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.