HSC Chemistry 2021

(a) [2 marks]. Organic liquids such as these are highly flammable. Address this by keeping them away from naked flames and other ignition sources, and heating only with a water bath or heating mantle in a well ventilated fume cupboard.

(b) [2 marks].

Flasks 2 and 3 react with the oxidant (the alkene is oxidised across its double bond; the primary alcohol oxidises to an acid). Of these, only propan-1-ol is miscible with water, so flask 3 is propan-1-ol and flask 2 is hex-1-ene. Of the non-reacting pair, propanoic acid is water-miscible (flask 1) and hexane is not (flask 4).

(c) [2 marks]. Liquid: propanoic acid. Add a small amount of solid sodium hydrogen carbonate to the liquid. Effervescence (bubbles of carbon dioxide gas) confirms a carboxylic acid, so the liquid is propanoic acid.

Marker's note. Link the named risk (flammable) to a precaution you would actually use in a school laboratory, not a vague mention of PPE. For (b), reason from the data to each compound. For (c), pair a specific test with its expected positive observation.

[3 marks]. Two ways, each justified by Le Chatelier's principle:

1. Reduce the hydrogen ion concentration (for example by adding a base that consumes H+). The system partly counteracts this by shifting left to replace the H+ removed, which increases the concentration of yellow chromate ions.

2. Increase the temperature. The forward reaction is exothermic, so heating adds energy that the system relieves by shifting in the endothermic (reverse) direction, again favouring the yellow chromate ions.

Either change moves the solution from orange towards yellow.

Marker's note. Pick changes that genuinely shift the position left and explain how the system counteracts each disturbance, rather than just citing Le Chatelier. Remember that adding water dilutes both sides and that pressure does not affect this aqueous system.

(a) [2 marks].

(b) [2 marks]. The salt, potassium methanoate, is basic. The methanoate ion (HCOO-) is the conjugate base of the weak acid methanoic acid, so it hydrolyses in water, accepting a proton and releasing hydroxide ions:

This raises the hydroxide concentration above neutral, giving a pH above 7. The potassium ion comes from a strong base and does not affect the pH.

Marker's note. Use the correct organic prefix (methan- for the one-carbon acid). Tie the basic pH to the methanoate ion being the conjugate base of a weak acid, ideally with the hydrolysis equation.

[4 marks]. For an alkane $C_nH_{2n+2}$, set $12.011n + 1.008(2n + 2) = 72.146$, which gives $n = 5$. The molecule is pentane, and its chain isomers are 2-methylbutane and 2,2-dimethylpropane.

• Pentane: the straight five-carbon chain CH3-CH2-CH2-CH2-CH3.
• 2-methylbutane: a four-carbon chain with a methyl branch on carbon 2.
• 2,2-dimethylpropane: a central carbon bonded to four methyl groups.

Each structural formula should be drawn in expanded form showing every carbon and hydrogen atom.

Marker's note. Confirm pentane from the molar mass using CnH2n+2, then find both branched isomers. Draw expanded structures with all hydrogen atoms shown, and follow IUPAC naming exactly.

[4 marks]. Fermentation of glucose:

The reaction has finished when the volume stops rising, so use the final volume 1006 mL = 1.006 L of CO2. At 25 degrees C and 100 kPa the molar volume is 24.79 L/mol:

The equation gives a 1:1 ratio of ethanol to carbon dioxide, so $n(\text{C}_2\text{H}_5\text{OH}) = 0.04058$ mol.

Marker's note. Use the final (plateau) volume of CO2, the molar volume at 25 degrees C and the correct 1:1 mole ratio. The yeast nutrient is not part of the equation. Keep full precision until the final answer and quote units.

(a) [4 marks]. Working from the final esterification backwards, the alcohol and acid that form methyl 2-methylpropanoate are methanol and 2-methylpropanoic acid.

• A (hydration of 2-methylprop-1-ene): 2-methylpropan-2-ol, the tertiary alcohol (OH on the central carbon), by Markovnikov addition.
• B (substitution with concentrated HCl and ZnCl2): 2-chloro-2-methylpropane, the chlorine replacing the hydroxyl on the tertiary carbon.
• C and D leading to the ester: the branch ends in 2-methylpropanoic acid, (CH3)2CHCOOH, the carboxylic acid that reacts with methanol. The primary alcohol 2-methylpropan-1-ol is the species that oxidises to this acid.

(b) [2 marks]. Two reasons for refluxing:

1. It heats the mixture continuously to speed up the slow esterification without boiling the volatile reactants away.
2. The vertical condenser returns evaporated methanol, acid and ester to the flask, so no volatile material is lost and the reaction can proceed to a higher yield.

Marker's note. Identify the esterification at the end and work backwards through the substitution and oxidation steps, drawing clear structures. For reflux, give reasons specific to the technique (continuous heating and returning volatiles), not generic safety reasons.

(a) [4 marks]. Dissolution: $\text{Li}_3\text{PO}_4\text{(s)} \rightleftharpoons 3\text{Li}^+\text{(aq)} + \text{PO}_4^{3-}\text{(aq)}$, so the solubility product is

Precipitation begins between sample 3 (no precipitate) and sample 4 (precipitate), so Ksp lies between the ionic products of these two samples.

So Ksp lies between $3.4 \times 10^{-5}$ and $3.4 \times 10^{-4}$ (to two significant figures).

(b) [2 marks]. Prepare more phosphate solutions with concentrations between 0.010 and 0.10 mol/L (for example 0.02, 0.04, 0.06, 0.08). The precipitate would first appear over a much smaller concentration step, narrowing the gap between the two bracketing ionic products and bringing the estimated Ksp closer to the true value.

Marker's note. Cube the lithium term in the Ksp expression and identify that precipitation starts between samples 3 and 4. For (b), link finer concentration increments directly to a narrower, more accurate Ksp range.

[4 marks]. Let the alkali metal be X. The precipitate is copper(II) hydroxide:

Moles of precipitate, with MM of Cu(OH)2 = 97.57 g/mol:

Each formula unit of precipitate uses two OH-, so $n(\text{XOH}) = 2 \times 0.04725 = 0.09450$ mol. The molar mass of XOH is

Subtracting OH (17.01 g/mol) leaves the metal at 56.08 - 17.01 = 39.08 g/mol, which matches potassium. The hydroxide is potassium hydroxide (KOH).

Marker's note. Write a balanced precipitation equation with correct formulae and states, use the 2:1 mole ratio of hydroxide to copper hydroxide, then locate the metal in the periodic table from its molar mass. Show every step.

[7 marks]

Pentane-1,5-diamine is H2N-CH2-CH2-CH2-CH2-CH2-NH2, a symmetric molecule with an amino group at each end.

Infrared spectrum

The broad absorption near 3300 to 3400 cm^-1 is the N-H stretch of the primary amine groups, confirming the two amino groups in the structure.

Mass spectrum

The highlighted fragment at m/z = 30 corresponds to CH2NH2+ (12 + 2 + 14 + 2 = 30), formed by cleavage next to a nitrogen. This is the characteristic terminal -CH2NH2 group at each end of the molecule.

Carbon-13 NMR

Although there are five carbons, only three signals appear, because the molecule is symmetric: carbons 1 and 5 are in identical environments, as are carbons 2 and 4, while carbon 3 is unique. The signals near 24 and 33 ppm are the -CH2- carbons, and the signal near 42 ppm is the carbon bonded to the deshielding nitrogen.

Proton NMR

The triplet (4H) is the two equivalent -CH2-NH2 groups, each next to one CH2. The quintet (4H) is the carbon-2 and carbon-4 protons, each next to four neighbouring hydrogens. The overlapping 6H multiplet arises because protons in two different but similar environments have nearly the same chemical shift, so their signals overlap.

Marker's note. Relate each highlighted feature to a specific part of the molecule across all four spectra. Distinguish the base fragment at m/z = 30 from the molecular ion at 102, attribute the larger carbon-13 shift to nitrogen deshielding, and link the symmetry to the reduced number of signals.

[5 marks]

The sodium chloride test is uninformative. Sodium does not precipitate either possible anion, and chloride does not precipitate any of barium, calcium or magnesium, so no reaction was always going to occur. This test contributes nothing.

The silver nitrate test identifies the anion well

Silver acetate is soluble, but silver hydroxide is an insoluble brown solid that dissolves in acid. The brown precipitate that redissolves therefore shows the anion is hydroxide, not acetate. Usefully, this also rules out magnesium, since magnesium hydroxide is insoluble and the original solution was clear.

The sodium sulfate test cannot distinguish the cation

Both barium sulfate and calcium sulfate are insoluble, so a white precipitate appears for either, and the test cannot tell barium from calcium.

Judgement

the procedure correctly identifies the anion as hydroxide but fails to identify the cation, so overall it is an inadequate method. A flame test (a green flame for barium, brick-red for calcium) would be needed to complete the identification.

Marker's note. Use solubility rules to interpret each test, outline both useful and unhelpful results, then make an informed overall judgement. State clearly that the sulfate test cannot separate barium from calcium and that a flame test is required.

[4 marks]. Increasing ammonia by 0.050 mol means the forward reaction advances by 0.025 mol of reaction extent, consuming 0.025 mol N2 and 0.075 mol H2. Let $x$ be the moles of N2 added. New equilibrium amounts and concentrations in the 10.0 L vessel:

The equilibrium expression must still equal 748:

Rearranging for the nitrogen concentration:

So $4.475 + x = 5.775$, giving $x = 1.3$ mol. About 1.3 moles of nitrogen must be added.

Marker's note. Use the mole ratio to find how much N2 and H2 change when ammonia rises by 0.050 mol, build the new concentrations, then substitute into the correct Keq expression (using concentrations, not moles, with the right exponents) and solve for the added nitrogen.

[4 marks]. HCl and HNO3 are strong acids, so they are fully ionised in solution. Their neutralisation with KOH is, in every case, the same net ionic reaction:

Because the actual chemical change is identical, the energy released is the same (-57.6 kJ/mol).

HCN is a weak acid and is only partially ionised, existing in equilibrium: $\text{HCN(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{CN}^-\text{(aq)}$. As the hydrogen ions are removed by the base, this ionisation equilibrium shifts right and more HCN ionises. Breaking the H-CN bond to ionise the remaining acid is endothermic, absorbing some of the energy released by forming water. The net result is a smaller, but still exothermic, enthalpy of -12.0 kJ/mol.

Marker's note. Discuss the chemistry, not just acid strength: identify the common net ionic equation for the strong acids, then explain that ionising the weak acid is an endothermic step that reduces the overall energy released.

(a) [2 marks]. Read the two values off the graph at 300 K: delta H is about -93 kJ/mol and T.delta S is about -78 kJ/mol. Using the data-sheet relationship:

(Answers within the marking centre's accepted reading range are acceptable.)

(b) [4 marks]. Across the whole range delta H is negative (exothermic) and, since T.delta S is negative while T is positive, delta S is also negative. As temperature rises, the -T.delta S term grows more positive, so delta G increases.

• At T1, T.delta S is smaller in magnitude than delta H, so delta G is negative: the reaction is spontaneous.
• At T2, the two lines cross, so delta H = T.delta S and delta G = 0: the system is at equilibrium.
• At T3, T.delta S is larger in magnitude than delta H, so delta G is positive: the reaction is non-spontaneous.

Marker's note. Read delta H and T.delta S accurately with a ruler, apply delta G = delta H - T.delta S, and link delta G relative to zero at each temperature to spontaneity, equilibrium (delta G = 0) and non-spontaneity using correct terms.

[5 marks]. At t0. Pure water has pH 7. Solutions X and Y start near pH 5, so both are mildly acidic; together with their later behaviour this identifies them as buffers made of a weak acid and its conjugate base.

At t1. As HCl is bubbled in, it dissolves and fully ionises, raising the hydronium concentration. In water there is nothing to absorb the added acid, so the pH falls sharply. In X and Y the conjugate base mops up the added protons, so the pH falls only slightly:

At t2. Eventually the conjugate base in the buffer is used up, so further HCl is no longer neutralised and the pH of X and Y begins to drop more steeply. The pH of X falls before Y because X is the less concentrated buffer, so its conjugate base is exhausted sooner.

Marker's note. Distinguish the complete ionisation of strong HCl (single arrow) from the buffer equilibrium, include a balanced equation, and explain the order of the pH drops by relating buffer capacity to the concentration of conjugate base.

[7 marks]. Discard titre 1 (29.9 mL) as an outlier and average the concordant titres:

Thiosulfate to iodine

$n(\text{S}_2\text{O}_3^{2-}) = 0.900 \times 0.02857 = 0.02571$ mol. The ratio I2 : S2O3 is 1 : 2, so $n(\text{I}_2) = 0.01285$ mol.

Iodine to excess dichromate

The ratio Cr2O7 : I2 is 1 : 3, so the excess dichromate is $n = 0.01285/3 = 0.004285$ mol.

Dichromate that reacted with ethanol

Initial dichromate is $0.500 \times 0.0200 = 0.0100$ mol, so the amount used by ethanol is $0.0100 - 0.004285 = 0.005715$ mol.

Ethanol

The ratio Cr2O7 : ethanol is 2 : 3, so $n(\text{ethanol}) = \frac{3}{2}\times 0.005715 = 0.008573$ mol in the 25.0 mL aliquot.

Scale up by the 1.00 L to 25.0 mL dilution to reach the undiluted 25.0 mL sample:

The sample is about 80% v/v ethanol, which is below 85%, so the sample does not meet the manufacturer's requirement.

Marker's note. Reject the outlier, then carry the mole ratios in sequence (thiosulfate to iodine to excess dichromate, then dichromate reacted with ethanol). Do not round until the end, convert mass to volume with the density, and give a v/v judgement that matches the calculation.

[5 marks]. Convert each pKa to a Ka:

The target reaction is the sum of the two ionisation steps, so its equilibrium constant is the product of their constants. Writing out the expressions (water is omitted as it is the aqueous solvent):

Marker's note. Convert the pKa values to Ka, recognise that adding the two equilibria multiplies their constants, exclude water from the expressions, and show every step rather than introducing numbers without working.