2025 Cambridge IAL Chemistry (9701) Practice Paper with Answers

First, the contraction after shaking with NaOH is due to the absorption of CO2, meaning 30 cm^{3} of CO2 was produced. Since 10 cm^{3} of X was used, each molecule of X contains 3 carbon atoms (30 / 10 = 3). Next, the reaction equation is: C_{3}H_{y} + (3 + y/4) O2 -> 3 CO2 + (y/2) H2O. The initial volume of reacting gases is 10 cm^{3} of C_{3}H_{y} + 10(3 + y/4) cm^{3} of O2. The final volume of reacting gases (excluding liquid water) is 30 cm^{3} of CO2. The contraction in volume is (10 + 30 + 2.5y) - 30 = 10 + 2.5y = 25 cm^{3}. Solving for y gives 2.5y = 15, so y = 6. Thus, the molecular formula is C_{3}H_{6}.

1 mark for the correct option B. (Method: Deduce carbon number from CO2 volume = 3. Use volume contraction formula to calculate hydrogen number = 6. Combine to get C3H6.)

Propan-2-ol is first oxidized to propanone using acidified potassium dichromate(VI) under reflux. Propanone then undergoes nucleophilic addition with HCN (in the presence of NaCN catalyst) to form 2-hydroxy-2-methylpropanenitrile. Finally, acid hydrolysis of the nitrile group using dilute hydrochloric acid under reflux yields 2-hydroxy-2-methylpropanoic acid.

1 mark for the correct option B. (Method: Identify propanone as intermediate. Recognize HCN addition to form a hydroxynitrile, followed by acid hydrolysis to carboxylic acid.)

Ammonia is a stronger ligand than water in the spectrochemical series and causes a larger d-orbital splitting (larger \Delta E). Since \Delta E = hf = hc/\lambda, a larger energy gap corresponds to the absorption of light with a shorter wavelength (higher frequency). The complementary color transmitted shifts from light blue to a deep royal blue.

1 mark for the correct option B. (Method: Relate ligand strength of NH3 vs H2O to the magnitude of d-orbital splitting and the resulting change in the wavelength of absorbed light.)

Initial moles of HA = 0.0500 \times 0.100 = 0.00500 mol. Moles of NaOH added = 0.0250 \times 0.120 = 0.00300 mol. The reaction is HA + OH^- -> A^- + H2O. Moles of A^- formed = 0.00300 mol. Moles of HA remaining = 0.00500 - 0.00300 = 0.00200 mol. Using the Henderson-Hasselbalch equation: pH = pK_{a} + \log_{10}([A^-]/[HA]). pK_{a} = -\log_{10}(1.35 \times 10^{-5}) = 4.87. pH = 4.87 + \log_{10}(0.00300 / 0.00200) = 4.87 + 0.18 = 5.05.

1 mark for the correct option C. (Method: Calculate moles of acid and base, determine excess acid and conjugate base concentrations, apply Henderson-Hasselbalch equation to get pH = 5.05.)

A reaction is thermodynamically feasible when \Delta G^\ominus < 0. Using \Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus, we set \Delta H^\ominus - T\Delta S^\ominus < 0, which gives T > \Delta H^\ominus / \Delta S^\ominus. Convert \Delta H^\ominus to J mol^{-1}: 178 \times 10^{3} J mol^{-1}. Thus, T > (178 \times 10^{3}) / 160 = 1112.5 K. This rounds to T > 1113 K.

1 mark for the correct option B. (Method: Set up Gibbs free energy equation for feasibility, convert units of enthalpy to J mol^{-1}, and solve for T to find T > 1113 K.)

The molecular ion peak at m/z = 172 corresponds to CH2(^{79}Br)(^{79}Br)^{+}. The peak at m/z = 174 corresponds to CH2(^{79}Br)(^{81}Br)^{+} and CH2(^{81}Br)(^{79}Br)^{+}. The peak at m/z = 176 corresponds to CH2(^{81}Br)(^{81}Br)^{+}. Since the probability of choosing each isotope is 0.5: P(172) = 0.5 \times 0.5 = 0.25; P(174) = 2 \times (0.5 \times 0.5) = 0.50; P(176) = 0.5 \times 0.5 = 0.25. The ratio is 0.25 : 0.50 : 0.25, which simplifies to 1 : 2 : 1.

1 mark for the correct option B. (Method: Establish the isotopic combinations for each m/z peak, calculate the binomial probabilities based on a 1:1 ratio, and simplify to the simplest integer ratio.)

Reaction of phenylamine with nitrous acid below 10 ^\circ C forms the benzenediazonium ion, C6H5N2^{+}. When coupled with phenol, the electrophilic diazonium ion attacks the electron-rich benzene ring of phenol. The -OH group is a strong ortho/para director, and because of steric hindrance at the ortho-positions, substitution occurs predominantly at the para-position, yielding 4-hydroxyphenylazobenzene.

1 mark for the correct option A. (Method: Identify diazonium ion as the intermediate, recognize para-coupling on phenol as the dominant orientation due to directing effects and steric hindrance.)

Methyl 2-methylbutanoate is an ester. Heating an ester under reflux with an aqueous acid catalyst (such as dilute H2SO4) results in acid hydrolysis, which is the reverse of esterification. This splits the ester back into its constituent carboxylic acid (2-methylbutanoic acid) and alcohol (methanol).

1 mark for the correct option B. (Method: Identify the ester reactants, recognize that acid hydrolysis yields the corresponding carboxylic acid and alcohol, and name them correctly.)

In Step 1, the alkyl side chain of 4-nitromethylbenzene is oxidised to a carboxylic acid group using alkaline potassium manganate(VII), \(\text{KMnO}_4\), heated under reflux, followed by acidification. K2Cr2O7 is not a strong enough oxidising agent to cleave the benzene side chain under standard reflux conditions. In Step 2, the nitro group (\(\text{-NO}_2\)) is selectively reduced to an amine group (\(\text{-NH}_2\)) using tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)) heated under reflux, which leaves the carboxylic acid group intact. (LiAlH4 would reduce the carboxylic acid group as well, which is undesired). In Step 3, the carboxylic acid group of 4-aminobenzoic acid is esterified with ethanol in the presence of a concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) catalyst under heat.

Award 1 mark for the correct option A.
- Reject B because K2Cr2O7 and LiAlH4 are inappropriate reagents for these specific selective transformations.
- Reject C because NaBH4 does not reduce aromatic nitro groups and nucleophilic substitution with bromoethane is inappropriate.
- Reject D because CrO3 in pyridine is used for mild alcohol oxidation, not alkyl side-chain oxidation.

Let the hydrocarbon be \(\text{C}_x\text{H}_y\).
1. The initial volume of hydrocarbon = 10.0 cm\(^3\).
2. After cooling to room temperature, water is liquid and has negligible volume. The remaining 50.0 cm\(^3\) of gas consists of produced \(\text{CO}_2\) and unreacted \(\text{O}_2\).
3. Passing the gas through aqueous \(\text{NaOH}\) absorbs \(\text{CO}_2\). The volume decreases by 50.0 - 20.0 = 30.0 cm\(^3\). Thus, the volume of \(\text{CO}_2\) produced is 30.0 cm\(^3\).
4. 10.0 cm\(^3\) of \(\text{C}_x\text{H}_y\) produces 30.0 cm\(^3\) of \(\text{CO}_2\), so \(x = 3\).
5. The remaining 20.0 cm\(^3\) of gas is excess \(\text{O}_2\). Therefore, the volume of \(\text{O}_2\) that reacted is 70.0 - 20.0 = 50.0 cm\(^3\).
6. The combustion equation is \(\text{C}_3\text{H}_y + (3 + y/4)\text{O}_2 \rightarrow 3\text{CO}_2 + (y/2)\text{H}_2\text{O}\).
7. Ratio of reacted \(\text{O}_2\) to hydrocarbon is 50.0 / 10.0 = 5.0.
8. Therefore, \(3 + y/4 = 5.0 \implies y/4 = 2.0 \implies y = 8\). The molecular formula is \(\text{C}_3\text{H}_8\).

Award 1 mark for the correct option B.
- Award 0 marks for incorrect stoichiometric analysis or wrong molecular formula calculation.

In transition metal complexes, the presence of ligands causes the d-orbitals of the transition metal ion to split into two groups of different energy levels. Electrons in the lower d-orbitals can absorb a photon of specific wavelength from the visible light spectrum to transition to a higher d-orbital (d-d transition). The light complementary to the absorbed wavelength is transmitted. Since copper(II) ions in water appear light blue, they absorb light in the complementary red-orange region of the spectrum.

Award 1 mark for the correct option A.
- Reject B because light is absorbed to promote electrons, not emitted.
- Reject C because this colour is due to d-d transitions, not ligand-to-metal charge transfer.
- Reject D because d-d orbital transitions within the split 3d levels are responsible, not excitation to the 4s orbital.

1. Moles of propanoic acid (\(\text{HA}\)) initially = \(0.0500 \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 0.00500\) mol.
2. Moles of \(\text{NaOH}\) added = \(0.0500 \text{ dm}^3 \times 0.0500 \text{ mol dm}^{-3} = 0.00250\) mol.
3. The reaction is: \(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\).
4. Since \(\text{OH}^-\) is the limiting reagent, all \(0.00250\) mol reacts. Moles of \(\text{HA}\) remaining = \(0.00500 - 0.00250 = 0.00250\) mol. Moles of propanoate ions (\(\text{A}^-\)) produced = \(0.00250\) mol.
5. Since the remaining moles of weak acid (\(\text{HA}\)) equal the moles of conjugate base (\(\text{A}^-\)) in the same final volume of 100.0 cm\(^3\), \([\text{HA}] = [\text{A}^-]\).
6. Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{pK}_a + \log_{10}([\text{A}^-]/[\text{HA}])\). Since \([\text{A}^-] = [\text{HA}]\), \(\text{pH} = \text{pK}_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

Award 1 mark for the correct option C.
- Reject A (calculates pH of the unreacted weak acid solution without buffer mixture).
- Reject B (incorrect ratio due to failure to account for the neutralization reaction).
- Reject D (inverted buffer ratio calculation).

The combustion reaction equation for propane is:
\(\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\)

According to Hess's Law:
\(\Delta H_c^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})\)
\(\Delta H_c^\ominus = [3 \times \Delta H_f^\ominus(\text{CO}_2(\text{g})) + 4 \times \Delta H_f^\ominus(\text{H}_2\text{O}(\text{l}))] - [\Delta H_f^\ominus(\text{C}_3\text{H}_8(\text{g}))]\)
\(\Delta H_c^\ominus = [3 \times (-393.5) + 4 \times (-285.8)] - [-103.8]\)
\(\Delta H_c^\ominus = [-1180.5 - 1143.2] + 103.8 = -2323.7 + 103.8 = -2219.9\) kJ mol\(^{-1}\).

Award 1 mark for the correct option B.
- Reject A (calculated as \(-2323.7 - 103.8\), a sign error in subtracting the reactant).
- Reject C (fails to multiply the product values by the stoichiometric coefficients).
- Reject D (reversed products and reactants calculation resulting in positive value).

The mass of the \(\text{CH}_2\) fragment is \(12 + 2(1) = 14\).
Possible isotopic combinations for \(\text{CH}_2\text{ClBr}\) molecular ions are:
1. \(^{35}\text{Cl}\) and \(^{79}\text{Br}\): \(14 + 35 + 79 = 128\). Probability = \(0.758 \times 0.507 = 0.384\).
2. \(^{37}\text{Cl}\) and \(^{79}\text{Br}\): \(14 + 37 + 79 = 130\). Probability = \(0.242 \times 0.507 = 0.123\).
3. \(^{35}\text{Cl}\) and \(^{81}\text{Br}\): \(14 + 35 + 81 = 130\). Probability = \(0.758 \times 0.493 = 0.374\).
4. \(^{37}\text{Cl}\) and \(^{81}\text{Br}\): \(14 + 37 + 81 = 132\). Probability = \(0.242 \times 0.493 = 0.119\).

Thus, peaks appear at \(m/z\) 128, 130, and 132. The peak at \(m/z\) 130 has two possible combinations, giving a combined probability of \(0.123 + 0.374 = 0.497\) (approx 50%), which is the most intense.

Award 1 mark for the correct option B.
- Reject A because the combined probability of the isotopes makes 130 more intense than 128.
- Reject C because hydrogen and carbon isotopes are negligible/specified as fixed, so there are no intermediate integer peaks like 129 or 131 of high intensity.
- Reject D because it misses the mixed isotope combination.

A stability constant (\(K_{\text{stab}}\)) is an equilibrium constant for the formation of a complex ion. The much larger value of \(K_{\text{stab}}\) for Reaction 2 (\(1.0 \times 10^{20}\)) compared to Reaction 1 (\(1.2 \times 10^{13}\)) shows that the ethylenediamine complex is thermodynamically much more stable. This is largely driven by the chelate effect (favourable increase in entropy when bidentate ligands displace monodentate ligands). Therefore, adding 'en' to a solution containing \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\) will cause a ligand exchange reaction.

Award 1 mark for the correct option A.
- Reject B because thermodynamics (indicated by the huge difference in \(K_{\text{stab}}\)) strongly favours the formation of the 'en' complex, making displacement by ammonia extremely unfavourable.
- Reject C because a higher stability constant is associated with stronger overall binding/stability of the complex.
- Reject D because stability depends on the nature of the ligands and chelate effect, not just coordination number.

In Step 1, dehydration of propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) by heating with concentrated sulfuric acid at 180 °C produces propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)) as alkene Z. In Step 2, mild oxidation of propene with cold, dilute, acidified \(\text{KMnO}_4\) adds two hydroxyl groups across the double bond to form a diol. The product is propane-1,2-diol (\(\text{CH}_3\text{CH(OH)CH}_2\text{OH}\)).

Award 1 mark for the correct option A.
- Reject B because propane-1,3-diol cannot be formed directly via addition across the double bond of propene.
- Reject C and D because these are the products of oxidation of propan-1-ol directly (using potassium dichromate), not via dehydration and diol formation.

Let the hydrocarbon be \(\text{C}_x\text{H}_y\). The equation for combustion is: \(\text{C}_x\text{H}_y(g) + \left(x + \frac{y}{4}\right)\text{O}_2(g) \rightarrow x\text{CO}_2(g) + \frac{y}{2}\text{H}_2\text{O}(l)\). Since water is liquid at room temperature, it does not contribute to the gas volume.

1. Total contraction on cooling = 30 cm\(^3\).
Initial gas volume = \(10 + V(\text{O}_2)\)
Final gas volume = \(V(\text{CO}_2) + \text{excess }\text{O}_2 = 10x + [V(\text{O}_2) - 10(x + y/4)] = V(\text{O}_2) - 2.5y\)
Decrease in volume = \(\text{Initial} - \text{Final} = 10 + 2.5y = 30 \implies 2.5y = 20 \implies y = 8\).

2. Carbon dioxide volume absorption by NaOH = 40 cm\(^3\).
\(10x = 40 \implies x = 4\).

Therefore, the molecular formula is \(\text{C}_4\text{H}_8\).

[1 mark] Correctly deduces the number of hydrogen atoms (y = 8) from the initial contraction of gas volume and the number of carbon atoms (x = 4) from the contraction on adding sodium hydroxide to find the formula C4H8.

The stability constant represents the equilibrium constant for the formation of a complex from the aqua complex. For the ligand displacement reaction: \([\text{Fe}(L^1)_6]^{3+} + 6L^2 \rightleftharpoons [\text{Fe}(L^2)_6]^{3+} + 6L^1\), we can subtract the formation reaction of \([\text{Fe}(L^1)_6]^{3+}\) from the formation reaction of \([\text{Fe}(L^2)_6]^{3+}\). Thus, the equilibrium constant \(K_c\) is: \(K_c = \frac{K_{\text{stab2}}}{K_{\text{stab1}}} = \frac{3.4 \times 10^{12}}{1.2 \times 10^5} = 2.8 \times 10^7\). Since \(K_c \gg 1\), the statement in option C is correct.

[1 mark] Correctly identifies that the equilibrium constant for the displacement reaction is the ratio of the stability constants, which is significantly greater than 1, meaning the forward reaction is highly favored.

Step 1: Gentle oxidation of propan-1-ol to propanal. Acidified potassium dichromate(VI) with immediate distillation prevents further oxidation to propanoic acid.
Step 2: Nucleophilic addition of HCN to propanal. Using \(\text{HCN}\) in the presence of \(\text{NaCN}\) (which provides the \(\text{CN}^-\)\ nucleophile catalyst) yields 2-hydroxybutanenitrile, lengthening the carbon chain by one carbon.
Step 3: Acid hydrolysis of the nitrile. Heating under reflux with dilute hydrochloric acid hydrolyzes the nitrile group to a carboxylic acid group, forming 2-hydroxybutanoic acid.

[1 mark] Correctly identifies the three-step sequence involving partial oxidation of the primary alcohol to an aldehyde, nucleophilic addition to form a hydroxynitrile, and subsequent acid-catalyzed hydrolysis of the nitrile.

Determine initial moles:
Moles of propanoic acid = \(0.0250 \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 2.50 \times 10^{-3} \text{ mol}\).
Moles of \(\text{NaOH}\) = \(0.0150 \text{ dm}^3 \times 0.0800 \text{ mol dm}^{-3} = 1.20 \times 10^{-3} \text{ mol}\).

Since \(\text{NaOH}\) is the limiting reactant, it reacts completely with the propanoic acid:
Moles of propanoate ions formed = \(1.20 \times 10^{-3} \text{ mol}\).
Moles of propanoic acid remaining = \(2.50 \times 10^{-3} - 1.20 \times 10^{-3} = 1.30 \times 10^{-3} \text{ mol}\).

Using the Henderson-Hasselbalch equation:
\(pH = pK_a + \log_{10}\left(\frac{[\text{salt}]}{[\text{acid}]}\right)\)
\(pK_a = -\log_{10}(1.35 \times 10^{-5}) = 4.870\)
\(pH = 4.870 + \log_{10}\left(\frac{1.20 \times 10^{-3}}{1.30 \times 10^{-3}}\right) = 4.870 - 0.035 = 4.835 \approx 4.84\).

[1 mark] Calculates the remaining moles of acid and the moles of salt formed, uses the Henderson-Hasselbalch equation with the correct pKa, and performs the arithmetic correctly to find pH = 4.84.

Lattice energy is proportional to the electrostatic attraction between cations and anions, which is governed by \(\frac{q_+ q_-}{r_+ + r_-}\). The charge of the cation is identical (\(\text{Ca}^{2+}\)), and both anions have a \(-1\) charge. However, the ionic radius of the fluoride ion is smaller than that of the chloride ion. This means the distance between the centres of the ions (\(r_+ + r_-\)) is smaller in \(\text{CaF}_2\), resulting in stronger electrostatic attraction and a more exothermic lattice energy.

[1 mark] Identifies that the smaller ionic radius of the fluoride ion compared to the chloride ion leads to a shorter distance between ion centers and stronger electrostatic forces, making the lattice energy more exothermic.

The relative atomic mass (\(A_r\)) is the weighted average of the masses of the isotopes based on their relative abundances:
\(A_r = \frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100} = \frac{1895.76 + 250.00 + 286.26}{100} = \frac{2432.02}{100} = 24.3202\). Rounding to two decimal places gives 24.32.

[1 mark] Calculates the weighted average using the given abundances and isotope masses, rounding correctly to 24.32 (two decimal places).

Basicity is determined by the availability of the lone pair on the nitrogen atom to accept a proton.
- Phenylamine (1) is the weakest base because the nitrogen lone pair is delocalized into the \(\pi\) ring system of the benzene ring.
- Ammonia (3) is more basic than phenylamine as there is no ring delocalization.
- Ethylamine (2) is a primary aliphatic amine; the electron-donating ethyl group increases electron density on the nitrogen atom via the inductive effect.
- Diethylamine (4) is a secondary aliphatic amine with two electron-donating ethyl groups, making it the most basic.

Thus, the correct order is 1 < 3 < 2 < 4.

[1 mark] Correctly ranks the basicity based on electronic effects (delocalization in phenylamine vs. electron-donation by alkyl groups in primary/secondary amines).

For a species \(R\) to reduce an oxidant \(Ox\) spontaneously under standard conditions, the cell potential must be positive: \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} > 0\). This requires that \(E^\ominus(\text{oxidant}) > E^\ominus(\text{reducing agent})\).

1. To reduce \(\text{Fe}^{3+}\) to \(\text{Fe}^{2+}\) (\(E^\ominus = +0.77\text{ V}\)), the reducing agent must have a standard electrode potential less than \(+0.77\text{ V}\).
2. To NOT reduce \(\text{I}_2\) to \(\text{I}^-\rightleftharpoons\) (\(E^\ominus = +0.54\text{ V}\)), the reducing agent must have a standard electrode potential greater than \(+0.54\text{ V}\).

The standard electrode potential for the \(\text{O}_2 / \text{H}_2\text{O}_2\) couple is \(+0.68\text{ V}\). Since \(+0.54\text{ V} < +0.68\text{ V} < +0.77\text{ V}\), \(\text{H}_2\text{O}_2(aq)\) can reduce \(\text{Fe}^{3+}\) (\(E^\ominus_{\text{cell}} = +0.77 - 0.68 = +0.09\text{ V}\)) but cannot reduce \(\text{I}_2\) (\(E^\ominus_{\text{cell}} = +0.54 - 0.68 = -0.14\text{ V}\)).

[1 mark] Deduces that the reducing agent must have a standard electrode potential between +0.54 V and +0.77 V and identifies H2O2 as meeting this condition.

Acid hydrolysis of an ester yields a carboxylic acid and an alcohol. Let X be \(\text{R}_1\text{COOR}_2\). Hydrolysis yields \(\text{R}_1\text{COOH}\) (Z) and \(\text{R}_2\text{OH}\) (Y). Since Y can be oxidized to Z, the alcohol Y must have the same number of carbons as the carboxylic acid Z, and Y must be a primary alcohol. Since X has 4 carbons in total, both Y and Z must contain 2 carbons. Thus, Y is ethanol and Z is ethanoic acid. The original ester X must be ethyl ethanoate.

1 mark: Correctly identifies B as ethyl ethanoate.
0 marks: Any other option.

The sequence is as follows:
1. Benzene is nitrated to nitrobenzene.
2. Nitrobenzene is reduced using tin and concentrated hydrochloric acid to form phenylamine (aniline).
3. Phenylamine reacts with nitrous acid (nitric(III) acid) at low temperature (under \(10\ ^\circ\text{C}\)) to form benzenediazonium chloride.
4. Benzenediazonium chloride couples with phenol. The electrophilic attack by the diazonium ion occurs at the electron-rich para-position (4-position) of phenol, yielding 4-(phenyldiazenyl)phenol.

1 mark: Correctly identifies B as 4-(phenyldiazenyl)phenol.
0 marks: Any other option.

Calculate mass of Carbon:
\(m(\text{C}) = 0.300\text{ g} \times \frac{12.01}{44.01} = 0.08187\text{ g}\)
\(n(\text{C}) = \frac{0.08187}{12.01} = 0.006817\text{ mol}\)

Calculate mass of Hydrogen:
\(m(\text{H}) = 0.123\text{ g} \times \frac{2 \times 1.008}{18.016} = 0.01378\text{ g}\)
\(n(\text{H}) = \frac{0.01378}{1.008} = 0.01367\text{ mol}\)

Calculate mass of Oxygen:
\(m(\text{O}) = 0.150\text{ g} - (0.08187 + 0.01378) = 0.05435\text{ g}\)
\(n(\text{O}) = \frac{0.05435}{16.00} = 0.003397\text{ mol}\)

Divide by the smallest number of moles:
\(\text{C}: \frac{0.006817}{0.003397} \approx 2\)
\(\text{H}: \frac{0.01367}{0.003397} \approx 4\)
\(\text{O}: \frac{0.003397}{0.003397} = 1\)

The empirical formula is \(\text{C}_2\text{H}_4\text{O}\).

1 mark: Correctly calculates moles of C, H, and O to find empirical formula C2H4O (C).
0 marks: Any other option.

Let the hydrocarbon be \(\text{C}_x\text{H}_y\).
1. Volume of hydrocarbon = \(10.0\text{ cm}^3\).
2. The remaining gas after combustion consists of excess \(\text{O}_2\) and product \(\text{CO}_2\) (since water is liquid at room temperature and its volume is negligible). Total remaining gas = \(60.0\text{ cm}^3\).
3. Passing the gas through aqueous \(\text{NaOH}\) removes \(\text{CO}_2\), leaving only the excess \(\text{O}_2\), which is \(20.0\text{ cm}^3\).
4. Therefore, the volume of \(\text{CO}_2\) produced = \(60.0 - 20.0 = 40.0\text{ cm}^3\).
From \(10.0\text{ cm}^3\) of \(\text{C}_x\text{H}_y\), we get \(40.0\text{ cm}^3\) of \(\text{CO}_2\), so \(x = 4\).
5. The volume of \(\text{O}_2\) reacted = Initial \(\text{O}_2\) - Excess \(\text{O}_2\) = \(80.0 - 20.0 = 60.0\text{ cm}^3\).
6. The combustion equation is:
\(\text{C}_x\text{H}_y + (x + y/4)\text{O}_2 \rightarrow x\text{CO}_2 + y/2\text{H}_2\text{O}\)
Thus, the ratio of reacted \(\text{O}_2\) to hydrocarbon is \(x + y/4 = 60.0 / 10.0 = 6\).
Since \(x = 4\):
\(4 + y/4 = 6 \implies y/4 = 2 \implies y = 8\).
The molecular formula is \(\text{C}_4\text{H}_8\).

1 mark: Correctly deduces the volumes of CO2 produced and O2 reacted to determine C4H8.
0 marks: Any other option.

The complex ion \([\text{Co(en)}_2\text{Cl}_2]^+\) exhibits both geometric and optical isomerism:
1. Trans-isomer: The two chlorine ligands are opposite to each other (180 degrees). This isomer has a plane of symmetry and is optically inactive. (1 isomer)
2. Cis-isomer: The two chlorine ligands are adjacent to each other (90 degrees). This isomer lacks a plane of symmetry (is chiral) and exists as a pair of non-superimposable mirror images (enantiomers). (2 isomers)
Total number of stereoisomers = 1 (trans) + 2 (cis) = 3.

1 mark: Correctly identifies 3 stereoisomers.
0 marks: Any other option.

1. Moles of propanoic acid (\(\text{HA}\)) originally present:
\(n(\text{HA}) = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\)
2. Moles of sodium hydroxide (\(\text{NaOH}\)) added:
\(n(\text{OH}^-) = 0.0250\text{ dm}^3 \times 0.120\text{ mol dm}^{-3} = 0.00300\text{ mol}\)
3. After reaction, the moles of propanoate ion (\(\text{A}^-\)) formed is equal to the moles of \(\text{NaOH}\) added:
\(n(\text{A}^-) = 0.00300\text{ mol}\)
4. The moles of remaining unreacted propanoic acid is:
\(n(\text{HA}) = 0.00500 - 0.00300 = 0.00200\text{ mol}\)
5. Use the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\)
\(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87\)
\(\text{pH} = 4.87 + \log_{10}\left(\frac{0.00300}{0.00200}\right) = 4.87 + \log_{10}(1.5) = 4.87 + 0.18 = 5.05\)

1 mark: Correctly calculates the remaining acid, formed salt, and uses the Henderson-Hasselbalch equation to find 5.05.
0 marks: Any other option.

Using the Born-Haber cycle equation:
\(\Delta H_f^\ominus = \Delta H_{at}^\ominus(\text{Na}) + IE_1(\text{Na}) + \Delta H_{at}^\ominus(\text{Cl}) + EA_1(\text{Cl}) + \Delta H_{latt}^\ominus(\text{NaCl})\)

Substitute the given values into the equation:
\(-411 = +107 + 496 + 121 + (-349) + \Delta H_{latt}^\ominus\)
\(-411 = 375 + \Delta H_{latt}^\ominus\)
\(\Delta H_{latt}^\ominus = -411 - 375 = -786\text{ kJ mol}^{-1}\)

1 mark: Correctly sets up the Born-Haber cycle summation and calculates -786 kJ/mol.
0 marks: Any other option.

Biodegradable polymers contain functional groups like esters or amides in their backbones, which are susceptible to chemical breakdown. Poly(lactic acid) (PLA) is a condensation polyester. The ester linkages can undergo alkaline hydrolysis to form the salt of lactic acid. In contrast, poly(ethene), poly(phenylethene), and poly(tetrafluoroethene) are addition polymers with inert C-C backbones that are resistant to hydrolysis.

1 mark: Correctly identifies poly(lactic acid) as biodegradable and hydrolyzable.
0 marks: Any other option.

Let's track the volume changes step-by-step:
1. The initial volume of hydrocarbon is \(10\text{ cm}^3\).
2. Concentrated aqueous sodium hydroxide absorbs carbon dioxide gas. The decrease in volume from \(55\text{ cm}^3\) to \(25\text{ cm}^3\) indicates that the volume of \(CO_2\) produced is:
\(55\text{ cm}^3 - 25\text{ cm}^3 = 30\text{ cm}^3\).
Since \(10\text{ cm}^3\) of \(C_xH_y\) yields \(30\text{ cm}^3\) of \(CO_2\), each molecule must contain 3 carbon atoms (i.e., \(x = 3\)).
3. The remaining \(25\text{ cm}^3\) of gas is unreacted oxygen. Therefore, the volume of oxygen that reacted is:
\(70\text{ cm}^3 - 25\text{ cm}^3 = 45\text{ cm}^3\).
4. The combustion equation for the hydrocarbon is:
\(C_3H_y + (3 + \frac{y}{4})O_2 \rightarrow 3CO_2 + \frac{y}{2}H_2O\)
The mole ratio of hydrocarbon to reacting oxygen is \(10 : 45 = 1 : 4.5\).
Thus, \(3 + \frac{y}{4} = 4.5\), which gives \(\frac{y}{4} = 1.5 \Rightarrow y = 6\).

Therefore, the molecular formula of the hydrocarbon is \(C_3H_6\).

1 mark for the correct option B.
- Award 1 mark for calculating the volume of CO2 (30 cm³) and O2 reacted (45 cm³), setting up the stoichiometric ratio, and identifying the correct molecular formula.

1. Compound X has the molecular formula \(C_4H_8O_2\) and undergoes acid hydrolysis, which means X is an ester.
2. Acid hydrolysis of ester X yields a carboxylic acid (Z) and an alcohol (Y).
3. Since Y can be oxidized to Z, Z must have the exact same carbon skeleton and number of carbon atoms as Y.
4. Since the total number of carbon atoms is 4, both Y and Z must contain 2 carbon atoms each. Thus, Y is ethanol (\(CH_3CH_2OH\)) and Z is ethanoic acid (\(CH_3COOH\)).
5. This means X must be ethyl ethanoate (\(CH_3COOCH_2CH_3\)).

Let's evaluate the options:
- A is incorrect: Ethyl ethanoate does not contain a chiral carbon, so it does not exhibit optical isomerism.
- B is incorrect: Esters do not react with sodium carbonate to produce carbon dioxide (only carboxylic acids do).
- C is correct: Compound X is indeed ethyl ethanoate.
- D is incorrect: Methyl propanoate would hydrolyze to methanol and propanoic acid, and methanol cannot be oxidized to propanoic acid.

1 mark for the correct option C.
- Award 1 mark for identifying the ester as ethyl ethanoate based on the oxidation of its alcohol fragment into its acid fragment.

The stability constant for the ammonia complex (\(K_{\text{stab}} = 2.1 \times 10^{13}\)) is significantly larger than that for the chloro complex (\(K_{\text{stab}} = 4.2 \times 10^5\)). This means that ammonia is a much stronger ligand than chloride and will successfully displace the chloride ligands from the copper(II) coordination sphere.

When concentrated ammonia is added, the yellow-green tetrachlorocuprate(II) complex, \([CuCl_4]^{2-}\), is converted into the highly stable deep-blue tetraamminediaquacopper(II) complex, \([Cu(NH_3)_4(H_2O)_2]^{2+}\).

1 mark for the correct option B.
- Award 1 mark for comparing the K_stab values to determine relative ligand strength and predicting the resulting color change.

To find the ratio of salt to acid in the buffer, we use the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\text{salt}]}{[\text{acid}]} \right)\)

First, calculate \(\text{p}K_a\):
\(\text{p}K_a = -\log_{10}(K_a) = -\log_{10}(1.75 \times 10^{-5}) = 4.757\)

Now substitute the known values into the equation:
\(4.80 = 4.757 + \log_{10} \left( \frac{[\text{sodium ethanoate}]}{[\text{ethanoic acid}]} \right)\)

\(\log_{10} \left( \frac{[\text{sodium ethanoate}]}{[\text{ethanoic acid}]} \right) = 4.80 - 4.757 = 0.043\)

Taking the antilog of both sides:
\(\frac{[\text{sodium ethanoate}]}{[\text{ethanoic acid}]} = 10^{0.043} \approx 1.10\)

1 mark for the correct option C.
- Award 1 mark for calculating pKa correctly and using the buffer equation to solve for the salt-to-acid ratio.

By applying Hess's Law to a Born-Haber cycle for \(CaO(s)\):

\(\Delta H_{\text{f}}^{\ominus}[CaO(s)] = \Delta H_{\text{at}}^{\ominus}[Ca(s)] + \text{IE}_1[Ca(g)] + \text{IE}_2[Ca(g)] + \Delta H_{\text{at}}^{\ominus}[O_2(g)] + \text{EA}_1[O(g)] + \text{EA}_2[O(g)] + \Delta H_{\text{latt}}^{\ominus}[CaO(s)]\)

Substitute the given values into the equation:

\(-635 = 178 + 590 + 1145 + 249 + (-141) + 798 + \Delta H_{\text{latt}}^{\ominus}\)

\(-635 = 2819 + \Delta H_{\text{latt}}^{\ominus}\)

\[\Delta H_{\text{latt}}^{\ominus} = -635 - 2819 = -3454 \text{ kJ mol}^{-1}\]

1 mark for the correct option A.
- Award 1 mark for correctly setting up the Born-Haber cycle calculation and computing the lattice energy.

1. Compound P is an acyl chloride because it has the formula \(C_4H_7ClO\) (corresponding to \(R-COCl\)) and reacts rapidly with water at room temperature to form dense white fumes of \(HCl\).
2. Since P is a branched-chain acyl chloride with 4 carbon atoms, its structure must be 2-methylpropanoyl chloride, \((CH_3)_2CHCOCl\).
3. Hydrolysis of 2-methylpropanoyl chloride with water produces 2-methylpropanoic acid (compound Q) and hydrogen chloride:

\((CH_3)_2CHCOCl + H_2O \rightarrow (CH_3)_2CHCOOH + HCl\)

Thus, compound Q is methylpropanoic acid (or 2-methylpropanoic acid).

1 mark for the correct option B.
- Award 1 mark for identifying the functional group as an acyl chloride, using the branching condition to identify the structural formula, and determining the hydrolyzed product.

The relative abundance ratio of \(^{35}Cl\) to \(^{37}Cl\) is \(3:1\). Let the probability of finding a \(^{35}Cl\) atom be \(p = \frac{3}{4}\) and a \(^{37}Cl\) atom be \(q = \frac{1}{4}\).

For a molecule containing two chlorine atoms, the probabilities of the isotope combinations are given by the expansion of \((3 + 1)^2\):
- \(M\) peak (containing two \(^{35}Cl\) atoms): \(3 \times 3 = 9\)
- \(M+2\) peak (containing one \(^{35}Cl\) and one \(^{37}Cl\) atom): \(2 \times (3 \times 1) = 6\)
- \(M+4\) peak (containing two \(^{37}Cl\) atoms): \(1 \times 1 = 1\)

Therefore, the ratio of the peak heights for \(M : M+2 : M+4\) is \(9:6:1\).

1 mark for the correct option D.
- Award 1 mark for using the binomial expansion (3+1)^2 or probability math to deduce the correct 9:6:1 ratio for a dichloro-compound.

1. Titanium has the electronic configuration \([Ar]3d^2 4s^2\). The \(Ti^{3+}\) ion has the configuration \([Ar]3d^1\).
2. Scandium has the electronic configuration \([Ar]3d^1 4s^2\). The \(Sc^{3+}\) ion has the configuration \([Ar]3d^0\).
3. In the presence of water ligands, the d-orbitals split into two sets of different energy levels.
4. \(Ti^{3+}\) has one electron in its split d-subshell. It can absorb a specific frequency of visible light to promote this electron from a lower d-orbital to a higher d-orbital (d-d transition). The unabsorbed frequencies of light are transmitted, making the solution colored.
5. \(Sc^{3+}\) has an empty d-subshell (\(3d^0\)). There are no d-electrons available to undergo d-d transitions, so no visible light is absorbed and the solution remains colorless.

1 mark for the correct option A.
- Award 1 mark for identifying the electronic configurations of Ti3+ (3d1) and Sc3+ (3d0) and correlating this with the capability to undergo d-d transitions.

(a)(i) \(\text{Mass of water lost} = 2.371\text{ g} - 1.723\text{ g} = 0.648\text{ g}\).

(a)(ii) \(\text{Percentage of water by mass} = \frac{0.648}{2.371} \times 100 = 27.3\%\).

(b)(i) \(\text{Mass of SO}_4^{2-} = 1.723\text{ g} - 0.353\text{ g} - 0.216\text{ g} = 1.154\text{ g}\).

(b)(ii) Convert masses to moles:
- \(\text{Moles of Co}^{2+} = \frac{0.353}{58.9} = 0.00599\text{ mol}\)
- \(\text{Moles of NH}_4^+ = \frac{0.216}{18.0} = 0.0120\text{ mol}\)
- \(\text{Moles of SO}_4^{2-} = \frac{1.154}{96.1} = 0.0120\text{ mol}\)

Divide by the smallest value (0.00599):
- \(\text{Co}^{2+} = 1\)
- \(\text{NH}_4^+ = 2.00\)
- \(\text{SO}_4^{2-} = 2.00\)
Therefore, the empirical formula of the anhydrous salt is \(\text{(NH}_4\text{)}_2\text{Co(SO}_4\text{)}_2\).

(c) The molar mass of \(\text{(NH}_4\text{)}_2\text{Co(SO}_4\text{)}_2\) is:
\(M_\text{r} = 2(18.0) + 58.9 + 2(96.1) = 287.1\text{ g mol}^{-1}\).
From Procedure 1:
- \(\text{Moles of anhydrous salt} = \frac{1.723}{287.1} = 0.00600\text{ mol}\).
- \(\text{Moles of water} = \frac{0.648}{18.0} = 0.0360\text{ mol}\).
Ratio: \(w = \frac{0.0360}{0.00600} = 6\).

(d) Formula: \(\text{(NH}_4\text{)}_2\text{Co(SO}_4\text{)}_2 \cdot 6\text{H}_2\text{O}\).
Equation: \(\text{(NH}_4\text{)}_2\text{Co(SO}_4\text{)}_2 \cdot 6\text{H}_2\text{O(s)} \rightarrow \text{(NH}_4\text{)}_2\text{Co(SO}_4\text{)}_2\text{(s)} + 6\text{H}_2\text{O(g)}\)

(a)(i) 0.648 g [1]
(a)(ii) Correct calculation showing 0.648/2.371 * 100 [1]; final answer 27.3% (must be 3 s.f.) [1]
(b)(i) 1.154 g (or 1.153 g) [1]
(b)(ii) Calculates moles of cobalt and ammonium [1]; calculates moles of sulfate [1]; finds simplest mole ratio of 1 : 2 : 2 [1]; states empirical formula (NH4)2Co(SO4)2 [1]
(c) Calculates Mr of anhydrous salt as 287.1 [1]; calculates moles of anhydrous salt (0.00600 mol) and moles of water (0.0360 mol) [1]; states w = 6 [1]
(d) Correct balanced equation with state symbols not required [1]

(a)(i) Reaction type: Dehydration / Elimination [1].
Reagent: Concentrated sulfuric acid, \(\text{H}_2\text{SO}_4\) (or conc. \(\text{H}_3\text{PO}_4\)) [1].
Conditions: Heat (approx. 170 °C) [1]. (Alternative: \(\text{Al}_2\text{O}_3\) catalyst and heat at 300 °C).

(a)(ii) Mechanism steps:
- Curly arrow from the double bond of propene pointing to the \(\text{H}\) of \(\text{H}^\delta\text{+}-\text{Br}^\delta\text{-}\) [1].
- Curly arrow from the \(\text{H}-\text{Br}\) bond to the \(\text{Br}\) atom, with correct partial charges [1].
- Correct structure of the secondary carbocation intermediate (\(\text{CH}_3\text{C}^+\text{HCH}_3\)) and the bromide ion (\(\text{Br}^-\)) [1].
- Curly arrow from the lone pair of the \(\text{Br}^-\) ion pointing to the carbocation carbon [1].

(b)(i) Reagent: Aqueous sodium hydroxide, \(\text{NaOH(aq)}\) [1/2]. Conditions: Heat under reflux [1/2]. Mechanism type: Nucleophilic substitution [1].

(b)(ii) Reagent: Acidified potassium dichromate(VI) / \(\text{H}^+/\text{Cr}_2\text{O}_7^{2-}\) [1]. Observation: Orange solution turns green [1].

(c) Add aqueous sodium carbonate/hydrogencarbonate (or magnesium ribbon) [1/2]. Propanoic acid produces effervescence/bubbles, while propanone shows no reaction [1/2].

(a)(i) Elimination / Dehydration [1]; conc. H2SO4 / conc. H3PO4 [1]; heat [1]
(a)(ii) Arrow from double bond to H and dipole on H-Br [1]; arrow from H-Br bond to Br [1]; structure of secondary carbocation [1]; arrow from lone pair of Br- to C+ [1]
(b)(i) NaOH(aq) / KOH(aq) and heat/reflux [1]; nucleophilic substitution [1]
(b)(ii) Acidified K2Cr2O7 [1]; orange to green [1]
(c) Add NaHCO3 / Na2CO3 / Mg: propanoic acid fizzes, propanone does not [1]

(a) Standard enthalpy change of combustion is the enthalpy change when one mole of a substance [1] is completely burned in excess oxygen under standard conditions (298 K, 100 kPa) [1].

(b)(i) \(q = m \cdot c \cdot \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 16.7\text{ K} = 10470.9\text{ J} = 10.47\text{ kJ}\) (Accept 10.5 kJ).

(b)(ii) \(M_\text{r}(\text{CH}_3\text{OH}) = 32.0\text{ g mol}^{-1}\). \(n = \frac{0.850}{32.0} = 0.02656\text{ mol}\) (Accept 0.0266 mol).

(b)(iii) \(\Delta H_\text{c} = -\frac{10.4709}{0.02656} = -394.2\text{ kJ mol}^{-1}\). Rounded to 3 s.f.: \(-394\text{ kJ mol}^{-1}\).

(b)(iv) Heat loss to the surroundings/calorimeter [1]; incomplete combustion of methanol [1].

(c) \(\text{C(s)} + 2\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}\)
Using Hess's Law:
\(\Delta H_\text{f}^\ominus = \Delta H_\text{c}^\ominus[\text{C(s)}] + 2 \times \Delta H_\text{c}^\ominus[\text{H}_2\text{(g)}] - \Delta H_\text{c}^\ominus[\text{CH}_3\text{OH(l)}]\)
\(\Delta H_\text{f}^\ominus = (-393.5) + 2(-285.8) - (-726.0)\)
\(\Delta H_\text{f}^\ominus = -393.5 - 571.6 + 726.0 = -239.1\text{ kJ mol}^{-1}\).

(a) Enthalpy change when 1 mole of substance [1] is completely burned in oxygen under standard conditions [1]
(b)(i) 10.5 kJ (or 10.47 kJ) [1]
(b)(ii) 0.0266 mol (or 0.02656 mol) [1]
(b)(iii) Correct calculation dividing heat by moles [1]; final answer with negative sign and units: -394 kJ mol-1 [1]
(b)(iv) Heat loss to surroundings [1]; incomplete combustion (or evaporation of fuel) [1]
(c) Writes balanced equation or draws correct Hess's cycle [1]; uses 2 x enthalpy of combustion of hydrogen [1]; correct algebraic expression [1]; final answer -239.1 kJ mol-1 [1]

(a)(i) A Brønsted–Lowry acid is a proton (\(\text{H}^+\)) donor [1].

(a)(ii) Acid: \(\text{HNO}_2\), Base: \(\text{HCOO}^-\), Conjugate Base: \(\text{NO}_2^-\), Conjugate Acid: \(\text{HCOOH}\). (All 4 correct [2], 2 or 3 correct [1]).

(b)(i) A strong acid completely dissociates/ionises in aqueous solution [1].

(b)(ii) \(\text{pH} = -\log_{10}(0.0400) = 1.40\) [1].

(b)(iii) At pH = 2.00, \([\text{H}^+] = 10^{-2.00} = 0.0100\text{ mol dm}^{-3}\) [1].
Using dilution formula: \(c_1 V_1 = c_2 V_2 \Rightarrow 0.0400 \times 25.0 = 0.0100 \times V_2 \Rightarrow V_2 = 100.0\text{ cm}^3\) [1].
Volume of water to be added = \(100.0 - 25.0 = 75.0\text{ cm}^3\) [1].

(c)(i) \(K_\text{a} = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}\) [1].

(c)(ii) \([\text{H}^+] = \sqrt{K_\text{a} \cdot c} = \sqrt{1.78 \times 10^{-4} \times 0.150} = 5.167 \times 10^{-3}\text{ mol dm}^{-3}\) [1].
\(\text{pH} = -\log_{10}(5.167 \times 10^{-3}) = 2.29\) [1].
Assumption: Dissociation of methanoic acid is negligible (so \([\text{HCOOH}]_\text{equilibrium} \approx [\text{HCOOH}]_\text{initial}\)) OR dissociation of water is negligible [1].

(a)(i) Proton donor [1]
(a)(ii) HNO2 is acid and HCOOH is conjugate acid [1]; HCOO- is base and NO2- is conjugate base [1]
(b)(i) Fully ionised/dissociated in water [1]
(b)(ii) 1.40 [1]
(b)(iii) [H+] at pH 2.00 is 0.0100 mol dm-3 [1]; final volume of 100 cm3 [1]; volume of water added is 75.0 cm3 [1]
(c)(i) Ka = [H+][HCOO-] / [HCOOH] [1]
(c)(ii) Correct [H+] calculation [1]; pH = 2.29 [1]; states valid assumption [1]

(a)(i)
- \(\text{Cu}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1\) [1]
- \(\text{Cu}^{2+}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9\) [1]

(a)(ii) A transition element forms at least one stable ion with an incomplete d-subshell. Copper forms \(\text{Cu}^{2+}\) with an incomplete d-subshell (\(3\text{d}^9\)) [1], but zinc only forms \(\text{Zn}^{2+}\) which has a completely filled d-subshell (\(3\text{d}^{10}\)) [1].

(b)(i) Geometry: Octahedral [1]. Principal bond angle: \(90^\circ\) (or \(180^\circ\)) [1].

(b)(ii) Equation: \(\text{[Cu(H}_2\text{O)}_6\text{]}^{2+} + 4\text{Cl}^- \rightleftharpoons \text{[CuCl}_4\text{]}^{2-} + 6\text{H}_2\text{O}\) [1].
Type of reaction: Ligand substitution (or ligand exchange) [1].

(c)
1. Ligands split the five degenerate d-orbitals into two groups of different energy levels [1].
2. d-electrons absorb specific frequencies/wavelengths of visible light [1].
3. This absorption promotes an electron from a lower d-orbital to a higher d-orbital (d-d electron transition) [1].
4. The non-absorbed light (complementary color) is transmitted/observed [1].

(a)(i) Cu configuration (3d10 4s1) [1]; Cu2+ configuration (3d9) [1]
(a)(ii) Identifies transition elements have incomplete d-subshells in stable ions [1]; links to Cu2+ having 3d9 and Zn2+ having 3d10 [1]
(b)(i) Octahedral [1]; 90 degrees [1]
(b)(ii) Correct equation [1]; states ligand substitution [1]
(c) Ligands split d-orbitals [1]; d-electrons absorb energy/visible light [1]; d-d electron transition [1]; complementary color is observed/transmitted [1]

**Step-by-step Solution:**

**(a) Titration Table:**

| Titration | Rough | 1 | 2 |
| :--- | :---: | :---: | :---: |
| Final reading / \(\text{cm}^3\) | 25.60 | 25.00 | 49.90 |
| Initial reading / \(\text{cm}^3\) | 0.00 | 0.00 | 25.00 |
| Titre / \(\text{cm}^3\) | 25.60 | 25.00 | 24.90 |

Concordant titres are within \(0.10\text{ cm}^3\) of each other: Titration 1 (\(25.00\text{ cm}^3\)) and Titration 2 (\(24.90\text{ cm}^3\)).

Average titre = \(\frac{25.00 + 24.90}{2} = 24.95\text{ cm}^3\).

**(b) Moles of sodium thiosulfate:**
\[\text{Moles of } \text{S}_2\text{O}_3^{2-} = 0.100\text{ mol dm}^{-3} \times \frac{24.95}{1000}\text{ dm}^3 = 2.495 \times 10^{-3}\text{ mol}\]

**(c) Ionic equations:**
(i) \(2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} \rightarrow 2\text{CuI(s)} + \text{I}_2\text{(aq)}\)
(ii) \(\text{I}_2\text{(aq)} + 2\text{S}_2\text{O}_3^{2-}\text{(aq)} \rightarrow 2\text{I}^-\text{(aq)} + \text{S}_4\text{O}_6^{2-}\text{(aq)}\)

**(d) Moles of \(\text{Cu}^{2+}\) in \(250.0\text{ cm}^3\):**
From the equations in (c), \(2\text{Cu}^{2+} \equiv \text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-}\).
Therefore, \(1\text{ mol of } \text{Cu}^{2+} \equiv 1\text{ mol of } \text{S}_2\text{O}_3^{2-}\).
Moles of \(\text{Cu}^{2+}\) in \(25.0\text{ cm}^3\) = \(2.495 \times 10^{-3}\text{ mol}\).
Moles of \(\text{Cu}^{2+}\) in \(250.0\text{ cm}^3\) = \(2.495 \times 10^{-3} \times 10 = 2.495 \times 10^{-2}\text{ mol}\).

**(e) Molar mass and value of \(x\):**
\(M_r(\text{CuSO}_4 \cdot x\text{H}_2\text{O}) = \frac{\text{mass}}{\text{moles}} = \frac{6.24\text{ g}}{2.495 \times 10^{-2}\text{ mol}} = 250.1\text{ g mol}^{-1}\).
\(M_r(\text{CuSO}_4) = 63.5 + 32.1 + (16.0 \times 4) = 159.6\text{ g mol}^{-1}\).
Mass of water of crystallization, \(x\text{H}_2\text{O} = 250.1 - 159.6 = 90.5\text{ g mol}^{-1}\).
\(x = \frac{90.5}{18.0} = 5.03 \approx 5\).

**(f) Percentage uncertainty:**
Weighing by difference requires two balance readings, each with an uncertainty of \(\pm 0.01\text{ g}\).
Total uncertainty = \(2 \times 0.01\text{ g} = 0.02\text{ g}\).
Percentage uncertainty = \(\frac{0.02}{6.24} \times 100\% = 0.32\%\).

**Marking Scheme (Total: 13.33 marks)**

- **(a) Titration table & mean titre (4 marks):**
- [1 mark] Headings clearly stated with appropriate units (e.g. \(/\text{cm}^3\)).
- [1 mark] All raw titration readings recorded to \(0.05\text{ cm}^3\).
- [1 mark] Correctly identifies concordant titres (within \(0.10\text{ cm}^3\) of each other) and shows how the mean titre is calculated.
- [1 mark] Correct calculation of average titre: \(24.95\text{ cm}^3\).

- **(b) Moles of sodium thiosulfate (2 marks):**
- [1 mark] Uses \(c \times V / 1000\).
- [1 mark] Correct final answer: \(2.495 \times 10^{-3}\text{ mol}\) (accept \(2.50 \times 10^{-3}\)).

- **(c) Ionic equations (2 marks):**
- [1 mark] Correct equation for reaction (i): \(2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI} + \text{I}_2\).
- [1 mark] Correct equation for reaction (ii): \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\).

- **(d) Moles of \(\text{Cu}^{2+}\) (2 marks):**
- [1 mark] Correctly relates moles to thiosulfate (\(1:1\) ratio) and multiplies by \(10\).
- [1 mark] Correct moles of \(\text{Cu}^{2+}\) in \(250.0\text{ cm}^3\): \(2.495 \times 10^{-2}\text{ mol}\).

- **(e) Molar mass & \(x\) (2.33 marks):**
- [1 mark] Correctly divides mass (6.24) by moles from (d) to find \(M_r \approx 250\text{ g mol}^{-1}\).
- [1 mark] Subtracts 159.6 from molar mass and divides by 18.0 to solve for \(x\).
- [0.33 mark] Correct final integer value: \(x = 5\).

- **(f) Percentage uncertainty (1 mark):**
- [1 mark] \(\frac{0.02}{6.24} \times 100\% = 0.32\%\) (Accept \(0.16\%\) if reasoned that a single weighing was used, but reject if no working is shown).

**Step-by-step Solution:**

**(a) Graph Extrapolation and \(\Delta T\):**
- Plot temperature on the y-axis and time on the x-axis.
- Draw a horizontal line through the initial temperature points from \(t = 0.0\) to \(3.0\text{ min}\) at \(21.0\text{ °C}\).
- Draw a straight line of best fit through the cooling points from \(t = 5.0\) to \(10.0\text{ min}\). The temperature drops by \(0.5\text{ °C}\) per minute.
- Extrapolate this cooling line back to the time of mixing, \(t = 4.0\text{ min}\).
- The extrapolated temperature at \(t = 4.0\text{ min}\) is \(33.5 + 0.5 = 34.0\text{ °C}\).
- Corrected temperature change, \(\Delta T = T_{\text{max}} - T_{\text{initial}} = 34.0 - 21.0 = 13.0\text{ °C}\).

**(b) Heat energy released, \(q\):**
\[\text{Mass of solution } m = 50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g}\]
\[q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 13.0\text{ K} = 2717\text{ J}\]

**(c) Moles of \(\text{CuSO}_4\):**
\[\text{Moles of } \text{CuSO}_4 = 0.500\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0250\text{ mol}\]

**(d) Enthalpy change of reaction, \(\Delta H\):**
\[\Delta H = -\frac{q}{\text{moles}} = -\frac{2717\text{ J}}{0.0250\text{ mol}} = -108,680\text{ J mol}^{-1} = -108.7\text{ kJ mol}^{-1} \approx -109\text{ kJ mol}^{-1}\]
(Since the reaction is exothermic, \(\Delta H\) is negative.)

**(e) Reasons & Improvements:**
- **Reasons:**
1. Polystyrene is an excellent thermal insulator, which minimizes heat loss to the surroundings during the reaction.
2. Polystyrene has a much lower heat capacity than glass, meaning it absorbs less heat from the solution.
- **Improvement:** Place a lid on the polystyrene cup to prevent convective heat loss, or place the cup inside another polystyrene cup / beaker to create a surrounding insulating layer of air.

**Marking Scheme (Total: 13.33 marks)**

- **(a) Graph Extrapolation & \(\Delta T\) (4 marks):**
- [1 mark] Explains drawing a flat line at \(21.0\text{ °C}\) for initial values.
- [1 mark] Explains drawing a line of best fit for points from \(t = 5.0\) to \(10.0\text{ min}\) and extrapolating back to \(t = 4.0\text{ min}\).
- [1 mark] Identifies the extrapolated maximum temperature as \(34.0\text{ °C}\).
- [1 mark] Correctly calculates \(\Delta T = 13.0\text{ °C}\).

- **(b) Heat energy calculation (2 marks):**
- [1 mark] Correctly calculates mass of solution (50.0 g).
- [1 mark] Correctly calculates \(q = 2717\text{ J}\) (accept \(2.72\text{ kJ}\)).

- **(c) Moles of \(\text{CuSO}_4\) (2 marks):**
- [1 mark] Uses volume and concentration correctly.
- [1 mark] Correct final moles: \(0.0250\text{ mol}\).

- **(d) Enthalpy change of reaction (3 marks):**
- [1 mark] Correctly divides \(q\) by moles of \(\text{CuSO}_4\) and converts to \(\text{kJ mol}^{-1}\).
- [1 mark] Applies negative sign (exothermic reaction).
- [1 mark] Gives answer to 3 significant figures: \(-109\text{ kJ mol}^{-1}\) (or \(-108.7\text{ kJ mol}^{-1}\) if using exact value of 2717 J).

- **(e) Discussion of apparatus (2.33 marks):**
- [1 mark] First reason: Polystyrene is a good thermal insulator / reduces heat loss.
- [1 mark] Second reason: Polystyrene has a low heat capacity / absorbs negligible heat.
- [0.33 mark] Suggestion: Use a plastic/styrofoam lid, or double cup.

**Step-by-step Solution:**

**(a) Order with respect to \(\text{S}_2\text{O}_8^{2-}\):**
- Compare Experiment 1 and Experiment 2.
- The volume of \(\text{KI}\), \(\text{Na}_2\text{S}_2\text{O}_3\), and total volume are kept constant. Therefore, the concentration of \(\text{I}^-\) is constant.
- The volume of \(\text{K}_2\text{S}_2\text{O}_8\) is halved (from \(20.0\text{ cm}^3\) to \(10.0\text{ cm}^3\)), which halves its concentration.
- The time taken, \(t\), doubles from \(48\text{ s}\) to \(96\text{ s}\). Since \(\text{rate} \propto 1/t\), the rate is halved.
- Since halving the concentration of \(\text{S}_2\text{O}_8^{2-}\) halves the rate, the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) is **1** (first-order).

**(b) Order with respect to \(\text{I}^-\):**
- Compare Experiment 1 and Experiment 3.
- The volume of \(\text{K}_2\text{S}_2\text{O}_8\), \(\text{Na}_2\text{S}_2\text{O}_3\), and total volume are kept constant. Therefore, the concentration of \(\text{S}_2\text{O}_8^{2-}\) is constant.
- The volume of \(\text{KI}\) is halved (from \(20.0\text{ cm}^3\) to \(10.0\text{ cm}^3\)), which halves its concentration.
- The time taken, \(t\), doubles from \(48\text{ s}\) to \(96\text{ s}\). Since \(\text{rate} \propto 1/t\), the rate is halved.
- Since halving the concentration of \(\text{I}^-\right.\) halves the rate, the order of reaction with respect to \(\text{I}^-\right.\) is **1** (first-order).

**(c) Rate equation:**
\[\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\]
Overall order = \(1 + 1 = 2\).

**(d) Initial concentrations in Experiment 1:**
- \([\text{S}_2\text{O}_8^{2-}]_{\text{initial}} = 0.100\text{ mol dm}^{-3} \times \frac{20.0\text{ cm}^3}{50.0\text{ cm}^3} = 0.0400\text{ mol dm}^{-3}\).
- \([\text{I}^-]_{\text{initial}} = 0.100\text{ mol dm}^{-3} \times \frac{20.0\text{ cm}^3}{50.0\text{ cm}^3} = 0.0400\text{ mol dm}^{-3}\).

**(e) Change in concentration and initial rate:**
- Moles of \(\text{S}_2\text{O}_3^{2-}\) added = \(0.010\text{ mol dm}^{-3} \times \frac{10.0}{1000}\text{ dm}^3 = 1.0 \times 10^{-4}\text{ mol}\).
- From the stoichiometry of the reactions:
\[1\text{ mol of } \text{I}_2 \equiv 2\text{ mol of } \text{S}_2\text{O}_3^{2-}\]
\[1\text{ mol of } \text{S}_2\text{O}_8^{2-} \equiv 1\text{ mol of } \text{I}_2\]
Therefore, \(1\text{ mol of } \text{S}_2\text{O}_8^{2-} \equiv 2\text{ mol of } \text{S}_2\text{O}_3^{2-}\).
- Moles of \(\text{S}_2\text{O}_8^{2-}\) reacted = \(\frac{1.0 \times 10^{-4}}{2} = 5.0 \times 10^{-5}\text{ mol}\).
- Change in concentration, \(\Delta [\text{S}_2\text{O}_8^{2-}] = \frac{5.0 \times 10^{-5}\text{ mol}}{0.0500\text{ dm}^3} = 1.00 \times 10^{-3}\text{ mol dm}^{-3}\).
- Rate = \frac{\Delta [\text{S}_2\text{O}_8^{2-}]}{\Delta t} = \frac{1.00 \times 10^{-3}\text{ mol dm}^{-3}}{48\text{ s}} = 2.08 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\).

**(f) Why total volume is kept constant:**
- To ensure that the volume of any added reactant is directly proportional to its concentration in the final mixture. This allows direct comparison between the volumes of reactants used and their relative concentrations.

**Marking Scheme (Total: 13.33 marks)**

- **(a) Order with respect to peroxodisulfate (2.5 marks):**
- [1 mark] States comparison between Exp 1 and Exp 2.
- [1 mark] Mentions halving \([\text{S}_2\text{O}_8^{2-}]\) while keeping other concentrations constant halves the rate (time doubles).
- [0.5 mark] Correctly concludes order is 1.

- **(b) Order with respect to iodide (2.5 marks):**
- [1 mark] States comparison between Exp 1 and Exp 3.
- [1 mark] Mentions halving \([\text{I}^-]\) while keeping other concentrations constant halves the rate (time doubles).
- [0.5 mark] Correctly concludes order is 1.

- **(c) Rate equation & overall order (1.5 marks):**
- [1 mark] Correct rate equation: \(\text{Rate} = k [\text{S}_2\text{O}_8^{2-}][\text{I}^-]\).
- [0.5 mark] Correct overall order: 2.

- **(d) Initial concentrations (2 marks):**
- [1 mark] Correctly applies dilution factor \(20.0 / 50.0 = 0.4\) to concentrations.
- [1 mark] Correctly obtains \(0.0400\text{ mol dm}^{-3}\) for both species.

- **(e) Concentration change & rate (3.33 marks):**
- [1 mark] Calculates moles of \(\text{S}_2\text{O}_3^{2-}\) correctly as \(1.0 \times 10^{-4}\text{ mol}\).
- [1 mark] Correctly uses stoichiometry (\(1:2\) ratio) to find moles of \(\text{S}_2\text{O}_8^{2-}\) reacted as \(5.0 \times 10^{-5}\text{ mol}\).
- [1 mark] Calculates concentration change of \(\text{S}_2\text{O}_8^{2-}\) correctly as \(1.00 \times 10^{-3}\text{ mol dm}^{-3}\).
- [0.33 mark] Correct rate value with units: \(2.08 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\).

- **(f) Volume constant explanation (1.5 marks):**
- [1.5 marks] Explains that keeping total volume constant ensures that the volume of a reactant is directly proportional to its initial concentration in the reaction mixture (so that change in volume directly reflects the change in concentration).

(a) Stability constant is the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules. (b) \(K_{\text{stab}} = \frac{[[Cu(NH_3)_4(H_2O)_2]^{2+}]}{[[Cu(H_2O)_6]^{2+}][NH_3]^4}\). (c) When ethylenediamine (a bidentate ligand) replaces monodentate water ligands, there is an increase in the number of particles in solution: \([Cu(H_2O)_6]^{2+} + 2\text{en} \rightleftharpoons [Cu(\text{en})_2(H_2O)_2]^{2+} + 4H_2O\) (3 particles become 5). This leads to an increase in disorder and a positive entropy change (\(\Delta S^\theta > 0\)). Since \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), the positive entropy change makes \(\Delta G^\theta\) more negative, increasing the stability of the complex (the chelate effect). (d) Let \([Cu^{2+}]_{\text{free}} = x\). Since \(K_{\text{stab}}\) is very large, almost all \(Cu^{2+}\) complexes with \(NH_3\). Thus, \([[Cu(NH_3)_4]^{2+}] \approx 0.010\text{ mol dm}^{-3}\). The concentration of remaining free \(NH_3 = 1.00 - (4 \times 0.010) = 0.96\text{ mol dm}^{-3}\). Substitute into the expression: \(1.2 \times 10^{13} = \frac{0.010}{x \times (0.96)^4}\). \(x = \frac{0.010}{1.2 \times 10^{13} \times 0.8493} = 9.81 \times 10^{-16}\text{ mol dm}^{-3}\).

(a) 2 marks: 1 mark for equilibrium constant of complex formation, 1 mark for stating it is from constituent ions/molecules in solvent. (b) 1 mark: Correct expression including charges and concentration of reactants and products. (c) 3 marks: 1 mark for stating increase in number of particles, 1 mark for stating increase in entropy/disorder, 1 mark for linking to \(\Delta G^\theta\) becoming more negative. (d) 5.11 marks: 1 mark for establishing \([[Cu(NH_3)_4]^{2+}] \approx 0.010\text{ mol dm}^{-3}\), 1.11 marks for calculating remaining \([NH_3] = 0.96\text{ mol dm}^{-3}\), 2 marks for setting up the equation, 1 mark for correct calculation of final concentration with units.

(a) Step 1: \(CH_3Cl\) and anhydrous \(AlCl_3\) (catalyst), warm. Step 2: Alkaline \(KMnO_4\) heated under reflux, followed by dilute acid (e.g., \(HCl\) or \(H_2SO_4\)). Step 3: \(Br_2\) and halogen carrier catalyst (e.g., \(FeBr_3\) or \(AlBr_3\)), dry/warm. (b) The methyl group (\(-CH_3\)) is electron-donating by positive inductive effect, increasing electron density at positions 2 and 4, making them more reactive (2,4-directing). The carboxylic acid group (\(-COOH\)) is strongly electron-withdrawing, decreasing electron density across the ring, but making position 3 relatively the most electron-rich/reactive (3-directing). (c) Electrophile generation: \(Br_2 + FeBr_3 \rightarrow Br^+ + FeBr_4^-\). Curly arrow from benzene ring to \(Br^+\) electrophile to form the arenium carbocation intermediate (with correct horseshoe charge representation). Curly arrow from \(C-H\) bond at position 3 back into the ring to restore aromaticity and release \(H^+\).

(a) 4 marks: 1 mark for Step 1 reagents and catalyst, 1.5 marks for Step 2 reagents and acid workup, 1.5 marks for Step 3 reagents and catalyst. (b) 3.11 marks: 1.51 marks for explaining the activating/2,4-directing effect of the methyl group, 1.6 marks for explaining the deactivating/3-directing effect of the carboxyl group. (c) 4 marks: 1 mark for electrophile generation, 1 mark for curly arrow from ring to \(Br^+\), 1 mark for correct arenium intermediate structure, 1 mark for curly arrow restoring ring stability.

(a) \(Pt(s) | Fe^{2+}(aq), Fe^{3+}(aq) || Ag^{+}(aq) | Ag(s)\). (b) \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.80\text{ V} - (+0.77\text{ V}) = +0.03\text{ V}\). (c) Using the Nernst equation for each half-cell: \(E(Fe^{3+}/Fe^{2+}) = 0.77 + 0.059 \log\frac{[Fe^{3+}]}{[Fe^{2+}]} = 0.77 + 0.059 \log\frac{0.050}{0.80} = 0.77 + 0.059(-1.204) = 0.77 - 0.071 = 0.699\text{ V}\). \(E(Ag^{+}/Ag) = 0.80 + 0.059 \log[Ag^+] = 0.80 + 0.059 \log(0.010) = 0.80 + 0.059(-2) = 0.80 - 0.118 = 0.682\text{ V}\). \(E_{\text{cell}} = E(Ag^{+}/Ag) - E(Fe^{3+}/Fe^{2+}) = 0.682 - 0.699 = -0.017\text{ V}\). (d) Since \(E_{\text{cell}}\) is negative, the cell reaction is not spontaneous in the forward direction. The spontaneous reaction is the reverse process: \(Fe^{3+}(aq) + Ag(s) \rightarrow Fe^{2+}(aq) + Ag^+(aq)\).

(a) 2 marks: 1 mark for correct reactants and phases in each half-cell, 1 mark for correct platinum electrode and cell structure. (b) 1 mark: Correct standard cell potential. (c) 5.11 marks: 2 marks for calculating non-standard \(E(Fe^{3+}/Fe^{2+}) = 0.699\text{ V}\), 2 marks for calculating non-standard \(E(Ag^+/Ag) = 0.682\text{ V}\), 1.11 marks for final difference of \(-0.017\text{ V}\). (d) 3 marks: 1 mark for identifying that the potential is negative, 2 marks for writing the spontaneous direction of the chemical reaction.

(a) Doubling \([CH_3COCH_3]\) doubles rate \(\rightarrow\) order = 1. Doubling \([I_2]\) does not change rate \(\rightarrow\) order = 0. Doubling \([H^+]\) doubles rate \(\rightarrow\) order = 1. Rate Equation: \(\text{Rate} = k[CH_3COCH_3][H^+]\). (b) \(k = \frac{\text{Rate}}{[CH_3COCH_3][H^+]} = \frac{1.2 \times 10^{-5}}{0.20 \times 0.050} = 1.2 \times 10^{-3}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\). (c) Step 1 (fast): \(CH_3COCH_3 + H^+ \rightleftharpoons [CH_3C(OH)CH_3]^+\). Step 2 (slow, rate-determining step): \([CH_3C(OH)CH_3]^+ + H_2O \rightarrow CH_3C(OH)=CH_2 + H_3O^+\) (enol formation). Step 3 (fast): \(CH_3C(OH)=CH_2 + I_2 \rightarrow CH_3COCH_2I + H^+ + I^-\). Since \(I_2\) is involved only after the rate-determining step (Step 2), it is zero order. Propanone and \(H^+\) are both involved up to and including the rate-determining step, which matches the overall rate equation.

(a) 3 marks: 1 mark for orders of reactants, 1 mark for order of catalyst, 1 mark for correct overall rate equation. (b) 3.11 marks: 2 marks for calculation of \(k\), 1.11 marks for correct units (\(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)). (c) 5 marks: 2 marks for writing steps where propanone and hydrogen ions react, 1 mark for identifying the slow step, 1 mark for including a fast step involving iodine, 1 mark for explanation relating steps to the rate equation.

(a) A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. (b) Moles of propanoic acid, \(HA = 0.0500 \times 0.150 = 7.50 \times 10^{-3}\text{ mol}\). Moles of propanoate, \(A^- = 0.0300 \times 0.200 = 6.00 \times 10^{-3}\text{ mol}\). \(pH = pK_a + \log\frac{[A^-]}{[HA]} = -\log(1.35 \times 10^{-5}) + \log\frac{6.00 \times 10^{-3}}{7.50 \times 10^{-3}} = 4.870 - 0.097 = 4.77\). (c) Moles of added \(H^+ = 0.0020 \times 0.10 = 2.0 \times 10^{-4}\text{ mol}\). The added \(H^+\) reacts with propanoate ions: \(A^- + H^+ \rightarrow HA\). New moles of \(A^- = 6.00 \times 10^{-3} - 2.0 \times 10^{-4} = 5.80 \times 10^{-3}\text{ mol}\). New moles of \(HA = 7.50 \times 10^{-3} + 2.0 \times 10^{-4} = 7.70 \times 10^{-3}\text{ mol}\). New \(pH = 4.870 + \log\frac{5.80 \times 10^{-3}}{7.70 \times 10^{-3}} = 4.870 - 0.123 = 4.75\). pH change = \(4.75 - 4.77 = -0.02\).

(a) 2 marks: 1 mark for resisting pH change, 1 mark for adding small amounts of acid or base. (b) 4.11 marks: 2 marks for moles of components, 1 mark for Henderson-Hasselbalch substitution, 1.11 marks for correct pH. (c) 5 marks: 1 mark for moles of added \(H^+\), 2 marks for updating moles of acid and salt, 1 mark for calculating new pH, 1 mark for correct pH change.

(a) Lattice energy is the enthalpy change when 1 mole of an ionic crystal lattice is formed from its constituent gaseous ions under standard conditions. (b) The Born-Haber cycle should show: \(Mg(s) + Cl_2(g)\) at baseline, rising to \(Mg(g) + Cl_2(g)\) (atomisation of Mg), rising to \(Mg^+(g) + Cl_2(g) + e^-\) (1st IE of Mg), rising to \(Mg^{2+}(g) + Cl_2(g) + 2e^-\) (2nd IE of Mg), rising to \(Mg^{2+}(g) + 2Cl(g) + 2e^-\) (atomisation of chlorine, note: \(2 \times 121 = 242\text{ kJ mol}^{-1}\)), falling to \(Mg^{2+}(g) + 2Cl^-(g)\) (electron affinity of chlorine, note: \(2 \times -349 = -698\text{ kJ mol}^{-1}\)), falling to \(MgCl_2(s)\) (lattice energy), with the direct arrow from \(Mg(s) + Cl_2(g)\) to \(MgCl_2(s)\) representing formation (\(-641\text{ kJ mol}^{-1}\)). (c) Using Hess's law: \(\Delta H^\theta_f = \Delta H^\theta_{at}(Mg) + IE_1(Mg) + IE_2(Mg) + 2 \times \Delta H^\theta_{at}(Cl) + 2 \times EA(Cl) + \Delta H^\theta_{latt}\). \(-641 = 148 + 738 + 1450 + 242 - 698 + \Delta H^\theta_{latt}\). \(-641 = 1880 + \Delta H^\theta_{latt}\). \(\Delta H^\theta_{latt} = -2521\text{ kJ mol}^{-1}\).

(a) 2 marks: 1 mark for formation of solid ionic lattice from gaseous ions, 1 mark for 1 mole under standard conditions. (b) 3 marks: 1 mark for correct upward arrows and species labels, 1 mark for downward arrows (EA and lattice energy) and labels, 1 mark for correct stoichiometric factors (e.g. 2 Cl). (c) 6.11 marks: 2 marks for setting up Hess's Law equation, 1.11 marks for using \(2 \times \Delta H^\theta_{at}(Cl)\) and \(2 \times EA(Cl)\), 2 marks for arithmetic, 1 mark for correct units (\(\text{kJ mol}^{-1}\)) and negative sign.

(a) Zwitterion structure: \(CH_3CH(NH_3^+)COO^-\). At pH 1: \(CH_3CH(NH_3^+)COOH\). At pH 12: \(CH_3CH(NH_2)COO^-\). (b) The isoelectric point is the pH value at which an amino acid has no net electric charge and exists primarily in its zwitterionic form. (c) At a buffered pH of 6.0: Alanine is at its isoelectric point, so its net charge is zero. It will remain stationary/not migrate. Lysine (isoelectric point 9.7) is in a medium below its isoelectric point, meaning it gains protons and has a net positive charge. It will migrate towards the negative electrode (cathode). Aspartic acid (isoelectric point 2.8) is in a medium above its isoelectric point, meaning it loses protons and has a net negative charge. It will migrate towards the positive electrode (anode).

(a) 3 marks: 1 mark for each correct structure showing precise ionization states of the amine and carboxylic groups. (b) 2 marks: 1 mark for the pH of no net electrical charge, 1 mark for mentioning zwitterion dominance. (c) 6.11 marks: 2 marks for describing alanine behavior (zero net charge, stationary), 2 marks for describing lysine behavior (net positive charge, cathode migration), 2.11 marks for describing aspartic acid behavior (net negative charge, anode migration).

(a) \(n(MnO_4^-) = 0.0250\text{ dm}^3 \times 0.00400\text{ mol dm}^{-3} = 1.00 \times 10^{-4}\text{ mol}\). The redox stoichiometry is \(5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O\). \(n(Fe^{2+}) = 5 \times n(MnO_4^-) = 5.00 \times 10^{-4}\text{ mol}\). (b) \(M_r(BaSO_4) = 137.3 + 32.1 + (4 \times 16.0) = 233.4\text{ g mol}^{-1}\). \(n(SO_4^{2-}) = n(BaSO_4) = \frac{0.233}{233.4} = 1.00 \times 10^{-3}\text{ mol}\). (c) Ratio of \(Fe^{2+} : SO_4^{2-} = 5.00 \times 10^{-4} : 1.00 \times 10^{-3} = 1 : 2\). Thus, \(y = 1\) and \(z = 2\). To balance the electrical charge: \(2 \times SO_4^{2-}\) carries a \(-4\) charge. \(1 \times Fe^{2+}\) carries a \(+2\) charge. Therefore, \(2 \times NH_4^+\) is needed to balance the remaining charge, giving \(x = 2\). (d) Total moles of salt in \(250\text{ cm}^3 = 10 \times 5.00 \times 10^{-4} = 5.00 \times 10^{-3}\text{ mol}\). \(M_r\) of hydrated salt = \frac{1.96}{5.00 \times 10^{-3}} = 392\text{ g mol}^{-1}\). \(M_r\) of anhydrous \((NH_4)_2Fe(SO_4)_2 = (2 \times 18.0) + 55.8 + (2 \times 96.1) = 36.0 + 55.8 + 192.2 = 284\text{ g mol}^{-1}\). Mass of water of crystallisation per mole = \(392 - 284 = 108\text{ g mol}^{-1}\). \(w = \frac{108}{18.0} = 6\).

(a) 2 marks: 1 mark for moles of manganate, 1 mark for moles of iron(II) using 5:1 stoichiometry. (b) 2 marks: 1 mark for molar mass of \(BaSO_4\), 1 mark for calculating correct moles of sulfate. (c) 4.11 marks: 1.11 marks for finding the 1:2 ratio, 1 mark for \(y = 1\), 1 mark for \(z = 2\), 1 mark for using electrical neutrality to find \(x = 2\). (d) 3 marks: 1 mark for total moles of salt in flask, 1 mark for calculating the total \(M_r\) as 392, 1 mark for correctly determining \(w = 6\).

(a) The standard electrode potential is the electromotive force (emf) of a half-cell coupled to a standard hydrogen electrode (SHE) measured under standard conditions of 298 K, 1 atm pressure, and 1.00 mol dm^-3 concentration of all solutions. (b) The standard cell potential is calculated as: \( E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.34 - (-0.76) = +1.10 \text{ V} \). The overall ionic equation is: \( \text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)} \). (c) Using the Nernst equation: \( E = E^\theta + \frac{0.059}{z}\log[\text{Cu}^{2+}] \). Substituting the values: \( E = +0.34 + \frac{0.059}{2}\log(0.0050) = +0.34 + (0.0295 \times -2.301) = +0.34 - 0.0679 = +0.272 \text{ V} \) (or +0.27 V). (d) The non-standard cell potential is: \( E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = +0.272 - (-0.76) = +1.032 \text{ V} \) (or +1.03 V). (e) Adding concentrated aqueous ammonia to the copper half-cell: (i) The concentration of free \( \text{Cu}^{2+}(\text{aq}) \) ions decreases significantly because \( \text{Cu}^{2+} \) reacts with \( \text{NH}_3 \) to form the stable complex ion \( [\text{Cu(NH}_3)_4]^{2+}(\text{aq}) \). (ii) The decrease in \( [\text{Cu}^{2+}(\text{aq})] \) shifts the equilibrium position of \( \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu(s)} \) to the left, causing the electrode potential of the copper half-cell to decrease (become less positive / more negative).

Part (a): [2 marks] 1 mark for mentioning 'potential difference/emf compared to a standard hydrogen electrode'. 1 mark for specifying 'standard conditions: 298 K, 1 atm (101 kPa), and 1.00 mol dm^-3 concentration'. Part (b): [2 marks] 1 mark for \( E^\theta_{\text{cell}} = +1.10 \text{ V} \). 1 mark for correct equation: \( \text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)} \) (state symbols not strictly required but must be balanced). Part (c): [3 marks] 1 mark for correct substitution of values into the equation: \( E = +0.34 + \frac{0.059}{2}\log(0.0050) \). 1 mark for correct log term calculation: \( \log(0.0050) = -2.301 \) (or equivalent intermediate). 1 mark for the final correct value: \( +0.27 \text{ V} \) or \( +0.272 \text{ V} \). Part (d): [1 mark] 1 mark for correct deduction of non-standard cell potential: \( E_{\text{cell}} = +0.272 - (-0.76) = +1.03 \text{ V} \) (accept +1.03 V to +1.04 V depending on rounding of part c). Part (e): [3 marks] 1 mark for stating that \( [\text{Cu}^{2+}] \) decreases due to the formation of the complex ion \( [\text{Cu(NH}_3)_4]^{2+} \). 1 mark for stating that the equilibrium \( \text{Cu}^{2+} + 2\text{e}^- \rightleftharpoons \text{Cu} \) shifts to the left. 1 mark for stating that the electrode potential decreases / becomes less positive.

(a)
- Moles of \(HX\) required: \(n = C \times V = 0.0500 \text{ mol dm}^{-3} \times 0.2500 \text{ dm}^3 = 0.0125 \text{ mol}\)
- Mass required: \(m = 0.0125 \text{ mol} \times 122.1 \text{ g mol}^{-1} = 1.53 \text{ g}\) (or \(1.526 \text{ g}\))
- Practical steps:
1. Weigh \(1.53 \text{ g}\) of \(HX\) in a weighing boat / beaker using a balance accurate to 2 decimal places.
2. Dissolve the solid in approximately \(100 \text{ cm}^3\) of distilled water in a beaker, stirring with a glass rod.
3. Quantitatively transfer the solution (including the washings of the beaker and stirring rod) to a \(250.0 \text{ cm}^3\) volumetric flask.
4. Add distilled water up to the graduation mark (using a dropper near the line so the bottom of the meniscus touches the line), stopper, and invert several times to mix.

(b) Table calculation values:
- Mixture 1: \(\frac{2.0}{50.0 - 4.0} = 0.0435\), \(\log_{10}(0.0435) = -1.36\)
- Mixture 2: \(\frac{5.0}{50.0 - 10.0} = 0.125\), \(\log_{10}(0.125) = -0.90\)
- Mixture 3: \(\frac{8.0}{50.0 - 16.0} = 0.235\), \(\log_{10}(0.235) = -0.63\)
- Mixture 4: \(\frac{11.0}{50.0 - 22.0} = 0.393\), \(\log_{10}(0.393) = -0.41\)
- Mixture 5: \(\frac{14.0}{50.0 - 28.0} = 0.636\), \(\log_{10}(0.636) = -0.20\)
- Mixture 6: \(\frac{17.0}{50.0 - 34.0} = 1.063\), \(\log_{10}(1.063) = +0.03\)
- Mixture 7: \(\frac{20.0}{50.0 - 40.0} = 2.000\), \(\log_{10}(2.000) = +0.30\)
- Mixture 8: \(\frac{23.0}{50.0 - 46.0} = 5.750\), \(\log_{10}(5.750) = +0.76\)

(c)
- Plot \(\text{pH}\) on the y-axis against \(\log_{10} \left( \frac{V}{50.0 - 2V} \right)\) on the x-axis.
- Draw a straight line of best fit. The \(\text{p}K_a\) value is read directly from the y-intercept (the point where the x-axis value is 0).
- Mixture 4 is the anomalous point. The measured pH (4.15) is lower than the expected value from the linear trend.
- Possible practical mistake: The student may have added less than the required \(11.0 \text{ cm}^3\) of \(NaOH\) (e.g. undershooting the burette reading), resulting in more unneutralised \(HX\) and thus a lower, more acidic pH.

(d)
- (i) Adding \(H^+\): \(X^-(aq) + H^+(aq) \rightarrow HX(aq)\)
- (ii) Adding \(OH^-\): \(HX(aq) + OH^-(aq) \rightarrow X^-(aq) + H_2O(l)\)

(e)
- Each reading of a burette has an uncertainty of \(\pm 0.05 \text{ cm}^3\).
- Since a delivered volume is calculated from two readings (initial and final), the overall uncertainty in the volume delivered is \(2 \times 0.05 = 0.10 \text{ cm}^3\).
- Maximum percentage error = \(\frac{0.10}{2.0} \times 100\% = 5.0\%\).

- (a) Mass calculation (1.53 g): 1 mark
- (a) Weighing and dissolving steps: 1 mark
- (a) Quantitative transfer with washings: 1 mark
- (a) Making up to the mark in a 250 cm3 volumetric flask and mixing: 1 mark
- (b) All ratio column values calculated correctly: 1 mark
- (b) All log10 column values calculated to 2 d.p. correctly: 1 mark
- (b) General precision and consistent decimal formatting of the table: 1 mark
- (c) Plotting description (pH on y-axis, log ratio on x-axis): 1 mark
- (c) Identifying y-intercept as pKa: 1 mark
- (c) Identification of Mixture 4 as anomalous: 1 mark
- (c) Plausible error for Mixture 4 (e.g. too little NaOH added): 1 mark
- (d) Correct ionic equation for added H+: 1 mark
- (d) Correct ionic equation for added OH-: 1 mark
- (e) Correct uncertainty calculation (0.10 cm3) and percentage error (5.0%): 2 marks (Award 1 mark if 2.5% is calculated based on a single reading error of 0.05 cm3)

(a)
- Moles of \(NaL\) needed: \(n = C \times V = 0.0200 \text{ mol dm}^{-3} \times 0.1000 \text{ dm}^3 = 0.00200 \text{ mol}\)
- Mass required: \(m = 0.00200 \text{ mol} \times 144.0 \text{ g mol}^{-1} = 0.288 \text{ g}\)
- Preparation: Weigh \(0.288 \text{ g}\) of \(NaL\) using a balance. Dissolve in a beaker with a small volume of distilled water. Transfer quantitatively with washings into a \(100.0 \text{ cm}^3\) volumetric flask (key apparatus). Add distilled water up to the mark and invert to mix.

(b)
- Colorimeter is more quantitative and objective than visual inspection, which is subjective and susceptible to human error/fatigue.
- Calibration/Configuration: Place distilled water (or a blank solution) in a cuvette, insert into the colorimeter, and calibrate to 0 absorbance. Set the colorimeter filter to green (or choose wavelength near \(540 \text{ nm}\)) because the purple/red complex absorbs its complementary color (green) most strongly.

(c)
(i)
- Independent variable: Volume of ligand, \(V_L\) (or mole fraction of ligand, \(x_L\)).
- Dependent variable: Absorbance (at \(540 \text{ nm}\)).

(ii)
- For Tube 8: \(x_L = \frac{7.5}{2.5 + 7.5} = 0.750\)
- For Tube 10: \(x_L = \frac{9.0}{1.0 + 9.0} = 0.900\)

(iii)
- Sketch: The graph should show two straight lines intersecting at a peak. The line rises from \(x_L = 0\) (absorbance = 0) to the peak, then falls from the peak back to \(x_L = 1.0\) (absorbance = 0).
- Explanation: The intersection point of the two straight lines of best fit corresponds to the exact stoichiometric ratio of the reactants.
- Calculation: The peak occurs at \(x_L = 0.75\).
- Ratio of \(L^- : M^{2+} = \frac{x_L}{1 - x_L} = \frac{0.75}{0.25} = 3\). Hence, \(n = 3\).

(iv)
- Anomalous tube: Tube 9.
- Explanation: The measured absorbance is \(0.62\), but the expected value (from the limiting metal ion line) is \(0.72\).
- Practical error: The cuvette used for Tube 9 may have had droplets of distilled water inside before adding the mixture, diluting the complex and lowering the measured absorbance.

- (a) Mass calculation (0.288 g): 1 mark
- (a) Quantitative transfer and dilution steps described: 1 mark
- (a) Use of a 100 cm3 volumetric flask: 1 mark
- (b) Quantitative/objective reason for using colorimeter: 1 mark
- (b) Calibration with blank (distilled water) and selection of green filter/540 nm: 1 mark
- (c)(i) Both independent and dependent variables correctly identified: 1 mark
- (c)(ii) Correct mole fractions for Tube 8 (0.75) and Tube 10 (0.90): 1 mark
- (c)(iii) Graph sketch with rising and falling straight lines: 1 mark
- (c)(iii) Explaining that the intersection represents the stoichiometric ratio: 1 mark
- (c)(iii) Correct calculation of n = 3: 2 marks
- (c)(iv) Identifying Tube 9 as anomalous: 1 mark
- (c)(iv) Suggesting a valid practical error (e.g., dilution of mixture due to water droplets inside the cuvette): 2 marks