marked-2022-merewether-trial-paper-criteria-sample-answers-64460dddc4290

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Question 21 (9 marks) Phosgene (COCl2) is a colourless gas that has an odour of freshly cut grass. It decomposes into chlorine gas and carbon monoxide through an endothermic reaction that establishes an equilibrium. a) Write a chemical equation for this reaction including an appropriate scientific representation that it is endothermic. (2 marks) b) Write an equilibrium expression for this reaction. (1 mark) Phosgene, chlorine and carbon monoxide were mixed in a sealed container and allowed to reach equilibrium. A change was imposed at time T and the equilibrium re-established. The concentration of each gas was plotted against time and are represented with the letters X, Y and Z. c) Identify the gas that is represented by the letter X on the graph. (1 mark) d) Identify the change that occurred at time T and explain using Le Chatelier's Principle why the equilibrium re-established as shown in the graph. (4 marks) e) Use the graph to calculate the equilibrium constant for the reaction before time T. (1 mark)
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Question 22 (4 marks) The heat of neutralisation for a reaction between a strong acid and a strong base is: ΔH = -57.62 kJ/mol. A student mixed 50.0 mL of 0.100 M hydrochloric acid with 50.0 mL of 0.200 M sodium hydroxide. They determined the mass of the final solution was 98.03 g. The teacher instructed the student to assume that the specific heat capacity of the solution would be the same as pure water. Calculate the theoretical change in temperature of the reaction mixture. (4 marks) Question 23 (5 marks) Magnesium chloride is an ionic compound which is highly soluble in water. a) Write the equation, including state symbols, for the process of forming a saturated solution of magnesium chloride in water (1 mark) b) The table below shows some thermochemical data related to the dissolving of magnesium chloride solid in water. Use these data to calculate the standard enthalpy of solution of magnesium chloride. (2 marks) c) Use your answer to part (b) to deduce how the solubility of MgCl2 changes as the temperature is increased. Justify your answer. (2 marks) Question 24 (5 marks) Propan-1-amine has a boiling point of 47º C while N,N-dimethylmethanamine has a boiling point of just 3º C. Draw the structural formula for each compound and explain why there is a difference in boiling point. (5 marks)
Question 25 (6 marks) A student wanted to determine the concentration of a solution of potassium hydroxide. They did this by reacting a 25 mL aliquot with a standard solution of hydrochloric acid while measuring the conductivity of the solution. 10.00 mL volumes of 0.1012 M hydrochloric acid were incrementally added using a burette and the conductivity was measured after each addition. The student recorded the following data. a) Explain the trend shown in the graph. (3 marks) b) Calculate the concentration of the potassium hydroxide solution. (3 marks)
Question 26 (8 marks) Linseed oil is a mixture of triesters. It is used for many purposes including as a protective layer for wooden furniture and as an ingredient in some cuisines. The diagram below shows the structure of a typical triester found in linseed oil. When the triester shown above undergoes the process of de-esterification (this process is the opposite of esterification), three fatty acids are produced: ● palmitic acid ● linoleic acid ● oleic acid which correspond to the three long chains in the diagram listed from top to bottom. a) Why are compounds like the one depicted above called triesters? (1 mark) b) In addition to the three acids listed above, ONE other product will form, and ONE other reactant will be consumed when the compound undergoes de-esterification. Identify the product and reactant and state the number of moles that would be produced/consumed from the complete de-esterification of one mole of the compound shown. (2 marks) c) Castor oil is another natural product that may find a use as a component in biofuel. It releases 39.5 MJ kg−1 of energy whe n fully combusted. When 2.00 L of castor oil was combusted in a home-made engine 59.3 MJ of energy was released. Given that the density of castor oil is 0.96 g mL−1, calculate percentage efficiency of the home - made engine. (3 marks) d) The picture below shows a laboratory reflux apparatus using a heating mantle.
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e) Explain how TWO risks associated with heating volatile organic compounds are reduced by using a reflux apparatus with a heating mantle. (2 marks) Question 27 (3 marks) Sodium dihydrogen phosphate is an amphiprotic salt. Explain what is meant by the term ‘amphiprotic’. Support your answer with chemical equations. (3 marks) Question 28 (5 marks) A student obtained the following spectroscopic data to help to identify an unknown organic compound, labelled Compound X.
Identify Compound X, using BOTH spectra to justify your identification. (5 marks) Question 29 (8 marks) A solution of 4-chlorobutanoic acid was titrated against standardised sodium hydroxide sol ution. Small volumes of sodium hydroxide solution were added, and the pH was measured after each addition. The data collected are shown in the table below. a) Draw the structural formula of 4-chlorobutanoic acid. (1 mark) b) Plot the values of the pH of the solution against the volume of NaOH and draw a smooth curve of best fit, ignoring possible outliers. (4 marks) Use own graph paper… c) Estimate the pH at the equivalence point in this titr ation. (1 mark) d) Use your graph to estimate a value of Ka for 4-chlorobutanoic acid. (2 marks)
Question 30 (8 marks) The following table shows four acids which were tested for relative conductivity and pH: a) Explain why the pH for glacial acetic acid is unable to be measu red. (2 marks) b) Explain the differences in the pH values for the 0.1 mol L -1 acids that were tested. Include relevant calculations. (6 marks) Exam continues on next page
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Question 31 (7 marks) The flow chart shows a series of reactions that would allow a chemist to synthesise propyl propanoate from propene.
a) Identify reagent 1, reagent 2, reagent 3, compound 3, compound 5 and catalyst 1 (5 marks) b) Identify the following reaction types: (2 marks) i) Compound 1 → Compound 2 ii)Compound 3 → Compound 4 iii)Compound 4 + Compound 5 → Compound 6 Question 32 (5 marks) a) Given the molar solubility of MnCO3 is 4.20 x 10-6 M, calculate the solubility product for MnCO3. (2 marks) b) A 0.200g sample of solid barium sulfate was added to 500mL of 0.010M sodium sulfate. Will all the barium sulfate sample dissolve? Justify your answer. (3 marks) Question 33 (7 marks) 100.0 mL of 0.200 M hydrochloric acid was combined with 100.0 mL of 0.200 M ammonia solution. Calculate the pH of the final solution once the reaction has gone through to completion. (7 marks)
10 Question 21 (9 marks) Marks Phosgene (COCl 2 ) is a colourless gas that has an odour of freshly cut grass. It decomposes into chlorine gas and carbon monoxide through an endothermic reaction that establishes an equilibrium. a) Write a chemical equation for this reaction including an appropriate scientific representation that it is endothermic. Marking Criteria Marks ● Correct chemical equation including states ● Indicates that the enthalpy change is positive 1 2 Sample Answer Markers Comments Generally well done. Common mistakes were to leave out states and to include heat as a reactant. Also, some students had +ΔH along with the other products which represents it as a product. This is not how you represent an endothermic reaction. You should h ave ΔH= or ΔH > 0 separate to the equation as it is not made of atoms, it is a representation of the energy change. b) Write an equilibrium expression for this reaction. Marking Criteria Marks ● Correct equilibrium expression 1 Sample Answer Markers Comments Well done by most students. A very small number of student had it upside down (the reactant on top) and some did not include K=. This is an important part of the equilibrium expression. Phosgene, chlorine and carbon monoxide were mixed in a sealed container and allowed to reach equilibrium. A change was imposed at time T and the equilibrium re-established. The 1 1
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11 concentration of each gas was plotted against time and are represented with the letters X, Y and Z. Question 21 continued c) Identify the gas that is represented by the letter X on the graph. Marking Criteria Marks ● Correctly identifies phosgene (COCl 2 ) 1 Markers Comments More than 90% of the cohort answered this question correctly. d) Identify the change that occurred at time T and explain using Le Chatelier's Principle why the equilibrium re-established as shown in the graph. Marking Criteria Marks ● Identifies the change as an increase in volume or that gas had been removed. ● States that this results in a decrease in pressure ● Explains that the equilibrium position shifts to the products to counteract the change ● Because this increases the number of gas particles, therefore increase pressure 4 4 1
12 ● One mistake or step missing from above 3 ● Explains why the shift in equilibrium position counteracts an identified change (does not have to be pressure) 2 ● States a change and predicts a shift in equilibrium position 1 Markers Comments The graph for this question showed a sharp almost instantaneous drop in both reactant and products, with all decreasing by the same proportion. There were only two possible changes that could have caused this. Either the volume of the container increased or gas (made up of the reaction mixture) was released. Both of these results in a pressure decrease . Students had to identify the change and that it resulted in a pressure increase to achieve full marks. Students then needed to explain that shifting the equilibrium position to the right will counteract the drop in pressure because it will create more gas particles, therefore, increase the pressure. Common mistakes x Identifying a drop in pressure as the change without identifying what must have caused the drop e.g. volume increase. x Identifying an increase in volume or a decrease in concentration of all chemical species as the disturbance that the equilibrium was counteracting. The equilibrium cannot shift to decrease the volume. The equilibrium shifting to the products does not counteract the change of everything decreasing in concentration. It shifted to the products to counteract the change in pressure. If students did not explain the shift in terms of pressure, they could only achieve a maximum of two marks. x Only stating that shifting to the right counteracts the change without explain how producing more gas particles increases pressure. Sample Answer
13 e) Use the graph to calculate the equilibrium constant for the reaction before time T. Marking Criteria Marks ● Calculates the equilibrium constant (0.024) 1 Markers Comments Well done by most students. Some students had their equation upside down and some used incorrect data from the graph. Question 22 (4 marks) Marks The heat of neutralisation for a reaction between a strong acid and a strong base is: ΔH = -57.62 kJ/mol. A student mixed 50.0 mL of 0.100 M hydrochloric acid with 50.0 mL of 0.200 M sodium hydroxide. They determined the mass of the final solution was 98.03 g. The teacher instructed the student to assume that the specific heat capacity of the solution would be the same as pure water. Calculate the theoretical change in temperature of the reaction mixture. 4 Marking Criteria Marks ● Justifies HCl as the limiting reagent ● Calculates q (288.1 J) with working ● Calculates the change in temperature (0.703° C) with working and 4
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14 units ● Give the final answer to 3 significant figures ● One mistake or step missing from above 3 ● Two mistakes or steps missing from above 2 ● Attempts at least one of the above step 1 Markers Comments Most students understood the general method needed to answer this question, however, there were many simple mistakes and a lack of understanding of units and setting out that resulted in marks being lost. Common mistakes x The question clearly stated that 98.03 g was the mass of the solution. Some students used this as a mass of NaCl, which is impossible given the moles of HCl you started with. Some other students subtracted this mass from 100 g. Don’t get locked into always using the volume of the solution to find the mass. The most accurate method for these types of question will measure the mass of the substance being heated or cooled by the reaction. x ΔH and q are not the same thing, a good understanding of the units in this question will make this clear. You must remember that ΔH= -q/n, and the units kJ/mol should help you to remember this. This was the most common major mistake. x Be very careful with the units you are using. You should always use J for q and then decide if you want to use C from the data sheet, in which case your mass must be in kg or 4.18 J/g/K in which case your mass must be g. This caused a number of students to be out by either x1000 or x1000000. A good way to check your working is to write all of your units in your data list and make sure they agree. x There was some generally poor setting out, missing equations and not substituting. Don’t be lazy or you will inevitably drop marks you shouldn’t. x Students did not identify and justify HCl as the limiting reagent. This must be done by comparing the number of moles. Too many students added the moles of HCl and NaOH together. I’m not sure where this came from, think of any simple molar calculation you do, no never add the moles together, they are telling you what will react and what will not. x Finally, be careful with you - ve and +ve numbers and check you final answer. If it’s an exothermic reaction, the temperature must go up. Sample Answer
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15 Question 23 (5 marks) Marks Magnesium chloride is an ionic compound which is highly soluble in water. a) Write the equation, including state symbols, for the process of forming a saturated solution of magnesium chloride in water. Marking Criteria Marks ● Correct chemical equation with states 1 Sample Answer 1
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16 Markers Comments This should have been a simple question, but many students lost this mark. There were many answers that had water as a reactant and product. This should indicate to you it shouldn’t be there. The equation should simply show solid magnesium chloride becoming aqueous ions. Some students had the reactant as aqueous; this would mean the reaction had already happened. Some did not use equilibrium arrows even though this is a large part of the equilibrium module. And others showed magnesium reacting with water to form magnesium o xide and hydrochloric acid… b) The table below shows some thermochemical data related to the dissolving of magnesium chloride solid in water. Energy required to overcome the lattice forces to form free Mg 2+ and Cl - ions. +2493 kJ mol – 1 Energy released upon hydration of Mg 2+ ions – 1920 kJ mol – 1 Energy released upon hydration of Cl - ions – 364 kJ mol – 1 Use these data to calculate the standard enthalpy of solution of magnesium chloride. Marking Criteria Marks ● Calculates the standard enthalpy ● Including units and working 2 ● Attempts to combine standard energies 1 Markers Comments To answer this question students had to apply the energy per mole data to the equation from part a. Energy required to overcome the lattice forces to form free Mg 2+ and Cl - ions is the energy required to break up the solid MgCl 2 and is kJ/mol, meaning that’s how much energy it would take if you had one mole of MgCl 2 . The next two is the energy released when each ion is hydrated by water molecules and is also per mol. From the reaction in part a, it is clear that the energy would be double for Cl - as there are 2 mole produced. All of the energies per mole then needed to added together according to the molar ratio of the reaction. To achieve full marks you must have shown you working. The most common mistake was not doubling the value for Cl - . Sample Answer 2 2
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17 c) Use your answer to part (b) to deduce how the solubility of MgCl 2 changes as the temperature is increased. Justify your answer. Marking Criteria Marks ● Explains that an increase in temperature will cause the equilibrium position to shift to the reactant. ● Links the shift left to reduced solubility If the previous question was a positive enthalpy, the opposite relationship was required. 2 ● Explains how an increase in temperature will cause equilibrium position to shift. 1 Markers Comments This question was best answered using LCP. As the enthalpy from the previous question was negative the reaction is exothermic. This means heating the reaction will cause the equilibrium position to shift to the reactants which is the solid MgCl 2 decreasing the concentration of dissolved ions and therefore solubility. If students had an incorrect positive value for the previous question, they needed to answer accordingly. Some students explained that the higher temperature would give of the MgCl 2 the correct activation energy to dissolve increasing solubility. This neglects that the reverse reaction would also have more ions with the correct activation energy to react in the reverse direction. It is all about the equilibrium position and it is equilibrium position that determines solubility. No marks were given to simply saying the solubility increased or decrease as this is a 50-50 guess. There must be some justification. Sample Answer Marks
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18 Question 24 (5 marks) Propan-1-amine has a boiling point of 47º C while N,N-dimethylmethanamine has a boiling point of just 3º C. Draw the structural formula for each compound in the space provided and explain why there is a difference in boiling point. Marking Criteria Marks ● Correct structures (one mark each) 2 5 Marking Criteria Marks ● Identifies the major intermolecular force for Propan-1-amine as hydrogen bonding ● Identifies the major intermolecular force for N,N- dimethylmethanamine as dipole-dipole interactions ● Explains that hydrogen bond require more energy to over come than dispersion forces, therefore, Propan-1-amine has a higher boiling point 3 ● Correctly identifies the major intermolecular force one of the compounds. ● Relates the strength of intermolecular forces to boiling point 2 ● Links intermolecular forces to boiling point 1 Markers Comments About 60% of students gave correct structures indicating that a significant number of you need to review naming and organic structures. The key to answering this question lies in intermolecular forces. As soon as you see a question about comparing physical properties like boiling point or solubility etc. you should go straight to intermolecular forces. In this case there is a key difference between the two molecules. Propan-1-amine will form hydrogen bonds between molecules because it has two hydrogens attached to its nitrogen atom. N,N-dimethylmethanamine cannot form hydrogen bonds because it has only carbons attached to its nitrogen. It is also asymmetric; therefore the electronegative nitrogen will form a partially negative pole and allow dipole-dipole interactions to form between molecules. However, dipole-dipole interactions are weaker than hydrogen bonds. It is this difference in intermolecular forces that account for the large difference in boiling points. Although the branching and linear nature of the two carbon chains will have a small effect on boiling point it is not the major effect in this case. Students had to correctly identify the major intermolecular force for each molecule and explain why this caused the difference in boiling points. There were no carried errors for an incorrect structure Common mistakes Marks
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19 x Stating that N,N-dimethylmethanamine forms hydrogen bonds because it has a nitrogen atom and hydrogen atoms. This is incorrect because the hydrogens were not attached to the nitrogen. x Incorrectly stating that N,N-dimethylmethanamine does not produce dipole-dipole interactions between molecules missing that this molecule is asymmetrical. x Correctly stating the intermolecular forces at play but not explaining why this causes the difference in boiling point. Sample Answer Question 25 (6 marks) A student wanted to determine the concentration of a solution of potassium hydroxide. They did this by reacting a 25 mL aliquot with a standard solution of hydrochloric acid while measuring the conductivity of the solution. 10.00 mL volumes of 0.1012 M hydrochloric acid were incrementally added using a burette and the conductivity was measured after each addition. The student recorded the following data.
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20 Question 25 continued a) Explain the trend shown in the graph. Marking Criteria Marks ● Explains that the decrease in conductivity it because of the hydroxide ions concentration decreasing. ● Identifies the lowest point of conductivity as the equivalence point. ● Explains that the increase in conductivity is because of the hydrogen ion concentration increasing 3 ● Correctly links either the decrease or increase in conductivity to the increase or decrease of hydrogen or hydroxide ions ● Identifies the lowest point of conductivity as the equivalence point. 2 ● Links conductivity to the concentration ions in solution 1 Markers Comments A lot of students achieved two marks for this question with the major mistake not focussing on the key ions that caused the change in conductivity. The initial decrease in conductivity was solely due to the OH - concentration decreasing. If you did not state that you could not achieve full marks. Common mistakes here were to say that the HCl neutralised the KOH decreasing the number of ions. Firstly this isn’t true, the number of ions remained const ant because for every OH - that was removed it was replaced with a Cl - . Another was to say that both the OH - and K + were removed, this was incorrect because the K + remained as free ions. If the OH - and K + were treated equally, only a max of 2 marks was possible. Better answers described the lowest point in conductivity as the equivalence point and explained that only the less conductive K + and Cl - were present at this point. The increase in conductivity must have been explained by an increase in H + . The Cl - were also increasing, but they had been increasing the whole time, therefore it is the H + that are the major cause for the increase in conductivity. Students that did not specifically write about any of the ions that effected conductivity could only achieve a maximum of 1 mark. Some students describe the equ Sample Answer Marks 3 3
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21 b) Calculate the concentration of the potassium hydroxide solution. Marking Criteria Marks ● Selects the correct volume of HCl (54-56 mL) ● Calculates the number of moles of HCl ● Calculates the concentration of KOH 3 ● One mistake of the above 2 ● Attempts of the above steps 1 Markers Comments This question was well done by the vast majority of students. Most students correctly identified the volume of HCl at the equivalence point, I accepted 54-56 mL. The most common way students lost marks was for poor setting out and not including steps they took in their answer. If you are using an equation write it down and substitute it. Some students added the titre to the aliquot volume when finding the final concentration. You never do this for a titration question as you are finding the concentration of the original solution, if you include the addition of the titre at the equivalence point then the concentration will be zero as
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22 it will have fully reacted. It was also important to state the molar ratio, even though it is 1:1 this can not be assumed. Some students used the equation C 1 V 1 =C 2 V 2 as this was a 1:1 molar ratio it works for this question, however, in a previous HSC this method was not accepted. So, I would advise that you use C=n/V to answer titration calculation questions. Sample Answer
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23
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10 Question 26 (8 marks) Marked by Mr Bell Linseed oil is a mixture of triesters. It is used for many purposes including as a protective layer for wooden furniture and as an ingredient in some cuisines. The diagram below shows the structure of a typical triester found in linseed oil. When the triester shown above undergoes the process of de-esterification (this process is the opposite of esterification), three fatty acids are produced: ● palmitic acid ● linoleic acid ● oleic acid which correspond to the three long chains in the diagram listed from top to bottom. a) Why are compounds like the one depicted above called triesters? Marking Criteria Marks ● Identifies that there are three ester functional groups in the compound 1 Marker’s Comments This question was completed well by approximately 70% of the cohort. The marker accepted answers that indicated that there were 3 ester functional groups in the compound. Disappointingly, some students stated that the compound was formed by a reaction between 3 esters; the question states that the process of de-esterification ‘d ecomposes ’ the compound, so it would imply that the process of esterification (a reaction between an alcohol + carboxylic acids as stated in the stimulus) would produce the compound. Sample Answer
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11 b) In addition to the three acids listed above, ONE other product will form, and ONE other reactant will be consumed when the compound undergoes de-esterification. Identify the product and reactant and state the number of moles that would be produced/consumed from the complete de-esterification of one mole of the compound shown. Marking Criteria Marks ● Identifies the reactant as water and the product as propane-1,2,3- triol (i.e. glycerin, glycerol, a triol) ● States that 3 moles of water are consumed and 1 mole of propane- 1,2,3-triol is produced 2 ● Correctly identifies a product or reactant 1 Note – ½ marks were awarded for the identification of the reactant + product and the mole ratio of each. Marker’s Comments Many of the cohort were able to gain full marks for this question as they were able to understand the de-esterification process would form carboxylic acids (or fatty acids as stated in the stimulus) and an alcohol. The marker accepted a variety of responses for the naming of the product, as stated in the marking criteria, and was pleased that a significant proportion of the cohort were able to realise that the terms were appropriate for this question. Please note that the numbering of the 3 hydroxyl groups should be within the name and not out the front of the name (i.e. 1,2,3- propanetriol). The use of the common names for the triol, glycerin and glycerol suggest that some students have studied the process of saponification. Technically, the process of saponification is considered extension knowledge for this course and is not mentioned in the syllabus; however, the knowledge of the outcomes of the process is. Additionally, despite saponification being a special example of de-esterification using NaOH (basic hydrolysis of an ester), this question is not an example of saponification (i.e. glycerin would be formed along with 3 carboxylic salts or soap molecule) since the 3 fatty acids (along with glycerin) are formed and are named carboxylic acids in the stimulus material. This example is an acidic hydrolysis reaction using water as a reactant in the presence of an acid to form 3 different acids and glycerin. Therefore, NaOH was not considered to be correct as the reactant in this case. The process of saponification is considered extension knowledge for this course. The majority of the cohort were able to gain the marks for the correct mole ratio of a reactant and product regardless of the reactant or product identified. Sample Answer
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12 c) Castor oil is another natural product that may find a use as a component in biofuel. It releases 39.5 MJ kg − 1 of energy when fully combusted. When 2.00 L of castor oil was combusted in a home-made engine 59.3 MJ of energy was released. Given that the density of castor oil is 0.96 g mL − 1 , calculate percentage efficiency of the home-made engine. Marking Criteria Marks Shows full working for the following calculations: ● Uses the density of the oil to determine the mass of 2 L sample ● Determines the theoretical energy release of energy from the 2L sample ● Determines the % efficiency of the engine. 3 ● TWO of the above criteria correct 2 ● ONE of the above criteria correct Or ● Some relevant information 1 Marker’s Comments There were several ways students in the cohort approached this problem and examples of the most popular ways have been included as sample answers below. This was a question about units ; pleasingly, many in the cohort made use of the units and manipulated them to calculate what the question was asking. Better responses showed good planning, setting out their calculations in sequential order, identifying what each calculation was, showing full working and clearly indicated the units for each calculation. A point to note for many students is recording your final answer in significant figures with many stating their answer to 3 sig figs whereas the data has a limit of 2 sig figs (i.e. 0.96 g mL −1 ). No marks were awarded here for sig figs this time, but any question could have a mark allocated to them in the HSC. Some other points students had difficulty with were: x All components of a calculation must be included (e.g. x 100 for % calculations) to gain full marks ( ½ marks were awarded if there were missing components) x Students who did not allow for the density of the oil but went on to calculate efficiency could only gain a maximum of 2 marks. x Students with a correct final answer, which was gained from calculated data that was incorrect only gain a maximum of 2 marks.
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13 x Some relevant information was appropriate if it was not originally mentioned in the stem of the question. Sample Answers d) The picture below shows a laboratory reflux apparatus using a heating mantle.
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14 Explain how TWO risks associated with heating volatile organic compounds are reduced by using a reflux apparatus with a heating mantle. Marking Criteria Marks ● Identifies TWO specific risks associated with heating volatile organic compounds and how they are minimised using a heating mantle with a reflux apparatus or ● Identifies ONE specific risk associated with heating volatile organic compounds and how it is minimised using a heating mantle with a reflux apparatus and ● Identifies ONE specific risk associated with heating volatile organic compounds and how it is minimised using a reflux apparatus 2 ● Identifies ONE specific risk associated with heating volatile organic compounds and how it is minimised using a heating mantle with a reflux apparatus or ● Identifies ONE specific risk associated with heating volatile organic compounds and how it is minimised using a reflux apparatus 1 Marker’s Comments This question was really designed to ask for 2 risks that were minimised using a heating mantle during reflux procedures. Since most of the cohort did not do this, the marking scheme was modified to include 1 other risk + minimisation. A better response for this question included the identification of 2 appropriate risks and relevant minimisations when using the apparatus shown in the diagram (Note – a least 1 of the risks + minimisation had to be associated with the use of the heating mantle). Unfortunately, only 35% of the cohort did this. Some difficulties students experienced when formulating a response were: x Identifying a risk but not clearly stating how the use of apparatus minimises the risk; many students implied the minimisation or risk a technique that does not gain marks in any HSC Science exam. ( ½ marks were awarded in these cases) x Many students claimed that the apparatus is an example of a closed system. If this were so, an increase in pressure would be a significant problem since volatile compounds are being heated. For this reason, it is an open system with the condenser minimising the release of volatile compounds x Misinterpreting what a heating mantle is and does. A heating mantle uses electricity and to heat thermal wool which is placed around the reaction vessel. This provides even heating of the compounds at a controlled temperature. Therefore, the use of the mantle avoids: o Vigorous boiling of the volatile compound (i.e. known as ‘bumping’) risks producing spills of hot liquid and/or toxic vapours. o No naked flames are used preventing the ignition of any vapour present Sample Answers
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15 Question 27 (3 marks) Marked by Mr Bell Sodium dihydrogen phosphate is an amphiprotic salt. Explain what is meant by the term ‘ amphiprotic ’. Support your answer with chemical equations. Marking Criteria Marks ● Explains that amphiprotic compounds can: o donate and accept a proton. or o act as a Bronsted-Lowry acid and base ● Supports the explanation using at least TWO fully correct equations (including states of matter) showing appropriate example(s) of an amphiprotic compound that is clearly acting as a: o Bronsted-Lowry acid (donating a proton) and o Bronsted-Lowry base (accepting a proton) 3 ● ONE component of the above criteria is missing or incorrect. 2 ● TWO components of the above criteria are missing or incorrect. 1 Marker’s Comments Many in the cohort had difficulty with this question; responses showed significant gaps in knowledge and understanding of the amphiprotic concept. To make it clear for many students, amphiprotic compounds can act as a Bronsted-Lowry acid or base since they can donate (acidic nature) or accept (basic nature) protons (i.e. H + ions). Additionally, when asked to support responses with equations amphiprotic substances act as an acid when placed with a strong base
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16 (e.g. OH - ) and as a base when placed with an strong acid (e.g. H + ) (Note - it is easier to balance the equation if the spectator ions are removed). Better responses were: x concise, providing a clear definition of the term. x acknowledged that this concept was an application of Bronsted-Lowry definition of acid and bases. x used the example (sodium dihydrogen phosphate - NaH 2 PO 3 ) provided to write balanced equations (including states of matter) demonstrating its acidic and basic capacity. (Most responses removed the spectator ions to balance the equation). Some responses used H + and OH - in these reactions while other showed the interaction with H 2 O (also being an amphiprotic substance). Some issues students experienced were: x confusion with the definition of the term as a significant number of students thought the question was referring to the concept of polyprotic acids; they defined the term and went on to explain the multiple ionisation reactions to support their response. (No marks awarded). x equations were written with no context to the question; often no explanation (e.g. how the amphiprotic compound was acting as an acid or base donating or accepting a proton) was given for the reaction. Remember if your response does not CLEARLY describe the action of the amphiprotic compound to the marker, then marks cannot be awarded to the response. x reactions were written without states of matter. It is expected at this stage of the course that students include states of matter in ALL equations. (Very minor errors were award 1/2 marks). Sample Answers
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18 Question 28 (5 marks) Marked by Mr Bell A student obtained the following spectroscopic data to help to identify an unknown organic compound, labelled Compound X. Question 28 continues over the page
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19 Question 28 continued Identify Compound X, using BOTH spectra to justify your identification. Marking Criteria Marks To identify Compound X, students: x Identify the peak at around 1715 cm -1 from the IR spectrum and links this to the presence of a carbonyl group x Identify the molecular ion peak at 100 from the mass spectrum and links this to the molar mass of the compound x Identify at least ONE other mass spectrum peak that corresponds to a fragment that could only form from Compound X. x Identify additional evidence from EITHER spectra that supports the identification of Compound X. x Use data, in a logical way, to justify the identity (i.e. name) of Compound X. 5 ● ONE of the above criteria is missing or incorrect. 4 ● TWO of the above criteria are missing or incorrect. 3 ● THREE of the above criteria are missing or incorrect. 2 ● Some relevant information taken from EITHER spectra 1 Marker’s Comments Very few of the cohort gained full marks for this question despite the marker accepting multiple answers for the identification of the compound (i.e. hexanal, hexan-2-one, hexan-3-one). Some key features of better responses were: x Clear use and correct interpretation of the spectra provided + the Infrared data from the HSC Data Sheet provided with the task. x Logical sequence used to justify the identification of the compound. x Structure of the compound was supported using fragment data from the Mass Spectrum x Clear description of the origin of the interpreted data from each spectrum (e.g. the molecular ion peak from the mass spectrum was 100 cm -1 which equates to the compound molecular mass 100 gmol -1 ) x Correct name of the compound x Correct terms used when describing the features of the spectra. (e.g. last peak vs molecular ion peak) Some issues students experienced were: x A significant number of students (approximately 70%) misinterpreted the features of the infrared spectrum; o the 2985 peak is due to the H-C stretch as it is narrow and not the presence of an acidic O-H bond which would be very broad. Unfortunately, this led students to believe that the molecule was an alcohol or carboxylic acid. o the 1175 peak was due to presence of C-O bond. Unfortunately, this led students to believe that the molecule was an ester. NOTE – the fingerprint region of the IR spectrum (i.e. data below 1500 cm -1 ) is ignored as it can cause confusion and its interpretation is outside the scope of this course.
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20 x used incorrect terms when describing features of the spectra. Eg. ‘beard’, ‘tongue’ vs narrow or broad peaks. x did not use multiple fragment data from the mass spectrum or did not apply the data to support their justification of the structure of the identified compound. x ignored discrepancies in the mass data when comparing fragments from compound. e.g. the identified molecule had a mass of 102gmol -1 when they stated the mass of the compound was 100gmol -1 . x Incorrect naming of the compound. This compound could be identified as a; o ketone - hexan-3-one is not hex-3-one as the ketone is unsaturated and needs to be included in the name. o aldehyde/alkynal – hexanal is not 1-hexanal as the carbonyl group is always on the C1 or hexaldehyde. Note: x To allow students who misinterpreted the spectra data to access a maximum of 3 marks if used a logic interpretation of their data to identify a compound. x fragments from a mass spectroscopy have a positive charge and this should be included on the fragments drawn in responses. x ticks did not equate to marks allocated. x no ½ marks were awarded Sample Answers
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23 Question 29 (8 marks) Marked by Mr Bell A solution of 4-chlorobutanoic acid was titrated against standardised sodium hydroxide solution. Small volumes of sodium hydroxide solution were added, and the pH was measured after each addition. The data collected are shown in the table below. Volume of NaOH added (mL) 0 5 10 15 18 19 20 21 22 25 30 pH 3.8 4.3 4.6 4.5 4.7 5.0 12.1 12.5 13.3 13.0 13.0 a) Draw the structural formula of 4-chlorobutanoic acid. Marking Criteria Marks ● Correct structural formula 1 Marker’s Comments This question was completed well by the cohort. A few students did not complete the diagram leaving off the hydrogen atoms and this was deemed to be an incomplete structure (i.e.0 mark). Sample Answer b) Plot the values of the pH of the solution against the volume of NaOH and draw a smooth curve of best fit, ignoring possible outliers. Marking Criteria Marks ● Correct labels with units ● Linear scales that use more than half the graph ● Correct data marked with X (may have outliers excluded) ● Smooth LOBF (outliers not included) 4 ● ONE of the above criteria is missing or incorrect. 3 ● TWO of the above criteria are missing or incorrect. 2 ● THREE of the above criteria are missing or incorrect. 1 Marker’s Comment Most of the cohort were able to achieve 3-4 marks for this question. The major issues student experienced were: x Not drawing a smooth LOBF x Including outliers (highlighted in the table) x Not drawing a vertical line through the equivalence point A few students did not even successfully draw a “typical’ titration graph and this was disturbing at this point of the course.
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24 Sample Answers Graph 1 - This graph shows the student not including the outliers that they identified. The information on this graph has been used for later sample answers. Graph 2 - This graph shows the student including the outliers that they identified but not including them in the plot.
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25 c) Estimate the pH at the equivalence point in this titration. Marking Criteria Marks ● Selects the pH that is in the middle of the vertical increase on their graph. 1 Marker ’s Comments Most students were able to achieve 1 mark for this question if their answer coincided with the mid-point of the vertical section of their graph. Better responses made it easier for the marker by marking/highlighting the point on their graphs. Sample Answer Varied with the students LOBF; however, answer had to coincide with the mid-point of the vertical section of their graph. A typical answer was pH 8.4. Sample answer taken from graph 1. d) The p K a of a weak acid is equal to the pH of the solution in which half of the weak acid has been neutralised. Use your graph to estimate a value of Ka for 4-chlorobutanoic acid. Marking Criteria Marks ● Uses the graph (i.e. clearly indicates on the graph) to select the volume of base that is half of the volume at the equivalence point (around 10 mL) Or x Uses the graph to calculate the volume of base that is half of the volume at the equivalence point ● Selects the pH at this volume and states that this is equal to the pKa. (pKa = 4.5) 2 ● Selects a pH and states that this is equal to the pKa 1 Marker’s Comments Many students were able to determine the pH from their graph directly and showed this value on the graph using a vertical and horizontal line draw on the grid (this is recommended to do in questions where it says “Use your graph”). Some students calculated the volume by halving the volume at the equivalence point and then reading the pH value from the graph. This was considered by the marker as ‘using the graph’. Again, the values for the pH varied due to their interpretation of the LOBF. A typical value was pH 4.5. Some students struggled with the calculation of Ka and they should review this calculation before the HSC (i.e. Ka = 10 -pKa ). The calculation does involve an antilog which is the 10 x button on the calculator (often associated with the log button). Sample Answer Sample answer taken from graph 1
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26 Question 30 (8 marks) Marked by Mr Bell The following table shows four acids which were tested for relative conductivity and pH: Acid tested Relative Conductivity pH Concentrated CH 3 COOH (100% pure) known as glacial acetic acid Does not conduct Unable to measure 0.1 mol L -1 HCl solution high 1.0 0.1 mol L -1 CH 3 COOH solution moderate 3.3 0.1 mol L -1 H 2 SO 4 solution high 0.7
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27 a) Explain why the pH for glacial acetic acid is unable to be measured. Marking Criteria Marks ● Identifies for an acid to conduct it must be in a solution of ions. (i.e. the acid dissociates in water to form ions) ● State that glacial acetic acid is not a solution of ions, since no water is present, and cannot conduct 2 ● ONE of the above criteria is missing or incorrect. 1 Marker’s Comments Approximately 60% of the cohort were awarded full marks, while the rest received 1 mark. The main issue students had responding to the question was that they did not state that water was essential to for ions and glacial acetic acid contains no water. Some students referred to a solvent implying water (i.e. ½ mark). Sample Answers b) Explain the differences in the pH values for the 0.1 mol L -1 acids that were tested. Include relevant calculations. Marking Criteria Marks For EACH 0.1 molL -1 acid (i.e. HCl, H 2 SO 4 , CH 3 COOH), students: x Explain thoroughly the differences in pH values in terms of: o relative strength of the acid o number of protons present (i.e. monoprotic, diprotic) x Support their response with an appropriate calculation (e.g. [H + ]) x Clearly link ALL the highlighted relationships above. 6 For EACH 0.1 molL -1 acid (i.e. HCl, H 2 SO 4 , CH 3 COOH), students: x Explain generally the differences in pH values in terms of: o relative strength of the acid o number of protons present (i.e monoprotic, diprotic) ● Support their response with an appropriate calculation (e.g. [H + ]) 4-5
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28 OR For EACH 0.1 molL -1 acid (i.e. HCl, H 2 SO 4 , CH 3 COOH), students: x Explain thoroughly the differences in pH values in terms of: o relative strength of the acid o number of protons present (i.e monoprotic, diprotic) x Include an appropriate calculation (e.g. [H + ]) Note - a least ONE appropriate calculation in the response For the 0.1 molL -1 acids (i.e. HCl and/or H 2 SO 4 and/or CH 3 COOH), students: x Identify the differences in pH values in terms of: o relative strength of the acid and/or o number of protons present (i.e. monoprotic, diprotic) x May include an appropriate calculation (e.g. [H + ]) 2 - 3 x Some relevant information 1 Marker’s Comments Despite the marker approaching this question holistically, only 3 students in the cohort attained full marks (see below in the sample answers) and several other students attained 5 marks; congratulations to those students as the marking criteria was very demanding and the marker required quality answers with no errors at all. Note – ticks on a response does note equate to marks and no ½ marks were awarded. The average for the cohort was 3/4 marks as students: x Missed the link that the original concentration of the acids was the same (i.e. 0.1 M) x gave general descriptions often with no calculations (or appropriate ones) to support their response. x did not address the concept of monoprotic/diprotic acids. x made errors such as; o HCl was a weak acid o Acids have high pH values x focused on the conductivity of the solutions rather than pH. x did not include units in their calculations x writing an equation such as pH = -log[H + ] and substituting the pH value from the table and then considering that this was an appropriate calculation to support their responses; this was not considered to be a calculation. What was pleasing was that some students (and hence better responses): x Calculated % ion dissociation and approached their explanation that way; better responses used the calculation + the concept of the second ionisation reaction for the diprotic acid to explain the lower pH. x Also included dissociation equations to explain the concept of monoprotic/diprotic acids; however, note that equations showing the dissociation of weak acids equations have a double arrow and dissociation of strong acids have a single arrow. x Sample Answers Note – the following sample answers were awarded full marks as the marker had paper copies of the student’s response; however, notice that the sc anned copy of the response has writing ‘cut off’. In the HSC, markers receive a scanned copy of a student’s response, so the scans
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29 below would not receive full marks as ‘vital’ information is missing. THEREFORE, FOR THE HSC, write in the lines available and ask for extra writing paper!
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31
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32 Question 31 (7 marks) Marks The flow chart shows a series of reactions that would allow a chemist to synthesise propyl propanoate from propene.
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33 Question 31 continues over the page Question 31 continued a) Identify the following substances from the flow chart. Marking Criteria Marks Five completely correct identifications 5 Four completely correct identifications, OR Three completely correct and two substantially correct identifications 4 Three completely correct identifications, OR Two completely correct and two substantially correct identifications 3 Two completely correct identifications, OR One completely correct and two substantially correct identifications 2 One completely correct identification 1 Marker’s Comments It was apparent that many students had not revised organic reactions sufficiently. Common errors include: x Leaving out ‘concentrated’ for Catalyst 1 x Leaving out ‘dilute’ for Reagent 2 x Misinterpreting “reagent“ as “reactant” – especially for Reagent 2. Many students identified the reactant, H 2 O, when the reagent (substance necessary for the reaction to occur) is dilute NaOH or KOH (which includes the water). x Including heat / reflux as a catalyst. Only substances can be catalysts, not energy or conditions. This did not cost marks this time. Sample Answer Mark s 5 2
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34 b) Identify each of the following reaction types. Marking Criteria Marks Three correct reaction types 2 One correct reaction type 1 Marker’s Comments Answers accepted were: x Addition, hydrohalogenation, hydrochlorination, hydrogen halide addition o Better answers identified the specific type of addition reaction. o Halogenation was not accepted, as this is the addition of X 2 not HX. x Substitution, nucleophilic substitution, substitution (hydration), substitution with HX o Note that ‘hydration’ alone was not accepted here, as this is a term used for addition of water to unsaturated hydrocarbons. x Esterification o Condensation was not accepted; although this is a condensation reaction, esterification is the term for this specific reaction. Sample Answer
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35 Question 32 (5 marks) Marks a) Given the molar solubility of MnCO 3 is 4.20 x 10 -6 M, calculate the solubility product for MnCO 3 . Marking Criteria Marks Writes correct K sp expression AND calculates K sp as 1.76 x 10 - 11 2 Writes correct K sp expression OR calculates K sp as 1.76 x 10 -11 1 Marker’s Comments Mostly well answered, but some simple errors cost students marks: x Not recognising that solubility product meant K sp , and attempting to calculate something else. x Including the solid MnCO 3 in the K sp expression. As the solid compound is pure, a “concentration” of it is a meaningless expression. You should not include it an d then justify getting rid of it; just don’t put it in. x Using incorrect formulae for the manganese and carbonate ions. This was a common reason for getting ½ marks despite a mathematically correct answer. x Considering 4.20 x 10 -6 as K sp then using it to calculate a molar solubility (x). Better answers included an equation for the dissolution of manganese (II) carbonate and justified x = [Mn 2+ ] = [CO 3 2- ] using a mole ratio. Sample Answer 2 3
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36 b) A 0.200g sample of solid barium sulfate was added to 500mL of 0.010M sodium sulfate. Will all the barium sulfate sample dissolve? Justify your answer. Marking Criteria Marks States that it will not dissolve or most will not dissolve, AND Justifies this by: x Calculating the molar solubility of BaSO 4 AND x Calculating the concentration of BaSO 4 required OR n BaSO4 present 3 Calculates the molar solubility of BaSO 4 (1.08 x 10 -8 ), OR Calculates the concentration of BaSO 4 required (1.7 x 10 -3 M) 2 Writes a K sp expression in terms of molar solubility (x), OR Calculates n BaSO4 present, OR Explains the common ion effect in general terms 1 Marker’s Comments Many students really struggled to get to grips with what to do in this question, and this was evidenced by a lot of poor setting out and attempts to create a chemical reaction or identify a limiting reagent in the mixed solution. The first challenge is to recognise that this is a common ion effect scenario, as both salts contain sulfate ions. Note that the only way to justify your conclusion for a question like this is via calculations which demonstrate (not just assert) that the quantity of barium sulfate provided cannot dissolve in the volume of sodium sulfate solution present; words are not enough. It was also critical to recall that a list of K sp (solubility product) values are given on the Data Sheet, and barium sulfate is one of them. Several suitable approaches were used by students: x Calculating the number of moles of BaSO 4 in 0.200g and comparing it to the calculated number of moles (or mass) of BaSO 4 that could be dissolved in 500mL of Na 2 SO 4 solution. x Calculating the molar solubility (x) of BaSO 4 in Na 2 SO 4 solution and comparing it to the concentration that a solution of BaSO 4 would have if 0.200g could be fully dissolved in 500mL. x Calculating a Q sp (assuming all the BaSO 4 could dissolve in the Na 2 SO 4
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37 solution) and comparing it to K sp for BaSO 4 . Sample Answers
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39 Question 33 (7 marks) Marks
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40 100.0 mL of 0.200 M hydrochloric acid was combined with 100.0 mL of 0.200 M ammonia solution. Calculate the pH of the final solution once the reaction has gone through to completion. Marking Criteria Marks ● Calculates number of moles of NH 4 + ● Calculates concentration of NH 4 + ● Calculates K a of NH 4 + ● Substituted equilibrium expression ● Justifies assumption that equilibrium concentration of NH 4 + is equal to initial concentration or correctly calculates x as a quadratic ● Calculates H + ion concentration ● Calculates pH 1 1 1 1 1 1 1 7 Ammonium pKa = 9.25 Hydrochloric acid pKa = -5.9
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41 End of paper
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10 Multiple Choice Answers 1 D 11 B 2 C 12 A 3 C 13 A 4 D 14 C 5 B 15 A 6 A 16 A 7 A 17 C 8 A 18 D 9 A 19 NA 10 D 20 A Question 21 (9 marks) Marks Phosgene (COCl 2 ) is a colourless gas that has an odour of freshly cut grass. It decomposes into chlorine gas and carbon monoxide through an endothermic reaction that establishes an equilibrium. a) Write a chemical equation for this reaction including an appropriate scientific representation that it is endothermic. Marking Criteria Marks ● Correct chemical equation including states ● Indicates that the enthalpy change is positive 1 2 Sample Answer Markers Comments Generally well done. Common mistakes were to leave out states and to include heat as a reactant. Also, some students had +ΔH along with the other products which represents it as a product. This is not how you represent an endothermic reaction. You should have ΔH= or ΔH > 0 separate to the equation as it is not made of atoms, it is a representation of the energy change. b) Write an equilibrium expression for this reaction. Marking Criteria Marks ● Correct equilibrium expression 1 1
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11 Sample Answer Markers Comments Well done by most students. A very small number of student had it upside down (the reactant on top) and some did not include K=. This is an important part of the equilibrium expression. Phosgene, chlorine and carbon monoxide were mixed in a sealed container and allowed to reach equilibrium. A change was imposed at time T and the equilibrium re-established. The concentration of each gas was plotted against time and are represented with the letters X, Y and Z. Question 21 continued c) Identify the gas that is represented by the letter X on the graph. Marking Criteria Marks ● Correctly identifies phosgene (COCl 2 ) 1 Markers Comments 1 4 1
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12 d) Identify the change that occurred at time T and explain using Le Chatelier's Principle why the equilibrium re-established as shown in the graph. Marking Criteria Marks ● Identifies the change as an increase in volume or that gas had been removed. ● States that this results in a decrease in pressure ● Explains that the equilibrium position shifts to the products to counteract the change ● Because this increases the number of gas particles, therefore increase pressure 4 ● One mistake or step missing from above 3 ● Explains why the shift in equilibrium position counteracts an identified change (does not have to be pressure) 2 ● States a change and predicts a shift in equilibrium position 1 Markers Comments Sample Answer
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13 e) Use the graph to calculate the equilibrium constant for the reaction before time T. Marking Criteria Marks ● Calculates the equilibrium constant (0.024) 1 Sample Answer Markers Comments Question 22 (4 marks) Marks The heat of neutralisation for a reaction between a strong acid and a strong base is: ΔH = -57.62 kJ/mol. A student mixed 50.0 mL of 0.100 M hydrochloric acid with 50.0 mL of 0.200 M sodium hydroxide. They determined the mass of the final solution was 98.03 g. The teacher instructed the student to assume that the specific heat capacity of the solution would be the same as pure water.
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14 Calculate the theoretical change in temperature of the reaction mixture. 4 Marking Criteria Marks ● Justifies HCl as the limiting reagent ● Calculates q (288.1 J) with working ● Calculates the change in temperature (0.703° C) with working and units ● Give the final answer to 3 significant figures 4 ● One mistake or step missing from above 3 ● Two mistakes or steps missing from above 2 ● Attempts at least one of the above step 1 Markers Comments Sample Answer
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15 Question 23 (5 marks) Marks Magnesium chloride is an ionic compound which is highly soluble in water. a) Write the equation, including state symbols, for the process of forming a saturated solution of magnesium chloride in water. Marking Criteria Marks ● Correct chemical equation with states 1 Sample Answer Markers Comments b) The table below shows some thermochemical data related to the dissolving of magnesium chloride solid in water. Energy required to overcome the lattice forces to form free Mg 2+ and Cl - ions. +2493 kJ mol – 1 Energy released upon hydration of Mg 2+ ions – 1920 kJ mol – 1 Energy released upon hydration of Cl - ions – 364 kJ mol – 1 Use these data to calculate the standard enthalpy of solution of magnesium chloride. Marking Criteria Marks ● Calculates the standard enthalpy ● Including units and working 2 ● Attempts to combine standard energies 1 1 2 2
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16 Markers Comments Sample Answer c) Use your answer to part (b) to deduce how the solubility of MgCl 2 changes as the temperature is increased. Justify your answer. Marking Criteria Marks ● Explains that an increase in temperature will cause the equilibrium position to shift to the reactant. ● Links the shift left to reduced solubility If the previous question was a positive enthalpy, the opposite relationship was required. 2 ● Explains how an increase in temperature will cause equilibrium position to shift. 1 Markers Comments Sample Answer Question 24 (5 marks) Marks Propan-1-amine has a boiling point of 47º C while N,N-dimethylmethanamine has a boiling point of just 3º C. Draw the structural formula for each compound in the space provided and explain why there is a difference in boiling point. Marking Criteria Marks ● Correct structures (one mark each) 2 5
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17 Marking Criteria Marks ● Identifies the major intermolecular force for Propan-1-amine as hydrogen bonding ● Identifies the only intermolecular force for N,N- dimethylmethanamine as dispersion forces ● Explains that hydrogen bond require more energy to over come than dispersion forces, therefore, Propan-1-amine has a higher boiling point 3 ● Correctly identifies the major intermolecular force one of the compounds. ● Relates the strength of intermolecular forces to boiling point 2 ● Links intermolecular forces to boiling point 1 Sample Answer Marks
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18 Question 25 (6 marks) A student wanted to determine the concentration of a solution of potassium hydroxide. They did this by reacting a 25 mL aliquot with a standard solution of hydrochloric acid while measuring the conductivity of the solution. 10.00 mL volumes of 0.1012 M hydrochloric acid were incrementally added using a burette and the conductivity was measured after each addition. The student recorded the following data.
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19 Question 25 continued a) Explain the trend shown in the graph. Marking Criteria Marks ● Explains that the decrease in conductivity it because of the hydroxide ions concentration decreasing. ● Identifies the lowest point of conductivity as the equivalence point. ● Explains that the increase in conductivity is because of the hydrogen ion concentration increasing 3 ● Correctly links either the decrease or increase in conductivity to the increase or decrease of hydrogen or hydroxide ions ● Identifies the lowest point of conductivity as the equivalence point. 2 Marks 3
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20 ● Links conductivity to the concentration ions in solution 1 Sample Answer b) Calculate the concentration of the potassium hydroxide solution. Marking Criteria Marks ● Selects the correct volume of HCl (54-56 mL) ● Calculates the number of moles of HCl ● Calculates the concentration of KOH 3 ● One mistake of the above 2 ● Attempts of the above steps 1 Sample Answer 3
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23 Question 26 (8 marks) Marks Linseed oil is a mixture of triesters. It is used for many purposes including as a protective layer for wooden furniture and as an ingredient in some cuisines. The diagram below shows the structure of a typical triester found in linseed oil. When the triester shown above undergoes the process of de-esterification (this process is the opposite of esterification), three fatty acids are produced: ● palmitic acid ● linoleic acid ● oleic acid which correspond to the three long chains in the diagram listed from top to bottom. a) Why are compounds like the one depicted above called triesters? Marking Criteria Marks ● Identifies that there are three ester functional groups 1 b) In addition to the three acids listed above, ONE other product will form, and ONE other reactant will be consumed when the compound undergoes de-esterification. Identify the product and reactant and state the number of moles that would be produced/consumed from the complete de-esterification of one mole of the compound shown. Marking Criteria Marks ● Identifies the reactant as water and the product as propane-1,2,3- triol ● States that 3 moles of water are consumed and 1 mole of propane- 1,2,3-triol is produced 2 1 2 Marks
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24 ● Correctly identifies one product or reactant 1 Question 26 continues over the page Question 26 continued c) Castor oil is another natural product that may find a use as a component in biofuel. It releases 39.5 MJ kg − 1 of energy when fully combusted.When 2.00 L of castor oil was combusted in a home-made engine 59.3 MJ of energy was released. Given that the density of castor oil is 0.96 g mL − 1 , calculate percentage efficiency of the home-made engine. Marking Criteria Marks ● Calculates the mass of 2 L as 1.920 Kg ● Calculates the theoretical energy release as 75.84 MJ ● Calculates 78.19… % efficiency 3 ● One mistake or step missing from above 2 ● Two mistakes or steps missing from above 1 d) The picture below shows a laboratory reflux apparatus using a heating mantle. Explain how TWO risks associated with heating volatile organic compounds are reduced by using a reflux apparatus with a heating mantle. Marking Criteria Marks ● Identifies two specific risks of using a naked flame to heat a reflux apparatus and explain how these are reduced using a heating mantle 3 2 Marks
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25 ● Identifies one specific risk 2 Question 27 (3 marks) Sodium dihydrogen phosphate is an amphiprotic salt. Explain what is meant by the term ‘ amphiprotic ’. Support your answer with chemical equations. Marking Criteria Marks ● Explains that amphiprotic compounds can donate and accept a proton. ● Correct equation showing Sodium dihydrogen phosphate donating a proton. ● Correct equation showing Sodium dihydrogen phosphate accepting a proton. 3 ● One mistake or step missing from above 2 ● Two mistakes or steps missing from above 1 . 3
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26 Question 28 (5 marks) Marks A student obtained the following spectroscopic data to help to identify an unknown organic compound, labeled Compound X.
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27 Question 28 continues over the page Question 28 continued Identify Compound X, using BOTH spectra to justify your identification. Marking Criteria Marks ● Identifies the compound as hexan-3-one (structure or name) ● Identifies the peak at around 1650 cm -1 from the IR and links this to the presence of a carbonyl group ● Identifies the molecular ion peak at 100 from the mass spec and links this to the molar mass of the compound ● Identifies at least one mass spec peak that corresponds to a fragment that could only be from hexan-3-one. E.g. 57 for CH 3 CH 2 CO + ● Uses this data to justify the identification of hexan-3-one 5 ● All of the above with one mistake or the fragment selected does not conclusively identifies hexan-3-one 4 ● Identifies the compound any of the possible chain isomers of hexan-3-one (structure or name) ● Identifies the peak at around 1650 cm -1 from the IR and links this to the presence of a carbonyl group ● Identifies the molecular ion peak at 100 from the mass spec and links this to the molar mass of the compound 3 Marks 5
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28 ● Identifies the peak at around 1650 cm -1 from the IR and links this to the presence of a carbonyl group ● Identifies the molecular ion peak at 100 from the mass spec and links this to the molar mass of the compound 2 ● A relevant reference to either of the spectra 1 Question 29 (8 marks) Marks A solution of 4-chlorobutanoic acid was titrated against standardised sodium hydroxide solution. Small volumes of sodium hydroxide solution were added, and the pH was measured after each addition. The data collected are shown in the table below. Volume of NaOH added (mL) 0 5 10 15 18 19 20 21 22 25 30 pH 3.8 4.3 4.6 4.5 4.7 5.0 12.1 12.5 13.3 13.0 13.0 a) Draw the structural formula of 4-chlorobutanoic acid. Marking Criteria Marks ● Correct structural formula 1 1
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29 b) Plot the values of the pH of the solution against the volume of NaOH and draw a smooth curve of best fit, ignoring possible outliers. Marking Criteria Marks ● Correct labels with units ● Linear scales that use more than half the graph ● Correct data marked with X ● Smooth LOBF 4 ● One mistake or step missing from above 3 ● Two mistakes or steps missing from above 2 ● Attempt at graphing data 1 Question 29 continues over the page Question 29 continued c) Estimate the pH at the equivalence point in this titration. Marking Criteria Marks ● Selects the pH that is in the middle of the vertical increase on their graph. Around pH = 8.5 (I think) 1 d) The p K a of a weak acid is equal to the pH of the solution in which half of the weak acid has been neutralised. Use your graph to estimate a value of Ka for 4-chlorobutanoic acid. Marking Criteria Marks ● Uses the graph to select the volume of base that is half of the 2 4 Marks 1 2
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30 volume at the equivalence point (around 10 mL) ● Selects the pH at this volume and states that this is equal to the pKa. (roughly pH = 4.5 therfore pKa = 4.5) ● Selects a pH and states that this is equal to the pKa 1 Question 30 (8 marks) Marks The following table shows four acids which were tested for relative conductivity and pH: Acid tested Relative Conductivity pH Concentrated CH 3 COOH (100% pure) known as glacial acetic acid Does not conduct Unable to measure 0.1 mol L -1 HCl solution high 1.0 0.1 mol L -1 CH 3 COOH solution moderate 3.3 0.1 mol L -1 H 2 SO 4 solution high 0.7
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31 a) Explain why the pH for glacial acetic acid is unable to be measured. …………………………………………………………………………………………...…… ………………………………………………………………………………………...……… ……………………………………………………………………………………...………… …………………………………………………………………………………...…………… b) Explain the differences in the pH values for the 0.1 mol L -1 acids that were tested. Include relevant calculations. …………………………………………………………………………………………...…… ………………………………………………………………………………………...……… ……………………………………………………………………………………...………… …………………………………………………………………………………...…………… ………………………………………………………………………………...……………… ……………………………………………………………………………...………………… …………………………………………………………………………...…………………… ………………………………………………………………………...……………………… ……………………………………………………………………...………………………… ………………………………………………………………… ................................................. 2 6 Question 31 (7 marks) Marks The flow chart shows a series of reactions that would allow a chemist to synthesise propyl propanoate from propene.
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32 Question 31 continues over the page
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33 Question 31 continued a) Identify the following substances from the flow chart. Marking Criteria Marks Five completely correct identifications 5 Four completely correct identifications, OR Three completely correct and two substantially correct identifications 4 Three completely correct identifications, OR Two completely correct and two substantially correct identifications 3 Two completely correct identifications, OR One completely correct and two substantially correct identifications 2 One completely correct identification 1 Marker’s Comments It was apparent that many students had not revised organic reactions sufficiently. Common errors include: x Leaving out ‘concentrated’ for Catalyst 1 x Leaving out ‘dilute’ for Reagent 2 x Misinterpreting “reagent” as “reactant” – especially for Reagent 2. Many students identified the reactant, H 2 O, when the reagent (substance necessary for the reaction to occur) is dilute NaOH or KOH (which includes the water). x Including heat / reflux as a catalyst. Only substances can be catalysts, not energy or conditions. This did not cost marks this time. Sample Answer Mark s 5 2
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34 b) Identify each of the following reaction types. Marking Criteria Marks Three correct reaction types 2 One correct reaction type 1 Marker’s Comments Answers accepted were: x Addition, hydrohalogenation, hydrochlorination, hydrogen halide addition o Better answers identified the specific type of addition reaction. o Halogenation was not accepted, as this is the addition of X 2 not HX. x Substitution, nucleophilic substitution, substitution (hydration), substitution with HX o Note that ‘hydration’ alone was not accepted here, as this is a term used for addition of water to unsaturated hydrocarbons. x Esterification o Condensation was not accepted; although this is a condensation reaction, esterification is the term for this specific reaction. Sample Answer
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35 Question 32 (5 marks) Marks a) Given the molar solubility of MnCO 3 is 4.20 x 10 -6 M, calculate the solubility product for MnCO 3 . Marking Criteria Marks Writes correct K sp expression AND calculates K sp as 1.76 x 10 -11 2 Writes correct K sp expression OR calculates K sp as 1.76 x 10 -11 1 Marker’s Comments Mostly well answered, but some simple errors cost students marks: x Not recognising that solubility product meant K sp , and attempting to calculate something else. x Including the solid MnCO 3 in the K sp expression. As the solid compound is pure, a “concentration” of it is a meaningless expression. You should not include it and then justify getting rid of it; just don’t put it in. x Using incorrect formulae for the manganese and carbonate ions. This was a common reason for getting ½ marks despite a mathematically correct answer. x Considering 4.20 x 10 -6 as K sp then using it to calculate a molar solubility (x). Better answers included an equation for the dissolution of manganese (II) carbonate and justified x = [Mn 2+ ] = [CO 3 2- ] using a mole ratio. Sample Answer 2 3
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36 b) A 0.200g sample of solid barium sulfate was added to 500mL of 0.010M sodium sulfate. Will all the barium sulfate sample dissolve? Justify your answer. Marking Criteria Marks States that it will not dissolve or most will not dissolve, AND Justifies this by: x Calculating the molar solubility of BaSO 4 AND x Calculating the concentration of BaSO 4 required OR n BaSO4 present 3 Calculates the molar solubility of BaSO 4 (1.08 x 10 -8 ), OR Calculates the concentration of BaSO 4 required (1.7 x 10 -3 M) 2 Writes a K sp expression in terms of molar solubility (x), OR Calculates n BaSO4 present, OR Explains the common ion effect in general terms 1 Marker’s Comments Many students really struggled to get to grips with what to do in this question, and this was evidenced by a lot of poor setting out and attempts to create a chemical reaction or identify a limiting reagent in the mixed solution. The first challenge is to recognise that this is a common ion effect scenario, as both salts contain sulfate ions. Note that the only way to justify your conclusion for a question like this is via calculations which demonstrate (not just assert) that the quantity of barium sulfate provided cannot dissolve in the volume of sodium sulfate solution present; words are not enough. It was also critical to recall that a list of K sp (solubility product) values are given on the Data Sheet, and barium sulfate is one of them. Several suitable approaches were used by students: x Calculating the number of moles of BaSO 4 in 0.200g and comparing it to the calculated number of moles (or mass) of BaSO 4 that could be dissolved in 500mL of Na 2 SO 4 solution. x Calculating the molar solubility (x) of BaSO 4 in Na 2 SO 4 solution and comparing it to the concentration that a solution of BaSO 4 would have if
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37 0.200g could be fully dissolved in 500mL. x Calculating a Q sp (assuming all the BaSO 4 could dissolve in the Na 2 SO 4 solution) and comparing it to K sp for BaSO 4 . Sample Answers
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39 Question 33 (7 marks) Marks
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40 100.0 mL of 0.200 M hydrochloric acid was combined with 100.0 mL of 0.200 M ammonia solution. Calculate the pH of the final solution once the reaction has gone through to completion. Marking Criteria Marks Justifies the correct answer with full working shown 7 Calculates the correct answer with incomplete working, OR Calculates the incorrect answer (one error allowed) with correct complete working 6 Provides two relevant pieces of information, AND Calculates n(NH 4 + ), AND Calculates K a (NH 4 + ), AND Writes the K a expression for the dissociation of NH 4 + 5 Provides two relevant pieces of information, AND Calculates n(NH 4 + ), AND Calculates K a (NH 4 + ) 4 Provides two relevant pieces of information, AND Calculates n(NH 4 + ) OR calculates K a (NH 4 + ) 3 Provides two relevant pieces of information 2 Provides some relevant information 1 Marker’s Comments Unsurprisingly, this question was not well answered by most students, who spent a lot of valuable time and pen ink wandering in a chemical wilderness. Five students in the cohort scored full marks. I marked this question quite generously in the 1 – 4mk range to give access to many students. This was definitely not a question to just start plugging numbers into likely- looking equations! To successfully approach this question, it was most important to recognise that: x the reaction described was a neutralisation that had already gone to completion x the stoichiometric ratios of the acid (HCl) and base (NH 3 ) involved meant that ONLY A SALT AND WATER WERE PRESENT at this point. This was at its heart, therefore, an acidic/basic/neutral salt question, as only ammonium chloride and water are present when the neutralisation reaction has finished. The K a value provided for HCl was a distractor, as HCl is not only a strong 7 Ammonium pKa = 9.25 Hydrochloric acid pKa = -5.9
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41 acid, but no longer present after neutralisation. Its conjugate base (Cl - ) does not react with water and therefore did not influence the solution pH; in this situation, only the ammonium ion reacts with water. Disappointingly many students are still not distinguishing ammonia (NH 3 , a weak base) from ammonium (NH 4 + , its conjugate, and weak, acid). This impacted on the capacity to write correct equilibrium expressions. K a is an acid dissociation constant, therefore the acid in question is going to be a reactant in the equilibrium expression: ?? 4 + + ? 2 ? ↔ ?? 3 + ? 3 ? + K a = [NH 3 ][H 3 O + ] [NH 4 + ] It was necessary to take into account the dilution factor of the solution (50:50 when mixed), which halved [NH 4 + ] i to 0.100M. An ICE table was required to calculate [NH 4 + ] eq before substitution into the K a expression. This necessitated a quadratic solution, OR a clearly stated justification for using [NH 4 + ] i instead (x << 0.100 for a weak acid with K a = 5.62 x 10 -10 ). This was not the significant figures question, but the answer should correctly be quoted to 2sf (note that the logarithmic function changes the way sig figs are applied and thus K a for NH 4 + has 2sf; also note that K a for HCl was not used, and thus the answer is not limited by its 1sf. Sample Answer
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42 End of paper
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