.73 PO2 = 2.75 0.5 PSO3= × 10 = 1.83 2.75 ch em ac ti 2SO3 2 0.5 20 31. (a) (i) fraction of the total pressure generated by a gas or or pressure gas would generate if it alone occupied the volume or Ptotal × mol fraction (1) Kp = p(CO) × p(H 2 ) 3 (1) p(CH 4 ) × p(H 2 O) 1 1 (ii) (iii) not [ ] m Increase in total pressure will result in less product molecules in the equilibrium mixture / equilibrium moves to left (1) because more molecules on product side of the equilibrium than on left (1) No change (1) KP increase (1) No change (1) 2 1 1 1 1 1 (b) (i) (ii) (iii) (ii) (iii) equilibrium has moved left in favour of gas (1) exothermic going left to right/in the forward direction / as written (1) Stand alone Answer yes or no with some sensible justification (1) e.g. No the costs would not justify the amount produced em ac ti 9.87 × 10–3 kPa–1/ 9.87 × 10−6 Pa−1 consequential on (i) (1) Allow 3 – 5 sig fig ve (c) (i) Kp = 1 (1) p(CH 4 ) .co 20 30 40 50 Volume of alkali added / cm 3 2 1 [12] (iv) (b) 14 12 pH 10 8 6 4 (c) 2 0 10 NT Exampro ww w. Maintains an almost constant pH / resists change in pH (1) with the addition of small amounts of acid or alkali (1) ch 32. (a) Few molecules dissociate (into protons) / partially dissociated / ionised (1) Not fully dissociated scores zero 1 2 21 starting pH (1) at 2.8 endpoint (1) vertical between 6 and 11 including 7-10 vertical (1) at 25 cm3 general shape (1) finish above 12 (d) (e) Almost horizontal area marked on graph (1) (i) 4 1 Ka = or [CH 3 COO − ][H + ] (1) [CH 3 COOH] [CH 3 COO − ][H 3 O + ] (1) [CH 3 COOH] Ka = (ii) 1 .co 2 H2 O pH = pKa at half way to neutralisation point = 12.5 cm3 (1) This could be shown on the graph because pH = pKa when [CH3COO−] = [CH3COOH] (1) m 2 [11] 33. (a) VO2+ V3+ yellow green VO2+ V2+ blue Lavender/mauve/lilac/purple/violet em ac ti ve 2 All four correct 2 marks – any 2 correct 1 mark (b) (i) (ii) 2VO2+ + Zn + 4H+ → 2VO2+ + 2H2O + Zn2+ species (1) balance (1) 2 (c) (i) w. V3+ or V3+ and VO2+ (1) allow [V(H2O)6]3+ E value for both reduction reactions positive so feasible (1) (further) reduction (to V2+) not feasible / E value negative (1) Disproportionation requires the original oxidation states to be able to both rise and fall (1) or In disproportionation a species must be able to be reduced and to be oxidised thus needs 3 oxidation states (1) ch 3 1 (ii) 2VO2+ → VO2+ + V3+ species (1) balance (1) ww 2 [10] 34. (a) (i) Equation (1) OH + H2 O 2 O + (1) OH O There are several ways of answering the rest of this section Alternative 1 E = + 2.47V (1) This is positive, therefore reaction feasible (1) NT Exampro 3 22 Alternative 2 Both E values are positive, so sum of E Therefore reaction is feasible (1) Number / fraction of molecules (with energy E) is positive (1) (ii) m Ea Ea cat (E) Do not award third mark if either of the E values is on left hand side of max on hump em ac ti Explanation area under graph to right of Ecat> area to right of Ea (1) greater number/ fraction of molecules or particles have enough energy to react on collision or greater number of effective / successful collisions (1) If draw two graphs showing different temperature – ignore ve .co 5 23 Maxwell-Boltzmann graph - 1 mark for shape of graph 1 mark for correct axes plotted and labelled 1 mark for Ea and Eacat NT Exampro ww w. ch (iii) The reactant/ 1,4-dihydroxybenzene (and the product/ quinone both) have delocalised ring (or resonance) systems/ or described delocalisation (or resonance) (1) these average bond energies are for localised bonds / do not apply to benzene ring compounds / compounds with delocalised or resonance systems. (1) 2 (b) 20 dm3 oxygen is 20/24 = 0.833 mol (1) amount peroxide in 1 dm3 = 0.833 mol × 2 (1) = 1.67 mol mass 1.67 mol × 34 g mol−1 = 57 / 56.6 / 56.7 / 56.8 g (dm–3) (1) (i) (ii) H2O2 → O2+ 2H+ + 2e− (1) 2MnO4− + 6H+ + 5 H2O2 → 2Mn2+ + 8H2O + 5O2 species on correct side of equation (1) balance (1) MnO4– + 8H+ + 5e– → Mn2+ + 4H2O If the overall equation, is not correct allow 1 mark for this equation if correct ignore 3 1 2 (c) (iii) ve Higher concentration increases collision frequency / more collisions per unit of time (1) therefore causes increase in reaction rate (1) More successful collisions therefore faster gets 1 mark (no reference to higher concentration) .co m 2 [18] 35. (a) The marks are for: • substituting correctly • calculating p(SO3) • writing the expression for K • correct generation of the ratio Ratio of SO3 = (b) (i) ww pSO32 = 150 pSO3 = 12.25 (1) w. Kp = pSO32/ pSO22 × pO2 (= 3.00 × 104) (1) 3.00 × 104 = pSO32 / 0.1 × 0.1 × 0.5 (1) if no expression for Kp is given this correct substitution can score 2 marks 12.25 × 100% (1) = 95% (1) (12.25 + 0.1 + 0.5) ch • calculation of the ratio to give answer which rounds to 95 t em ac ti 5 The marks are for • Recognizing the existence of hydrogen bonds ( between molecules) (1) • That each molecule can form more than one hydrogen bond because of the two OH (and two S=O groups) / or a description of hydrogen bonds in this case / or a diagram showing the hydrogen bonds (1) • That hydrogen bonds make for strong intermolecular forces (and hence high boiling temperature) which requires higher energy to break / separate molecules (1) 3 NT Exampro 24 (ii) If water is added to acid heat generated boils and liquid spits out (1) if acid added to water the large volume of water absorbs the heat generated (and the mixture does not boil) (1) pH = –log10 (0.200) = 0.70 (1) allow 0.7 or 0.699 realising that the first ionisation / dissociation of sulphuric and that of HCl are both complete (1) that the second ionisation of sulphuric is suppressed by the H+ from the first (1) little contribution from 2nd ionisation so reduces the pH very little / increases the [H+] very little (1) 2 1 (c) (i) (ii) 3 ve Lead(IV) oxide equations 2 marks PbO2 + H2SO4+ 2H+ + 2e– → PbSO4 + 2H2O or PbO2 +SO42– + 4H+ + 2e− → PbSO4 + 2H2O+ Species (1) balancing (1) (ii) PbO2 + Pb + 2H2SO4 → 2PbSO4 + 2H2O (1) .co 3 1 [18] 36. (a) (i) (ii) pH = −log10 [H+] / pH = –lg [H+] (1) em ac ti m 1 1 1 1 2 1 3 25 (d) (i) Lead equations 1 mark Pb + H2SO4 → PbSO4 + 2H+ + 2e– (1) or Pb + SO42− → PbSO4 + 2e– KW = [H+] [OH−] or KW = [H3+O] [OH−] (1) 0.70 (or 0.699) (1) (b) (c) fully ionised / fully dissociated / almost completely ionised (1) [H+] = KW / [OH−] = 1.25 × 10–14 (1) pH = 13.9 or 13.90 (1) Ka = (ii) (d) (i) (ii) [H+] = √ (Ka × [HA]) (1) = 0.00474 (1) pH = 2.32 / 2.33 (1) NT Exampro ww [H + ][A – ] (1) [HA] allow [H3+O] w. ch (i) (e) K a [HA] (1) [A – ] [H+] = (5.62 × 10−5 × 0.3) / 0.6 = 0.0000281 / 2.81 × 10−5 (1) pH = 4.55 (1) Or [A – ] pH = pKa + log [HA] [0.600] = −log10 (5.62 × 10−5) + log10 = 4.55 [0.300] [H+] = If initial error in statement of [H+] or Henderson equation max 1 3 [13] 37. (a) (i) (ii) Kc = [SO3]2 / [SO2]2 [O2] (1) m ve .co 1 0.1 1.8 60 60 1.67 × 10–3 0.03 (1) (0.03) 2 Kc = = 4860or 4.86 × 10−4 (1) (3.33 × 10 −3 ) 2 × 1.67 × 10 −3 mol−1 dm3 (1) Kc decreases (1) no effect (1) no effect (1) 0.2 30 = 3.33 × 10−3 3 1 1 1 1 1 2 1 (b) (i) (ii) shifts to left / in reverse (1) (c) (i) (ii) (d) (i) (ii) (iii) (iv) Kp = pSO32 / pSO22 × pO2 (1) penalise square brackets Total number of moles (1) consequential on a (ii) SO2 = 0.0952(4); O2 = 0.0476 (2); SO3 = 0.857 (1) (1) Partial pressures: SO2 = 0. 190 (5) atm; O2 = 0.0952 (4) atm; SO3 = 1.71(4) atm (1) i.e. multiply answer in (ii) by 2 w. 1.7142 / 0.19052 × 0.09524 = 850 (1) atm−1 (1) ch em ac ti 2 [14] 38. (a) (b) (i) (i) [Ar]3d6 (1) allow 1s2 etc ww 1 • Zn / Iron (1) not the zinc ion or iron ion • more negative potential than -0.28V (1) NB this mark must show evidence of use of the data gives +E for reduction reaction (1) consequential on second mark • rate too slow / activation energy too high / kinetically stable / allow oxide layer if metal electrode specified (1) • non-standard conditions (1) 3 (ii) 2 NT Exampro 26 (c) (i) (ii) Co(H2O)62+ (1) Example: Co(H2O)62+ + 4Cl− → CoCl42− + 6 H2O Any valid equation that shows a ligand exchange but begins with Co(H2O)62+ (1) ligand exchange correctly balanced (1) 1 2 [9] 39. (a) (i) A B C CH3CH2CH2OH → CH3CH2COOH → CH3CHCICOOH (1) (1) 1 .co 186 100% D (ii) substitution (1) free radical (1) m Or propan-l-ol, propanoic acid, 2-chloropropanoic acid (3) 3 2 (iii) 122 100% C ww Comment that vapour richer in more volatile component (1) Link via diagram tie-lines to repeated distillation (1) D left behind (1) (b) 0.8 × 0.8 × 0.8 = 0.51 (1) 1 mol propanol = 60g 1 mol D = 90g (1) therefore 0.5 × 90 = 46.1 g (1) (i) pH = 2.04 [H+] = 9.12 ×10−3 (1) [H + ] 2 Ka = (1) [acid] Ka = (9.12 × 10−3)2/0.1 = 8.32 (mol dm−3) (1) w. diagram (2) Details axes labelled and bp in right order (1) vapour, liquid lines or areas labelled (1) ch em ac ti liquid % composition Boiling temperature ve 5 3 3 vapour (c) NT Exampro 27 (ii) [Consequential on the answer to part (i)] • Propanoic acid is a weaker acid than D and would produce less energy than D in its reaction with NaOH (1) as more energy would be needed to dissociate the acid into ions (1) 2 [18] 40. (a) The pressure which the gas exerts if it alone occupies the same volume at the same temperature (1) or partial pressure of a gas in a mixture = mole fraction of gas × total pressure (1) (i) Kp = p(PCl3)p(Cl2)/p(PCl5) (1) no square brackets Or Kp = (ii) 1 (b) mark consequentially on (i) PCl5 mols at start mols at eqm mol fraction partial pressures 1 0.60 0.60 1.40 0.60 ×2 1.40 = 0.857 .co ve (1) (1) (1) (1) (1) PCl3 0 0.40 0.40 1.40 0.40 ×2 1.40 = 0.571 PPCl3 × PCl2 PPCl5 m + Cl2 0 0.40 0.40 1.40 0.40 ×2 1.40 = 0.571 1 em ac ti (1) (1) (1) Kp = (0.571)2/0.857 = 0.38 or 0.381 (1) atm (1) but 2-4 sf acceptable i.e. mols at eqm mols fraction partial pressures (ie. multiply x2) substitute in Kp + answer units ch (c) (i) (ii) (d) (i) Kp increases (1) ww Endothermic (1) conditional on increase of temperature moves equilibrium in direction that absorbs heat (1) w. 5 2 1 Kp = P(CO2) (1) or Kp = PCO2 1 1 [12] (ii) 16 (atm) (1) ignore units. Consequential on (d)(i) NT Exampro 28 41. (a) (i) Pairs up CH3CH2COOH and CH3CH2COO– and H2O/H3O+(1); correct identification of which is acid and which base (1) Ka = [CH3CH2COO–] [H3O+] / [CH3CH2COOH] (1) [H+] is acceptable. [H+] = (Ka[HA])½ or √Ka[HA] (1) = (1. 3 × 10–5 × 0.10) ½ = 1.14 × 10–3 mol dm–3 (1) pH = 2.9 or 2.94, i.e. to 1 or 2 d.p. (1) Consequential on the value of [H+] provided the pH resulting is between 0 and 7. [H+][OH–] = 10–14 (1) = 1.14 × 10–3 [OH–] Thus [OH–] =10–14 × 10–3 (1) = 8.77 (8.8) × 10–12 mol dm–3 (1) units needed (2 or 3 sf) Consequential on the answer to (iii) for [H+] Allow 8.71 × 10–12 if solved using pH + pOH and pH = 2.94; 7.9 × 10–12 if solved using pH + pOH and pH = 2.9. 2 1 (ii) (iii) 3 (iv) .co m 3 2 2 3 [16] 29 (b) (c) (i) (ii) Solution that maintains almost constant pH (1) for small addition of acid or alkali (1) pH = pKa + lg [salt]/[acid] (1) = 4.9 + Ig (0.05)/(0.025) (1) for dividing by 2 = 5.19 or 5.2 (1). If the Henderson equation is wrong but concs are divided by 2 then 1/3 max. Or pH = 5.19 or 5.2 (1) If the concns are twice what they should be, ie. candidate does not spot the volume increase, then max (2). The pH is still 5.2, so care is needed. NT Exampro ww 1.30 × 10 -5 × 0.025 = (1) 0.050 w. [H ] = Ka[acid] (1) [salt] + ch em ac ti or CH3CH2COONa + H2O → CH3CH2COOH + NaOH (1) Explanation then must comment that acid is weak/not fully ionised ve CH3CH2COO– + H2O CH3CH2COOH + OH (1) Hydroxide ions make the solution alkaline (1) or propanoate ion deprotonates the water 42. (a) (i) P 24.6 / 31} (1) 0.794 / 0.794 = 1 } EF is PF5 (1) F 75.4 /19} 3.97 / 0.794 = 5 } Mr of EF = 126 (1) Therefore MF = EF = PF5 (1) There must be some use of the data of 126 g mol–1 OR Mass of phosphorus in I mole = 126 × 24.6/100 = 31 (1) Mass of fluorine in 1 mole = 126 × 75.4/100 = 95 (1) Moles of phosphorus in 1 mole compound = 31/31 = 1 Mole of fluorine in 1 mole compound = 95/19 = 5 (1) MF = PF5 (1) 4 F (ii) F Angles drawn on diagram of 90° (1) and 120° (1) (iii) (–) F F F P F F 90º F ve (–) F F P F F F (1) note: there must be an attempt at a 3-D drawing (i.e. one wedge and one dotted line) 3 F (b) HF has intermolecular hydrogen bonding (but others do not) (1) Because F atom is very small / other halogen atoms / chlorine etc. radii are too large (1) Hydrogen bonding is stronger than IMF/vdW/dipole-dipole/induced dipoledipole/dispersion forces and so more energy required (to boil) (1) Do not give any marks if the candidate answers in terms of strength of covalent bonds. Do not give all 3 marks unless the candidate has expressed their ideas clearly. ww w. Name stated as octahedral (1) Angle marked / stated as 90° (1) ch (1) note: again it must be 3-D (again wedges and dotted lines) 3 em ac ti F OR F .co 90º F 120º P F F OR F P F F m 3 30 NT Exampro (c) Ka = [H + ][F − ] [HF] (1) allow H + or H 3 O + [H+] = 10-2.04 (1) = 0.009120 mol dm-3 [H+] = [F-] or Ka = [H+]2 / [HF] (0.009120) 2 (0.009120) 2 or = (1) = (0.150 − 0.009120) 0.0150 5 = 5.90 × 10-4 (1) mol dm-3 (1) or = 5.55 (or 5.54) × 10-4 the unit mark can be given in the expression for K. [18] Kc = [CO 2 ] eq [H 2 ] eq = 1.0 (or 1 or 1.00) (1) mol2 dm-6 (1) (b) w. [CH 4 ] eq [H 2 O] eq ch 4 2 Equilibrium concentrations: above values ÷ 4 dm3 (1) (mark consequently) [CH4] = 0.03125 mol dm-3 [H2O] = 0.500 mol dm-3 [CO2] = 0.125 mol dm-3 [H2] = 0.50 mol dm-3 = 0.125 × 0.50 4 0.03125 × 0.500 2 em ac ti Equilibrium amounts: CH4 = 0.625 – ¼ × 2.0 = 0. 1 25 mol (1) H2O = 3.0 – ½ 2.0 = 2.0 mol (1) CO2 = 0.0 + ¼ × 2.0 = 0.500 mol (1) H2 = 2.0 mol (given) ve 8 3 Starting amounts: CH 4 = 10g / 16 g mol −1 = 0.625 mol (1) −1 H 2 O = 54 g / 18g mol = 3.0 mol ∆H = (-394) – [(-76) + (2 × – 242)] (2) 1 mark for × 2, 1 mark for signs and values = + 166 (1) kJ mol–1 166 scores zero NT Exampro ww .co 31 43. (a) Kc = [CH 4 ][H 2 O]2 [CO 2 ][H 2 ] 4 (1) m (c) A catalyst (of nickel) is used because the reaction, even at 750°C, is too slow / to speed up the reaction (1) Then any six of the following eight points: a temperature of 750°C is used: • as the reaction is endothermic (1) • a high temperature increases the value of the equilibrium constant (1) • and so increases the equilibrium yield (1) • a high temperature also favours a fast rate (1) • but a temperature > 750 would be too expensive / cause engineering problems (1) Temperature could score up to 5 but max 6 for T and P combined. If their calculation in (b) gives an exothermic answer, mark consequentially [exothermic (1), decreases Kp (1), decreases yield (1) but faster rate(1), so 750 is compromise of fast rate and lower yield (1)] A pressure of 30 atm is used • even though the reaction goes from 3 to 5 gas moles / more gas moles of right of equation (1) • causing a decrease in equilibrium yield (1) • but a moderately high pressure is needed to push the gases through the 7 plant (1) Ignore any reference to rate Pressure could score up to 3 but max 6 for T and P combined. Do not give all 7 marks unless the candidate has expressed their ideas clearly em ac ti ve .co m [18] NT Exampro ww w. ch 32