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HSC Chemistry 2021 Exam Solutions (1 Viewer)

jazz519

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Hi, here are my answers to this year's chemistry exam.

I will post them progressively over today as I am not currently at home, so check back for updates later today.

Comment by another user:
just for convenience - this is the paper from the thread posted by @Life'sHard:
https://drive.google.com/file/d/14QFLQcYAv8UkAY-qj-V7QCi5bWElL4mU/view


Multiple choice:
1) A
The definition of a dynamic equilibrium is the rate of forward and reverse reactions being equal. Concentrations of reactants and products are CONSTANT not equal

2) B
Only one that will form a precipitate is Cl- to form AgCl precipitate

3) A
Propanamide molecule which has one of the H atoms in an amine group replaced with a methyl group

4) C
Ethyl pentanoate ester synthesis:
Ethanol + pentanoic acid <--> ethyl pentanoate + water

5) B
Rinsing the conical flask with the acetic acid will mean that more moles of acid are present than without this step. This means more volume of base from the burette will be added and so a greater moles will be obtained for the acid through a molar ratio. The greater the amount of moles, the higher the concentration for the acetic acid

6) C
Weak acid has the following equilibrium:
HA(aq) + H2O(l) <--> A-(aq) + H3O+(aq)
If it is diluted by adding water the equilibrium will shift to the right and so ionisation increases. The Ka value is only changed by temperature and not concentration changes like in a dilution so it remains the same

7) A
Methanol when it reacts with HBr forms bromomethane. This molecule has a lower boiling point because it only has dipole dipole interactions which are weaker than the hydrogen bonding in Methanol.

8) D
The non-polar hydrocarbon tail of soap forms dispersion forces with oil molecules (inwards the droplet). The polar head group forms ion-dipole forces with polar water molecules (head pointing outwards).

9) D
The solvent should not have an absorption spectra that overlaps with the paracetamol. This is because if we try to measure the sample solution, the absorption will not be due to our paracetamol sample but rather a combination of the solvent and the paracetamol.

10) B
The polymer formed is a condensation polymer as there is a C-O-C linkage. This could be formed using the monomer B.

11) C
Adding solids to an equilibrium will not cause any shift

12) A
Mass spectrum: 98 m/z molecular ion peak. Therefore molecular mass of the substance is approximately 98. This fits with either A or D.

Based on the C-13 NMR: there are 4 signals indicating 4 carbon environments. A fits the correct number of carbon environments, while D has 6.

13) C
Hydration is an addition reaction which happens with alkenes and not alkanes.
X = prop-1-ene

Y = Propan-2-ol as there is no other option for the alkene start in this case.

Z = propanone. Oxidation of a secondary alcohol makes a ketone

14) D
Absorbance = 0.30 gives [Ni2+] = 0.0022 mol/L
n(Ni2+) = cv = (0.0022)(0.250) = 0.00055 moles
This is the moles in the 10.0 mL solution. However, the initial solution was 50.0 mL so times moles by 5
n(Ni2+) in 50 mL = 0.00055 x 5 = 0.00275 moles
m(Ni2+) = n x MM = (0.00275)(58.69)
m(Ni2+) = 0.16g

15) B
HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)
n(HCl) = cv = (0.20)(20/1000) = 0.004 moles ==> n(H+) = 0.004 moles
n(NaOH) = cv = (0.50)(20/1000) = 0.01 moles ==> n(OH-) = 0.01 moles

OH- is excess
n(OH-) excess = 0.01 - 0.004
n(OH-) excess = 0.006 moles
[OH-] excess = n/v = 0.006 / (40/1000) = 0.15 mol/L
pOH = -log(0.15) = 0.823...
pH = 14-0.823 = 13.2

16) B
2 equivalance points and therefore 2H+ number

17) D
n((NH4)3PO4.12MoO3) = m/MM = 24.21 / 1877 = 0.01289... moles
n(Na3PO4) = n((NH4)3PO4.12MoO3) (this is because every one of the precipitate molecules has 1 PO43- and the moles of PO4 3- is equal to the moles of Na3PO4)

n(Na3PO4) = 0.01289... moles
m(Na3Po4) = n x MM = (0.01289...)(3x22.99 + 30.97 + 4x16)
m(Na3PO4) = 2.115 g

18) D
8 hydrogen in total, which only works with the B and D formulas. 3 hydrogen environments for both. D has a 3H, singlet. This is not present in B

19) C
2AgNO3(aq) + K2SO4(aq) --> Ag2SO4(s) + 2KNO3(aq)
[SO4 2-] = 0.100 mol/L as each K2SO4 has 1 sulfate ion

It's a Ksp question because it's asking about when will the precipitation start
Ag2SO4(s) < -- > 2Ag+(aq) + SO4 2-(aq)
Ksp = [Ag+]^2 [SO4 2-]

From data sheet Ksp = 1.20 x 10^-5

1.20 x 10^-5 = [Ag+]^2 [0.100]
[Ag+] = 0.001095... mol/L
[Ag+] = [AgNO3]

n(AgNO3) = cv = (0.001095...)(0.250) = 0.000273... moles
m(AgNO3) = nMM = (0.000273..)(169.9)
m(AgNO3) = 0.0465 g

20) C
Ka = [(CH3)3N] [H3O+] / [(CH3)3NH)+]
1.55 x 10^-10 = [(CH3)3N] [H3O+] / [(CH3)3NH)+]

Ksp equation:
[(CH3)3NH]Cl(s) < -- > (CH3)3NH)+(aq) + Cl-(aq)
Ksp = [(CH3)3NH)+] [Cl-]

pH = 4.46
[H3O+] = 10^-4.46

therefore using the first Ka one and a RICE table:
1.55 x 10^-10 = (10^-4.46)^2 / [(CH3)3NH)+]
[(CH3)3NH)+] = 7.7565... mol/L

At the Ksp the concentration of the [Cl-] and [(CH3)3NH)+] are equal

therefore:
Ksp = (7.7565...) (7.7565...)
Ksp = 60.2
 
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jazz519

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Short answer:
21)
a) not the only possible answer
- flammable substances as they are all fuels
- address by keeping away open flames or ignition sources that may lead to a fire

b)
Flask 1: propanoic acid (polar so soluble in water and not able to oxidised)
Flask 2: hex-1-ene (non-polar so insoluble in water. Alkenes can be oxidised)
Flask 3: propan-1-ol (polar so soluble in water and oxidised as it is a primary alcohol)
Flask 4: hexane (insoluble with water and doesn't react with the acidified permanganate)

c) many possible answers here such as:
- hex-1-ene use bromine water where an addition reaction occurs and the colour changes from orange to colourless
- propanoic acid use Na2CO3 where a neutralisation reaction will occur resulting in bubbles forming
- propan-1-ol use acidified dichromate and the colour will change from green to orange through an oxidation reaction

There may be other possibilities for this one than the above

22) possible ways to shift equilibrium left:
Add dichromate ions: this will increase the concentration of dichromate ions and so disturbs the equilibrium. According to Le Chateliers Principle, an equilibrium that is disturbed will undergo a chemical reaction to minimise the disturbance. Therefore the equilibrium shifts left leading to an increase in the reactant concentrations and changing the colour.
- remove H+ by adding a base such as NaOH: this will reduce the [H+] through an acid-base reaction. Similarly like above the equilibrium shifts left.
- increase the temperature: for an exothermic reaction when temperature is increased the equilibrium will shift towards the endothermic side and therefore a left shift in this case.
- dilute the solution by adding water
- some form of a precipitation reaction that when adding a reagent leads to formation of a precipitate with the chromate ions. This will remove them from the system

23)
a) HCOOH(aq) + KOH(aq) --> KHCOO(aq) + H2O(l)

b) KHCOO(aq) --> K+(aq) + HCOO-(aq)
The K+ is a weak conjugate acid as it comes from a strong base KOH and so does not react with water. The HCOO- is a moderately strong conjugate base as it comes from a weak acid HCOOH and so reacts with water:
HCOO-(aq) + H2O(l) <--> HCOOH(aq) + OH-(aq). As hydroxide ions are formed the pH will be basic.

24) to find the chemical formula of the alkane:
mm(CnH2n+2) = 72.146
12n + 1(2n+2) = 72.146
14n = 70.146
n = 5

Therefore it is pentane

Isomers:
2-methylbutane
2,2-dimethylpropane

25)
Fermentation equation: C6H12O6(aq) --> 2CO2(g) + 2C2H5OH(aq)
v(CO2) = 1006 mL = 1.006 L
n(CO2) = v / Vm (you can use PV=nRT here too)
n(CO2) = 1.006 / 24.79 = 0.04058... moles
n(C2H5OH) = n(CO2) (1:1 ratio)
n(C2H5OH) = 0.04058... moles
m(C2H5OH) = nMM = (0.04058..)(46.068)
m(C2H5OH) = 1.8694.... g
m(C2H5OH) = 1.869 g (4sf)

26)
a)
A = 2-methylpropan-2-ol
B = 2-methylpropan-1-ol (has to be a primary alcohol as it makes the carboxylic acid to be used in the esterification step)
C = 2-chloro-2-methylpropane (replaces OH with Cl)
D = 2-methylpropanoic acid (oxidation of primary alcohol first makes an aldehyde and then a carboxylic acid)

b) refluxing has 2 purposes:
- heating the reaction mixture which leads to an increase in reaction rate. This is because at a higher temperature a greater proportion of molecules will have the collisional energy to overcome the activation energy barrier
- heating leads to formation of vapours that may be lost as the substances are volatile. Refluxing uses a cooling condenser to convert the vapours back to liquids and return them to the reaction mixture

27)
a) Li3PO4(s) < -- > 3Li+(aq) + PO4 3-(aq)
Ksp = [Li3+]^3 [PO4 3-]

For the range we should compare between just before no precipitate is made and when the precipitate is first made. This is when the solution first reaches saturation

Using Sample 3:
Ksp = (0.15)^3 (0.010) = 3.375 x 10^-5

Using sample 4:
Ksp = (0.15)^3 (0.10) = 3.375 x 10^-4

Therefore Ksp is between the range of: 3.375 x 10^-5 - 3.375 x 10^-4

b)
- Add more samples between the 0.010 and 0.10 mol/L for the [PO4 3-] this will make the range smaller and therefore more accurate
There are probably other acceptable ways too than this answer

28)
The precipitate that will form is copper hydroxide.

m(Cu(OH)2) = 4.61 g
n(Cu(OH)2) = m / MM = 4.61 / (97.561) = 0.04725... moles

Alkali metals have a +1 charge as they belong to group of the periodic table

Therefore the equation will be:
2XOH(aq) + Cu(NO3)2(aq) --> Cu(OH)2(s) + 2XNO3(aq)
where X = alkali metal

n(XOH) = 2n(Cu(OH)2)
n(XOH) = 2(0.04725...)
n(XOH) = 0.0945... moles

n = m/MM
MM = m / n
MM = 5.30 / 0.0945...
MM(XOH) = 56.08 g/mol

The OH contributes 16 + 1.008 = 17.008 g/mol to the MM
therefore X = 56.08 - 17.008 = 39.07 g/mol

This is consistent with potassium (K)

Therefore, the alkali metal hydroxide is KOH

29)
IR: this techniques is used to find the key functional groups in a molecule. The two signals are 3300 cm-1 that are weak and sharp arise from the amine (NH2) group in the structure

MS: indicates the molecular mass and fragments of a molecule. The 30 m/z signal arises due to the fragmentation of the molecule during the ionisation process which involves high energy electron bombardment which can lead to the breaking of bonds. This signal arises from the NH2CH2+ fragment which has a MM = 14+2x1 +12 +2x1 = 30

13C NMR: there are 3 signals in the spectra as there are 3 carbon environments in the molecule. The signal at 45 ppm is due to a RCNH2. The 25 and 35 ppm signals are due to C-C bonds with the 35 ppm carbon being closer to the amine group due to dieshielding of the carbon environment as result of being closer to the electron withdrawing amine group.

1H NMR: 6H integration is coming from the middle three CH2 groups. There are 2 unique hydrogen environments in this location. One is a CH2 while the other is a 2×CH2. A quintet arises because each hydrogen environment has 4H atom neighbours adjacent to them. The chemical shifts of these 2 hydrogen environments are lower than the other 2 signals because the hydrogen atoms in these 2 environments are less dieshielded as they are further away from the electronegative amine groups

30) Step 1: is not necessary for the student to conduct because none of the cations or anions possibly present will form a precipitate with the NaCl reagent being added. Therefore this step does not help in identifying the presence of the possible ions.
(I would include some equations here to show your understanding)

Step 2: is a valid method for testing for the presence of the hydroxide ion. The silver nitrate will react with the OH- ion forming an AgOH precipitate (give the equation for this). This precipitate can be dissolved by adding an acid which will cause the solubility of the precipitate to increase. This test is suitable to use because the acetate forms a different colour precipitate which can be visually distinguished

Step 3: invalid to use this as a test alone. Sodium sulfate forms a precipitate with both the barium and calcium ions (give the equations). Both of these precipitates are white and therefore the identity of the precipitate being due to calcium or barium cannot be confirmed with the method used by the student. Instead a confirmatory test such as adding NaF or doing a flame test where a red flame will be found for calcium and green flame for barium is required.

Also the method does not include a way to test for the presence of magnesium ions such as filtering the solution from step 3 and then adding sodium hydroxide and if a precipitate forms magnesium ions are present.

Evaluation: valid for step 2, but the rest of the method can not accurately determine which of the possible ions are present

31)
K = [NH3]^2 / [N2] [H2]^3

Find the concentrations first because it's a 10.0 L vessel:
[N2] = 4.50 / 10 = 0.45 mol/L
[H2] = 1.00 / 10 = 0.10 mol/L
[NH3] = 5.80 / 10 = 0.58 mol/L

these are the initial concentrations (bar the nitrogen where we are adding more moles at the beginning)
We want to increase [NH3] by 0.050 moles i.e. 0.050/10 = 0.005 mol/L
N2H2NH3
Ratio132
initial conc0.45+x0.100.58
change conc-0.0025-0.0075+0.005
equil conc0.4475+x0.09250.585
Sub into K exp:




748 = (0.585)^2 / (0.4475+x) (0.0925)^3
1.729... = 1/(0.4475+x)
1/1.729... = 0.4475 + x
0.578... = 0.4475 + x
x = 0.1305... mol/L

To find moles use n = cv
n = 0.1305 x 10 = 1.3 mol (2sf)

32)
The first two enthalpy of neutralisation values are for reactions between strong acids and strong bases. These types of acid/base ionise completely in solution when added to water:
HCl(aq) + H2O(l) --> Cl-(aq) + H3O+(aq)
NaOH(aq) --> Na+(aq) + OH-(aq)
HNO3(aq) + H2O(l) --> NO3-(aq) + H3O+(aq)
KOH(aq) --> K+(aq) + OH-(aq)

Therefore as they are dissociated fully the Cl-, Na+ are spectator ions in the first equation and the NO3- and K+ are spectator ions in the second equation. Spectator ions are not involved in the bond breaking and forming processes in a chemical reaction and so therefore do not affect the enthalpy value.

Both these reactions have the same net ionic equation: H+(aq) + OH-(aq) --> H2O(l), and therefore the enthalpy value that is obtained is the same for both of them.

In the reaction 3, the HCN is a weak acid which only partially ionises: HCN(aq) + H2O(l) < -- > CN-(aq) + H3O+(aq). As bond breaking is an endothermic process some of the energy involved in this chemical reaction will be consumed in the endothermic bond breaking process and therefore the overall deltaH value will be less exothermic.
 
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jazz519

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33)
a) deltaG = deltaH - TdeltaS
deltaH = -95 kJ/mol at 300 K
TdeltaS = -80 kJ/mol at 300 K
deltaG = -95 - (-80)
deltag = -15 kJ/mol

b) The deltaH for this reaction is negative and so therefore this reaction is exothermic (releases heat to the surroundings). The TdeltaS term is also less negative and therefore this means in the deltaG equation, the magnitude of the deltaH and TdeltaS will dictate where the reaction has a deltaG < 0 (i.e. spontaneous) and where the deltaG > 0 (i.e. non-spontaneous).

At T1: the deltaH term is more negative than the TdeltaS and so therefore DeltaG < 0. This means the reaction will be spontaneous at this temperature.

At T2: the deltaH term is equal to the TdeltaS term and so therefore DeltaG = 0. This means the system will be at equilibrium at this temperature.

At T3: the deltaH term is less negative than the TdeltaS term and so DeltaG > 0. This means the system will be non-spontaneous at this temperature.

34)
HCl(aq) + H2O(l) --> Cl-(aq) + H3O+(aq)
HCl is an acid which dissociates in water to form the hydronium ion.

Water: At t0 the pH of water is = 7 as this substance is neutral due to the [H+] = 10^-7 mol/L. As the HCl is bubbled through it begins dissolving like the equation above releasing hydronium ions. The pH lowers as it is inversely proportional to the [H3O+]. This decrease is steep for the time between t0 and t1 as initially a big change in [H3O+] is occurring. Between t1 and t2, the decrease becomes more gradual. This is because pH is calcualted on a log10 scale where a 10x change in [H3O+] translates to a 1 pH unit change. In this case as more [H3O+] is added the x10 factor grows larger for the next 1 pH unit decrease as shown in the graph.

X and Y: the pH decreases gradually due to the addition of HCl. This indicates these two solutions are buffers (mixture of a weak acid/base and its conjugate in equimolar concentrations. They resist changes in pH when acids and bases are added). HA(aq) + H2O(l) < -- > A-(aq) + H3O+(aq). These buffers are likely acidic buffers made from a weak acid because pH starts at 5 which is a weakly acidic solution. When the [H3O+] increases due to the addition of HCl, the equilibrium is disturbed and by LCP it shifts to the left to minimise the disturbance. This reduces [H3O+] and therefore the pH change is minimised as shown in the t0 to t1 time. At t2 the buffers begin to fail as they have reached their buffer capacity. The sudden drop in pH of X indicates that it has a lower buffer capacity in comparison to Y. This could be due to a higher concentration of the weak acid and its conjugate (salt).

35)
v(Na2S2O3) avg = (28.7 + 28.4 + 28.6) / 3 = 28.5666... mL
The Time 1 is an outlier and so isn't included in the average

n(Na2S2O3) = cv = (0.900) (28.5666.../1000)
n(Na2S2O3) = 0.02571 moles

n(S2SO3 2-) = 0.02571 moles (1 S2O3 2- in the Na2S2O3)

n(I2) = 1/2 n(S2O3 2-)
n(I2) = 1/2 (0.02571) = 0.012855 moles

this moles is the same as the one in equation 2
Compare moles of I2 to Cr2O7 2-
n(Cr2O7 2-) = 1/3 x n(I2)
n(Cr2O7 2-) excess = 1/3 (0.012855) = 0.004285 moles

n(Cr2O7 2-) initial = cv = (0.500)(20/1000)
n(Cr2O7 2-) initial = 0.01 moles

n(Cr2O7 2-) reacted with ethanol = 0.01 - 0.004285
n(Cr2O7 2-) reacted with ethanol = 0.005715 moles

n(C2H5OH) = 3/2 n(Cr2O7 2-)
n(C2H5OH) = 3/2 (0.005715)
n(C2H5OH) = 0.0085725 moles
m(C2H5OH) = nMM = (0.0085725)(46.068)
m(C2H5OH) = 0.3949... g

This however is just the ethanol mass in the 25 mL aliquot. The 1.00 L solution is 40x as much ethanol mass

m(C2H5OH) in original = 0.3949... x 40 = 15.796... g

D = m/v
0.789 = 15.796... / v
v = 20.021... mL

%v/v = v(ethanol) / v(sample) x 100
%v/v = 20.021... / 25.0 x 100
%v/v = 80.1% (3sf)

Therefore doesn't meet the manufacturer's requirement as it is lower than 85%


36) lucky last question :)
Ka1 = [HSO3-] [H3O+] / [H2SO3]

Ka2 = [SO3 2-] [H3O+] / [HSO3-]

Keq = [SO3 2-] [H3O+]^2 / [H2SO3]

Keq can be derived by doing Ka1 x Ka2

Therefore if we find the individual Ka1 and Ka2 we can get the Keq

Ka1 = 10^(-1.82)
Ka2 = 10^(-7.17)

therefore Keq = 10^(-1.82) x 10^(-7.17)
Keq = 1.02 x 10^-9
 
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CM_Tutor

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Do you think q21(b) is fair, in that oxidation of alkenes is not covered in the syllabus... and yet, the answer can also be deduce by a process of elimination?
 

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Do you think q21(b) is fair, in that oxidation of alkenes is not covered in the syllabus... and yet, the answer can also be deduce by a process of elimination?
Yeah that was a bit odd they put that one in but I guess it's fair then since you can do it by elimination and they didn't ask for a reason. I think it would be problematic though if they asked to explain it and then I doubt 99% of students would know. Maybe a select few who did something like Olympiad would know
 

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Yeah that was a bit odd they put that one in but I guess it's fair then since you can do it by elimination and they didn't ask for a reason. I think it would be problematic though if they asked to explain it and then I doubt 99% of students would know. Maybe a select few who did something like Olympiad would know
I have seen a couple of trial papers with MCQ that couldn't be answered without knowing that alkenes can be oxidised by permanganate (not requiring knowledge of the products, however)... it just makes me wonder. They got creative with equilibrium constants, including combining them (which I have been anticipating will happen at some point), I guess 2022 papers will have a lot more exploration of those ideas.
 

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Solutions finished :D (bar the 33b which I will do). Hopefully these were useful to you after your exam. Keep in mind though the level of answer here may be higher than what is needed for a full mark since I am writing this from a perspective of someone who has tutored for 5 years and completed a degree at university in Chemistry.

Try not to dwell too much if you lost marks and focus on other exams and if you are finished your exams, congratulations! on completing the HSC
 

jazz519

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I have seen a couple of trial papers with MCQ that couldn't be answered without knowing that alkenes can be oxidised by permanganate (not requiring knowledge of the products, however)... it just makes me wonder. They got creative with equilibrium constants, including combining them (which I have been anticipating will happen at some point), I guess 2022 papers will have a lot more exploration of those ideas.
Yeah that last question honestly is actually very easy. But students who over analyse the question too much probably would've got stumped by it before even attempting it because it looks very complicated. Those students who just stuck to the basics such as soon as they see a K (equilibrium constant) type of question they write down the Ka in variable form and the Keq one they could have easily spotted the answer.

This is a handy tip for future HSC students reading this forum thread. If you are ever stuck on a calculation question don't stare at it and try to do all of it in one go in your head. No matter how complicated the calculation it will always come back to some of the basics you have learnt. So start by writing down things such as:
- formulas that might be useful
- assigning values in the question to variables
- writing K constant expression out like you would in a simple calculate the K question
- writing chemical equation out

These things I teach to my tutoring students and is the only way you can master higher level questions if you haven't been exposed to them like a tutor or teacher who has taught the syllabus for multiple years and so therefore knows the method before doing the question since they may have seen similar questions in the past. Using these type of troubleshoot methods in the exam will help you to gain valuable marks. Maybe you can't get to the final answer, that's okay, but at least you will have gained some marks for the working of the initial steps and if you're lucky something might click in your brain after you write out the basic steps and the pathway to the answer becomes more clearer
 

CM_Tutor

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These things I teach to my tutoring students and is the only way you can master higher level questions if you haven't been exposed to them like a tutor or teacher who has taught the syllabus for multiple years and so therefore knows the method before doing the question since they may have seen similar questions in the past. Using these type of troubleshoot methods in the exam will help you to gain valuable marks. Maybe you can't get to the final answer, that's okay, but at least you will have gained some marks for the working of the initial steps and if you're lucky something might click in your brain after you write out the basic steps and the pathway to the answer becomes more clearer
Yes, it was gratifying to me that some students who took the 2021 BoS paper picked up ideas it explored that were applicable in this HSC paper. Exposure to enough challenging questions definitely helps students to cope with unfamiliar challenges when they appear on assessments. I used pentanedioic acid as an example for 1H NMR with one of my students recently, and I know that he found that helped him handle the 1,5-diaminopentane question. Transfer of learning between different but related questions can be very helpful.
 

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Yes, it was gratifying to me that some students who took the 2021 BoS paper picked up ideas it explored that were applicable in this HSC paper. Exposure to enough challenging questions definitely helps students to cope with unfamiliar challenges when they appear on assessments. I used pentanedioic acid as an example for 1H NMR with one of my students recently, and I know that he found that helped him handle the 1,5-diaminopentane question. Transfer of learning between different but related questions can be very helpful.
Yeah good to see that the exam helped them
 

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Guys, I've got a query about the equilibrium constant (approx 1 x 10^-9) in the last question. I agree that it was very easy to algebraically come up with that solution, but I'm trying to understand the utility (usefulness) of that overall equilibrium equation. I'm probably doing something dumb here, but to me it appears to predict neither the [H+] nor the [SO3^(2-)] concentrations correctly.

My understanding of polyprotic acids was that, if one Ka was much larger than the others then the pH could be estimated by considering only that one. In other cases it can be solved using simultaneous equations (like using "x" for the [HSO3-] in the first eqn, and "y" for the [SO3^2-] in the second equation, and (x+y) for the [H+] in both equations).

If for example I take the initial [H2SO3] as 1M, then when I solve the two equations using the Ka1>>Ka2 approximation I get the following:

[H+] = 0.11569
[SO3^2-] = 6.7608 x 10^-8

And if I get my computer to solve the two simultaneous equations exactly, then I get the same as the above to at least 6 significant digits.

However, if I use the overall equation with the calculated overall Keq of 1x10^-9 then I get vastly different results :

[H+] = 0.00126
[SO3^2-] = 6.30 x 10^-4

Can someone tell me what's going on here. What am I doing wrong?
 
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uart

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I think that the problem with the equation given in Q37 is that the SO3^(2-) just wont exist in aqueous equilibrium with H+, not without also forming HSO3-.

I'm starting to think that maybe the correct answer to the last question is that an equilibrium cannot be established with only the species listed in that equation?

Not a very satisfying answer I know, but I just can't see how that equilibrium equation can work. It certainly seems to give the wrong values for the concentrations.
 
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janeway

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33)
a) deltaG = deltaH - TdeltaS
deltaH = -95 kJ/mol at 300 K
TdeltaS = -80 kJ/mol at 300 K
deltaG = -95 - (-80)
deltag = -15 kJ/mol

b) The deltaH for this reaction is negative and so therefore this reaction is exothermic (releases heat to the surroundings). The TdeltaS term is also less negative and therefore this means in the deltaG equation, the magnitude of the deltaH and TdeltaS will dictate where the reaction has a deltaG < 0 (i.e. spontaneous) and where the deltaG > 0 (i.e. non-spontaneous).

At T1: the deltaH term is more negative than the TdeltaS and so therefore DeltaG < 0. This means the reaction will be spontaneous at this temperature.

At T2: the deltaH term is equal to the TdeltaS term and so therefore DeltaG = 0. This means the system will be at equilibrium at this temperature.

At T3: the deltaH term is less negative than the TdeltaS term and so DeltaG > 0. This means the system will be non-spontaneous at this temperature.

34)
HCl(aq) + H2O(l) --> Cl-(aq) + H3O+(aq)
HCl is an acid which dissociates in water to form the hydronium ion.

Water: At t0 the pH of water is = 7 as this substance is neutral due to the [H+] = 10^-7 mol/L. As the HCl is bubbled through it begins dissolving like the equation above releasing hydronium ions. The pH lowers as it is inversely proportional to the [H3O+]. This decrease is steep for the time between t0 and t1 as initially a big change in [H3O+] is occurring. Between t1 and t2, the decrease becomes more gradual. This is because pH is calcualted on a log10 scale where a 10x change in [H3O+] translates to a 1 pH unit change. In this case as more [H3O+] is added the x10 factor grows larger for the next 1 pH unit decrease as shown in the graph.

X and Y: the pH decreases gradually due to the addition of HCl. This indicates these two solutions are buffers (mixture of a weak acid/base and its conjugate in equimolar concentrations. They resist changes in pH when acids and bases are added). HA(aq) + H2O(l) < -- > A-(aq) + H3O+(aq). These buffers are likely acidic buffers made from a weak acid because pH starts at 5 which is a weakly acidic solution. When the [H3O+] increases due to the addition of HCl, the equilibrium is disturbed and by LCP it shifts to the left to minimise the disturbance. This reduces [H3O+] and therefore the pH change is minimised as shown in the t0 to t1 time. At t2 the buffers begin to fail as they have reached their buffer capacity. The sudden drop in pH of X indicates that it has a lower buffer capacity in comparison to Y. This could be due to a higher concentration of the weak acid and its conjugate (salt).

35)
v(Na2S2O3) avg = (28.7 + 28.4 + 28.6) / 3 = 28.5666... mL
The Time 1 is an outlier and so isn't included in the average

n(Na2S2O3) = cv = (0.900) (28.5666.../1000)
n(Na2S2O3) = 0.02571 moles

n(S2SO3 2-) = 0.02571 moles (1 S2O3 2- in the Na2S2O3)

n(I2) = 1/2 n(S2O3 2-)
n(I2) = 1/2 (0.02571) = 0.012855 moles

this moles is the same as the one in equation 2
Compare moles of I2 to Cr2O7 2-
n(Cr2O7 2-) = 1/3 x n(I2)
n(Cr2O7 2-) excess = 1/3 (0.012855) = 0.004285 moles

n(Cr2O7 2-) initial = cv = (0.500)(20/1000)
n(Cr2O7 2-) initial = 0.01 moles

n(Cr2O7 2-) reacted with ethanol = 0.01 - 0.004285
n(Cr2O7 2-) reacted with ethanol = 0.005715 moles

n(C2H5OH) = 3/2 n(Cr2O7 2-)
n(C2H5OH) = 3/2 (0.005715)
n(C2H5OH) = 0.0085725 moles
m(C2H5OH) = nMM = (0.0085725)(46.068)
m(C2H5OH) = 0.3949... g

This however is just the ethanol mass in the 25 mL aliquot. The 1.00 L solution is 40x as much ethanol mass

m(C2H5OH) in original = 0.3949... x 40 = 15.796... g

D = m/v
0.789 = 15.796... / v
v = 20.021... mL

%v/v = v(ethanol) / v(sample) x 100
%v/v = 20.021... / 25.0 x 100
%v/v = 80.1% (3sf)

Therefore doesn't meet the manufacturer's requirement as it is lower than 85%


36) lucky last question :)
Ka1 = [HSO3-] [H3O+] / [H2SO3]

Ka2 = [SO3 2-] [H3O+] / [HSO3-]

Keq = [SO3 2-] [H3O+]^2 / [H2SO3]

Keq can be derived by doing Ka1 x Ka2

Therefore if we find the individual Ka1 and Ka2 we can get the Keq

Ka1 = 10^(-1.82)
Ka2 = 10^(-7.17)

therefore Keq = 10^(-1.82) x 10^(-7.17)
Keq = 1.02 x 10^-9
Hi 🙂 Thank you for taking the time to do these solutions
have you assumed that the 2 [H3O+] are the same? Does the answer need to include a comment about limiting assumption for full marks?
Thanks for your thoughts
 

jazz519

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Hi 🙂 Thank you for taking the time to do these solutions
have you assumed that the 2 [H3O+] are the same? Does the answer need to include a comment about limiting assumption for full marks?
Thanks for your thoughts
Which question are you referring to?
 

jazz519

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There's no limiting assumption like K is small. The solution was reached by multiplying Ka1 and Ka2 they gave, so none of those were individually calculated
 

Velocifire

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the hard adv maths test paid off for question 36 with the logs
 

janeway

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There's no limiting assumption like K is small. The solution was reached by multiplying Ka1 and Ka2 they gave, so none of those were individually calculated
Hi .. hard to see where 5 marks can be allocated
 

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Hi .. hard to see where 5 marks can be allocated
Probably 2 in the writing the K constants of the two reactions and the final one

One in the final answer being correct

Two in the working and rearranging

Probably should be a 4 mark question but you can't really give more working than that there, if the question doesn't require it
 

janeway

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Probably 2 in the writing the K constants of the two reactions and the final one

One in the final answer being correct

Two in the working and rearranging

Probably should be a 4 mark question but you can't really give more working than that there, if the question doesn't require it
Hi- thanks - I agree that it is worth less than 5 marks - think 3 marks max ... so curious as to the way the marking centre will appropach this. :)
 

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