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Advanced Inorganic & Organic Chemistry Core Practicals

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  Edexcel IAL Chemistry — Core Practicals

Advanced Inorganic & Organic
Chemistry Core Practicals

The Big Idea: Five practicals, one skill underneath all of them — using colour changes, precipitates, and precise measurements to work out exactly how much of something is present, or exactly what something is, by controlling every variable so the only thing that changes is the chemistry you're trying to observe.

Summary — What This Chapter Covers

  • Core Practical 13a: Redox titration between iron(II) and manganate(VII) — self-indicating, no indicator needed.
  • Core Practical 13b: Redox titration between iodine and thiosulfate — uses starch as an indicator near the end point.
  • Core Practical 12: Preparing a transition metal complex — tetraamminecopper(II) sulfate crystals — and calculating percentage yield.
  • Core Practical 15: Qualitative analysis — identifying unknown positive ions (Group 2 metals, ammonium), negative ions (halides, hydroxide, carbonate, sulfate), and organic functional groups (alkenes, alcohols, carbonyls, carboxylic acids).
  • Core Practical 16: Preparing aspirin from salicylic acid, then purifying it by recrystallisation and checking its purity with a melting point test.
Core Practical 13a

Redox Titration — Iron(II) & Manganate(VII)

Think of a titration as a chemical "weighing scale." You know the exact concentration of one solution (the one in the burette), and you slowly add it to a known volume of the other solution until the reaction is exactly complete. The volume you needed tells you how much of the unknown substance was there.

In a redox titration, instead of an acid neutralising a base, you have an oxidising agent reacting with a reducing agent — electrons are being transferred from one species to the other. The clever part is that many transition metal ions change colour when they change oxidation state, so you often don't even need to add a separate indicator — the reaction indicates itself.

Why Sulfuric Acid, and Nothing Else?

In this titration, manganate(VII) ions (MnO₄⁻) are the oxidising agent — they get reduced to Mn²⁺. Iron(II) is the reducing agent — it gets oxidised to Fe³⁺. This reaction only works in acidic conditions, so excess acid is added to the iron(II) solution before you even start titrating.

But you can't just grab any acid off the shelf. The acid has one job: provide H⁺ ions, and stay completely out of the redox chemistry. Dilute sulfuric acid is the only one that does this cleanly:

AcidWhy it's rejected
Hydrochloric acidChloride ions get oxidised to chlorine by the manganate(VII) — it reacts when it shouldn't.
Nitric acidIt's an oxidising agent itself — it could oxidise the iron(II) directly, ruining the titre.
Ethanoic acidIt's a weak acid — not enough H⁺ ions are released to keep the solution acidic enough.
Concentrated sulfuric acidConcentrated (not dilute) sulfuric acid can act as an oxidising agent too.
The pattern to remember A "safe" acid for a redox titration must be (1) strong enough to be fully acidic, and (2) redox-inert — meaning its own ions can't be oxidised or reduced under the reaction conditions. Dilute sulfuric acid ticks both boxes; nothing else on a typical exam's list does.

Spotting the End Point

Potassium manganate(VII) is intensely purple. As you drip it into the (acidified) iron(II) solution, it reacts instantly and is decolourised — the Mn²⁺ produced is such a pale pink that the solution just looks colourless. This continues until every Fe²⁺ ion has reacted. The very next drop of manganate(VII) has nothing left to react with, so it survives, and the whole flask turns a persistent pale pink. That's your end point.

Practical detail examiners love to test Use a burette with white numbering, not black. Reading black numbers through a deep purple potassium manganate(VII) solution is genuinely difficult — this is a real, examinable practical skill point, not just trivia.

Building and Balancing the Equation

You're given two half-equations and asked to combine them. The rule is always the same: multiply each half-equation so the number of electrons matches, then add them together and cancel the electrons.

Half Equations MnO₄⁻(aq) + 5e⁻ + 8H⁺(aq) → Mn²⁺(aq) + 4H₂O(l)
Fe²⁺(aq) → Fe³⁺(aq) + e⁻   (× 5 to balance electrons)
Overall Ionic Equation MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)

Notice the mole ratio buried in that equation: 1 mole of MnO₄⁻ reacts with 5 moles of Fe²⁺. Every calculation in this practical hinges on that 1:5 ratio.

Worked Example — Analysing an Iron Tablet

Question: An iron tablet weighing 0.960 g was dissolved in dilute sulfuric acid. An average titre of 28.50 cm³ (approximated as 25.0 cm³ in the working below to match the source data) of 0.0180 mol dm⁻³ potassium manganate(VII) solution was needed to reach the end point. What is the percentage by mass of iron in the tablet?

Step 1 — moles of MnO₄⁻ used:
moles = (0.0180 × 25.0) / 1000 = 5.13 × 10⁻⁴ mol

Step 2 — use the 1:5 ratio to get moles of Fe²⁺:
moles of Fe²⁺ = 5 × 5.13 × 10⁻⁴ = 2.565 × 10⁻³ mol

Step 3 — convert moles to mass using Mᵣ(Fe) = 56.0:
mass of Fe = 56.0 × 2.565 × 10⁻³ = 0.14364 g

Step 4 — express as a percentage of the tablet's total mass:
% by mass = (0.14364 / 0.960) × 100 = 15.0%

Practice Question 1

Why must the iron(II) solution be acidified before the titration begins, rather than during it?

Practice Question 2

A student uses hydrochloric acid instead of sulfuric acid to acidify their iron(II) solution. Explain the error this introduces and predict the effect on the calculated iron content.

Core Practical 13b

Redox Titration — Thiosulfate & Iodine

This is a two-stage idea, and it's easy to get lost if you only look at the final titration equation. Here's the full logic chain:

  1. You have an oxidising agent of unknown concentration (e.g. chlorate(I) ions in bleach).
  2. You react it with an excess of iodide ions — the oxidising agent converts iodide into iodine (I₂). The amount of iodine produced is directly linked to how much oxidising agent was present.
  3. You then titrate that iodine against a sodium thiosulfate solution of known concentration. The volume of thiosulfate needed tells you exactly how much iodine was made — and working backwards, how much oxidising agent you started with.
The Titration Reaction 2S₂O₃²⁻(aq) + I₂(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)

Reading the Colour Changes

Iodine solution starts off light brown/yellow. As thiosulfate is added from the burette, the iodine is steadily converted to colourless iodide ions, and the solution gets paler and paler. Once it fades to a pale straw colour, that's your cue to add starch indicator — not at the start, because starch forms such an intense blue-black complex with iodine that it would mask the gradual colour change and make it impossible to judge when you're getting close to the end point.

What you're watching for After starch is added, the solution turns deep blue-black. As you continue adding thiosulfate dropwise, the very last trace of iodine reacts and the blue-black colour disappears suddenly and completely — that sharp, sudden loss of colour is the end point.

Worked Example — Chlorate(I) in Household Bleach

Setup: 10.0 cm³ of bleach was made up to 250.0 cm³. A 25.0 cm³ portion of this diluted solution had 10.0 cm³ of 1.0 mol dm⁻³ potassium iodide added, then was acidified with 1.0 mol dm⁻³ HCl:

ClO⁻(aq) + 2I⁻(aq) + 2H⁺(aq) → Cl⁻(aq) + I₂(aq) + H₂O(l)

This was titrated against 0.05 mol dm⁻³ sodium thiosulfate, average titre = 25.20 cm³:

2S₂O₃²⁻(aq) + I₂(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)

Step 1 — moles of thiosulfate used:
(0.05 × 25.20) / 1000 = 1.26 × 10⁻³ mol

Step 2 — moles of I₂ (and therefore ClO⁻) in the 25.0 cm³ sample — the S₂O₃²⁻ : I₂ ratio is 2 : 1, and the ClO⁻ : I₂ ratio is 1 : 1, so:
1.26 × 10⁻³ / 2 = 6.30 × 10⁻⁴ mol

Step 3 — scale up to the full 250.0 cm³:
6.30 × 10⁻⁴ × 10 = 6.30 × 10⁻³ mol of ClO⁻

Step 4 — scale back to the original 10.0 cm³ of undiluted bleach, then to 1 dm³:
10 cm³ bleach contains 6.30 × 10⁻³ mol ClO⁻ → 1.0 dm³ contains 0.630 mol ClO⁻

Answer: concentration of ClO⁻ in the bleach = 0.630 mol dm⁻³

The universal 7-step method for redox titration calculations 1. Write the half-equations for oxidant and reductant.
2. Combine into the overall equation.
3. Calculate moles of the titrant (manganate/dichromate/thiosulfate) used.
4. Read off the mole ratio from the overall equation.
5. Calculate moles of the substance you're analysing, in the sample solution.
6. Scale up (or down) to the moles in the original solution, if it was diluted.
7. Convert to a final concentration or percentage as the question asks.
Practice Question 1

Why is starch added only once the solution has turned pale straw-coloured, rather than at the very start of the titration?

Practice Question 2

A student forgets to dilute their 10.0 cm³ bleach sample to 250.0 cm³ and instead titrates the iodine produced from the concentrated sample directly. What practical problem would they run into, and why does the dilution step matter for the maths?

Core Practical 12

Preparing a Transition Metal Complex

This practical makes crystals of tetraamminecopper(II) sulfate-1-water, Cu(NH₃)₄SO₄·H₂O, starting from ordinary copper(II) sulfate. The whole thing is really a story about solubility — you dissolve something, add a reagent to build a new, different compound in solution, and then force it back out as a solid by making the solvent "unfriendly" to it.

The Method, Step by Step

  1. Weigh 1.4–1.6 g of copper(II) sulfate accurately (weigh test tube empty, then with the solid — the difference is the mass).
  2. Dissolve it in 4 cm³ of water, warming gently in a water bath.
  3. In the fume cupboard, wearing gloves, add 2 cm³ of concentrated ammonia solution while stirring — the solution turns deep blue as the [Cu(NH₃)₄(H₂O)₂]²⁺ complex ion forms.
  4. Pour this into 6 cm³ of ethanol and cool in an ice bath — dark blue Cu(NH₃)₄SO₄·H₂O crystals precipitate out.
  5. Filter using a Büchner funnel (vacuum filtration — much faster than gravity filtration), wash with cold ethanol.
  6. Dry the crystals between filter paper, then weigh them.
Why does adding ethanol make crystals appear? Ethanol is less polar than water. Ionic compounds like Cu(NH₃)₄SO₄·H₂O are far less soluble in a less polar solvent. As ethanol is mixed in, the solubility of the complex drops sharply, so it has no choice but to come out of solution as solid crystals. This is essentially the same principle behind recrystallisation later in the chapter — change the solvent, change the solubility.
Why cold ethanol, specifically, for washing? Cold ethanol is used because the product is barely soluble in it — the crystals stay solid while impurities wash away. If hot ethanol were used instead, the product itself would start to dissolve and you'd lose yield.

Calculating Percentage Yield

This is a classic "actual vs. theoretical" calculation. You work out the maximum mass you could have made (assuming a perfect 1:1 reaction with no losses), then compare it to what you actually collected.

Percentage Yield % yield = (actual mass obtained ÷ theoretical mass) × 100

Worked Example

Reaction: CuSO₄·5H₂O + 4NH₃ → Cu(NH₃)₄SO₄·H₂O + 4H₂O

A student used 1.5 g of CuSO₄·5H₂O and obtained 1.2 g of dry product.

Step 1 — relative formula masses:
Mᵣ(CuSO₄·5H₂O) = 63.5 + 32.1 + (4×16.0) + 5×((2×1.0)+16.0) = 249.6
Mᵣ(Cu(NH₃)₄SO₄·H₂O) = 63.5 + 4×(14.0+3×1.0) + 32.1 + (4×16.0) + (2×1.0) + 16.0 = 245.6

Step 2 — moles of CuSO₄·5H₂O used:
1.5 / 249.6 = 0.00601 mol

Step 3 — theoretical moles of product (the equation shows a 1:1 molar ratio):
0.00601 mol of Cu(NH₃)₄SO₄·H₂O expected

Step 4 — theoretical mass:
0.00601 × 245.6 = 1.48 g

Step 5 — percentage yield:
(1.2 / 1.48) × 100 = 81%

The classic "over 100% yield" trap If a student's crystals aren't fully dried, leftover water adds extra mass, making the actual yield look bigger than it should be — sometimes pushing the calculated percentage yield above 100%, which is physically impossible. If you ever see this happen in your own results, incomplete drying is almost always the reason.

Safety in This Practical

HazardWhy it matters
Concentrated ammoniaCorrosive and dangerous to the environment — must be used in the fume cupboard, wearing gloves. It also releases toxic ammonia gas.
Copper saltsHarmful and dangerous to the environment.
EthanolFlammable — keep away from naked flames.
Practice Question 1

Explain, in terms of solubility, why cooling the mixture in an ice bath after adding ethanol improves the yield of crystals.

Practice Question 2

Suggest two reasons why a student's percentage yield in this experiment might be lower than 100%, other than incomplete drying.

Core Practical 15

Qualitative Analysis of Inorganic & Organic Unknowns

This practical is a toolkit of "detective tests." You're handed an unknown substance, and each test gives you a clue — a colour change, a precipitate, a smell, a gas — that lets you rule ions and functional groups in or out. The trick to mastering this section isn't memorising every single line; it's understanding why each reagent produces the result it does, so you can reconstruct the table under exam pressure even if your memory blanks.

Testing for Positive Ions: Group 2 Metals

Add sodium hydroxide (or ammonia solution, or sulfuric acid) dropwise, in excess, to a solution of the unknown metal ion, and watch for a precipitate.

ReagentMg²⁺Ca²⁺Sr²⁺Ba²⁺
Ammonia solutionWhite ppt — Mg(OH)₂No changeNo changeNo change
Excess NaOHWhite ppt — Mg(OH)₂White ppt — Ca(OH)₂Slight white ppt — Sr(OH)₂No change
Excess H₂SO₄Colourless (no ppt)Slight white ppt — CaSO₄White ppt — SrSO₄White ppt — BaSO₄
The pattern hiding in this table Notice the trends run in opposite directions down Group 2: hydroxide solubility decreases down the group (so precipitates get less obvious going Mg → Ba with NaOH), while sulfate solubility also decreases down the group (so sulfate precipitates get more obvious going Mg → Ba with sulfuric acid). If you remember the direction of each trend, you can rebuild this whole table from scratch.

Testing for Ammonium Ions (NH₄⁺)

Add sodium hydroxide to the unknown solution, then gently warm it in a water bath. Ammonium ions react with hydroxide to release ammonia gas: NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l). Hold a piece of damp red litmus paper near (not in) the mouth of the tube using tongs.

Positive Result Damp red litmus paper turns blue — confirms ammonia gas, and therefore ammonium ions.

Testing for Negative Ions

Halide Ions (Cl⁻, Br⁻, I⁻)

Add dilute nitric acid (to remove interference from other ions like carbonate), then add silver nitrate solution.

Halide presentPrecipitate colourFormulaBehaviour with ammonia
Chloride, Cl⁻WhiteAgClDissolves in dilute ammonia
Bromide, Br⁻CreamAgBrDissolves only in concentrated ammonia
Iodide, I⁻YellowAgIDoes not dissolve, even in concentrated ammonia
Why the ammonia follow-up test matters Cream and pale yellow can be hard to tell apart by eye alone. The ammonia solubility test is a second, independent check that removes any ambiguity — if you're not sure whether you're looking at AgBr or AgCl, dilute ammonia settles it instantly.

Hydroxide Ions (OH⁻)

Test the pH directly with red litmus paper or universal indicator. A positive result: red litmus turns blue, or the universal indicator paper reads clearly alkaline.

Carbonate Ions (CO₃²⁻)

Add dilute hydrochloric acid, then immediately attach a bung and delivery tube leading into a second test tube containing limewater (calcium hydroxide solution).

Positive Result Effervescence in the first tube (CO₂ gas evolved) + the limewater in the second tube turns cloudy/milky as insoluble CaCO₃ forms.

Sulfate Ions (SO₄²⁻)

Acidify the sample with dilute hydrochloric acid first, then add a few drops of aqueous barium chloride.

Reaction Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)   (white precipitate)
Why acidify with HCl before adding barium chloride? Without the acid, barium ions would also form a precipitate with any carbonate ions present (BaCO₃), giving a false positive for sulfate. The HCl reacts with and removes any carbonate first, so any precipitate that forms afterwards can only be barium sulfate — a clean, unambiguous result.

Testing for Organic Functional Groups

Alkenes / Unsaturation — Bromine Water Test

Shake the unknown compound with orange/yellow bromine water. If a carbon–carbon double bond is present, an addition reaction occurs and the bromine is used up.

Positive Result Orange bromine water is decolourised (turns colourless).

Alcohols — PCl₅ Test

Add solid phosphorus(V) chloride. If an −OH group is present, a vigorous reaction occurs at room temperature, with no heating needed.

Positive Result Steamy white fumes of HCl gas are evolved.

Distinguishing 1° / 2° Alcohols from Tertiary Alcohols — Oxidation Test

Warm the unknown with acidified potassium dichromate(VI), or acidified potassium manganate(VII).

Alcohol typeBehaviour
Primary / secondaryGets oxidised (to an aldehyde/ketone). Dichromate(VI): orange → green. Manganate(VII): purple → colourless.
TertiaryCannot be oxidised this way — no colour change at all.

Carbonyls (General) — 2,4-DNPH Test

2,4-dinitrophenylhydrazine (2,4-DNPH) undergoes a condensation reaction with the carbonyl group (C=O) in both aldehydes and ketones.

Positive Result A deep-orange precipitate forms. It can then be recrystallised and its melting point compared against data book values to identify the exact aldehyde/ketone.

Distinguishing Aldehydes from Ketones

Both Tollens' reagent and Fehling's solution rely on the same underlying idea: aldehydes can be oxidised further (to a carboxylic acid); ketones cannot.

ReagentWith an aldehydeWith a ketone
Tollens' reagent (ammoniacal AgNO₃)Silver mirror forms — Ag⁺ is reduced to Ag metal as the aldehyde is oxidised.No reaction — no silver mirror.
Fehling's/Benedict's solutionBrick-red precipitate of Cu₂O forms — Cu²⁺ reduced to Cu⁺.No reaction — stays clear blue.

Carboxylic Acids

Add solid sodium carbonate, or aqueous sodium hydrogen carbonate.

Positive Result Effervescence — bubbles of CO₂ gas are evolved.
Practice Question 1

A student adds excess sodium hydroxide to an unknown Group 2 metal chloride solution and sees no precipitate at all. They then test with excess sulfuric acid and see a dense white precipitate immediately. Identify the metal ion and justify your answer.

Practice Question 2

An unknown liquid gives a positive result with 2,4-DNPH (orange precipitate) but no silver mirror with Tollens' reagent, and no colour change when warmed with acidified potassium dichromate(VI). What type of compound is it, and how do you know?

Core Practical 16

Aspirin Preparation

This practical brings together several separate skills you've built up across the chapter: accurate measurement, controlled heating, purification by recrystallisation, and using a melting point to check purity. It's really two experiments glued together — first you make the aspirin, then you purify and verify it.

Stage 1 — Synthesis

  1. Add 6.0 g of salicylic acid to a conical flask with 10 cm³ of ethanoic anhydride and 5 drops of concentrated sulfuric acid (a catalyst).
  2. Swirl and hold in a warm water bath at around 60°C for about 20 minutes.
  3. Cool the flask, then pour the contents into 75 cm³ of cold water — the aspirin crystallises out.
  4. Recover the crude aspirin by Büchner (vacuum) filtration, and leave to dry.
Why 60°C, not higher? This reaction needs gentle, controlled heating — hot enough to speed the reaction up, but not so hot that side reactions or decomposition of reagents/products become a problem. This is a common exam-style question: always link a specific temperature choice back to reaction rate vs. avoiding unwanted side effects.

Stage 2 — Recrystallisation (Purification)

The crude aspirin you've collected is impure — it contains left-over reactants and by-products. Recrystallisation is the standard technique for purifying an organic solid, and the logic is elegant:

  1. Dissolve the impure solid in the minimum volume of hot solvent (here, ethanol) needed to fully dissolve it.
  2. If any insoluble impurities remain, filter the hot solution (a "hot filtration").
  3. Let the solution cool slowly to room temperature — the desired product crystallises out, but the impurities (present in much smaller amounts) stay dissolved in the solvent.
  4. Filter again (typically Büchner filtration, since it's faster — filtration under reduced pressure) to collect the pure crystals.
  5. Wash with a small amount of fresh, cold solvent to remove any remaining surface impurities, then dry.
The Core Principle of Recrystallisation Choose a solvent in which the product is soluble hot but only slightly soluble cold — so cooling forces the pure product out while impurities (present in smaller quantities) stay dissolved.
The minimum-solvent rule — a favourite exam trap Using too much hot solvent is a genuine mistake, not just a technicality: if there's more solvent than necessary, more of your product will still be dissolved even after cooling, and you'll lose yield through the filter. Always add just enough hot solvent to dissolve the solid — no more.

Stage 3 — Melting Point Analysis

The melting point of a solid is a fingerprint for both its identity and its purity. Pure aspirin melts sharply at 135°C.

ObservationWhat it tells you
Melting point matches the literature value closelyProduct is likely to be pure aspirin.
Melting point is lower than the literature valueImpurities are present — impurities characteristically lower a substance's melting point.
Melting occurs over a wide temperature rangeThe sample is impure — pure substances melt sharply over a very narrow range; impure ones melt gradually over a broad range.

Practical Skills for a Good Melting Point Test

  • The sample must be completely dry and finely powdered — crush it with the back of a spatula on filter paper or a white tile to absorb any remaining moisture.
  • Do a quick first run to find the approximate melting range, heating fairly fast.
  • Repeat with a much slower heating rate for an accurate reading — heating too fast means the thermometer reading lags behind the true temperature of the sample, giving you an inaccurately high value.
  • Take repeat readings (three is standard) and quote the result as a range, ideally referenced against a data book value.
Recrystallisation trade-off worth remembering You can recrystallise a sample more than once for extra purity, but every repeat loses some product to the solvent — so purity and yield genuinely trade off against each other. Similarly, cooling the solution slowly (rather than quickly) produces larger, well-defined crystals that are easier to filter and dry.
Practice Question 1

A student measures the melting point of their recrystallised aspirin as 128–133°C, compared to the literature value of 135°C. What does this tell you about the purity of their product, and why does an impure solid melt over a range rather than at one exact temperature?

Practice Question 2

Explain why the crystals are washed with cold ethanol (rather than hot ethanol, or water) as the final step of recrystallisation.


What to Memorise

Manganate(VII) titration equation

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ (mole ratio MnO₄⁻:Fe²⁺ = 1:5)

Thiosulfate–iodine equation

2S₂O₃²⁻ + I₂ → 2I⁻ + S₄O₆²⁻ (mole ratio S₂O₃²⁻:I₂ = 2:1)

Why sulfuric acid, not others

HCl gets oxidised; HNO₃ is itself an oxidiser; ethanoic acid is too weak; conc. H₂SO₄ can oxidise too.

Halide precipitate colours

Cl⁻ → white (AgCl); Br⁻ → cream (AgBr); I⁻ → yellow (AgI). Test with AgNO₃ after acidifying with HNO₃.

Sulfate test

Acidify with HCl first (removes carbonate), then add BaCl₂ → white BaSO₄ precipitate.

Carbonate test

Add dilute HCl → CO₂ gas evolved → turns limewater milky/cloudy.

Alkene test

Shake with bromine water → orange decolourises if C=C double bond present.

Alcohol tests

PCl₅ → steamy HCl fumes confirms −OH. Dichromate(VI)/manganate(VII): 1°/2° alcohols oxidised (colour change); 3° alcohols unaffected.

Aldehyde vs ketone

Tollens': aldehyde gives silver mirror, ketone gives none. Fehling's: aldehyde gives brick-red ppt, ketone gives none.

Carboxylic acid test

Add Na₂CO₃ or NaHCO₃ → effervescence (CO₂ gas evolved).

Percentage yield formula

% yield = (actual mass obtained ÷ theoretical mass) × 100

Recrystallisation core principle

Dissolve in minimum hot solvent → hot filter if needed → cool slowly to crystallise → filter → wash with cold solvent → dry.


Concepts Checklist


Exam Tips — Common Mistakes & Mark-Scheme Traps

Mistake: forgetting to use the mole ratio Students often calculate moles of titrant correctly, then forget to scale by the reaction's mole ratio (e.g. 1:5 for manganate/iron, or 2:1 for thiosulfate/iodine) before converting to a final answer. Always write the balanced equation first and underline the ratio before you touch a calculator.
Mistake: vague acid/reagent explanations "It reacts with it" is never enough for the marks. Examiners want the specific redox behaviour — e.g. "chloride ions would be oxidised to chlorine by the manganate(VII) ions," not just "HCl would interfere." Always name the species and the type of reaction (oxidised/reduced) explicitly.
Mistake: mixing up which colour change belongs to which test Dichromate(VI) goes orange → green. Manganate(VII) goes purple → colourless. Fehling's gives a brick-red precipitate. Tollens' gives a silver mirror. These four are commonly swapped under exam pressure — write them out as a table during revision and test yourself blind.
Mistake: not linking impurities to melting point in enough detail Don't just say "impurities lower the melting point" — for full marks, explain why: impurities disrupt the regular lattice structure, weakening the intermolecular/ionic forces holding it together, which is why the solid starts melting at a lower temperature and over a wider range.
Mistake: describing recrystallisation vaguely A full-marks answer names the specific steps in order: dissolve in minimum hot solvent, hot filter if needed, cool slowly, filter (often specifying Büchner/vacuum filtration), wash with a small volume of cold solvent, dry. Missing "minimum" or "slowly" is a very common way to drop marks.
Typical question patterns to expect
  • "Explain why [acid X] is unsuitable for this titration" — always link back to redox interference, not just general reactivity.
  • Full titration calculations requiring the 7-step method (moles → ratio → scale-up → final answer).
  • "Identify the ion(s) present" from a table of test results — work through each test's logic, don't guess.
  • "Suggest why the percentage yield is greater than 100%" — always link to incomplete drying/impure product.
  • "Describe how you would purify this compound and check its purity" — full method plus melting point interpretation.
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Also in the full note
  • Redox Titration — Iron(II) & Manganate(VII)
  • Redox Titration — Thiosulfate & Iodine
  • Qualitative Analysis of Inorganic & Organic Unknowns
  • Exam Tips — Common Mistakes & Mark-Scheme Traps
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