HSC Chemistry 2019

(a) [2 marks]. A valid structural isomer is butan-1-ol: a straight four-carbon chain, CH3CH2CH2CH2OH, with the OH on carbon 1. It has the same molecular formula (C4H10O) as 2-methylpropan-2-ol but a different arrangement of atoms.

(b) [2 marks]. Both compounds have the same molecular formula (C3H6O) but contain different functional groups. Isomer A (propanone) has a ketone group (C=O on an interior carbon), while Isomer B (propanal) has an aldehyde group (C=O on a terminal carbon). Isomers that differ in functional group are functional group isomers.

(c) [3 marks]. Use the Tollens' test (ammoniacal silver nitrate). The aldehyde (Isomer B) is readily oxidised to a carboxylic acid and reduces the silver(I) ions to metallic silver, depositing a silver mirror on the inside of the tube. The ketone (Isomer A) is not oxidised, so it gives no reaction. The presence or absence of the silver mirror distinguishes the two. (Acidified dichromate, which turns from orange to green only with the aldehyde, is equally acceptable.)

Marker's note. Draw and name a genuine structural isomer (not just restate the given structure), and be clear it is a functional group isomer, not merely a structural one. The test must be a chemical test, not a physical one, with a named reagent, the expected observation, and the chemistry behind it.

[4 marks]. The indicator colour did not change because the pH barely changed: the buffer resists pH change. The buffer contains the weak acid and its conjugate base in the equilibrium

When acid (extra H3O+) is added, the added H3O+ reacts with the acetate ions and the position of equilibrium shifts to the left, consuming most of the added H3O+. When base (OH-) is added, the OH- removes H3O+ to form water, and the equilibrium shifts to the right as acetic acid ionises to replace it. In both cases the hydronium ion concentration, and therefore the pH, stays nearly constant, so the indicator colour does not change.

Marker's note. Explain the reason for the shift, not just its direction, and use the proper equilibrium arrow with the hydronium ion. Strong answers state that a buffer is a weak acid on one side and its conjugate base on the other, and use the specific effect of added OH- and H3O+ rather than a vague appeal to Le Chatelier's principle.

(a) [4 marks]. Heat absorbed by the water:

Moles of ethanol (C2H5OH, M = 46.08 g/mol):

Molar enthalpy of combustion:

(b) [2 marks]. The experimental value is too low (less exothermic) mainly because heat is lost to the surrounding air rather than going into the water. A change that improves accuracy is to move the spirit burner closer to the beaker (or add a draught shield), so more of the heat released reaches the water, giving a value closer to the accepted enthalpy of combustion.

Marker's note. Use the mass of the water (105 g), not the ethanol, in mc(delta)T, and use the correct molar mass of ethanol from the periodic table. Quote the final answer to the correct significant figures with units. In (b) the change must be linked to the reason the value was low (heat loss).

(a) [3 marks]. Before equivalence the curve falls: each added OH- reacts with the highly mobile H3O+ ions of the acid to form water, removing the most conductive ions and replacing them with the less mobile Ba2+ and Cl- ions, so conductivity drops. At the equivalence point the conductivity is at its minimum (but not zero), because the solution then contains only Ba2+ and Cl- ions, which conduct, just far less than H3O+ or OH- do. Past equivalence the curve rises again, because further Ba(OH)2 adds excess, highly mobile OH- ions back into the solution.

(b) [4 marks]. Balanced equation:

Moles of HCl:

From the 2 : 1 ratio, $n(\text{Ba(OH)}_2) = \tfrac{1}{2} n(\text{HCl}) = 1.3125 \times 10^{-4}\ \text{mol}.$

Marker's note. Explain why the curve has its shape in terms of the specific ions (H3O+ removed, then Ba2+ and Cl- at equivalence, then excess OH-), and note conductivity is non-zero at equivalence. In the calculation use volumes in litres, apply the 2 : 1 mole ratio from a balanced equation, and avoid c1v1 = c2v2 (it is for dilutions, not this titration).

(a) [2 marks]. Immediately after T, CO drops suddenly (the species removed) while H2O, CO2 and H2 are unchanged at that instant. After T the system re-establishes equilibrium: CO2 and H2 (products) fall, while H2O (the other reactant) rises, and CO gradually rises back toward (but below) its original level. All four level off at new equilibrium concentrations. The product curves fall by the same number of moles per litre as each other (1 : 1 stoichiometry), and CO and H2O rise by the same amount.

(b) [3 marks]. Removing CO lowers its concentration, so there are fewer collisions between CO and H2O molecules and the rate of the forward reaction decreases. The reverse reaction rate is momentarily unchanged, so it is now greater than the forward rate; the reverse reaction dominates, consuming CO2 and H2 to re-form CO and H2O. As CO and H2O build up, the forward rate increases again and the product rates fall, until the forward and reverse rates become equal once more and a new equilibrium is established.

Marker's note. The removal of CO does not cause a sudden change in the other species (only a gradual one as the system re-equilibrates). Link the lowered CO concentration to fewer collisions and a lower forward rate, then compare forward and reverse rates. Le Chatelier's principle is a consequence of collision theory, so explain via collisions and rates.

(a) [5 marks]. The compound is propanoic acid, CH3CH2COOH.

• Reactivity: it fizzes with sodium carbonate, releasing CO2, which shows it is a carboxylic acid (an organic acid).
• Mass spectrum: the molecular ion (parent) peak at m/z = 74 gives a molar mass of 74 g/mol, matching C3H6O2 (propanoic acid).
• 13C NMR: three peaks show three carbon environments (CH3, CH2 and COOH), and the peak near 180 ppm confirms a carboxylic acid carbon (C=O).
• 1H NMR: three proton environments with splitting and integration consistent with a CH3 (triplet), a CH2 (quartet) and the acidic OH proton.

Together these fix the structure as CH3CH2COOH.

(b) [3 marks]. A single technique gives only partial information, so combining techniques builds a complete and reliable structure. For example, mass spectrometry gives the molar mass (from the molecular ion) and characteristic fragment masses, but not how the atoms are bonded. 1H NMR then reveals the number of distinct hydrogen environments, their relative numbers (integration) and neighbouring hydrogens (splitting), letting you build the carbon-hydrogen skeleton and distinguish isomers of that same molar mass. Using both together confirms a structure that either alone would leave ambiguous.

Marker's note. Use values read from the spectra and relate them to the data sheet (for example m/z = 74, the 180 ppm carboxyl carbon), and link the sodium carbonate reactivity to a carboxylic acid. For (b), state what key information each named technique provides and explain how the techniques work together, rather than describing them in isolation.

(a) [1 mark].

(b) [4 marks]. Set up an ICE table for the hydrolysis, with x the concentration of OH- formed:

Because Kb is very small, the ionisation is slight, so [OCl-] stays essentially 0.20 mol/L. Solving:

Marker's note. Use Kw from the data sheet and keep Ka, Kb and Kw straight. Justify the assumption that [OCl-] stays at 0.20 (the degree of ionisation is small), build a clear ICE table, and do not confuse pH with pOH at the end.

[5 marks]. The Bronsted-Lowry model defines an acid as a proton (H+) donor and a base as a proton acceptor, judged by proton transfer rather than by producing OH- in water.

Advantage. It classifies substances the older Arrhenius model cannot. For example:

Here ammonia accepts a proton (a base) and hydrogen chloride donates one (an acid), even though there is no water and ammonia produces no OH- ions, so Arrhenius theory fails but Bronsted-Lowry succeeds.

Limitation. The model still needs a proton to transfer, so it cannot classify acid-base reactions with no H+ transfer, such as between an acidic oxide and a basic oxide:

This is an acid-base reaction (a basic oxide neutralising an acidic oxide), yet no proton moves, so Bronsted-Lowry cannot describe it; the broader Lewis model is needed.

Judgement. The Bronsted-Lowry model is highly useful for the great majority of acid-base reactions, especially proton transfers in solution and in the gas phase, but it is limited because it cannot handle reactions without proton transfer.

Marker's note. Address the key word "assess": give a clear advantage and a clear limitation, each tied to a correctly balanced equation, then make an overall judgement. Avoid stating everything known about the models; link each point to its equation and use proper terminology.

(a) [2 marks]. Calcium hydroxide is slightly soluble, so adding it raises the hydroxide ion concentration in the stormwater. Copper(II) hydroxide and lead(II) hydroxide are very insoluble (very small Ksp values), so the added OH- precipitates the metal ions out of solution, removing them:

(The analogous equation for Pb(OH)2 is equally acceptable.)

(b) [2 marks]. Each metal absorbs light strongly only at wavelengths characteristic of that element (matching its electron transitions). By using a lamp tuned to copper's characteristic wavelength, the instrument measures only the copper absorbance, and switching to lead's wavelength measures only lead. So the two ions can be quantified independently in the same solution without interfering with each other.

(c) [7 marks]. Plot two calibration curves (absorbance against concentration in 10^-5 mol/L) for Cu2+ and Pb2+, with labelled axes, a key, a sensible scale, and a ruled line of best fit for each set of standards.

Read the sample concentrations off the curves by interpolation, then convert to mg/L (multiply by the molar mass in g/mol and by 1000):

Sample conversion for post-treatment copper:

Conclusion. The copper concentration falls to 0.20 mg/L, comfortably below the 1.0 mg/L limit, so treatment succeeds for copper. The lead concentration falls from 9.84 to 1.8 mg/L, but 1.8 mg/L still exceeds the limit. The treatment is therefore only partially effective: acceptable for copper, but further treatment is needed to bring lead below 1.0 mg/L.

Marker's note. Plot points accurately and draw ruled lines of best fit (not dot-to-dot), then interpolate the unknowns. Convert mol/L to mg/L carefully (watch the powers of ten and the decimal point), compare each metal to the 1.0 mg/L limit, and reach a judgement that copper passes while lead does not. Use specific AAS terminology and Ksp values rather than general statements.

[3 marks]. For both salts, dissolving is exothermic (negative DsolH) and moves toward greater order (negative DsolS), so the entropy term -TDsolS is positive and opposes dissolving in each case.

For magnesium chloride, DsolH is strongly negative (-160 kJ/mol), large enough to outweigh the unfavourable -TDsolS contribution (+34.2 kJ/mol), giving a negative DsolG (-125 kJ/mol). Its dissolving is enthalpy driven, so it dissolves spontaneously.

For magnesium fluoride, DsolH is only slightly negative (-7.81 kJ/mol), much smaller than the unfavourable -TDsolS contribution (+66.4 kJ/mol), so DsolG is positive (+58.6 kJ/mol) and it does not dissolve spontaneously.

So in both salts entropy opposes solubility; the difference is that the enthalpy release in magnesium chloride is large enough to overcome it, while in magnesium fluoride it is not.

Marker's note. Explain what the values mean in terms of energy released and order increasing, rather than just citing how large the numbers are. Identify which salt is soluble using DsolG, and make sure no part of the answer contradicts another part.

[4 marks]. Let x be the concentration of Hg2+ (and of CuCl4^2-) formed. The ICE table gives equilibrium concentrations 0.100 - x for each reactant and x for each product:

Because Keq is tiny, x is very small, so 0.100 - x is approximately 0.100. Taking the square root of both sides:

So $[\text{Hg}^{2+}] = 6.75 \times 10^{-7}$ mol/L.

Marker's note. Recognise the products start at zero, set up the ICE table, and either take the square root (the expression is a perfect square) or use the x-is-small approximation rather than wrestling with a full quadratic. Substitute carefully and watch for calculator errors.

[4 marks]. Three patterns appear.

1. Both series rise with chain length. Adding carbons increases the number of electrons and the surface area of each molecule, so the dispersion forces between molecules strengthen and more energy is needed to separate them, raising the boiling point.
2. The alcohol always boils higher than the matching thiol. The O-H group of an alcohol forms hydrogen bonds, which are stronger than the dispersion forces (and weak interactions) between thiols, whose S-H group does not hydrogen bond effectively. So at each chain length the alcohol needs more energy to boil.
3. The gap between the two series narrows as the chain lengthens. As the chains get longer, dispersion forces grow and become the dominant intermolecular force in both series, so the fixed contribution of hydrogen bonding in the alcohol makes up a smaller fraction of the total, shrinking the difference.

Marker's note. Identify and explain each of the three trends, naming the specific intermolecular force responsible (dispersion forces and hydrogen bonding). Distinguish a hydroxyl group from a hydroxide ion, and explain what controls the strength of dispersion forces.

[4 marks]. Balanced equation:

Moles of each reactant (M of Al(OH)3 = 78.00 g/mol):

From the 1 : 3 ratio, 0.0150 mol Al(OH)3 needs 0.0450 mol HCl. The HCl is in excess:

Marker's note. Write the balanced equation, find the limiting reagent through the 1 : 3 mole ratio, and use the excess HCl to find [H+]. Use exact values throughout, the molar mass from the periodic table, and show all working.

[7 marks]

Step 1 (chloroethane to ethanol)

Heat chloroethane with dilute aqueous sodium hydroxide (or potassium hydroxide) under reflux; the halogen is substituted by OH (a substitution reaction). Identify the product: ethanol gives an O-H stretch in the infrared spectrum (broad, around 3230 to 3550 cm^-1) that chloroethane lacks; it also turns acidified potassium dichromate from orange to green, while chloroethane does not react.

Step 2 (ethanol to ethanoic acid)

Heat ethanol under reflux with an acidified strong oxidant, for example acidified potassium permanganate (or acidified potassium dichromate); the alcohol is oxidised to a carboxylic acid. Identify the product: ethanoic acid releases bubbles of CO2 with sodium carbonate and turns blue litmus red, whereas ethanol does neither; a C=O signal also appears in the 13C NMR spectrum (160 to 185 ppm).

Step 3 (ethanoic acid to ethyl ethanoate)

Heat ethanoic acid with ethanol and a few drops of concentrated sulfuric acid (the catalyst) under reflux; this esterification produces ethyl ethanoate. Identify the product: the ester has no broad O-H peak in its infrared spectrum (present in both ethanol and ethanoic acid) and has a sweet, fruity smell; its molecular ion appears at m/z = 88.

Marker's note. Address all parts: reagents, conditions and a way to identify each product. Use acidified reagents in Step 2, name the colour change of any indicator or oxidant used, and recall that concentrated sulfuric acid is the catalyst for esterification. Annotate the scheme with the names of the compounds and reagents before writing.