A-level Chemistry (7405/1) - PDF Free Download

A-level Chemistry (7405/) Paper : Inorganic and Physical Chemistry Specimen 205 v0.5 Session 2 hours Materials For this paper you must have: the Data Booklet, provided as an insert a ruler a calculator. Instructions Answer all questions. Show all your working. Information The maximum mark for this paper is 05. Please write clearly, in block capitals, to allow character computer recognition. Centre number Candidate number Surname Forename(s) Candidate signature Barcode v0.5 7405/

2 Answer all questions. 0. Explain how the electron pair repulsion theory can be used to deduce the shape of, and the bond angle in, PF 3 [6 marks] Barcode Typesetter code

3 0. 2 State the full electron configuration of a cobalt(ii) ion. [ mark] 0. 3 Suggest one reason why electron pair repulsion theory cannot be used to predict the shape of the [CoCl 4 ] 2 ion. [ mark] 0. 4 Predict the shape of, and the bond angle in, the complex rhodium ion [RhCl 4 ] 2 [2 marks] Shape Bond angle Turn over for the next question Barcode Typesetter code Turn over

4 0 2. Explain why the atomic radii of the elements decrease across Period 3 from sodium to chlorine. [2 marks] 0 2. 2 Explain why the melting point of sulfur (S 8 ) is greater than that of phosphorus (P 4 ). [2 marks] 0 2. 3 Explain why sodium oxide forms an alkaline solution when it reacts with water. [2 marks] Barcode Typesetter code

5 0 2. 4 Write an ionic equation for the reaction of phosphorus(v) oxide with an excess of sodium hydroxide solution. [ mark] Turn over for the next question Barcode Typesetter code Turn over

6 3 Fuel cells are an increasingly important energy source for vehicles. Standard electrode potentials are used in understanding some familiar chemical reactions including those in fuel cells. Table contains some standard electrode potential data. Table Electrode half-equation E Ɵ / V F 2 + 2e 2F +2.87 Cl 2 + 2e 2Cl +.36 O 2 + 4H + + 4e 2H 2 O +.23 Br 2 + 2e 2Br +.07 I 2 + 2e 2I +0.54 O 2 + 2H 2 O + 4e 4OH +0.40 SO 2 4 + 4H + + 2e SO 2 + 2H 2 O +0.7 2H + + 2e H 2 0.00 4H 2 O + 4e 4OH + 2H 2 0.83 0 3. A salt bridge was used in a cell to measure electrode potential. Explain the function of the salt bridge. [2 marks] 0 3. 2 Use data from Table to deduce the halide ion that is the weakest reducing agent. [ mark] Barcode Typesetter code

7 0 3. 3 Use data from Table to justify why sulfate ions should not be capable of oxidising bromide ions. [ mark] 0 3. 4 Use data from Table to calculate a value for the EMF of a hydrogen oxygen fuel cell operating under alkaline conditions. [ mark] EMF = V 0 3. 5 There are two ways to use hydrogen as a fuel for cars. One way is in a fuel cell to power an electric motor, the other is as a fuel in an internal combustion engine. Suggest the major advantage of using the fuel cell. [ mark] Turn over for the next question Barcode Typesetter code Turn over

8 4 Many chemical processes release waste products into the atmosphere. Scientists are developing new solid catalysts to convert more efficiently these emissions into useful products, such as fuels. One example is a catalyst to convert these emissions into methanol. The catalyst is thought to work by breaking a H H bond. An equation for this formation of methanol is given below. CO 2 (g) + 3H 2 (g) CH 3 OH(g) + H 2 O(g) ΔH = 49 kj mol Some mean bond enthalpies are shown in Table 2. Table 2 Bond C=O C H C O O H Mean bond enthalpy / kj mol 743 42 360 463 0 4. Use the enthalpy change for the reaction and data from Table 2 to calculate a value for the H H bond enthalpy. [3 marks] H H bond enthalpy = kj mol 0 4. 2 A data book value for the H H bond enthalpy is 436 kj mol. Suggest one reason why this value is different from your answer to Question 4.. [ mark] Barcode Typesetter code

9 0 4. 3 Suggest one environmental advantage of manufacturing methanol fuel by this reaction. [ mark] 0 4. 4 Use Le Chatelier's principle to justify why the reaction is carried out at a high pressure rather than at atmospheric pressure. [3 marks] 0 4. 5 Suggest why the catalyst used in this process may become less efficient if the carbon dioxide and hydrogen contain impurities. [ mark] Question 4 continues on the next page Barcode Typesetter code Turn over

0 0 4. 6 In a laboratory experiment to investigate the reaction shown in the equation below,.0 mol of carbon dioxide and 3.0 mol of hydrogen were sealed into a container. After the mixture had reached equilibrium, at a pressure of 500 kpa, the yield of methanol was 0.86 mol. CO 2 (g) + 3H 2 (g) CH 3 OH(g) + H 2 O(g) Calculate a value for K p Give your answer to the appropriate number of significant figures. Give units with your answer. [7 marks] K p = Units = Barcode Typesetter code

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2 5 Table 3 contains some entropy data relevant to the reaction used to synthesise methanol from carbon dioxide and hydrogen. The reaction is carried out at a temperature of 250 C. Table 3 Substance CO 2 (g) H 2 (g) CH 3 OH(g) H 2 O(g) Entropy (S Ɵ ) / J K mol 24 3 238 89 CO 2 (g) + 3H 2 (g) CH 3 OH(g) + H 2 O(g) H = 49 kj mol 0 5. Use this enthalpy change and data from Table 2 to calculate a value for the free-energy change of the reaction at 250 C. Give units with your answer. [4 marks] Free-energy change = Units = Barcode Typesetter code

3 0 5. 2 Calculate a value for the temperature when the reaction becomes feasible. [2 marks] Temperature = K 0 5. 3 Gaseous methanol from this reaction is liquefied by cooling before storage. Draw a diagram showing the interaction between two molecules of methanol. Explain why methanol is easy to liquefy. [4 marks] Diagram Explanation Barcode Typesetter code Turn over

4 6 Ammonium chloride, when dissolved in water, can act as a weak acid as shown by the following equation. NH 4 + (aq) NH 3 (aq) + H + (aq) Figure shows a graph of data obtained by a student when a solution of sodium hydroxide was added to a solution of ammonium chloride. The ph of the reaction mixture was measured initially and after each addition of the sodium hydroxide solution. 4 3 2 0 Figure ph 9 8 7 6 5 4 0 2 3 4 5 6 7 8 9 0 2 3 Volume of NaOH added / cm 3 0 6. Suggest a suitable piece of apparatus that could be used to measure out the sodium hydroxide solution. Explain why this apparatus is more suitable than a pipette for this purpose. [2 marks] Apparatus Explanation Barcode Typesetter code

5 0 6. 2 Use information from the curve in Figure to explain why the end point of this reaction would be difficult to judge accurately using an indicator. [2 marks] 0 6. 3 The ph at the end point of this reaction is.8 Use this ph value and the ionic product of water, K w =.0 0 4 mol 2 dm 6, to calculate the concentration of hydroxide ions at the end point of the reaction. [3 marks] Concentration = mol dm 3 Question 6 continues on the next page Barcode Typesetter code Turn over

6 0 6. 4 The expression for the acid dissociation constant for aqueous ammonium ions is KK = NN3 [N + ] [NN 4 + ] The initial concentration of the ammonium chloride solution was 2.00 mol dm 3. Use the ph of this solution, before any sodium hydroxide had been added, to calculate a value for K a [3 marks] K a = mol dm 3 0 6. 5 A solution contains equal concentrations of ammonia and ammonium ions. Use your value of K a from Question 6.4 to calculate the ph of this solution. Explain your working. (If you were unable to calculate a value for K a you may assume that it has the value 4.75 0 9 mol dm 3. This is not the correct value.) [2 marks] ph= Barcode Typesetter code

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8 7 Table 4 shows some successive ionisation energy data for atoms of three different elements X, Y and Z. Elements X, Y and Z are Ca, Sc and V but not in that order. Table 4 First Second Third Fourth Fifth Sixth X 648 370 2870 4600 6280 2 400 Y 590 50 4940 6480 820 0 496 Z 632 240 2390 70 8870 0 720 For questions 7. and 7.2, only one answer per question is allowed. For each answer, completely fill in the circle alongside the appropriate answer. CORRECT METHOD WRONG METHODS If you want to change your answer you must cross out your original answer as shown. If you wish to return to an answer previously crossed out, ring the answer you now wish to select as shown. 0 7. Which element is calcium? [ mark] X Y Z 0 7. 2 Which element is vanadium? [ mark] X Y Z Barcode Typesetter code

9 0 7. 3 Justify your choice of vanadium in Question 7.2 [ mark] 0 7. 4 An acidified solution of NH 4 VO 3 reacts with zinc. Explain how observations from this reaction show that vanadium exists in at least two different oxidation states. [2 marks] Question 7 continues on the next page Barcode Typesetter code Turn over

20 0 7. 5 The vanadium in 50.0 cm 3 of a 0.800 mol dm 3 solution of NH 4 VO 3 reacts with 506 cm 3 of sulfur(iv) oxide gas measured at 20.0 C and 98.0 kpa. Use this information to calculate the oxidation state of the vanadium in the solution after the reduction reaction with sulfur(iv) oxide. Explain your working. The gas constant R = 8.3 J K mol. [6 marks] Oxidation state = Barcode Typesetter code

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22 0 8. A co-ordinate bond is formed when a transition metal ion reacts with a ligand. Explain how this co-ordinate bond is formed. [2 marks] 0 8. 2 Describe what you would observe when dilute aqueous ammonia is added dropwise, to excess, to an aqueous solution containing copper(ii) ions. Write equations for the reactions that occur. [4 marks] Barcode Typesetter code

23 0 8. 3 When the complex ion [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ reacts with,2-diaminoethane, the ammonia molecules but not the water molecules are replaced. Write an equation for this reaction. [ mark] 0 8. 4 Suggest why the enthalpy change for the reaction in Question 8.3 is approximately zero. [2 marks] 0 8. 5 Explain why the reaction in Question 8.3 occurs despite having an enthalpy change that is approximately zero. [2 marks] Turn over for the next question Barcode Typesetter code Turn over

24 9 A 5.00 g sample of potassium chloride was added to 50.0 g of water initially at 20.0 C. The mixture was stirred and as the potassium chloride dissolved, the temperature of the solution decreased. 0 9. Describe the steps you would take to determine an accurate minimum temperature that is not influenced by heat from the surroundings. [4 marks] 0 9. 2 The temperature of the water decreased to 4.6 C. Calculate a value, in kj mol, for the enthalpy of solution of potassium chloride. You should assume that only the 50.0 g of water changes in temperature and that the specific heat capacity of water is 4.8 J K g. Give your answer to the appropriate number of significant figures. [4 marks] Enthalpy of solution = kj mol Barcode Typesetter code

25 0 9. 3 The enthalpy of solution of calcium chloride is 82.9 kj mol. The enthalpies of hydration for calcium ions and chloride ions are 650 and 364 kj mol, respectively. Use these values to calculate a value for the lattice enthalpy of dissociation of calcium chloride. [2 marks] Lattice enthalpy of dissociation = kj mol 0 9. 4 Explain why your answer to Question 9.3 is different from the lattice enthalpy of dissociation for magnesium chloride. [2 marks] Barcode Typesetter code Turn over

26 0 Table 5 shows observations of changes from some test-tube reactions of aqueous solutions of compounds Q, R and S with five different aqueous reagents. The initial colours of the solutions are not given. Table 5 Q R S BaCl 2 + HCl no change observed no change observed white precipitate AgNO 3 + HNO 3 pale cream precipitate white precipitate no change observed NaOH Na 2 CO 3 HCl (conc) white precipitate white precipitate, dissolves in excess of NaOH brown precipitate white precipitate white precipitate, bubbles of a gas brown precipitate, bubbles of a gas no change observed no change observed yellow solution 0. Identify each of compounds Q, R and S. You are not required to explain your answers. Identity of Q [6 marks] Identity of R Identity of S Barcode Typesetter code

27 0. 2 Write ionic equations for each of the positive observations with S. [4 marks] END OF QUESTIONS Barcode Typesetter code Turn over

AQA Qualifications A-level Chemistry Paper (7405/): Inorganic and Physical Chemistry Mark scheme 7405 Specimen paper Version 0.5

MARK SCHEME A-level Chemistry Specimen paper Question Marking guidance Mark AO Comments 0. This question is marked using levels of response. Refer to the Mark Scheme Instructions for Examiners for guidance on how to mark this question. Level 3 All stages are covered and the explanation of each stage is generally correct and virtually complete. 5 6 marks Level 2 3 4 marks Level 2 marks Answer is communicated coherently and shows a logical progression from stage to stage 2 then stage 3. All stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies OR two stages are covered and the explanations are generally correct and virtually complete. Answer is mainly coherent and shows progression from stage to stage 3. Two stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies, OR only one stage is covered but the explanation is generally correct and virtually complete. 6 2 AOa 2 AO2a 2 AO2b Indicative chemistry content Stage : Electrons round P P has 5 electrons in the outside shell With 3 electrons from 3 fluorine, there are a total of 8 electrons in outside shell so 3 bond pairs, non-bond pair Stage 2: Electron pair repulsion theory Electron pairs repel as far as possible Lone pair repels more than bonding pairs Stage 3: Conclusions Therefore, tetrahedral / trigonal pyramidal shape With angle of 09(.5) decreased to 07 Level 0 0 marks Answer includes isolated statements but these are not presented in a logical order or show confused reasoning. Insufficient correct chemistry to gain a mark. 2 of 5

MARK SCHEME A-level Chemistry Specimen paper 0.2 s 2 2s 2 2p 6 3s 2 3p 6 3d 7 AOa Allow correct numbers that are not superscripted 0.3 Too many electrons in d sub-shell / orbitals AO3 b 0.4 Tetrahedral (shape) 09.5 AO2a AO2a Allow 09 3 of 5

MARK SCHEME A-level Chemistry Specimen paper Question Marking guidance Mark AO Comments 02. The number of protons increases (across the period) / nuclear charge increases AOa Therefore, the attraction between the nucleus and electrons increases AOa Can only score M2 if M is correct 02.2 S 8 molecules are bigger than P 4 molecules Therefore, van der Waals / dispersion / London forces between molecules are stronger in sulfur AOa AOa Allow sulfur molecules have bigger surface area and sulfur molecules have bigger M r 02.3 Sodium oxide contains O 2 ions These O 2 ions react with water forming OH ions AO2c AO2c O 2 + H 2 O 2OH scores M and M2 02.4 P 4 O 0 + 2OH 4PO 4 3 + 6H2 O AO2d 4 of 5

MARK SCHEME A-level Chemistry Specimen paper Question Marking Guidance Mark AO Comments 03. The ions in the ionic substance in the salt bridge move through the salt bridge To maintain charge balance / complete the circuit AOb AOb 03.2 F AO3 a 03.3 E Ɵ SO 4 2 / SO 2 < E Ɵ Br 2 /Br AO3 a Allow correct answer expressed in words, eg electrode potential for sulfate ions / sulfur dioxide is less than that for bromine / bromide 03.4.23 (V) AO2d 03.5 A fuel cell converts more of the available energy from combustion of hydrogen into kinetic energy of the car / an internal combustion engine wastes more (heat) energy AO3 b 5 of 5

MARK SCHEME A-level Chemistry Specimen paper Question Marking guidance Mark AO Comments 04. Bonds broken = 2(C=O) + 3(H H) = 2 743 + 3 H H Bonds formed = 3(C H) +(C O) + 3(O H) = 3 42 + 360 + 3 463 AOb Both required 49 = [2 743 + 3 (H H)] [3 42 + 360 + 3 463] 3(H H) = 49 2 743 + [3 42 + 360 + 3 463] = 450 AOb Both required H H = 483 (kj mol ) AOb Allow 483.3(3) 04.2 Mean bond enthalpies are not the same as the actual bond enthalpies in CO 2 (and/or methanol and/or water) AOb 04.3 The carbon dioxide (produced on burning methanol) is used up in this reaction AO3 b 04.4 4 mol of gas form 2 mol At high pressure the position of equilibrium moves to the right to lower the pressure / oppose the high pressure This increases the yield of methanol AO2f AO3 b AO3 b 04.5 Impurities (or sulfur compounds) block the active sites AOb Allow catalyst poisoned 6 of 5

MARK SCHEME A-level Chemistry Specimen paper 04.6 Stage : moles of components in the equilibrium mixture CO 2 (g) + 3H 2 (g) CH 3 OH(g) + H 2 O(g) Extended response question Initial moles.0 3.0 0 0 Eqm moles ( 0.86) = 0.4 (3-3 0.86) = 0.42 0.86 0.86 AO2f Stage 2: Partial pressure calculations Total moles of gas = 2.28 AO2f Partial pressures = mol fraction p total p CO2 = mol fraction p total p H2 = mol fraction p total = 0.4 500/2.28 = 30.7 kpa = 0.42 500/2.28 = 92. kpa AO2f M3 is for partial pressures of both reactants Alternative M3 = pp CO2 = 0.064 500 pp H2 = 0.842 500 p CH3OH = mol fraction p total = 0.86 500/2.28 = 88.6 kpa p H2O = mol fraction p total = 0.86 500/2.28 = 88.6 kpa AO2f M4 is for partial pressures of both products Alternative M4 = Stage 3: Equilibrium constant calculation K p = p CH3OH p H2O / p CO2 (p H2 ) 3 AO2f pp CH3OH = 0.3772 500 pp H2O = 0.3772 500 Hence K p = 88.6 88.6 / 30.7 (92.) 3 =.483 0 3 =.5 0 3 AOb Answer must be to 2 significant figures Units = kpa 2 AO2f 7 of 5

MARK SCHEME A-level Chemistry Specimen paper Question Marking guidance Mark AO Comments 05. S = 238 + 89 24 3 3 = 80 J K mol AOb G = H T S AOa = 49 523 ( 80) 000 AOb = +45. kj mol AOb Units essential 05.2 When G = 0, H = T S therefore T = H/ S = 49 000/ 80 = 272 (K) AOb AOb Mark consequentially to S in 5. 8 of 5

MARK SCHEME A-level Chemistry Specimen paper 05.3 Diagram marks δ- δ+ Diagram of a molecule showing O H bond and two lone pairs on each oxygen AO2a Labels on diagram showing δ+ and δ- charges AO2a Allow explanation of position of δ+ and δ- charges on H and O Diagram showing δ+ hydrogen on one molecule attracted to lone pair on a second molecule AO2a Explanation mark Hydrogen bonding (the name mentioned) is a strong enough force (to hold methanol molecules together in a liquid) AO2a 9 of 5

MARK SCHEME A-level Chemistry Specimen paper Question Marking guidance Mark AO Comments 06. Burette AO3 b Because it can deliver variable volumes AO2g 06.2 The change in ph is gradual / not rapid at the end point AO3 a An indicator would change colour over a range of volumes of sodium hydroxide AO3 a Allow indicator would not change colour rapidly / with a few drops of NaOH 06.3 [H + ] = 0 ph =.58 0 2 K w = [H + ] [OH ] therefore [OH ] = K w / [H + ] Therefore, [OH ] = 0 4 /.58 0 2 = 6.33 0 3 (mol dm 3 ) AO2h AO2h AO2h Allow 6.3 6.33 0 3 (mol dm 3 ) 06.4 At this point, [NH 3 ] = [H + ] Therefore K a = [H + ] 2 [NH + 4 ] [H + ] = 0 4.6 = 2.5 0 5 K a = (2.5 0 5 ) 2 /2 = 3.5 0 0 (mol dm 3 ) AO2f AO2f AO2f Allow 3.5 3.6 0 0 (mol dm 3 ) 06.5 When [NH 3 ] = [NH 4 + ], K a = [H + ] therefore log K a = log [H + ] AO2h Answer using alternative value Therefore ph = log 0 (3.5 0 0 ) = 9.50 AO2h M2 ph = log 0 (4.75 0 9 ) = 8.32 Allow consequential marking based on answer from 6.4 0 of 5

MARK SCHEME A-level Chemistry Specimen paper Question Marking guidance Mark AO Comments 07. Y AO3 a 07.2 X AO3 a 07.3 Jump in trend of ionisation energies after removal of fifth electron Fits with an element with 5 outer electrons (4s 2 3d 3 ) like V AO2b 07.4 Explanation: Two different colours of solution are observed AO2g Because each colour is due to vanadium in a different oxidation state AO2g of 5

MARK SCHEME A-level Chemistry Specimen paper 07.5 Extended response Stage : mole calculations in either order Moles of vanadium = 50.0 0.800/000 = 4.00 x 0 2 Moles of SO 2 = pv/rt = (98 000 506 0 6 )/(8.3 293) = 2.04 0 2 AO2d AO2d Maximum of 5 marks for answers which do not show a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. Stage 2: moles of electrons added to NH 4 VO 3 When SO 2 (sulfur(iv) oxide) acts as a reducing agent, it is oxidised to sulfate(vi) ions so this is a two electron change AO2d Moles of electrons released when SO 2 is oxidised = 2.04 0 2 2 = 4.08 0 2 AO2d Stage 3 : conclusion But in NH 4 VO 3 vanadium is in oxidation state 5 AO2b 4.00 0 2 mol vanadium has gained 4.08 0 2 mol of electrons therefore mol vanadium has gained 4.08 0 2 / 4.00 0 2 = mol of electrons to the nearest integer, so new oxidation state is 5 =4. AO2d 2 of 5

MARK SCHEME A-level Chemistry Specimen paper Question Marking guidance Mark AO Comments 08. An electron pair on the ligand Is donated from the ligand to the central metal ion 08.2 Blue precipitate Dissolves to give a dark blue solution [Cu(H 2 O) 6 ] 2+ + 2NH 3 + Cu(H 2 O) 4 (OH) 2 + 2NH 4 Cu(H 2 O) 4 (OH) 2 + 4NH 3 [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ + 2OH + 2H 2 O AOa AOa AOb AOb AO2d AO2d 08.3 [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ + 2H 2 NCH 2 CH 2 NH 2 AO2b [Cu(H 2 NCH 2 CH 2 NH 2 ) 2 (H 2 O) 2 ] 2+ + 4NH 3 08.4 Cu N bonds formed have similar enthalpy / energy to Cu N bonds broken AO3 b And the same number of bonds broken and made AO3 b 08.5 3 particles form 5 particles / disorder increases because more particles are formed / entropy change is positive AO2e Therefore, the free-energy change is negative AO2e M2 can only be awarded if M is correct 3 of 5

MARK SCHEME A-level Chemistry Specimen paper Question Marking guidance Mark AO Comments 09. Start a clock when KCl is added to water AO3 2b Record the temperature every subsequent minute for about 5 minutes Plot a graph of temperature vs time Extrapolate back to time of mixing = 0 and determine the temperature AO3 2b AO3 2a AO3 2a Allow record the temperature at regular time intervals until some time after all the solid has dissolved for M2 09.2 Heat taken in = m c T = 50 4.8 5.4 = 28.6 J AO2h Max 2 if 4.6 C used as T Moles of KCl = 5.00/74.6 = 0.0670 AO2h Enthalpy change per mole = +28.6/0.0670 = 6 839 J mol AO2h = +6.8 (kj mol ) AOb Answer must be given to this precision 09.3 H solution = H lattice + H(hydration of calcium ions) + 2 H(hydration of chloride ions) H lattice = H solution H(hydration of calcium ions) 2 H(hydration of chloride ions) H lattice = 82.9 ( 650 + 2 364) = +2295 (kj mol ) AO2f AO2f 09.4 Magnesium ion is smaller than the calcium ion AO2a Therefore, it attracts the chloride ion more strongly / stronger ionic bonding AO2a 4 of 5

MARK SCHEME A-level Chemistry Specimen paper Question Marking guidance Mark AO Comments 0. Q is calcium or magnesium AO3 b Mark this question independently bromide AO3 b R is aluminium AO3 b chloride AO3 b S is iron(iii) AO3 b sulfate AO3 b 0.2 Ba 2+ + SO 4 2 BaSO 4 AOa [Fe(H 2 O) 6 ] 3+ + 3OH Fe(H 2 O) 3 (OH) 3 + 3H 2 O AOa 2[Fe(H 2 O) 6 ] 3+ + 3CO 3 2 [Fe(H 2 O) 6 ] 3+ + 4Cl 2Fe(H 2 O) 3 (OH) 3 + 3H 2 O + 3CO 2 [FeCl 4 ] + 6H 2 O AOa AOa 5 of 5