Parramatta Library HSC Chemistry Trial Discussion (1 Viewer)

strawberrye

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Feel free to comment and ask questions about this paper on this thread.

As promised, here is the chemistry paper for you to have a go:) Please leave a comment on this thread about how you found the paper:)

View attachment Parramatta Chemistry exam-Meihua(1).pdf

Firstly, I correct my initial point about the first question, the answer is that the weak acid will require the same amount of base as the strong acid. The reason being that the dissociation of the weak acid doesn't change the concentration of it, which is what is important in a weak-acid base titration.

Queston 9 was taken out for the actual session in Parramatta Library, but a replacement question has been included in this online version.

Just some interesting statistics. The highest mark for multiple choice was 16/19. Most people are averaging around 11/19 mark, which demonstrates that this indeed was a very hard exam:)
 
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iStudent

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That first MCQ question reminds me of HSC 2011 Q15 :)
 

strawberrye

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If you know how to calculate pH, you should definitely know how to calculate pOH...there are skills that are not explicitly stated in syllabus but that I think will be assumed for the HSC
 

ManDarren

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Organic chemistry is difficult. Those who study it have alkynes of trouble.
 
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In the preceding days I will be uploading answers and general comments. Feel free to comment and ask questions about this paper on this thread.

Firstly, I correct my initial point about the first question, the answer is that the weak acid will require the same amount of base as the strong acid. The reason being that the dissociation of the weak acid doesn't change the concentration of it, which is what is important in a weak-acid base titration.

I have also decided to take out question 9 because as some people have kindly pointed out, I didn't correctly balanced the equation I had in mind which subsequently meant there was no correct answer in the multiple choice. Once again, I apologise for this mistake.

Just some interesting statistics. The highest mark for multiple choice was 16/19. Most people are averaging around 11/19 mark, which demonstrates that this indeed was a very hard exam:)
This makes me want to do the paper. Is there any chance you could upload it or?
 

iStudent

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Can we find out our individual marks for each section + rank?(without going to parra library to collect it)
 
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M_Francis

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In the preceding days I will be uploading answers and general comments. Feel free to comment and ask questions about this paper on this thread.

Firstly, I correct my initial point about the first question, the answer is that the weak acid will require the same amount of base as the strong acid. The reason being that the dissociation of the weak acid doesn't change the concentration of it, which is what is important in a weak-acid base titration.

I have also decided to take out question 9 because as some people have kindly pointed out, I didn't correctly balanced the equation I had in mind which subsequently meant there was no correct answer in the multiple choice. Once again, I apologise for this mistake.

Just some interesting statistics. The highest mark for multiple choice was 16/19. Most people are averaging around 11/19 mark, which demonstrates that this indeed was a very hard exam:)
When will our papers be available for collection?
 

strawberrye

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Please understand marking paper,particularly with only one person is a very time consuming process. I would not be giving out ranks-but I will give a broad description of distribution of marks so you will have a rough idea how you went. It will be available for collection either late this week or next-i need to double mark to make sure it is consistent. Unfortunately at this stage I don't think I will be uploading my paper since it is not mistake free and the purpose of me slaving countless hours on this paper was to make people come and do it under exam conditions. I apologise for this inconvenience but I might consider uploading a few extended response questions to give a taste to what the exam was like
 

iStudent

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Can you explain why Q11 is C?
I thought it would be B because increase in pressure -> indicates decrease in volume -> decrease in volume = increase concentration of all species (explaining the spike in CO2 concentration). -> equilibrium shifts to the right due to Le Chatelier's -> hence, concentration decreases afterwards

Also, how do you collect the results? Who do we ask?
:)
 

strawberrye

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I will be posting answers by the end of this week so you will get a detailed explanation of why...just go to level 3 at the desk/reception and ask them for your paper
 

strawberrye

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Can we find out our individual marks for each section + rank?(without going to parra library to collect it)
You actually got the highest mark in the written section of the paper, I can say that you got one of the top three marks in the exam (as to what that is, you will have to come to the library to find out for yourself:))

I've changed my mind about not posting marks, I will annoymously post the marks below (there are some people whom I will not include in the marks distribution for various reasons, such as one person was a year 11 girl attempting the paper, so inclusion of extreme outliers would not serve as an accurate reflection of the average), in no particular order the marks were: (they are out of 74 since I took out a multiple choice)

37.5, 42.5,49,26.5,34,37.5,24.5,33.5, 35.5,41,32.5,31.5,36.5,34.5,53.5,38,37,55.5


So just some answers and explanations-the missing answers will be posted up in the next two days-(I also start each question answer with some comments about people's performance in the question or things to look out for so that candidates can better understand how to improve for their HSC Chemistry exam). I am going to be focusing on the extended responses for now-section I part B. As with many of the following questions, there are various possible responses, and my responses just represents a model of what might have gotten you a band 6 response.

Multiple Choice Answers: (note question 9 was omitted)

1)C 2)A 3)B 4)B 5)C 6)C 7)C 8)D 10)C 11)B or D(the two possible answers are for the Parramatta session only, only one answer is correct for the online version, that is B) 12)B 13)D 14)A 15)B 16)A 17)B 18)C 19)C 20)B

Question 21: (This question was very well done by most candidates, however, it is notable that a significant amount of people ignored discussion of the impact of bases on society and environment with specific examples, most just focused on acid and acid rain, which a partial reply to the question resulted in a loss of half of the marks).

The use of weak acids as food additives have enabled food to be preserved for longer, suc as citric acid whichis used in canned foods to preent microbe growth. However the release of acidic oxides from the combustion of fossil fueld such as S+O2->SO2, when reacting with water vapour in the air can form acid rain, which detrimentally results in the termination of plant growth and damages vegetation, resulting in loss of crops. SO2+H2O-.H2SO3 (sulfurous acid). Furthermore, the use of weak bases have had an overall positive impact on society and environment, such as sodium hydrogen carbonate which is an ampiphrotic substance often used to clean up acid and base spills, effectively minimising the destructive impacts of these spills on the environment. Sodium hydrogen carbonate is also widely used in cooking and as an efficient cleaning agent, further assisting to enhance quality of life in society.

(With base) HCO3-+OH-->H2O+CO2-3
(with acid)HCO3-+H+->H2O+CO2

Q22)(Many people interpreted the question in various ways, most refer to AAS as a general method, however, my intention in this question, particularly with the phrase "Outline how this method can be used to detect the concentration of lead ions in a river is to aim for a specific procedural method using AAS in relation to the scenario, which only one or two people actually managed to do, in this instance, I did not deduct marks for the general method as I realised I could have worded the question a bit less ambiguously, furthermore, some people did not have a clearly labelled diagram, and this did not assist them to acquire more marks-for the diagram-refer to jacaranda chemistry textbook or conquering chemistry)

A potential procedure that can be used to detect metal ions in a river stream are as follows:
1)Collect water samples from different areas of the river

2)In a lab, construct a range of standard lead solutions by using appropriate dilutions and dissolutions, including a control solution which contained deionised water which can be used to calibrate the AAS machine. The standards can be made up by dissolving 10ml of 0.1M of lead nitrate and make it up to volumes of 100ml, 1L etc to get solutions of corresponding concentrations of 0.01M, 0.001M etc.

3)Calibrate the AAS machine using the range of prepared standard solutions of known concentrations, also can draw a calibration curve with absorbance on y-axis and concentration of lead ions on the x-axis.

4)For the water sample from the river, put in AAS and measure absorbance, interpolating the concentration of lead ions from the calibration curve.

One potential limitation is that AAS allows only for the measurement of one type of metal ion, one sample at a time. This can be a time consuming pocess, particularly for pollution monitoring an control over a large area, also AAS is extremely expensive and not portable, so not always the most convenient instrumentation for analysis of heavy metal pollutants.

Q23)(many people did not read the question properly and erroneously thought that standard cell notation involved writing half equations and calculating cell potential which was completely irrevelant, many did not correctly identify the standard cell conditions, some introduced absurd answers such as requirement of salt bridge, electrolyte, solutions of equal concentrations, cleaned electrodes which all kind of is relevant, but missed the mark to answering the question).

Standard cell notation involved Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) -(in this instance, the lack of inclusion of states which are very important in cell notations resulted in people losing a mark for this section of the question). Set of standard conditions include concentration of electrolytes at both anode and cathode at 1M, measured at STP and 25 degrees.

Question 24(almost all papers got this question right so I am not going to go into too much detail on the answer, but just one thing is for people to read the question very carefully, the question asked for a chemical equation that is composed of structural formulas, some people started to count hydrogens and oxygens to construct a condensed molecular formula equation which is not the requirement of the question, candidates are advised to revise basic nomenclature representations, some people erroneously thought the benzene ring had hydroxide ions all around it, the formula of the benzene ring in the methyl salicylate was C6H6.-knowing this would have resulted in drawing of correct diagrams, a notable amount of individuals did not include equilibrium arrows in the equation or ignore the formation of water as the other product-these candidates all lost a mark in this question).

Question 25-(Some people had very complicated flowchart diagrams, just one simple word of advice, the more simple your diagram, the more likely you are going to get the marks, markers are looking for answers, not looking for how much you know about cations and anions in general. Some people separated out the two ions when it was explicitly stated in the question both ions are in the sample, so a procedure where for the same sample, for one, you test for presence of carbonate-evolution of gas, for the other, no-evolution, than add barium nitrate is implicitly wrong-because it is impossible that the same sample will have different results for the same test in the same procedure-these candidates all lost marks for these mistakes).

Essentially, the flowchart involved adding nitric acid to the sample, gas evolving will indicate presence of carbonate, then nitric acid shouldbe added until effervescene of gas ceases(i.e. all carbonate is removed) before adding barium nitrate -wich results in a white precipitate of barium sulfate which confirm the presence of sulfate ions as well.

Question 26-(Many people failed to realise that using a table format would have been the most effective way of answering this question, some people wrote way too much and went off tangent, others did not effectively compare the two processes together-which again resulted in a loss of marks-equations in this instance was essential to establish an awareness that both carbonation and esterification was equilibrium process and as a result, factors such as temperature, pressure and the use of catalyst all came to influence the rate of reaction and yield of products formed).

The following information could be presented in a table format(and it was in my answer on paper).
Carbonation

Equation: CO2+H2O<->H2CO3 (include states)

Catalyst:Usually don't require a catalyst since other measures are more effective

Pressure:Increase in pressure, according to Le Chatelier's principle, equilibrium shift to the right to reduce pressure, causing increase in dissolution of carbon dioxide, and thus carbonation in soft drinks.

Temperature: Decrease in temperature since this reaction is exothermic, by Le-Chatelier's principle, the equlibrium will shift to the right, again increasing in dissolution of carbon dioxide gas.

Esterification

Equation: alkanol+alkanoic acid<->ester+water

Catalyst: Often small amounts of concentrated sulfuric acid or hydrochloric acid is used to speed the rate of reaction by lowering the activation energy of the reaction. Sulfuric acid also serves as a dehydrating agent, absorbs the product water and thus according to Le Chatelier's principle, again forces the equilibrium to the right to increase the yield of ester.

Pressure:Usually don't increase pressure because can be danergous in esterification reactions which can result in an explosion with heating in a pressurised environment.

Temperature:Heat is often used to speed up the reaction, boiling chips is often included in the reaction vessel to ensure even heating.

Q27:Generally responses for this question was extremely well answered. However, it is notable that many answers lost marks for a poor structure of the question, not including chemical equations when it was explicitly required in the question and not making a close address to the statement in the question.

Monitoring Water Quality:
High water quality is essential for human consumption as well as critical for agriculture developments. Portable technologies such as pH meters and nephelometers enables quantitative measurement of water pH and turbidity, as well as various ion-elective electrodes such as oxygen to measure dissolved oxygen in water, these technologies has allowed for rapid analysis and efficient/affordable ways to quickly assess water quality. The apt alleviation of poor water quality has been made possible with products such as AAS which allows monitoring of heavy metal poisoning in rivers which previously has been impossible the the limitations of sensitivity limits of traditional gravimetric analysis techniques. The use of new chemical products such as membrane filters has not been economically feasible and still inadequate in removal of heavy metal ions, but overall, the statement is highly valid. (can include a reaction involving floccuating agents to purify water)

CFC Alternatives and Monitoring Ozone Hole:
The ozone layer is vital for the filtration of harmful UVB radiation which has the potential to cause skin cancer in humans (genetic mutation change). The use and production of CFCs, although cheap, such as aerosols, coolants in fridges and air conditioners, have had a detrimental long term effect through increase in medical cost which impedes sustainable economic development. Many people were pretty good in writing the equations for the depletion of ozone layer, but not many managed to relate these equations back in their answers.

The chain reaction and cyclic process which leads to the regeneration of chlorine free radicals which decomposes ozone into oxygen thus causing a reduction/thinning of ozone layer is illustrated by the following equations:

CCl3F->Cl*+CCl2F*
Cl*+O3->ClO*+O2
ClO*+O*->O2+Cl*

The use of UV spectrophotometers on both ground-based and satellite based instruments such as TOMS have allowed complete ozone mapping profiles to be constructed as a function of altitude and latitude. However, alternatives such as HCFCS and HFCs which have more reactive C-H bonds have been critical to minimizing the depletion of the ozone layer and as part of the Montreal Protocol, it also ensures sustainable economic and environmental development. Hence, it is evident that the statement is fully valid for this area of science.

Q28:Many people failed to read the instructions of the question carefully which required drawing of the titration curves on the same graph. Many people also failed to label their axis or label them properly, such as the y-axis should have been labelled as pH and the x-axis should be the volume of sodium carbonate added (it can only be the volume of base since both graphs have to be drawn on the same graph).

Essentially, the titration graph for HCl should start from a low pH such as 1 or 2 and climb upwards, having one equivalence point around 6/7. The graph for citric acid was a bit more tricky, it was a tripotic acid, which means it actually had multiple equivalence points and multiple possible indicators for each equivalence point.The graph should look a bit like this image-but the pH starting point should be higher, say around 3/4 because tripotic acid is a weak acid.
http://www.bing.com/images/search?q...2DB2A876E5B3380F5E22DDC0E91E&selectedIndex=15

b)Equivalence points was mostly identified correctly for HCl, but not for the citric acid graph.
c)Potential indicators can include methy orange, phenophathalein etc

Q29)a)Many of the diagrams were well drawn, some people draw diagrams for all the experimental steps but it wasn't necessary to get full marks, one clear labelled diagram of the filtration/precipitation process would have been sufficient, many people did not write their procedure step by step, hence it was quite confusing to read a procedure not specific enough, and these candidates subsequently lost marks as a result. Candidates are strongly advised to plan their answers before writing them out.I will not include the experimental diagram since I have no way of drawing it using what I know on BOS.

1)Meausre a small amount of fertiliser in a clean beaker.
2)Dissolve about 50ml of 0.1M of HCl to dissolve as much of the solid as possible.
3)Filter off insoluble material
4)Heat solution to near boiling, while adding barium chloride solution in excess with continuous stirring until no more white precipitate is observed to be formed.
5)Allow mixture to cool and for precipitate to settle down the bottom of the beaker.
6)Filter and dry precipitate to constant mass, weigh precipitate on an electronic balance.
7)Calculate percentage of sulfate by the formula:

Percentage of sulfate/100=((molar mass of sulfate/molar mass of barium sulfate) times mass of sulfate precipitate))/(mass of fertiliser sample)

Relevant ionic equation: Ba2+ +SO2-4->BaSO4

b)AAS will likely likely to be ineffective to detect sulfate content because it is only useful for low concentrations of metal ions, unsuitable for the high percentage of sulfate content which is more appropriately analysed by the quantitative analytical method of gravimetric analysis.

Q30:Many people got the neutralisation reaction right, but almost all candidates fail to realise that sulfuric acid is a diprotic acid, so a lot of people failed to times the given concentration of the acid by two in the hydrogen ion concentration of the pH equation to calculate the correct moles.


Neutralisation equation: H2SO4+Na2CO3=>H2O+CO2+Na2SO4

Moles of hydrogen ion in sulfuric acid: n=cV=0.4times 2 times (12/1000)=9.6times 10^3 moles.
Moles of sodium carbonate: n=cV=(56/1000)times 4=0.224 moles
Excess moles(basic)=0.223-(9.6times10^3moles)=0.2144 moles

poH=-log(0.2144/(68/1000))=-0.50

Q31)Many people left out this question blank or gave a very general approach to making the required solution without giving specific example numbers that could make up to the required concentration. Most people neglected highlighting an analytical analysis where making a series of standard solutions would be useful. Again, candidates are advised to read the question carefully to make sure they have truly answered all parts of the question.


Making a series of standard solutions are beneficial in calibrating an AAS machine for AAS. For example, making 0.234 M of sodium carbonate(Na2CO3) would be done via the following two methods:

Method 1:
Molar mass of anhydrous sodium carbonate is 105.99g/mol (anhydrous simply means without water). so n=m/M, and substituting n=0.234 and M=105.99 in the formula, we get m as 24.8g. So to make 0.234M solution can just weigh 24.8g of Na2CO3, dissolve in some water, and wash the dissolved solution into a 1L volumetric flask, and add water until the meniscus is touching the calibration mark, stopper the flask, invert it several times, and you've got yourself a standard solution of 0.234M Na2CO3.

Method 2:
Can use the formula C1V1=C2V2, so say you have a 1L of 0.5M of sodium carbonate, can dilute to 2.14L to get a concentration of 0.234M.

Q32a): Many people made the mistake of thinking that the solution with pH 1.4 was a strong acid, very few people managed to recognise that both substances were weak, and there were many erroneous explanations and comparisons, such as saying the solution with pH 3.6 was a stronger acid than the solution with pH 1.4-this is technically NOT correct, you can either have a strong or weak acid, you can't really say there is a 'stronger' weak acid, since strong denotes 100% dissociation into ions in scientific terms. People also need to revise how to calculate degrees of ionisation since many people seems to either fail to notice this part of the question or did not know how to calculate it.

Strong acids denotes 100% dissociation into its respective ions in aqueous solutions, so for a solution of 0.2M, if it was a strong acid, pH=-log[H+]=-log(0.2)=0.7, since both solutions have a higher pH than this, this means both solutions are weak acids.

Degree of ionisation for solution of pH 1.4: =((10^1.4)/0.2)times 100=19.9%

Degree of ionisation for solution of pH 3.6:=((10^3.6)/0.2)times 100=0.13%

b)any of the following could have gotten full marks-the question did not ask for justification, many people justified unnecessarily: Cloud chamber, Geiger-Muller counter, scintillation counter

Q33a)(many people ignored the wording of the question which demanded the drawing of separate flow charts, many used only one flowchart and some had missing steps, and one person did not draw arrows between steps hence resulted in loss of marks, it is important for students once again to read the question extremely carefully to avoid the loss of unnecessary marks).

Physical process:
Screening out large debris (makes subsequent purifying steps easier)->coagulation to increase particle size->sedimentation to remove larger suspended particles->filtration to remove finer suspended particles.

Chemical process:
Aeration to oxidise soluble iron and manganese salts to insoluble compounds->oxidatione.g. by potassium permanganate of iron and manganese salts->lime softening to precipitate out possible high levels of calcium and magnesium ions->pH adjustment to neutralise the water fit for consumption->chlorination(to kill microbes)->fluoridation(assist in tooth decay prevention)

b)This was well done by most people. A simple procedure for testing dissolved solid would have been to use an electrical conductivity probe(assuming most of the dissolved solids were ionic) or through using the process of evaporation and then weighing the precipitate, dividing it by the mass of the water sample and times by 100 to get percentage of dissolved solids-but this latter method only works if the concentration of dissolved solids is adequately high enough. For testing hardness, a quantitative procedure would involve titration with an EDTA complex reagent.
 
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strawberrye

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All answers to the exams are up now, if you have any questions feel free to ask:)
 

strawberrye

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I should clarify, when I said enough likes, I meant the number of different peple, so far I could count 13 different people who have liked a comment I posted:) So 3 more people like my comments-I will be posting up the full paper:) Keep the likes coming and I wish everyone all the best for their HSC exams:)
 
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