HL readiness
IB Chemistry HL Readiness Check: Four Skills to Test Before Choosing
Four original mini-tasks and a pattern-based review of chemical explanation, quantitative setup, data reasoning and learning habits.
- Syllabus
- IB Chemistry · first assessment 2025
- Relevant to
- SL · HL
In this guide
This is not an entrance examination, and it does not automatically declare someone “an HL student” or “not an HL student”. Choosing Chemistry HL also depends on university routes, the other two HL subjects and the workload of the complete Diploma. These four tasks answer a narrower question: which foundations already work independently, and which require targeted repair?
Work on paper without notes or internet access. A calculator is allowed. Give each task no more than 8–10 minutes, then assess the reasoning as well as the final result.
Use a calm, ordinary day rather than immediately after a major test or late-night study. Record start and finish time and the point at which help would first have been requested. This boundary of independence is useful evidence beside the final answer. If a concept has not yet appeared in the student’s prior curriculum, mark it “not yet taught” rather than as a knowledge failure, then test the same skill with a more familiar example.
What this check can—and cannot—decide
The result is a snapshot. It can show whether the student:
- connects a particle-level model to an observable property;
- can set up an unfamiliar calculation without copying a memorised template;
- reasons from data instead of merely recalling course content;
- corrects an error and later retries independently.
It cannot by itself show which universities require HL, how three HL subjects interact, or how much progress is possible before the course begins. For those decisions, use the SL/HL guide that works backwards from university requirements.
Four readiness domains
Readiness is a pattern across four domains
Connect structure, particles and observed behaviour.
Use ratios, algebra, units and multi-step relationships.
Interpret unfamiliar tables, graphs and uncertainty.
Practise consistently, correct errors and retry independently.
Chemical explanation: the number of technical terms matters less than whether the response builds a clear causal chain.
Quantitative setup: can the student turn data into quantities, select a ratio, handle units and check an intermediate result?
Data reasoning: can the student separate observation from explanation and notice when the evidence is weaker than the apparent difference?
Learning habits: does the method change after an error? Across two years of HL, this matters at least as much as one early test result.
Four original mini-tasks
1. Chemical explanation
8–10 minutes
At approximately atmospheric pressure, CH3OH boils at about 64.7 °C and CH3SH at about 6.0 °C. These small molecules have molar masses of the same order, although they are not identical.
Explain the difference by connecting structure, intermolecular interactions and the energy needed to enter the gas phase.
Show worked answer and self-review
Both substances have London dispersion forces. Methanol's strongly polar O–H bond and oxygen lone pairs allow intermolecular hydrogen bonding; methanethiol does not form an equivalently strong hydrogen-bonded network. More energy is needed to overcome the overall attractions in liquid methanol, giving it the higher boiling point. Boiling does not break the covalent O–H bonds within its molecules.
Self-review: did you identify the structural cause, compare both substances and explain the energy consequence? ‘More polar’ alone is incomplete.
2. Quantitative setup
8–10 minutes
25.00 cm3 of H2SO4 of unknown concentration is titrated with 0.1200 mol dm−3 NaOH. Complete neutralisation at the equivalence point requires 18.60 cm3 NaOH. Determine the acid concentration, showing units and the stoichiometric ratio.
Show worked answer and self-review
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
- V(NaOH) = 0.01860 dm3; V(H2SO4) = 0.02500 dm3.
- n(NaOH) = 0.1200 × 0.01860 = 0.002232 mol.
- n(H2SO4) = 0.002232 / 2 = 0.001116 mol.
- c(H2SO4) = 0.001116 / 0.02500 = 0.04464 mol dm−3.
A 1:1 ratio would give 0.08928 mol dm−3, twice the correct result: complete neutralisation uses two moles of NaOH per mole of acid. Check the cm3-to-dm3 conversion, the 2:1 ratio for complete neutralisation and division by the acid volume. Four significant figures match the given data.
3. Data reasoning
8–10 minutes
Original illustrative data: each temperature has three repeats, summarised by the mean and half-range, (maximum − minimum)/2. Half-range is not standard deviation, standard error or a confidence interval.
| °C | Mean / mmol s−1 | Half-range / mmol s−1 |
|---|---|---|
| 20 | 1.20 | 0.10 |
| 25 | 1.28 | 0.12 |
| 35 | 1.86 | 0.11 |
Evaluate the evidence for an increase from 20 to 25 °C, then compare it with the change from 25 to 35 °C.
Show worked answer and self-review
The first mean difference is 0.08 mmol s−1, small relative to half-ranges of 0.10–0.12 mmol s−1; the repeat ranges overlap. The second difference is 0.58 mmol s−1, much larger than half-ranges of 0.12 and 0.11, giving a clearer pattern under the investigated conditions. Three repeats and half-ranges do not establish statistical significance or a law for all temperatures.
‘Temperature increases rate’ does not evaluate the strength of this evidence.
4. Learning habits — reflection
8–10 minutes, followed by later observation
This task examines your correction routine, rather than Chemistry knowledge. Choose a previously missed question and record:
- the first failed step;
- the smallest specific repair;
- a date to retry without help;
- a fresh similar task to test transfer.
Show worked answer and self-review
For example: ‘I ignored the acid's two protons. I will write and justify the complete neutralisation equation, then solve a reaction with a different ratio. In three days I will check method selection without notes.’ This is more specific than ‘be more careful’.
A convincing written plan is not evidence of an established habit. Observe for two weeks whether correction and delayed retries actually happen.
Answer key — separate from the task sheet
1. Chemical explanation
Both substances have London dispersion forces. Methanol's strongly polar O–H bond and oxygen lone pairs allow intermolecular hydrogen bonding; methanethiol does not form an equivalently strong hydrogen-bonded network. More energy is needed to overcome the overall attractions in liquid methanol, giving it the higher boiling point. Boiling does not break the covalent O–H bonds within its molecules.
Self-review: did you identify the structural cause, compare both substances and explain the energy consequence? ‘More polar’ alone is incomplete.
2. Quantitative setup
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
- V(NaOH) = 0.01860 dm3; V(H2SO4) = 0.02500 dm3.
- n(NaOH) = 0.1200 × 0.01860 = 0.002232 mol.
- n(H2SO4) = 0.002232 / 2 = 0.001116 mol.
- c(H2SO4) = 0.001116 / 0.02500 = 0.04464 mol dm−3.
A 1:1 ratio would give 0.08928 mol dm−3, twice the correct result: complete neutralisation uses two moles of NaOH per mole of acid. Check the cm3-to-dm3 conversion, the 2:1 ratio for complete neutralisation and division by the acid volume. Four significant figures match the given data.
3. Data reasoning
The first mean difference is 0.08 mmol s−1, small relative to half-ranges of 0.10–0.12 mmol s−1; the repeat ranges overlap. The second difference is 0.58 mmol s−1, much larger than half-ranges of 0.12 and 0.11, giving a clearer pattern under the investigated conditions. Three repeats and half-ranges do not establish statistical significance or a law for all temperatures.
‘Temperature increases rate’ does not evaluate the strength of this evidence.
4. Learning habits — reflection
For example: ‘I ignored the acid's two protons. I will write and justify the complete neutralisation equation, then solve a reaction with a different ratio. In three days I will check method selection without notes.’ This is more specific than ‘be more careful’.
A convincing written plan is not evidence of an established habit. Observe for two weeks whether correction and delayed retries actually happen.
Rate the pattern
Choose one state for each domain.
Scroll sideways if the table does not fit.
| State | Meaning | Evidence |
|---|---|---|
| Secure | Independent, correct and clearly communicated; checking is visible. | It also works on a fresh similar problem. |
| Developing | The central idea is present, but one or two steps need a prompt or correction. | Targeted practice produces quick improvement. |
| Not yet reliable | The opening step, model or method does not form without help. | The same failure returns in a new context. |
Do not calculate a total score. A secure explanation does not cancel a not-yet-reliable quantitative setup. The pattern shows where time should go first.
Keep the working when rating it. Secure does not mean instantly flawless; a small independent check is compatible with secure work. In a developing response, a brief prompt allows the student to continue their own route. Not yet reliable means the prompt would need to supply the model, opening step or central method. The three states therefore describe the amount of support required as well as correctness.
A parent or teacher observing should not give hints during the task. Instead, note whether the student returns to check a unit, recognises an unreasonable result and can name the source of an impasse. These observations add useful context; they are not validated predictions.
What the pattern means
Rate each area separately as secure, developing or not yet reliable. Do not decide HL suitability by counting secure labels. This is an original pedagogical self-review, not a standardised or psychometrically validated assessment.
One convincing account of study habits is not enough. Observe actual corrections and delayed retries over the following fortnight. University requirements and the Diploma workload need separate evidence; use the SL/HL decision guide to weigh those factors.
Two-week repair and retest
Choose no more than two developing or not-yet-reliable domains.
Days 1–3: relearn the missing connection and complete short direct tasks with immediate feedback.
Days 4–7: mix the same skill among different topics. Do not repeat only the original form.
Days 8–11: reduce support and record the first uncertain step.
Days 12–14: complete a new mini-task of comparable difficulty under a time limit. Rate the domain again using the same three states.
For the level decision, the most useful evidence is not whether the first check was perfect. It is whether targeted feedback produced independent, transferable improvement.
The retest should not merely replace the numbers. Preserve the skill while changing the molecule, data format or quantitative context. If the student has memorised only the original solution steps, the new setting will expose that quickly. For the final level decision, place the new pattern beside the university matrix and an actual weekly time log.
A ready-made IB retest covering these four domains is not currently published. The atomic and isotope resources below cover only part of the foundations; the SL bank questions are not an HL-placement test. Select fresh intermolecular, titration and data tasks from your school resources.
Strengthen the skills behind your result
Browse the current IB bank questions in English. The published SL atomic questions are a starting point, not a complete HL retest.
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