2021 BoS Chemistry Trial - Marking and Answers (2 Viewers)

[CM_Tutor]

First, let me start with an apology that this process has taken much longer than expected.

Second, here are the distributions of answers for the MCQ:

The most difficult questions, based on the rate at which the question was answered correctly, were questions 5, 14, 11, and 2. The easiest questions on this basis were questions 10, 17, and 1.

Third, for all those who took the paper, marking is complete. I have sent the final marks to @Trebla. I will upload each person's marked paper into the same forum as where you did the exam today.

The exam was seriously difficult / challenging, as it was meant to be, and so results do not predict HSC scores.

Raw marks, out of 100, ranged from 20 to 78 with an average of 53.

Adjusting for non-attempts - so, for example, taking a student who scored 44 marks but only answered 86 marks worth of questions as 51% on questions attempted - the range was 38% to 78% with an average of 60%.

 

[CM_Tutor]

Here is a box-and-whisker plot of the results:

On the left, in blue, is the plot based on raw marks / 100. We see a median slightly above the mean, only one outlier.

On the left, in orange, are percentages of marks obtained from questions attempted. So, as mentioned above, a candidate who attempted 86 marks worth of questions and was awarded 44 marks would be scored as 44 on the blue plot and 44 / 86 = 51% on the orange plot. This gives some measure of achievement and feedback in cases where a significant part of the paper was not attempted. We see the mean and median are both larger on this measure, and the mean is further below the median, indicating greater skew.

 

[CM_Tutor]

Question 21
This question proved to be quite challenging.

Part (a) averaged 0.875 / 1
Part (b) averaged 1.458 / 3

I had two answers in mind in relation to part (a) - that the BP gap for water and 2-propyl propanoate could be a problem, and that 2-propanol having the lowest of the BPs meant that an attempt to distill would extract the alcohol first, causing the equilibrium position to shift left and the yield of ester to be decreased. This second aspect is the major barrier to using distillation to extract a product from an equilibrium system, and one that I believe is not well understood. Almost no answers addressed this point, and if it was not understood then part (b) becomes more difficult.

With part (b), some responses misinterpreted the question as being about the reflux procedure and forming the ester, rather than about extracting a pure ester product from the reaction mixture. I will clarify the question in the final version of the paper to something like "... pure 2-propyl propanoate could be obtained from the product mixture after refluxing in good yield." Many answers showed an awareness of the steps involved (neutralising acids, separating the immiscible aqueous and organic layers, drying the organic layer, and distilling the ester from the alcohol), though the sequence was unclear and / or steps were omitted in some cases. The larger problem was that the question asked that you "justify" the the procedure rather than to describe it. It is clear that understanding of the reasons for the steps involved is an area of weakness for many. In short, they are:

1. Neutralise to remove the catalyst as the rate that the equilibrium position can shift is much slower without the catalyst being present. This also removes the unreacted carboxylic acid and the salt products shift predominantly to the aqueous layer, making separation from the organic layer convenient. The position of equilibrium will shift as a result, there no longer being a forward reaction possible, so stopping that shift as quickly as possible is important.
2. Having separated the organic layer, we dry it with something like anhydrous calcium sulfate. Removing the water from the organic layer prevents the reverse reaction from taking place.
3. After filtering, we now have an organic phase that is a mixture of ester and unreacted alcohol. This is not an equilibrium system any longer as we don't have both reactants needed for either the forward or reverse reactions. Distillation (with the necessary heating to vapourise substances) can no longer promote a shift in equilibrium position and the two substances have much more separated BPs, making it an appropriate purification method at this point.

 

[CM_Tutor]

Question 22

Part (a) averaged 0.917 / 2
Part (b) averaged 1.042 / 2

For 2 marks in part (a), it was necessary to note cover (n some way):

• An indicator has (at least) two forms, existing in equilibrium (such as HInd + H₂O <---in equilibrium---> H₃O⁺ + Ind^-), and that these forms have different colours (due to structural differences)
• Colour change occurs over a range where the equilibrium is shifting from the indicator being predominantly in one form to being predominantly in the other.
• This means it is shifting its equilibrium position around [HInd] = [Ind^-], which corresponds to pH = pK_a

The latter point proved difficult for some, but there were also answers that did not score full marks because they omitted something like that the indicators forms are coloured and different from one another.

For 2 marks in part (b), comment was needed to address both accuracy and validity:

• Since the indicator is itself a weak acid or base, the more of it that is present, the more it will alter the pH of the system and undergo reactions that effect when the end point will be observed. Further, since the colour change depends on it reacting with the excess of whichever compound is in the burette once the equivalence point has been past, having more of it will mean more reaction is needed for a clear change in colour to be observable and thus will move the end point further beyond the equivalence point. Either of these impacts on accuracy.
• Validity is not necessarily lost by the introduction of inaccuracy. A titration with five drops of indicator is less accurate than one with two drops, but that doesn't make it invalid. The property that you seek to measure about the unknown still causitively leads to differences in results in the titration, so the experiment is valid. Some noted that adding a huge excess of indicator could mean the experiment is invalid in that determining an end point is near-impossible or that the experiment is now measuring a property of a system that is not just the material being investigated, and that is a reasonable point.

The lactucin question on the 2021 CSSA paper offers an example where a direct titration would be invalid because the the reaction is so slow that conducting the experiment in the conventional way would be so slow and it would confusing to determine what a permanent colour change of the indicator actually was - which is why a back titration method was used. The question did not ask about this but it is an example of how validity can become an issue in a titration.

Note also that the question made no mention of reliability, so describing any impact on reliability was pointless.

 

[CM_Tutor]

Question 23
This question was handled well. Every response scored 2 or 3 out of 3, and the average was exactly 2.5.

There was no particular answer sought or points that needed to be raised, but there did need to be:

• both advantages and disadvantages considered
• some discussion - a list of five of each would not score 3 / 3, for example
• focus on the particular fuels mentioned - straying into alternatives was pointless and could not score credit
• the connection to societal implications needed to be explicit

The last point here probably caused the most difficulty, despite wide scope to consider societal implications that were economic, or social, or environmental.

 

[CM_Tutor]

Question 24

Part (a) averaged 0.708 / 1
Part (b) averaged 0.625 / 1
Part (c) averaged 1.375 / 2
Part (d) averaged 1 / 2

The system involved was 2 A (g) + C (g) <---in equilibrium---> 2 B (g). The possible mistakes were:

• not recognising that "synthesis" meant one product on the RHS, which then had to be B.
• taking the stoichiometric ratio from the vertical change at 3.5 min (a volume change) rather than from the shift in position change from 1.5 to 3 min

For part (b), the graph needed to show the [C] increasing from the value after its vertical drop, returning to a position that was below the start of that vertical drop, and the three moving in the correct ratio (or the one from an incorrect equation in (a). ALL THREE graphs needed to be completed.

Part (c) said "identify ... and justify", so both were needed. t = 1.5 min shows a shift to the left favouring the endothermic reverse reaction, and thus must have been due to a temperature increase because that is the change that favours an ENDO reaction, according to Le Chatelier's Principle. t = 3.5 min shows a proportionate increase in all concentrations, resulting from a volume decrease (proportionate in that each concentration increased by 50%, meaning the volume was reduced by 1/3 - though this fraction need not be identified). The resulting shift right occurs because the pressure increase will favour fewer gaseous molecules, from Le Chatelier's Principle, and here there are 3 gas molecules on the reactant side ad 2 gas molecules on the products side).

Part (d) was unusual in asking for a graph of Q. The temperature increase that led to a shift left produced a change in the value of the equilibrium constant. Equilibrium occurs when Q = K, so Q needed to decrease to a new K value. The subsequent shifts caused Q to shift away from that new K before returning to it. The change at t = 3.5 min decreased Q as it prompted a shift right, while the change at t = 5.5 min prompted a shift left and so must have caused Q to increase, along the lines of this illustration:

 

[CM_Tutor]

Question 25

Part (a) averaged 0.286 / 1 amongst those who attempted the question
Part (b) averaged 1.571 / 2 amongst those who attempted the question
Part (c) averaged 0.417 / 1 amongst those who attempted the question

Compounds A and B are isomers of C₃H₇Br, which means they can only be the two possible bromopropanes. C, the intermediate gas, is propene, so the process is carrying out a position isomerism via an elimination then addition sequence.

Most responses recognised that the higher BP of A made it the unbranched form, so the reaction was:

1-bromopropane ---KOH in ethanol (or other suitable solvent but NOT water)---> propene ---HBr---> 2-bromopropane (major product)

The solvent to used in the dehydrohalogenation was important as aqueous KOH will promote substitution to form 1-propanol, which is not a gas.

Part (b) required an explanation of why A had a higher BP than B and why C has a lower BP than both. The second part, that C has only dispersion forces between its molecules, compared with the dipole-dipole interactions in A and B, was much better handled than the difference between A and B. A difference in the strength of dipole-dipole interactions was only accepted if an explanation of why there is a difference was provided. The major reason is actually the difference in dispersion forces due to the greater distance between molecules of B due to less efficient packing of molecules with side chains.

Part (c) required a recognition that "in good yield" meant that the process won't run backwards because the major addition product from HBr and propene is the 2-bromo isomer, according to Markovnikov's Rule.

As an aside, there are methods to convert propene to 1-bromopropane using a process called hydroboration oxidation, but this is beyond the syllabus:

3 CH₃-CH=CH₂ + 3 BH₃ ---ether (solvent)---> B(CH₂CH₂CH₃)₃

B(CH₂CH₂CH₃)₃ + 3 H₂O₂ -----> B(OH)₃ + 3 CH₃CH₂CH₂OH

CH₃CH₂CH₂OH + HBr -----> CH₃CH₂CH₂Br + H₂O​

 

[CM_Tutor]

Question 26
Many students struggled here because they accepted what the maths said without considering the chemistry.

The average score was 1.458 / 3 in this question.

A common mistake was to take the pH = 10.54 solution and do a 1-in-10 dilution to conclude that [H⁺] = 10^-10.54 was diluted to [H⁺] = 10^-11.54, resulting in a solution of pH = 11.54. This carries out a mathematical procedure correctly but on a flawed premise. If this was true, the pOH would have fallen from 3.46 to 2.46, meaning that the concentration of base increased by a factor of 10. Where would all this extra base have come from?

Whether a solution is acidic or basic, dilution will result in a shift in pH towards neutrality, and the student's prediction of a pH of 9.54 is correct, as is the observation that this means the hydronium concentration has increased.

One approach is to note that the dilution predicts a change in [H₃O⁺] from 10^-10.54 M to 10^-11.54 M and a change in [OH^-] from 10^-3.46 M to 10^-4.46 M. Considering the autoionisation of water, this system is not at equilibrium as Q = [H₃O⁺][OH^-] = 10^-16 < K_w and so further ionisation will occur, resulting in a mixture very close to [H₃O⁺] = 10^-9.54 and a change in [OH^-] = 10^-4.46 M.

Another approach is to recognise that the flaw in the dilution formula is that its use rests on the assumption that the dilutant does not contain the species being diluted. Diluting with water does mean adding some extra H₃O⁺ and OH^- ions (each 10^-7 M). This can be ignored for diluting [OH^-] = 10^-3.46 M, but not when diluting [H₃O⁺] = 10^-10.54 M as the water contains more H₃O⁺ than is in the solution being diluted.

Answers needed to recognise that the pH would decrease on dilution from 10.54 to roughly 9.54 and then account for the [H₃O⁺] having increased as a consequence, supported with calculations or a qualitative theoretical explanation.

 

[someth1ng]

Validity is not necessarily lost by the introduction of inaccuracy. A titration with five drops of indicator is less accurate than one with two drops, but that doesn't make it invalid. The property that you seek to measure about the unknown still causitively leads to differences in results in the titration, so the experiment is valid. Some noted that adding a huge excess of indicator could mean the experiment is invalid in that determining an end point is near-impossible or that the experiment is now measuring a property of a system that is not just the material being investigated, and that is a reasonable point.

Everything you've said here is correct (from both an industrial and research perspective). I'm wondering whether this question is difficult for students to answer because validity and accuracy are not clearly defined at HSC-level beyond "is it a reasonable method?" and "how close is it?".

In industry, validity often includes accuracy, precision, range, linearity, repeatability (and reproducibility), range, robustness, and specificity (depending on the method). To be valid, you need to pass all the required tests. For each parameter, there's a clearly defined tolerance.

For example, in HPLC, accuracy is defined as the recovery of 95-105% of your sample (sometimes 90-110%). That is, you make a sample at 1 mg/mL (example only) concentration and find the area, then you "spike" the sample with another 0.5 mg/mL (example only) so the total concentration is 1.5 mg/mL and find the new area. If the area increased by 48%, then it went up by 96% of what you expected, so recovery/accuracy is 96%.

For a titration, it'd be different and I'm not sure what the normal tolerance is, but let's say it's 5%. In this case, you'd be allowed to add more indicator until it causes a 5% change in titrant volume. In this way, it's completely clear when the method becomes invalid.

Also, you might want to make your MC answer more than just C and D haha. I'm surprised 5 was the most poorly answered. It doesn't seem that hard.

 

[CM_Tutor]

Everything you've said here is correct (from both an industrial and research perspective). I'm wondering whether this question is difficult for students to answer because validity and accuracy are not clearly defined at HSC-level beyond "is it a reasonable method?" and "how close is it?".

In industry, validity often includes accuracy, precision, range, linearity, repeatability (and reproducibility), range, robustness, and specificity (depending on the method). To be valid, you need to pass all the required tests. For each parameter, there's a clearly defined tolerance.

For example, in HPLC, accuracy is defined as the recovery of 95-105% of your sample (sometimes 90-110%). That is, you make a sample at 1 mg/mL (example only) concentration and find the area, then you "spike" the sample with another 0.5 mg/mL (example only) so the total concentration is 1.5 mg/mL and find the new area. If the area increased by 48%, then it went up by 96% of what you expected, so recovery/accuracy is 96%.

For a titration, it'd be different and I'm not sure what the normal tolerance is, but let's say it's 5%. In this case, you'd be allowed to add more indicator until it causes a 5% change in titrant volume. In this way, it's completely clear when the method becomes invalid.

Also, you might want to make your MC answer more than just C and D haha. I'm surprised 5 was the most poorly answered. It doesn't seem that hard.

I strongly agree that a lack of definition is part of the problem, but it also doesn't help that most teachers have little experience in these concepts in an authentic research or industrial setting, so they lack the concrete experience to give these terms nuanced meaning. Plus, meaning is often contextual. Nevertheless, validity, reliability and accuracy are different concepts and students need to be able to distinguish between them. For example, I have seen answers stating that reliability is improved by repeating an experiment and averaging results, but repetition gives information on whether an experiment is reliable. It should not improve reliability unless the unreliability is due to poor technique that improves with skill. Repeating and averaging is a way to improve accuracy through minimising the effect of random error.

Significant inaccuracy is not proof of invalidity either. Take a heat of combustion experiment where the heat from a burning fuel is used to heat a fixed quantity of water. The calculated enthalpy change will likely be low by at least 40% as there is a large systematic error (heat loss to the environment). The accuracy can be greatly improved by calibrating the equipment on known samples to determine the heat loss and then make a correction for unknown samples. Since no correction can make an invalid experiment into a valid one, I consider the standard experiment moderately reliable (a whole class of students will get similar results) and highly inaccurate, but still valid.

On the MCQ, a set of answers was posted in another thread with comments, and I hope a set of answers will be posted, but I can't see me finishing writing one in time to be useful for students tomorrow. As for Q5, it was easy to get the copper concentration from the Beer-Lambert law (option A) and skip over the fact that the carbonate concentration was sought.

 

[someth1ng]

I strongly agree that a lack of definition is part of the problem, but it also doesn't help that most teachers have little experience in these concepts in an authentic research or industrial setting, so they lack the concrete experience to give these terms nuanced meaning. Plus, meaning is often contextual. Nevertheless, validity, reliability and accuracy are different concepts and students need to be able to distinguish between them. For example, I have seen answers stating that reliability is improved by repeating an experiment and averaging results, but repetition gives information on whether an experiment is reliable. It should not improve reliability unless the unreliability is due to poor technique that improves with skill. Repeating and averaging is a way to improve accuracy through minimising the effect of random error.

Yeah, I don't blame the teachers at all. It's a pain to become a teacher with X years of experience in industry or research. To be honest, I was taught in HS that you can improve reliability by repetition. In a sense, it's not completely wrong because if you repeat a method and get similar results, it does demonstrate some level of reliability but you're right, it's more about providing the information. More formally, reliability is the probability that the method will perform as intended, without failure.

Significant inaccuracy is not proof of invalidity either. Take a heat of combustion experiment where the heat from a burning fuel is used to heat a fixed quantity of water. The calculated enthalpy change will likely be low by at least 40% as there is a large systematic error (heat loss to the environment). The accuracy can be greatly improved by calibrating the equipment on known samples to determine the heat loss and then make a correction for unknown samples. Since no correction can make an invalid experiment into a valid one, I consider the standard experiment moderately reliable (a whole class of students will get similar results) and highly inaccurate, but still valid.

This is an example where you need a clear definition. Conceptually, it'd be correct under ideal conditions but in the real world, you could argue that it's no longer valid because it doesn't account for heat loss etc. Personally, I don't see why a correction can't turn an invalid experiment into a valid one. I'd argue that there's quite a lot of corrections made in science (e.g. baseline correction and phasing in NMR). Often, legitimate decisions need to be made with data analysis and if you're applying it evenly across your data, then it can still be valid.

 

[CM_Tutor]

Question 27

Average was 3.952 / 7 amongst those who attempted the question

This was an "assess" question, so it required a judgement on the question of the "validity of the conclusion that the zinc smelter is releasing high concentrations of cadmium pollution" that needs to be based on a balanced discussion of evidence. Though it was not stated directly, an assessment of the validity of a conclusion necessarily includes consideration of the validity of the evidence on which it is based. To this is added a discussion of improving ACME's investigation. The above comments from @someth1ng are timely, in that validity is not well defined in the syllabus. Consequently, I looked at the assessments from the perspective of the whole argument, even where I would disagree with the position being advanced.

This question was, in part, inspired by a question from a past HSC that asked about the validity of a conclusion. In that question, the conclusion reached was not supported by the evidence presented and was invalid, even though the conclusion may well have been correct. Looking at example answers across the band levels, difficulties arose when students focused on how to determine if the conclusion was correct by seeking additional evidence, rather than on whether the conclusion was supported by (or an inevitable consequence of) the evidence presented. A similar issue arose here. If I was answering this question in an exam, I would assess the conclusion as invalid on the grounds that nothing presented proves that the smelter is releasing cadmium pollution. There is good circumstantial evidence to strongly suspect that the smelter is releasing cadmium pollution, but nothing that proves it definitively. This nuance was identified explicitly by a few students, and implied by more, and was an essential part of the question, irrespective of how it was used in the answer.

Evidence and issues that are present in the question include:

• We are given the guideline that [Cd²⁺] > 0.9 nmol L^-1 indicates a terrestrial source of cadmium when found in coastal waters. Ideally, this is used to look at site E, which is the closest to coastal waters, though several students noted that a sample further out would be instructive. This is the only piece of information that is not in ppb, so for comparison purposes a conversion calculation is needed, which shows that 0.9 nmol L^-1 = 0.101 ppb. Alternatively, all of the readings can be converted to nmol L^-1, but that is time consuming, producing site A being 0.75 nmol L^-1 to site E being 170 nmol L^-1. Either conversion is acceptable. If the latter approach is taken, the 2.0 ppb limit on cadmium in drinking water also needed conversion to 19 nmol L^-1. Those students whose conversions were incorrect had the rest of their answers considered based on their incorrect conversions, so long as the error did not greatly simplify the question.
• Based on these conversions, site A provides water that meets drinking water guidelines in relation to cadmium, site B is slightly over the guideline, and sites C, D, and E are far above acceptable levels. Site E provides strong support for the inference of a terrestrial cadmium source, while only site A is below this guideline. These results strongly suggest a source downstream from A, though a far upstream source that is diluted by a long distance (think of a river like the Amazon) could be possible. Testing upstream of A is thus a potential improvement to discuss.
• The use of single samples with no repetition over time raises issues of reliability - could this be a one-off release of waste, or is it a regular pattern of pollution? Considerations here certainly provide options for potential improvements to the investigation.
• Sites B, C, and D have levels that are so high that the conclusion of there being large amounts of cadmium present can safely be drawn. Some discussed the techniques of AAS to make the point that it would be cadmium being detected and not some other pollutant, which is a relevant point.
• The calibration curve itself shows that 13 standard solutions of varying concentrations were used, which provide a strong basis for the linearity of concentration with absorbance over the range to 50 ppb. Sample D is calculated based on an extrapolation of 20% above this range to 60 ppb, while sample E requires extrapolation to triple this domain. ACME has made the assumption that linearity continues but no evidence is presented on this point. Some students suggested additional calibration to extend the domain to 150 ppb, though the simpler alternative of dilutions of samples D and E and re-measuring to test the linearity assumption was not suggested often. Given the problems with extrapolation, the accuracy of these measurements is open to question at very least... but even if they are overestimates by 50%, the conclusion that there is very substantial amounts of cadmium present in the river remains. In my view, this flaw in the investigation should be addressed but it does not threaten the conclusion that there is a cadmium source in the vicinity.
• As was noted in another thread, linearity at very low concentrations can also be problematic and so there is reason to be less confident in the accuracy of the site A - though this is a nuance that I would not expect any high school student to pick up.
• There is some noise in the calibration curve that might also be considered. The correlation coefficient is not given (though that concept is studied in both Standard 2 and Advanced Maths syllabi) and it is high, but the amount of scatter is atypical for AAS data - again, not something that was expected to be discussed, but another potential area that may be explored.
• The biggest threat to the validity of the conclusion that the zinc smelter is the source of the cadmium is that the evidence is circumstantial. There is a lot of cadmium near the smelter, and the concentration increases in its vicinity and is much lower just upstream of the smelter, but that is not proof that the smelter is the source. We know nothing of the vicinity so there could be an alternative source nearby - suppose that next to the smelter, there is a facility for recycling batteries that processes Nicad (nickel-cadmium) batteries. Unlikely, yes, but we don't know. Some students raised the possibility of alternative, or even additional, cadmium sources. Most students commented on the desirability of more sampling, some suggesting that it be taken from near both banks of the river, which would be informative. However, and surprisingly to me, not many commented that the question of sourcing can be addressed by sampling right at the outflow points from the smelter, or even better, from sources inside the smelter.
• One piece of circumstantial evidence that was mentioned in the question (but not raised by anyone, IIRC) was in the first sentence of the question - that zinc sulfide ores often contain cadmium. In other words, the ACME investigation is not only showing cadmium adjacent to the zinc smelter, but also offering a reason why a zinc smelter might be (in fact, typically does) produce cadmium as a by-product. Real-world zinc smelters collect and sell this cadmium if it is present in a high enough concentration in the ore.

Having said all of that, what was I looking for? Essentially, the following 7 things, sometimes modified for equivalent merit:

1. A calculation for conversion of units to all ppb or 0.9 nmol L^-1
2. The observation that the site A data meets drinking water guidelines, but the rest do not - and by a very substantial margin in the cases of C, D, and E
3. The evidence does support a terrestrial cadmium source
4. The calibration extrapolation raises issues of accuracy of the measurements (particularly of C) which should be addressed in some way
5. The evidence does support the conclusion that there is substantial cadmium pollution in the vicinity of the zinc smelter despite the above
6. The conclusion that the zinc smelter is the source may well be (in fact, probably is) true but the evidence presented does not validly support the conclusion. Conclusive evidence linking the smelter to the cadmium is needed.
7. Discussion of improvements relating to issues raised

I was surprised that several responses used only the terrestrial source from coastal waters or the drinking water guidelines, but not both, and in so doing created an easily-addressed weakness in their answer. No one scored 7 / 7 but there were a couple of 6 / 7 responses.

 

[CM_Tutor]

Question 28

Average on part (a) was 2.13 / 3 amongst those who attempted the question
Average on part (b) was 0.818 / 1 amongst those who attempted the question
Average on part (c) was 0.9 / 2 amongst those who attempted the question
Average on part (a) was 1.143 / 2 amongst those who attempted the question

Part (a) was a back titration calculation. The most common mistake was not reading the question and thus (correctly) calculating the chloride ion concentration and then not reporting it in ppm, as requested by the question. The answers were:

n(Ag⁺) that reacted with SCN^- in the titration = 1.8626... x 10^-3 mol

n(Ag⁺) added initially = 1.234 x 10^-2 mol

n(Ag⁺) reacted with Cl^- = 1.0477... x 10^-2 mol

[Cl^-] = 0.1396... mol L^-1 = 4953 ppm

Part (b) required recognising that the iron(III) nitrate was an indicator as the blood-red coloured complex FeSCN²⁺ is formed in much greater quantities once the Ag⁺ has reacted and the SCN^- concentration rises. An equation was useful for supporting the answered but not required. Some mention of why a colour change occurs or what it would be was required, however.

Part (c) required recognising a problem arising from a basic solution. The one I had in mind was the precipitation of iron(III) hydroxide, given its extremely low K_sp, but I also accepted the formation of silver(I) oxide or silver(I) hydroxide. The second mark required a suitable buffer to be suggested, which needed to be an acidic buffer and one that would not introduce a new problem. Carbonate, sulfate, or phosphate buffers all risk precipitation of the silver cations and so were unacceptable. The most common acceptable buffer suggested was an acetic acid / acetate buffer.

Part (d) required identifying why having AgCl (s) present during titration would be problematic and providing an explanation. Some students did not recognise the relevance of the quoted K_sp for silver thiocyanate or thought it referred to silver chloride (the K_sp of which was given on the data sheet). Identifying that extra thiocyanate would be reacted proved easier than explaining why this was the case. The clearest explanation, IMO, is to look at the direct reaction:

AgCl (s) + SCN^- (aq) <---in equilibrium---> AgSCN (s) + Cl^- (aq)​

and to note that it occurs to a significant extent by finding its equilibrium constant:

​

Explanations using Le Chatelier's Principle and competing equilbria, that the precipitation of AgSCN will reduce the concentration of Ag⁺ by so much that there will be significant redissolving of AgCl, were accepted. However, the interpretation that extra chloride would be present and thus determined was incorrect and thus not accepted. It was the extra silver in solution leading to more thiocyanate reacting that as the problem as this reducing the calculated silver consumed in the first part of the back titration and produced an underestimate of chloride content.

 

[CM_Tutor]

Question 29
This question was done well

Average on part (a) was 2.348 / 3 amongst those who attempted it
Average on part (b) was 1.625 / 3 amongst those who attempted it

For part (a), the species sought were Na⁺, HCO₃^-, CO₃^2-, H₂CO₃, H₂O, H₃O⁺, OH^-, resulting from dissolution of the sodium hydrogencarbonate, auto-ionisation of water, and the hydrogencarbonate ion being amphiprotic. Another possible species was CO₂ but this was not required. Suitable equations were needed to support the answer.

For part (b), any household acid or base was accepted. Some mistaken identifications were ignored. Examples include:

• Household acids: Hydrochloric acid (pool acid), acetic acid (vinegar), sulfuric acid (car batteries), citric acid (cooking)
• Household bases: Ammonia (cleaning), sodium hydroxide (oven / drain cleaner), bleach (sodium hypochlorite), baking powder (sodium carbonate)

Addressing spills with a solid is desirable to prevent spreading, and it is desirable to make a solid product that can be swept up and disposed of in a bin. Though I did not penalise it, some answers contemplated an over-the-top response... vinegar spilled on the kitchen floor is more likely wiped up than neutralised, and citric acid powder (or in a liquid like lemon juice) is likewise not going to be treated with sodium hydrogencabonate.

 

[CM_Tutor]

Question 30

Average for part (a) was 1.522 / 2 amongst those who attempted it
Average for part (b) was 0.565 / 1 amongst those who attempted it
Average for part (c) was 1.4 / 2 amongst those who attempted it
Average for part (d) was 0.526 / 1 amongst those who attempted it
Average for part (e) was 1 / 2 amongst those who attempted it

The biggest observation here is about graphing... graphs do not change from linear to suddenly being curves and then returning to linearity with some new trend. There was a linear trend here for the data from beakers 1 to 6, and a separate linear trend (constant) for beakers 7 to 9). These two trends each need to be extrapolated to find where they meet. The increasing trend is associated with the sulfuric acid being the limiting reagent. The constant trend is produced when the sulfuric acid is present in excess. The change from one trend to the other occurs at the point where the two lines meet and represents the two reagents being present in their stoichiometric ratio and neither being in excess. The 2020 HSC had a graph where there were similarly two distinct trends and the point of intersection represented neither reagent being in excess, about which a calculation is required. Conductivity curves in titrations also show this pattern, and there are many experiments that could be used as questions where something similar happens.

DO NOT, under any circumstances, plot a set of data where there is a clear line of best fit through part of the data and then a sudden change to a different trend by drawing a curve. Put a line through the linear parts, and extrapolate for where they meet.

The answer to part (a), from reaction of the solid sodium hydroxide with water, was -44.6 kJ mol^-1.

Gaining the mark in part (b) required noting that the entropy change must be positive, so that the Gibbs free energy is always negative and increasingly negative as temperature increases, resulting in increased solubility.

The graph for part (c) has two separate lines of best fit, one for when sulfuric acid is limiting, and one for when it is present in excess. Each trendline is extrapolated to find the point of intersection, representing neither reagent being in excess but rather both are present in their stoichiometric ratio.



The answer to part (d) depending on reading a suitable volume from the graph - I accepted anything over 50 mL up to about 55 mL - and then a mole calculation, giving 0.361 mol L^-1.

There were two ways to do part (e). The first was to use the maximum temperature change from 21.3 ^oC to find the delta H for the reaction of solid sodium hydroxide with acid and then use Hess' Law to remove the effect of sodium hydroxide dissolution to get the answer. The other approach was to look at the temperature change from 26.3 ^oC to isolate just the neutralisation of hydronium and hydroxide. Both approaches give the same answer of about -55 kJ mol^-1.

 

[CM_Tutor]

Question 31

Average for part (a) was 1.87 / 2 amongst those who attempted it
Average for part (b) was 1.818 / 3 amongst those who attempted it

Part (a) used the standard K_a calculation method on an atypical (non-BL) acid reaction. I was very pleased to see that most people handled this without problem, correctly deducing the pH was 5.21. Some responses used the assumption that [H₃O]⁺_eqm << [B(OH)₃]_i without explanation, which is dangerous on the "show all working" front and may make marks more difficult to earn in the event of a mistake in the calculation. The data all used 2 significant figures so the pH in the answer should have been given to 2 decimal places, though I did not penalise this.

Part (b) showed that respondents have a strong knowledge of Lowry-Bronsted theory and how it has improved our understanding of aqueous acids and bases. A variety of (mostly correct) statements were made about non-aqueous systems, which was not capable of gaining credit as the question specifically limited the context to aqueous solutions. These comments were not penalised, of course, but they did waste time and space, potentially creating an impression of a thorough answer and hiding the weakness that remained. The weakness was that the question said "assess" and that means that a judgement is required (mostly done well) based on evidence (also well done) that was balanced (here was the problem). An assessment of LB theory needs some recognition of its limitations. The question gave a big hint by having you calculate the pH of an acid based on a reaction that does not fit LB theory as the acidity of boric acid arises from hydroxide acceptance rather than proton donation, so a suitable example of a limitation was immediately in front of you. There was no requirement that this example be used, but some comment on limitations was necessary. An alternative example used by some is the formation of hydrogencarbonate ions when carbon dioxide (an acidic gas) dissolves in a basic solution like sodium hydroxide, which is also an example of hydroxide acceptance. The inclusion of a limitation also meant that the assessment needed some nuance, like that LB theory improved understanding over A theory greatly, but does not account for all cases and so can be improved / is still limited but to a much lesser extent than A theory. The assessment offered needed to be supported by the evidence presented.

 

[CM_Tutor]

Question 32

Average for part (a) was 1 / 1 amongst those who attempted it
Average for part (b) was 1.476 / 2 amongst those who attempted it
Average for part (c) was 1.474 / 2 amongst those who attempted it
Average for part (d) was 1.056 / 3 amongst those who attempted it

Part (a) was the easiest question in the short answer section - every student who provided an answer correctly identified the two compounds as functional group isomers.

In part (b), I was looking for the shift left from heating, in line with Le Chatelier's Principle, and the fact that reaction rates would be higher (owing to greater average kinetic energy / greater rate of effective collisions). I also accepted, as the second effect, that the equilibrium constant would decrease on heating. The major cause of lost marks here was that two effects were identified, as required, but not justified. Reasons needed to be given to explain why each stated effect would occur.

Part (c) was a calculation that was unfamiliar and caused a few problems. The easiest solution was:

​

The solution can also be found using an ICE table. The same numerical answer is found by K^-1 and multiplying by 100 (to get a percentage) but this was not accepted as K^-1 is actually the equilibrium constant for the reverse process acetaldehyde <---in equilibrium---> vinyl alcohol. Partial credit could be earned by an estimate supported with a suitable explanation.

Part (d) proved to be challenging. Many students noted that vinyl acetate could be hydrolysed to vinyl alcohol and then tried to explain why this can be immediately polymerised without the rearrangement being problematic. None were successful because the rearrangement is the exact reason why polyvinyl alcohol cannot be formed by the addition polymerisation of vinyl alcohol. Hydrolysis was the process needed but the sequence needed to be reversed. The addition polymerisation of vinyl acetate gives polyvinyl acetate (a few did this step and then incorrectly asserted that the product was polyvinyl alcohol). Hydrolysis of polyvinyl acetate gives polyvinyl alcohol and a lot of acetic acid. Organic chemistry is about functional groups, and so the ester functional group on a polymer can be hydrolysed in same way as the monomer can be hydrolysed. However, by forming the polymer backbone with the acetate side chains in place, the result of hydrolysis cannot rearrange as vinyl alcohol would because we never have the hydroxyl functional group bound to a carbon-carbon double bond. The equation required was the sequence:

vinyl acetate ---heat / initator---> polyvinyl acetate ---hydrolysis---> polyvinyl alcohol + acetic acid (acid hydrolysis) or acetate salt (base hydrolysis)​

 

[CM_Tutor]

Question 33

Average on part (a) was 4.05 / 7 amongst those who attempted it
Average on part (b) was 1.267 / 2 amongst those who attempted it
Average on part (c) was 1.133 / 3 amongst those who attempted it
Average on part (d) was 1.083 / 3 amongst those who attempted it

Half of all those who took the exam did not respond to all of part of this question.

For part (a), as in the earlier 7 mark question, marks were lost by not accounting for evidence that provided justification for parts of the answer. The information provided and which needed to be presented is:

• W oxidises to X which further oxidises to Y. X can be oxidised in another way to Z, the salt of Y. Based on HSC chemistry and all of the compounds containing C, H, and O only, we have W = primary alcohol, X = aldehyde, Z = carboxylic acid. The fact that W is a primary alcohol is confirmed by its reaction with sodium metal and its O-H stretch in the IR. The fact that W does not react with sulfuric acid is therefore strange and requires comment.
• The percentage composition data gave that Y has an empirical formula of C₃H₆O.
• The parent ion mass-to-charge ratio gives the parent ion at m/z = 116, meaning the molar mass is 116 g mol^-1, from which we can deduce that the molecular formula is C₆H₁₂O₂.
• The triplet / quartet combination in the ¹H NMR, accounting for 5 hydrogens, is strong evidence of an ethyl group that is isolated (i.e. its neighbours do not bear any H atoms).
• The 6H singlet in the ¹H NMR must also be isolated, and being 6H must be either two identical methyls, three identical methylenes (CH₂), or six identical methines (CH). Given we already have an ethyl and a carboxylic acid (so 3C accounted for), it can't be six methines and three methylenes is also impossible if they are to be isolated, so we must have two methyls. The proton from the carboxylic acid is missing from the ¹H NMR as its signal will be way above 3 ppm.
• The ¹H NMR plus knowledge of the carboxyl group gives the following fragments: -COOH, -CH₃ x 2 (identical and isolated), -CH₂CH₃ (isolated), which account for C₅H₁₂O₂, and so we have one more carbon atom - which is good news, as we can use it to get the required isolations. Piecing these together, there is only one possibility: Y is 2,2-dimethylbutanoic acid.
• Looking at this structure, the reason for the non-reaction of W with sulfuric acid becomes clear. Primary alcohols should undergo dehydration but W is 2,2-dimethyl-1-butanol, which cannot dehydrate as C2, the carbon adjacent to the carbon bearing the hydroxyl group (C1) is tertiary and bears no H atom, making elimination (and the consequent formation of a C1 to C2 double carbon-carbon bond) impossible. This non-reaction thus provides evidence that confirms the structure of Y. This information could also have been used in the reverse way: if W is a primary alcohol but does not undergo elimination, the reason is likely the carbon skeleton is HO-C-C(C)₂-C, and thus we have the likely arrangement of five of the six carbon atoms. In fact, we have all six as, whichever branch we lengthen, we end up with a carbon attached to a CH₂OH, two one-carbon chains, and a two-carbon chain.

For part (b), it was important to note that Z is the salt of Y and so the only difference is the carboxyl group being present as -COOH or -CO₂^-. The ¹³C NMR for all but carbon 1 should be very similar (four environments, one with double intensity, all below 50 ppm) while C1 of Z will experience greater electron density as an anion and so be shifted from its position near 185 ppm shown for Y. I did not penalise comments that predicted the direction of the shift incorrectly, the point was to recognise that there would be a shift.

Part (c) was meant to be two relatively straight-forward marks and one difficult mark - identifying the bonds in the IR by position (easy) and then explaining why A had an unusual appearance. The question even identified A as unusual, identifying it as the one that was different for its type. However, it is evident that many took this as meaning A wasn't due to a hydroxyl (despite its position), which implies the hydroxyl (a large and prominent bond stretch) was somewhere else. The stretches were:
A = O-H
B = C-H
C = C=O
D = C-O
This is entirely consistent with every other IR spectrum presented in the HSC. The unusual feature is that A, the hydroxyl stretch, looks nothing like it usually would - very broad and intense. The reason for the difference is that this is a gas-phase IR spectrum. What we know about gases includes that intermolecular interactions are absent, so this spectrum is unaffected by hydrogen bonding, which is the reason that hydroxyl groups are usually broad. I know no HSC student will have done IR in gas phase or likely seen one, but in discussing IR everyone would learn about the O-H stretch being so intense and broad and different from virtually every other IR signal, and I would hope most teachers / books would mention that broadening is due to hydrogen bonding.

Part (d) contained a typo. It was meant to ask the name of Z, not Y, to test whether the cation that would be in the salt was identified correctly. The broader point in (d) was how questions can cross topics, as this was actually based in the module 8 content of cation and anion testing. Most did not recognise this, and the marks awarded were mostly for the name of Y, where the correct answer was credited along with correct names of incorrect Y structures from part (a). The answer was meant to be sodium 2,2-dimethylbutanoate, the sodium cation coming from the sodium hydroxide used to precipitate the silver nitrate, described in the preamble to the question.

The chemistry of part (d) involved the formation of silver oxide:

2 AgNO₃ (aq) + 2 NaOH (aq) -----> Ag₂O (s) + 2 NaNO₃ (aq) + H₂O (l)

or

2 Ag⁺ (aq) + 2 OH^- (aq) -----> Ag₂O (s) + H₂O (l)​

which is then re-dissolved in ammonia:

Ag₂O (s) + 4 NH₃ (aq) + H₂O (l) -----> [Ag(NH₃)₂]⁺ (aq) + 2 OH^- (aq)​

The oxidant was the diamminesilver(I) cation, [Ag(NH₃)₂]⁺, that resulted from these reactions (and hence the answer to (d)(ii). The answer to (d)(iii) was then the above equations.

The question did not ask for the redox reaction in the silver mirror test, but if it had, the half-equations would be:

[Ag(NH₃)₂]⁺ (aq) + e^- -----> Ag (s) + 2 NH₃ (aq)

and

CH₃-CH₂-C(CH₃)₂-C(=O)H (aq) + H₂O (l) -----> CH₃-CH₂-C(CH₃)₂-CO₂^- (aq) + 3 H⁺ (aq) + 2e^-​

for an overall reaction of:

CH₃-CH₂-C(CH₃)₂-C(=O)H (aq) + 2 [Ag(NH₃)₂]⁺ (aq) + H₂O (l) -----> CH₃-CH₂-C(CH₃)₂-CO₂^- (aq) + 2 Ag (s) + 3 NH₄⁺ (aq) + NH₃ (aq)
​

The Ag (s) produced adheres to the inner surface of the reaction vessel producing the "silver mirror". Note that this equation accounts for the reaction of hydrogen ions with ammonia, explaining the ammonium ions in the product mixture. Consequently, the salt Z could also be ammonium 2,2-dimethylbutanoate. An equation like this written with a carboxylate and ammonia, both weak bases in the presence of hydrogen ions would be unacceptable.

On part (d), it was disappointing to see mistakes over the valence of silver, which exists in the +I oxidation state in compounds.

 

[Loquaar]

I just wanted to thank all those who contributed to the paper particularly CM_Tutor and his solutions. Whilst the paper itself was difficult from my perspective it tested my abilities in various ways exposing me to questions I would never have seen before. However I believe it is the BOS exam that I sat which really allowed me to answer more of the difficult calculations and theory questions in the 2021 HSC. This is particulary evident through the penguinone question in the multiple choice where we had to deduce the No. of carbon environments in a cyclical molecule which the hsc also had. The question on Hertmans Solution and the Volhards titration involving the manipulation of the equilibrium constant also provided me with the skills needed to answer the last question which was similar in theory. I also wanted to make a special mention of the spectra question in the trial, due to its unique nature and difficulty I was able to improve my skills in its analysis and took on the feedback from CM_Tutor proving to be crucial in this years HSC question which was unique in its concept. Once again, thanks and the BOS trial certainly proved to be helpful.

 

[CM_Tutor]

Thank you very much, @Loquaar, it is gratifying to hear that the paper was helpful for your HSC exam!