- Tell strong electrolytes, weak electrolytes and non-electrolytes apart, and write ionisation equations
- Write and check net ionic equations (what is split, conservation of atoms and charge)
- Decide whether ions react or can coexist in a solution, including hidden conditions (colour, acidity, redox)
- Choose the reagent and the order of tests for Cl⁻, SO₄²⁻, CO₃²⁻, NH₄⁺, Fe³⁺ and Fe²⁺, and read flame colours
Add a drop of silver nitrate to tap water and it turns faintly cloudy: chloride ions are there. Pour hydrochloric acid on marble and it fizzes. Almost everything that happens in a school test tube is a reaction between ions, and the CSCA syllabus names it directly: “Ionic reactions and testing methods”. The questions ask you to write or judge an ionic equation, to decide whether a group of ions can exist together, and to identify an ion from what is seen. What happens to the electrons in reactions such as Fe + Cu²⁺ is in the lesson “Redox reactions”; the equilibrium of weak electrolytes and hydrolysis are in “Theories of electrolyte solutions”.
Electrolytes: what is really in the solution
电解质, 非电解质)An electrolyte is a compound that conducts electricity when dissolved in water or when molten, because it releases free ions: acids, bases, salts, metal oxides and water. A non-electrolyte is a compound that conducts in neither state: sucrose, ethanol, most organic compounds, and also CO₂, SO₂ and NH₃, whose solutions conduct only because the products H₂CO₃, H₂SO₃ and NH₃·H₂O ionise. Elements (Cu, Cl₂) and mixtures (NaCl solution) are neither.
Strong electrolytes (强电解质) ionise completely in water: strong acids (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄), strong bases (NaOH, KOH, Ba(OH)₂, Ca(OH)₂) and almost all salts. Even insoluble BaSO₄ is strong, because the little that dissolves is fully ionised. Weak electrolytes (弱电解质) ionise only partly: weak acids (CH₃COOH, H₂CO₃, H₂S, HClO, HF), the weak base NH₃·H₂O and water. How far a weak electrolyte ionises is the subject of the lesson “Theories of electrolyte solutions”; here we need only the list. Conductivity depends on the concentration and charge of the free ions, not on whether the electrolyte is strong: a very dilute HCl solution can conduct worse than a concentrated acetic acid solution.
- →complete ionisation (strong electrolyte)
- ⇌partial, reversible ionisation (weak electrolyte)
The charges on the right add up to zero. Acid salts: NaHCO₃ → Na⁺ + HCO₃⁻ (HCO₃⁻ is not split), but NaHSO₄ → Na⁺ + H⁺ + SO₄²⁻ in solution. Polyprotic weak acids ionise step by step: first H₂CO₃ ⇌ H⁺ + HCO₃⁻.
1) Write the ionisation equations of Ba(OH)₂, NaHCO₃, NaHSO₄ (in solution) and NH₃·H₂O.
2) Find c(Fe³⁺) and c(SO₄²⁻) in 0.1 mol/L Fe₂(SO₄)₃ solution (ignore hydrolysis).
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2) Fe₂(SO₄)₃ → 2Fe³⁺ + 3SO₄²⁻: c(Fe³⁺) = 2 · 0.1 = 0.2 mol/L, c(SO₄²⁻) = 3 · 0.1 = 0.3 mol/L. The concentration of an ion does not depend on the volume taken: 10 mL and 1 L of this solution have the same c(SO₄²⁻).
Which list gives a strong electrolyte, a weak electrolyte and a non-electrolyte, in this order?
A) NaCl, CH₃COOH, C₂H₅OH B) HCl, BaSO₄, SO₂ C) Cu, NH₃·H₂O, sucrose D) KNO₃, H₂O, Cl₂
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B) BaSO₄ is insoluble but a strong electrolyte. C) Cu is an element, neither electrolyte nor non-electrolyte. D) Cl₂ is an element too (water is correctly weak). The trap is to judge by solubility or by “does the solution conduct?”.
Net ionic equations
An ionic equation (离子方程式) shows what really reacts. When BaCl₂ solution is poured into Na₂SO₄ solution, Na⁺ and Cl⁻ stay in solution unchanged — they are spectator ions — and the whole change is Ba²⁺ + SO₄²⁻ → BaSO₄↓. That is why one net equation describes a whole family of reactions: any soluble barium salt with any soluble sulfate. Chinese textbooks teach four steps, 写、拆、删、查 (write, split, delete, check).
- 1Write (
写)Write the balanced molecular equation: BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl.
- 2Split (
拆)Rewrite as ions only the strong electrolytes that are dissolved: strong acids, strong bases, soluble salts. Keep as formulas: precipitates, gases, weak electrolytes (water, CH₃COOH, NH₃·H₂O), oxides and elements.
- 3Delete (
删)Cross out the ions that appear unchanged on both sides (here 2Na⁺ and 2Cl⁻).
- 4Check (
查)Atoms and total charge must be equal on both sides, the coefficients must be the smallest whole numbers, and the equation must match the real reaction.
| Written as ions | Kept as the formula |
|---|---|
| strong acids: HCl, HNO₃, dilute H₂SO₄ | weak acids: CH₃COOH, H₂CO₃, H₂S, HClO |
| strong bases in solution: NaOH, KOH, Ba(OH)₂, clear limewater Ca(OH)₂ | the weak base NH₃·H₂O; milk of lime (solid Ca(OH)₂); water |
| soluble salts: NaCl, CuSO₄, AgNO₃, Na₂CO₃ | precipitates (AgCl, BaSO₄, CaCO₃, Cu(OH)₂), gases, oxides, elements |
Write the net ionic equations: 1) AgNO₃ + KCl; 2) CaCO₃ (marble) + hydrochloric acid; 3) acetic acid + NaOH solution; 4) dilute H₂SO₄ + Ba(OH)₂ solution.
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2) CaCO₃ + 2H⁺ → Ca²⁺ + CO₂↑ + H₂O (marble is a solid, so it is not split).
3) CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O (acetic acid is weak).
4) Ba²⁺ + 2OH⁻ + 2H⁺ + SO₄²⁻ → BaSO₄↓ + 2H₂O. Two products form at once, and the 1 : 2 ratios of the formulas must be kept: “Ba²⁺ + OH⁻ + H⁺ + SO₄²⁻ → BaSO₄↓ + H₂O” is wrong. As the acid is added, the ions disappear from the solution: a bulb in the circuit goes almost dark at exact neutralisation and glows again when the acid is in excess.
- zthe charge of each ion multiplied by its coefficient
Together with the balance of atoms. In a redox ionic equation the electrons lost must also equal the electrons gained; if you check only the atoms, many wrong options look right.
Correct the equations: 1) Al + 2H⁺ → Al³⁺ + H₂↑; 2) Fe²⁺ + Cl₂ → Fe³⁺ + 2Cl⁻; 3) Zn + Fe³⁺ → Zn²⁺ + Fe²⁺ (little zinc).
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2) Left +2, right +3 − 2 = +1. 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻ (+4 = +6 − 2).
3) Left +3, right +4. Zn gives 2e⁻ and each Fe³⁺ takes 1e⁻: Zn + 2Fe³⁺ → Zn²⁺ + 2Fe²⁺ (+6 = +2 + 4).
Which ionic equation is correct?
A) Copper with dilute sulfuric acid: Cu + 2H⁺ → Cu²⁺ + H₂↑
B) Magnesium hydroxide with hydrochloric acid: H⁺ + OH⁻ → H₂O
C) Copper(II) oxide with dilute sulfuric acid: CuO + 2H⁺ → Cu²⁺ + H₂O
D) Sodium with CuSO₄ solution: 2Na + Cu²⁺ → 2Na⁺ + Cu
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A) Copper is after H in the activity series and does not release hydrogen from dilute acid: no reaction. B) Mg(OH)₂ is a precipitate and must be written as a formula: Mg(OH)₂ + 2H⁺ → Mg²⁺ + 2H₂O. D) Sodium reacts with the water first; then Cu²⁺ + 2OH⁻ → Cu(OH)₂↓, and no copper forms. A and D are balanced but do not match what really happens.
When do ions react? Coexistence of ions
An ion-exchange reaction in solution goes to completion only if the products remove ions from the solution: a precipitate (Ba²⁺ + SO₄²⁻, Ag⁺ + Cl⁻, Ca²⁺ + CO₃²⁻, Cu²⁺ + 2OH⁻), a gas (2H⁺ + CO₃²⁻ → CO₂↑ + H₂O, 2H⁺ + SO₃²⁻ → SO₂↑ + H₂O, and on warming NH₄⁺ + OH⁻ → NH₃↑ + H₂O) or a weak electrolyte (H⁺ + OH⁻ → H₂O, H⁺ + CH₃COO⁻ → CH₃COOH, NH₄⁺ + OH⁻ → NH₃·H₂O). Mixing KNO₃ and NaCl solutions only gives a solution of four ions: no reaction.
Coexistence of ions (离子共存). Ions can be present together in large amounts only if no pair of them forms a precipitate, a gas or a weak electrolyte, and no pair reacts by redox. Questions hide extra conditions in the wording:
- colourless: no Cu²⁺ (blue), Fe³⁺ (yellow-brown), Fe²⁺ (pale green) or MnO₄⁻ (purple);
- acidic (pH < 7, litmus turns red): add H⁺ to every group; it removes OH⁻, CO₃²⁻, HCO₃⁻, SO₃²⁻, S²⁻, CH₃COO⁻ and ClO⁻;
- basic (pH > 7, phenolphthalein turns pink): add OH⁻; it removes H⁺, NH₄⁺, Mg²⁺, Cu²⁺, Fe²⁺, Fe³⁺, Al³⁺ and HCO₃⁻ (HCO₃⁻ survives in neither medium: HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O);
- redox pairs: NO₃⁻ together with H⁺ (dilute nitric acid) oxidises Fe²⁺, I⁻, SO₃²⁻ and S²⁻; MnO₄⁻ oxidises Fe²⁺ and I⁻; Fe³⁺ oxidises I⁻ and S²⁻ (see the lesson “Redox reactions”).
Which group of ions can be present in large amounts in a colourless solution with pH = 13?
A) Na⁺, NH₄⁺, SO₄²⁻, Cl⁻ B) K⁺, Cu²⁺, SO₄²⁻, NO₃⁻ C) K⁺, Na⁺, CO₃²⁻, NO₃⁻ D) Na⁺, K⁺, HCO₃⁻, Cl⁻
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A) NH₄⁺ + OH⁻ → NH₃·H₂O. B) Cu²⁺ is blue and gives Cu(OH)₂↓. D) HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O. In every group, first check for a coloured ion and for the hidden OH⁻.
Tests for ions
An ion test relies on a sign that only this ion gives under the chosen conditions. So every test has two parts: a reagent that gives a clear sign (a precipitate of a certain colour, a gas, a colour) and a way to exclude the ions that could give the same sign.
| Ion | What to do | What you see | Net ionic equation |
|---|---|---|---|
| Cl⁻ | dilute HNO₃, then AgNO₃ solution | white precipitate that does not dissolve in HNO₃ | Ag⁺ + Cl⁻ → AgCl↓ |
| SO₄²⁻ | dilute HCl first (no precipitate, no gas), then BaCl₂ solution | white precipitate | Ba²⁺ + SO₄²⁻ → BaSO₄↓ |
| CO₃²⁻ | CaCl₂ (or BaCl₂) solution, then dilute HCl on the precipitate | white precipitate that dissolves, giving a colourless, odourless gas that turns limewater milky | Ca²⁺ + CO₃²⁻ → CaCO₃↓ |
| NH₄⁺ | NaOH solution, warm | gas turns moist red litmus paper blue; white smoke near a rod dipped in concentrated HCl | NH₄⁺ + OH⁻ → NH₃↑ + H₂O |
| Fe³⁺ | KSCN solution (or NaOH) | blood-red solution (red-brown precipitate with NaOH) | Fe³⁺ + 3SCN⁻ ⇌ Fe(SCN)₃ |
| Fe²⁺ | KSCN, no colour, then chlorine water (or H₂O₂); or NaOH; or K₃[Fe(CN)₆] | red only after oxidation; with NaOH a white precipitate that turns grey-green, then red-brown; blue precipitate with K₃[Fe(CN)₆] | 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻ |
Why the acid first? In the Cl⁻ test, dilute HNO₃ dissolves Ag₂CO₃ (and other silver salts of weak acids), so only AgCl remains. In the SO₄²⁻ test, HCl added first removes CO₃²⁻ and SO₃²⁻ (as gases) and Ag⁺ (as AgCl, seen at once); a white precipitate that appears only after BaCl₂ is then BaSO₄. Nitric acid is not used here: it would oxidise SO₃²⁻ to SO₄²⁻ and create sulfate that was not there. For CO₃²⁻, the first step with CaCl₂ tells it from HCO₃⁻, which gives no precipitate with dilute CaCl₂, and the odourless gas tells it from SO₃²⁻, whose SO₂ also clouds limewater but smells sharp. When several ions are present, remove the disturbing one first: to test for Cl⁻ in a sulfate solution, add excess Ba(NO₃)₂ (not BaCl₂, which brings Cl⁻ itself), filter, then use AgNO₃ and HNO₃.
NaOH solution sorts cations by colour: Mg²⁺ gives a white precipitate, Cu²⁺ blue, Fe³⁺ red-brown, Fe²⁺ white that turns grey-green and then red-brown in air (4Fe(OH)₂ + O₂ + 2H₂O → 4Fe(OH)₃), and NH₄⁺ no precipitate but NH₃ on warming. How iron’s two ions turn into each other is in the lesson “Common metals and their compounds”.
焰色试验)Some metals colour a flame. The colour belongs to the element — sodium metal, NaCl and NaOH all give the same yellow — and the process is a physical change: electrons excited by the heat give out light as they fall back. Older Chinese textbooks called it 焰色反应 (“flame reaction”).
- 1Clean the wire
Dip a platinum (or clean iron) wire into dilute hydrochloric acid and heat it in the outer flame until the flame shows its original colour. HCl is used because chlorides are volatile; sulfates are not.
- 2Pick up the sample
Touch the wire to the solution or the solid and hold it in the outer flame.
- 3Observe the colour
Look at potassium through blue cobalt glass: the glass absorbs the strong yellow light of any sodium traces.
- 4Clean again
Before the next sample, clean the wire with HCl and heat it again.
| Element | Flame colour |
|---|---|
| lithium | crimson |
| sodium | yellow |
| potassium | violet (through blue cobalt glass) |
| calcium | brick-red |
| barium | yellow-green |
| copper | green |
A colourless solution contains one salt. 1) Warming a sample with NaOH solution gives a gas that turns moist red litmus blue. 2) Another sample shows no change with dilute HCl, and then gives a white precipitate with BaCl₂. What is the salt? Write the net ionic equations.
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2) No change with HCl, so no CO₃²⁻, SO₃²⁻ or Ag⁺; then BaSO₄: Ba²⁺ + SO₄²⁻ → BaSO₄↓ ⇒ SO₄²⁻.
The salt is (NH₄)₂SO₄.
Which single reagent can tell apart the four colourless solutions NH₄Cl, (NH₄)₂SO₄, NaCl and Na₂SO₄ (warming is allowed)?
A) NaOH solution B) BaCl₂ solution C) Ba(OH)₂ solution D) AgNO₃ solution
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A) NaOH gives NH₃ with both ammonium salts and nothing with both sodium salts. B) BaCl₂ only separates the sulfates from the chlorides. D) AgNO₃ gives white precipitates with all four (Ag₂SO₄ is only slightly soluble). The idea: combine two tests in one reagent, Ba²⁺ for SO₄²⁻ and OH⁻ for NH₄⁺.
1) A solution contains Fe³⁺. How can you show whether it also contains Fe²⁺?
2) A flame test of a solid gives a yellow flame; through blue cobalt glass the flame looks violet. What does the solid contain?
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2) The yellow shows sodium; the cobalt glass absorbs the yellow light, and the violet seen through it shows potassium as well. Without the glass, the strong yellow of even traces of sodium hides the violet of potassium.
How CSCA asks about this
- Electrolytes: classify substances; BaSO₄ is strong, CO₂ and NH₃ are non-electrolytes, Cu and NaCl solution are neither.
- Ionic equations: pick the correct one by checking the facts, the splitting, the atoms, the charge and the ratios; watch reactions that depend on amounts (CO₂ with NaOH, acid with Na₂CO₃).
- Which reaction a given net equation represents (H⁺ + OH⁻ → H₂O, Ba²⁺ + SO₄²⁻ → BaSO₄↓).
- Coexistence: find the group that can coexist, or the reason why one cannot (precipitate, gas, weak electrolyte, redox), with colour and pH clues.
- Tests: the right reagent and order, what is observed, why an acid goes in first, one reagent that tells several solutions apart, what a set of observations proves and what it leaves open.
- Conductivity: a bulb in Ba(OH)₂ solution goes almost dark as H₂SO₄ is added and glows again when the acid is in excess.
- Time-savers (about 75 s per item): in coexistence items, first strike out options with a coloured ion or an obvious pair (Ba²⁺/SO₄²⁻, Ag⁺/Cl⁻, H⁺/CO₃²⁻, NH₄⁺/OH⁻); in equation items, check the charge first, it takes two seconds.
The test can be taken in English or in Chinese. These are the terms of this lesson in both languages:
| Term | 中文 | Pinyin |
|---|---|---|
| electrolyte | 电解质 | diànjiězhì |
| non-electrolyte | 非电解质 | fēi diànjiězhì |
| strong electrolyte | 强电解质 | qiáng diànjiězhì |
| weak electrolyte | 弱电解质 | ruò diànjiězhì |
| ionisation (dissociation) | 电离 | diànlí |
| ionic equation | 离子方程式 | lízǐ fāngchéngshì |
| precipitate | 沉淀 | chéndiàn |
| white precipitate | 白色沉淀 | báisè chéndiàn |
| coexistence of ions | 离子共存 | lízǐ gòngcún |
| test for ions | 离子检验 | lízǐ jiǎnyàn |
| flame test | 焰色试验 | yànsè shìyàn |
| blue cobalt glass | 蓝色钴玻璃 | lánsè gǔ bōli |
| limewater | 澄清石灰水 | chéngqīng shíhuīshuǐ |
| moist red litmus paper | 湿润的红色石蕊试纸 | shīrùn de hóngsè shíruǐ shìzhǐ |
| dilute nitric acid | 稀硝酸 | xī xiāosuān |
Key points
- Strong electrolytes (strong acids, alkalis, salts, even BaSO₄) ionise completely, weak ones (CH₃COOH, NH₃·H₂O, water) partly; CO₂ and NH₃ are non-electrolytes.
- Write, split, delete, check: only dissolved strong electrolytes are written as ions; atoms and charge must balance.
- An ionic reaction runs when a precipitate, a gas or a weak electrolyte forms; ions that make such a pair cannot coexist, and neither can redox pairs.
- Cl⁻: HNO₃ + AgNO₃; SO₄²⁻: HCl first, then BaCl₂; CO₃²⁻: CaCl₂, then acid and limewater; NH₄⁺: NaOH, warm, moist red litmus.
- Fe³⁺: blood-red with KSCN; Fe²⁺: no colour with KSCN but red after oxidation, or a blue precipitate with K₃[Fe(CN)₆]. Flames: Na yellow, K violet (through cobalt glass), Ca brick-red.
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