كيف يحسب الرقم الهيدروجيني pH للمحاليل؟ How to calculate pH and pOH?

يحسب الرقم الهيدروجيني بالعلاقة: pH = -log[H⁺]، وللمحاليل المائية عند 25°C: pH + pOH = 14. إذا كان pH 7 فالمحلول قاعدي. [الصيغة الرمزية الموزونة: pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴]

1gPrinciples of Chemistry

Covalent bonding — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions

5 min read 11 parts Lesson 9 of 25 1 quiz

📚 Edexcel Linear - OL (Chemistry) Hub / Topical Revision

Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution

pH = -log[H⁺]  |  pOH = -log[OH⁻]  |  pH + pOH = 14  |  [H⁺][OH⁻] = 10⁻¹⁴

Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14

Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.

0%read

Lesson summary

Covalent bonding Edexcel O Level Linear Chemistry notes: know that a covalent bond is formed between atoms by the sharing of a pair of electronsEach lone pair reduces bond angle by about 2.5°.Hydrogen bonding explains high bp of water and low density of ice.σ bonds form by head-o…

Covalent bonding Edexcel O Level Linear Chemistry: what to do next
Ionic bond
Dative bond
Metallic bond
Electronegativity
van der Waals (id–id)

Section 1 of 11Process & sequence

Structure types

Giant ionic NaCl – high mp, conducts molten

Simple molecular I₂ – low mp

Giant covalent diamond, SiO₂ – very high mp

Giant metallic Cu – conducts, malleable

Lone pair–lone pair
>
lone pair–bond pair
>
bond pair–bond pair

Section 2 of 11Keep going

Covalent bonding Edexcel O Level Linear Chemistry: what to do next

These Covalent bonding Edexcel O Level Linear Chemistry notes covered the key ideas, worked examples and exam technique for this topic. Check your Covalent bonding Edexcel O Level Linear Chemistry understanding with the practice questions below, then reread any Covalent bonding Edexcel O Level Linear Chemistry point you were unsure about before moving on.

Practice: Covalent bonding Edexcel O Level Linear Chemistry practice questions

Previous lesson: 1f Ionic bonding — Notes

Next lesson: 2a Group 1 (alkali metals) - lithium, sodium and potassium — Notes

Course: Edexcel Linear – OL (chemistry) — all units

Official syllabus: Pearson Edexcel International GCSEs

Section 3 of 11Giant covalent structures

Giant covalent structures

Diamond and graphite are forms of carbon with giant covalent structures. They contain networks of atoms held by many strong covalent bonds, not separate carbon molecules.

Both have high melting points because many strong covalent bonds must be broken to melt their structures. Breaking those bonds requires a large energy input.

Section 4 of 11Diamond

Four covalent bonds

In diamond, every carbon atom forms four covalent bonds to other carbon atoms in a three-dimensional network. The network resists scratching, making diamond very hard.

Diamond does not conduct electricity because its outer electrons are used in bonds and there are no mobile charged particles to carry current. A hard material can still be brittle: a sufficiently strong impact can fracture diamond.

Section 5 of 11Graphite

Three covalent bonds

In graphite, each carbon atom forms three covalent bonds within a flat layer of hexagons. The layers are held together by weak forces, so they can slide over one another. This makes graphite soft and slippery.

Delocalised

One electron from each carbon atom is delocalised and can move within the layers. These mobile electrons allow graphite to conduct electricity and help transfer heat.

Section 6 of 11Compare and explain

To explain a property, name the structural feature responsible: diamond's four-bond network gives hardness; graphite's weak forces between layers allow sliding; graphite's delocalised electrons enable conduction; strong covalent bonds give both high melting points.

Check:

Check Why is graphite a conductor but diamond is not? Answer: Graphite has mobile delocalised electrons, whereas diamond has no mobile charged particles.

Section 7 of 11Compare

Shapes of molecules (VSEPR)
Bonding pairs	Lone pairs	Shape	Bond angle	Example	Polar?
2	0	Linear	180°	CO₂	No
3	0	Trigonal planar	120°	BF₃	No
4	0	Tetrahedral	109.5°	CH₄	No
3	1	Pyramidal	107°	NH₃	Yes
2	2	Non-linear (bent)	104.5°	H₂O	Yes
6	0	Octahedral	90°	SF₆	No

Section 8 of 11Key terms

Section 9 of 11Syllabus checklist

know that a covalent bond is formed between atoms by the sharing of a pair of electrons

understand covalent bonds in terms of electrostatic attractions

understand how to use dot-and-cross diagrams to represent covalent bonds in

- diatomic molecules
including hydrogen, oxygen
nitrogen
halogens and hydrogen halides - inorganic molecules including water
ammonia and carbon dioxide - organic molecules containing up to two carbon atoms
including methane, ethane
ethene and those containing halogen atoms

explain why substances with a simple molecular structures are gases or liquids, or solids with low melting and boiling points the term intermolecular forces of attraction can be used to represent all forces between molecules

explain why the melting and boiling points of substances with simple molecular structures increase, in general, with increasing relative molecular mass

explain why substances with giant covalent structures are solids with high melting and boiling points

explain how the structures of diamond, graphite and C60 fullerene influence their physical properties, including electrical conductivity and hardness

know that covalent compounds do not usually conduct electricity

Section 10 of 11Command terms

Command terms

Give / State / Name

Recall a fact; one idea per mark.

Define

Give the exact meaning using the specification keywords.

Describe

Give an account of what happens / the pattern in data, quoting values; no reasons needed.

Explain

Give reasons with linked statements (cause → effect). Each linked step is usually a separate marking point.

Suggest

Use knowledge in a new context; more than one answer may be accepted.

Compare

Similarities and differences; each point must refer to both items.

Calculate

Show working; include units; check significant figures. Method marks are given even if the final answer is wrong.

Discuss

Explore the issue from different angles and link the points.

Evaluate / Assess

Weigh strengths and weaknesses and reach a supported judgement.

Deduce

Draw a conclusion from the information given.

Justify

Support a statement/choice with evidence.

Section 11 of 11Exam technique

Keywords the mark scheme looks for in this lesson

Ionic bond · Covalent bond · Dative bond · Metallic bond · Electronegativity · VSEPR · Hydrogen bond · van der Waals (id–id)

Learn each definition in the Key words tab exactly; a missing keyword in a definition usually loses the mark.

How to get full marks (examiner guidance)

Use specification keywords exactly (e.g. electronegativity, nucleophile, dynamic equilibrium, activation energy).

Link ideas

Link ideas for “explain” each marking point is usually a cause → consequence step; unlinked facts often score nothing.

Use the information in the question

“discuss/evaluate” answers must use the data provided and match the number of marks.

Convert units carefully (mg → g, cm³ → dm³) and show working for calculations.

Avoid vague words

“it”, “stuff”, “things”, “goes up”, “energy is made”. Name the substance and the process.

Do not extrapolate a graph line beyond the data or to zero unless asked.

Six-mark extended responses are marked on content AND a logical, linked line of reasoning.

Last updated: 11 October 2026