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1. Find the oxidation state of sulfur in the thiosulfate ion, \(\text{S}_2\text{O}_3^{2-}\):
\(2(S) + 3(-2) = -2 \Rightarrow 2(S) = +4 \Rightarrow S = +2\).

2. Find the oxidation state of sulfur in the tetrathionate ion, \(\text{S}_4\text{O}_6^{2-}\):
\(4(S) + 6(-2) = -2 \Rightarrow 4(S) = +10 \Rightarrow S = +2.5\).

Therefore, the change in the average oxidation state of sulfur is from +2 to +2.5.

[1] A - from +2 to +2.5.
Award 1 mark for the correct option. Reject all other options.

Benzene-1,4-dicarboxylic acid (terephthalic acid) is highly symmetrical:
- The two carboxylic carbon atoms (\(-\text{COOH}\)) are equivalent (1 peak).
- The two aromatic carbon atoms attached directly to the carboxylic groups are equivalent (1 peak).
- The remaining four aromatic carbon atoms bonded to hydrogen are equivalent (1 peak).
This gives a total of 3 peaks in the \(^{13}\text{C}\) NMR spectrum.

[1] B - 3 peaks.
Award 1 mark for the correct answer. Reject all other options.

1. Calculate the moles of \(\text{AgCl}\) precipitate:
\(n(\text{AgCl}) = \frac{2.87\text{ g}}{143.4\text{ g mol}^{-1}} = 0.0200\text{ mol}\).

2. Deduce moles of \(\text{NaCl}\):
Since \(1\text{ mol}\) of \(\text{NaCl}\) yields \(1\text{ mol}\) of \(\text{AgCl}\), \(n(\text{NaCl}) = 0.0200\text{ mol}\).

3. Calculate mass of \(\text{NaCl}\):
\(m(\text{NaCl}) = 0.0200\text{ mol} \times 58.5\text{ g mol}^{-1} = 1.17\text{ g}\).

4. Calculate percentage by mass:
\(\% = \left(\frac{1.17\text{ g}}{2.00\text{ g}}\right) \times 100 = 58.5\%\).

[1] B - 58.5%.
Award 1 mark for the correct calculation path leading to the percentage. Reject other percentages.

1. Convert wavelength to meters:
\(\lambda = 580 \times 10^{-9}\text{ m}\).

2. Calculate the energy of a single photon:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{580 \times 10^{-9}\text{ m}} = 3.4293 \times 10^{-19}\text{ J}\).

3. Calculate energy per mole in kJ mol\(^{-1}\):
\(E_{\text{mol}} = 3.4293 \times 10^{-19}\text{ J} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 2.064 \times 10^5 \text{ J mol}^{-1} = 206\text{ kJ mol}^{-1}\).

[1] B - 206 kJ mol\(^{-1}\).
Award 1 mark for correct use of Planck's equation and molar conversion.

Magnesium ions (\(\text{Mg}^{2+}\)) have a much smaller ionic radius than barium ions (\(\text{Ba}^{2+}\)) whilst carrying the same 2+ charge. Consequently, the magnesium ion has a higher charge density. This allows it to polarise (distort) the carbonate ion's electron cloud more strongly, which weakens the carbon-oxygen bonds in the carbonate anion, making it easier to break apart on heating.

[1] B - Magnesium ions have a higher charge density and polarise the carbonate ion more.
Award 1 mark for selecting the option relating ionic radius/charge density and polarizing power.

1. Calculate the \(\text{p}K_a\) of ethanoic acid:
\(\text{p}K_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74\).

2. Use the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\)
\(\text{pH} = 4.74 + \log_{10}\left(\frac{0.200}{0.100}\right)\)
\(\text{pH} = 4.74 + \log_{10}(2) = 4.74 + 0.30 = 5.04\).

[1] C - 5.04.
Award 1 mark for correct calculation of pH using the buffer equation.

A redox reaction is thermodynamically feasible if the standard cell potential, \(E^\theta_{\text{cell}}\), is positive (\(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} > 0\)).

- Option C: \(\text{Fe}^{3+}\) acts as the oxidising agent (reduced) and \(\text{I}^-\) acts as the reducing agent (oxidised).
\(E^\theta_{\text{cell}} = E^\theta(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\theta(\text{I}_2/\text{I}^-) = +0.77 - (+0.54) = +0.23\text{ V}\).
Since \(E^\theta_{\text{cell}} > 0\), this reaction is feasible.

[1] C - Fe^{3+}(aq) oxidising I^{-}(aq).
Award 1 mark for identifying the correct feasible combination of half-cells.

1. Find the energy required to break one \(\text{O=O}\) bond:
\(E = \frac{498 \times 10^3\text{ J mol}^{-1}}{6.02 \times 10^{23}\text{ mol}^{-1}} = 8.272 \times 10^{-19}\text{ J}\).

2. Use the relation \(E = \frac{hc}{\lambda}\) to find wavelength:
\(\lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{8.272 \times 10^{-19}\text{ J}} = 2.404 \times 10^{-7}\text{ m}\).

3. Convert meters to nanometers:
\(\lambda = 2.404 \times 10^{-7}\text{ m} \times 10^9 = 240\text{ nm}\).

[1] A - 240 nm.
Award 1 mark for correct use of energy conversion per molecule and the wave equation to get 240 nm.

1. Calculate the amount in moles of thiosulfate used:
\(n(S_2O_3^{2-}) = 0.0225\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.25 \times 10^{-3}\text{ mol}\)

2. From the second equation, \(1\text{ mol}\) of \(I_2\) reacts with \(2\text{ mol}\) of \(S_2O_3^{2-}\):
\(n(I_2) = \frac{1}{2} \times 2.25 \times 10^{-3}\text{ mol} = 1.125 \times 10^{-3}\text{ mol}\)

3. From the first equation, \(1\text{ mol}\) of \(IO_3^-\)\ produces \(3\text{ mol}\) of \(I_2\):
\(n(IO_3^-) = \frac{1}{3} \times 1.125 \times 10^{-3}\text{ mol} = 3.75 \times 10^{-4}\text{ mol}\)

4. Calculate the concentration of \(IO_3^-\)\ in the \(10.0\text{ cm}^3\) sample:
\(c(IO_3^-) = \frac{3.75 \times 10^{-4}\text{ mol}}{0.0100\text{ dm}^3} = 0.0375\text{ mol dm}^{-3}\)

[1] Award 1 mark for the correct option (A).
[0] For any other answer.

Notes: Common errors include failing to apply the correct stoichiometry, leading to distractors B, C, or D.

To show only two singlets with relative areas of 3:2:
- The molecule must contain two types of proton environments (highly symmetrical structure).
- The ratio of the number of protons in these two environments must simplify to 3:2.

Let us analyze Dimethyl terephthalate (dimethyl benzene-1,4-dicarboxylate):
- The six methyl ester protons (\(2 \times -OCH_3\)) are equivalent and have no neighboring protons, so they yield a singlet of area 6.
- The four aromatic ring protons are chemically equivalent due to the para-symmetry of the benzene ring. They do not couple with each other because they are equivalent, yielding a singlet of area 4.
- The ratio of the areas of these two singlets is \(6 : 4 = 3 : 2\).

Other options, such as dimethyl glutarate or adipate, contain multiple methylene groups in different environments, resulting in more than two proton signals with complex splitting patterns.

[1] Award 1 mark for the correct option (A).
[0] For any other answer.

1. Determine the amount of \(Cl_2\) in moles:
\(n(Cl_2) = \frac{1.50\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.0625\text{ mol}\)

2. Use the stoichiometric ratio from the balanced equation:
\(1\text{ mol}\) of \(MnO_2\) reacts to produce \(1\text{ mol}\) of \(Cl_2\).
Therefore, \(n(MnO_2) = 0.0625\text{ mol}\).

3. Calculate the mass of \(MnO_2\):
\(m(MnO_2) = 0.0625\text{ mol} \times 86.9\text{ g mol}^{-1} = 5.43\text{ g}\)

[1] Award 1 mark for the correct option (C).
[0] For any other answer.

1. Calculate the energy absorbed by one photon:
\(\Delta E = h \nu = 6.63 \times 10^{-34}\text{ J s} \times 6.10 \times 10^{14}\text{ s}^{-1} = 4.0443 \times 10^{-19}\text{ J}\)

2. Calculate the energy change per mole of molecules:
\(\Delta E_{\text{mole}} = 4.0443 \times 10^{-19}\text{ J} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 2.4347 \times 10^5\text{ J mol}^{-1}\)

3. Convert to kilojoules per mole:
\(\Delta E_{\text{mole}} = 243\text{ kJ mol}^{-1}\)

[1] Award 1 mark for the correct option (C).
[0] For any other answer.

Let's review the trends down Group 2:
- A: Carbonate thermal stability increases down the group because larger cations have a lower charge density, making them less able to polarize (and weaken the C-O bonds of) the carbonate anion.
- B: Hydroxide solubility increases down the group. Both lattice enthalpy and cation hydration enthalpy become less exothermic down the group, but the lattice enthalpy decreases more rapidly than the hydration enthalpy of the cations due to the relatively small size of the hydroxide ion. This is correct.
- C: First ionization energy decreases down the group due to increased shielding and atomic radius.
- D: Reactivity with water increases down the group as outer electrons are more easily lost.

[1] Award 1 mark for the correct option (B).
[0] For any other answer.

1. Write the reaction equation:
\(HA + OH^- \rightarrow A^- + H_2O\)

2. Calculate the initial amounts in moles:
\(n(HA)_{\text{initial}} = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol}\)
\(n(OH^-)_{\text{initial}} = 0.0500\text{ dm}^3 \times 0.050\text{ mol dm}^{-3} = 0.00250\text{ mol}\)

3. Determine the moles after reaction:
Since \(OH^-\)\ is the limiting reactant, all of it reacts:
\(n(HA)_{\text{final}} = 0.00500 - 0.00250 = 0.00250\text{ mol}\)
\(n(A^-)_{\text{final}} = 0.00250\text{ mol}\)

4. Use the Henderson-Hasselbalch equation:
Since \([HA] = [A^-]\), the pH is equal to \(\text{p}K_a\):
\(\text{pH} = \text{p}K_a = -\log_{10}(1.30 \times 10^{-5}) = 4.89\)

[1] Award 1 mark for the correct option (C).
[0] For any other answer.

A reaction is thermodynamically feasible under standard conditions if the cell potential \(E^\theta_{\text{cell}}\)\ is positive.
Let's evaluate each mixture:
- A: \(Fe^{2+}(aq)\) and \(I_2(aq)\). Reaction: \(2Fe^{2+} + I_2 \rightarrow 2Fe^{3+} + 2I^-\).
\(E^\theta_{\text{cell}} = E^\theta(\text{reduction}) - E^\theta(\text{oxidation}) = 0.54 - 0.77 = -0.23\text{ V}\) (Not feasible).
- B: \(Cu^{2+}(aq)\) and \(I_2(aq)\). Both are oxidizing agents (the oxidized species in their half-equations), so no redox reaction can occur.
- C: \(Fe^{3+}(aq)\) and \(I^-(aq)\). Reaction: \(2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2\).
\(E^\theta_{\text{cell}} = E^\theta(\text{reduction}) - E^\theta(\text{oxidation}) = 0.77 - 0.54 = +0.23\text{ V}\) (Feasible).
- D: \(Fe^{2+}(aq)\) and \(Cu^{2+}(aq)\). Reaction: \(2Fe^{2+} + Cu^{2+} \rightarrow 2Fe^{3+} + Cu(s)\).
\(E^\theta_{\text{cell}} = 0.34 - 0.77 = -0.43\text{ V}\) (Not feasible).

[1] Award 1 mark for the correct option (C).
[0] For any other answer.

1. Recall the definition of atom economy:
\(\text{Atom Economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100\%\)

2. Calculate the formula mass of the desired product (\(Br_2\)):
\(M_r(Br_2) = 2 \times 79.9 = 159.8\)

3. Calculate the sum of the formula masses of the reactants:
\(M_r(Cl_2) = 2 \times 35.5 = 71.0\)
\(2 \times M_r(KBr) = 2 \times (39.1 + 79.9) = 2 \times 119.0 = 238.0\)
\(\text{Total reactant mass} = 71.0 + 238.0 = 309.0\)

4. Calculate the atom economy:
\(\text{Atom Economy} = \frac{159.8}{309.0} \times 100\% = 51.7\%\)

[1] Award 1 mark for the correct option (B).
[0] For any other answer.

The reaction equations are:
1) \(\text{ClO}^- + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{Cl}^- + \text{I}_2 + \text{H}_2\text{O}\)
2) \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\)
From these equations, the reacting ratio of \(\text{ClO}^- : \text{S}_2\text{O}_3^{2-}\) is \(1 : 2\).
Moles of \(\text{S}_2\text{O}_3^{2-}\) used = \(0.0240\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.40 \times 10^{-3}\text{ mol}\).
Moles of \(\text{ClO}^-\approx 1.20 \times 10^{-3}\text{ mol}\).
Concentration of \(\text{ClO}^- = 1.20 \times 10^{-3}\text{ mol} / 0.0100\text{ dm}^3 = 0.120\text{ mol dm}^{-3}\).

1 mark for the correct calculation of moles and final concentration leading to option B.

In the presence of \(\text{D}_2\text{O}\), labially labile protons on the hydroxyl group (\(\text{-OH}\)) undergo rapid exchange with deuterium and do not appear in the proton NMR spectrum. The remaining non-exchanging protons are: the ester methyl group (\(\text{-OCH}_3\)) which has no adjacent protons and appears as a singlet; the methyl group on the carbon adjacent to the CH (\(\text{-CH}_3\)) which is split by one adjacent CH proton to give a doublet; and the CH proton itself which is split by three adjacent methyl protons to give a quartet.

1 mark for identifying the correct splitting patterns and the effect of D2O exchange leading to option A.

Atom economy is calculated as: \((\text{mass of desired product} / \text{total mass of reactants}) \times 100\).
Molar mass of reactants: \(\text{Cl}_2 = 71.0\text{ g mol}^{-1}\), \(2\text{KBr} = 2 \times 119.0 = 238.0\text{ g mol}^{-1}\).
Total reactant mass = \(71.0 + 238.0 = 309.0\text{ g mol}^{-1}\).
Desired product: \(\text{Br}_2 = 159.8\text{ g mol}^{-1}\).
Atom economy = \((159.8 / 309.0) \times 100 = 51.7\%\).

1 mark for the correct calculation of relative formula masses and percentage atom economy leading to option C.

The energy of one photon is given by \(E = hc / \lambda\).
\(E = (6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}) / (580 \times 10^{-9}\text{ m}) = 3.429 \times 10^{-19}\text{ J}\).
For one mole of photons: \(E_{\text{mol}} = 3.429 \times 10^{-19}\text{ J} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 2.064 \times 10^5\text{ J mol}^{-1} = 206\text{ kJ mol}^{-1}\).

1 mark for correctly applying the equation and converting the units to kJ/mol to find option B.

The atomic number of manganese is 25, with electron configuration \([\text{Ar}] 3\text{d}^5 4\text{s}^2\). Formation of the \(\text{Mn}^{2+}\) ion involves losing the two \(4\text{s}\) electrons, leaving the configuration as \([\text{Ar}] 3\text{d}^5\). \(\text{Cr}^{3+}\) is \(3\text{d}^3\); \(\text{Fe}^{2+}\) is \(3\text{d}^6\); \(\text{Co}^{3+}\) is \(3\text{d}^6\).

1 mark for the correct deduction of the transition metal ion's d-electron configuration leading to option B.

For a buffer, \([\text{H}^+] = K_a \times ([\text{HA}] / [\text{A}^-])\).
Since equal volumes are mixed, the concentration ratio of acid to conjugate base is unchanged: \([\text{HA}] / [\text{A}^-] = 0.20 / 0.10 = 2\).
\([\text{H}^+] = 1.8 \times 10^{-5} \times 2 = 3.6 \times 10^{-5}\text{ mol dm}^{-3}\).
\(\text{pH} = -\log_{10}(3.6 \times 10^{-5}) = 4.44\).

1 mark for the correct calculation of buffer pH leading to option B.

Since doubling \([\text{X}]\) quadruples the rate, the order with respect to \(\text{X}\) is 2. Since doubling \([\text{Y}]\) has no effect, the order with respect to \(\text{Y}\) is 0. The rate equation is \(\text{Rate} = k[\text{X}]^2\).

1 mark for the correct determination of reaction order and the resulting rate equation leading to option C.

Alkaline hydrolysis of a dipeptide cleaves the peptide amide linkage. Under alkaline conditions, carboxylic acid groups are deprotonated to form carboxylate anions, while amine groups remain unprotonated. Thus, the products are the anions \(\text{H}_2\text{NCH}_2\text{COO}^-\)\ and \(\text{H}_2\text{NCH(CH}_3\text{)COO}^-\).

1 mark for identifying the correct ionic forms of the hydrolysis products in alkaline solution leading to option B.

In \(\text{S}_2\text{O}_3^{2-}\), the average oxidation state of sulfur is +2. In \(\text{S}_4\text{O}_6^{2-}\), the average oxidation state of sulfur is +2.5. Thus, the average oxidation state of sulfur increases from +2 to +2.5, which represents oxidation. Iodine is reduced from 0 to -1, so it acts as an oxidizing agent. Each iodine atom gains one electron.

1 mark: Correct option chosen (B).

For methyl 2-hydroxypropanoate (\(\text{CH}_3\text{CH(OH)COOCH}_3\)), the singlet at \(\delta = 3.7\text{ ppm}\) (3H) corresponds to the \(\text{-COOCH}_3\) group. The doublet at \(\delta = 1.4\text{ ppm}\) (3H) corresponds to the \(\text{-CH}_3\) adjacent to the CH. The quartet at \(\delta = 4.3\text{ ppm}\) (1H) corresponds to the \(\text{-CH-}\) split by the neighboring \(\text{-CH}_3\). The singlet at \(\delta = 3.1\text{ ppm}\) (1H) is the labile \(\text{-OH}\) proton.

1 mark: Correct option chosen (A).

1. Moles of \(\text{S}_2\text{O}_3^{2-} = 0.0200\text{ mol dm}^{-3} \times 0.02250\text{ dm}^3 = 4.50 \times 10^{-4}\text{ mol}\). 2. Since \(2\text{ mol}\) of \(\text{S}_2\text{O}_3^{2-}\) reacts with \(1\text{ mol}\) of \(\text{I}_2\), which is produced from \(1\text{ mol}\) of \(\text{Cl}_2\), the moles of \(\text{Cl}_2 = \frac{4.50 \times 10^{-4}}{2} = 2.25 \times 10^{-4}\text{ mol}\). 3. Concentration of \(\text{Cl}_2 = \frac{2.25 \times 10^{-4}\text{ mol}}{0.0500\text{ dm}^3} = 4.50 \times 10^{-3}\text{ mol dm}^{-3}\).

1 mark: Correct option chosen (B).

1. Energy of a single photon: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{480 \times 10^{-9}} = 4.14375 \times 10^{-19}\text{ J}\). 2. Energy per mole of photons: \(\Delta E = 4.14375 \times 10^{-19} \times 6.02 \times 10^{23} = 249453.75\text{ J mol}^{-1}\). 3. Convert to \(\text{kJ mol}^{-1}\): \(\Delta E = 249.45 \approx 250\text{ kJ mol}^{-1}\).

1 mark: Correct option chosen (C).

Thermal stability of Group 2 carbonates increases down the group because the cation size increases, reducing the polarization of the carbonate ion. Solubility of hydroxides increases down the group, and solubility of sulfates decreases down the group. The first ionisation energy decreases down the group.

1 mark: Correct option chosen (C).

1. Moles of propanoic acid, \(n(\text{HA}) = 0.100 \times 0.0250 = 2.50 \times 10^{-3}\text{ mol}\). 2. Moles of propanoate, \(n(\text{A}^-) = 0.0600 \times 0.0250 = 1.50 \times 10^{-3}\text{ mol}\). 3. \([\text{H}^+] = K_{\text{a}} \times \frac{[\text{HA}]}{[\text{A}^-]} = 1.30 \times 10^{-5} \times \frac{2.50 \times 10^{-3}}{1.50 \times 10^{-3}} = 2.167 \times 10^{-5}\text{ mol dm}^{-3}\). 4. \(\text{pH} = -\log_{10}(2.167 \times 10^{-5}) = 4.66\).

1 mark: Correct option chosen (B).

### Part a)
i) The ionic equation is:
\( \text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O} \)

ii) Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced.
In this reaction, chlorine (oxidation state 0) is reduced to chlorine in \( \text{Cl}^- \) (oxidation state -1) and oxidised to chlorine in \( \text{ClO}^- \) (oxidation state +1).

### Part b)
1. Calculate moles of \( \text{S}_2\text{O}_3^{2-} \) used in the titration:
\( n(\text{S}_2\text{O}_3^{2-}) = c \times V = 0.120 \text{ mol dm}^{-3} \times 0.02250 \text{ dm}^3 = 2.70 \times 10^{-3} \text{ mol} \)

2. From Equation 2, find moles of \( \text{I}_2 \):
\( n(\text{I}_2) = \frac{1}{2} \times n(\text{S}_2\text{O}_3^{2-}) = \frac{2.70 \times 10^{-3}}{2} = 1.35 \times 10^{-3} \text{ mol} \)

3. From Equation 1, find moles of \( \text{ClO}^- \) in the 25.0 cm³ aliquot:
\( n(\text{ClO}^-) = n(\text{I}_2) = 1.35 \times 10^{-3} \text{ mol} \)

4. Scale up to the 250.0 cm³ volumetric flask:
\( n(\text{ClO}^-)_{\text{total}} = 1.35 \times 10^{-3} \text{ mol} \times \frac{250.0}{25.0} = 1.35 \times 10^{-2} \text{ mol} \)

5. Calculate concentration of \( \text{ClO}^- \) in the original 10.0 cm³ bleach sample:
\( c(\text{ClO}^-)_{\text{original}} = \frac{n}{V} = \frac{1.35 \times 10^{-2} \text{ mol}}{0.0100 \text{ dm}^3} = 1.35 \text{ mol dm}^{-3} \)

### Part c)
i) The colour change at the end point is from blue-black to colourless.
ii) Starch indicator is added late in the titration, when the solution becomes pale yellow straw-coloured. If added too early, the starch forms a strong, stable, insoluble complex with a high concentration of iodine, which prevents an accurate and sharp end point.

### Part d)
i) \( E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = E^\theta(\text{ClO}^-/\text{Cl}^-) - E^\theta(\text{I}_2/2\text{I}^-) = +0.89\text{ V} - (+0.54\text{ V}) = +0.35\text{ V} \).
ii) A positive \( E^\theta_{\text{cell}} \) indicates that the reaction is thermodynamically feasible.

### Part e)
Chlorine has fewer shells / a smaller atomic radius and less shielding than iodine. Therefore, there is a stronger electrostatic attraction between the nucleus of a chlorine atom and an incoming electron, meaning chlorine gains electrons more readily (is a stronger oxidising agent).

a) i) Correct equation with correct species and balancing: [1 mark]
ii) Definition: simultaneous oxidation and reduction of the same element [1 mark].
Identification of oxidation states: 0 in Cl2 to -1 in Cl- (reduction) and +1 in ClO- (oxidation) [1 mark].

b)
- Moles of thiosulfate = 2.70 x 10^-3 mol [1 mark]
- Moles of iodine = 1.35 x 10^-3 mol [1 mark]
- Moles of ClO- in 25 cm3 = 1.35 x 10^-3 mol [1 mark]
- Moles of ClO- in 250 cm3 = 1.35 x 10^-2 mol [1 mark]
- Original concentration = 1.35 (mol dm^-3) [1 mark]
(Allow ECF for calculation errors, but penalize missing dilution factors).

c)
i) Blue-black to colourless (reject 'clear') [1 mark].
ii) Added when the solution is pale yellow/straw-coloured [1 mark].
Explanation: Iodine forms an insoluble complex with starch if concentration is high / prevents sharp colour change [1 mark].

d)
i) +0.35 V (accept with or without sign, units optional but must be correct if present) [1 mark].
ii) E-theta cell is positive (or > 0) [1 mark].

e)
- Chlorine has smaller atomic radius / fewer electron shells [1 mark].
- Chlorine has less electron shielding [1 mark].
- Stronger attraction between nucleus and incoming electron [1 mark].

### Part a)
1. Relative moles of each element:
- \( \text{C}: \frac{40.45}{12.0} = 3.371 \)
- \( \text{H}: \frac{7.87}{1.0} = 7.87 \)
- \( \text{N}: \frac{15.73}{14.0} = 1.124 \)
- \( \text{O}: \frac{35.95}{16.0} = 2.247 \)

2. Divide by the smallest value (1.124):
- \( \text{C}: \frac{3.371}{1.124} = 3.00 \)
- \( \text{H}: \frac{7.87}{1.124} = 7.00 \)
- \( \text{N}: \frac{1.124}{1.124} = 1.00 \)
- \( \text{O}: \frac{2.247}{1.124} = 2.00 \)

Therefore, the empirical formula of **Y** is \( \text{C}_3\text{H}_7\text{NO}_2 \).

3. Empirical formula mass:
\( M = (3 \times 12.0) + (7 \times 1.0) + 14.0 + (2 \times 16.0) = 89.0 \text{ g mol}^{-1} \).
Since the molecular ion peak in the mass spectrum has \( m/z = 89 \), the molecular formula is also \( \text{C}_3\text{H}_7\text{NO}_2 \).

### Part b)
i)
1. Carboxylic acid \( \text{O–H} \) bond corresponding to the broad absorption at 2500–3300 cm⁻¹.
2. Carbonyl \( \text{C=O} \) bond corresponding to the strong sharp absorption at 1715 cm⁻¹.
3. Primary or secondary amine \( \text{N–H} \) bond corresponding to the sharp absorption at 3350 cm⁻¹.

ii) When \( \text{D}_2\text{O} \) is added, the acidic/labile proton of the \( \text{-COOH} \) group (and the amine group) undergoes rapid chemical exchange with deuterium:
\( \text{R-COOH} + \text{D}_2\text{O} \rightleftharpoons \text{R-COOD} + \text{HOD} \)
Deuterons (\( ^2\text{H} \)) do not absorb in the proton \( ^1\text{H} \) NMR range, causing the signal to disappear.

### Part c)
The formula and functional groups show **Y** is 2-aminopropanoic acid (alanine), \( \text{CH}_3\text{CH(NH}_2)\text{COOH} \).

Assignment of NMR signals:
- \( \delta = 11.0 \text{ ppm} \) (singlet, area 1): Assigned to the carboxylic acid proton (\( \text{-COOH} \)). It is a singlet because there are no adjacent protons, and it is highly de-shielded.
- \( \delta = 3.6 \text{ ppm} \) (quartet, area 1): Assigned to the \( \text{-CH-} \) proton. It is split into a quartet by the 3 adjacent protons of the neighbouring methyl group (using the \( n+1 \) rule, \( 3 + 1 = 4 \)).
- \( \delta = 5.5 \text{ ppm} \) (broad singlet, area 2): Assigned to the two protons of the primary amine group (\( \text{-NH}_2 \)). It is a broad singlet due to rapid proton exchange and lacks coupling.
- \( \delta = 1.4 \text{ ppm} \) (doublet, area 3): Assigned to the three protons of the methyl group (\( \text{-CH}_3 \)). It is split into a doublet by the single adjacent \( \text{-CH-} \) proton (using the \( n+1 \) rule, \( 1 + 1 = 2 \)).

### Part d)
i) The structure of the zwitterion is:
\( \text{H}_3\text{N}^+\text{-CH(CH}_3\text{)-COO}^- \)
ii) In the solid state, there are strong ionic electrostatic attractions between the oppositely charged parts of the zwitterions (ionic bonds), which require much more thermal energy to overcome than weak intermolecular forces like hydrogen bonding. This causes the compound to have a high melting point.

a)
- Correct relative moles of each element: C (3.37), H (7.87), N (1.12), O (2.25) [1 mark].
- Divide by smallest to show C3H7NO2 [1 mark].
- Calculation of empirical mass = 89 [1 mark].
- Deduction that molecular formula is C3H7NO2 because m/z of M+ peak = 89 [1 mark].

b) i)
- O-H carboxylic acid: 2500-3300 cm-1 [1 mark].
- C=O carbonyl/acid: 1715 cm-1 [1 mark].
- N-H amine: 3350 cm-1 [1 mark].
ii) Labile proton exchanges with deuterium / D2O, which does not resonate in proton frequency range [1 mark].

c)
- Correct structure drawn or named as 2-aminopropanoic acid / alanine [1 mark].
- Assignment of 11.0 ppm to carboxylic acid H (-COOH) [1 mark].
- Assignment of 1.4 ppm to methyl protons (-CH3) and explanation of doublet (adjacent to 1 H) [1 mark].
- Assignment of 3.6 ppm to CH proton and explanation of quartet (adjacent to 3 H) [1 mark].
- Assignment of 5.5 ppm to NH2 protons [1 mark].
- Explanation of integration areas (3H:1H:2H:1H ratio matches CH3:CH:NH2:COOH) [1 mark].

d)
- i) Correct zwitterion structure: H3N+-CH(CH3)-COO- [1 mark].
- ii) Strong ionic attractions / electrostatic forces between opposite charges of zwitterions require a lot of energy to break [1 mark].

### Part a)
i) The ionic equation is:
\( \text{Cl}_2(\text{g}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq}) \)

ii) Chlorine gas (\( \text{Cl}_2 \)) is the oxidizing agent. It causes bromide ions to lose electrons and is itself reduced by gaining electrons:
\( \text{Cl}_2 + 2\text{e}^- \rightarrow 2\text{Cl}^- \).

### Part b)
i)
- In \( \text{SO}_2 \): Oxygen is \(-2\), so \( \text{S} + 2(-2) = 0 \Rightarrow \text{S} = +4 \).
- In \( \text{H}_2\text{SO}_4 \): Hydrogen is \(+1\) and oxygen is \(-2\), so \( 2(+1) + \text{S} + 4(-2) = 0 \Rightarrow 2 + \text{S} - 8 = 0 \Rightarrow \text{S} = +6 \).

ii)
1. Convert masses to moles:
- Moles of \( \text{Br}_2 = \frac{160 \times 10^3 \text{ g}}{159.8 \text{ g mol}^{-1}} = 1001.25 \text{ mol} \)
- Moles of \( \text{SO}_2 = \frac{80.0 \times 10^3 \text{ g}}{64.1 \text{ g mol}^{-1}} = 1248.05 \text{ mol} \)

2. Determine limiting reagent:
According to Equation 3, the molar ratio of reaction is \( 1:1 \).
Since we have fewer moles of \( \text{Br}_2 \) (1001.25 mol) than \( \text{SO}_2 \) (1248.05 mol), **\( \text{Br}_2 \) is the limiting reagent**.

3. Calculate theoretical yield of \( \text{HBr} \):
- Moles of \( \text{HBr} = 2 \times 1001.25 \text{ mol} = 2002.5 \text{ mol} \)
- Mass of \( \text{HBr} = 2002.5 \text{ mol} \times 80.9 \text{ g mol}^{-1} = 162002 \text{ g} \approx 162 \text{ kg} \).

### Part c)
i) Percentage yield:
\( \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% = \frac{121.5 \text{ kg}}{162.0 \text{ kg}} \times 100\% = 75.0\% \)

ii) Any valid chemical reason, such as: the reaction is reversible and does not go to completion, some gases escape/dissolve, or physical losses occur when transferring and concentrating the solutions.

### Part d)
i)
\( \% \text{ Atom Economy} = \frac{\text{Mass of Desired Products}}{\text{Total Mass of All Reactants}} \times 100\% \)
\( \text{Mass of desired product (2HBr)} = 2 \times 80.9 = 161.8 \)
\( \text{Total mass of reactants} = 159.8 (\text{Br}_2) + 64.1 (\text{SO}_2) + 2 \times 18.0 (\text{H}_2\text{O}) = 259.9 \)
\( \% \text{ Atom Economy} = \frac{161.8}{259.9} \times 100\% = 62.25\% \approx 62.3\% \)

ii) The sulfuric acid co-product can be recycled and used directly in Step 1 to acidify the incoming seawater, reducing the need to buy and transport raw sulfuric acid.

a) i) Correct equation: Cl2 + 2Br- -> 2Cl- + Br2 [1 mark]. Correct state symbols (g for Cl2, others aq) [1 mark].
ii) Oxidizing agent: Chlorine/Cl2 [1 mark] because it gains electrons / oxidation state of Cl decreases [1 mark].

b) i) SO2: +4 [1 mark]. H2SO4: +6 [1 mark].
ii)
- Moles of Br2 = 1001.25 mol [1 mark].
- Moles of SO2 = 1248.05 mol [1 mark].
- Correct deduction that Br2 is the limiting reagent based on 1:1 stoichiometry [1 mark].
- Moles of HBr theoretically formed = 2002.5 mol [1 mark].
- Mass of HBr = 162 kg (accept 161.8 to 162.2) [1 mark].

c) i) 75.0% (accept 75%) [1 mark].
ii) Incomplete reaction / reversible reaction / product remaining in solution [1 mark].

d) i)
- Total mass of desired product = 161.8 [1 mark].
- Correct percentage calculation: 62.3% (accept 62% or 62.25%) [1 mark].
ii) Recycled to acidify seawater / neutralised to make useful sulfates [1 mark].

### Part a)
The atomic number of Fe is 26. The ground-state electronic configuration of an Fe atom is \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 4\text{s}^2 \).
When the \( \text{Fe}^{2+} \) ion forms, the two electrons in the outer 4s subshell are lost first:
\( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 \) (or \( [\text{Ar}] 3\text{d}^6 \)).

### Part b)
i) The shape of \( \left[\text{Cu}(\text{H}_2\text{O})_6\right]^{2+} \) is octahedral.
- The central \( \text{Cu}^{2+} \) ion is bonded to 6 water ligands via dative covalent bonds.
- The drawing must show an octahedral arrangement with wedges (coming out) and dashes (going in).
- The bond angle between adjacent ligands is \( 90^\circ \).

ii) Explanation of colour:
1. The lone pairs on the oxygen atoms of the water ligands approach the copper ion, creating an electrostatic field.
2. This causes the five degenerate 3d orbitals of the \( \text{Cu}^{2+} \) ion to split into two different energy levels.
3. A d-electron in the lower energy level absorbs a photon of visible light and is promoted to the higher energy level (a d–d transition).
4. The energy of this photon corresponds to the equation \( \Delta E = h\nu \).
5. The absorbed light is in the red-orange region of the visible spectrum. The complementary wavelength (pale blue) is transmitted or reflected, which is the colour we see.

### Part c)
i) Ligand substitution equation:
\( \left[\text{Cu}(\text{H}_2\text{O})_6\right]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons \left[\text{CuCl}_4\right]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l}) \)

ii) Shape: Tetrahedral.
Explanation: Chloride ions (\( \text{Cl}^- \)) are larger and more negatively charged than neutral water molecules. Only four chloride ligands can fit around the copper(II) central ion due to steric hindrance and electrostatic repulsion between the ligands.

### Part d)
i) Energy change, \( \Delta E \):
\( \Delta E = \frac{hc}{\lambda} \)
\( \lambda = 790 \text{ nm} = 790 \times 10^{-9} \text{ m} = 7.90 \times 10^{-7} \text{ m} \)
\( \Delta E = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m s}^{-1})}{7.90 \times 10^{-7} \text{ m}} = 2.5177 \times 10^{-19} \text{ J} \approx 2.52 \times 10^{-19} \text{ J} \) per ion.

ii) Converting to \( \text{kJ mol}^{-1} \):
\( \Delta E_{\text{mol}} = 2.5177 \times 10^{-19} \text{ J} \times (6.02 \times 10^{23} \text{ mol}^{-1}) = 151566 \text{ J mol}^{-1} \)
\( \Delta E_{\text{mol}} = 151.57 \text{ kJ mol}^{-1} \approx 152 \text{ kJ mol}^{-1} \).

a)
- Correct electron configuration: 1s2 2s2 2p6 3s2 3p6 3d6 (or [Ar] 3d6) [2 marks].
- Award [1 mark] for configuration ending in 3d4 4s2 or showing 4s electrons retained.

b) i)
- Diagram showing octahedral geometry with wedges/dashes and labels [1 mark].
- Bond angle of 90 degrees correctly labelled [1 mark].
ii)
- Ligands donate electron pairs / split the 3d orbitals into two different energy levels [1 mark].
- An electron absorbs energy/photon of visible light [1 mark] and is promoted to a higher d-orbital / undergoes d-d transition [1 mark].
- Energy gap corresponds to a specific frequency/wavelength (Delta E = hv) [1 mark].
- Remaining wavelengths/complementary colour is transmitted [1 mark].

c) i) Balanced equation with correct formulas: [Cu(H2O)6]2+ + 4Cl- -> [CuCl4]2- + 6H2O [2 marks]. (deduct 1 mark if state symbols are requested and missing, but here they are optional. Deduct 1 mark for incorrect balancing).
ii) Shape: Tetrahedral [1 mark].
Reason: Chloride ligands are larger than water molecules, so fewer can fit around the central ion [1 mark].

d) i)
- Wavelength converted to meters (7.90 x 10^-7 m) [1 mark].
- Correct calculation: 2.52 x 10^-19 J (allow 2.5 x 10^-19) [1 mark].
ii)
- Multiplication by Avogadro constant and division by 1000: 152 kJ mol-1 [1 mark].

### Part a)
i) Dissociation equation:
\( \text{CH}_3\text{CH(OH)COOH}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{CH}_3\text{CH(OH)COO}^-(\text{aq}) + \text{H}_3\text{O}^+(\text{aq}) \)
(Accept \( \text{CH}_3\text{CH(OH)COOH}(\text{aq}) \rightleftharpoons \text{CH}_3\text{CH(OH)COO}^-(\text{aq}) + \text{H}^+(\text{aq}) \))

ii) \( K_{\text{a}} = \frac{[\text{H}^+][\text{CH}_3\text{CH(OH)COO}^-]}{[\text{CH}_3\text{CH(OH)COOH}]} \) or \( K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \)

### Part b)
Using the weak acid approximation:
\( K_{\text{a}} \approx \frac{[\text{H}^+]^2}{[\text{HA}]} \)
\( [\text{H}^+]^2 = K_{\text{a}} \times [\text{HA}]_{\text{initial}} \)
\( [\text{H}^+] = \sqrt{1.38 \times 10^{-4} \text{ mol dm}^{-3} \times 0.120 \text{ mol dm}^{-3}} = \sqrt{1.656 \times 10^{-5}} = 4.069 \times 10^{-3} \text{ mol dm}^{-3} \)
\( \text{pH} = -\log_{10}(4.069 \times 10^{-3}) = 2.39 \)

Assumptions:
1. \( [\text{H}^+] \approx [\text{A}^-] \) (ionisation of water is negligible).
2. \( [\text{HA}]_{\text{equilibrium}} \approx [\text{HA}]_{\text{initial}} \) (dissociation of lactic acid is negligible).

### Part c)
i) Action of the buffer:
- When acid (\( \text{H}^+ \)) is added, it reacts with the conjugate base (lactate ions) to form undissociated lactic acid:
\( \text{H}^+(\text{aq}) + \text{A}^-(\text{aq}) \rightarrow \text{HA}(\text{aq}) \)
- When alkali (\( \text{OH}^- \)) is added, the hydroxide ions react with the weak acid (lactic acid) to form water and lactate ions:
\( \text{OH}^-(\text{aq}) + \text{HA}(\text{aq}) \rightarrow \text{A}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) \)

ii)
1. Target \( \text{pH} = 4.00 \Rightarrow [\text{H}^+] = 1.00 \times 10^{-4} \text{ mol dm}^{-3} \).
2. Rearrange the \( K_{\text{a}} \) expression to find the required ratio:
\( K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \Rightarrow \frac{[\text{A}^-]}{[\text{HA}]} = \frac{K_{\text{a}}}{[\text{H}^+]} = \frac{1.38 \times 10^{-4}}{1.00 \times 10^{-4}} = 1.38 \)
3. Calculate the required concentration of lactate, \( [\text{A}^-] \):
\( [\text{A}^-] = 1.38 \times [\text{HA}] = 1.38 \times 0.150 \text{ mol dm}^{-3} = 0.207 \text{ mol dm}^{-3} \)
4. Calculate moles of sodium lactate in 500 cm³ (0.500 dm³):
\( n(\text{sodium lactate}) = 0.207 \text{ mol dm}^{-3} \times 0.500 \text{ dm}^3 = 0.1035 \text{ mol} \)
5. Calculate mass of sodium lactate:
\( \text{Mass} = 0.1035 \text{ mol} \times 112.0 \text{ g mol}^{-1} = 11.592 \text{ g} \approx 11.6 \text{ g} \).

### Part d)
i) The chiral carbon is the central carbon bonded to four different groups: \( \text{-H} \), \( \text{-CH}_3 \), \( \text{-OH} \), and \( \text{-COOH} \).
The two 3D tetrahedral structures must be drawn as non-superimposable mirror images of each other, using wedge-and-dash representation.

a)
- i) Correct dissociation equation with reversible arrow: [1 mark]
- ii) Ka expression: [1 mark]

b)
- Correct equation/expression setup: [1 mark]
- Correct [H+] = 4.07 x 10^-3 mol dm-3: [1 mark]
- pH = 2.39: [1 mark]
- Two assumptions stated correctly: [2 marks] (1 mark per assumption).

c) i)
- Equation for addition of H+: [1 mark]
- Equation for addition of OH-: [1 mark]
- Explanation that the buffer maintains pH by keeping [H+] and [OH-] relatively constant: [1 mark]
ii)
- Calculation of [H+] = 1.00 x 10^-4: [1 mark]
- Ratio [A-]/[HA] = 1.38: [1 mark]
- Moles of sodium lactate = 0.1035 mol: [1 mark]
- Mass of sodium lactate = 11.6 g (allow 11.59 g): [1 mark]

d) i)
- Identification of chiral carbon / 4 different groups: [1 mark]
- Correctly drawn mirror images with 3D tetrahedral representation (wedges and dashes): [1 mark]

(a) Combining the half-equations:
\(\text{MnO}_2(s) + 4\text{H}^+(aq) + 2e^- \rightarrow \text{Mn}^{2+}(aq) + 2\text{H}_2\text{O}(l)\) and
\(2\text{I}^-(aq) \rightarrow \text{I}_2(aq) + 2e^-\)
gives: \(\text{MnO}_2(s) + 4\text{H}^+(aq) + 2\text{I}^-(aq) \rightarrow \text{Mn}^{2+}(aq) + \text{I}_2(aq) + 2\text{H}_2\text{O}(l)\).

(b) The oxidation number of manganese decreases from +4 in \(\text{MnO}_2\) to +2 in \(\text{Mn}^{2+}\), indicating reduction. The oxidation number of iodine increases from -1 in \(\text{I}^-\) to 0 in \(\text{I}_2\), indicating oxidation. Since both oxidation and reduction occur simultaneously, it is a redox reaction.

(c)(i) \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = 1.36 - 0.54 = +0.82\text{ V}\).

(c)(ii) The standard electrode potential for the chlorine system (\(E^\theta = +1.36\text{ V}\)) is more positive than that for the iodine system (\(E^\theta = +0.54\text{ V}\)). This means chlorine gas is a stronger oxidising agent than iodine, so chlorine is easily reduced to chloride while driving the oxidation of iodide to iodine.

(d)(i) In \(\text{SO}_2\), the oxidation state of \(\text{S}\) is +4. In \(\text{SO}_4^{2-}\), the oxidation state of \(\text{S}\) is +6.

(d)(ii) The reducing agent is \(\text{SO}_2\) (or sulfur) because its oxidation state increases from +4 to +6, meaning it has lost electrons (donated electrons to iodine).

(e)(i) The colour change at the end-point is blue-black to colourless.

(e)(ii)
1. Moles of \(\text{S}_2\text{O}_3^{2-}\) reacted = \(0.0500 \times \frac{18.40}{1000} = 9.20 \times 10^{-4}\text{ mol}\).
2. From the titration equation: \(1\text{ mol } \text{I}_2 \equiv 2\text{ mol } \text{S}_2\text{O}_3^{2-}\).
Therefore, moles of \(\text{I}_2 = 4.60 \times 10^{-4}\text{ mol}\).
3. Since \(2\text{I}^-\) ions produce \(1\text{ mol } \text{I}_2\), the original moles of \(\text{I}^-\) in the \(25.0\text{ cm}^3\) sample = \(9.20 \times 10^{-4}\text{ mol}\).
4. Concentration of \(\text{I}^-\) in \(\text{mol dm}^{-3} = \frac{9.20 \times 10^{-4}}{0.0250} = 0.0368\text{ mol dm}^{-3}\).
5. Concentration of \(\text{I}^-\) in \(\text{g dm}^{-3} = 0.0368 \times 126.9 = 4.67\text{ g dm}^{-3}\).

(e)(iii) Adding the starch indicator too early when the iodine concentration is high causes starch-iodine complexes to form irreversibly, preventing a sharp end-point. This is minimised by adding starch only near the end-point when the solution turns a pale straw-yellow color.

(a)
- 1 mark: Correct reactants and products: \(\text{MnO}_2 + 4\text{H}^+ + 2\text{I}^- \rightarrow \text{Mn}^{2+} + \text{I}_2 + 2\text{H}_2\text{O}\).
- 1 mark: Fully balanced species and charges.

(b)
- 1 mark: Identifies Mn decreases from +4 to +2 (reduced).
- 1 mark: Identifies I increases from -1 to 0 (oxidised).
- 1 mark: Explicitly links both to the definition of a redox reaction.

(c)(i)
- 1 mark: \(+0.82\text{ V}\) (must include sign).

(c)(ii)
- 1 mark: States that the \(\text{Cl}_2/\text{Cl}^-\) system has a more positive standard electrode potential.
- 1 mark: Explains that chlorine is a stronger oxidising agent / more readily reduced than iodine.

(d)(i)
- 1 mark: \(+4\) in \(\text{SO}_2\).
- 1 mark: \(+6\) in \(\text{SO}_4^{2-}\).

(d)(ii)
- 1 mark: Identifies sulfur / \(\text{SO}_2\) as the reducing agent.
- 1 mark: Explains that its oxidation number increases / it loses electrons.

(e)(i)
- 1 mark: Blue-black to colourless (reject 'clear').
- Reject: Purple to colourless.

(e)(ii)
- 1 mark: Moles of \(\text{S}_2\text{O}_3^{2-} = 9.20 \times 10^{-4}\text{ mol}\).
- 1 mark: Moles of \(\text{I}_2 = 4.60 \times 10^{-4}\text{ mol}\).
- 1 mark: Moles of \(\text{I}^- = 9.20 \times 10^{-4}\text{ mol}\).
- 1 mark: Concentration of \(\text{I}^- = 0.0368\text{ mol dm}^{-3}\).
- 1 mark: Concentration in \(\text{g dm}^{-3} = 4.67\text{ g dm}^{-3}\) (accept range 4.66-4.68; must be to 3 significant figures).

(e)(iii)
- 1 mark: Starch added too early binds irreversibly to iodine / gives an inaccurate endpoint.
- 1 mark: Add starch only when the mixture becomes pale yellow / straw-coloured.

(a)(i) Lactic acid: Skeletal structure consists of a three-carbon chain. Carbon-1 has a double-bonded oxygen and a single-bonded oxygen (carboxylic acid). Carbon-2 has a single-bonded hydroxyl group (\(\text{-OH}\)).

(a)(ii) Glycolic acid: Skeletal structure consists of a two-carbon chain. Carbon-1 has a carboxylic acid group (\(\text{=O}\) and \(\text{-OH}\)). Carbon-2 has a single-bonded hydroxyl group (\(\text{-OH}\)).

(b)(i) Copolymerisation of lactic acid and glycolic acid:
\(\text{CH}_3\text{CH(OH)COOH} + \text{HOCH}_2\text{COOH} \rightarrow \text{-[O-CH(CH_3)-CO-O-CH_2-CO]-} + 2\text{H}_2\text{O}\)
(or represented as a repeating unit showing alternating ester linkages).

(b)(ii) The polymer linkage formed is an ester linkage.

(c) Glycolic acid (\(\text{HO-CH}_2\text{-COOH}\)) has 3 proton environments:
1. Carboxylic acid proton (\(\text{-COOH}\)): chemical shift \(\delta \approx 10.0-12.0\text{ ppm}\); splitting pattern = singlet (1H).
2. Alkyl protons (\(\text{-CH}_2-\)): chemical shift \(\delta \approx 3.0-4.5\text{ ppm}\) (deshielded by adjacent oxygen and carbonyl); splitting pattern = singlet (2H).
3. Alcohol proton (\(\text{-OH}\)): chemical shift \(\delta \approx 2.0-6.0\text{ ppm}\); splitting pattern = singlet (1H).

(d) Lactic acid (\(\text{CH}_3\text{-CH(OH)-COOH}\)) has 3 distinct carbon environments, yielding 3 peaks in the \(^{13}\text{C}\) NMR spectrum:
1. Carbonyl carbon (\(\text{-COOH}\)) at \(\delta \approx 160-185\text{ ppm}\).
2. Carbon bonded to oxygen (\(\text{-CH(OH)-}\)) at \(\delta \approx 50-90\text{ ppm}\).
3. Methyl carbon (\(\text{-CH}_3\)) at \(\delta \approx 5-40\text{ ppm}\).

(e) In the monomer mixture, there are intense, broad absorption bands for the carboxylic acid \(\text{O-H}\) stretch (\(2500-3300\text{ cm}^{-1}\)) and alcohol \(\text{O-H}\) stretch (\(3200-3600\text{ cm}^{-1}\)). In the fully polymerised PLGA, these peaks will be absent or significantly reduced (only remaining at the chain ends).

(f)
- The peak at \(m/z = 45\) is due to the carboxyl cation: \(\text{[COOH]}^+\) (or \(\text{[HCO}_2\text{]}^+\)).
- The peak at \(m/z = 31\) is due to the hydroxymethyl cation: \(\text{[CH}_2\text{OH]}^+\).

(g)
1. At regular time intervals, take small aliquots of the reaction mixture and spot them onto a silica gel TLC plate along with reference spots of pure lactic acid and glycolic acid.
2. Place the plate in a beaker containing a suitable polar solvent (eluent) and allow the solvent front to run.
3. Because monomers are colourless and do not absorb UV strongly, visualise the spots using an iodine chamber, a UV lamp, or a suitable chemical stain (such as bromocresol green/indicator spray or basic permanganate).
4. Compare the \(R_f\) values of the hydrolysed mixture spots with the reference monomer spots to confirm their identity.

(a)
- (i) 1 mark: Correct skeletal structure of lactic acid.
- (ii) 1 mark: Correct skeletal structure of glycolic acid.

(b)
- (i) 1 mark: Correct structure of both monomers.
- (i) 1 mark: Correct alternating polymer repeating unit showing open bonds at the ends and elimination of water molecules.
- (ii) 1 mark: Ester.

(c)
- 1 mark: Identifies 3 proton environments / 3 peaks.
- 1 mark: \(\text{-COOH}\) proton peak at \(10-12\text{ ppm}\) as a singlet.
- 1 mark: \(\text{-CH}_2-\) proton peak at \(3.0-4.5\text{ ppm}\) as a singlet.
- 1 mark: Alcohol \(\text{-OH}\) proton peak at \(2.0-6.0\text{ ppm}\) (or \(0.5-5.0\text{ ppm}\)) as a singlet.

(d)
- 1 mark: Identifies 3 peaks.
- 1 mark: Assigns one peak to the \(\text{C=O}\) carbon at \(160-185\text{ ppm}\).
- 1 mark: Assigns other peaks to \(\text{C-O}\) (\(50-90\text{ ppm}\)) and \(\text{C-C}\) (\(5-40\text{ ppm}\)).

(e)
- 1 mark: Disappearance / significant reduction of alcohol \(\text{O-H}\) stretch (\(3200-3600\text{ cm}^{-1}\)).
- 1 mark: Disappearance / significant reduction of carboxylic acid \(\text{O-H}\) stretch (\(2500-3300\text{ cm}^{-1}\)).

(f)
- 1 mark: \(\text{COOH}^+\) (charge must be present).
- 1 mark: \(\text{CH}_2\text{OH}^+\) (charge must be present).

(g)
- 1 mark: Spot the sample alongside reference standards (lactic acid and glycolic acid).
- 1 mark: Run using a suitable developing solvent.
- 1 mark: Visualise spots using iodine vapor / basic potassium permanganate stain / pH indicator spray.
- 1 mark: Match \(R_f\) values of hydrolysed products to standards.

(a)(i) Atom economy for Reaction 1:
$$\text{Atom economy} = \frac{M_r(\text{desired product})}{\sum M_r(\text{reactants})} \times 100$$
Desired product = \(\text{Ca(IO}_3)_2\) (\(M_r = 389.9\)).
Reactants = \(\text{CaCO}_3 + 2\text{HIO}_3\) = \(100.1 + 2 \times 175.9 = 451.9\text{ g mol}^{-1}\).
$$\text{Atom economy} = \frac{389.9}{451.9} \times 100 = 86.28\% \approx 86.3\%$$

(a)(ii) Atom economy for Reaction 2:
Reactants = \(6\text{Ca(OH)}_2 + 6\text{I}_2\) = \(6 \times 74.1 + 6 \times 253.8 = 444.6 + 1522.8 = 1967.4\text{ g mol}^{-1}\).
$$\text{Atom economy} = \frac{389.9}{1967.4} \times 100 = 19.82\% \approx 19.8\%$$

(b)(i)
1. Moles of \(\text{I}_2\) starting material = \frac{15.0}{253.8} = 0.05910\text{ mol}\).
2. According to Reaction 2, \(6\text{ mol } \text{I}_2\) produces \(1\text{ mol } \text{Ca(IO}_3)_2\).
Therefore, moles of \(\text{Ca(IO}_3)_2\) = \(\frac{0.05910}{6} = 9.850 \times 10^{-3}\text{ mol}\).
3. Maximum theoretical mass = \(9.850 \times 10^{-3} \times 389.9 = 3.84\text{ g}\).

(b)(ii)
$$\text{Percentage yield} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100 = \frac{2.15}{3.84} \times 100 = 55.99\% \approx 56.0\%$$

(c)(i) \(\text{Ca(IO}_3)_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{IO}_3^-(aq)\)
$$K_{sp} = [\text{Ca}^{2+}][\text{IO}_3^-]^2$$

(c)(ii) Let \(s = 2.40 \times 10^{-3}\text{ mol dm}^{-3}\).
\([\text{Ca}^{2+}] = s = 2.40 \times 10^{-3}\text{ mol dm}^{-3}\).
\([\text{IO}_3^-] = 2s = 4.80 \times 10^{-3}\text{ mol dm}^{-3}\).
$$K_{sp} = (2.40 \times 10^{-3}) \times (4.80 \times 10^{-3})^2 = 5.5296 \times 10^{-8} \approx 5.53 \times 10^{-8}$$
Units: \((\text{mol dm}^{-3})^3 = \text{mol}^3\text{ dm}^{-9}\).

(d)(i) From the first equation, \(1\text{ mol}\) of \(\text{IO}_3^-\). reacts to form \(3\text{ mol}\) of \(\text{I}_2\).
From the second equation, each mole of \(\text{I}_2\) reacts with \(2\text{ mol}\) of \(\text{S}_2\text{O}_3^{2-}\).
Therefore, \(3\text{ mol}\) of \(\text{I}_2\) will react with \(3 \times 2 = 6\text{ mol}\) of \(\text{S}_2\text{O}_3^{2-}\).
Hence, \(1\text{ mol}\) of \(\text{IO}_3^-\). is chemically equivalent to \(6\text{ mol}\) of \(\text{S}_2\text{O}_3^{2-}\).

(d)(ii)
1. Moles of \(\text{S}_2\text{O}_3^{2-}\) in titration = \(0.100 \times \frac{22.50}{1000} = 2.25 \times 10^{-3}\text{ mol}\).
2. Moles of \(\text{IO}_3^-\) in the \(25.0\text{ cm}^3\) aliquot = \(\frac{2.25 \times 10^{-3}}{6} = 3.75 \times 10^{-4}\text{ mol}\).
3. Concentration of \(\text{IO}_3^-\) = \(\frac{3.75 \times 10^{-4}}{0.0250} = 0.0150\text{ mol dm}^{-3}\).

(a)(i)
- 1 mark: Calculation of total reactant mass = 451.9.
- 1 mark: Correct percentage atom economy = 86.3% (accept 86.28%).

(a)(ii)
- 1 mark: Calculation of total reactant mass = 1967.4.
- 1 mark: Correct percentage atom economy = 19.8% (accept 19.82%).

(b)(i)
- 1 mark: Moles of \(\text{I}_2 = 0.0591\text{ mol}\).
- 1 mark: Moles of \(\text{Ca(IO}_3)_2 = 9.85 \times 10^{-3}\text{ mol}\).
- 1 mark: Final mass = \(3.84\text{ g}\) (accept 3.83-3.85 g).

(b)(ii)
- 1 mark: Correct process (actual/theoretical ratio).
- 1 mark: \(56.0\%\) (must match working from b(i)).

(c)(i)
- 1 mark: \(K_{sp} = [\text{Ca}^{2+}][\text{IO}_3^-]^2\) (must include state symbols or brackets; do not accept round brackets).

(c)(ii)
- 1 mark: Correct concentrations substituted: \([\text{Ca}^{2+}] = 2.40 \times 10^{-3}\) and \([\text{IO}_3^-] = 4.80 \times 10^{-3}\).
- 1 mark: Correct numerical value of \(5.53 \times 10^{-8}\).
- 1 mark: Correct units: \(\text{mol}^3\text{ dm}^{-9}\).

(d)(i)
- 1 mark: Connects \(1\text{ mol of } \text{IO}_3^-\) to \(3\text{ mol of } \text{I}_2\).
- 1 mark: Connects \(3\text{ mol of } \text{I}_2\) to \(6\text{ mol of } \text{S}_2\text{O}_3^{2-}\).

(d)(ii)
- 1 mark: Moles of \(\text{S}_2\text{O}_3^{2-} = 2.25 \times 10^{-3}\text{ mol}\).
- 2 marks: Moles of \(\text{IO}_3^- = 3.75 \times 10^{-4}\text{ mol}\) (1 mark if divide-by-6 ratio is missed but carried forward).
- 2 marks: Concentration of \(\text{IO}_3^- = 0.0150\text{ mol dm}^{-3}\) (1 mark for correct division of moles by volume).

(a) The azo group consists of two nitrogen atoms connected by a double bond: \(\text{-N=N-}\).

(b)
- Both dyes contain an extensive delocalised \(\pi\) electron system (conjugation) that spans across the aromatic rings and the double-bonded nitrogen atoms in the azo group.
- In non-conjugated molecules, the energy gap between the Highest Occupied Molecular Orbital (HOMO) and Lowest Unoccupied Molecular Orbital (LUMO) is large, meaning they only absorb high-energy ultraviolet light.
- As the length of the conjugated system increases, the molecular orbital energy levels become closer together, significantly lowering the HOMO-LUMO energy gap (\(\Delta E\)).
- This smaller energy gap allows \(\pi \rightarrow \pi^*\) electronic transitions to occur by absorbing lower-energy photons in the visible light spectrum (\(400-700\text{ nm}\)).
- The colour observed is the complementary colour to the wavelength of light absorbed.

(c)
1. Convert wavelength to meters:
\(\lambda = 504\text{ nm} = 504 \times 10^{-9}\text{ m}\).
2. Calculate the energy of a single photon:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{504 \times 10^{-9}\text{ m}} = 3.946 \times 10^{-19}\text{ J}\).
3. Calculate the energy per mole of photons:
\(E_{\text{mole}} = 3.946 \times 10^{-19}\text{ J} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 2.376 \times 10^5\text{ J mol}^{-1}\).
4. Convert to \(\text{kJ mol}^{-1}\):
\(E_{\text{mole}} = 238\text{ kJ mol}^{-1}\) (to 3 significant figures).

(d)(i) Tartrazine (\(427\text{ nm}\)) has a larger HOMO-LUMO energy gap than Allura Red AC (\(504\text{ nm}\)). Since \(E = hc/\lambda\), energy is inversely proportional to wavelength. A shorter wavelength of absorption corresponds to a larger energy transition.

(d)(ii) When white light passes through the dye, some wavelengths are absorbed while the unabsorbed (complementary) wavelengths are transmitted/reflected to be seen by the eye.
- Allura Red AC absorbs green/blue-green light (around \(504\text{ nm}\)) and transmits/reflects red light.
- Tartrazine absorbs blue/violet light (around \(427\text{ nm}\)) and transmits/reflects yellow light.

(e)(i) Reagents: Sodium nitrate(III) (\(\text{NaNO}_2\)) and dilute hydrochloric acid (\(\text{HCl}\)) (to form nitrous acid, \(\text{HNO}_2\)).
Condition: Temperature kept between \(0\ ^\circ\text{C}\) and \(5\ ^\circ\text{C}\).

(e)(ii) The benzenediazonium chloride intermediate formed is highly unstable. If the temperature rises above \(5\ ^\circ\text{C}\), it rapidly decomposes via a nucleophilic substitution reaction with water, producing phenol and releasing nitrogen gas.

(a)
- 1 mark: Correct structure of azo group showing double bond: \(\text{-N=N-}\).

(b)
This is a starred question designed to evaluate the Quality of Extended Response (QER).
- Level 3 (5-6 marks): Comprehensive explanation linking conjugation, HOMO-LUMO energy gap reduction, electron transitions (\(\pi \rightarrow \pi^*\)), and visible light absorption to complementary colours. Wording is clear and technically precise.
- Level 2 (3-4 marks): Explains that conjugation decreases the HOMO-LUMO gap allowing absorption of lower energy light, but explanation of the mechanism of colour observation or transition types lacks detail.
- Level 1 (1-2 marks): Mentions that visible light is absorbed because of the azo/conjugated group, but fails to clearly define HOMO-LUMO gaps or complementary colours.

(c)
- 1 mark: Conversion of wavelength to \(504 \times 10^{-9}\text{ m}\).
- 1 mark: Energy of single photon = \(3.95 \times 10^{-19}\text{ J}\).
- 1 mark: Multiplies by Avogadro constant.
- 1 mark: Correct final answer: \(238\text{ kJ mol}^{-1}\) (accept 237-238, must show correct units and 3 significant figures).

(d)(i)
- 1 mark: Identifies Tartrazine as having the larger gap.
- 1 mark: Justifies using \(E \propto 1/\lambda\) or states shorter wavelength means higher energy.

(d)(ii)
- 1 mark: Mentions the concept of complementary colours.
- 1 mark: Links absorption of green/blue-green (\(504\text{ nm}\)) in Allura Red to red appearance.
- 1 mark: Links absorption of blue/violet (\(427\text{ nm}\)) in Tartrazine to yellow appearance.

(e)(i)
- 1 mark: Reagents: nitrous acid / \(\text{NaNO}_2\) and \(\text{HCl}\).
- 1 mark: Temperature range: \(0-5\ ^\circ\text{C}\).

(e)(ii)
- 1 mark: Explains that benzenediazonium salt/ion is unstable at higher temperatures.
- 1 mark: Identifies the products of decomposition (phenol and nitrogen gas).

(a) The acid dissociation constant is:
$$K_a = \frac{[\text{H}^+][\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}$$

(b)
$$pK_a = -\log_{10}(K_a) = -\log_{10}(1.60 \times 10^{-7}) = 6.80$$

(c) Using the Henderson-Hasselbalch equation:
$$\text{pH} = pK_a + \log_{10}\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right)$$
$$7.20 = 6.80 + \log_{10}\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right)$$
$$\log_{10}\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right) = 0.40$$
$$\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} = 10^{0.40} = 2.51$$

(d)(i) Added hydrogen ions react with the conjugate base, hydrogenphosphate:
$$\text{HPO}_4^{2-}(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{PO}_4^-(aq)$$
Because the added free \(\text{H}^+\) ions are converted into the weak acid, the concentration of free \(\text{H}^+\) (and therefore pH) remains nearly constant.

(d)(ii) Added hydroxide ions react with the weak acid, dihydrogenphosphate:
$$\text{H}_2\text{PO}_4^-(aq) + \text{OH}^-(aq) \rightarrow \text{HPO}_4^{2-}(aq) + \text{H}_2\text{O}(l)$$
Because the added free \(\text{OH}^-\) ions are consumed to produce water, the pH remains nearly constant.

(e)(i)
Total final volume of mixture = \(50.0 + 150 = 200.0\text{ cm}^3 = 0.200\text{ dm}^3\).
Moles of \(\text{H}_2\text{PO}_4^-\). added = \(0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\).
Final concentration of \(\text{H}_2\text{PO}_4^- = \frac{7.50 \times 10^{-3}\text{ mol}}{0.200\text{ dm}^3} = 0.0375\text{ mol dm}^{-3}\).
Moles of \(\text{HPO}_4^{2-}\). added = \(0.150\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 1.50 \times 10^{-2}\text{ mol}\).
Final concentration of \(\text{HPO}_4^{2-} = \frac{1.50 \times 10^{-2}\text{ mol}}{0.200\text{ dm}^3} = 0.0750\text{ mol dm}^{-3}\).

(e)(ii)
$$[\text{H}^+] = K_a \times \frac{[\text{H}_2\text{PO}_4^-]}{[\text{HPO}_4^{2-}]} = 6.20 \times 10^{-8} \times \frac{0.0375}{0.0750} = 3.10 \times 10^{-8}\text{ mol dm}^{-3}$$
$$\text{pH} = -\log_{10}(3.10 \times 10^{-8}) = 7.51$$

(f)
- A buffer solution is a solution that minimises pH changes when small amounts of an acid or base are added.
- A mixture of a strong acid (like \(\text{HCl}\)) and its conjugate base (\(\text{Cl}^-\)) cannot buffer because:
1. The chloride ion, \(\text{Cl}^-\), is an exceptionally weak conjugate base with no basic properties in water; it will not react with added \(\text{H}^+\) to form undissociated \(\text{HCl}\).
2. A strong acid is fully dissociated, so there is no reservoir of undissociated acid molecules to release more hydrogen ions when an alkali is added.

(a)
- 1 mark: Correct expression for \(K_a = \frac{[\text{H}^+][\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\) (reject round brackets).

(b)
- 1 mark: \(6.80\) (must be to 2 decimal places).

(c)
- 1 mark: Setting up pH equation with correct values: \(7.20 = 6.80 + \log_{10}(\text{ratio})\).
- 1 mark: Rearranging to find \(\log_{10}(\text{ratio}) = 0.40\).
- 1 mark: Correct ratio = \(2.51\) (or \(2.5\)).

(d)(i)
- 1 mark: Equation: \(\text{HPO}_4^{2-} + \text{H}^+ \rightarrow \text{H}_2\text{PO}_4^-\).
- 1 mark: Explains that \(\text{H}^+\) is removed from solution, keeping pH stable.

(d)(ii)
- 1 mark: Equation: \(\text{H}_2\text{PO}_4^- + \text{OH}^- \rightarrow \text{HPO}_4^{2-} + \text{H}_2\text{O}\).
- 1 mark: Explains that \(\text{OH}^-\). is neutralised, keeping pH stable.

(e)(i)
- 1 mark: Total volume of \(0.200\text{ dm}^3\).
- 1 mark: Correct concentration of \([\text{H}_2\text{PO}_4^-] = 0.0375\text{ mol dm}^{-3}\).
- 1 mark: Correct concentration of \([\text{HPO}_4^{2-}] = 0.0750\text{ mol dm}^{-3}\).

(e)(ii)
- 1 mark: Correct rearrangement of \(K_a\) to solve for \([\text{H}^+]\).
- 1 mark: Substitution of concentrations and Ka value to get \([\text{H}^+] = 3.10 \times 10^{-8}\text{ mol dm}^{-3}\).
- 1 mark: Calculation of pH = \(7.51\).
- 1 mark: Appropriate rounding (2 decimal places for pH).

(f)
- 1 mark: Defines buffer solution (minimises change in pH on addition of small amounts of acid/alkali).
- 1 mark: Explains that \(\text{Cl}^-\). has negligible basic strength and cannot absorb added \(\text{H}^+\).
- 2 marks: Explains that \(\text{HCl}\). is fully dissociated, so adding alkali directly changes \([\text{H}^+]\) without any reservoir to replace lost \(\text{H}^+\) ions.

(a) Pipette exactly 25.0 cm3 of the commercial bleach using a safety pipette filler into a clean 250.0 cm3 volumetric flask. Add distilled water to the flask until the level is just below the graduation mark, then add water dropwise using a teat pipette until the bottom of the meniscus is aligned with the graduation mark at eye level. Stopper the flask securely and invert it several times to mix the solution thoroughly to ensure it is homogeneous.

(b) (i) \(ClO^- + 2I^- + 2H^+ \rightarrow Cl^- + I_2 + H_2O\)
(ii) \(I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}\)

(c) Titrate the mixture with sodium thiosulfate until the brown iodine colour fades to a pale straw-yellow colour. Add a few drops of starch indicator solution at this point, which will turn the mixture blue-black. Continue adding sodium thiosulfate dropwise until the blue-black colour just disappears to leave a completely colourless solution.

(d) Moles of thiosulfate used: \(n(S_2O_3^{2-}) = 0.100 \times \frac{26.35}{1000} = 2.635 \times 10^{-3}\text{ mol}\).
From the equations, \(1\text{ mol of } ClO^-\ \equiv 1\text{ mol of } I_2 \equiv 2\text{ mol of } S_2O_3^{2-}\).
Moles of \(ClO^-\ \text{ in } 25.0\text{ cm}^3\ \text{ aliquot} = \frac{2.635 \times 10^{-3}}{2} = 1.3175 \times 10^{-3}\text{ mol}\).
Moles of \(ClO^-\ \text{ in } 250.0\text{ cm}^3\ \text{ diluted solution} = 1.3175 \times 10^{-2}\text{ mol}\), which is also the amount in \(25.0\text{ cm}^3\) of original bleach.
Mass of \(NaClO\ \text{ in } 25.0\text{ cm}^3\ \text{ of original bleach} = 1.3175 \times 10^{-2} \times 74.5 = 0.9815\text{ g}\).
Mass of \(25.0\text{ cm}^3\ \text{ of original bleach} = 25.0 \times 1.08 = 27.0\text{ g}\).
Percentage by mass of \(NaClO = \frac{0.9815}{27.0} \times 100 = 3.635\%\), which rounds to \(3.64\%\).

(a) [Total: 4 marks]
- Award 1 mark for pipetting 25.0 cm3 of commercial bleach into a 250.0 cm3 volumetric flask.
- Award 1 mark for adding distilled water dropwise until the bottom of the meniscus is aligned with the graduation mark (at eye level).
- Award 1 mark for inserting the stopper.
- Award 1 mark for inverting the flask multiple times to ensure the solution is homogeneous.

(b) [Total: 2 marks]
- Award 1 mark for: \(ClO^- + 2I^- + 2H^+ \rightarrow Cl^- + I_2 + H_2O\) (ignore state symbols).
- Award 1 mark for: \(I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}\).

(c) [Total: 4 marks]
- Award 1 mark for titrating until the solution is a pale straw/yellow colour.
- Award 1 mark for adding starch indicator (turns blue-black).
- Award 1 mark for adding thiosulfate dropwise near the end point.
- Award 1 mark for identifying the correct end point colour change (blue-black to colourless).

(d) [Total: 5 marks]
- Award 1 mark for calculating thiosulfate moles: \(2.635 \times 10^{-3}\text{ mol}\).
- Award 1 mark for calculating \(ClO^-\ \text{ moles in aliquot}: 1.3175 \times 10^{-3}\text{ mol}\).
- Award 1 mark for calculating \(NaClO\ \text{ mass in original bleach}: 0.9815\text{ g}\).
- Award 1 mark for calculating mass of bleach sample: \(27.0\text{ g}\).
- Award 1 mark for the final percentage: \(3.64\%\) (must be 3 s.f.). Accept \(3.63\%\) to \(3.65\%\).

(a) The diagram must show a heating source, a flask (round-bottomed or pear-shaped) containing the reaction mixture, and a vertical Liebig condenser with cold water entering at the bottom and leaving at the top. The top of the condenser must remain unstoppered. Reflux is used because it allows the reaction mixture to be heated to its boiling point for a prolonged period without the loss of volatile reactants, products, or solvent. Volatile vapours condense in the vertical condenser and drop back into the reaction flask.

(b) (i) \(R_f\) of Spot 1 = \(\frac{3.2\text{ cm}}{8.0\text{ cm}} = 0.40\), which corresponds to Glycine. \(R_f\) of Spot 2 = \(\frac{5.6\text{ cm}}{8.0\text{ cm}} = 0.70\), which corresponds to Alanine. Therefore, dipeptide **X** is made of Glycine and Alanine.
(ii) To make the spots visible, spray the TLC plate with ninhydrin and heat (or expose to UV light). Amino acids separate because of their different relative solubilities in the mobile phase and their different adsorption onto the stationary phase.

(c) The amino acid corresponding to Spot 2 is Alanine (structure: \(H_2N-CH(CH_3)-COOH\) or protonated form). The doublet at \(\delta = 1.4\text{ ppm}\) (integrating to 3H) corresponds to the three equivalent protons on the methyl group (\(-CH_3\)) split by the single adjacent \(-CH-\) proton into a doublet (\(n+1 = 1+1 = 2\)). The quartet at \(\delta = 3.9\text{ ppm}\) (integrating to 1H) corresponds to the single proton on the \(-CH-\) carbon split by the three adjacent methyl protons into a quartet (\(n+1 = 3+1 = 4\)).

(a) [Total: 5 marks]
- Award 1 mark for a clearly drawn round-bottomed/pear-shaped flask with a vertical Liebig condenser attached.
- Award 1 mark for correct labeling of the Liebig condenser showing water entering at the bottom and exiting at the top.
- Award 1 mark for showing the top of the condenser is open/unstoppered.
- Award 1 mark for explaining that reflux prevents loss of volatile substances/solvent during heating.
- Award 1 mark for explaining that condensed vapours return to the reaction mixture, allowing prolonged high-temperature reaction.

(b) [Total: 4 marks]
- Award 1 mark for calculating \(R_f\) of Spot 1 = \(0.40\) and identifying Glycine.
- Award 1 mark for calculating \(R_f\) of Spot 2 = \(0.70\) and identifying Alanine.
- Award 1 mark for stating that ninhydrin spray (and heating) or UV light is used to visualize the spots.
- Award 1 mark for explaining separation in terms of relative partition between the mobile phase and stationary phase.

(c) [Total: 6 marks]
- Award 1 mark for identifying the amino acid as Alanine.
- Award 1 mark for assigning the doublet at \(\delta = 1.4\text{ ppm}\) to the methyl (\(-CH_3\)) protons.
- Award 1 mark for stating that the 3H integration confirms the three protons of the methyl group.
- Award 1 mark for explaining that the doublet splitting is caused by the adjacent single \(-CH-\) proton.
- Award 1 mark for assigning the quartet at \(\delta = 3.9\text{ ppm}\) to the \(-CH-\) proton with a 1H integration.
- Award 1 mark for explaining that the quartet splitting is caused by the three adjacent methyl protons.

(a) 'Heating to constant mass' means heating the sample, cooling it in a desiccator, and weighing it. This cycle is repeated until successive mass measurements are identical, which ensures all water of crystallisation is completely driven off.

(b) (i) Mass of water lost = \(21.66\text{ g} - 20.94\text{ g} = 0.72\text{ g}\). Mass of anhydrous residue = \(20.94\text{ g} - 18.25\text{ g} = 2.69\text{ g}\).
(ii) Moles of \(H_2O = \frac{0.72}{18.0} = 0.040\text{ mol}\). Moles of \(CuCl_2 = \frac{2.69}{134.5} = 0.020\text{ mol}\). Ratio of \(H_2O : CuCl_2 = \frac{0.040}{0.020} = 2 : 1\). Therefore, \(x = 2\).

(c) (i) \(Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)\)
(ii) Moles of \(AgCl = \frac{1.434}{143.4} = 0.0100\text{ mol}\). Since each mole of \(AgCl\) contains one mole of \(Cl^-\), moles of \(Cl^-\ \text{ in } 25.0\text{ cm}^3\ \text{ aliquot} = 0.0100\text{ mol}\). The total volume is \(100.0\text{ cm}^3\), so total moles of \(Cl^-\ \text{ in } 2.69\text{ g} = 0.0100 \times \frac{100.0}{25.0} = 0.0400\text{ mol}\).

(d) Mass of chlorine in the anhydrous sample = \(0.0400\text{ mol} \times 35.5\text{ g mol}^{-1} = 1.42\text{ g}\). Experimental percentage of chlorine = \(\frac{1.42}{2.69} \times 100 = 52.79\%\). Theoretical percentage of chlorine in pure anhydrous \(CuCl_2 = \frac{2 \times 35.5}{134.5} \times 100 = 52.79\%\). The experimental percentage is identical to the theoretical value, showing that the anhydrous copper halide residue is pure \(CuCl_2\).

(a) [Total: 2 marks]
- Award 1 mark for explaining that the process involves repeating the heating, cooling, and weighing cycle.
- Award 1 mark for stating that it is finished when consecutive mass readings are constant to ensure all water of crystallisation has been removed.

(b) [Total: 4 marks]
- Award 1 mark for correct masses of water lost (0.72 g) and anhydrous residue (2.69 g).
- Award 1 mark for calculating moles of water (0.040 mol) and moles of anhydrous copper halide (0.020 mol).
- Award 1 mark for ratio calculation and concluding that \(x = 2\).

(c) [Total: 5 marks]
- Award 1 mark for the precipitation equation with state symbols: \(Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)\).
- Award 1 mark for calculating moles of \(AgCl = 0.0100\text{ mol}\).
- Award 1 mark for equating moles of chloride in aliquot to moles of \(AgCl\) (\(0.0100\text{ mol}\)).
- Award 2 marks for scaling up by a factor of 4 to find total moles of chloride = \(0.0400\text{ mol}\) (Award 1 mark if scaling factor is omitted).

(d) [Total: 4 marks]
- Award 1 mark for calculating experimental mass of chlorine = \(1.42\text{ g}\).
- Award 1 mark for finding experimental chlorine percentage = \(52.8\%\) (or \(52.79\%\)).
- Award 1 mark for calculating theoretical percentage of chlorine in pure \(CuCl_2\) = \(52.8\%\).
- Award 1 mark for comparing the values and concluding that the sample is pure anhydrous \(CuCl_2\).

(a) Use the dilution formula: \(V_1 \times C_1 = V_2 \times C_2 \implies V_1 \times 0.100 = 100 \times 0.040 \implies V_1 = 40.0\text{ cm}^3\). Measure exactly \(40.0\text{ cm}^3\) of the stock copper(II) solution using a volumetric pipette (or a burette) and transfer it into a clean \(100\text{ cm}^3\) volumetric flask. Add distilled water up to the graduation mark, and then invert the stoppered flask several times to mix. The volumetric flask is critical because it has a very low uncertainty (is highly precise) and ensures the final volume is exactly \(100\text{ cm}^3\), which is necessary for preparing accurate standard solutions.

(b) (i) The student should record the absorbance of a standard copper(II) solution across a range of wavelengths to determine the wavelength of maximum absorption (\(\lambda_{\max}\)). Alternatively, a complementary orange/red filter should be selected because a blue copper(II) solution absorbs light in the orange-red region of the spectrum, providing maximum sensitivity for the measurements.
(ii) Distilled water (the solvent used to prepare the standard solutions) is used as the blank.

(c) (i) Using the calibration equation: \([Cu^{2+}(aq)] = \frac{\text{Absorbance}}{7.50} = \frac{0.54}{7.50} = 0.0720\text{ mol dm}^{-3}\).
(ii) Moles of \(Cu^{2+}\) in \(250.0\text{ cm}^3\ \text{ of solution} = 0.0720\text{ mol dm}^{-3} \times 0.250\text{ dm}^3 = 0.0180\text{ mol}\). Since each mole of copper metal dissolves to produce one mole of copper(II) ions, moles of copper in the coin = \(0.0180\text{ mol}\). Mass of copper in the coin = \(0.0180\text{ mol} \times 63.5\text{ g mol}^{-1} = 1.143\text{ g}\). Percentage of copper by mass in the coin = \(\frac{1.143}{1.20} \times 100 = 95.25\%\), which rounds to \(95.3\%\) (three significant figures).

(a) [Total: 5 marks]
- Award 1 mark for showing dilution calculation: \(40.0\text{ cm}^3\) of stock solution is needed.
- Award 1 mark for specifying a volumetric pipette (or burette) to measure the \(40.0\text{ cm}^3\) portion.
- Award 1 mark for transferring to a \(100\text{ cm}^3\) volumetric flask and filling to the mark with distilled water.
- Award 1 mark for mixing the solution by inversion.
- Award 1 mark for explaining that a volumetric flask has high precision/low uncertainty which is critical for preparing accurate standard solutions.

(b) [Total: 3 marks]
- Award 1 mark for identifying the need to measure absorbance across a range of wavelengths to find \(\lambda_{\max}\) or choosing a complementary filter (orange/red).
- Award 1 mark for explaining that this choice maximizes the sensitivity of the colorimetric analysis.
- Award 1 mark for identifying distilled water as the blank to set the colorimeter to zero.

(c) [Total: 7 marks]
- Award 2 marks for calculating the concentration of \(Cu^{2+}\) = \(0.0720\text{ mol dm}^{-3}\) (Award 1 mark if formula is inverted or calculation error is present).
- Award 1 mark for calculating moles of \(Cu^{2+}\) in \(250.0\text{ cm}^3\) = \(0.0180\text{ mol}\).
- Award 1 mark for equating moles of copper metal to moles of copper(II) ions.
- Award 1 mark for calculating the mass of copper in the coin = \(1.143\text{ g}\).
- Award 1 mark for finding the percentage = \(95.25\%\).
- Award 1 mark for rounding the final percentage to three significant figures: \(95.3\%\).