Cambridge IAL · Thinka-original Practice Paper

2023 Cambridge IAL Chemistry (9701) Practice Paper with Answers

Thinka Jun 2023 (V3) Cambridge International A Level-Style Mock — Chemistry (9701)

270 marks465 mins2023
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V3) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.

Paper 13 (Multiple Choice)

Answer all forty questions on the multiple choice answer sheet. Each correct answer scores one mark.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
For a reaction \(2A + B + 2C \rightarrow \text{products}\), initial rate data were obtained at constant temperature:

* Experiment 1: \([A] = 0.10\text{ mol dm}^{-3}\), \([B] = 0.10\text{ mol dm}^{-3}\), \([C] = 0.10\text{ mol dm}^{-3}\); initial rate = \(1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)
* Experiment 2: \([A] = 0.20\text{ mol dm}^{-3}\), \([B] = 0.10\text{ mol dm}^{-3}\), \([C] = 0.10\text{ mol dm}^{-3}\); initial rate = \(2.4 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)
* Experiment 3: \([A] = 0.10\text{ mol dm}^{-3}\), \([B] = 0.20\text{ mol dm}^{-3}\), \([C] = 0.10\text{ mol dm}^{-3}\); initial rate = \(4.8 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)
* Experiment 4: \([A] = 0.10\text{ mol dm}^{-3}\), \([B] = 0.10\text{ mol dm}^{-3}\), \([C] = 0.20\text{ mol dm}^{-3}\); initial rate = \(1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)

What are the units and the numerical value of the rate constant, \(k\), for this reaction?
  1. A.Numerical value = \(1.2\); units = \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)
  2. B.Numerical value = \(1.2\); units = \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
  3. C.Numerical value = \(12\); units = \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)
  4. D.Numerical value = \(12\); units = \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
Show answer & marking scheme

Worked solution

First, determine the order of reaction with respect to each reactant:
1. Comparing Experiment 1 and 2, when \([A]\) is doubled while \([B]\) and \([C]\) are kept constant, the initial rate doubles (\(1.2 \times 10^{-3}\) to \(2.4 \times 10^{-3}\)). Thus, the order with respect to \(A\) is 1.
2. Comparing Experiment 1 and 3, when \([B]\) is doubled while \([A]\) and \([C]\) are kept constant, the initial rate quadruples (\(1.2 \times 10^{-3}\) to \(4.8 \times 10^{-3}\)). Thus, the order with respect to \(B\) is 2.
3. Comparing Experiment 1 and 4, when \([C]\) is doubled while \([A]\) and \([B]\) are kept constant, the initial rate remains unchanged. Thus, the order with respect to \(C\) is 0.

The rate equation is: \(\text{Rate} = k[A]^1[B]^2[C]^0 = k[A][B]^2\).

To find the value of \(k\), substitute values from Experiment 1:
\(1.2 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} = k \times (0.10\text{ mol dm}^{-3}) \times (0.10\text{ mol dm}^{-3})^2\)
\(1.2 \times 10^{-3} = k \times 1.0 \times 10^{-3}\)
\(k = 1.2\)

Units of \(k\):
\(k = \frac{\text{Rate}}{[A][B]^2} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Marking scheme

1 mark: Correctly identifies the order of reaction for each reactant, calculates the value of the rate constant, and determines the correct units.
Question 2 · multiple-choice
1 marks
The stability constants, \(K_{\text{stab}}\), for three different copper(II) complexes in aqueous solution at \(298\text{ K}\) are given:

1. \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O} \quad K_{\text{stab}} = 4.2 \times 10^5\text{ dm}^{12}\text{ mol}^{-4}\)
2. \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O} \quad K_{\text{stab}} = 1.2 \times 10^{13}\text{ dm}^{12}\text{ mol}^{-4}\)
3. \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 2\text{en} \rightleftharpoons [\text{Cu}(\text{en})_2(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O} \quad K_{\text{stab}} = 1.6 \times 10^{20}\text{ dm}^6\text{ mol}^{-2}\)
(where 'en' represents 1,2-diaminoethane)

Which statement is correct?
  1. A.Addition of excess 1,2-diaminoethane to a solution of \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\) will result in the displacement of ammonia ligands because \([\text{Cu}(\text{en})_2(\text{H}_2\text{O})_2]^{2+}\) is thermodynamically more stable.
  2. B.The complex \([\text{CuCl}_4]^{2-}\) is the most stable of the three complexes because its stability constant is the smallest.
  3. C.The value of \(K_{\text{stab}}\) indicates that the rate of ligand exchange is fastest for the formation of \([\text{Cu}(\text{en})_2(\text{H}_2\text{O})_2]^{2+}\).
  4. D.The addition of a high concentration of chloride ions to \([\text{Cu}(\text{en})_2(\text{H}_2\text{O})_2]^{2+}\) will completely displace the 1,2-diaminoethane ligands under standard conditions.
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Worked solution

A larger stability constant value, \(K_{\text{stab}}\), indicates a more stable complex. Since \(K_{\text{stab}}\) for the 1,2-diaminoethane complex (\(1.6 \times 10^{20}\)) is significantly larger than that for the ammonia complex (\(1.2 \times 10^{13}\)), \([\text{Cu}(\text{en})_2(\text{H}_2\text{O})_2]^{2+}\) is thermodynamically much more stable. Thus, adding 'en' to a solution of the ammonia complex will displace ammonia ligands. Note that stability constants are thermodynamic parameters and do not indicate the rates of reactions.

Marking scheme

1 mark: Correctly identifies that a higher K_stab value corresponds to greater thermodynamic stability, leading to ligand displacement.
Question 3 · multiple-choice
1 marks
The amino acid aspartic acid has the structure \(\text{HOOC-CH}_2\text{-CH(NH}_2\text{)-COOH}\). It has three \(pK_a\) values:

* \(pK_{a1} = 2.0\) (for the \(\alpha\)-carboxyl group)
* \(pK_{a2} = 3.9\) (for the side-chain carboxyl group)
* \(pK_{a3} = 9.8\) (for the \(\alpha\)-ammonium group)

What is the predominant ionic structure of aspartic acid in a buffer solution at \(\text{pH} = 6.0\)?
  1. A.\({}^-\text{OOC-CH}_2\text{-CH(NH}_3^+)\text{-COO}^-\)
  2. B.\({}^-\text{OOC-CH}_2\text{-CH(NH}_2)\text{-COO}^-\)
  3. C.\(\text{HOOC-CH}_2\text{-CH(NH}_3^+)\text{-COOH}\)
  4. D.\(\text{HOOC-CH}_2\text{-CH(NH}_3^+)\text{-COO}^-\)
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Worked solution

At \(\text{pH} = 6.0\):
1. Since \(\text{pH} > pK_{a1}\) (6.0 > 2.0), the \(\alpha\)-carboxyl group is deprotonated to form \({}^-\text{COO}^-\).
2. Since \(\text{pH} > pK_{a2}\) (6.0 > 3.9), the side-chain carboxyl group is also deprotonated to form \({}^-\text{COO}^-\).
3. Since \(\text{pH} < pK_{a3}\) (6.0 < 9.8), the \(\alpha\)-amino group remains protonated as \(-\text{NH}_3^+\).

Combining these yields the structure: \({}^-\text{OOC-CH}_2\text{-CH(NH}_3^+)\text{-COO}^-\).

Marking scheme

1 mark: Correctly identifies the protonation/deprotonation state of all three functional groups based on the relative pH and pKa values.
Question 4 · multiple-choice
1 marks
A sample of \(2.32\text{ g}\) of an anhydrous metal carbonate, \(\text{MCO}_3\), is heated strongly until it completely decomposes according to the following equation:

\[\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})\]

The carbon dioxide gas produced occupies a volume of \(480\text{ cm}^3\) at room temperature and pressure (r.t.p.).
[Assume 1 mole of gas occupies \(24.0\text{ dm}^3\) at r.t.p.]

What is the identity of the metal, \(\text{M}\)?
  1. A.Calcium (\(A_r = 40.1\))
  2. B.Iron (\(A_r = 55.8\))
  3. C.Copper (\(A_r = 63.5\))
  4. D.Zinc (\(A_r = 65.4\))
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Worked solution

First, calculate the moles of \(\text{CO}_2\) produced:
\(\text{Moles of CO}_2 = \frac{480\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.020\text{ mol}\).

Since the molar ratio of \(\text{MCO}_3 : \text{CO}_2\) is \(1:1\), the moles of \(\text{MCO}_3\) decomposed is also \(0.020\text{ mol}\).

Calculate the relative formula mass (\(M_r\)) of \(\text{MCO}_3\):
\(M_r(\text{MCO}_3) = \frac{\text{mass}}{\text{moles}} = \frac{2.32\text{ g}}{0.020\text{ mol}} = 116.0\text{ g mol}^{-1}\).

Subtract the relative formula mass of the carbonate group (\(\text{CO}_3^{2-}\)) to find the relative atomic mass (\(A_r\)) of \(\text{M}\):
\(M_r(\text{CO}_3) = 12.0 + (3 \times 16.0) = 60.0\text{ g mol}^{-1}\).
\(A_r(\text{M}) = 116.0 - 60.0 = 56.0\text{ g mol}^{-1}\).

This relative atomic mass matches iron (\(A_r = 55.8\)) very closely.

Marking scheme

1 mark: Correctly calculates moles of CO2, relates it to the moles of carbonate, determines the molar mass of the compound, and identifies iron.
Question 5 · multiple-choice
1 marks
An aqueous solution contains \(4.0\text{ g}\) of an organic compound \(X\) dissolved in \(100\text{ cm}^3\) of water. The partition coefficient, \(K_{pc}\), of compound \(X\) between ether and water is defined as:

\[K_{pc} = \frac{[X]_{\text{ether}}}{[X]_{\text{water}}} = 4.0\]

If the aqueous solution is shaken with a single \(50\text{ cm}^3\) portion of ether, what mass of compound \(X\) is extracted into the ether layer?
  1. A.\(1.33\text{ g}\)
  2. B.\(2.00\text{ g}\)
  3. C.\(2.67\text{ g}\)
  4. D.\(3.20\text{ g}\)
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Worked solution

Let \(x\) be the mass (in grams) of \(X\) extracted into the ether layer. The mass of \(X\) remaining in the water layer is \((4.0 - x)\text{ g}\).

Concentrations in each layer:
\([X]_{\text{ether}} = \frac{x}{50}\text{ g cm}^{-3}\)
\([X]_{\text{water}} = \frac{4.0 - x}{100}\text{ g cm}^{-3}\)

Substitute into the partition coefficient expression:
\(K_{pc} = \frac{x / 50}{(4.0 - x) / 100} = 4.0\)

Simplify and solve for \(x\):
\(\frac{x}{50} \times \frac{100}{4.0 - x} = 4.0\)
\(\frac{2x}{4.0 - x} = 4.0\)
\(2x = 4.0(4.0 - x)\)
\(2x = 16.0 - 4x\)
\(6x = 16.0\)
\(x = 2.67\text{ g}\).

Marking scheme

1 mark: Correctly formulates the equilibrium partition expression and solves the algebra to find the extracted mass.
Question 6 · multiple-choice
1 marks
Which statement correctly describes and explains the trend in properties of Group 2 elements and their compounds from magnesium to barium?
  1. A.The solubility of Group 2 hydroxides increases down the group because the lattice energy of the hydroxides decreases more rapidly than the hydration energy of the cations.
  2. B.The solubility of Group 2 sulfates increases down the group because the hydration energy of the cations decreases less rapidly than the lattice energy.
  3. C.The thermal stability of Group 2 carbonates decreases down the group because the charge density of the cation increases.
  4. D.The first ionisation energy of the elements increases down the group because of increased shielding of the outer electrons.
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Worked solution

As we go down Group 2, the solubility of the hydroxides increases. This is because both lattice energy and the hydration energy of the cation decrease down the group, but due to the small size of the hydroxide ion, the lattice energy decreases much more rapidly. This makes the overall enthalpy of solution increasingly exothermic (or less endothermic), increasing the solubility. For sulfates, solubility decreases down the group. Thermal stability of carbonates increases down the group because the larger cations have a lower charge density and polarize the carbonate ion less. First ionisation energy decreases down the group due to increased shielding and distance.

Marking scheme

1 mark: Identifies the correct trend and the scientific rationale regarding lattice and hydration energies for Group 2 hydroxides.
Question 7 · multiple-choice
1 marks
Aqueous solutions of transition metal complexes are often coloured. Which statement correctly explains the origin of the colour in the complex ion \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\)?
  1. A.d-orbitals split into two energy levels; d-electrons are promoted to a higher energy level by absorbing a photon of visible light, and the complementary colour is transmitted.
  2. B.Electrons are promoted from the 3d orbital to the 4s orbital by absorbing visible light.
  3. C.Light is emitted when excited electrons fall back from a higher energy d-orbital to a lower energy d-orbital.
  4. D.The ligands absorb visible light, causing vibrations that emit light of a specific wavelength.
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Worked solution

When ligands like \(\text{H}_2\text{O}\) approach the transition metal ion \(\text{Cu}^{2+}\), the five degenerate 3d orbitals split into two energy levels separated by an energy difference, \(\Delta E\). d-electrons in the lower energy level absorb a photon of visible light and are promoted to the higher energy level (d-d transition). The complementary colour (the wavelength not absorbed) is transmitted, which is the colour observed. Energy is not emitted as visible light in this process.

Marking scheme

1 mark: Identifies that splitting of d-orbitals, absorption of a photon of visible light during d-d excitation, and transmission of the complementary color are responsible for the color of transition complexes.
Question 8 · multiple-choice
1 marks
The decomposition of hydrogen peroxide, \(\text{H}_2\text{O}_2\), is first-order with respect to \(\text{H}_2\text{O}_2\):

\[2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})\]

In an experiment, the concentration of a sample of \(\text{H}_2\text{O}_2\) decreases from \(1.60\text{ mol dm}^{-3}\) to \(0.10\text{ mol dm}^{-3}\) in exactly \(120\text{ minutes}\).

What is the half-life of this reaction under these conditions?
  1. A.\(15\text{ minutes}\)
  2. B.\(30\text{ minutes}\)
  3. C.\(40\text{ minutes}\)
  4. D.\(60\text{ minutes}\)
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Worked solution

For a first-order reaction, the half-life is constant. Track the concentration halves over successive half-lives:

* Initial concentration = \(1.60\text{ mol dm}^{-3}\)
* After 1 half-life = \(0.80\text{ mol dm}^{-3}\)
* After 2 half-lives = \(0.40\text{ mol dm}^{-3}\)
* After 3 half-lives = \(0.20\text{ mol dm}^{-3}\)
* After 4 half-lives = \(0.10\text{ mol dm}^{-3}\)

Thus, exactly 4 half-lives (\(4 \times t_{1/2}\)) have passed in \(120\text{ minutes}\).
\(t_{1/2} = \frac{120\text{ minutes}}{4} = 30\text{ minutes}\).

Marking scheme

1 mark: Correctly determines the number of half-lives that have elapsed and divides the total time to get the correct half-life.
Question 9 · Multiple Choice
1 marks
In an investigation of the reaction \(2\text{P} + \text{Q} \rightarrow \text{R} + \text{S}\), the following initial rates were obtained at a constant temperature. Experiment 1: \([\text{P}] = 0.050\text{ mol dm}^{-3}\), \([\text{Q}] = 0.100\text{ mol dm}^{-3}\), Initial rate = \(1.50 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Experiment 2: \([\text{P}] = 0.100\text{ mol dm}^{-3}\), \([\text{Q}] = 0.100\text{ mol dm}^{-3}\), Initial rate = \(6.00 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Experiment 3: \([\text{P}] = 0.100\text{ mol dm}^{-3}\), \([\text{Q}] = 0.200\text{ mol dm}^{-3}\), Initial rate = \(1.20 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). What is the correct value and unit of the rate constant, \(k\), for this reaction?
  1. A.\(0.60\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)
  2. B.\(0.60\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
  3. C.\(1.20\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)
  4. D.\(1.20\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
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Worked solution

Comparing Experiment 1 and Experiment 2: \([\text{Q}]\) is constant, \([\text{P}]\) doubles, and the rate increases by a factor of 4 (from \(1.50 \times 10^{-4}\) to \(6.00 \times 10^{-4}\)). Therefore, the reaction is second order with respect to \(\text{P}\). Comparing Experiment 2 and Experiment 3: \([\text{P}]\) is constant, \([\text{Q}]\) doubles, and the rate doubles (from \(6.00 \times 10^{-4}\) to \(1.20 \times 10^{-3}\)). Therefore, the reaction is first order with respect to \(\text{Q}\). The rate equation is: \(\text{Rate} = k[\text{P}]^2[\text{Q}]\). Substituting values from Experiment 1: \(1.50 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} = k \times (0.050\text{ mol dm}^{-3})^2 \times (0.100\text{ mol dm}^{-3})\). This simplifies to \(1.50 \times 10^{-4} = k \times 2.50 \times 10^{-4}\), which gives \(k = 0.600\). The overall order is 3, so the units of \(k\) are \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Marking scheme

Award 1 mark for the correct answer A. Method: Deduce the order with respect to P is 2 and with respect to Q is 1. Set up the rate equation. Substitute experimental values to solve for the value of k (0.60). Determine the units based on the third-order overall rate equation.
Question 10 · Multiple Choice
1 marks
The table below lists some thermodynamic data for a Group 2 metal, \(\text{M}\), and fluorine. Enthalpy change of formation of \(\text{MF}_2\text{(s)}\) = \(-1100\text{ kJ mol}^{-1}\); Enthalpy change of atomisation of \(\text{M}\text{(s)}\) = \(+150\text{ kJ mol}^{-1}\); First ionisation energy of \(\text{M}\text{(g)}\) = \(+550\text{ kJ mol}^{-1}\); Second ionisation energy of \(\text{M}\text{(g)}\) = \(+1150\text{ kJ mol}^{-1}\); Enthalpy change of atomisation of fluorine, \(\Delta H_{at}^\ominus[\text{F}\text{(g)}]\) = \(+80\text{ kJ mol}^{-1}\); First electron affinity of fluorine, \(\text{EA}[\text{F}\text{(g)}]\) = \(-330\text{ kJ mol}^{-1}\). What is the lattice energy, \(\Delta H_{latt}^\ominus\), of \(\text{MF}_2\text{(s)}\)?
  1. A.\(-2450\text{ kJ mol}^{-1}\)
  2. B.\(-2370\text{ kJ mol}^{-1}\)
  3. C.\(-2700\text{ kJ mol}^{-1}\)
  4. D.\(-3770\text{ kJ mol}^{-1}\)
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Worked solution

The Born-Haber cycle for \(\text{MF}_2\) is: \(\Delta H_f^\ominus = \Delta H_{at}^\ominus[\text{M}\text{(s)}] + \text{IE}_1[\text{M}] + \text{IE}_2[\text{M}] + 2\Delta H_{at}^\ominus[\text{F}\text{(g)}] + 2\text{EA}[\text{F}] + \Delta H_{latt}^\ominus\). Note that 2 moles of gaseous F atoms are formed from the element, so we must double the enthalpy change of atomisation of fluorine: \(2 \times (+80) = +160\text{ kJ mol}^{-1}\). Similarly, 2 moles of \(\text{F}^-\text{(g)}\) are formed, so we must double the electron affinity of fluorine: \(2 \times (-330) = -660\text{ kJ mol}^{-1}\). Substituting the values: \(-1100 = 150 + 550 + 1150 + 160 + (-660) + \Delta H_{latt}^\ominus\), which simplifies to \(-1100 = 1350 + \Delta H_{latt}^\ominus\). Therefore, \(\Delta H_{latt}^\ominus = -1100 - 1350 = -2450\text{ kJ mol}^{-1}\).

Marking scheme

Award 1 mark for the correct answer A. Method: Identify and apply Hess's Law to the Born-Haber cycle for a Group 2 dihalide. Remember to double both the atomisation enthalpy and the electron affinity of fluorine since two moles of fluoride ions are present per mole of formula unit. Solve for the lattice enthalpy.
Question 11 · Multiple Choice
1 marks
The stability constant, \(K_{stab}\), for the complex ion \([\text{Fe(SCN)(H}_2\text{O)}_5]^{2+}\) is represented by the equilibrium: \([\text{Fe(H}_2\text{O)}_6]^{3+}\text{(aq)} + \text{SCN}^-\text{(aq)} \rightleftharpoons [\text{Fe(SCN)(H}_2\text{O)}_5]^{2+}\text{(aq)} + \text{H}_2\text{O(l)}\). The value of \(K_{stab}\) is \(8.9 \times 10^2\text{ dm}^3\text{ mol}^{-1}\). In an equilibrium mixture, the concentration of uncomplexed \([\text{Fe(H}_2\text{O)}_6]^{3+}\text{(aq)}\) is \(2.0 \times 10^{-3}\text{ mol dm}^{-3}\) and the concentration of \([\text{Fe(SCN)(H}_2\text{O)}_5]^{2+}\text{(aq)}\) is \(4.0 \times 10^{-3}\text{ mol dm}^{-3}\). What is the equilibrium concentration of \(\text{SCN}^-\text{(aq)}\)?
  1. A.\(2.2 \times 10^{-3}\text{ mol dm}^{-3}\)
  2. B.\(4.5 \times 10^2\text{ mol dm}^{-3}\)
  3. C.\(5.6 \times 10^{-4}\text{ mol dm}^{-3}\)
  4. D.\(1.1 \times 10^{-3}\text{ mol dm}^{-3}\)
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Worked solution

The stability constant expression is: \(K_{stab} = \frac{[[\text{Fe(SCN)(H}_2\text{O)}_5]^{2+}]}{[[\text{Fe(H}_2\text{O)}_6]^{3+}][\text{SCN}^-]}\). Substituting the given values: \(8.9 \times 10^2 = \frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3} \times [\text{SCN}^-]}\). This simplifies to \(8.9 \times 10^2 = \frac{2.0}{[\text{SCN}^-]}\), which gives \([\text{SCN}^-] = \frac{2.0}{8.9 \times 10^2} = 2.25 \times 10^{-3}\text{ mol dm}^{-3}\). To two significant figures, this is \(2.2 \times 10^{-3}\text{ mol dm}^{-3}\).

Marking scheme

Award 1 mark for the correct answer A. Method: Write down the correct stability constant expression, omitting water. Rearrange the expression to solve for the concentration of the thiocyanate ligand. Calculate the numerical value.
Question 12 · Multiple Choice
1 marks
Lysine is an amino acid with the formula \(\text{H}_2\text{N(CH}_2)_4\text{CH(NH}_2)\text{COOH}\). Which structure represents the predominant ionic species present in a solution of lysine at \(\text{pH} = 1.0\)?
  1. A.\({}^+\text{H}_3\text{N(CH}_2)_4\text{CH(NH}_3^+)\text{COOH}\)
  2. B.\({}^+\text{H}_3\text{N(CH}_2)_4\text{CH(NH}_3^+)\text{COO}^-\)
  3. C.\({}^+\text{H}_3\text{N(CH}_2)_4\text{CH(NH}_2)\text{COO}^-\)
  4. D.\(\text{H}_2\text{N(CH}_2)_4\text{CH(NH}_2)\text{COO}^-\)
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Worked solution

At a highly acidic pH of 1.0, there is a very high concentration of hydrogen ions. Both of the basic amine groups (one on the side chain and one on the alpha-carbon) are protonated to carry positive charges as \(-\text{NH}_3^+\). The carboxylic acid group remains protonated as \(-\text{COOH}\). Thus, the predominant species has a net positive charge of \(+2\), represented by \({}^+\text{H}_3\text{N(CH}_2)_4\text{CH(NH}_3^+)\text{COOH}\).

Marking scheme

Award 1 mark for the correct answer A. Method: Identify that at pH 1.0, both amine groups of lysine are protonated and the carboxylic acid group remains un-ionised. Select the corresponding structure.
Question 13 · Multiple Choice
1 marks
An organic compound, X, is soluble in both water and ether. The partition coefficient, \(K_{pc}\), of X between ether and water is defined as: \(K_{pc} = \frac{[\text{X}]_{\text{ether}}}{[\text{X}]_{\text{water}}} = 4.0\). An aqueous solution contains \(6.0\text{ g}\) of X dissolved in \(100\text{ cm}^3\) of water. This solution is shaken with a single \(50\text{ cm}^3\) portion of ether at a constant temperature. What mass of X remains in the aqueous layer after equilibrium has been established?
  1. A.\(1.2\text{ g}\)
  2. B.\(2.0\text{ g}\)
  3. C.\(4.0\text{ g}\)
  4. D.\(5.3\text{ g}\)
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Worked solution

Let \(m\) be the mass of X that is extracted into the ether layer. The mass of X remaining in the water layer is \(6.0 - m\). The concentration of X in the ether layer is \(\frac{m}{50}\text{ g cm}^{-3}\), and the concentration in the water layer is \(\frac{6.0 - m}{100}\text{ g cm}^{-3}\). Using the partition coefficient expression: \(K_{pc} = \frac{m / 50}{(6.0 - m) / 100} = 4.0\). This simplifies to \(\frac{2m}{6.0 - m} = 4.0\), leading to \(2m = 24.0 - 4.0m\), which gives \(6.0m = 24.0\) and \(m = 4.0\text{ g}\). Therefore, \(4.0\text{ g}\) is extracted and the mass remaining in the water layer is \(6.0 - 4.0 = 2.0\text{ g}\).

Marking scheme

Award 1 mark for the correct answer B. Method: Set up the algebraic equation for the partition coefficient using mass and volume. Solve for the extracted mass of X, then subtract it from the initial mass to find the mass remaining in the aqueous layer.
Question 14 · Multiple Choice
1 marks
When butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\), reacts with \(\text{HCN}\) in the presence of \(\text{NaCN}\) catalyst, an organic product is formed. Which statement about this reaction and its product is correct?
  1. A.The reaction is a nucleophilic addition that produces a racemic mixture of enantiomers.
  2. B.The reaction is an electrophilic addition that produces a single chiral enantiomer.
  3. C.The rate of reaction is increased by adding a strong mineral acid to increase the concentration of \(\text{H}^+\) ions.
  4. D.The organic product has the molecular formula \(\text{C}_5\text{H}_{11}\text{NO}\).
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Worked solution

The reaction is a nucleophilic addition of HCN to the carbonyl group. The carbonyl carbon in butanone is planar, allowing the \(\text{CN}^-\right.\) nucleophile to attack with equal probability from above or below the plane. This yields equal amounts of the two optical isomers, creating a racemic mixture. Since the carbon atom in the product is bonded to four different groups (methyl, ethyl, hydroxyl, and nitrile), it is chiral. Adding acid reduces the concentration of \(\text{CN}^-\right.\) nucleophile, slowing the rate. The molecular formula of the product is \(\text{C}_5\text{H}_9\text{NO}\).

Marking scheme

Award 1 mark for the correct answer A. Method: Identify the mechanism of the reaction between a ketone and HCN as nucleophilic addition. Recognise that planar reactant carbonyl geometry leads to equal rates of attack from both sides, producing a racemic mixture of the chiral product. Verify that other options are incorrect.
Question 15 · Multiple Choice
1 marks
A section of a synthetic polymer is shown below: \(\text{...-CO-CH}_2\text{CH}_2\text{-CO-NH-CH}_2\text{CH}_2\text{CH}_2\text{-NH-CO-CH}_2\text{CH}_2\text{-CO-NH-CH}_2\text{CH}_2\text{CH}_2\text{-NH-...}\). Which pair of monomers can be used to prepare this polymer?
  1. A.\(\text{HOOC(CH}_2)_2\text{COOH}\) and \(\text{H}_2\text{N(CH}_2)_3\text{NH}_2\)
  2. B.\(\text{HOOC(CH}_2)_3\text{COOH}\) and \(\text{H}_2\text{N(CH}_2)_2\text{NH}_2\)
  3. C.\(\text{HOOC(CH}_2)_2\text{COOH}\) and \(\text{H}_2\text{N(CH}_2)_2\text{NH}_2\)
  4. D.\(\text{HOOC(CH}_2)_3\text{COOH}\) and \(\text{H}_2\text{N(CH}_2)_3\text{NH}_2\)
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Worked solution

The synthetic polymer is a polyamide. Splitting the polyamide chain at the amide linkages (\(\text{-CO-NH-}\)) reveals the constituent monomers: the dicarboxylic acid portion has the carbon skeleton \(\text{-CO-CH}_2\text{CH}_2\text{-CO-}\), which corresponds to the monomer \(\text{HOOC-CH}_2\text{CH}_2\text{-COOH}\), or \(\text{HOOC(CH}_2)_2\text{COOH}\). The diamine portion has the carbon skeleton \(\text{-NH-CH}_2\text{CH}_2\text{CH}_2\text{-NH-}\), which corresponds to the monomer \(\text{H}_2\text{N-CH}_2\text{CH}_2\text{CH}_2\text{-NH}_2\), or \(\text{H}_2\text{N(CH}_2)_3\text{NH}_2\).

Marking scheme

Award 1 mark for the correct answer A. Method: Identify the amide repeating unit, separate the dicarboxylic acid and diamine segments, and count the number of carbon atoms in each segment to find the correct molecular formulas of the monomers.
Question 16 · Multiple Choice
1 marks
Two different Group 2 anhydrous nitrates, \(\text{X(NO}_3)_2\) and \(\text{Y(NO}_3)_2\), are heated under identical conditions. It is observed that \(\text{X(NO}_3)_2\) decomposes at a lower temperature than \(\text{Y(NO}_3)_2\). Which statement about \(\text{X}\) and \(\text{Y}\) is correct?
  1. A.The ionic radius of \(\text{X}^{2+}\) is greater than that of \(\text{Y}^{2+}\).
  2. B.The first ionisation energy of \(\text{X}\) is lower than that of \(\text{Y}\).
  3. C.The sulfate of \(\text{X}\) is more soluble in water than the sulfate of \(\text{Y}\).
  4. D.The hydroxide of \(\text{X}\) is more soluble in water than the hydroxide of \(\text{Y}\).
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Worked solution

Thermal stability of Group 2 nitrates increases down the group. Since \(\text{X(NO}_3)_2\) decomposes at a lower temperature than \(\text{Y(NO}_3)_2\), \(\text{X}\) must be located higher in Group 2 than \(\text{Y}\). Moving down the group: the ionic radius increases (making \(\text{X}^{2+}\) smaller than \(\text{Y}^{2+}\)), first ionisation energy decreases (making first IE of \(\text{X}\) higher than \(\text{Y}\)), sulfate solubility decreases (making the sulfate of \(\text{X}\) more soluble than that of \(\text{Y}\)), and hydroxide solubility increases (making the hydroxide of \(\text{X}\) less soluble than that of \(\text{Y}\)). Thus, statement C is correct.

Marking scheme

Award 1 mark for the correct answer C. Method: Use the decomposition temperatures to determine that X is higher in Group 2 than Y. Recall and apply the Group 2 trends for ionic radius, first ionisation energy, sulfate solubility, and hydroxide solubility to identify the correct statement.
Question 17 · Multiple Choice
1 marks
Which gaseous ion has the greatest number of unpaired d-electrons?
  1. A.\(\text{V}^{2+}\)
  2. B.\(\text{Fe}^{2+}\)
  3. C.\(\text{Ni}^{2+}\)
  4. D.\(\text{Cu}^{2+}\)
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Worked solution

To determine the number of unpaired d-electrons, we first look at the electronic configurations of the transition metal gaseous ions:

- \(\text{V}^{2+}\): Vanadium has the atomic number 23. The configuration of \(\text{V}\) is \([\text{Ar}] 3\text{d}^3 4\text{s}^2\). Removing two electrons to form \(\text{V}^{2+}\) gives \([\text{Ar}] 3\text{d}^3\). There are 3 unpaired d-electrons.
- \(\text{Fe}^{2+}\): Iron has the atomic number 26. The configuration of \(\text{Fe}\) is \([\text{Ar}] 3\text{d}^6 4\text{s}^2\). Removing two electrons to form \(\text{Fe}^{2+}\) gives \([\text{Ar}] 3\text{d}^6\). Following Hund's rule, the 5 d-orbitals are filled with 6 electrons, resulting in 1 paired orbital and 4 unpaired d-electrons.
- \(\text{Ni}^{2+}\): Nickel has the atomic number 28. The configuration of \(\text{Ni}\) is \([\text{Ar}] 3\text{d}^8 4\text{s}^2\). Removing two electrons to form \(\text{Ni}^{2+}\) gives \([\text{Ar}] 3\text{d}^8\). This results in 3 paired orbitals and 2 unpaired d-electrons.
- \(\text{Cu}^{2+}\): Copper has the atomic number 29. The configuration of \(\text{Cu}\) is \([\text{Ar}] 3\text{d}^{10} 4\text{s}^1\). Removing two electrons to form \(\text{Cu}^{2+}\) gives \([\text{Ar}] 3\text{d}^9\). This results in 4 paired orbitals and 1 unpaired d-electron.

Thus, \(\text{Fe}^{2+}\) has the greatest number of unpaired d-electrons (4).

Marking scheme

1 mark for identifying the correct electronic configurations of each ion and determining that \(\text{Fe}^{2+}\) has the maximum number of unpaired d-electrons (4).
Question 18 · Multiple Choice
1 marks
The table shows the initial rate data for the reaction: \(2\text{A} + \text{B} + \text{C} \rightarrow \text{products}\)

* Experiment 1: \([\text{A}] = 0.10 \text{ mol dm}^{-3}\), \([\text{B}] = 0.10 \text{ mol dm}^{-3}\), \([\text{C}] = 0.10 \text{ mol dm}^{-3}\); Initial Rate = \(1.2 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}\)
* Experiment 2: \([\text{A}] = 0.20 \text{ mol dm}^{-3}\), \([\text{B}] = 0.10 \text{ mol dm}^{-3}\), \([\text{C}] = 0.10 \text{ mol dm}^{-3}\); Initial Rate = \(4.8 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}\)
* Experiment 3: \([\text{A}] = 0.10 \text{ mol dm}^{-3}\), \([\text{B}] = 0.20 \text{ mol dm}^{-3}\), \([\text{C}] = 0.10 \text{ mol dm}^{-3}\); Initial Rate = \(1.2 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}\)
* Experiment 4: \([\text{A}] = 0.10 \text{ mol dm}^{-3}\), \([\text{B}] = 0.10 \text{ mol dm}^{-3}\), \([\text{C}] = 0.30 \text{ mol dm}^{-3}\); Initial Rate = \(3.6 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}\)

What are the overall order of reaction and the correct units for the rate constant, \(k\)?
  1. A.Overall order: 3; Units of \(k\): \(\text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}\)
  2. B.Overall order: 3; Units of \(k\): \(\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\)
  3. C.Overall order: 4; Units of \(k\): \(\text{dm}^9 \text{ mol}^{-3} \text{ s}^{-1}\)
  4. D.Overall order: 4; Units of \(k\): \(\text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}\)
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Worked solution

1. Determine the order with respect to each reactant:
- Comparing Experiment 1 and 2: When \([\text{A}]\) is doubled (while \([\text{B}]\) and \([\text{C}]\) are kept constant), the rate increases by a factor of 4 (\(\frac{4.8 \times 10^{-3}}{1.2 \times 10^{-3}} = 4\)). Therefore, the reaction is second order with respect to \([\text{A}]\).
- Comparing Experiment 1 and 3: When \([\text{B}]\) is doubled (while \([\text{A}]\) and \([\text{C}]\) are kept constant), the rate does not change. Therefore, the reaction is zero order with respect to \([\text{B}]\).
- Comparing Experiment 1 and 4: When \([\text{C}]\) is tripled (while \([\text{A}]\) and \([\text{B}]\) are kept constant), the rate triples (\(\frac{3.6 \times 10^{-3}}{1.2 \times 10^{-3}} = 3\)). Therefore, the reaction is first order with respect to \([\text{C}]\).

2. Overall order of reaction = \(2 + 0 + 1 = 3\).

3. Determine the units of the rate constant, \(k\):
Rate equation: \(\text{Rate} = k[\text{A}]^2[\text{C}]\)
\(k = \frac{\text{Rate}}{[\text{A}]^2[\text{C}]} = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^2(\text{mol dm}^{-3})} = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{\text{mol}^3 \text{ dm}^{-9}} = \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}\).

Marking scheme

1 mark for calculating individual orders, finding the overall order is 3, and correctly deriving the units of the third-order rate constant as \(\text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}\).
Question 19 · Multiple Choice
1 marks
A \(10.0 \text{ cm}^3\) sample of a gaseous hydrocarbon was mixed with excess oxygen (\(80.0 \text{ cm}^3\)) and exploded. After cooling to room temperature, the total volume of gaseous mixture remaining was \(65.0 \text{ cm}^3\). When this remaining gas was passed through concentrated aqueous potassium hydroxide (which absorbs carbon dioxide), the volume decreased to \(25.0 \text{ cm}^3\). All gas volumes were measured at room temperature and pressure. What is the molecular formula of the hydrocarbon?
  1. A.\(\text{C}_4\text{H}_{10}\)
  2. B.\(\text{C}_4\text{H}_8\)
  3. C.\(\text{C}_4\text{H}_6\)
  4. D.\(\text{C}_3\text{H}_8\)
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Worked solution

Let the hydrocarbon be \(\text{C}_x\text{H}_y\).

1. The volume decrease upon passing through aqueous KOH corresponds to the volume of \(\text{CO}_2\) produced:
\(V(\text{CO}_2) = 65.0 - 25.0 = 40.0 \text{ cm}^3\).
Since \(10.0 \text{ cm}^3\) of hydrocarbon produced \(40.0 \text{ cm}^3\) of \(\text{CO}_2\), by Avogadro's hypothesis, 1 mole of hydrocarbon contains 4 moles of carbon atoms. Thus, \(x = 4\).

2. The remaining \(25.0 \text{ cm}^3\) of gas is unreacted oxygen. Thus, the volume of oxygen reacted is:
\(V(\text{O}_2\text{ reacted}) = 80.0 - 25.0 = 55.0 \text{ cm}^3\).

3. The equation for the complete combustion of the hydrocarbon \(\text{C}_4\text{H}_y\) is:
\(\text{C}_4\text{H}_y\text{(g)} + \left(4 + \frac{y}{4}\right)\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + \frac{y}{2}\text{H}_2\text{O}\text{(l)}\)

Using the volume ratios:
\(\frac{V(\text{O}_2\text{ reacted})}{V(\text{C}_4\text{H}_y)} = 4 + \frac{y}{4} = \frac{55.0}{10.0} = 5.5\)
\(4 + \frac{y}{4} = 5.5 \implies \frac{y}{4} = 1.5 \implies y = 6\).

Thus, the molecular formula of the hydrocarbon is \(\text{C}_4\text{H}_6\).

Marking scheme

1 mark for establishing the volume of carbon dioxide and oxygen reacted, and using mole ratios to deduce \(x=4\) and \(y=6\).
Question 20 · Multiple Choice
1 marks
Aspartic acid has the structural formula \(\text{HOOC-CH}_2\text{-CH(NH}_2\text{)-COOH}\). Its three acid-dissociation constants (\(\text{p}K_a\)) are approximately 2.0 (for the \(\alpha\)-carboxyl group), 3.9 (for the side-chain carboxyl group), and 9.8 (for the \(\alpha\)-amino group). Which structure represents the predominant species of aspartic acid present in a buffer solution maintained at \(\text{pH} = 1.0\)?
  1. A.\(\text{HOOC-CH}_2\text{-CH(NH}_3^+\text{)-COOH}\)
  2. B.\(\text{^{-}OOC-CH}_2\text{-CH(NH}_3^+\text{)-COO^{-}}\)
  3. C.\(\text{HOOC-CH}_2\text{-CH(NH}_2\text{)-COO^{-}}\)
  4. D.\(\text{^{-}OOC-CH}_2\text{-CH(NH}_3^+\text{)-COOH}\)
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Worked solution

At \(\text{pH} = 1.0\), the solution is more acidic than the \(\text{p}K_a\) of all three acidic groups present on aspartic acid (\(2.0, 3.9, 9.8\)).

When \(\text{pH} < \text{p}K_a\), the protonated form of each group is the dominant species:
- The two carboxylic acid groups will exist in their protonated (neutral, non-ionised) form: \(-\text{COOH}\).
- The amino group will exist in its protonated (cationic, ionised) form: \(-\text{NH}_3^+\).

Therefore, the overall structure is \(\text{HOOC-CH}_2\text{-CH(NH}_3^+\)-COOH\).

Marking scheme

1 mark for identifying that at \(\text{pH} = 1.0\) (lower than all \(\text{p}K_a\) values), both carboxyl groups are neutral (protonated) and the amino group is positively charged (protonated).
Question 21 · Multiple Choice
1 marks
The table lists thermodynamic data for calcium and bromine.

* Enthalpy of formation of \(\text{CaBr}_2\text{(s)}\) = \(-683 \text{ kJ mol}^{-1}\)
* Enthalpy of atomisation of \(\text{Ca(s)}\) = \(+178 \text{ kJ mol}^{-1}\)
* First ionisation energy of \(\text{Ca(g)}\) = \(+590 \text{ kJ mol}^{-1}\)
* Second ionisation energy of \(\text{Ca(g)}\) = \(+1145 \text{ kJ mol}^{-1}\)
* Enthalpy of atomisation of \(\text{Br}_2\text{(l)}\) = \(+112 \text{ kJ mol}^{-1}\)
* Lattice energy of \(\text{CaBr}_2\text{(s)}\) = \(-2176 \text{ kJ mol}^{-1}\)

Using these data, what is the value of the first electron affinity of bromine?
  1. A.\(-322 \text{ kJ mol}^{-1}\)
  2. B.\(-161 \text{ kJ mol}^{-1}\)
  3. C.\(-644 \text{ kJ mol}^{-1}\)
  4. D.\(-532 \text{ kJ mol}^{-1}\)
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Worked solution

According to the Born-Haber cycle for \(\text{CaBr}_2\text{(s)}\):
\(\Delta H_f^\ominus = \Delta H_{at}^\ominus(\text{Ca}) + \text{IE}_1(\text{Ca}) + \text{IE}_2(\text{Ca}) + 2\Delta H_{at}^\ominus(\text{Br}) + 2\text{EA}(\text{Br}) + \Delta H_{latt}^\ominus(\text{CaBr}_2)\)

Note that 2 moles of gaseous bromine atoms are formed, requiring twice the enthalpy of atomisation, and 2 moles of bromide ions are formed, requiring twice the first electron affinity.

Substitute the known values:
\(-683 = 178 + 590 + 1145 + 2(112) + 2\text{EA}(\text{Br}) - 2176\)
\(-683 = 2137 - 2176 + 2\text{EA}(\text{Br})\)
\(-683 = -39 + 2\text{EA}(\text{Br})\)
\(2\text{EA}(\text{Br}) = -644\)
\(\text{EA}(\text{Br}) = -322 \text{ kJ mol}^{-1}\).

Marking scheme

1 mark for correctly applying the Born-Haber cycle equation, using correct multipliers for bromine processes, and calculating the final value as \(-322 \text{ kJ mol}^{-1}\).
Question 22 · Multiple Choice
1 marks
Four separate samples of Period 3 chlorides, \(\text{NaCl}\), \(\text{MgCl}_2\), \(\text{AlCl}_3\), and \(\text{SiCl}_4\), are added to separate beakers of water. Which of the following lists the resulting mixtures in order of decreasing pH (from highest pH to lowest pH)?
  1. A.\(\text{NaCl} > \text{MgCl}_2 > \text{AlCl}_3 > \text{SiCl}_4\)
  2. B.\(\text{SiCl}_4 > \text{AlCl}_3 > \text{MgCl}_2 > \text{NaCl}\)
  3. C.\(\text{MgCl}_2 > \text{NaCl} > \text{AlCl}_3 > \text{SiCl}_4\)
  4. D.\(\text{NaCl} > \text{AlCl}_3 > \text{MgCl}_2 > \text{SiCl}_4\)
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Worked solution

- \(\text{NaCl}\) dissolves in water to form a neutral solution (pH ~ 7.0) because the sodium ion is of relatively low charge density and does not hydrolyse water.
- \(\text{MgCl}_2\) undergoes very slight hydrolysis due to the moderate charge density of the magnesium ion, resulting in a weakly acidic solution (pH ~ 6.5).
- \(\text{AlCl}_3\) contains the high charge density \(\text{Al}^{3+}\) ion, which undergoes significant hydrolysis to produce an acidic solution (pH ~ 3.0).
- \(\text{SiCl}_4\) reacts violently and completely with water to yield \(\text{SiO}_2\) and \(\text{HCl}\), forming a strongly acidic solution (pH ~ 1–2).

Therefore, the order of decreasing pH is: \(\text{NaCl} > \text{MgCl}_2 > \text{AlCl}_3 > \text{SiCl}_4\).

Marking scheme

1 mark for correctly ordering the Period 3 chlorides from those that undergo no/minor hydrolysis to those that undergo complete acidic hydrolysis.
Question 23 · Multiple Choice
1 marks
An organic compound \(X\) has a partition coefficient, \(K_{pc}\), of 4.0 between ethoxyethane (ether) and water, defined as:

\(K_{pc} = \frac{[X\text{ in ethoxyethane}]}{[X\text{ in water}]}\)

An aqueous solution containing \(5.0\text{ g}\) of \(X\) dissolved in \(100\text{ cm}^3\) of water is shaken with a single \(50\text{ cm}^3\) portion of ethoxyethane. What mass of \(X\) is extracted into the ethoxyethane layer?
  1. A.\(1.67\text{ g}\)
  2. B.\(2.50\text{ g}\)
  3. C.\(3.33\text{ g}\)
  4. D.\(4.00\text{ g}\)
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Worked solution

Let \(x\) be the mass of \(X\) extracted into the ethoxyethane layer (in grams).
The mass of \(X\) remaining in the aqueous layer is \((5.0 - x)\text{ g}\).

Calculate concentrations:
- Concentration in ethoxyethane: \([X\text{ in ethoxyethane}] = \frac{x}{50} \text{ g cm}^{-3}\)
- Concentration in water: \([X\text{ in water}] = \frac{5.0 - x}{100} \text{ g cm}^{-3}\)

Use the partition coefficient equation:
\(K_{pc} = \frac{x/50}{(5.0 - x)/100} = 4.0\)
\(\frac{100x}{50(5.0 - x)} = 4.0\)
\(\frac{2x}{5.0 - x} = 4.0\)
\(2x = 4.0(5.0 - x)\)
\(2x = 20.0 - 4x\)
\(6x = 20.0\)
\(x = 3.33\text{ g}\).

Marking scheme

1 mark for setting up the algebraic equation for the partition coefficient correctly, using the volumes given, and solving for the mass extracted.
Question 24 · Multiple Choice
1 marks
Which carbonyl compound reacts with hydrogen cyanide, \(\text{HCN}\), in the presence of a catalytic amount of \(\text{NaCN}\), to produce a reaction mixture that does not rotate the plane of plane-polarised light, even though the reaction product itself is a chiral molecule?
  1. A.propanone
  2. B.methanal
  3. C.pentan-3-one
  4. D.propanal
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Worked solution

Nucleophilic addition of \(\text{HCN}\) to a planar carbonyl group produces a 50:50 racemic mixture of enantiomers if a chiral carbon is created. This racemic mixture is optically inactive (does not rotate the plane of plane-polarised light).

We need to identify which reactant gives a chiral product (a carbon with 4 different groups attached):
- **Propanone** (\(\text{CH}_3\text{COCH}_3\)) produces 2-hydroxy-2-methylpropanenitrile. The central carbon has two identical methyl groups attached, so it is achiral.
- **Methanal** (\(\text{HCHO}\)) produces hydroxyacetonitrile. The central carbon has two hydrogen atoms attached, so it is achiral.
- **Pentan-3-one** (\(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\)) produces 2-ethyl-2-hydroxybutanenitrile. The central carbon has two identical ethyl groups attached, so it is achiral.
- **Propanal** (\(\text{CH}_3\text{CH}_2\text{CHO}\)) produces 2-hydroxybutanenitrile, \(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\). The central carbon is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CN}\), and \(-\text{CH}_2\text{CH}_3\). Thus, the product is a chiral molecule, but the reaction forms a racemic mixture which is optically inactive.

Marking scheme

1 mark for determining that only the nucleophilic addition of HCN to propanal yields a chiral product, which is produced as an optically inactive racemic mixture.
Question 25 · Multiple Choice Questions
1 marks
In an aqueous solution, the copper(II) ion forms various complex ions. When concentrated hydrochloric acid is added to aqueous copper(II) sulfate, the hexaaquacopper(II) ion, \([Cu(H_2O)_6]^{2+}\), is converted to the tetrachlorocuprate(II) ion, \([CuCl_4]^{2-}\).

The stability constants, \(K_{stab}\), for these complexes are:
\([Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O\) \(K_{stab} = 4.0 \times 10^5\text{ dm}^{12}\text{ mol}^{-4}\)

If another ligand \(L\) forms a complex \([CuL_4]^{2+}\) with a stability constant of \(2.5 \times 10^{11}\text{ dm}^{12}\text{ mol}^{-4}\), what is the equilibrium constant for the reaction:
\([CuCl_4]^{2-} + 4L \rightleftharpoons [CuL_4]^{2+} + 4Cl^-\text{?}\)
  1. A.\(1.6 \times 10^{-6}\)
  2. B.\(6.25 \times 10^5\)
  3. C.\(1.0 \times 10^{17}\)
  4. D.\(6.25 \times 10^{16}\)
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Worked solution

The equilibrium constant, \(K\), for the replacement reaction of \([CuCl_4]^{2-}\) by ligand \(L\) can be calculated from the two stability constants:

\(K = \frac{K_{stab}([CuL_4]^{2+})}{K_{stab}([CuCl_4]^{2-})} = \frac{2.5 \times 10^{11}}{4.0 \times 10^5} = 6.25 \times 10^5\text{ dm}^{12}\text{ mol}^{-4}\)

Marking scheme

Award 1 mark for the correct calculation showing the division of the two stability constants to arrive at the correct value of 6.25 x 10^5.
Question 26 · Multiple Choice Questions
1 marks
For the reaction \(2A + B \rightarrow C\), the rate equation is determined to be:

\(\text{rate} = k[A]^2[B]\)

If the concentration of \(A\) is halved and the concentration of \(B\) is tripled, by what factor is the initial rate of the reaction multiplied?
  1. A.\(0.375\)
  2. B.\(0.75\)
  3. C.\(1.5\)
  4. D.\(3.0\)
Show answer & marking scheme

Worked solution

Let the initial rate be \(\text{rate}_1 = k[A]^2[B]\).

When the concentration of \(A\) is halved and \(B\) is tripled, the new rate is:

\(\text{rate}_2 = k(0.5[A])^2(3[B])\)
\(\text{rate}_2 = 0.25 \times 3 \times k[A]^2[B] = 0.75 \times \text{rate}_1\)

Therefore, the rate is multiplied by a factor of 0.75.

Marking scheme

Award 1 mark for substituting the scale factors into the rate equation and solving for the overall multiplier (0.5^2 * 3 = 0.75).
Question 27 · Multiple Choice Questions
1 marks
A 2.30 g sample of an anhydrous metal carbonate, \(MCO_3\), was dissolved in excess dilute hydrochloric acid. The carbon dioxide gas evolved was collected and measured to be \(480\text{ cm}^3\) at room temperature and pressure (rtp).

[Assume 1 mole of gas occupies \(24.0\text{ dm}^3\) at rtp.]

What is the identity of the metal \(M\)?
  1. A.Calcium
  2. B.Manganese
  3. C.Nickel
  4. D.Copper
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Worked solution

1. Find the moles of \(CO_2\) gas evolved:
\(n(CO_2) = \frac{480\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.020\text{ mol}\)

2. The equation for the reaction is:
\(MCO_3(s) + 2HCl(aq) \rightarrow MCl_2(aq) + CO_2(g) + H_2O(l)\)

The mole ratio of \(MCO_3\) to \(CO_2\) is 1:1, so \(n(MCO_3) = 0.020\text{ mol}\).

3. Calculate the molar mass of \(MCO_3\):
\(M_r(MCO_3) = \frac{\text{mass}}{\text{moles}} = \frac{2.30\text{ g}}{0.020\text{ mol}} = 115\text{ g mol}^{-1}\)

4. Calculate the atomic mass of \(M\):
\(M_r(M) + M_r(CO_3^{2-}) = 115\)
\(M_r(M) + 12.0 + 3(16.0) = 115\)
\(M_r(M) + 60.0 = 115 \implies M_r(M) = 55.0\text{ g mol}^{-1}\)

This relative atomic mass corresponds to manganese (\(Mn = 54.9\)).

Marking scheme

Award 1 mark for calculating the moles of gas, deducing the molar mass of the metal carbonate as 115 g/mol, and subtracting the mass of the carbonate group (60 g/mol) to identify the metal as Manganese.
Question 28 · Multiple Choice Questions
1 marks
Aspartic acid has the structure \(HOOC-CH(NH_2)-CH_2-COOH\).

Which structure represents the predominant ionic species present in a highly alkaline solution of aspartic acid at pH 12?
  1. A.\(^-OOC-CH(NH_2)-CH_2-COO^-\)
  2. B.\(^-OOC-CH(NH_3^+)-CH_2-COO^-\)
  3. C.\(HOOC-CH(NH_3^+)-CH_2-COOH\)
  4. D.\(^-OOC-CH(NH_2)-CH_2-COOH\)
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Worked solution

In highly alkaline solutions (high pH, such as pH 12), the concentration of hydroxide ions is high, causing proton-donating groups to lose their protons. Both carboxylic acid groups (\(-COOH\)) lose a proton to become carboxylate anions (\(-COO^-\)). The amine group (\(-NH_2\)) remains unprotonated as the neutral free base. Therefore, the overall species formed has a charge of 2- with the structural formula \(^-OOC-CH(NH_2)-CH_2-COO^-\).

Marking scheme

Award 1 mark for identifying the correct ionic species where both carboxylic acid groups are deprotonated and the amine group is in its neutral amine form.
Question 29 · Multiple Choice Questions
1 marks
Consider the following thermodynamic data for magnesium fluoride, \(MgF_2\):

Enthalpy change of formation of \(MgF_2(s) = -1124\text{ kJ mol}^{-1}\)
Enthalpy change of atomisation of \(Mg(s) = +148\text{ kJ mol}^{-1}\)
First ionisation energy of \(Mg(g) = +738\text{ kJ mol}^{-1}\)
Second ionisation energy of \(Mg(g) = +1451\text{ kJ mol}^{-1}\)
Bond dissociation enthalpy of \(F_2(g) = +158\text{ kJ mol}^{-1}\)
First electron affinity of \(F(g) = -328\text{ kJ mol}^{-1}\)

What is the lattice energy, \(\Delta H_{latt}^\ominus\), of \(MgF_2(s)\)?
  1. A.\(-2477\text{ kJ mol}^{-1}\)
  2. B.\(-2963\text{ kJ mol}^{-1}\)
  3. C.\(-3121\text{ kJ mol}^{-1}\)
  4. D.\(-3447\text{ kJ mol}^{-1}\)
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Worked solution

Using a Born-Haber cycle and Hess's Law:

\(\Delta H_f^\ominus = \Delta H_{at}^\ominus(Mg) + IE_1(Mg) + IE_2(Mg) + E(F-F) + 2 \times EA_1(F) + \Delta H_{latt}^\ominus(MgF_2)\)

Note that 1 mole of \(F_2(g)\) dissociates to form 2 moles of \(F(g)\) atoms, so we use the full bond enthalpy of \(F_2\) (\(+158\text{ kJ mol}^{-1}\)).

\(-1124 = 148 + 738 + 1451 + 158 + 2(-328) + \Delta H_{latt}^\ominus\)
\(-1124 = 2495 - 656 + \Delta H_{latt}^\ominus\)
\(-1124 = 1839 + \Delta H_{latt}^\ominus\)
\(\Delta H_{latt}^\ominus = -1124 - 1839 = -2963\text{ kJ mol}^{-1}\)

Marking scheme

Award 1 mark for the correct mathematical application of Hess's Law to the Born-Haber cycle, taking into account the factor of 2 for the electron affinity of fluorine.
Question 30 · Multiple Choice Questions
1 marks
Which statement best explains why the first ionisation energy of sulfur is lower than that of phosphorus?
  1. A.Phosphorus has a larger nuclear charge than sulfur.
  2. B.The outer electron in sulfur is in a d-orbital, whereas in phosphorus it is in a p-orbital.
  3. C.Spin-pair repulsion occurs between the two electrons in one of the 3p orbitals of sulfur.
  4. D.The 3p subshell in phosphorus is closer to the nucleus than the 3p subshell in sulfur.
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Worked solution

Phosphorus has the outer electronic configuration \(3s^2 3p^3\) (with three singly occupied p-orbitals), while sulfur has the configuration \(3s^2 3p^4\) (with one doubly occupied p-orbital and two singly occupied p-orbitals). The spin-pair repulsion between the two electrons in the shared 3p orbital of sulfur makes it easier to remove one of these electrons compared to a single electron in a half-filled p-orbital of phosphorus, resulting in a lower first ionisation energy.

Marking scheme

Award 1 mark for identifying that the paired electrons in the 3p orbital of sulfur experience spin-pair repulsion, making them easier to remove.
Question 31 · Multiple Choice Questions
1 marks
An organic compound, \(X\), has a partition coefficient, \(K_{pc}\), of 4.0 between diethyl ether and water:

\(K_{pc} = \frac{[X]_{\text{ether}}}{[X]_{\text{water}}} = 4.0\)

An aqueous solution containing 5.00 g of \(X\) in \(100\text{ cm}^3\) of water is shaken with one \(50\text{ cm}^3\) portion of diethyl ether.

What mass of \(X\) remains in the aqueous layer?
  1. A.\(1.25\text{ g}\)
  2. B.\(1.67\text{ g}\)
  3. C.\(2.50\text{ g}\)
  4. D.\(3.33\text{ g}\)
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Worked solution

Let \(w\) be the mass of \(X\) remaining in the aqueous layer. The mass of \(X\) extracted into the diethyl ether layer is \(5.00 - w\).

Concentration in ether = \(\frac{5.00 - w}{50\text{ cm}^3}\)
Concentration in water = \(\frac{w}{100\text{ cm}^3}\)

Using the partition coefficient equation:
\(K_{pc} = \frac{(5.00 - w)/50}{w/100} = 4.0\)
\(\frac{5.00 - w}{50} \times \frac{100}{w} = 4.0\)
\(2 \times \frac{5.00 - w}{w} = 4.0\)
\(10.0 - 2w = 4.0w\)
\(6.0w = 10.0\)
\(w = 1.67\text{ g}\)

Marking scheme

Award 1 mark for setting up the partition coefficient expression with the appropriate volumes and solving for the mass in the aqueous layer.
Question 32 · Multiple Choice Questions
1 marks
Compound \(Y\) has the molecular formula \(C_4H_8O\). It reacts with 2,4-dinitrophenylhydrazine reagent to form an orange precipitate, but does not form a red precipitate when heated with Fehling's solution.

What is the structural formula of \(Y\)?
  1. A.\(CH_3CH_2CH_2CHO\)
  2. B.\(CH_3COCH_2CH_3\)
  3. C.\(CH_3CH(CH_3)CHO\)
  4. D.\(CH_3CH_2CH_2CH_2OH\)
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Worked solution

1. Reaction with 2,4-DNPH indicates that \(Y\) is a carbonyl compound (either an aldehyde or a ketone). This rules out alcohols like butan-1-ol (option d).
2. No reaction with Fehling's solution indicates that \(Y\) is not an aldehyde (aldehydes are easily oxidised and give a red precipitate of \(Cu_2O\) with Fehling's, whereas ketones are not oxidised).
3. Therefore, \(Y\) must be a ketone. The only ketone with four carbon atoms is butanone, \(CH_3COCH_2CH_3\).

Marking scheme

Award 1 mark for identifying that the positive 2,4-DNPH test confirms a carbonyl group and the negative Fehling's test indicates a ketone rather than an aldehyde.
Question 33 · multiple-choice
1 marks
The reaction \(2\text{A} + \text{B} + \text{C} \rightarrow \text{D} + \text{E}\) was investigated at constant temperature. The following initial rate data were obtained:

- Exp 1: \([\text{A}] = 0.10 \text{ mol dm}^{-3}\), \([\text{B}] = 0.10 \text{ mol dm}^{-3}\), \([\text{C}] = 0.10 \text{ mol dm}^{-3}\), Initial rate = \(1.2 \times 10^{-3} \text{ mol dm}^{-3}\text{ s}^{-1}\)
- Exp 2: \([\text{A}] = 0.20 \text{ mol dm}^{-3}\), \([\text{B}] = 0.10 \text{ mol dm}^{-3}\), \([\text{C}] = 0.10 \text{ mol dm}^{-3}\), Initial rate = \(4.8 \times 10^{-3} \text{ mol dm}^{-3}\text{ s}^{-1}\)
- Exp 3: \([\text{A}] = 0.10 \text{ mol dm}^{-3}\), \([\text{B}] = 0.20 \text{ mol dm}^{-3}\), \([\text{C}] = 0.10 \text{ mol dm}^{-3}\), Initial rate = \(2.4 \times 10^{-3} \text{ mol dm}^{-3}\text{ s}^{-1}\)
- Exp 4: \([\text{A}] = 0.10 \text{ mol dm}^{-3}\), \([\text{B}] = 0.10 \text{ mol dm}^{-3}\), \([\text{C}] = 0.20 \text{ mol dm}^{-3}\), Initial rate = \(1.2 \times 10^{-3} \text{ mol dm}^{-3}\text{ s}^{-1}\)

What are the units of the rate constant, \(k\), for this reaction?
  1. A.\(\text{dm}^{3}\text{ mol}^{-1}\text{ s}^{-1}\)
  2. B.\(\text{dm}^{6}\text{ mol}^{-2}\text{ s}^{-1}\)
  3. C.\(\text{dm}^{9}\text{ mol}^{-3}\text{ s}^{-1}\)
  4. D.\(\text{s}^{-1}\)
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Worked solution

Comparing Exp 1 and Exp 2: when \([\text{A}]\) doubles (while \([\text{B}]\) and \([\text{C}]\) are constant), the rate increases by a factor of 4. Therefore, the reaction is second order with respect to \([\text{A}]\). Comparing Exp 1 and Exp 3: when \([\text{B}]\) doubles (while \([\text{A}]\) and \([\text{C}]\) are constant), the rate increases by a factor of 2. Therefore, the reaction is first order with respect to \([\text{B}]\). Comparing Exp 1 and Exp 4: when \([\text{C}]\) doubles, the rate remains unchanged. Therefore, the reaction is zero order with respect to \([\text{C}]\). The overall rate equation is:

\(\text{Rate} = k[\text{A}]^2[\text{B}]\)

Solving for the units of the rate constant, \(k\):

\(k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2 \times \text{mol dm}^{-3}} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Marking scheme

Award 1 mark for the correct option B. Deduce that the reaction is second order with respect to A, first order with respect to B, and zero order with respect to C, then calculate the units of k.
Question 34 · multiple-choice
1 marks
When aqueous chromium(III) chloride is treated with excess aqueous ammonia, a violet-purple solution of a complex ion, \([\text{Cr}(\text{NH}_3)_6]^{3+}\), is formed. When treated with excess aqueous sodium hydroxide, a green solution of the complex ion \([\text{Cr}(\text{OH})_6]^{3-}\) is formed instead. Which statement best explains why these two complexes have different colours?
  1. A.Ammonia is a neutral ligand, while hydroxide is an anionic ligand, which affects the oxidation state of the central chromium ion.
  2. B.The different ligands produce different splittings of the d-orbitals (\(\Delta E\)), leading to the absorption of different wavelengths of visible light.
  3. C.The \([\text{Cr}(\text{NH}_3)_6]^{3+}\) complex is octahedral, whereas the \([\text{Cr}(\text{OH})_6]^{3-}\) complex is tetrahedral, resulting in different crystal field arrangements.
  4. D.Ammonia has a weaker ligand field strength than hydroxide, causing a larger d-d transition energy difference.
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Worked solution

In both complex ions, chromium has an oxidation state of +3 and both complexes are octahedral. Different ligands (ammonia vs hydroxide) have different ligand field strengths. Ammonia has a stronger field strength than hydroxide, creating a larger crystal field splitting energy, \(\Delta E\), in the d-orbitals of the chromium(III) ion. Since \(\Delta E = h\nu\), different frequencies/wavelengths of visible light are absorbed by the d-d electronic transitions, resulting in different transmitted complementary colours.

Marking scheme

Award 1 mark for selecting B. Recognize that ligand field strength determines d-orbital splitting, which affects the complementary color observed.
Question 35 · multiple-choice
1 marks
A \(1.20\text{ g}\) sample of an impure mixture of calcium carbonate (\(\text{CaCO}_3\), \(M_{\text{r}} = 100.1\)) and barium carbonate (\(\text{BaCO}_3\), \(M_{\text{r}} = 197.3\)) was completely dissolved in excess dilute hydrochloric acid. The carbon dioxide gas evolved was collected and measured to be \(192\text{ cm}^3\) at room temperature and pressure (r.t.p.). Assuming the impurities do not react with acid to produce gas, and that one mole of gas occupies \(24.0\text{ dm}^3\) at r.t.p., what is the total number of moles of carbonate ions present in the \(1.20\text{ g}\) sample?
  1. A.\(4.00 \times 10^{-3}\text{ mol}\)
  2. B.\(8.00 \times 10^{-3}\text{ mol}\)
  3. C.\(1.20 \times 10^{-2}\text{ mol}\)
  4. D.\(1.60 \times 10^{-2}\text{ mol}\)
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Worked solution

The general reaction equation for a metal carbonate reacting with acid is:

\(\text{MCO}_3(\text{s}) + 2\text{H}^+(\text{aq}) \rightarrow \text{M}^{2+}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)

One mole of carbonate ions (\(\text{CO}_3^{2-}\)) yields exactly one mole of carbon dioxide gas (\(\text{CO}_2\)).

Moles of \(\text{CO}_2\) collected = \frac{192\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 8.00 \times 10^{-3}\text{ mol}\).

Since \(n(\text{CO}_3^{2-}) = n(\text{CO}_2)\), there are \(8.00 \times 10^{-3}\text{ mol}\) of carbonate ions present in the sample.

Marking scheme

Award 1 mark for the correct answer B. Correctly identify the 1:1 mole ratio between carbonate ions and carbon dioxide gas, then calculate the moles of gas.
Question 36 · multiple-choice
1 marks
The structural formula of the amino acid glutamic acid is \(\text{HOOC-CH}_2\text{-CH}_2\text{-CH}(\text{NH}_2)\text{-COOH}\). Which species is the major component of a solution of glutamic acid at a very high pH (\(\text{pH} = 13\))?
  1. A.\(\text{HOOC-CH}_2\text{-CH}_2\text{-CH}(\text{NH}_3^+)\text{-COOH}\)
  2. B.\({^-\text{OOC-CH}}_2\text{-CH}_2\text{-CH}(\text{NH}_3^+)\text{-COO}^-\)
  3. C.\({^-\text{OOC-CH}}_2\text{-CH}_2\text{-CH}(\text{NH}_2)\text{-COO}^-\)
  4. D.\(\text{HOOC-CH}_2\text{-CH}_2\text{-CH}(\text{NH}_2)\text{-COO}^-\)
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Worked solution

At a highly alkaline pH of 13, carboxylic acid groups are fully deprotonated to form carboxylate anions (\(-\text{COO}^-\)), and amino groups remain as the basic, unprotonated form (\(-\text{NH}_2\)). Glutamic acid has two carboxylic acid groups and one amino group, so both carboxylic acid groups will be deprotonated (forming two \(-\text{COO}^-\)) and the amine remains unprotonated. This gives the species \({^-\text{OOC-CH}}_2\text{-CH}_2\text{-CH}(\text{NH}_2)\text{-COO}^-\).

Marking scheme

Award 1 mark for selecting C. Understand that under high pH, carboxyl groups are deprotonated and amino groups remain unprotonated.
Question 37 · multiple-choice
1 marks
The Born-Haber cycle for the formation of magnesium oxide, \(\text{MgO}\)(s), involves several energy terms. Which reaction corresponds to the enthalpy change of atomisation of oxygen, \(\Delta H_{\text{at}}^{\ominus}[\text{O}]\)?
  1. A.\(\text{O}_2\text{(g)} \rightarrow 2\text{O(g)}\)
  2. B.\(\frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{O(g)}\)
  3. C.\(\text{O(g)} \rightarrow \text{O}^-\text{(g)}\)
  4. D.\(\text{O}_2\text{(g)} \rightarrow \text{O}_2\text{(l)}\)
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Worked solution

The standard enthalpy change of atomisation (\(\Delta H_{\text{at}}^{\ominus}\)) is the enthalpy change that occurs when 1 mole of gaseous atoms is formed from the element in its standard state under standard conditions. Oxygen exists as diatomic molecules, \(\text{O}_2\)(g), in its standard state. The equation representing the formation of 1 mole of \(\text{O}\)(g) is \(\frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{O(g)}\).

Marking scheme

Award 1 mark for selecting B. Correctly apply the definition of standard enthalpy change of atomisation to diatomic oxygen.
Question 38 · multiple-choice
1 marks
An organic compound \(\text{X}\) is distributed between water and an organic solvent, dichloromethane. The partition coefficient, \(K_{\text{pc}}\), is defined as:

\[K_{\text{pc}} = \frac{[\text{X}\text{(dichloromethane)}]}{[\text{X}\text{(aqueous)}]} = 4.00\]

A student has \(100\text{ cm}^3\) of an aqueous solution containing \(2.50\text{ g}\) of \(\text{X}\). Which extraction procedure will remove the greatest mass of \(\text{X}\) from the aqueous solution?
  1. A.A single extraction using \(40\text{ cm}^3\) of dichloromethane.
  2. B.Two successive extractions, each using \(20\text{ cm}^3\) of dichloromethane.
  3. C.A single extraction using \(50\text{ cm}^3\) of dichloromethane.
  4. D.Two successive extractions, each using \(25\text{ cm}^3\) of dichloromethane.
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Worked solution

Let \(w\) be the mass of solute remaining in water. The relationship is:
\(K_{\text{pc}} = \frac{(2.50 - w)/V_{\text{org}}}{w/100} \implies w = 2.50 \times \frac{100}{100 + 4V_{\text{org}}}\).

- Option A (single extraction, \(40\text{ cm}^3\)): \(w = 2.50 \times \frac{100}{260} = 0.962\text{ g}\); mass extracted = \(1.538\text{ g}\).
- Option B (two extractions, \(20\text{ cm}^3\) each): After 1st: \(w_1 = 2.50 \times \frac{100}{180} = 1.389\text{ g}\); after 2nd: \(w_2 = 1.389 \times \frac{100}{180} = 0.772\text{ g}\); mass extracted = \(1.728\text{ g}\).
- Option C (single extraction, \(50\text{ cm}^3\)): \(w = 2.50 \times \frac{100}{300} = 0.833\text{ g}\); mass extracted = \(1.667\text{ g}\).
- Option D (two extractions, \(25\text{ cm}^3\) each): After 1st: \(w_1 = 2.50 \times \frac{100}{200} = 1.250\text{ g}\); after 2nd: \(w_2 = 1.250 \times \frac{100}{200} = 0.625\text{ g}\); mass extracted = \(1.875\text{ g}\).

Option D yields the greatest extracted mass of \(\text{X}\).

Marking scheme

Award 1 mark for the correct answer D. Show by calculations that two successive extractions with 25 cubic centimeters of solvent are more efficient than the other single or double extraction scenarios.
Question 39 · multiple-choice
1 marks
An organic compound \(\text{Y}\) has the molecular formula \(\text{C}_5\text{H}_{10}\text{O}\).

- It gives a yellow precipitate when warmed with alkaline aqueous iodine.
- It does not react with Tollens' reagent or Fehling's solution.
- When reduced with \(\text{NaBH}_4\), it forms a secondary alcohol.

Which structure is consistent with this information?
  1. A.Pentan-3-one
  2. B.3-Methylbutan-2-one
  3. C.Pentanal
  4. D.2,2-Dimethylpropanal
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Worked solution

The negative Tollens' and Fehling's tests rule out aldehydes (pentanal and 2,2-dimethylpropanal). Thus, compound Y must be a ketone. Reduction of ketones with \(\text{NaBH}_4\) gives secondary alcohols. The positive iodoform test (yellow precipitate of \(\text{CHI}_3\) with alkaline aqueous iodine) indicates the presence of a methyl ketone (\(\text{CH}_3\text{C=O}\) group). Among the ketones, pentan-3-one, \(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\), does not contain a methyl ketone group, whereas 3-methylbutan-2-one, \(\text{CH}_3\text{COCH(CH}_3)_2\), is a methyl ketone and yields a positive iodoform test.

Marking scheme

Award 1 mark for selecting B. Identify the functional group requirements (ketone and methyl ketone) based on the qualitative analysis descriptions.
Question 40 · multiple-choice
1 marks
Which statement correctly describes the trend in properties of the Group 2 elements, magnesium to barium, or their compounds down the group?
  1. A.The thermal stability of the carbonates decreases down the group because the cationic radius decreases.
  2. B.The solubility of the hydroxides increases down the group because the lattice energy of the hydroxides decreases more rapidly than the hydration energy of the cations.
  3. C.The solubility of the sulfates increases down the group because the hydration energy of the cations decreases less rapidly than the lattice energy.
  4. D.The first ionisation energy increases down the group because the nuclear charge increases.
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Worked solution

As Group 2 is descended, the solubility of Group 2 hydroxides increases. This is because both the lattice energy and the hydration energy decrease, but because the hydroxide ion is relatively small, the lattice energy decreases more rapidly than the hydration energy of the metal cations, causing dissolution to become more thermodynamically favorable. Decreasing cation charge density down the group decreases polarizing power, which increases thermal stability of carbonates (ruling out A). Sulfates become less soluble down the group (ruling out C). First ionization energies decrease down the group due to increased atomic radius and shielding (ruling out D).

Marking scheme

Award 1 mark for selecting B. Explain the solubility of Group 2 hydroxides correctly in terms of lattice and hydration enthalpies.

Paper 23 (AS Level Structured)

Answer all questions in the spaces provided on the question paper. Show all working and use appropriate units.
6 Question · 60 marks
Question 1 · Structured
10 marks
A sample of anhydrous Group 2 nitrate, \( \text{M}(\text{NO}_3)_2 \), of mass 1.483 g is heated strongly. It decomposes completely according to the following equation:
\( 2\text{M}(\text{NO}_3)_2(\text{s}) \rightarrow 2\text{MO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g}) \)

The gaseous products are passed through an excess of concentrated aqueous sodium hydroxide to absorb all the nitrogen dioxide. The remaining oxygen gas is collected and measured at r.t.p. and has a volume of 120 cm\(^3\).

(a) Define thermal decomposition. [1]
(b) (i) Calculate the number of moles of oxygen gas collected at r.t.p. (Assume 1.00 mol of gas occupies 24.0 dm\(^3\) at r.t.p.). [1]
(ii) Use the equation to determine the number of moles of \( \text{M}(\text{NO}_3)_2 \) that decomposed. [2]
(iii) Calculate the relative formula mass of \( \text{M}(\text{NO}_3)_2 \) and identify metal M. [2]
(c) The solid residue of \( \text{MO} \) (where M is the identified metal) is reacted with acid.
(i) Write a balanced equation for the reaction of \( \text{MO} \) with dilute hydrochloric acid. [1]
(ii) The residue of \( \text{MO} \) from the decomposition of 1.483 g of \( \text{M}(\text{NO}_3)_2 \) is dissolved in 50.0 cm\(^3\) of 1.00 mol dm\(^{-3}\) hydrochloric acid (an excess). Calculate the concentration of the unreacted hydrochloric acid remaining in the solution, assuming the volume remains 50.0 cm\(^3\). [3]
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Worked solution

(a) Thermal decomposition is the breakdown of a chemical compound into two or more substances by heating.

(b)(i) \( n(\text{O}_2) = \frac{120 \text{ cm}^3}{24000 \text{ cm}^3 \text{ mol}^{-1}} = 0.00500 \text{ mol} \).

(ii) From the stoichiometry of the equation, the mole ratio of \( \text{M}(\text{NO}_3)_2 : \text{O}_2 \) is 2:1. Therefore, moles of \( \text{M}(\text{NO}_3)_2 \) decomposed = \( 2 \times 0.00500 \text{ mol} = 0.0100 \text{ mol} \).

(iii) \( M_{\text{r}}(\text{M}(\text{NO}_3)_2) = \frac{\text{mass}}{\text{moles}} = \frac{1.483 \text{ g}}{0.0100 \text{ mol}} = 148.3 \text{ g mol}^{-1} \).
Since \( M_{\text{r}}(\text{M}(\text{NO}_3)_2) = A_{\text{r}}(\text{M}) + 2 \times [14.0 + (3 \times 16.0)] = A_{\text{r}}(\text{M}) + 124.0 \),
\( A_{\text{r}}(\text{M}) = 148.3 - 124.0 = 24.3 \).
Looking at the Periodic Table, the metal M is Magnesium (Mg).

(c)(i) \( \text{MgO}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) \).

(ii) Moles of \( \text{MgO} \) produced = moles of \( \text{Mg}(\text{NO}_3)_2 \) decomposed = 0.0100 mol.
Moles of \( \text{HCl} \) reacted = \( 2 \times 0.0100 \text{ mol} = 0.0200 \text{ mol} \).
Initial moles of \( \text{HCl} \) in solution = \( 0.0500 \text{ dm}^3 \times 1.00 \text{ mol dm}^{-3} = 0.0500 \text{ mol} \).
Moles of \( \text{HCl} \) remaining = \( 0.0500 - 0.0200 = 0.0300 \text{ mol} \).
Concentration of remaining \( \text{HCl} \) = \( \frac{0.0300 \text{ mol}}{0.0500 \text{ dm}^3} = 0.600 \text{ mol dm}^{-3} \).

Marking scheme

(a) 1 mark: breakdown of a compound into other substances using heat.
(b)(i) 1 mark: 0.00500 mol.
(b)(ii) 1 mark: states ratio of 2:1; 1 mark: 0.0100 mol.
(b)(iii) 1 mark: 148.3; 1 mark: identifies Mg from calculated Ar of 24.3 (or 24.0).
(c)(i) 1 mark: balanced equation with correct formulas.
(c)(ii) 1 mark: moles of HCl reacted = 0.0200 mol; 1 mark: moles of HCl remaining = 0.0300 mol; 1 mark: final concentration = 0.600 mol dm^-3.
Question 2 · Structured
10 marks
The rate of reaction between reactant A and reactant B was studied at constant temperature.
\( 2\text{A} + \text{B} \rightarrow \text{C} + \text{D} \)

The following initial rate data were obtained:
- Experiment 1: \( [\text{A}] = 0.100 \text{ mol dm}^{-3} \), \( [\text{B}] = 0.100 \text{ mol dm}^{-3} \), Initial Rate = \( 1.20 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1} \)
- Experiment 2: \( [\text{A}] = 0.200 \text{ mol dm}^{-3} \), \( [\text{B}] = 0.100 \text{ mol dm}^{-3} \), Initial Rate = \( 4.80 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1} \)
- Experiment 3: \( [\text{A}] = 0.200 \text{ mol dm}^{-3} \), \( [\text{B}] = 0.200 \text{ mol dm}^{-3} \), Initial Rate = \( 4.80 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1} \)

(a) Determine the order of reaction with respect to A and with respect to B, showing your reasoning. [4]
(b) Write the rate equation for this reaction. [1]
(c) Using the data from Experiment 1, calculate the value of the rate constant, k, and state its units. [3]
(d) Suggest a mechanism for this reaction consisting of two steps, indicating which step is the rate-determining step. [2]
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Worked solution

(a) Comparison of Expt 1 and Expt 2: \( [\text{B}] \) is constant while \( [\text{A}] \) doubles. The rate increases by a factor of \( \frac{4.80 \times 10^{-4}}{1.20 \times 10^{-4}} = 4.0 \). Since \( 2^2 = 4 \), the reaction is second order with respect to A.
Comparison of Expt 2 and Expt 3: \( [\text{A}] \) is constant while \( [\text{B}] \) doubles. The rate remains unchanged (ratio of 1.0). Therefore, the reaction is zero order with respect to B.

(b) \( \text{Rate} = k[\text{A}]^2 \)

(c) From Expt 1: \( 1.20 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1} = k (0.100 \text{ mol dm}^{-3})^2 \)
\( k = \frac{1.20 \times 10^{-4}}{0.0100} = 1.20 \times 10^{-2} \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1} \) (or 0.0120).
Units: \( \text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1} \).

(d) Since B is zero-order and A is second-order, B cannot be involved in the rate-determining step. A possible mechanism is:
Step 1 (slow): \( \text{A} + \text{A} \rightarrow \text{A}_2 \) (rate-determining step)
Step 2 (fast): \( \text{A}_2 + \text{B} \rightarrow \text{C} + \text{D} \)

Marking scheme

(a) 2 marks for A: 1 mark for noting rate increases 4x as [A] doubles, 1 mark for deducing second order. 2 marks for B: 1 mark for noting rate is constant as [B] doubles, 1 mark for deducing zero order.
(b) 1 mark: Rate = k[A]^2 (accept [B]^0 included).
(c) 1 mark: correct calculation of value (0.0120 or 1.20 x 10^-2); 1 mark: correct units (dm^3 mol^-1 s^-1); 1 mark: working shown.
(d) 1 mark: Rate-determining step involving 2A (slow step); 1 mark: fast step involving intermediate and B to give products, overall equation balances.
Question 3 · Structured
10 marks
Cobalt is a transition element that forms various complex ions with interesting properties.

(a) A pink solution containing hexaaquacobalt(II) ions, \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} \), is mixed with excess concentrated hydrochloric acid to form a blue tetrahedral complex, \( \mathbf{X} \).
(i) State the formula and coordination number of complex \( \mathbf{X} \). [2]
(ii) Write the equilibrium equation for the ligand exchange reaction and describe the color change. [2]
(b) When aqueous ammonia is added dropwise to \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} \), a blue-green precipitate \( \mathbf{Y} \) forms first. Upon adding excess ammonia, \( \mathbf{Y} \) dissolves to give a pale brown solution containing the octahedral complex \( \mathbf{Z} \).
(i) Give the formula of precipitate \( \mathbf{Y} \). [1]
(ii) Write an equation for the conversion of \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} \) to \( \mathbf{Z} \) in excess ammonia. [1]
(iii) When the solution of \( \mathbf{Z} \) is exposed to air, it is oxidised. Write the formula of the cobalt(III) complex ion formed. [1]
(c) Chromium forms a complex with the formula \( [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^{+} \).
(i) Draw the 3D structures of the two stereoisomers of this complex, labeling them as cis or trans. [3]
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Worked solution

(a)(i) Complex \( \mathbf{X} \) is \( [\text{CoCl}_4]^{2-} \). The coordination number is 4.
(ii) Equation: \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O} \).
Color change: pink to blue.

(b)(i) Precipitate \( \mathbf{Y} \) is cobalt(II) hydroxide, \( \text{Co}(\text{OH})_2 \) (or \( [\text{Co}(\text{H}_2\text{O})_4(\text{OH})_2] \)).
(ii) Equation: \( [\text{Co}(\text{H}_2\text{O})_6]^{2+} + 6\text{NH}_3 \rightarrow [\text{Co}(\text{NH}_3)_6]^{2+} + 6\text{H}_2\text{O} \).
(iii) The cobalt(III) complex is \( [\text{Co}(\text{NH}_3)_6]^{3+} \).

(c)(i) Cis-isomer: Cl- ligands are drawn at adjacent positions (90 degrees apart), e.g., one axial and one equatorial. Trans-isomer: Cl- ligands are drawn at opposite positions (180 degrees apart), e.g., both axial.

Marking scheme

(a)(i) 1 mark: [CoCl4]2-; 1 mark: 4.
(ii) 1 mark: correct equation with correct charges; 1 mark: pink to blue.
(b)(i) 1 mark: Co(OH)2 or [Co(H2O)4(OH)2].
(ii) 1 mark: [Co(H2O)6]2+ + 6NH3 -> [Co(NH3)6]2+ + 6H2O.
(iii) 1 mark: [Co(NH3)6]3+ (charge must be 3+).
(c)(i) 1 mark: 3D representation of octahedral cis-isomer with Cl adjacent; 1 mark: 3D representation of trans-isomer with Cl opposite; 1 mark: correct labeling of cis and trans.
Question 4 · Structured
10 marks
Amino acids such as serine, \( \text{HO-CH}_2\text{-CH}(\text{NH}_2)\text{-COOH} \), exhibit both acidic and basic properties.

(a) Draw the structural formula of serine in:
(i) strongly acidic solution (pH 1) [1]
(ii) strongly alkaline solution (pH 13) [1]
(iii) its zwitterionic form at its isoelectric point. [1]
(b) Serine can be esterified by reacting it with methanol in the presence of concentrated sulfuric acid catalyst.
(i) Draw the structural formula of the protonated organic product formed. [1]
(ii) State the name of the new organic functional group formed. [1]
(c) Two different amino acids, alanine (Ala) and serine (Ser), react to form two different dipeptides.
(i) Draw the structural formulas of the two dipeptides, Ala-Ser and Ser-Ala. Circle the peptide bond in one of your drawn structures. [3]
(d) Explain, in terms of structure and bonding, why amino acids have relatively high melting points compared to other organic compounds of similar molecular mass. [2]
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Worked solution

(a)(i) pH 1: \( \text{HO-CH}_2\text{-CH}(\text{NH}_3^+)\text{-COOH} \)
(ii) pH 13: \( \text{HO-CH}_2\text{-CH}(\text{NH}_2)\text{-COO}^- \)
(iii) Zwitterion: \( \text{HO-CH}_2\text{-CH}(\text{NH}_3^+)\text{-COO}^- \)

(b)(i) \( \text{HO-CH}_2\text{-CH}(\text{NH}_3^+)\text{-COOCH}_3 \) (under highly acidic conditions, the amine remains protonated).
(ii) Ester.

(c)(i) Structure of Ala-Ser (alanine at N-terminus, serine at C-terminus):
\( \text{CH}_3\text{-CH}(\text{NH}_2)\text{-CONH-CH}(\text{CH}_2\text{OH})\text{-COOH} \)
Structure of Ser-Ala (serine at N-terminus, alanine at C-terminus):
\( \text{HO-CH}_2\text{-CH}(\text{NH}_2)\text{-CONH-CH}(\text{CH}_3)\text{-COOH} \)
The peptide bond is the \( \text{-CONH-} \) amide link, which must be circled in one of the structures.

(d) In the solid state, amino acids exist as zwitterions. There are strong electrostatic forces of attraction (ionic bonds) between the oppositely charged groups (\( -\text{NH}_3^+ \) and \( -\text{COO}^- \)) of adjacent zwitterions. A large amount of thermal energy is required to overcome these strong forces, resulting in high melting points.

Marking scheme

(a)(i) 1 mark: HO-CH2-CH(NH3+)-COOH.
(ii) 1 mark: HO-CH2-CH(NH2)-COO-.
(iii) 1 mark: HO-CH2-CH(NH3+)-COO-.
(b)(i) 1 mark: HO-CH2-CH(NH3+)-COOCH3 (allow unprotonated HO-CH2-CH(NH2)-COOCH3).
(ii) 1 mark: ester.
(c)(i) 1 mark: Ala-Ser structure; 1 mark: Ser-Ala structure; 1 mark: peptide bond (C=O connected to N-H) clearly circled.
(d) 1 mark: mention of ionic bonding / electrostatic attraction between zwitterions (NH3+ and COO-); 1 mark: requiring a large amount of energy to break.
Question 5 · Structured
10 marks
The Born-Haber cycle is used to determine lattice energies of ionic compounds.

(a) Define the term *lattice energy*. [2]
(b) Consider the determination of the lattice energy of calcium chloride, \( \text{CaCl}_2(\text{s}) \), using the following data:
- Enthalpy change of formation of \( \text{CaCl}_2(\text{s}) = -796 \text{ kJ mol}^{-1} \)
- Enthalpy change of atomisation of calcium = \( +178 \text{ kJ mol}^{-1} \)
- First ionisation energy of calcium = \( +590 \text{ kJ mol}^{-1} \)
- Second ionisation energy of calcium = \( +1145 \text{ kJ mol}^{-1} \)
- Enthalpy change of atomisation of chlorine = \( +121 \text{ kJ mol}^{-1} \)
- Electron affinity of chlorine = \( -349 \text{ kJ mol}^{-1} \)

(i) Write the chemical equation, including state symbols, that represents the lattice energy of calcium chloride. [1]
(ii) Use the data above to calculate the lattice energy of \( \text{CaCl}_2(\text{s}) \). Show your working clearly. [4]
(c) Explain how the lattice energy of calcium chloride compares with:
(i) the lattice energy of calcium bromide, \( \text{CaBr}_2 \). [1]
(ii) the lattice energy of magnesium chloride, \( \text{MgCl}_2 \). [2]
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Worked solution

(a) Lattice energy is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions.

(b)(i) \( \text{Ca}^{2+}(\text{g}) + 2\text{Cl}^-(\text{g}) \rightarrow \text{CaCl}_2(\text{s}) \)

(ii) By Hess's Law:
\( \Delta H_{\text{f}}^\ominus = \Delta H_{\text{at}}^\ominus(\text{Ca}) + 1\text{st IE}(\text{Ca}) + 2\text{nd IE}(\text{Ca}) + 2 \times \Delta H_{\text{at}}^\ominus(\text{Cl}) + 2 \times \text{EA}(\text{Cl}) + \Delta H_{\text{latt}}^\ominus \)
Substituting the given values:
\( -796 = 178 + 590 + 1145 + 2(121) + 2(-349) + \Delta H_{\text{latt}}^\ominus \)
\( -796 = 178 + 590 + 1145 + 242 - 698 + \Delta H_{\text{latt}}^\ominus \)
\( -796 = 1457 + \Delta H_{\text{latt}}^\ominus \)
\( \Delta H_{\text{latt}}^\ominus = -796 - 1457 = -2253 \text{ kJ mol}^{-1} \).

(c)(i) The lattice energy of \( \text{CaBr}_2 \) is less exothermic (less negative) because the bromide ion is larger than the chloride ion, which reduces the electrostatic forces of attraction between the ions.
(ii) The lattice energy of \( \text{MgCl}_2 \) is more exothermic (more negative) because the magnesium ion has a smaller ionic radius than the calcium ion. This results in a higher charge density and stronger electrostatic forces of attraction between the ions.

Marking scheme

(a) 1 mark: enthalpy change when 1 mole of ionic solid is formed; 1 mark: from its gaseous ions (under standard conditions).
(b)(i) 1 mark: Ca^2+(g) + 2Cl^-(g) -> CaCl2(s) (state symbols essential).
(ii) 1 mark: correctly doubling enthalpy of atomisation of Cl (242) and electron affinity of Cl (-698); 1 mark: correct algebraic expression of Hess's law; 1 mark: calculation showing -796 - 1457; 1 mark: final value of -2253 kJ mol^-1 (must include correct sign and units).
(c)(i) 1 mark: CaBr2 is less exothermic because bromide is larger than chloride.
(ii) 1 mark: MgCl2 is more exothermic; 1 mark: because magnesium ion is smaller than calcium ion / has higher charge density.
Question 6 · Structured
10 marks
An organic acid, \( \mathbf{W} \), partitions between ethoxyethane (ether) and water. The partition coefficient, \( K_{\text{pc}} \), is defined as:
\( K_{\text{pc}} = \frac{[\mathbf{W}]_{\text{ether}}}{[\mathbf{W}]_{\text{water}}} = 4.0 \)

(a) Define the term *partition coefficient*. [2]
(b) A solution is prepared containing 3.00 g of \( \mathbf{W} \) in 100 cm\(^3\) of water.
(i) Calculate the mass of \( \mathbf{W} \) extracted when the solution is shaken with a single 50 cm\(^3\) portion of ether. [3]
(ii) Calculate the total mass of \( \mathbf{W} \) extracted when the original solution (3.00 g of \( \mathbf{W} \) in 100 cm\(^3\) of water) is shaken with two successive 25 cm\(^3\) portions of ether. [4]
(c) State, with a reason, which extraction method is more efficient. [1]
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Worked solution

(a) The partition coefficient is the ratio of the concentrations of a solute in two immiscible solvents at equilibrium at a specified temperature.

(b)(i) Let \( x \) be the mass of \( \mathbf{W} \) extracted into ether (in g).
\( [\mathbf{W}]_{\text{ether}} = \frac{x}{50} \)
\( [\mathbf{W}]_{\text{water}} = \frac{3.00 - x}{100} \)
\( K_{\text{pc}} = 4.0 = \frac{x/50}{(3.00 - x)/100} = \frac{2x}{3.00 - x} \)
\( 4.0(3.00 - x) = 2x \Rightarrow 12.0 - 4.0x = 2x \Rightarrow 6x = 12.0 \Rightarrow x = 2.00 \text{ g} \).

(ii) First extraction with 25 cm\(^3\) of ether:
Let \( y \) be the mass of \( \mathbf{W} \) extracted.
\( K_{\text{pc}} = 4.0 = \frac{y/25}{(3.00 - y)/100} = \frac{4y}{3.00 - y} \)
\( 4.0(3.00 - y) = 4y \Rightarrow 12.0 - 4y = 4y \Rightarrow 8y = 12.0 \Rightarrow y = 1.50 \text{ g} \).
Mass of \( \mathbf{W} \) remaining in water = \( 3.00 - 1.50 = 1.50 \text{ g} \).
Second extraction with 25 cm\(^3\) of ether:
Let \( z \) be the mass of \( \mathbf{W} \) extracted.
\( K_{\text{pc}} = 4.0 = \frac{z/25}{(1.50 - z)/100} = \frac{4z}{1.50 - z} \)
\( 4.0(1.50 - z) = 4z \Rightarrow 6.0 - 4z = 4z \Rightarrow 8z = 6.0 \Rightarrow z = 0.75 \text{ g} \).
Total mass of \( \mathbf{W} \) extracted = \( 1.50 + 0.75 = 2.25 \text{ g} \).

(c) Shaking with two successive smaller portions of solvent is more efficient because a greater total mass of \( \mathbf{W} \) is extracted (2.25 g compared to 2.00 g) using the same total volume of solvent (50 cm\(^3\)).

Marking scheme

(a) 1 mark: ratio of concentrations of a solute in two immiscible solvents; 1 mark: at equilibrium.
(b)(i) 1 mark: sets up correct expression \( 4.0 = \frac{x/50}{(3.00-x)/100} \); 1 mark: algebraic manipulation leading to \( 6x = 12.0 \); 1 mark: final mass of 2.00 g.
(b)(ii) 1 mark: calculation of 1.50 g extracted in the first step; 1 mark: shows mass remaining in water is 1.50 g; 1 mark: calculation of 0.75 g extracted in the second step; 1 mark: final total mass of 2.25 g.
(c) 1 mark: states that successive extraction is more efficient because a larger mass of solute is extracted (2.25 g > 2.00 g).

Paper 33 (Advanced Practical Skills)

Answer all questions. Record all observations, burette readings, and temperatures with appropriate precision.
3 Question · 39.99 marks
Question 1 · Practical Tasks & Calculations
13.33 marks
You are required to determine the water of crystallisation, \( x \), in hydrated copper(II) sulfate, \( \text{CuSO}_4 \cdot x\text{H}_2\text{O} \), using iodometric titration.

**Procedure**
1. Weigh out accurately \( 6.20\text{ g} \) of the hydrated copper(II) sulfate sample and dissolve it in distilled water. Transfer the solution quantitatively to a \( 250.0\text{ cm}^3 \) volumetric flask and make it up to the mark with distilled water. Label this solution **FA 1**.
2. Pipette a \( 25.00\text{ cm}^3 \) aliquot of **FA 1** into a conical flask.
3. Add \( 10\text{ cm}^3 \) of \( 1.0\text{ mol dm}^{-3} \) potassium iodide, \( \text{KI} \) (an excess), to the flask. A brown mixture containing a precipitate of copper(I) iodide and dissolved iodine is formed.
4. Titrate the liberated iodine with \( 0.100\text{ mol dm}^{-3} \) sodium thiosulfate solution, **FA 2**, until the brown colour fades to a pale yellow.
5. Add a few drops of starch indicator. The mixture will turn blue-black.
6. Continue titrating dropwise with **FA 2** until the blue-black colour completely disappears, leaving a white/off-white precipitate.
7. Record your results in a suitable table showing initial and final burette readings, and the volume of **FA 2** added for a rough titration and two suitable concordant titrations.

**Data and Results**
Assume the average concordant titre of **FA 2** obtained is \( 24.85\text{ cm}^3 \).

(a) Calculate the number of moles of thiosulfate ions, \( \text{S}_2\text{O}_3^{2-} \), used in your average titre.
(b) Using the equations below, determine the number of moles of \( \text{Cu}^{2+} \) ions present in the \( 25.00\text{ cm}^3 \) sample of **FA 1**:
\( 2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq}) \)
\( \text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq}) \)
(c) Calculate the number of moles of \( \text{Cu}^{2+} \) in the original \( 250.0\text{ cm}^3 \) of solution **FA 1**.
(d) Calculate the molar mass of \( \text{CuSO}_4 \cdot x\text{H}_2\text{O} \).
(e) Determine the value of \( x \), showing your working. (\( A_r \): \( \text{Cu} = 63.5 \), \( \text{S} = 32.1 \), \( \text{O} = 16.0 \), \( \text{H} = 1.0 \))
Show answer & marking scheme

Worked solution

1. **Moles of \( \text{S}_2\text{O}_3^{2-} \)**:
\( n(\text{S}_2\text{O}_3^{2-}) = \frac{24.85}{1000} \text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.485 \times 10^{-3}\text{ mol} \)

2. **Moles of \( \text{Cu}^{2+} \) in \( 25.00\text{ cm}^3 \)**:
From the stoichiometry:
\( 2\text{Cu}^{2+} \equiv 1\text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-} \)
Therefore, \( n(\text{Cu}^{2+}) = n(\text{S}_2\text{O}_3^{2-}) = 2.485 \times 10^{-3}\text{ mol} \)

3. **Moles of \( \text{Cu}^{2+} \) in \( 250.0\text{ cm}^3 \)**:
\( n(\text{Cu}^{2+})_{\text{total}} = 2.485 \times 10^{-3}\text{ mol} \times 10 = 2.485 \times 10^{-2}\text{ mol} \)

4. **Molar mass of \( \text{CuSO}_4 \cdot x\text{H}_2\text{O} \)**:
\( M_r = \frac{\text{mass}}{\text{moles}} = \frac{6.20\text{ g}}{2.485 \times 10^{-2}\text{ mol}} = 249.50\text{ g mol}^{-1} \)

5. **Value of \( x \)**:
\( M_r(\text{CuSO}_4) = 63.5 + 32.1 + (4 \times 16.0) = 159.6\text{ g mol}^{-1} \)
Mass of water of crystallisation per mole = \( 249.50 - 159.6 = 89.9\text{ g mol}^{-1} \)
\( x = \frac{89.9}{18.0} = 4.99 \approx 5 \)

Marking scheme

**Table of Titration Results (3 marks)**:
- 1 mark for structured table with appropriate headings, units (cm\(^3\)), and recording burette readings to 2 decimal places (ending in .00 or .05).
- 1 mark for obtaining concordant titres within \( 0.10\text{ cm}^3 \).
- 1 mark for correctly calculating the average titre using concordant results.

**Calculations (7.33 marks)**:
- 1.33 marks for calculating moles of \( \text{S}_2\text{O}_3^{2-} \) correctly to 3 or 4 significant figures: \( 2.485 \times 10^{-3}\text{ mol} \).
- 1.5 marks for deducing that moles of \( \text{Cu}^{2+} \) in \( 25.00\text{ cm}^3 \) is equal to moles of \( \text{S}_2\text{O}_3^{2-} \), giving \( 2.485 \times 10^{-3}\text{ mol} \).
- 1.5 marks for scaling up by a factor of 10 to find total moles in \( 250.0\text{ cm}^3 \): \( 2.485 \times 10^{-2}\text{ mol} \).
- 1.5 marks for calculating the molar mass of the hydrate correctly: \( 249.5\text{ g mol}^{-1} \).
- 1.50 marks for calculating \( x = 5 \) (accept values from 4.9 to 5.1 rounded to the nearest integer).

**Evaluating Errors (3 marks)**:
- 1.5 marks for identifying the percentage error in the pipette (\( \pm 0.06\text{ cm}^3 \) on \( 25.00\text{ cm}^3 \)): \( \frac{0.06}{25.00} \times 100 = 0.24\% \).
- 1.5 marks for explaining the effect of using a wet volumetric flask (no effect, as water is added to the mark anyway).
Question 2 · Practical Tasks & Calculations
13.33 marks
You are investigating the reaction between peroxodisulfate ions, \( \text{S}_2\text{O}_8^{2-} \), and iodide ions, \( \text{I}^- \):
\( \text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq}) \)

The reaction rate is studied using an iodine clock method. A small volume of sodium thiosulfate solution, \( \text{Na}_2\text{S}_2\text{O}_3 \), and starch indicator are added to the reaction mixture. The liberated iodine reacts instantly with thiosulfate:
\( \text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq}) \)

Once all the thiosulfate is consumed, any further iodine produced reacts with starch to form a blue-black complex. The time taken, \( t \), for the blue-black colour to appear is measured.

Three experiments were conducted at a constant temperature of \( 298\text{ K} \) with varying initial concentrations of the reactants. The total volume of each reaction mixture was kept constant at \( 50.0\text{ cm}^3 \) by adding distilled water.

The experimental data is shown below:

- **Experiment 1**:
- Volume of \( 0.100\text{ mol dm}^{-3}\ \text{KI}(\text{aq}) \): \( 20.0\text{ cm}^3 \)
- Volume of \( 0.100\text{ mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3(\text{aq}) \): \( 5.0\text{ cm}^3 \)
- Volume of distilled water: \( 15.0\text{ cm}^3 \)
- Volume of \( 0.100\text{ mol dm}^{-3}\ \text{K}_2\text{S}_2\text{O}_8(\text{aq}) \): \( 10.0\text{ cm}^3 \)
- Time taken for blue-black colour to appear, \( t_1 \): \( 44\text{ s} \)

- **Experiment 2**:
- Volume of \( 0.100\text{ mol dm}^{-3}\ \text{KI}(\text{aq}) \): \( 10.0\text{ cm}^3 \)
- Volume of \( 0.100\text{ mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3(\text{aq}) \): \( 5.0\text{ cm}^3 \)
- Volume of distilled water: \( 25.0\text{ cm}^3 \)
- Volume of \( 0.100\text{ mol dm}^{-3}\ \text{K}_2\text{S}_2\text{O}_8(\text{aq}) \): \( 10.0\text{ cm}^3 \)
- Time taken for blue-black colour to appear, \( t_2 \): \( 88\text{ s} \)

- **Experiment 3**:
- Volume of \( 0.100\text{ mol dm}^{-3}\ \text{KI}(\text{aq}) \): \( 20.0\text{ cm}^3 \)
- Volume of \( 0.100\text{ mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3(\text{aq}) \): \( 5.0\text{ cm}^3 \)
- Volume of distilled water: \( 20.0\text{ cm}^3 \)
- Volume of \( 0.100\text{ mol dm}^{-3}\ \text{K}_2\text{S}_2\text{O}_8(\text{aq}) \): \( 5.0\text{ cm}^3 \)
- Time taken for blue-black colour to appear, \( t_3 \): \( 88\text{ s} \)

(a) Deduce the order of reaction with respect to:
(i) \( \text{I}^- \) ions. Show your reasoning by comparing appropriate experiments.
(ii) \( \text{S}_2\text{O}_8^{2-} \) ions. Show your reasoning by comparing appropriate experiments.

(b) Construct the rate equation for this reaction.

(c) Calculate the initial rate of reaction for Experiment 1 in \( \text{mol dm}^{-3}\text{ s}^{-1} \).
*(Hint: First calculate the concentration of \( \text{S}_2\text{O}_8^{2-} \) that has reacted during the time \( t_1 \), which is stoichiometrically related to the moles of \( \text{S}_2\text{O}_3^{2-} \) consumed.)*

(d) Calculate the rate constant, \( k \), for the reaction, stating its units.
Show answer & marking scheme

Worked solution

(a)
(i) Order with respect to \( \text{I}^- \):
Comparing Experiment 1 and Experiment 2:
- \( [\text{S}_2\text{O}_8^{2-}] \) is kept constant (the volume of \( \text{K}_2\text{S}_2\text{O}_8 \) is \( 10.0\text{ cm}^3 \)).
- \( [\text{I}^-] \) is halved (volume of \( \text{KI} \) decreases from \( 20.0\text{ cm}^3 \) to \( 10.0\text{ cm}^3 \)).
- The time taken, \( t \), doubles from \( 44\text{ s} \) to \( 88\text{ s} \), which means the rate (\( \propto 1/t \)) is halved.
- Since halving the concentration halves the rate, the reaction is **first order** with respect to \( \text{I}^- \).

(ii) Order with respect to \( \text{S}_2\text{O}_8^{2-} \):
Comparing Experiment 1 and Experiment 3:
- \( [\text{I}^-] \) is kept constant (the volume of \( \text{KI} \) is \( 20.0\text{ cm}^3 \)).
- \( [\text{S}_2\text{O}_8^{2-}] \) is halved (volume of \( \text{K}_2\text{S}_2\text{O}_8 \) decreases from \( 10.0\text{ cm}^3 \) to \( 5.0\text{ cm}^3 \)).
- The time taken, \( t \), doubles from \( 44\text{ s} \) to \( 88\text{ s} \), which means the rate is halved.
- Since halving the concentration halves the rate, the reaction is **first order** with respect to \( \text{S}_2\text{O}_8^{2-} \).

(b) Rate equation:
\( \text{Rate} = k[\text{I}^-][\text{S}_2\text{O}_8^{2-}] \)

(c) Initial rate of reaction for Experiment 1:
- Moles of \( \text{S}_2\text{O}_3^{2-} \) added = \( \frac{5.0}{1000} \times 0.100 = 5.0 \times 10^{-4}\text{ mol} \).
- Since \( 2\text{S}_2\text{O}_3^{2-} \equiv 1\text{I}_2 \), moles of \( \text{I}_2 \) produced before the blue-black colour appears = \( 2.5 \times 10^{-4}\text{ mol} \).
- Since \( 1\text{S}_2\text{O}_8^{2-} \equiv 1\text{I}_2 \), the amount of \( \text{S}_2\text{O}_8^{2-} \) reacted is \( 2.5 \times 10^{-4}\text{ mol} \).
- The concentration of \( \text{S}_2\text{O}_8^{2-} \) reacted in the \( 50.0\text{ cm}^3 \) total volume is:
\( \Delta[\text{S}_2\text{O}_8^{2-}] = \frac{2.5 \times 10^{-4}\text{ mol}}{0.050\text{ dm}^3} = 5.0 \times 10^{-3}\text{ mol dm}^{-3} \).
- Initial Rate = \( \frac{\Delta[\text{S}_2\text{O}_8^{2-}]}{t} = \frac{5.0 \times 10^{-3}\text{ mol dm}^{-3}}{44\text{ s}} = 1.136 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} \) (or \( 1.14 \times 10^{-4} \)).

(d) Rate constant, \( k \):
In Experiment 1:
- \( [\text{I}^-] = 0.100 \times \frac{20.0}{50.0} = 0.0400\text{ mol dm}^{-3} \).
- \( [\text{S}_2\text{O}_8^{2-}] = 0.100 \times \frac{10.0}{50.0} = 0.0200\text{ mol dm}^{-3} \).
- \( k = \frac{\text{Rate}}{[\text{I}^-][\text{S}_2\text{O}_8^{2-}]} = \frac{1.136 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}}{0.0400\text{ mol dm}^{-3} \times 0.0200\text{ mol dm}^{-3}} = 0.142\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1} \) (allow \( 0.14 \) to \( 0.143 \)).

Marking scheme

**Order of Reaction (4 marks)**:
- 1 mark for identifying that comparing Exp 1 and Exp 2 involves halving \( [\text{I}^-] \) while keeping \( [\text{S}_2\text{O}_8^{2-}] \) constant.
- 1 mark for deducing first order for \( \text{I}^- \) because the rate is halved.
- 1 mark for identifying that comparing Exp 1 and Exp 3 involves halving \( [\text{S}_2\text{O}_8^{2-}] \) while keeping \( [\text{I}^-] \) constant.
- 1 mark for deducing first order for \( \text{S}_2\text{O}_8^{2-} \).

**Rate Equation (1.33 marks)**:
- 1.33 marks for the correct rate equation: \( \text{Rate} = k[\text{I}^-][\text{S}_2\text{O}_8^{2-}] \) (do not accept if \( k \) is missing).

**Rate Calculation (4 marks)**:
- 1 mark for calculating the moles of \( \text{S}_2\text{O}_3^{2-} \) as \( 5.0 \times 10^{-4}\text{ mol} \).
- 1 mark for showing that moles of \( \text{S}_2\text{O}_8^{2-} \) reacted is half of that: \( 2.5 \times 10^{-4}\text{ mol} \).
- 1 mark for converting this to a change in concentration: \( 5.0 \times 10^{-3}\text{ mol dm}^{-3} \).
- 1 mark for dividing by the time (44 s) to get \( 1.14 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} \) (accept \( 1.1 \times 10^{-4} \) to \( 1.15 \times 10^{-4} \)).

**Rate Constant & Units (4 marks)**:
- 1 mark for calculating correct concentrations of \( [\text{I}^-] = 0.040\text{ mol dm}^{-3} \) and \( [\text{S}_2\text{O}_8^{2-}] = 0.020\text{ mol dm}^{-3} \) in the mixture.
- 2 marks for calculating \( k = 0.142 \) (allow \( 0.14 \) to \( 0.143 \)).
- 1 mark for the correct units: \( \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1} \).
Question 3 · Practical Tasks & Calculations
13.33 marks
You are provided with two solutions: **FA 3** (containing one transition metal cation and one halide anion) and **FA 4** (containing one transition metal cation and one Group 16 oxyanion).

You are required to perform the tests described in the table below and record all of your observations. From these observations, you will identify the ions present and write balanced ionic equations.

| Test | Observation |
|---|---|
| **Test 1**: To a \( 1.0\text{ cm}^3 \) portion of **FA 3** in a test-tube, add aqueous sodium hydroxide dropwise until in excess. | A red-brown precipitate is formed, which is insoluble in excess sodium hydroxide. |
| **Test 2**: To a \( 1.0\text{ cm}^3 \) portion of **FA 3** in a test-tube, add \( 1\text{ cm}^3 \) of dilute nitric acid followed by a few drops of aqueous silver nitrate. | A thick white precipitate is formed. |
| **Test 3**: To a \( 1.0\text{ cm}^3 \) portion of **FA 4** in a test-tube, add aqueous sodium hydroxide dropwise until in excess. | A grey-green precipitate is formed. The precipitate dissolves in excess sodium hydroxide to give a dark green solution. |
| **Test 4**: To a \( 1.0\text{ cm}^3 \) portion of **FA 4** in a test-tube, add aqueous ammonia dropwise until in excess. | A grey-green precipitate is formed, which is insoluble in excess ammonia. |
| **Test 5**: To a \( 1.0\text{ cm}^3 \) portion of **FA 4** in a test-tube, add a few drops of aqueous barium nitrate, followed by dilute nitric acid. | A white precipitate is formed, which does not dissolve when dilute nitric acid is added. |

(a) Identify the ions present in **FA 3** and **FA 4**:
- Cation in **FA 3**
- Anion in **FA 3**
- Cation in **FA 4**
- Anion in **FA 4**

(b) Write a balanced ionic equation for the reaction that occurs when aqueous sodium hydroxide is added to **FA 3** in Test 1. Include state symbols.

(c) Write a balanced ionic equation showing the dissolution of the precipitate formed in Test 3 when excess sodium hydroxide is added.
Show answer & marking scheme

Worked solution

(a) Identifying the ions:
- **FA 3 Cation**: The red-brown precipitate with \( \text{NaOH} \) that is insoluble in excess is characteristic of the **iron(III)** ion, \( \text{Fe}^{3+} \).
- **FA 3 Anion**: The thick white precipitate with silver nitrate that forms in acidic conditions (nitric acid) is characteristic of the **chloride** ion, \( \text{Cl}^- \).
- **FA 4 Cation**: The grey-green precipitate with \( \text{NaOH} \) that dissolves in excess to give a green solution, and with \( \text{NH}_3 \) forms a grey-green precipitate insoluble in excess, is characteristic of the **chromium(III)** ion, \( \text{Cr}^{3+} \).
- **FA 4 Anion**: The white precipitate with barium nitrate that is insoluble in dilute nitric acid is characteristic of the **sulfate** ion, \( \text{SO}_4^{2-} \).

(b) Ionic equation for Test 1:
\( \text{Fe}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Fe}(\text{OH})_3(\text{s}) \)

(c) Ionic equation for the dissolution in excess \( \text{NaOH} \):
The precipitate is chromium(III) hydroxide, \( \text{Cr}(\text{OH})_3 \).
In excess hydroxide, it forms the soluble hexahydroxochromate(III) complex ion:
\( \text{Cr}(\text{OH})_3(\text{s}) + 3\text{OH}^-(\text{aq}) \rightarrow [\text{Cr}(\text{OH})_6]^{3-}(\text{aq}) \)
*(Alternative acceptable equations: \( \text{Cr}^{3+}(\text{aq}) + 6\text{OH}^-(\text{aq}) \rightarrow [\text{Cr}(\text{OH})_6]^{3-}(\text{aq}) \) or \( \text{Cr}(\text{OH})_3(\text{s}) + \text{OH}^-(\text{aq}) \rightarrow [\text{Cr}(\text{OH})_4]^-(\text{aq}) \).)*

Marking scheme

**Identification of Ions (6 marks)**:
- 1.5 marks for identifying the cation in **FA 3** as \( \text{Fe}^{3+} \) (accept iron(III)).
- 1.5 marks for identifying the anion in **FA 3** as \( \text{Cl}^- \) (accept chloride).
- 1.5 marks for identifying the cation in **FA 4** as \( \text{Cr}^{3+} \) (accept chromium(III)).
- 1.5 marks for identifying the anion in **FA 4** as \( \text{SO}_4^{2-} \) (accept sulfate).

**Ionic Equation for Test 1 (3.33 marks)**:
- 2 marks for the correct equation: \( \text{Fe}^{3+} + 3\text{OH}^- \rightarrow \text{Fe}(\text{OH})_3 \).
- 1.33 marks for correct state symbols: (aq) for reactants and (s) for product.

**Ionic Equation for Test 3 dissolution (4 marks)**:
- 2.5 marks for writing the correct species: \( \text{Cr}(\text{OH})_3(\text{s}) \) and \( \text{OH}^-(\text{aq}) \) reacting to form \( [\text{Cr}(\text{OH})_6]^{3-}(\text{aq}) \) (or tetrahydroxochromate(III), \( [\text{Cr}(\text{OH})_4]^- \)).
- 1.5 marks for balancing the equation and ensuring correct charges.

Paper 43 (A Level Structured)

Answer all questions. Show your working and use appropriate units where necessary.
8 Question · 100 marks
Question 1 · A Level Structured
12.5 marks
Cobalt(III) complexes with bidentate ligands show significant thermodynamic stability and interesting stereoisomerism.

(a) Define the term *bidentate ligand*.

(b) An aqueous solution of cobalt(II) ions is oxidised in the presence of 1,2-diaminoethane (en) to form the octahedral complex ion \([Co(en)_3]^{3+}\).
(i) Draw the 3D stereoisomers of \([Co(en)_3]^{3+}\). State the type of isomerism shown.
(ii) Suggest why the coordination of three bidentate 'en' ligands to a single metal ion results in a highly thermodynamically stable complex compared to coordination with monodentate ammonia ligands, referring to entropy.

(c) The stability constant, \(K_{stab}\), for the equilibrium:
\[[Co(H_2O)_6]^{2+} + 3en \rightleftharpoons [Co(en)_3]^{2+} + 6H_2O\]
is \(1.5 \times 10^{14}\ \text{dm}^9\ \text{mol}^{-3}\).
(i) Write the expression for \(K_{stab}\).
(ii) Calculate the concentration of free \([Co(H_2O)_6]^{2+}\) remaining in a solution where the equilibrium concentration of \([Co(en)_3]^{2+}\) is \(0.050\ \text{mol\ dm}^{-3}\) and the equilibrium concentration of free \(en\) ligand is \(2.0 \times 10^{-4}\ \text{mol\ dm}^{-3}\).
Show answer & marking scheme

Worked solution

(a) A bidentate ligand is a species that donates two lone pairs of electrons to a central metal ion to form two coordinate (dative covalent) bonds.

(b) (i) The stereoisomers are non-superimposable mirror images of each other (enantiomers). They show optical isomerism. The octahedral structure should be drawn with three bidentate loops (each representing \(\text{H}_2\text{N-CH}_2\text{-CH}_2\text{-NH}_2\)) connecting adjacent cis-positions on the octahedron.
(ii) Replacing monodentate ligands with bidentate ligands (the chelate effect) increases the total number of independent particles in the system (e.g., from 4 reactant particles to 7 product particles if replacing 6 monodentate ligands with 3 bidentate ligands). This results in a highly positive entropy change of the system (\(\Delta S^\ominus > 0\)). Since \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\), a positive entropy change makes \(\Delta G^\ominus\) significantly more negative, resulting in a highly stable complex.

(c) (i) \(K_{stab} = \frac{[[Co(en)_3]^{2+}]}{[[Co(H_2O)_6]^{2+}][en]^3}\)
(ii) Rearranging for \([[Co(H_2O)_6]^{2+}]\):
\[[[Co(H_2O)_6]^{2+}] = \frac{[[Co(en)_3]^{2+}]}{K_{stab} \times [en]^3}\]
\[[[Co(H_2O)_6]^{2+}] = \frac{0.050}{(1.5 \times 10^{14}) \times (2.0 \times 10^{-4})^3}\]
\[[[Co(H_2O)_6]^{2+}] = \frac{0.050}{(1.5 \times 10^{14}) \times (8.0 \times 10^{-12})}\]
\[[[Co(H_2O)_6]^{2+}] = \frac{0.050}{1200} = 4.17 \times 10^{-5}\ \text{mol\ dm}^{-3}\]

Marking scheme

(a) Award 1.5 marks for defining bidentate ligand:
- Donates two lone pairs of electrons [0.75 marks]
- Forms two coordinate/dative covalent bonds to a central metal ion [0.75 marks]

(b)(i) Award 4 marks total:
- Correct 3D representations of both isomers showing mirror symmetry (using wedge/dash lines) [2 marks]
- Identification of optical isomerism / enantiomers [1 mark]
- Correctly representation of bidentate 'en' loops connecting cis positions [1 mark]

(b)(ii) Award 3 marks total:
- Replacement of monodentate ligands with bidentate ligands increases the number of species/particles in solution [1 mark]
- Leading to an increase in entropy / disorder (\(\Delta S^\ominus > 0\)) [1 mark]
- Making \(\Delta G^\ominus\) more negative (and thus complex more thermodynamically stable) [1 mark]

(c)(i) Award 2 marks:
- Correct formula for \(K_{stab}\) with products on top and reactants on bottom [1.5 marks]
- Correct power of 3 for \([en]\) [0.5 marks]

(c)(ii) Award 2 marks:
- Correct substitution and rearranging of formula [1 mark]
- Final answer: \(4.17 \times 10^{-5}\) (or \(4.2 \times 10^{-5}\)) with correct units [1 mark]
Question 2 · A Level Structured
12.5 marks
The kinetics of the reaction between three reactants, A, B, and C, were investigated at 298 K:
\[2A + B + C \rightarrow D + E\]
The initial rate data obtained from four experiments are shown below:

| Experiment | Initial [A] / \(\text{moldm}^{-3}\) | Initial [B] / \(\text{moldm}^{-3}\) | Initial [C] / \(\text{moldm}^{-3}\) | Initial Rate / \(\text{moldm}^{-3}\text{s}^{-1}\) |
| :--- | :--- | :--- | :--- | :--- |
| 1 | 0.10 | 0.10 | 0.10 | \(1.2 \times 10^{-3}\) |
| 2 | 0.20 | 0.10 | 0.10 | \(2.4 \times 10^{-3}\) |
| 3 | 0.10 | 0.30 | 0.10 | \(1.08 \times 10^{-2}\) |
| 4 | 0.10 | 0.10 | 0.20 | \(1.2 \times 10^{-3}\) |

(a) Deduce the order of reaction with respect to A, B, and C. Explain your reasoning clearly.

(b) Write the rate equation for this reaction.

(c) Calculate the rate constant, \(k\), using the data from Experiment 1, and state its units.

(d) Describe and explain how the rate constant, \(k\), changes, if at all, when:
(i) the temperature is increased.
(ii) a catalyst is added at constant temperature.
Show answer & marking scheme

Worked solution

(a)
- **Order with respect to A**:
Comparing Experiments 1 and 2, the concentrations of B and C are kept constant, while the concentration of A is doubled (from 0.10 to 0.20 \(\text{moldm}^{-3}\)). The initial rate doubles (from \(1.2 \times 10^{-3}\) to \(2.4 \times 10^{-3}\ \text{moldm}^{-3}\text{s}^{-1}\)). Therefore, the reaction is **first order** with respect to A.
- **Order with respect to B**:
Comparing Experiments 1 and 3, the concentrations of A and C are kept constant, while the concentration of B is tripled (from 0.10 to 0.30 \(\text{moldm}^{-3}\)). The initial rate increases by a factor of 9 (from \(1.2 \times 10^{-3}\) to \(1.08 \times 10^{-2}\), which is \(9 \times 1.2 \times 10^{-3}\)). Since \(3^2 = 9\), the reaction is **second order** with respect to B.
- **Order with respect to C**:
Comparing Experiments 1 and 4, the concentrations of A and B are kept constant, while the concentration of C is doubled (from 0.10 to 0.20 \(\text{moldm}^{-3}\)). The initial rate remains unchanged (\(1.2 \times 10^{-3}\ \text{moldm}^{-3}\text{s}^{-1}\)). Therefore, the reaction is **zero order** with respect to C.

(b) The rate equation is:
\[\text{Rate} = k[A][B]^2\]

(c) Using the data from Experiment 1:
\[1.2 \times 10^{-3} = k (0.10) (0.10)^2\]
\[1.2 \times 10^{-3} = k (1.0 \times 10^{-3})\]
\[k = \frac{1.2 \times 10^{-3}}{1.0 \times 10^{-3}} = 1.2\]
Units of \(k\):
\[\text{Units} = \frac{\text{mol\ dm}^{-3}\ \text{s}^{-1}}{(\text{mol\ dm}^{-3})(\text{mol\ dm}^{-3})^2} = \text{dm}^6\ \text{mol}^{-2}\ \text{s}^{-1}\]

(d) (i) When temperature is increased, the rate constant \(k\) increases because a higher fraction of reactant molecules have kinetic energy greater than or equal to the activation energy (\(E \ge E_a\)), resulting in a higher frequency of successful collisions.
(ii) When a catalyst is added, the rate constant \(k\) increases because the catalyst provides an alternative pathway with a lower activation energy, meaning more molecules have sufficient energy to react at the same temperature.

Marking scheme

(a) Award 4.5 marks total (1.5 marks for each order & explanation):
- First order with respect to A + explanation [1.5 marks]
- Second order with respect to B + explanation [1.5 marks]
- Zero order with respect to C + explanation [1.5 marks]

(b) Award 1 mark:
- Rate = k[A][B]^2 [1 mark]

(c) Award 3 marks:
- Correct calculation of k value (1.2) [1.5 marks]
- Correct units (dm^6 mol^-2 s^-1) [1.5 marks]

(d)(i) Award 2 marks:
- Rate constant k increases [1 mark]
- More particles have energy >= Ea / greater frequency of successful collisions [1 mark]

(d)(ii) Award 2 marks:
- Rate constant k increases [1 mark]
- Alternative pathway with lower activation energy [1 mark]
Question 3 · A Level Structured
12.5 marks
Lattice energy measurements provide crucial insight into ionic bonding and structure.

(a) Define the term *lattice energy*.

(b) Construct a fully labelled Born-Haber cycle for the formation of solid calcium chloride, \(CaCl_2(s)\), from its constituent elements in their standard states. Include state symbols and enthalpy labels.

(c) Use the data below to calculate the lattice energy (\(\Delta H^\ominus_{latt}\)) of \(CaCl_2(s)\).
- Standard enthalpy of formation of \(CaCl_2(s) = -796\ \text{kJ\ mol}^{-1}\)
- Enthalpy change of atomisation of \(Ca(s) = +178\ \text{kJ\ mol}^{-1}\)
- First ionisation energy of \(Ca(g) = +590\ \text{kJ\ mol}^{-1}\)
- Second ionisation energy of \(Ca(g) = +1145\ \text{kJ\ mol}^{-1}\)
- Bond enthalpy of \(Cl_2(g) = +242\ \text{kJ\ mol}^{-1}\)
- First electron affinity of \(Cl(g) = -349\ \text{kJ\ mol}^{-1}\)

(d) Explain why the lattice energy of calcium fluoride, \(CaF_2\), is significantly more exothermic than that of calcium chloride, \(CaCl_2\).
Show answer & marking scheme

Worked solution

(a) Lattice energy is the enthalpy change when one mole of an ionic compound is formed from its constituent gaseous ions under standard conditions.

(b) The Born-Haber cycle should show the following steps:
- Enthalpy of formation path: \(Ca(s) + Cl_2(g) \rightarrow CaCl_2(s)\)
- Atomisation of Calcium: \(Ca(s) \rightarrow Ca(g)\)
- First Ionisation of Calcium: \(Ca(g) \rightarrow Ca^+(g) + e^-\)
- Second Ionisation of Calcium: \(Ca^+(g) \rightarrow Ca^{2+}(g) + e^-\)
- Atomisation of Chlorine: \(Cl_2(g) \rightarrow 2Cl(g)\) (equal to the bond enthalpy of \(Cl_2\))
- Electron Affinity of Chlorine: \(2Cl(g) + 2e^- \rightarrow 2Cl^-(g)\) (equal to \(2 \times\) first electron affinity of \(Cl\))
- Lattice Energy: \(Ca^{2+}(g) + 2Cl^-(g) \rightarrow CaCl_2(s)\)

(c) Using Hess's law:
\[\Delta H^\ominus_f = \Delta H^\ominus_{at}(Ca) + IE_1(Ca) + IE_2(Ca) + E_{bond}(Cl_2) + 2 \times EA_1(Cl) + \Delta H^\ominus_{latt}\]
\[-796 = +178 + 590 + 1145 + 242 + 2(-349) + \Delta H^\ominus_{latt}\]
\[-796 = +2155 - 698 + \Delta H^\ominus_{latt}\]
\[-796 = 1457 + \Delta H^\ominus_{latt}\]
\[\Delta H^\ominus_{latt} = -796 - 1457 = -2253\ \text{kJ\ mol}^{-1}\]

(d) The fluoride ion (\(F^-\)) has a smaller ionic radius than the chloride ion (\(Cl^-\)). Since the charge density of the fluoride ion is higher, there is a stronger electrostatic attraction between the calcium ion (\(Ca^{2+}\)) and fluoride ions compared to chloride ions. Consequently, more energy is released when the ionic lattice of \(CaF_2\) is formed, making its lattice energy more exothermic.

Marking scheme

(a) Award 2 marks:
- Enthalpy change when 1 mole of solid ionic compound is formed [1 mark]
- From its gaseous ions under standard conditions [1 mark]

(b) Award 4.5 marks:
- Correct elements/compounds with correct state symbols on all levels [1.5 marks]
- Correct arrows and directions with associated thermodynamic labels [1.5 marks]
- Double chlorine atomisation and electron affinity steps shown with appropriate coefficient '2' [1.5 marks]

(c) Award 3 marks:
- Correct expression of Hess's Law [1 mark]
- Correct multiplication of first electron affinity by 2 (i.e. -698) [1 mark]
- Correct calculation to yield -2253 kJ mol^-1 with sign and units [1 mark]

(d) Award 3 marks:
- Fluoride ion is smaller than chloride ion [1 mark]
- Stronger electrostatic attraction between Ca2+ and F- than Ca2+ and Cl- [1 mark]
- More energy released upon lattice formation [1 mark]
Question 4 · A Level Structured
12.5 marks
The partition coefficient of an organic carboxylic acid, X, between ethoxyethane and water is 8.0 (meaning X is more soluble in ethoxyethane).
\[K_{partition} = \frac{[X]_{ethoxyethane}}{[X]_{water}} = 8.0\]
An aqueous solution contains \(4.50\ \text{g}\) of solute X dissolved in \(100\ \text{cm}^3\) of water.

(a) Define the term *partition coefficient*.

(b) Calculate the mass of X extracted if the \(100\ \text{cm}^3\) aqueous solution is shaken with a single \(50.0\ \text{cm}^3\) portion of ethoxyethane.

(c) Alternatively, the same aqueous solution of X is extracted using two successive \(25.0\ \text{cm}^3\) portions of ethoxyethane.
(i) Calculate the total mass of X extracted after these two successive extractions.
(ii) Compare the efficiency of a single extraction versus successive extractions based on your calculations.
Show answer & marking scheme

Worked solution

(a) The partition coefficient is the ratio of the concentration of a solute in two immiscible solvents at equilibrium at a constant temperature.

(b) Let \(x\) be the mass of X extracted into ethoxyethane.
The mass of X remaining in water is \(4.50 - x\).
\[K_{partition} = 8.0 = \frac{[X]_{ethoxyethane}}{[X]_{water}} = \frac{x / 50.0}{(4.50 - x) / 100}\]
\[8.0 = \frac{x \times 100}{50.0 \times (4.50 - x)}\]
\[8.0 = \frac{2x}{4.50 - x}\]
\[8.0(4.50 - x) = 2x\]
\[36.0 - 8.0x = 2x\]
\[10.0x = 36.0 \implies x = 3.60\ \text{g}\]

(c) (i)
- **First Extraction** with \(25.0\ \text{cm}^3\):
Let \(y_1\) be the mass extracted into the first portion of ethoxyethane.
\[8.0 = \frac{y_1 / 25.0}{(4.50 - y_1) / 100} = \frac{4y_1}{4.50 - y_1}\]
\[8.0(4.50 - y_1) = 4y_1\]
\[36.0 - 8.0y_1 = 4y_1 \implies 12y_1 = 36.0 \implies y_1 = 3.00\ \text{g}\]
Mass remaining in water = \(4.50 - 3.00 = 1.50\ \text{g}\).

- **Second Extraction** with \(25.0\ \text{cm}^3\):
Let \(y_2\) be the mass extracted from the remaining \(1.50\ \text{g}\).
\[8.0 = \frac{y_2 / 25.0}{(1.50 - y_2) / 100} = \frac{4y_2}{1.50 - y_2}\]
\[8.0(1.50 - y_2) = 4y_2\]
\[12.0 - 8.0y_2 = 4y_2 \implies 12y_2 = 12.0 \implies y_2 = 1.00\ \text{g}\]
Total mass extracted = \(y_1 + y_2 = 3.00 + 1.00 = 4.00\ \text{g}\).

(ii) Shaking with two successive portions of \(25.0\ \text{cm}^3\) is more efficient because a greater total mass of X is extracted (\(4.00\ \text{g}\) compared to \(3.60\ \text{g}\) from a single extraction) using the exact same total volume of solvent (\(50.0\ \text{cm}^3\)).

Marking scheme

(a) Award 2 marks:
- Ratio of concentration of solute in two immiscible solvents [1 mark]
- At equilibrium / constant temperature [1 mark]

(b) Award 4 marks:
- Correct algebraic setup of equation using volumes [1.5 marks]
- Rearranging equation correctly [1.5 marks]
- Final mass of 3.60 g [1 mark]

(c)(i) Award 4.5 marks:
- Mass extracted in first stage is 3.00 g [1.5 marks]
- Correct calculation of remaining mass (1.50 g) [0.5 marks]
- Mass extracted in second stage is 1.00 g [1.5 marks]
- Total mass is 4.00 g [1 mark]

(c)(ii) Award 2 marks:
- Successive extractions are more efficient [1 mark]
- Extracted 4.00 g vs 3.60 g with same total solvent volume [1 mark]
Question 5 · A Level Structured
12.5 marks
The behavior of amino acids in aqueous solution depends significantly on pH.

(a) Draw the structural formula of aspartic acid, \(\text{H}_2\text{N-CH(CH}_2\text{COOH)-COOH}\), in:
(i) its zwitterionic form at its isoelectric point.
(ii) strongly acidic conditions (\(\text{pH} = 1\)).
(iii) strongly alkaline conditions (\(\text{pH} = 13\)).

(b) A mixture containing aspartic acid (isoelectric point = 2.8), alanine (isoelectric point = 6.0), and lysine (isoelectric point = 9.7) is subjected to electrophoresis in a buffer solution maintained at \(\text{pH} = 6.0\).
(i) State which amino acid(s) will migrate towards the anode (+), which will migrate towards the cathode (-), and which will remain stationary at the origin.
(ii) Explain your answer to (b)(i) by describing the net charge on each amino acid at \(\text{pH} = 6.0\).

(c) Draw the structure of the dipeptide formed when the carboxyl group of alanine reacts with the amino group of aspartic acid.
Show answer & marking scheme

Worked solution

(a) (i) Zwitterion of aspartic acid at its isoelectric point:
\(\text{H}_3\text{N}^+\text{-CH(CH}_2\text{COO}^-\text{)-COOH}\) or \(\text{H}_3\text{N}^+\text{-CH(CH}_2\text{COOH)-COO}^-\)
(ii) Under strongly acidic conditions (pH = 1):
\(\text{H}_3\text{N}^+\text{-CH(CH}_2\text{COOH)-COOH}\)
(iii) Under strongly alkaline conditions (pH = 13):
\(\text{H}_2\text{N-CH(CH}_2\text{COO}^-\text{)-COO}^-\)

(b) (i)
- Migrate towards the anode (+): Aspartic acid
- Migrate towards the cathode (-): Lysine
- Remain stationary at the origin: Alanine
(ii)
- At pH 6.0, alanine is at its isoelectric point, so it exists as a neutral zwitterion with no net charge and does not move.
- At pH 6.0, the environment is more basic than the isoelectric point of aspartic acid (2.8), so both carboxyl groups are deprotonated, giving aspartic acid a net negative charge, causing it to migrate to the positive anode.
- At pH 6.0, the environment is more acidic than the isoelectric point of lysine (9.7), so its basic side-chain and amino group are protonated, giving lysine a net positive charge, causing it to migrate to the negative cathode.

(c) Dipeptide structure (alanyl-aspartic acid):
\(\text{H}_2\text{N-CH(CH}_3\text{)-CO-NH-CH(CH}_2\text{COOH)-COOH}\)

Marking scheme

(a)(i) Award 1.5 marks for correct zwitterion structure (net neutral charge).
(a)(ii) Award 1.5 marks for correct structure with both carboxyl groups protonated and amine protonated (+1 charge).
(a)(iii) Award 1.5 marks for correct structure with both carboxyl groups deprotonated and amine neutral (-2 charge).

(b)(i) Award 3 marks:
- Aspartic acid to anode (+) [1 mark]
- Lysine to cathode (-) [1 mark]
- Alanine stationary [1 mark]

(b)(ii) Award 3 marks:
- Alanine is at its isoelectric point, net charge is zero [1 mark]
- Aspartic acid has a net negative charge (deprotonated) because pH 6.0 > pI 2.8 [1 mark]
- Lysine has a net positive charge (protonated) because pH 6.0 < pI 9.7 [1 mark]

(c) Award 2 marks:
- Correct peptide linkage (-CO-NH-) shown [1 mark]
- Correct alignment of residue structures (alanine amino end, aspartic acid carboxyl end) [1 mark]
Question 6 · A Level Structured
12.5 marks
Propanone undergoes nucleophilic addition with hydrogen cyanide.

(a) Propanone reacts with hydrogen cyanide, \(HCN\), in the presence of a catalyst, \(NaCN\), to form 2-hydroxy-2-methylpropanenitrile.
(i) Write the chemical equation for this reaction, showing the structural formulas of both the reactant and the organic product.
(ii) Outline the detailed, step-by-step mechanism for this reaction. Include curly arrows to show the movement of electron pairs, and show any relevant dipoles and lone pairs.
(iii) Explain why \(NaCN\) (or a base like \(NaOH\)) is added to initiate the reaction, rather than using pure \(HCN\) alone.

(b) Describe a chemical test (reagents and observations) that can be used to distinguish between:
(i) Propanal and propanone.
(ii) Propanone and propan-2-ol.
Show answer & marking scheme

Worked solution

(a) (i) Equation:
\(\text{CH}_3\text{COCH}_3 + HCN \rightarrow \text{CH}_3\text{C(OH)(CN)CH}_3\)

(ii) Mechanism:
- **Step 1**: The nucleophile \(:\text{CN}^-\)\ is attracted to the electron-deficient carbonyl carbon. Show the dipole \(\text{C}^{\delta+}=\text{O}^{\delta-}\) on propanone.
- Draw a curly arrow from the lone pair on the carbon of \(:\text{CN}^-\)\ to the carbonyl carbon.
- Draw a curly arrow from the \(\pi\) bond of the \(\text{C}=\text{O}\) double bond to the oxygen atom.
- This forms the intermediate anion: \(\text{(CH}_3)_2\text{C(O}^-\text{)(CN)}\).
- **Step 2**: The intermediate anion is protonated.
- Draw a curly arrow from the lone pair on the negative oxygen atom (\(\text{O}^-\)) to the hydrogen of \(\text{H-CN}\) (or \(\text{H}^+\)).
- Draw a curly arrow breaking the \(\text{H-CN}\) bond, regenerating the catalyst \(\text{CN}^-\).
- The final product is 2-hydroxy-2-methylpropanenitrile.

(iii) \(HCN\) is a very weak acid and partially dissociates, meaning the concentration of the nucleophile \(\text{CN}^-\)\ in pure \(HCN\) is too low for the reaction to proceed. Adding \(NaCN\) (or \(NaOH\) which reacts with \(HCN\)) provides a high concentration of \(\text{CN}^-\)\ to initiate the nucleophilic attack.

(b) (i) Tollens' reagent (or Fehling's solution):
- **Propanal**: Silver mirror formed (or red precipitate with Fehling's).
- **Propanone**: No visible change.
(ii) Acidified potassium dichromate(VI) (\(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+\)):
- **Propan-2-ol**: Orange solution turns green.
- **Propanone**: No visible change (remains orange).

Marking scheme

(a)(i) Award 2 marks:
- Correct structures of propanone and 2-hydroxy-2-methylpropanenitrile [2 marks]

(a)(ii) Award 5 marks total:
- Correct dipole \(\text{C}^{\delta+}=\text{O}^{\delta-}\) and curly arrow from \(\text{CN}^-\)\ to carbonyl carbon [1.5 marks]
- Curly arrow from C=O bond to O [1 mark]
- Structure of intermediate alkoxide ion with correct charge on oxygen [1 mark]
- Curly arrow from O- to proton donor (HCN or H+) [1 mark]
- Regeneration of CN- showing bond cleavage [0.5 marks]

(a)(iii) Award 2 marks:
- HCN is a weak acid and dissociates poorly (very low concentration of CN-) [1 mark]
- NaCN/NaOH provides/generates a high concentration of the CN- nucleophile [1 mark]

(b)(i) Award 2 marks:
- Tollens' / Fehling's reagent [1 mark]
- Positive result for aldehyde / negative result for ketone [1 mark]

(b)(ii) Award 1.5 marks:
- Acidified potassium dichromate(VI) [0.75 marks]
- Orange to green for propan-2-ol, no change for propanone [0.75 marks]
Question 7 · A Level Structured
12.5 marks
Electrode potentials provide quantitative predictions regarding redox reactions, particularly for transition metals.

(a) An electrochemical cell is set up using the standard \(Fe^{3+}/Fe^{2+}\) half-cell and the standard \(Cr^{3+}/Cr^{2+}\) half-cell.
Given standard electrode potentials:
\[Fe^{3+}(aq) + e^- \rightleftharpoons Fe^{2+}(aq) \quad E^\ominus = +0.77\ \text{V}\]
\[Cr^{3+}(aq) + e^- \rightleftharpoons Cr^{2+}(aq) \quad E^\ominus = -0.41\ \text{V}\]
(i) Write the overall ionic equation for the spontaneous cell reaction.
(ii) Calculate the standard cell potential, \(E^\ominus_{cell}\).
(iii) Deduce which species is the strongest reducing agent under standard conditions.

(b) The concentration of ions in the chromium half-cell is changed.
The Nernst equation is given by:
\[E = E^\ominus + \frac{0.059}{z} \log_{10} \frac{[\text{oxidised species}]}{[\text{reduced species}]}\]
Calculate the electrode potential, \(E\), of the chromium half-cell at \(298\ \text{K}\) when \([Cr^{3+}(aq)] = 0.015\ \text{mol\ dm}^{-3}\) and \([Cr^{2+}(aq)] = 0.35\ \text{mol\ dm}^{-3}\).

(c) Explain, with reference to the \(d\)-orbital splitting, why transition metal complexes are typically coloured.
Show answer & marking scheme

Worked solution

(a) (i) Overall spontaneous reaction:
\[Fe^{3+}(aq) + Cr^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cr^{3+}(aq)\]
(ii) \(E^\ominus_{cell} = E^\ominus_{cathode} - E^\ominus_{anode} = +0.77 - (-0.41) = +1.18\ \text{V}\)
(iii) Strongest reducing agent is \(Cr^{2+}\) because it is the reduced species of the half-cell with the most negative electrode potential.

(b) For the half-cell reaction \(Cr^{3+}(aq) + e^- \rightleftharpoons Cr^{2+}(aq)\):
- \(z = 1\)
- Oxidised species = \(Cr^{3+}(aq)\)
- Reduced species = \(Cr^{2+}(aq)\)
\[E = -0.41 + \frac{0.059}{1} \log_{10} \left(\frac{0.015}{0.35}\right)\]
\[E = -0.41 + 0.059 \log_{10}(0.042857)\]
\[E = -0.41 + 0.059 \times (-1.368)\]
\[E = -0.41 - 0.0807 = -0.491\ \text{V}\ \text{(or } -0.49\ \text{V)}\]

(c)
- In a transition metal complex, the approach of surrounding ligands causes the degenerate \(d\)-orbitals of the transition metal ion to split into two sets of non-degenerate energy levels (lower and higher energy).
- When white light passes through, an electron in a lower energy \(d\)-orbital absorbs a specific wavelength of light and is promoted to a higher energy \(d\)-orbital (a \(d-d\) transition).
- The energy of the absorbed photon is given by \(\Delta E = h\nu\).
- The remaining wavelengths of light are transmitted (not absorbed) and are seen as the complementary colour of the complex.

Marking scheme

(a)(i) Award 1.5 marks:
- Correct balanced overall equation, including state symbols [1.5 marks]

(a)(ii) Award 1 mark:
- E_cell = +1.18 V [1 mark]

(a)(iii) Award 1 mark:
- Cr2+ identified [1 mark]

(b) Award 4 marks:
- Correctly identifying z = 1 [1 mark]
- Correctly substituted values into Nernst equation [1.5 marks]
- Value of log term (-1.368) [0.5 marks]
- Final calculated potential of -0.49 V (or -0.491 V) [1 mark]

(c) Award 5 marks:
- Splitting of d-orbitals into lower and higher energy levels by ligands [2 marks]
- Absorption of light promotes an electron from lower to higher level / d-d transition [1 mark]
- Energy absorption relates to specific wavelength / \(\Delta E = h\nu\) [1 mark]
- Color seen is the complementary color transmitted [1 mark]
Question 8 · A Level Structured
12.5 marks
Polymers can be synthesized by different mechanisms, yielding distinct physical properties and environmental lifetimes.

(a) Kevlar is a highly strong synthetic polyamide. It is synthesized from benzene-1,4-dicarboxylic acid and benzene-1,4-diamine.
(i) Draw the repeat unit of Kevlar, clearly showing the amide linkage.
(ii) Explain, in terms of intermolecular forces, why Kevlar has such high mechanical strength and thermal stability.

(b) Poly(lactic acid), PLA, is a biodegradable polyester made from 2-hydroxypropanoic acid (lactic acid), \(\text{CH}_3\text{CH(OH)COOH}\).
(i) Draw the repeat unit of PLA.
(ii) Explain why PLA is biodegradable whereas addition polymers like poly(ethene) are not.
(iii) State the type of reaction that breaks down PLA in the environment.
Show answer & marking scheme

Worked solution

(a) (i) Repeat unit of Kevlar:
\(-[\text{CO}-\text{C}_6\text{H}_4-\text{CO}-\text{NH}-\text{C}_6\text{H}_4-\text{NH}]-\)
(where \(-\text{C}_6\text{H}_4-\) represents the 1,4-substituted benzene ring).
(ii) Kevlar chains are planar and contain polar amide groups (\(\text{C}=\text{O}\) and \(\text{N}-\text{H}\)). This allows extensive hydrogen bonding to form between adjacent polymer chains. Additionally, the flat aromatic rings can align closely to form strong van der Waals forces (\(\pi-\pi\) stacking) between parallel sheets, making the material highly crystalline and strong.

(b) (i) Repeat unit of PLA:
\(-[\text{O}-\text{CH(CH}_3)-\text{CO}]-\)
(ii) PLA is a condensation polymer containing polar ester linkages (\(\text{C}-\text{O}\) and \(\text{C}=\text{O}\)). These polar bonds are susceptible to chemical attack and can be cleaved by water (or micro-organisms/enzymes) in the environment. In contrast, poly(ethene) is an addition polymer containing only non-polar \(\text{C}-\text{C}\) and \(\text{C}-\text{H}\) bonds, which are highly inert and resistant to attack.
(iii) Hydrolysis

Marking scheme

(a)(i) Award 2.5 marks:
- Correct structure of the amide link (-CO-NH-) [1 mark]
- Correct 1,4-linkage on benzene rings [1 mark]
- Extension bonds on both ends of repeat unit [0.5 marks]

(a)(ii) Award 3 marks:
- Presence of hydrogen bonding between N-H and C=O of adjacent chains [1.5 marks]
- Flat aromatic rings permit close packing / strong van der Waals forces [1.5 marks]

(b)(i) Award 2 marks:
- Correct repeat unit of PLA with extension bonds [2 marks]

(b)(ii) Award 3 marks:
- PLA has polar ester links which are open to attack [1.5 marks]
- Poly(ethene) contains only non-polar C-C/C-H bonds and is inert [1.5 marks]

(b)(iii) Award 2 marks:
- Hydrolysis [2 marks]

Paper 53 (Planning, Analysis and Evaluation)

Answer all questions. Show your working, plot graphs with a sharp pencil, and use appropriate coordinates.
3 Question · 30 marks
Question 1 · Analytical & Planning Questions
10 marks
A student plans to determine the partition coefficient, \(K_{pc}\), of butanoic acid between water and trichloromethane at \(298\text{ K}\). In each of four experiments, a known mass of butanoic acid was dissolved in \(50.0\text{ cm}^3\) of water. This solution was shaken thoroughly with \(50.0\text{ cm}^3\) of trichloromethane in a separating funnel and allowed to reach equilibrium. A \(10.0\text{ cm}^3\) sample of the aqueous layer was then titrated against \(0.100\text{ mol dm}^{-3}\) aqueous sodium hydroxide.

Experimental Data:
- Experiment 1: Mass of butanoic acid initially dissolved = \(1.20\text{ g}\); Volume of \(0.100\text{ mol dm}^{-3}\) \(NaOH\) required = \(9.10\text{ cm}^3\).
- Experiment 2: Mass of butanoic acid initially dissolved = \(1.80\text{ g}\); Volume of \(0.100\text{ mol dm}^{-3}\) \(NaOH\) required = \(13.60\text{ cm}^3\).

[Molar mass of butanoic acid = \(88.1\text{ g mol}^{-1}\)]

(a) For Experiment 1, calculate:
(i) the concentration of butanoic acid remaining in the aqueous layer, \([\text{butanoic acid}]_{\text{aq}}\), in \\text{mol dm}^{-3}\\.
(ii) the concentration of butanoic acid extracted into the trichloromethane layer, \([\text{butanoic acid}]_{\text{org}}\), in \\text{mol dm}^{-3}\\.

(b) Calculate the partition coefficient, \(K_{pc} = \frac{[\text{butanoic acid}]_{\text{org}}}{[\text{butanoic acid}]_{\text{aq}}}\), for Experiment 1 and Experiment 2. Show that these values are consistent.

(c) Using an average \(K_{pc}\) value of \(2.00\), calculate the mass of butanoic acid that would remain in the aqueous layer if \(1.20\text{ g}\) of butanoic acid dissolved in \(50.0\text{ cm}^3\) of water was equilibrated with \(100\text{ cm}^3\) of trichloromethane.

(d) State and explain how the value of \(K_{pc}\) would change if the experiment was conducted at a higher temperature, given that the transfer of butanoic acid from water to trichloromethane is an exothermic process.
Show answer & marking scheme

Worked solution

(a)(i) Moles of \(NaOH\) used = \(9.10 \times 10^{-3} \times 0.100 = 9.10 \times 10^{-4}\text{ mol}\). Since 1 mole of butanoic acid reacts with 1 mole of \(NaOH\), the moles of butanoic acid in \(10.0\text{ cm}^3\) = \(9.10 \times 10^{-4}\text{ mol}\). Thus, \([\text{butanoic acid}]_{\text{aq}} = \frac{9.10 \times 10^{-4}}{0.0100} = 0.0910\text{ mol dm}^{-3}\).

(a)(ii) Total initial moles of butanoic acid = \\frac{1.20}{88.1} = 0.01362\text{ mol}\). Moles in the \(50.0\text{ cm}^3\) aqueous layer = \(0.0910 \times 0.0500 = 4.55 \times 10^{-3}\text{ mol}\). Moles in organic layer = \(0.01362 - 0.00455 = 9.07 \times 10^{-3}\text{ mol}\). Therefore, \([\text{butanoic acid}]_{\text{org}} = \frac{9.07 \times 10^{-3}}{0.0500} = 0.1814\text{ mol dm}^{-3}\).

(b) For Experiment 1, \(K_{pc} = \frac{0.1814}{0.0910} = 1.99\). For Experiment 2, initial moles = \\frac{1.80}{88.1} = 0.02043\text{ mol}\). \([\text{butanoic acid}]_{\text{aq}} = \frac{13.60 \times 10^{-4}}{0.0100} = 0.1360\text{ mol dm}^{-3}\). Moles in aqueous layer = \(0.1360 \times 0.0500 = 6.80 \times 10^{-3}\text{ mol}\). Moles in organic layer = \(0.02043 - 0.00680 = 0.01363\text{ mol}\). \([\text{butanoic acid}]_{\text{org}} = \frac{0.01363}{0.0500} = 0.2726\text{ mol dm}^{-3}\). \(K_{pc} = \frac{0.2726}{0.1360} = 2.00\). Both values are very close (1.99 and 2.00), demonstrating consistency.

(c) Let \(m\) be the mass of butanoic acid remaining in the \(50.0\text{ cm}^3\) aqueous layer. Mass in organic layer = \(1.20 - m\). \(K_{pc} = 2.00 = \frac{(1.20 - m)/100}{m/50.0}\). \(2.00 = \frac{1.20 - m}{2m} \Rightarrow 4.00m = 1.20 - m \Rightarrow 5.00m = 1.20 \Rightarrow m = 0.24\text{ g}\).

(d) Since the transfer process is exothermic, increasing the temperature shifts the equilibrium in the endothermic direction (backwards, favouring the aqueous layer). The concentration in the organic layer will decrease relative to the aqueous layer, so \(K_{pc}\) will decrease.

Marking scheme

Part (a): [3 marks total]
- Award 1 mark for correct calculation of aqueous concentration: 0.0910 mol dm^-3.
- Award 1 mark for calculating total initial moles (0.01362 mol) and moles remaining in the aqueous layer (0.00455 mol).
- Award 1 mark for correct organic concentration calculation: 0.1814 mol dm^-3 (allow ecf).

Part (b): [2 marks total]
- Award 1 mark for calculating K_pc for Experiment 1 (1.99) and Experiment 2 (2.00).
- Award 1 mark for stating that the values are very close / consistent.

Part (c): [3 marks total]
- Award 1 mark for setting up the correct algebraic expression incorporating volumes: K_pc = [(1.20 - m)/100] / [m/50.0].
- Award 1 mark for rearranging the equation to 5m = 1.20.
- Award 1 mark for the final mass value: 0.24 g.

Part (d): [2 marks total]
- Award 1 mark for identifying that K_pc decreases.
- Award 1 mark for explaining that increasing temperature shifts the exothermic process to the left / favouring the aqueous phase (Le Chatelier's principle).
Question 2 · Analytical & Planning Questions
10 marks
The kinetics of the reaction between peroxodisulfate ions and iodide ions was investigated using an iodine clock reaction:

\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq})

In each experiment, the reaction was timed at \(298\text{ K}\) until a blue-black colour suddenly appeared. This occurs when all the added sodium thiosulfate, \text{Na}_2\text{S}_2\text{O}_3, has reacted with the iodine produced.

Experimental Data (total volume of mixture in each run is exactly \(100.0\text{ cm}^3\), containing \(10.0\text{ cm}^3\) of \(0.0100\text{ mol dm}^{-3}\) \text{Na}_2\text{S}_2\text{O}_3 and starch indicator):

- Experiment 1: \(20.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) \text{K}_2\text{S}_2\text{O}_8, \(20.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) \text{KI}; time, \(t = 45.0\text{ s}\).
- Experiment 2: \(10.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) \text{K}_2\text{S}_2\text{O}_8, \(20.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) \text{KI}; time, \(t = 90.0\text{ s}\).
- Experiment 3: \(20.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) \text{K}_2\text{S}_2\text{O}_8, \(10.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) \text{KI}; time, \(t = 90.0\text{ s}\).

(a) Explain the chemical basis of this clock reaction, explaining why the blue-black colour only appears after a specific time interval.
(b) Deduce the order of reaction with respect to both \text{S}_2\text{O}_8^{2-} and \text{I}^-, showing your reasoning.
(c) For Experiment 1, calculate:
(i) the initial concentration of \text{S}_2\text{O}_8^{2-}(\text{aq}) in the mixture.
(ii) the initial rate of reaction, expressed as the rate of formation of \text{I}_2 in \\text{mol dm}^{-3}\\text{ s}^{-1}.
(d) Write the rate equation for the reaction, and calculate the value of the rate constant, \(k\), for Experiment 1, including its units.
Show answer & marking scheme

Worked solution

(a) Iodine produced in the main reaction reacts instantly with the added sodium thiosulfate: \(2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^-\). No free iodine is available to react with starch. Once all thiosulfate ions are completely consumed, any further iodine produced remains in the mixture and reacts with starch to form the dark blue-black complex.

(b) Since total volume is constant, reactant volumes are proportional to their initial concentrations. Rate \\propto 1/t.
- Comparing Expt 1 and Expt 2: \([\text{I}^-]\) is constant, \([\text{S}_2\text{O}_8^{2-}]\) is halved, and the time doubles (rate is halved). Thus, the order with respect to \([\text{S}_2\text{O}_8^{2-}]\) is 1.
- Comparing Expt 1 and Expt 3: \([\text{S}_2\text{O}_8^{2-}]\) is constant, \([\text{I}^-]\) is halved, and the time doubles (rate is halved). Thus, the order with respect to \([\text{I}^-]\) is 1.

(c)(i) \([\text{S}_2\text{O}_8^{2-}]_{\text{initial}} = \frac{20.0}{100.0} \times 0.100 = 0.0200\text{ mol dm}^{-3}\).

(c)(ii) Moles of \text{S}_2\text{O}_3^{2-} used = \(10.0 \times 10^{-3} \times 0.0100 = 1.00 \times 10^{-4}\text{ mol}\). Moles of \text{I}_2 produced to consume all thiosulfate = \(5.00 \times 10^{-5}\text{ mol}\). Concentration change of \text{I}_2, \\Delta[\text{I}_2] = \\frac{5.00 \\times 10^{-5}}{0.100} = 5.00 \\times 10^{-4}\text{ mol dm}^{-3}\). Rate = \\frac{\\Delta[\text{I}_2]}{\\Delta t} = \\frac{5.00 \\times 10^{-4}}{45.0} = 1.11 \\times 10^{-5}\text{ mol dm}^{-3}\\text{ s}^{-1}\).

(d) Rate equation: \text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\). Initial concentration of \text{I}^- = \\frac{20.0}{100.0} \\times 0.200 = 0.0400\text{ mol dm}^{-3}\). \(1.11 \times 10^{-5} = k (0.0200)(0.0400) \Rightarrow k = 0.0139\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

Marking scheme

Part (a): [2 marks total]
- Award 1 mark for stating that thiosulfate reacts with iodine as it forms, preventing it from reacting with starch.
- Award 1 mark for stating that when thiosulfate is completely exhausted, free iodine forms a blue-black complex with starch.

Part (b): [3 marks total]
- Award 1 mark for stating rate is proportional to 1/t.
- Award 1 mark for comparing Expt 1 & 2 to deduce first-order with respect to S2O8^2-.
- Award 1 mark for comparing Expt 1 & 3 to deduce first-order with respect to I^-.

Part (c): [3 marks total]
- Award 1 mark for (c)(i): 0.0200 mol dm^-3.
- Award 1 mark for calculating correct concentration of iodine produced: 5.00 x 10^-4 mol dm^-3.
- Award 1 mark for (c)(ii) rate calculation: 1.11 x 10^-5 mol dm^-3 s^-1.

Part (d): [2 marks total]
- Award 1 mark for rate equation and correct substitutions: 1.11 x 10^-5 = k (0.0200)(0.0400).
- Award 1 mark for correct value of k (0.0139) and correct units (dm3 mol-1 s-1).
Question 3 · Analytical & Planning Questions
10 marks
The formula of the purple complex formed between iron(III) ions and salicylate ligands is investigated using Job's method of continuous variation. In a series of 9 mixtures, the total volume of \(0.00200\text{ mol dm}^{-3}\) \text{Fe}^{3+}(\text{aq}) and \(0.00200\text{ mol dm}^{-3}\) salicylate, \text{Sal}^-(\text{aq}), was kept constant at \(10.0\text{ cm}^3\). The absorbance of each mixture was measured at \(525\text{ nm}\).

Data:
- Mixture 2: \(V_{\text{sal}} = 2.0\text{ cm}^3\), Absorbance = \(0.30\)
- Mixture 4: \(V_{\text{sal}} = 4.0\text{ cm}^3\), Absorbance = \(0.60\)
- Mixture 5: \(V_{\text{sal}} = 5.0\text{ cm}^3\), Absorbance = \(0.68\)
- Mixture 6: \(V_{\text{sal}} = 6.0\text{ cm}^3\), Absorbance = \(0.60\)
- Mixture 8: \(V_{\text{sal}} = 8.0\text{ cm}^3\), Absorbance = \(0.30\)

(a) Calculate the mole fraction of salicylate, \(x_{\text{sal}}\), for Mixtures 2, 4, 6, and 8.
(b) Explain how the formula of the complex can be determined by plotting absorbance against \(x_{\text{sal}}\), drawing two best-fit intersecting straight lines, and finding the stoichiometric intersection point. State the value of \(x_{\text{sal}}\) at this intersection.
(c) Use your answer to (b) to deduce the formula of the complex, including its overall charge (assume salicylate has a charge of -1 and iron is in the +3 oxidation state).
(d) Explain why the absorbance measured for Mixture 5 (the stoichiometric point) is lower than the value predicted by the intersection of the two straight lines.
(e) Suggest why a wavelength of \(525\text{ nm}\) (green light) was selected for this colorimetric measurement.
Show answer & marking scheme

Worked solution

(a) Since the total volume is always \(10.0\text{ cm}^3\), \(x_{\text{sal}} = \frac{V_{\text{sal}}}{10.0}\).
- Mixture 2: \(x_{\text{sal}} = 0.20\)
- Mixture 4: \(x_{\text{sal}} = 0.40\)
- Mixture 6: \(x_{\text{sal}} = 0.60\)
- Mixture 8: \(x_{\text{sal}} = 0.80\)

(b) Before the stoichiometric point, salicylate is limiting, so concentration of the complex (and hence absorbance) increases linearly with salicylate volume. Beyond the stoichiometric point, iron(III) is limiting, so absorbance decreases linearly. Intersecting the two straight lines eliminates the effects of equilibrium dissociation. The intersection occurs at \(x_{\text{sal}} = 0.50\).

(c) At \(x_{\text{sal}} = 0.50\), the ratio of \text{Fe}^{3+} : \text{Sal}^- is 1 : 1. The formula of the complex is therefore \([\text{Fe(Sal)}]^{2+}\).

(d) The formation of the complex is a reversible equilibrium reaction: \(\text{Fe}^{3+}(\text{aq}) + \text{Sal}^-(\text{aq}) \rightleftharpoons [\text{Fe(Sal)}]^{2+}(\text{aq})\). At the stoichiometric ratio, the reaction does not go to 100% completion; some complex dissociates back into free ions, lowering its concentration and resulting in a lower absorbance.

(e) A purple complex transmits purple (blue/red) light and absorbs its complementary color, green. Measuring at \(525\text{ nm}\) (green light) ensures maximum light absorbance, yielding the greatest experimental sensitivity and adherence to the Beer-Lambert law.

Marking scheme

Part (a): [2 marks total]
- Award 2 marks for all four mole fractions calculated correctly (0.20, 0.40, 0.60, 0.80). Award 1 mark if 2 or 3 are correct.

Part (b): [2 marks total]
- Award 1 mark for explaining that the straight lines represent the limiting reactant stoichiometry before and after the equivalence point.
- Award 1 mark for identifying the intersection at x_sal = 0.50.

Part (c): [2 marks total]
- Award 1 mark for deducing the 1:1 ratio of Fe3+ to salicylate.
- Award 1 mark for the correct formula including charge: [Fe(Sal)]^2+.

Part (d): [2 marks total]
- Award 1 mark for identifying that the reaction is an equilibrium/reversible process.
- Award 1 mark for explaining that dissociation of the complex lowers its equilibrium concentration below the theoretical maximum.

Part (e): [2 marks total]
- Award 1 mark for explaining that purple substances absorb green light (complementary color).
- Award 1 mark for stating that this wavelength provides the maximum absorbance/sensitivity.

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