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]

5.3Unit 5. Transition Metals and Organic Nitrogen Chemistry

Transition Metal Reactions — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 12 parts Lesson 3 of 43 1 quiz

📚 IAL Edexcel Linear (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.

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Lesson summary

5.3 Transition Metal Reactions — vanadium, chromium, aqua ions and catalysis — IAL EDEXCEL LINEAR (CHEMISTRY) notes: Describe the redox chemistry of vanadium and chromium, the reactions of aqueous ions with hydroxide and ammonia, and heterogeneous, homogeneous and autocatalysis.V…

Amphoteric hydroxide
Heterogeneous catalyst
Adsorption
Desorption
Advantages
Contact process

Part 1 of 10Vanadium oxidation states and colours

Vanadium shows four common oxidation states, from +2 to +5.

Oxidation state	Species	Colour
+5	VO₂+	Yellow
+4	VO₂+	Blue
+3	V3+	Green
+2	V2+	Violet (lilac)

Vanadium oxidation states and colours

Zinc reduces acidified ammonium vanadate(V) stepwise: yellow, then green (a yellow/blue mixture), blue, green and finally violet.

Take care not to confuse the two oxo ions, VO₂⁺ (+5) and VO²⁺ (+4).

Keep air out of V²⁺, which is readily oxidised.

Part 2 of 10Vanadium electrode potentials and feasibility

E⦵ values (V)

VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O, +1.00.
VO²⁺ + 2H⁺ + e⁻ → V³⁺ + H₂O, +0.34.
V³⁺ + e⁻ → V²⁺, −0.26.
V²⁺ + 2e⁻ → V, −1.18.
Zn²⁺ + 2e⁻ → Zn, −0.76.

The strongest oxidising agent is the reactant of the most positive half-equation (VO₂⁺); the strongest reducing agent is the product of the most negative one (V).

Use the half-equation with the more positive E⦵ as the reduction and reverse the other, then balance electrons.

2VO₂⁺ + 4H⁺ + Zn → 2VO²⁺ + Zn²⁺ + 2H₂O.
2VO²⁺ + 4H⁺ + Zn → 2V³⁺ + Zn²⁺ + 2H₂O.
2V³⁺ + Zn → 2V²⁺ + Zn²⁺.

Zn (−0.76 V) reduces vanadium down to V²⁺ but not to V metal, because V + Zn²⁺ → V²⁺ + Zn is not feasible.

Weaker reducing agents such as SO₂ or I⁻ stop at lower states whose E⦵ is above theirs.

Part 3 of 10Chromium redox chemistry

Chromium shows +2 (blue), +3 (green or violet) and +6 (orange dichromate, yellow chromate).

Dichromate is the strongest oxidising agent, E⦵ = +1.33 V; Zn is the strongest reducing agent, −0.76 V.

Zn in acid reduces Cr₂O₇²⁻ to Cr³⁺: Cr₂O₇²⁻ + 14H⁺ + 3Zn → 2Cr³⁺ + 7H₂O + 3Zn²⁺ (orange to green).

Further reduction gives Cr²⁺: 2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ (E⦵ Cr³⁺/Cr²⁺ = −0.41 V); Cr²⁺ is re-oxidised by air.

Alkaline H₂O₂ oxidises Cr(III) to yellow chromate: 2Cr(OH)₃ + 4OH⁻ + 3H₂O₂ → 2CrO₄²⁻ + 8H₂O.

Acidic conditions reduce dichromate; alkaline conditions give the oxidation of Cr(III).

Acidified dichromate(VI) also oxidises alcohols and Fe²⁺.

Part 4 of 10Chromate–dichromate equilibrium and reaction schemes

Chromate–dichromate equilibrium and reaction schemes

2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O

Acid shifts the equilibrium to orange dichromate; alkali shifts it to yellow chromate.
It is an acid–base reaction, not redox, because chromium is +6 in both.

A reaction scheme records for each species its formula and state, its colour, and the reagents and conditions to convert it into another.

[Cr(H₂O)₆]³⁺ (green) + 3OH⁻ → [Cr(H₂O)₃(OH)₃](s) (grey-green precipitate) + 3H₂O.

Part 5 of 10Aqua ions with hydroxide and ammonia

Transition metal aqua ions form precipitates with NaOH(aq) and NH₃(aq);
- the number of OH⁻ in the precipitate equals the charge on the ion.

Ion	NaOH (few drops)	Excess NaOH	Excess NH₃
Cr³⁺	Grey-green precipitate	Dark green [Cr(OH)₆]3-	Purple [Cr(NH₃)₆]3+
Fe²⁺	Green precipitate	No change	No change
Fe³⁺	Orange-brown precipitate	No change	No change
Co²⁺	Blue precipitate	No change	Yellow [Co(NH₃)₆]2+
Mn²⁺	Pale brown precipitate	No change	No change
Ni²⁺	Green precipitate	No change	Dark blue [Ni(NH₃)₆]2+
Cu²⁺	Blue precipitate	No change	Deep blue [Cu(NH₃)₄(H₂O)₂]2+
Zn²⁺	White precipitate	Colourless solution	Colourless [Zn(NH₃)₄]2+

Examples
- [Fe(H₂O)₆]²⁺ + 2OH⁻ → [Fe(H₂O)₄(OH)₂](s) + 2H₂O.
- [Fe(H₂O)₆]³⁺ + 3OH⁻ → [Fe(H₂O)₃(OH)₃](s) + 3H₂O.
- [Cu(H₂O)₆]²⁺ + 2NH₃ → Cu(H₂O)₄(OH)₂(s) + 2NH₄⁺.

Part 6 of 10Deprotonation and amphoteric hydroxides

Limited OH⁻ or NH₃ gives the same hydroxide precipitates by deprotonation acid–base reactions, not ligand substitution.

The hydroxide ions or ammonia remove H⁺ from water ligands, which become hydroxide ligands.

[M(H₂O)₆]³⁺ solutions are more acidic than [M(H₂O)₆]²⁺ because the greater polarising power of the metal ion makes water ligands release H⁺, e.g. [Fe(H₂O)₆]³⁺ ⇌ [Fe(H₂O)₅(OH)]²⁺ + H⁺.

Amphoteric hydroxide

Amphoteric hydroxide reacts with both acids and bases.
[Cr(H₂O)₃(OH)₃] + 3OH⁻ → [Cr(OH)₆]³⁻ + 3H₂O (acting as an acid).
[Cr(H₂O)₃(OH)₃] + 3H⁺ → [Cr(H₂O)₆]³⁺ (acting as a base).

Part 7 of 10Excess ammonia and ligand substitution

With excess NH₃, the precipitates of Cu, Co and Cr dissolve by ligand substitution.

NH₃ and H₂O are similar in size and uncharged, so the coordination number does not change for Co and Cr.

[Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(H₂O)₂(NH₃)₄]²⁺ + 4H₂O.

[Co(H₂O)₆]²⁺ + 6NH₃ → [Co(NH₃)₆]²⁺ + 6H₂O.

From the precipitate

[Cu(H₂O)₄(OH)₂] + 4NH₃ → [Cu(H₂O)₂(NH₃)₄]²⁺ + 2H₂O + 2OH⁻.

NH₃ acts as a Lewis base, donating an electron pair.

Part 8 of 10Heterogeneous catalysis

Heterogeneous catalyst

Heterogeneous catalyst in a different phase from the reactants, usually a solid with gaseous or dissolved reactants;
- reaction occurs at active sites on its surface.

Adsorption

Surface adsorption theory

Adsorption
- reactants attach to the surface, bonds weaken and the local concentration rises.
Reaction
- the weakened bonds react.
Desorption
- the product leaves the surface.

Strength of adsorption must be about right.

Tungsten adsorbs too strongly, so products are not released.
Silver adsorbs too weakly, so too few reactants adsorb.
Ni and Pt are about right and are most useful.

A larger surface area increases effectiveness; a support medium (e.g. Rh on a ceramic) saves cost.

Advantages

Advantages easy to filter off and suited to continuous processes.

Impurities adsorb on active sites, prevent bond weakening and reduce effectiveness (catalyst poisoning).

Transition metals are good catalysts because of variable oxidation states and the ability to form weak bonds using d orbitals.

Contact process

V₂O₅ catalyses 2SO₂ + O₂ ⇌ 2SO₃.
SO₂ + V₂O₅ → V₂O₄ + SO₃ (V +5 to +4).
O₂ + 2V₂O₄ → 2V₂O₅ (V +4 to +5, regenerating the catalyst).

The Haber process uses an iron catalyst.

Part 9 of 10Catalytic converters, homogeneous catalysis and autocatalysis

Catalytic converter

Catalytic converter finely divided platinum and rhodium on a ceramic honeycomb.
2NO + 2CO → N₂ + 2CO₂.
CO and NO adsorb, bonds weaken, the molecules react and the products desorb.
Spreading the catalyst thinly over an inert support maximises surface area and minimises cost.

Do not confuse adsorption (at a surface) with absorption (distributed throughout a substance).

Homogeneous catalyst

Homogeneous catalyst in the same phase as the reactants; it works by forming an intermediate.
S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂ is slow because the negative ions repel.
Fe²⁺ reduces S₂O₈²⁻: S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺.
Fe³⁺ then oxidises iodide: 2I⁻ + 2Fe³⁺ → I₂ + 2Fe²⁺, regenerating the catalyst.
Each step involves oppositely charged ions, so activation energy is lower; starch shows I₂ as blue-black.

Autocatalysis

Autocatalysis a product catalyses the reaction, so the rate increases as product forms.
2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O.
Mn²⁺ is not present at the start, so the reaction begins slowly and then accelerates.
Mn²⁺ cycles through Mn³⁺: 4Mn²⁺ + MnO₄⁻ + 8H⁺ → 5Mn³⁺ + 4H₂O and 2Mn³⁺ + C₂O₄²⁻ → 2CO₂ + 2Mn²⁺.
The concentration–time curve is S-shaped and can be followed by colorimetry.

Part 10 of 10More from the specification

Electrode potential
- A measure of how easily a species is reduced, relative to a reference half-cell.

Key terms

Term	Meaning
Variable oxidation state	The ability of transition metals to form ions in several oxidation states because 3d and 4s energies are close.
Heterogeneous catalyst	A catalyst in a different phase from the reactants, working at active sites on its surface.
Homogeneous catalyst	A catalyst in the same phase as the reactants, working through an intermediate.
Autocatalysis	Catalysis of a reaction by one of its own products.
Adsorption	The attachment of reactant molecules to the surface of a catalyst.
Catalyst poisoning	Blocking of active sites by impurities.
Amphoteric hydroxide	A hydroxide that reacts with both acids and bases.
Deprotonation	Loss of H⁺ from a coordinated water ligand.

Avoid these mistakes

Mixing up VO₂+ and VO₂+ — they are the +4 and +5 oxo ions.

Saying the chromate–dichromate change is a redox reaction — chromium is +6 in both; it is acid–base.

Saying hydroxide ions in the first precipitate step are ligand substitution — they remove H⁺ from water ligands.

Confusing absorption with adsorption — adsorption is at a surface.

Saying a catalyst is in the rate-determining step only in heterogeneous catalysis — homogeneous catalysts also form intermediates.

Last updated: 18 October 2026