Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
76 original questions across 15 selected topic areas. No account or payment required to practise.
5 questions in this topic. Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.
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Original English concept practice addressing selected identified gaps. Not complete syllabus coverage or practical competence.
Selected subtopics: Group 2 solution energetics · Born-Haber cycles · Nernst and Gibbs relations · Weak acids and Ksp · Partitioning and chromatography
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Question 1For a hypothetical Group 2 sulfate, lattice formation enthalpy is -2500 kJ mol^-1 and total gaseous-ion hydration enthalpy is -2480 kJ mol^-1. Calculate enthalpy of solution. Explain why sulfate solubility generally falls down Group 2, and why enthalpy alone cannot determine solubility at every temperature.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
For a hypothetical Group 2 sulfate, lattice formation enthalpy is -2500 kJ mol^-1 and total gaseous-ion hydration enthalpy is -2480 kJ mol^-1. Calculate enthalpy of solution. Explain why sulfate solubility generally falls down Group 2, and why enthalpy alone cannot determine solubility at every temperature.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Verified lattice-formation sign convention and +20 kJ/mol; hydration/lattice comparison specifically addresses the A-level Group-2 sulfate outcome. This is conceptual placement, not completion of the entire referenced subsection.
Reverse the stated formation lattice process to obtain separated gas ions, then hydrate them: this Hess path gives +20 kJ mol^-1.
The trend argument compares changes in two energy contributions, not merely ionic mass. It is qualitative: equilibrium solubility depends on the full free-energy change, not on one supplied enthalpy number.
Self-marking points
Dissociating the lattice needs +2500 kJ mol^-1.
deltaH_solution=2500-2480=+20 kJ mol^-1.
Down the group, hydration becomes less exothermic as the cation grows; the large sulfate ion makes the lattice-energy change relatively smaller, tending to reduce solubility.
Entropy and temperature also enter Gibbs energy, so an endothermic solution enthalpy alone is not an absolute solubility verdict.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 2For a hypothetical MX(s), use formation enthalpy -400, metal atomisation +100, metal first ionisation +500, half X-X dissociation +120 and first electron affinity -350 kJ mol^-1. Use Cambridge's lattice-formation convention for M+(g)+X-(g)->MX(s). Calculate lattice energy and explain the sign of electron affinity versus ionisation in this example.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
For a hypothetical MX(s), use formation enthalpy -400, metal atomisation +100, metal first ionisation +500, half X-X dissociation +120 and first electron affinity -350 kJ mol^-1. Use Cambridge's lattice-formation convention for M+(g)+X-(g)->MX(s). Calculate lattice energy and explain the sign of electron affinity versus ionisation in this example.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked Hess closure, pre-halved halogen term and definition of lattice formation in 23.1; no reversed sign convention silently imported. This is conceptual placement, not completion of the entire referenced subsection.
The atomisation and half-dissociation terms already generate one mole of each gas atom; do not halve 120 again. The affinity term already has its exothermic negative sign.
The direct formation route and indirect gas-ion route have the same endpoints. Subtract +370 from -400, giving -770; lattice dissociation would have the opposite sign.
Self-marking points
Non-lattice steps sum to 100+500+120-350=370 kJ mol^-1.
Hess law gives -400=370+deltaH_lattice.
Lattice formation energy is -770 kJ mol^-1.
Ionisation removes an electron and needs energy; the supplied first electron affinity releases energy when the electron is added.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 3At 298 K, a Cu2+/Cu electrode has Estandard=+0.340 V and [Cu2+]=0.0100 mol dm^-3. Use E=Estandard+(0.0590/2)log10([Cu2+]) with unit activity for the solid. Find E. Separately, a standard two-electron cell has Ecell=1.10 V; use F=96500 C mol^-1 to find standard deltaG and interpret its sign. Do not combine the nonstandard electrode with an unspecified second electrode.3 indicative marks
Calculate and explain3 indicative marksCalculator allowed
At 298 K, a Cu2+/Cu electrode has Estandard=+0.340 V and [Cu2+]=0.0100 mol dm^-3. Use E=Estandard+(0.0590/2)log10([Cu2+]) with unit activity for the solid. Find E. Separately, a standard two-electron cell has Ecell=1.10 V; use F=96500 C mol^-1 to find standard deltaG and interpret its sign. Do not combine the nonstandard electrode with an unspecified second electrode.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Recomputed Nernst logarithm and independent standard-cell free energy; explicit separation prevents mixing incompatible standard/nonstandard data. This is conceptual placement, not completion of the entire referenced subsection.
9701: 24.2: Electrode potentials and Nernst equation
The reduced species is solid Cu, whose activity is treated as one. The supplied dilute concentration model gives a less positive reduction potential on dilution.
The separate standard cell calculation uses two transferred electrons; retain the minus sign in deltaG=-nFE. Electrode potentials are not multiplied by balancing coefficients, although n enters the Gibbs relation.
Self-marking points
log10(0.0100)=-2, so E=0.340-0.0590=+0.281 V.
deltaGstandard=-nFEcell= -2 x 96500 x 1.10 = -212300 J mol^-1.
This is approximately -212 kJ mol^-1 and indicates favourable standard thermodynamic direction, not guaranteed rapid reaction.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 4A weak monoprotic acid has analytical concentration 0.100 mol dm^-3 and Ka=1.00 x 10^-5 mol dm^-3. Estimate [H+] and pH using small dissociation, then check that approximation. Separately AgCl has Ksp=1.80 x 10^-10 mol^2 dm^-6. Estimate [Ag+] in 0.0100 mol dm^-3 chloride, neglecting chloride from dissolution and complexes.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
A weak monoprotic acid has analytical concentration 0.100 mol dm^-3 and Ka=1.00 x 10^-5 mol dm^-3. Estimate [H+] and pH using small dissociation, then check that approximation. Separately AgCl has Ksp=1.80 x 10^-10 mol^2 dm^-6. Estimate [Ag+] in 0.0100 mol dm^-3 chloride, neglecting chloride from dissolution and complexes.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked weak-acid dissociation fraction and exact-root sanity check, and separate 1:1 Ksp dimensional/common-ion calculation; no Kb relation beyond syllabus required. This is conceptual placement, not completion of the entire referenced subsection.
9701: 25.1: Acid, buffer and solubility equilibria
Let x=[H+]=[A-]. Ka=x^2/(0.100-x); neglecting x only in the denominator yields x=0.00100 and pH=3.00. The exact positive root is about 0.000995, which confirms the estimate.
Use the supplied dominant chloride concentration for the separate salt problem: 1.80e-10/0.0100=1.80e-8. This is tiny compared with 0.0100, so dissolution scarcely changes the assumed chloride concentration.
Self-marking points
[H+] is approximately sqrt(Ka c)=1.00 x 10^-3 mol dm^-3; pH is approximately 3.00.
Dissociation is approximately 1.0% of analytical acid, supporting c-x approximately c.
For AgCl, Ksp=[Ag+][Cl-], giving [Ag+]=1.80 x 10^-8 mol dm^-3.
Added chloride suppresses silver-ion solubility by the common-ion effect under the stated model.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 5A neutral solute has partition coefficient Kpc=[solute]organic/[solute]water=4.00. Initially 1.00 mmol is in 20.0 cm^3 water; it equilibrates once with 10.0 cm^3 organic solvent. Find the amount remaining in water. Separately, a TLC spot travels 3.00 cm while solvent travels 5.00 cm, and a gas chromatogram has integrated areas 20 and 80 for two components with equal detector response. Find Rf and percentage composition.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
A neutral solute has partition coefficient Kpc=[solute]organic/[solute]water=4.00. Initially 1.00 mmol is in 20.0 cm^3 water; it equilibrates once with 10.0 cm^3 organic solvent. Find the amount remaining in water. Separately, a TLC spot travels 3.00 cm while solvent travels 5.00 cm, and a gas chromatogram has integrated areas 20 and 80 for two components with equal detector response. Find Rf and percentage composition.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Verified conservation over unequal solvent volumes, coefficient direction, Rf ratio and equal-response area assumption; these are data exercises, not hazardous extraction instructions. This is conceptual placement, not completion of the entire referenced subsection.
Both phases contain the same chemical form of the solute. The equation reduces to 4=2(1-x)/x, hence 6x=2. Equal concentrations are not implied when Kpc differs from one.
Distances for Rf use the same baseline, so length units cancel. For the gas chromatogram divide each area by their sum of 100; response equality is essential to using areas as composition.
Self-marking points
If x mmol remains in water, 4.00=[(1.00-x)/10.0]/[x/20.0].
Solving gives x=1/3 mmol, so 0.667 mmol transfers to organic solvent.
Rf=3.00/5.00=0.600, dimensionless.
Equal-response gas-chromatogram areas give 20% and 80%, not a percentage inferred from retention times.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
