One-to-one tutoring for Pearson Edexcel papers 1H and 2H, built on the board's own Higher-only content.
Book a Free ConsultationHigher-tier Pearson Edexcel GCSE Chemistry (1CH0) is the route examined on papers 1H and 2H, each lasting 1 hour and 45 minutes and worth 100 marks, with questions targeting grades 4 to 9. Leading Tuition provides expert one-to-one tutoring for that tier, built on the specification statements Pearson prints in bold as Higher-only and on three June series of Higher grade boundaries.
Our expert GCSE chemistry tutors help Higher-tier students on the Pearson Edexcel course, building each programme around papers 1H and 2H. Book a short call to talk through your child’s current grade and where marks are being lost.
Book a Free Consultation Message us on WhatsAppPearson states the Higher tier paper will target grades 4–9, and that students must complete all assessments for the qualification in the same tier. The qualification is graded on the total subject mark and individual papers are not graded, so the 100 marks on Paper 1H and the 100 marks on Paper 2H add to a single total out of 200. For a Higher entry the specification adds one rule at the lower edge: if the mark falls a smaller number of marks below the 4/3 grade boundary, a grade 3 may be awarded.
| Paper | Codes | Content assessed | Format |
|---|---|---|---|
| Paper 1 | 1CH0/1H | Topic 1 Key concepts in chemistry; Topic 2 States of matter and mixtures; Topic 3 Chemical changes; Topic 4 Extracting metals and equilibria; Topic 5 Separate chemistry 1 | 1 hour and 45 minutes; 100 marks; ten questions; calculators may be used |
| Paper 2 | 1CH0/2H | Topic 1 Key concepts in chemistry; Topic 6 Groups in the periodic table; Topic 7 Rates of reaction and energy changes; Topic 8 Fuels and Earth science; Topic 9 Separate chemistry 2 | 1 hour and 45 minutes; 100 marks; ten questions; calculators may be used |
Further figures from the specification shape how we plan. First, 27 marks of each paper are overlap questions that appear in both the foundation and higher tier papers, and a Higher entrant therefore meets them on both papers. Second, the board sets the assessment objectives at 20, 20 and 10 marks of each paper for AO1, AO2 and AO3, which is 40%, 40% and 20% of the qualification, give or take 3 points: applying knowledge (AO2) is weighted as heavily as knowing it (AO1).
The specification also fixes the mathematics. In its words, the minimum level of mathematics in the higher tier examination papers will be equivalent to foundation tier GCSE in Mathematics. Shaky algebra and ratio work shows first in the mole calculations, so Higher-tier Edexcel GCSE Maths tutoring can run alongside chemistry. For the whole course, see our Edexcel GCSE Chemistry overview; families still choosing a tier can read the Foundation-tier page as well.
What this changes in our programme. Every Higher student starts with both papers in view, because the grade comes from the 200-mark total. We map their first attempt onto the Higher-only list below.
Pearson’s notation, in the specification’s own words, is that text in bold indicates content that is for higher tier only. We read that marking from the typeface in the PDF itself, statement by statement, and list below only the statements Pearson prints in bold. Where a statement is plain type, it is outside the Higher-only marker, so we say what it covers and what the bold statement beside it adds. A “C” in a statement number means, in the specification’s words, content which is only in the GCSE in Chemistry and is not found in the GCSE in Combined Science.
| Bold statement (board’s wording) | Plain-type statements beside it | What Higher adds |
|---|---|---|
| 0.4 “Write balanced ionic equations” | 0.1 recalls the formulae of elements, simple compounds and ions; 0.3 writes balanced chemical equations, including the state symbols (s), (l), (g) and (aq) | Balanced ionic equations, which also underpin the half equations in 3.27 |
| Bold statement (board’s wording) | Plain-type statement beside it | What Higher adds |
|---|---|---|
| 1.12 “Calculate the relative atomic mass of an element from the relative masses and abundances of its isotopes” | 1.11 explains how isotopes mean some relative atomic masses are not whole numbers | The calculation itself, from relative masses and abundances |
| 1.50 “Recall that one mole of particles of a substance is defined as” the Avogadro constant number of particles (6.02 × 1023 atoms, molecules, formulae or ions) or a mass of ‘relative particle mass’ g | 1.48 calculates masses of reactants and products from balanced equations; 1.49 calculates concentration in g dm–3 | The definition of the mole and the Avogadro constant, which the plain-type calculations sit beside |
| 1.51 “Calculate the number of” moles of particles in a given mass and vice versa; particles in a given number of moles and vice versa; particles in a given mass and vice versa | 1.48 and 1.49 as above | Conversions between mass, moles and numbers of particles, both directions |
| 1.52 “Explain why, in a reaction, the mass of product formed is controlled by the mass of the reactant which is not in excess” | 1.48 gives the mass calculation from one known substance | The reasoning about which reactant limits the product |
| 1.53 “Deduce the stoichiometry of a reaction from the masses of the reactants and products” | 1.48 runs the other way, from an equation to masses | Working backwards from measured masses to the equation |
| Bold statement (board’s wording) | Plain-type statement beside it | What Higher adds |
|---|---|---|
| 3.4 “Recall that the higher the concentration of hydrogen ions in an acidic solution, the lower the pH”, and the higher the hydroxide ion concentration in an alkaline solution, the higher the pH | 3.1 acids are sources of hydrogen ions; 3.2 a neutral solution has a pH of 7 | The link between ion concentration and pH, in both directions |
| 3.5 “Recall that as hydrogen ion concentration in a solution increases by a factor of 10, the pH of the solution decreases by 1” | 3.2 a neutral solution has a pH of 7, with acidic solutions lower and alkaline solutions higher | The factor-of-10 rule |
| 3.7 “Explain the terms dilute and concentrated, with respect to amount of substances in solution” | 3.1 to 3.3 as above | Dilute and concentrated defined by amount of substance |
| 3.8 “Explain the terms weak and strong acids, with respect to the degree of dissociation into ions” | 3.1 as above | Weak and strong explained through dissociation |
| 3.27 “Write half equations for reactions occurring at the anode and cathode in electrolysis” | 3.24 movement of cations and anions; 3.26 predict products from molten binary ionic compounds | Half equations at each electrode |
| 3.28 “Explain oxidation and reduction in terms of loss or gain of electrons” | 3.24 as above | Oxidation and reduction as electron transfer |
| 3.29 “Recall that reduction occurs at the cathode and that oxidation occurs at the anode in electrolysis reactions” | 3.24 as above | Which process happens at which electrode |
Our expert GCSE chemistry tutors help Higher-tier students turn this list into a plan. Talk to us about papers 1H and 2H and where your child’s marks are going.
Book a Free Consultation Message us on WhatsApp| Bold statement (board’s wording) | Plain-type statement beside it | What Higher adds |
|---|---|---|
| 4.2 “Explain displacement reactions as redox reactions, in terms of gain or loss of electrons” | 4.1 deduces relative reactivity from reactions with water, acids and salt solutions; 4.3 explains the reactivity series | The electron-transfer description of a displacement reaction |
| 4.8 “Evaluate alternative biological methods of metal extraction (bacterial and phytoextraction)” | 4.7 explains why the extraction method relates to a metal’s position in the reactivity series | Evaluation of the biological alternatives |
| 4.17 “Predict how the position of a dynamic equilibrium is affected by changes in” temperature, pressure and concentration | 4.14 explains dynamic equilibrium; 4.16 recalls the Haber process conditions of 450 °C, 200 atmospheres and an iron catalyst | Prediction of how the position shifts under each change |
| Bold statement (board’s wording) | Plain-type statement beside it | What Higher adds |
|---|---|---|
| 5.8C “Calculate the concentration of solutions in mol dm–3 and convert concentration in g dm–3 into mol dm–3 and vice versa” | 1.49 gives concentration in g dm–3 | Concentration in moles per cubic decimetre and the conversion |
| 5.10C “Carry out simple calculations using the results of titrations to calculate an unknown concentration of a solution or an unknown volume of solution required” | 5.9C is the core practical: an accurate acid-alkali titration with burette, pipette and a suitable indicator | The calculation from titration results |
| 5.15C “Explain why a particular reaction pathway is chosen to produce a specified product, given appropriate data such as atom economy, yield, rate, equilibrium position and usefulness of by-products” | 5.11C calculates percentage yield; 5.14C calculates atom economy | Choosing a pathway from the data |
| 5.16C “Describe the molar volume, of any gas at room temperature and pressure”, as the volume occupied by one mole of molecules, with the molar volume provided as 24 dm3 or 24000 cm3 where required | No plain-type partner in this group; the nearest plain calculation is 1.48 | The molar volume idea, with the figure supplied |
| 5.17C “Use the molar volume and balanced equations in calculations involving the masses of solids and volumes of gases” | 1.48 as above | Mass-to-gas-volume calculations |
| 5.18C “Use Avogadro’s law to calculate volumes of gases involved in a gaseous reaction, given the relevant equation” | 1.48 as above | Gas-to-gas volume calculations |
| 5.20C “Predict how the rate of attainment of equilibrium is affected by” changes in temperature, pressure, concentration and use of a catalyst | 5.19C describes the Haber process as a reversible reaction between nitrogen and hydrogen to form ammonia | Prediction of the rate at which equilibrium is reached |
| 5.21C “Explain how, in industrial reactions, including the Haber process, conditions used are related to” the availability of raw materials and energy supplies, and the control of temperature, pressure and catalyst to produce an acceptable yield in an acceptable time | 4.16 recalls the Haber conditions | Explaining why the industrial conditions are chosen |
| Bold statement (board’s wording) | Plain-type statement beside it | What Higher adds |
|---|---|---|
| 6.12 “Explain why these displacement reactions are redox reactions in terms of gain and loss of electrons, identifying which of the substances are oxidised and which are reduced” | 6.11 describes the relative reactivity of chlorine, bromine and iodine through displacement reactions; 6.13 explains halogen reactivity through electronic configurations | The redox description with oxidised and reduced species identified |
| 7.14 “Calculate the energy change in a reaction given the energies of bonds (in kJ mol–1)” | 7.12 recalls that breaking bonds is endothermic and making bonds is exothermic; 7.13 links that to exothermic and endothermic reactions | The bond-energy calculation |
| 9.22C “Explain” why polyesters are condensation polymers, how a polyester forms from a monomer with two carboxylic acid groups and a monomer with two alcohol groups, and how a molecule of water is formed each time an ester link is formed | 9.18C describes addition polymerisation of ethene; 9.20C deduces monomer from addition polymer | Condensation polymerisation and the polyester link |
What this changes in our programme. We hand each student this list as a checklist and mark every statement as secure, shaky or untouched after a diagnostic Higher paper. Sessions then follow the gaps, not the order of the textbook. Statements 4.17 and 5.20C get paired practice, because Pearson separates what happens to the position of an equilibrium from how fast it is reached.
The Higher-only calculations form a ladder. The plain-type rungs come first: 1.43 to 1.49 cover relative formula mass, empirical formulae, conservation of mass, masses from balanced equations and concentration in g dm–3. The bold rungs follow: the mole and the Avogadro constant in 1.50 to 1.53, concentration in mol dm–3 and titration results in 5.8C and 5.10C, gas volumes in 5.16C to 5.18C, and bond energies in 7.14. The plain-type percentage yield (5.11C) and atom economy (5.14C) statements sit between them.
The specification also marks these use-of-mathematics bullets as Higher-only: calculations with numbers written in standard form when using the Avogadro constant, and converting units where appropriate, particularly from mass to moles. Pearson allows calculators in the examination, and it supplies the molar volume as 24 dm3 or 24000 cm3 where a question needs it, so the examined skill is the method around the number.
Pearson’s June 2025 report on Paper 1H shows where that method is tested. Of the atom economy calculation in Question 4(c)ii, it says the majority of candidates scored all 3 marks, and that the commonest errors were using only 1 mole of phosphoric acid or giving the total formula mass of reactants and products. In a six-mark titration answer, the report describes a full-marks response as one that states how to calculate the number of moles of acid used, the number of moles of alkali used and the final concentration of the alkali.
What this changes in our programme. We teach one fixed layout for every Higher calculation: write the balanced equation, convert to moles, apply the ratio, then convert back with units and the requested number of significant figures. The board’s own advice to Paper 1H candidates is to show all working and round correctly, so we mark practice answers on working before the final value.
Struggling with moles, titrations or gas volumes? Our expert GCSE chemistry tutors help Higher-tier students work through the calculation ladder one rung at a time. Book a short call about tutoring for papers 1H and 2H.
Book a Free Consultation Message us on WhatsAppPearson’s Paper 1H report introduces the paper as ten questions, six of which also form one of the GCSE Combined Science higher tier papers, with three of the ten also appearing on the equivalent foundation paper, and as targeted from grade 4 to grade 9. It adds that there is often confusion about command words and the incorrect use of science specific vocabulary, which leaves candidates showing some scientific knowledge without being able to write it in a creditworthy way.
The hardest item it names is Question 9, on ammonia and equilibrium, which it says proved the most challenging question on the paper overall, with a noticeable dip in performance throughout. Within it, approximately half of candidates could state that the forward and reverse rates are equal at dynamic equilibrium, the commonest error being to say the reactions happen at the same time rather than at the same rate. Most could say that reducing the pressure would decrease the yield or shift the equilibrium to the left, but only about half of those could explain why. Some contradicted themselves by saying the yield would fall because the position shifted to the right. Its closing advice for 1H includes focusing on reversible reactions, dynamic equilibrium and the factors that affect the position of a dynamic equilibrium, and making sure candidates have experience of all of the chemistry core practicals.
The Paper 2H report says its questions 2, 3 and 4 were common with the 1CH0 2F paper and questions 2, 3, 5, 6, 7 and 8 with the Combined Science higher tier paper, and that the paper targeted grades 4 to 9 with about half the marks targeting grades 4 and 5. It also noted that most candidates performed extremely well on the two 6-mark questions. Its advice to candidates lists balancing equations, ionic and half equations and polymer-forming reactions; the patterns of reactivity of groups 1, 7 and 0 explained in terms of electrons; drawing reaction profiles for exothermic and endothermic reactions; and learning the qualitative tests for cations and anions.
What this changes in our programme. Equilibrium gets its own block on Paper 1H preparation, taught as a definition (equal rates), a prediction (which way the position moves) and an explanation (why), because the report shows candidates losing marks at the third step. For Paper 2H we keep a running set of equations and reaction profiles, and we drill the command words “describe” and “explain” on past questions until the student answers the verb that was asked.
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Pearson’s grade boundary documents define a grade boundary as the minimum mark at which a numbered grade can be achieved. The Higher rows for 1CH0 are out of 200 marks, across papers 1H and 2H together. The table gives the boundary for each grade from 9 down to 3 in each of the three June series we use.
| Series | Grade 9 | Grade 8 | Grade 7 | Grade 6 | Grade 5 | Grade 4 | Grade 3 |
|---|---|---|---|---|---|---|---|
| June 2024 | 148 | 128 | 109 | 87 | 66 | 45 | 34 |
| June 2025 | 161 | 144 | 128 | 106 | 84 | 62 | 51 |
| June 2026 | 165 | 146 | 127 | 103 | 79 | 55 | 43 |
Reading the marks as shares of the 200-mark total gives a clearer picture of what a grade asks for.
| Series | Grade 4 boundary | Share of 200 | Grade 9 boundary | Share of 200 |
|---|---|---|---|---|
| June 2024 | 45 | 22.5% | 148 | 74.0% |
| June 2025 | 62 | 31.0% | 161 | 80.5% |
| June 2026 | 55 | 27.5% | 165 | 82.5% |
Each percentage is the boundary divided by 200: 45 ÷ 200 = 22.5%, 62 ÷ 200 = 31.0%, 55 ÷ 200 = 27.5%, and 148 ÷ 200 = 74.0%, 161 ÷ 200 = 80.5%, 165 ÷ 200 = 82.5%. The distance from the grade 4 boundary to the grade 9 boundary was 103 marks in the first series (148 minus 45), 99 in the second (161 minus 62) and 110 in the third (165 minus 55).
| Grade | Lowest boundary | Highest boundary | Spread (marks) |
|---|---|---|---|
| 9 | 148 | 165 | 17 |
| 8 | 128 | 146 | 18 |
| 7 | 109 | 128 | 19 |
| 6 | 87 | 106 | 19 |
| 5 | 66 | 84 | 18 |
| 4 | 45 | 62 | 17 |
| 3 | 34 | 51 | 17 |
Every grade moved across a spread of between 17 and 19 marks in these three series. A boundary is set after each series is marked, so no single figure is a promise about the next one.
What this changes in our programme. We set a Higher student’s target as a band rather than a point, and we plan the grade 4 target against the top of the range, 62 marks, not the lowest, 45. For a student aiming at grade 7 the band runs from 109 to 128 marks, and we use the upper figure when we judge whether a timed paper score is secure. The board has said individual papers are not graded, so we also total 1H and 2H marks to the 200-mark scale before reading any result.
Our expert GCSE chemistry tutors help Higher-tier students at every target grade from 4 to 9, working from the board’s own boundaries and Higher-only list. Book a short call and tell us which paper is losing your child the most marks.
Book a Free Consultation Message us on WhatsAppThe programme has four repeating steps. We begin with a diagnostic on a past Higher paper, marked against the Higher-only list so the first session already knows which rung of the calculation ladder is weakest. We then teach to the gaps, using the specification wording as the standard: if the statement says “predict”, the student practises predicting; if it says “evaluate”, they practise weighing both sides. Core practicals are revised alongside the content, because Pearson’s Paper 1H report advises candidates to have experience of all of the chemistry core practicals. Last, each cycle ends with a timed paper of 1 hour and 45 minutes, so exam stamina is trained against the same clock as papers 1H and 2H.
Our tutors work one to one. For students who also sit Higher Maths, we coordinate topics so that the algebra behind rearranging a concentration formula is secure before it appears in a titration question. You can meet the team on our tutors page or read about GCSE tutoring across subjects. The board’s full documents are on Pearson’s GCSE Chemistry specification.
Pearson says 27 marks of each paper are overlap questions that appear in both the foundation and higher tier papers. Its June 2025 examiners' report for Paper 1H says three of the ten questions also appear on the equivalent foundation paper. The Paper 2H report says questions 2, 3 and 4 were common with the 1CH0 2F paper and questions 2, 3, 5, 6, 7 and 8 with the Combined Science higher tier paper.
Yes, within a limit Pearson states: grades 4 to 9 are available at higher tier, but if the mark achieved is a smaller number of marks below the 4/3 boundary, a grade 3 may be awarded. The Higher grade 3 boundary was 34 marks in June 2024, 51 in June 2025 and 43 in June 2026, out of 200, so the marks that separate a 3 from a 4 move with each series.
No. Pearson prints the figure inside its Higher-only statement 5.16C: the molar volume will be provided as 24 dm3 or 24000 cm3 in calculations where it is required. What the student must still do is use it, with a balanced equation, to link masses of solids and volumes of gases (statement 5.17C), so our sessions drill the mole-ratio step rather than recall of the number.
Working from Pearson's Higher rows: 45 of 200 is 22.5% in June 2024, 62 of 200 is 31.0% in June 2025 and 55 of 200 is 27.5% in June 2026. The grade 4 boundary therefore moved across a 17-mark spread (62 minus 45) in three series. We plan each student's target against the top of that range rather than the lowest figure.
Question 9, on ammonia and equilibrium. The report says it proved the most challenging question on the paper overall, with a noticeable dip in performance throughout. For 9(b)i, approximately half of candidates stated that the forward and reverse rates are equal at dynamic equilibrium, and the commonest error was saying the reactions happen at the same time. For 9(b)iii only about half could explain why lower pressure reduces the yield.
Topic 1 is assessed on both papers, so statements 1.50 to 1.53, 1.12 and 0.4 can all be examined on either. Paper 1H alone assesses Topics 2 to 5, so 3.4, 3.5, 3.7, 3.8, 3.27 to 3.29, 4.2, 4.8, 4.17, 5.8C, 5.10C, 5.15C to 5.18C, 5.20C and 5.21C belong there. Paper 2H alone assesses Topics 6 to 9, so 6.12, 7.14 and 9.22C belong there. Pearson lists 0.4 among the topics common to Paper 1 and Paper 2.
No. Pearson's specification says students must complete all assessments for the qualification in the same tier, and that students are entered for individual papers at the same tier. A Higher entrant is entered for 1CH0/1H and 1CH0/2H, and the grade comes from the total subject mark out of 200, because individual papers are not graded. The specification lists the paper codes as 1CH0/1F or 1CH0/1H and 1CH0/2F or 1CH0/2H, and the same-tier rule means the two letters are never mixed across papers.
Step 1: the student attempts a Higher paper and we mark it against the Higher-only statements. Step 2: sessions start with the weakest rung of the calculation ladder, from the mole in 1.50 to gas volumes in 5.18C and bond energies in 7.14. Step 3: we rehearse equilibrium predictions and half equations in the board's command-word style. Step 4: timed 1 hour 45 minute papers close each cycle, with the core practicals revised alongside.
Our expert GCSE chemistry tutors help Higher-tier students on the Pearson Edexcel course. Book a free consultation and we will talk through papers 1H and 2H.
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