# 2021 BoS Chemistry Trial (1 Viewer)

#### someth1ng

##### Retired Nov '14
I agree that there are plenty of approaches, which actually makes marking more difficult to. What astounds me is that a teacher at HSC level would not see any of the problems with it, and nor did anyone notice in the checking process. I wouldn't let first year uni students near glacial acetic acid, let alone genuinely anhydrous acetic acid! So why would a teacher contemplate a question on it without considering "hmmm... this is too dangerous to let them near in reality, maybe I should ask something else". Maybe they considered oleum or anhydrous perchloric acid...
Yeah, good luck to the marker. You'd have to assess every single answer and consider whether it's valid or not. I could see each answer taking 5+ minutes to evaluate properly. It would be...highly varied.

#### CM_Tutor

##### Moderator
Moderator
These were the marking criteria:

Marking CriteriaMarks
• Labels bottles
• Flow chart is clear
• Uses micro test tubes or depression tiles
• Procedure uses minimal chemicals
• Correct procedure
4-5
2-3
• Provides some relevant information
1

I bet that would make marking the question easier for you!

#### someth1ng

##### Retired Nov '14
These were the marking criteria:

Marking CriteriaMarks
• Labels bottles
• Flow chart is clear
• Uses micro test tubes or depression tiles
• Procedure uses minimal chemicals
• Correct procedure
4-5
2-3
• Provides some relevant information
1
I bet that would make marking the question easier for you!
That's...bad. Really bad. I can imagine a lot of students giving completely valid answers and getting marked down for it.

#### CM_Tutor

##### Moderator
Moderator
Here was the sample answer:

The teacher appears unaware that sodium acetate is a weak base and so could cause a colour change with an indicator. And also that ethylamine is a gas.

#### someth1ng

##### Retired Nov '14
The teacher appears unaware that sodium acetate is a weak base and so could cause a colour change with an indicator. And also that ethylamine is a gas.
...and it's not even a flow chart.

#### idkkdi

##### Well-Known Member
That's...bad. Really bad. I can imagine a lot of students giving completely valid answers and getting marked down for it.
great summary of hsc chemistry.

#### Mina the HSC slayer

##### New Member
is there a solutions document?

#### CM_Tutor

##### Moderator
Moderator
is there a solutions document?
One will be released once the marking is complete and marks are announced.

In the meantime, if you have specific questions, feel free to ask.

#### CM_Tutor

##### Moderator
Moderator
I have received a question about q22(a) by DM, so am posting about it here in case others want to know.

An indicator for an acid-base titration will itself be a weak acid or base, so take it as having HInd as its acidic form. It is therefore present in solution as an equilibrium:

HInd (aq) + H2O (l) <---equilibrium---> H3O+ + Ind-

with the forms HInd and Ind- having different visible spectra, absorbing visible light in different patterns, and thus having different colours. The colour of the solution will therefore depend on the relative concentrations of these species and thus on the ratio of [HInd] : [Ind-]. There will be some value of this ratio (k : 1, say, where k > 1 is real and ) beyond which the colour of [HInd] will dominate and so will not be noticeably changing to an observer looking at the solution without any instrumentation (with the naked eye, say) - the solution will display the low pH colour of the indicator. Similarly, there will be some value (1 : k, say) beyond which the colour of [Ind-] will dominate and the solution will display the high pH colour of the indicator. Between these values there will be a mixture of the two colours and change will be visible to the observer. This change will occur around the ratio being 1 : 1, which means that

\bg_white \begin{align*} K_{\text{a(indicator)}} &= \cfrac{[\text{H}_3\text{O}^+][\text{Ind}^-]}{[\text{HInd}]} \\ &= [\text{H}_3\text{O}^+] \times \cfrac{[\text{Ind}^-]}{[\text{HInd}]} \\ &= [\text{H}_3\text{O}^+] \times 1 \\ &= [\text{H}_3\text{O}^+]\ \\ \text{pH} &= \text{p}K_{\text{a}} \end{align*}
It was a two mark question, so I was looking for:
• Indicator exists in equilibrium between two forms of different colours with colour dependent on the relative amounts of each​
• Colour change occurs around 1 : 1 mixture which happens when pH of the solution is equal to pKa of indicator​

#### dontbeagrumpmrtrump

##### New Member
I stumbled onto this excellent exam recently. CM_Tutor, some of the questions you have written are absolutely fantastic in how it requires deep knowledge of the syllabus as well as including many tricks to throw students off. Seeing the discussion here, I am going to throw my own hat into the ring.

Here are my solutions and thoughts on the MCQ portion of the exam, as I haven't yet had time to go through the SAQ in depth. There is every chance that there is some glaring error - please let me know and I will be more than happy to discuss.

1. B. Well written question, designed to trick students into picking D. Since H2O is gaseous, its concentration will contribute towards equilibrium and increasing [H2O] will shift equilibrium right. D is incorrect as although equilibrium moves right, by LCP the shift will not increase [CO2] to its concentration before the volume increase (remember that volume increases will decrease all concentrations)
2. D. Other choices all had tricks attached - C seems correct at first glance but the charges are incorrect,
3. C. 1-bromo-1-chloro, 1-bromo-2-chloro, 1-bromo-3-chloro, 2-bromo-1-chloro, 2-bromo-2-chloro. Question would be more fun with more carbons as then skeletal isomers would be in play.
4. C. By Avogadro’s Law, 2Cl2 + 5O2 -> 2Cl2O5. Then, Cl2O5, + H2O -> 2HCLO3.
5. D. Good question. Using Beer-Lambert Law, A = ecl. Sub in A = 0.437, e = 12.3, l = 1, we get c = 3.55 x 10-2. It is easy to stop here (and choose the very convenient option A), but the question is asking for carbonate concentration and not Cu2+. We must use the fact that the Ksp for CuCO3 is 1.4 x 10-10. As Ksp = [Cu2+][CO32-], we sub in 3.55 x 10-2 to get [CO32-] = 3.94 x 10-9,
6. C. It is necessary to consider Ksp expressions for all compounds, as the variable valencies means the magnitude of Ksp alone does not necessarily correlate with solubility. The fastest way probably involved rooting each Ksp by the power of the number of ions present in the ionic formula of each compound. We obtain C as the answer with concentrations of 7.55 x 10-11 and 1.15 x 10-4 for lowest and highest arsenate concentration respectively.
7. D. D is a “better” answer than B I believe, as it is more of an observation which the question stem necessitates.
8. B. Indicators are very often weakly acidic/basic (as the pKa in the question stem implies), and so will affect the pH. Kw is not affected by anything except temperature which, presumably, the adding of indicator does not affect,
9. C. This question was one of the easier ones, and should be fairly straightforward (I find that students generally do understand that acid strength does not affect volume of titrant required). I would suggest modifying the question stem to use “equivalence point” instead of endpoint to be less ambiguous. As the question stem mentions “appropriate indicator”, we are left to assume that endpoint = equivalence point. However this may never truly be the case - the (very slight) discrepancy between any endpoint and equivalence point means that B or D could be potentially correct if one chooses the right (wrong) indicators.
10. C. C is the only choice which would produce 2 hydrogen environments. It is unnecessary to use the table on page 34, or to be caught up in analysing each peak. I often teach that the number of environments is often the most important clue yet often overlooked - this question is a great reflection of this line of thought.
11. A. Very good question. It is good to know that any buffer is most effective at pH = pKa of the conjugate acid with an approx. +-1 pH range. B is incorrect as the acetic acid/acetate buffer would only really exist at pH below 7. A is the only other sensible choice, and would produce a buffer at pH < 7.
12. A. The trick was that students would number alkyl branches from left to right despite needing to prioritise the alkene. However, A is not possible to obtain from this dehydration reaction.
13. D. We need to refer to the data sheet and see that the Ksp of Fe(OH)3 is low. Consequently, we expect some Fe3+ to precipitate; this decreases [Fe3+] which causes equilibrium to shift the equation left. However, the shift will not completely counteract the initial disturbance. Hence, both [Fe3+] and [FeSCN2+] will decrease.
14. A. Good question. NMR would provide 5 peaks for A and 4 for B. Lucas’ reagent (Zn/HCl) will react more quickly with B than A as B is a tertiary alcohol whereas A is a secondary alcohol. Although I believe there would be subtle differences on IR, it would be unreasonable for students to determine the difference without reference spectra. Every other test would yield identical results for A and B.
15. C. Very interesting question, though not too difficult. Pyruvic acid was the dione, as the question stem stated that pyruvic acid was obtained from the oxidation of lactic acid. One then had to realise that the protonated form of lactic acid was on the RHS, whereas the protonated form of pyruvic acid on the LHS. Thus, it is necessary to divide by the Ka of pyruvic acid.
16. C. This question was simple as long as students identified which compound was lactic acid. As a fun aside, Hartmann’s may be used to treat mild acidosis (pH of blood being too low) as lactate will react with carbonic acid to form bicarbonate, lactate being a weak base. Many students (medical students included) mistake lactate for lactic acid and do not realise this fact. Hartmann’s is also more preferred than saline in many situations (anaesthetists tend to like them) as it better resembles blood plasma; however, it is not always used as it is more expensive.
17. D. This question is again fairly straightforward as temperature is always proportional to r.o.r. and the reaction is exothermic. It would be more interesting if the question referenced Maxwell-Boltzmann distributions. I assume the trick is that students would be put off by “decreased yield”.
18. D. Difficult question for students who did not know the ion tests inside out. Pb2+ is in the mixture due to the chloride salt not dissolving in ammonia. Cu2+ was present due to the formation of a dark blue solution. The reason the last ion is Mg2+ and not Ba2+.is because we expect barium to have precipitated with sulfate, whereas magnesium would have not.
19. C. I assume the author peruses the wikipedia page on List of chemical compounds with unusual names. There are 6 carbon environments but only 3 hydrogen ones. I’m glad the author chose penguinone and not a more difficult compound such as olympiadane.
20. D. This is a strong acid vs weak base titration with acid being added to base. I think that D is the best answer; however I believe that the graph is not entirely accurate. The initial equilibria before addition of acid would consist of B + H2O <-> BH+ + OH-. Addition of HA/H+ via a strong acid initially would briefly result in both H2O formation from OH-, and BH+ formation from B. This results in a net replacement of OH- by BH+ and also slightly inhibits further conversion of B into BH+ by LCP. As we know, OH- has a much higher conductance, which when combined with the inhibition in BH+ production, would likely outweigh the increase in conductance provided by A-. As further acid is added, eventually B is converted to BH+ substantially resulting in conductance increase. The shape of the conductance vs acid volume curve should thus decrease slightly before the increase.
If people want, I cam go through the SAQ also.

#### CM_Tutor

##### Moderator
Moderator
I stumbled onto this excellent exam recently. CM_Tutor, some of the questions you have written are absolutely fantastic in how it requires deep knowledge of the syllabus as well as including many tricks to throw students off. Seeing the discussion here, I am going to throw my own hat into the ring.

Here are my solutions and thoughts on the MCQ portion of the exam, as I haven't yet had time to go through the SAQ in depth. There is every chance that there is some glaring error - please let me know and I will be more than happy to discuss.
Thank you for your comments and the solutions. You would have scored 19 / 20.

Some specific comments / thoughts:
1. B. Well written question, designed to trick students into picking D. Since H2O is gaseous, its concentration will contribute towards equilibrium and increasing [H2O] will shift equilibrium right. D is incorrect as although equilibrium moves right, by LCP the shift will not increase [CO2] to its concentration before the volume increase (remember that volume increases will decrease all concentrations)
It did not trick many students, however... 83% of respondents chose (B).
2. D. Other choices all had tricks attached - C seems correct at first glance but the charges are incorrect,
Far too many did not go on to the second glance. Surprisingly, all four answers were selected at significant rates.
5. D. Good question. Using Beer-Lambert Law, A = ecl. Sub in A = 0.437, e = 12.3, l = 1, we get c = 3.55 x 10-2. It is easy to stop here (and choose the very convenient option A), but the question is asking for carbonate concentration and not Cu2+. We must use the fact that the Ksp for CuCO3 is 1.4 x 10-10. As Ksp = [Cu2+][CO32-], we sub in 3.55 x 10-2 to get [CO32-] = 3.94 x 10-9
You are correct that it was easy to stop at the copper(II) ion concentration, as 71% of respondents chose (A). This was the question with the lowest rate of correct answers in the MCQ.
9. C. This question was one of the easier ones, and should be fairly straightforward (I find that students generally do understand that acid strength does not affect volume of titrant required). I would suggest modifying the question stem to use “equivalence point” instead of endpoint to be less ambiguous. As the question stem mentions “appropriate indicator”, we are left to assume that endpoint = equivalence point. However this may never truly be the case - the (very slight) discrepancy between any endpoint and equivalence point means that B or D could be potentially correct if one chooses the right (wrong) indicators.
The wording of the question was modified shortly before the exam but the MCQ options were not modified to be consistent - an oversight on my part. Nevertheless, this question was answered incorrectly by about a third of respondents.
10. C. C is the only choice which would produce 2 hydrogen environments. It is unnecessary to use the table on page 34, or to be caught up in analysing each peak. I often teach that the number of environments is often the most important clue yet often overlooked - this question is a great reflection of this line of thought.
I agree that the table was not needed but it could be helpful and I didn't want people to get to the table and then jump back, feeling mislead by not knowing this information was available.
11. A. Very good question. It is good to know that any buffer is most effective at pH = pKa of the conjugate acid with an approx. +-1 pH range. B is incorrect as the acetic acid/acetate buffer would only really exist at pH below 7. A is the only other sensible choice, and would produce a buffer at pH < 7.
More than 60% went with (B), however, and this was the MCQ with the third lowest rate of correct response. PS: I think you mean that (A) will form a buffer at pH > 7
14. A. Good question. NMR would provide 5 peaks for A and 4 for B. Lucas’ reagent (Zn/HCl) will react more quickly with B than A as B is a tertiary alcohol whereas A is a secondary alcohol. Although I believe there would be subtle differences on IR, it would be unreasonable for students to determine the difference without reference spectra. Every other test would yield identical results for A and B.
The second most difficult question on the paper, according to rate of correct answers, and the only one you have incorrect:
• bromine water will not react with A or B, and so is unhelpful
• acidified potassium permanganate will oxidise A (a secondary alcohol) but not B (a tertiary alcohol), and will occur with a colour change, so is helpful
• sodium carbonate will give carbon dioxide gas with both A and B, and so is unhelpful
• zinc metal and hydrochloric acid (Lucas test) will give a near-instant cloudiness with B while the cloudiness will develop slowly over time with A, so useful
• IR spectra will differ but not in any way that it is reasonable for an HSC student to identify (assuming no reference spectra are provided), so not useful
• 1H NMR will differ in number of environments and appearance - two methyls will appear as a doublet in A but as a singlet in B, for example. So, useful.
With 3 useful techniques, the answer is (B)
16. C. This question was simple as long as students identified which compound was lactic acid. As a fun aside, Hartmann’s may be used to treat mild acidosis (pH of blood being too low) as lactate will react with carbonic acid to form bicarbonate, lactate being a weak base. Many students (medical students included) mistake lactate for lactic acid and do not realise this fact. Hartmann’s is also more preferred than saline in many situations (anaesthetists tend to like them) as it better resembles blood plasma; however, it is not always used as it is more expensive.
I thought this would be easier than it turned out to be - nearly half of respondents chose (A) or (B).
17. D. This question is again fairly straightforward as temperature is always proportional to r.o.r. and the reaction is exothermic. It would be more interesting if the question referenced Maxwell-Boltzmann distributions. I assume the trick is that students would be put off by “decreased yield”.
The second easiest / most successfully answered question, after question 10.
18. D. Difficult question for students who did not know the ion tests inside out. Pb2+ is in the mixture due to the chloride salt not dissolving in ammonia. Cu2+ was present due to the formation of a dark blue solution. The reason the last ion is Mg2+ and not Ba2+.is because we expect barium to have precipitated with sulfate, whereas magnesium would have not.
Yes, answered correctly but less than half of the candidates. (C) was the most popular wrong answer, but (A) and (B) were chosen by over 20%.

I will publish a set of solutions, etc, after the results are provided.

#### dontbeagrumpmrtrump

##### New Member
Thank you for your response CM_Tutor!

With question 14, of course! Clear oversight on my part - I did not consider the colour change. Thank you for the correction. You are also correct about question 11, was a typo on my part.

I personally think these questions are fantastic and challenging to even potential state rankers. How do you sit down and write these questions? I would be most interested to learn how to write questions of my own to challenge my own students. I have noticed the amount of care put into each question; from staying within syllabus scope, whilst also providing an original twist on essentially every MCQ question, to carefully curating the choices to mislead and deceive students who do not have a solid fundamental understanding. Do you come up with ideas/twists for questions spontaneously?

#### CM_Tutor

##### Moderator
Moderator
Some of the questions come from examples that I have seen confuse students before - questions 1, 2, 5, and 9, for example. It helps that I have quite a lot of experience at secondary and tertiary levels, and also have qualifications in science and education. It also helps to have specialised in extending highly able students, and so I can put together questions that require a genuinely deep understanding of the theory. I won't ask a MCQ without having a reason for each of the distractors / alternatives. The Hartman's solution question, for example, gives the correct names for lactate and pyruvate and then alternatives that do describe each structure, even if they are not in line with IUPAC nomenclature.

Writing the whole paper did take me quite a while, to be honest, and I did get feedback during the process from @jazz519 and @Trebla. Q27 gives plenty of space for a variety of approaches, any of which I would accept, but all needed to include an assessment of the validity of the suggested conclusion and a discussion of potential improvements. These higher order directing words are easy to mishandle. I also deliberately selected some contexts that are unfamiliar from regular questions, which demands a good understanding rather than just good knowledge / memory of questions that have been asked in the past.

#### jazz519

##### Moderator
Moderator
Thank you for your response CM_Tutor!

With question 14, of course! Clear oversight on my part - I did not consider the colour change. Thank you for the correction. You are also correct about question 11, was a typo on my part.

I personally think these questions are fantastic and challenging to even potential state rankers. How do you sit down and write these questions? I would be most interested to learn how to write questions of my own to challenge my own students. I have noticed the amount of care put into each question; from staying within syllabus scope, whilst also providing an original twist on essentially every MCQ question, to carefully curating the choices to mislead and deceive students who do not have a solid fundamental understanding. Do you come up with ideas/twists for questions spontaneously?
Adding onto @CM_Tutor comments as I wrote the 2020 BOS trial paper last year. The main aspect that helps I would say in being able to write these type of questions is an understanding of the content at a level that is far beyond the scope of the syllabus. If I had just come out of Year 12 I don't think it would be possible to write an exam like that because you may be unsure about your own understanding on the more difficult questions and so it would be hard to develop them with a correct answer and reasoning. This mainly for me came from doing a university degree with a major in chemistry and so a lot of areas in the syllabus like the NMR, buffers, organic chemistry part I have learnt at a university level and so that builds a very detailed knowledge of how these concepts work. Sometimes the type of questions I would include would be like a watered-down version of a uni question that is within the scope of the HSC syllabus. Being a tutor for 5 years also helps with that because naturally over that time I have gone through many past paper questions with students and so you pick up on hard questions you may have seen and get inspiration from those to remodel them into a different question that may be more difficult.

#### jazz519

##### Moderator
Moderator
One general way though to make questions more difficult from experience is by making questions that are linked to multiple areas of the syllabus across modules. I find that a lot of students have trouble for example being able to link module 7 organic chemistry reaction pathways with module 8 spectroscopy and spectrometry graphs. So that is a strategy you can use in some aspects to make harder types of questions.

#### CM_Tutor

##### Moderator
Moderator
One general way though to make questions more difficult from experience is by making questions that are linked to multiple areas of the syllabus across modules. I find that a lot of students have trouble for example being able to link module 7 organic chemistry reaction pathways with module 8 spectroscopy and spectrometry graphs. So that is a strategy you can use in some aspects to make harder types of questions.
Yes, the paper I wrote has brought multiple topics into single questions - equilibrium, organic, and polymers in q32, Ksp, titration, acids and bases in q28 - as other ways to make for challenging questions.

#### someth1ng

##### Retired Nov '14
The questions are also written in a way where if quickly form an answer without thinking, it's easy to get it wrong. There's also a bunch of questions where I'm like, lol, don't remember a single thing about this stuff (e.g. some of the solubility stuff).

#### CM_Tutor

##### Moderator
Moderator
The questions are also written in a way where if quickly form an answer without thinking, it's easy to get it wrong. There's also a bunch of questions where I'm like, lol, don't remember a single thing about this stuff (e.g. some of the solubility stuff).
Yes, I believe that exams where a strong performance is possible without use of the brain are not particularly worthwhile. Those who perform well should be demonstrating the ability to think like a scientist, to apply knowledge to solve unfamiliar problems, and to reason sensibly even if a solution is not evident. I do not feel that the ability to regurgitate a text book or to demonstrate rote-learned "skills" which lack meaning for the student are ingredients for becoming a scientifically literate individual, whatever career path might be chosen.

#### idkkdi

##### Well-Known Member
Yes, I believe that exams where a strong performance is possible without use of the brain are not particularly worthwhile. Those who perform well should be demonstrating the ability to think like a scientist, to apply knowledge to solve unfamiliar problems, and to reason sensibly even if a solution is not evident. I do not feel that the ability to regurgitate a text book or to demonstrate rote-learned "skills" which lack meaning for the student are ingredients for becoming a scientifically literate individual, whatever career path might be chosen.
any merit to regurgitating prac methods?
they make me cry. especially the ultra nuanced titration preparation.

#### CM_Tutor

##### Moderator
Moderator
any merit to regurgitating prac methods?
they make me cry. especially the ultra nuanced titration preparation.
If by "merit" you mean marks in an exam, marking can only be done based on what is written in responses, not on what an examiner might think the student understands, etc. So, a response that answers the question will always gain credit. Verbatim memorisation becomes much less necessary, however, with understanding. By understanding what a prac is about and how it works, much of the sequence of steps in a method becomes necessary consequences and so can be deduced. For example, washing techniques for a titration... once someone understands that the concentration of the solution in the pipette and burette must not be changed for accurate results, washing each with the solution they are to contain becomes clear - using anything else will cause a change in concentration by dilution or reaction or both. In the conical flask, by contrast, all of the reacting solutions must come from the pipette and the burette, so any residue from washing the CF must be water. This does dilute the substance in the CF but that doesn't matter as the chemical amount / moles is unchanged and its volume when it was in the pipette is known.

If you find yourself needing to memorise a procedure, I suggest you reflect on which steps you don't understand. If the reason for a step is clear, its position in the sequence should be clear. A good example is in purifying an ester product, where you should be able to answer:
• why is simple distillation inappropriate (2 reasons)?
• why neutralise with Na2CO3 or NaHCO3 (at least 2 reasons)?
• why can you use a separating funnel? What does it achieve?
• why anhydrous calcium sulfate (or equivalent)?
• why is distillation now appropriate (2 reasons, including describing what is different at this point than at the start)?
Being able to answer these questions should also explain why a different sequence would be problematic, meaning the order need not be memorised.