- Write dissociation equations of acids, bases and salts (including stepwise dissociation) and calculate amounts of ions from the degree of dissociation.
- Write molecular, full ionic and net ionic equations and decide whether an ion-exchange reaction goes to completion.
- Identify the ions Cl⁻, SO₄²⁻, CO₃²⁻, PO₄³⁻, NO₃⁻, NH₄⁺, Fe²⁺, Fe³⁺ and Al³⁺ by their test reactions.
- Use the composition of a salt to decide whether it hydrolyses and what medium (acidic, alkaline, neutral) its solution has.
Distilled water hardly conducts electricity, and neither does sugar solution. But drop a pinch of table salt into the water and the tester’s bulb lights up. So salt breaks up in water into charged particles — ions. In this lesson you will learn to speak the «language» of ions: dissociation equations, ionic equations, tests for ions and hydrolysis of salts. DİM likes this topic: in the 2026 group IV exam one of the written tasks was exactly to complete a table of molecular and full ionic equations.
Electrolytes and dissociation equations
Electrolytes are substances that break up into ions in aqueous solution or in the melt and therefore conduct electricity: acids, bases and salts. Non-electrolytes (sugar, alcohol, oxygen, most organic substances) do not form ions. The breaking up of an electrolyte into ions is called electrolytic dissociation.
The theory was put forward in 1887 by the Swedish scientist S. Arrhenius. In ionic substances (NaCl) the ions already exist in the crystal and water only pulls them apart; polar covalent molecules (HCl) are ionised by water. In solution the ions are surrounded by water molecules — they are hydrated (D. I. Mendeleev’s hydrate theory). In an electric field positive ions (cations) move to the cathode and negative ions (anions) to the anode, while the solution as a whole stays neutral.
- Acids → H⁺ cations and acid-residue anions: HNO₃ → H⁺ + NO₃⁻. The common properties of acids (sour taste, litmus turning red) come from the H⁺ ions.
- Alkalis → metal cations and OH⁻ ions: Ba(OH)₂ → Ba²⁺ + 2OH⁻.
- Salts → metal (or NH₄⁺) cations and acid-residue anions: Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻; an acid salt: NaHCO₃ → Na⁺ + HCO₃⁻.
- Stepwise dissociation: polybasic acids give off H⁺ ions one at a time, and each next step is much weaker: H₃PO₄ ⇌ H⁺ + H₂PO₄⁻; H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻; HPO₄²⁻ ⇌ H⁺ + PO₄³⁻. That is why a phosphoric acid solution contains the fewest PO₄³⁻ ions.
- n(electrolyte)amount of dissolved electrolyte, mol
- αdegree of dissociation (as a fraction); α = 1 for complete dissociation
- x + ynumber of ions from one formula unit: AₓBᵧ → xAᵐ⁺ + yBⁿ⁻
Total amount of ions formed on dissociation
1) Complete dissociation of 1 mol of which salt gives the most ions?
A) NaCl B) K₂SO₄ C) AlCl₃ D) K₃PO₄ E) Fe₂(SO₄)₃
2) A solution contains 0.4 mol of Fe₂(SO₄)₃, and the degree of dissociation is 75%. How many moles of Fe³⁺ ions, of SO₄²⁻ ions and of ions in total have formed?
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2) Fe₂(SO₄)₃ → 2Fe³⁺ + 3SO₄²⁻; the salt that dissociates: 0.75 · 0.4 = 0.3 mol.
n(Fe³⁺) = 2 · 0.3 = 0.6 mol; n(SO₄²⁻) = 3 · 0.3 = 0.9 mol.
Total: 0.75 · 0.4 · 5 = 1.5 mol of ions.
Degree of dissociation. Strong and weak electrolytes
- αdegree of dissociation, %
- n(dissociated)amount of molecules (formula units) that broke up into ions, mol
- n(dissolved)total amount of substance dissolved, mol
The degree of dissociation shows what percentage of the dissolved particles broke up into ions
For 0.1 mol/L solutions, substances with α > 30% are strong electrolytes, those with α < 3% are weak, and the ones in between are of medium strength. Dissociation of weak electrolytes is reversible, so ⇌ is written. Dilution and (in most cases) heating increase α, while adding an ion of the same kind decreases it: adding CH₃COONa to a CH₃COOH solution suppresses the dissociation of acetic acid.
| Substances | Strong electrolytes | Weak electrolytes |
|---|---|---|
| Acids | HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄ | H₂S, H₂CO₃, HF, HNO₂, CH₃COOH, H₂SiO₃ |
| Bases | alkalis: LiOH, NaOH, KOH, Ca(OH)₂, Ba(OH)₂ | NH₃·H₂O and insoluble bases: Cu(OH)₂, Fe(OH)₃ |
| Salts | almost all soluble salts | — |
1) A solution contains 0.5 mol of HF, and 0.04 mol of it has broken up into ions. Calculate the degree of dissociation (%) and the total amount (mol) of HF molecules and ions in the solution.
2) A solution contains 3 mol of MgCl₂ (α = 90%) and 2 mol of KCl (α = 85%). How many moles of Cl⁻ ions have formed?
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HF ⇌ H⁺ + F⁻: undissociated molecules 0.5 − 0.04 = 0.46 mol, ions 0.04 + 0.04 = 0.08 mol.
Total: 0.46 + 0.08 = 0.54 mol of particles.
2) MgCl₂ → Mg²⁺ + 2Cl⁻: n(Cl⁻) = 2 · 0.9 · 3 = 5.4 mol.
KCl → K⁺ + Cl⁻: n(Cl⁻) = 0.85 · 2 = 1.7 mol.
Total: 5.4 + 1.7 = 7.1 mol of Cl⁻.
Ion-exchange reactions and ionic equations
Reactions between electrolyte solutions are reactions between ions. An exchange reaction goes to completion when ions «leave» the solution: a precipitate (↓), a gas (↑) or a weakly dissociating substance (water, a weak acid) forms. Otherwise the ions just stay mixed: mixing KNO₃ and NaCl solutions gives no reaction. Precipitates are found from the solubility table.
- 1Molecular equation
Write the equation, balance it, mark the precipitate with ↓ and the gas with ↑.
- 2Full ionic equation
Write strong electrolytes (soluble salts, strong acids, alkalis) as ions; precipitates, gases, water, weak electrolytes, oxides and elements stay as formulas. The coefficient applies to the ions too: 3BaCl₂ → 3Ba²⁺ + 6Cl⁻.
- 3Net ionic equation
Cancel the ions that are the same on both sides — they take no part in the reaction (spectator ions).
- 4Check
Both the numbers of atoms and the sums of charges must be equal on the two sides.
Excess NaOH solution was added to 200 g of an 8% CuSO₄ solution. Write the molecular, full ionic and net ionic equations (1) and calculate the mass (g) of the precipitate (2). Mr(CuSO₄) = 160, Mr(Cu(OH)₂) = 98
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Full ionic: Cu²⁺ + SO₄²⁻ + 2Na⁺ + 2OH⁻ → Cu(OH)₂↓ + 2Na⁺ + SO₄²⁻
Net ionic: Cu²⁺ + 2OH⁻ → Cu(OH)₂↓ (blue precipitate). Charges: (+2) + 2 · (−1) = 0, and 0 on the right too.
(2) m(CuSO₄) = 200 · 0.08 = 16 g → n = 16 / 160 = 0.1 mol.
n(Cu(OH)₂) = n(CuSO₄) = 0.1 mol → m = 0.1 · 98 = 9.8 g.
Between the aqueous solutions of which substances does the reaction go to completion?
I. FeCl₃ and KOH
II. Na₂SO₄ and HNO₃
III. Na₂S and HCl
IV. KCl and NaNO₃
A) I, II B) I, III C) II, IV D) III, IV E) I, IV
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II. The possible products (NaNO₃, H₂SO₄) are soluble strong electrolytes — no reaction.
III. Na₂S + 2HCl → 2NaCl + H₂S↑ — a gas escapes, it goes.
IV. All ions stay in solution — no reaction.
Answer: B) I, III.
Tests for ions
A qualitative (test) reaction shows which ion is in the solution: it gives a visible sign — a coloured precipitate, a gas or a change of colour. One reagent can detect several ions (OH⁻ detects NH₄⁺, Fe²⁺, Fe³⁺ and Al³⁺), so remember the colour of each sign.
| Ion | Reagent | Sign | Net ionic equation |
|---|---|---|---|
| Cl⁻ | Ag⁺ (AgNO₃) | white curdy precipitate, insoluble in HNO₃ | Ag⁺ + Cl⁻ → AgCl↓ |
| SO₄²⁻ | Ba²⁺ (BaCl₂) | white precipitate, insoluble in acids | Ba²⁺ + SO₄²⁻ → BaSO₄↓ |
| CO₃²⁻ | H⁺ (an acid) | gas fizzes off and turns limewater milky | CO₃²⁻ + 2H⁺ → CO₂↑ + H₂O |
| PO₄³⁻ | Ag⁺ (AgNO₃) | yellow precipitate, soluble in HNO₃ | 3Ag⁺ + PO₄³⁻ → Ag₃PO₄↓ |
| NO₃⁻ | Cu + concentrated H₂SO₄, heating | brown gas (NO₂), the solution turns blue | Cu + 4H⁺ + 2NO₃⁻ → Cu²⁺ + 2NO₂↑ + 2H₂O |
| NH₄⁺ | OH⁻ (alkali), heating | smell of ammonia; wet red litmus paper turns blue | NH₄⁺ + OH⁻ → NH₃↑ + H₂O |
| Fe²⁺ | OH⁻; K₃[Fe(CN)₆] | greenish precipitate that browns in air; blue precipitate with potassium ferricyanide | Fe²⁺ + 2OH⁻ → Fe(OH)₂↓ |
| Fe³⁺ | OH⁻; KSCN | brown precipitate; blood-red colour with thiocyanate | Fe³⁺ + 3OH⁻ → Fe(OH)₃↓ |
| Al³⁺ | OH⁻ (a little) | white jelly-like precipitate that dissolves in excess alkali | Al³⁺ + 3OH⁻ → Al(OH)₃↓ |
Match the items.
1. Fe³⁺
2. NH₄⁺
3. SO₄²⁻
a. BaCl₂
b. AgNO₃
c. NaOH
d. KSCN
e. HCl
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2. NH₄⁺: ammonia is released when heated with NaOH → c.
3. SO₄²⁻: a white BaSO₄ precipitate with BaCl₂ → a.
AgNO₃ is the reagent for Cl⁻ and PO₄³⁻, and HCl for CO₃²⁻.
Answer: 1 – c, d; 2 – c; 3 – a.
Hydrolysis of salts and the medium of the solution
The reaction of salt ions with water that forms a weak electrolyte (or its ion) and leaves an excess of H⁺ or OH⁻ ions in the solution. The ion that comes from a weak acid or a weak base is the one that hydrolyses.
Split the salt into its «parents» — the base and the acid that form it. The strong parent decides the medium, and the ion of the weak parent hydrolyses. Hydrolysis is usually reversible and is written as the first step:
Na₂CO₃: CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ (Na₂CO₃ + H₂O ⇌ NaHCO₃ + NaOH) — the medium is alkaline.
AlCl₃: Al³⁺ + H₂O ⇌ AlOH²⁺ + H⁺ (AlCl₃ + H₂O ⇌ AlOHCl₂ + HCl) — the medium is acidic.
| Salt formed by | Examples | Hydrolysing ion | Medium | Litmus |
|---|---|---|---|---|
| strong base + weak acid | Na₂CO₃, K₂S, CH₃COONa, K₃PO₄, Na₂SiO₃ | anion | alkaline, pH > 7 | blue |
| weak base + strong acid | AlCl₃, ZnSO₄, CuCl₂, NH₄NO₃, FeCl₃ | cation | acidic, pH < 7 | red |
| strong base + strong acid | NaCl, KNO₃, Na₂SO₄, BaCl₂ | no hydrolysis | neutral, pH = 7 | violet |
| weak base + weak acid | CH₃COONH₄, Al₂S₃ | both ions | close to neutral; Al₂S₃ is completely decomposed by water | — |
Salts of a weak base and a weak volatile acid are decomposed by water completely and irreversibly: Al₂S₃ + 6H₂O → 2Al(OH)₃↓ + 3H₂S↑. That is why the solubility table shows «—» for Al₂S₃ and Al₂(CO₃)₃: when AlCl₃ and Na₂CO₃ solutions are mixed, Al(OH)₃ precipitates instead of a carbonate and CO₂ escapes: 2AlCl₃ + 3Na₂CO₃ + 3H₂O → 2Al(OH)₃↓ + 3CO₂↑ + 6NaCl. Hydrolysis is endothermic: heating and dilution increase it; to suppress the hydrolysis of salts such as FeCl₃, a little acid is added to the solution.
Match the items.
1. acidic medium
2. alkaline medium
3. neutral medium
a. K₂SiO₃
b. NH₄NO₃
c. Ba(NO₃)₂
d. Na₂S
e. Al₂(SO₄)₃
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NH₄NO₃ (weak NH₃·H₂O + HNO₃) and Al₂(SO₄)₃ (weak Al(OH)₃ + H₂SO₄): the cation hydrolyses, the medium is acidic.
Ba(NO₃)₂ is a salt of a strong base and a strong acid, so there is no hydrolysis.
Answer: 1 – b, e; 2 – a, d; 3 – c.
Key points
- Acids, alkalis and salts are electrolytes; polybasic acids dissociate stepwise, each step weaker than the one before.
- α = n(dissociated) / n(dissolved); the amount of ions is α · n · (x + y). In 0.1 mol/L solutions strong electrolytes have α > 30% and weak ones α < 3%.
- Ion exchange goes to completion when a precipitate, a gas or a weak electrolyte (water) forms; in ionic equations only soluble strong electrolytes are written as ions.
- Cl⁻ — Ag⁺ (white), SO₄²⁻ — Ba²⁺ (white), CO₃²⁻ — H⁺ (CO₂), PO₄³⁻ — Ag⁺ (yellow), NH₄⁺ — OH⁻ (NH₃), Fe³⁺ — SCN⁻ (blood-red); Fe²⁺, Fe³⁺ and Al³⁺ give coloured precipitates with alkali.
- The ion of the weak parent hydrolyses and the strong parent decides the medium: Na₂CO₃ — alkaline, AlCl₃ — acidic, NaCl — neutral.
Check yourself
12 questions. Every correct answer earns XP.