IGCSE Chemistry Topic 12: Experimental Techniques and Chemical Analysis

12.1 Apparatus & Experimental Design 12.2 Acid–Base Titrations 12.3 Chromatography 12.4 Separation, Purity & Identification

Hey Tara! Welcome to Topic 12 — the practical heart of IGCSE Chemistry. Four stories here. One: apparatus — what measures time, temperature, mass and volume, which instrument is the precise one, and how to collect and dry a gas without losing it. Two: the titration — a ritual with burette, pipette and indicator that examiners ask about every single year, right down to reading the burette to 0.05 cm³. Three: chromatography — pencil baselines, solvent fronts and the R_f value that is always less than 1. Four: the big one — separating mixtures, judging purity from melting and boiling points, and the full identification toolkit: flame colours, hydroxide precipitates, anion tests and gas tests. Learn the tables in section 12.4 cold and Paper 6 turns into a treasure hunt where you already have the map. Let's go!

12.1 Apparatus & Experimental Design ▼

The Big Idea: Every Measurement Needs the Right Instrument

Before any experiment can be trusted, four things usually have to be measured: time, temperature, mass and volume. The first three are easy — one instrument each. Volume is where the exam marks live, because there are three instruments for measuring the volume of a liquid, and choosing between them is a judgement about precision.

Quantity	Apparatus	Notes for the exam
Time	Stopwatch / stop clock	Read to the nearest second (digital ones to 0.01 s, but human reaction time makes that precision illusory)
Temperature	Thermometer	Typical laboratory thermometer reads to 1 °C or 0.5 °C; read at eye level
Mass	Electronic balance	Reads to 0.1 g or 0.01 g; remember to subtract the mass of the container (tare)
Volume of liquid	Burette, volumetric pipette, measuring cylinder	See the precision table below — this choice is a favourite exam question

The Three Volume Instruments — Precision Compared

Apparatus	What it does	Precision	When to choose it
Burette	Delivers a variable volume, drop by drop, through a tap	Read to the nearest 0.05 cm³ — very precise	Titrations — any time you need to add liquid gradually and know exactly how much went in
Volumetric pipette	Delivers one fixed volume (e.g. exactly 25.0 cm³) with high accuracy	Typically ±0.06 cm³ on a 25 cm³ pipette — very precise	Measuring out the fixed sample of solution for a titration; always used with a safety filler
Measuring cylinder	Measures an approximate variable volume	Read to about 0.5–1 cm³ — the least precise of the three	When the exact volume does not matter, e.g. adding an excess of acid

The Precision Question — How to Answer It

"Name the most suitable piece of apparatus to measure 23.7 cm³ of solution" — a variable, precise volume needs a burette. "Exactly 25.0 cm³, the same every time" — a pipette. "About 50 cm³ of dilute acid (an excess)" — a measuring cylinder is fine. The keywords are variable vs fixed and precise vs approximate. When reading any of them, read the bottom of the meniscus at eye level.

Collecting Gases: Three Questions Decide the Method

When a reaction makes a gas, you must collect it — and the collection method is chosen by asking: Is the gas soluble in water? Is it denser or less dense than air? Do I need to measure its volume? The four standard set-ups are below.

Supplement

Judging density from relative molecular mass

Air behaves as if its relative molecular mass were about 29 (a 78:21 mix of N₂ = 28 and O₂ = 32). A gas with M_r below 29 is less dense than air (H₂ = 2, NH₃ = 17, CH₄ = 16); above 29, denser (CO₂ = 44, SO₂ = 64, Cl₂ = 71). This single comparison answers every "which collection method?" question with a reason attached — and reasons are what the supplement marks pay for.

Drying a Gas

Gases collected from aqueous reactions carry water vapour with them. To dry a gas, bubble it through or pass it over a drying agent that does not react with it: concentrated sulfuric acid (for most gases, but never ammonia — the acid would react with the alkaline gas), anhydrous calcium chloride, or calcium oxide (the one to use for ammonia, since it is a base and will not react with it). Choosing a drying agent is a compatibility question: acid gas → do not use a basic drying agent; alkaline gas (NH₃) → do not use an acidic one.

The Language of Solutions

solute + solvent → solution

Solute — the substance that dissolves (e.g. the salt). Solvent — the liquid that does the dissolving (e.g. the water). Water is the most common solvent; ethanol and propanone are common non-aqueous solvents. Solution — the uniform mixture formed. A substance that dissolves is soluble; one that does not is insoluble. Saturated solution — a solution containing the maximum amount of dissolved solute at that temperature, in the presence of undissolved solute. The phrase "at that temperature" is a marking point — solubility of most solids increases with temperature.

Solubility Is Temperature's Servant

Most solids dissolve better in hot solvent. That single fact explains crystallisation (cool a hot saturated solution and crystals appear, because the cold solution cannot hold as much solute) and it explains why a "saturated at 60 °C" solution dumps solid when cooled to 20 °C. Gases are the opposite — they dissolve less in warm water — but for Topic 12 the solids rule is the one you use.

Worked Example 1 A student in Mumbai investigates the rate of reaction between marble chips and dilute hydrochloric acid by measuring the volume of carbon dioxide produced every 30 seconds. Name the apparatus needed to (a) measure 50 cm³ of acid, (b) measure the mass of marble chips, (c) time the reaction, (d) collect and measure the gas — and justify each choice. [4]

Step 1: The acid — how precise does it need to be?

The acid is in excess — its exact volume does not change the final volume of gas. So a measuring cylinder is sufficient. Choosing a burette here is not wrong chemistry, but the question asks for the most suitable apparatus, and precision you do not need is effort wasted.

Step 2: Mass and time — the easy pair

Mass of marble chips: electronic balance (tare the container first). Timing: stopwatch, started the moment the reactants meet.

Step 3: The gas — the word "measure" decides it

The question says measure the volume of gas, so a gas syringe is the best answer. (Collecting over water into an inverted measuring cylinder also works for CO₂ in rate experiments, though CO₂ is slightly soluble, so the syringe is superior.)

Step 4: Assemble the justifications

Every choice must carry its reason: measuring cylinder — approximate volume is acceptable for an excess; balance — mass; stopwatch — time; gas syringe — measures gas volume directly and works even for soluble gases.

(a) Measuring cylinder — the acid is in excess so an approximate volume is acceptable [1]. (b) Electronic balance [1]. (c) Stopwatch [1]. (d) Gas syringe — it measures the volume of gas collected directly (and does not lose soluble gas, unlike collection over water) [1].

Worked Example 2 State and explain the best method of collection for each gas: (a) ammonia, NH₃; (b) hydrogen, H₂; (c) chlorine, Cl₂. [6]

Step 1: Ask the two questions for each gas

Question one: is it soluble in water? If insoluble → collect over water. If soluble, question two: denser or less dense than air? Less dense → upward delivery (into an inverted, mouth-down tube); denser → downward delivery (into an upright, mouth-up tube).

Step 2: Ammonia

NH₃ is very soluble in water — collecting it over water would simply dissolve it. Its M_r = 17 < 29, so it is less dense than air: upward delivery.

Step 3: Hydrogen

H₂ is insoluble in water, so collection over water works beautifully and shows you how much has been collected. (Upward delivery also works — H₂ is the least dense gas of all — but over water is the standard answer because you can see the gas displacing the water.)

Step 4: Chlorine

Cl₂ is soluble in water (it forms chlorine water), so not over water. M_r = 71 >> 29: denser than air, so downward delivery — the gas sinks and fills an upright gas jar. (Do it in a fume cupboard: chlorine is toxic.)

(a) NH₃: upward delivery [1] — very soluble in water and less dense than air [1]. (b) H₂: over water [1] — insoluble in water (and the volume collected is visible) [1]. (c) Cl₂: downward delivery [1] — soluble in water and denser than air [1].

Worked Example 3 A student adds potassium nitrate to 100 cm³ of water at 60 °C, stirring, until solid remains undissolved at the bottom. She filters off the excess solid and lets the clear solution cool to 20 °C. Crystals form. Explain, using the terms solute, solvent, solution and saturated, everything that happened. [4]

Step 1: Name the actors

Potassium nitrate is the solute; water is the solvent; the mixture formed is the solution.

Step 2: Recognise saturation

When no more will dissolve and solid sits undissolved at the bottom, the solution is saturated at 60 °C — it holds the maximum possible solute at that temperature.

Step 3: What cooling does

Solubility of potassium nitrate decreases as temperature falls. At 20 °C the water can hold much less solute, so the solution becomes supersaturated for an instant and the excess solute crystallises out.

Step 4: Tie it together

This is exactly how crystallisation as a purification method works (section 12.4): hot saturated solution → cool → crystals of the pure salt, while soluble impurities in small amounts stay dissolved.

KNO₃ (the solute) dissolves in water (the solvent) to form a solution [1]. When undissolved solid remains, the solution is saturated — it contains the maximum solute at 60 °C [1]. On cooling, the solubility decreases [1], so the solution can no longer hold all the dissolved solute and the excess forms crystals [1].

Exam Tips for 12.1

1. Match precision to purpose. Burette = variable + precise (0.05 cm³). Pipette = one fixed volume, very accurate. Measuring cylinder = approximate. If the volume "doesn't need to be exact", say measuring cylinder and win the mark for judgement.

2. Learn the burette's precision number. Burettes are read to the nearest 0.05 cm³ — so a burette reading always has two decimal places, ending in 0 or 5 (e.g. 24.35, 24.40).

3. Gas collection is a two-question decision. Insoluble → over water. Soluble and lighter than air → upward delivery. Soluble and heavier than air → downward delivery. Need the volume → gas syringe.

4. "Upward delivery" = tube mouth pointing DOWN. The gas travels upward into an inverted tube. Downward delivery = tube mouth up. Name the method, and if asked to draw it, the tube orientation carries the mark.

5. Ammonia is the upward-delivery poster child. Too soluble for over-water collection, lighter than air. Chlorine, sulfur dioxide and hydrogen chloride are the downward-delivery trio.

6. Drying agents must not react with the gas. Concentrated H₂SO₄ dries most gases but NEVER ammonia; calcium oxide dries ammonia.

7. Saturated needs "at that temperature". A saturated solution contains the maximum dissolved solute at that temperature — omit the phrase, lose the mark.

8. Read all liquid levels at the bottom of the meniscus, at eye level. This sentence earns marks in apparatus questions across Papers 4 and 6.

🔬 Apply It: Real-World Chemistry

From a school lab in Chennai to a brewery in Munich, choosing the right instrument and the right collection method is the difference between data and guesswork.

During a practical exam in Chennai, a student needs exactly 25.0 cm³ of sodium hydroxide solution in a conical flask, then must add roughly 20 cm³ of distilled water to dilute it, and finally add acid gradually until the indicator changes. She has a pipette, a burette and a measuring cylinder on her bench.

Match each of the three volume tasks to the correct instrument, with reasons.

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Fixed, Variable-Precise, Approximate

Exactly 25.0 cm³, the same volume every time: volumetric pipette — built to deliver one fixed volume with high accuracy. Adding acid gradually while tracking exactly how much has gone in: burette — variable volume, read to 0.05 cm³, with a tap for drop-by-drop control. The dilution water: measuring cylinder — "roughly 20 cm³" needs no precision (adding water does not change the moles of NaOH present).

Why the Dilution Doesn't Need Precision

The titration measures amount of substance, not concentration in the flask. Extra water changes the volume but not the moles of NaOH, so the end-point volume of acid is unaffected. Recognising which measurement matters is the real skill being tested.

Chemistry Connection

This exact three-way choice is the most common apparatus question on Paper 6. The examiners' logic: precision costs time and effort, so a good experimenter is precise only where precision changes the answer.

A chemistry teacher in Leeds demonstrates the reaction of ammonium chloride with calcium hydroxide, which produces ammonia gas. A student suggests collecting the ammonia over water "like we did for hydrogen last week". The teacher smiles and instead clamps a dry test tube upside down over the delivery tube.

Why would the student's method have collected almost nothing? Why is the teacher's method correct, and why must the collecting tube be dry?

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Ammonia and Water Are Best Friends

Ammonia is extremely soluble in water — hundreds of volumes of gas dissolve in one volume of water. Bubbled through a trough, virtually all of it would dissolve before reaching the collecting tube. Collection over water only works for insoluble gases like H₂ and O₂.

Upward Delivery, Dry Glassware

NH₃ (M_r = 17) is less dense than air (~29), so it rises into the inverted tube and pushes the air out downwards — upward delivery. The tube must be dry because any film of water would start dissolving the gas you are trying to trap.

Chemistry Connection

The same solubility that ruins over-water collection is what makes the ammonia fountain demonstration so dramatic, and it is why the NH₃ gas test uses damp litmus — the gas must dissolve in the water film to act as an alkali. One property, three exam appearances.

At a water-testing laboratory in Singapore, a technician measures how much dissolved solid is in a sample by evaporating exactly 100.0 cm³ of water and weighing the residue on a balance reading to 0.001 g. Her trainee suggests speeding things up by estimating the water volume with a beaker's printed graduations and using the classroom balance that reads to 1 g.

The residue from typical samples is about 0.35 g. Explain why the trainee's shortcuts would destroy the measurement.

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When the Quantity Is Tiny, Precision Is Everything

A balance reading to 1 g cannot see 0.35 g at all — it would read 0 g or 1 g. The measurement needs an instrument whose precision is much smaller than the quantity measured. The 0.001 g balance gives 0.350 g — three meaningful figures.

The Volume Matters Just as Much

The final answer is "grams of solid per 100 cm³ of water", so an error in the water volume feeds straight into the result. Beaker graduations can be out by 5–10%; a pipette or burette delivers the volume to a fraction of a percent.

Chemistry Connection

Choosing apparatus is really about matching the instrument's precision to the size of the quantity and the purpose of the experiment — the exact judgement Paper 6 asks for when it says "suggest a more suitable piece of apparatus and explain why".

A pharmaceutical plant near Hyderabad synthesises a compound whose next reaction step is destroyed by moisture. The hydrogen chloride gas feeding the reactor is first bubbled through concentrated sulfuric acid. In the same plant, a different line dries ammonia for a separate process — but that line uses towers packed with calcium oxide instead.

Why do the two gas lines use different drying agents? What would go wrong if they were swapped?

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A Drying Agent Must Be a Bystander

A drying agent must absorb water only, without touching the gas. Concentrated sulfuric acid is an acid: fine for the acidic gas HCl, but it would react with and swallow the alkaline gas ammonia (forming ammonium sulfate).

Matching by Character

Calcium oxide is a base: it dries ammonia safely but would react with HCl. The rule: acidic gas → acidic (or neutral) drying agent; alkaline gas → basic drying agent. Anhydrous calcium chloride is the general-purpose neutral option for most gases.

Chemistry Connection

This is acid–base chemistry (Topic 7) hiding inside a techniques question. The exam version: "explain why concentrated sulfuric acid cannot be used to dry ammonia" — answer: it is an acid and would react with the alkaline ammonia. One sentence, one mark, pure logic.

In the Rann of Kutch, salt farmers flood shallow pans with seawater and let the fierce sun do the rest. As the water evaporates through the summer, nothing happens for weeks — then, quite suddenly, salt crystals begin to appear across the whole pan and keep growing until harvest.

Using the idea of a saturated solution, explain why the crystals appear only after weeks of evaporation, and all at once.

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Below Saturation, Nothing Can Crystallise

Seawater is a dilute solution — far less salt than the water could hold. As water evaporates, the same solute is dissolved in less and less solvent, so the solution becomes steadily more concentrated, but stays clear.

The Saturation Threshold

Crystals can only form once the solution is saturated — holding the maximum salt possible at that temperature. From that moment, every extra drop of water that evaporates leaves salt with nowhere to stay dissolved, so crystals form continuously across the pan.

Chemistry Connection

Salt pans are crystallisation-by-evaporation on an industrial scale — the same principle as the school practical where you evaporate a solution to the point of crystallisation and then let it cool. Saturation is the tipping point in both.

Practice Questions: 12.1

20 multiple choice questions. Click an option to check your answer.

Your Score 0 / 20

Question 1

Which piece of apparatus measures a variable volume of liquid to the nearest 0.05 cm³?

A Measuring cylinder

B Burette

C Volumetric pipette

D Conical flask

A burette delivers any chosen volume and is read to 0.05 cm³. A pipette is equally precise but delivers only one fixed volume; a measuring cylinder is approximate; a conical flask does not measure at all.

Question 2

A student needs exactly 25.0 cm³ of alkali, the same volume in every repeat. The best apparatus is a

A measuring cylinder

B beaker with graduations

C volumetric pipette

D gas syringe

A volumetric pipette is designed to deliver one fixed volume (e.g. exactly 25.0 cm³) with high accuracy — the standard way to measure the sample for a titration.

Question 3

Which gas is best collected by upward delivery (into an inverted tube)?

A Chlorine

B Carbon dioxide

C Ammonia

D Sulfur dioxide

Upward delivery suits gases less dense than air: ammonia (M_r = 17 < 29) rises into the inverted tube. Cl₂, CO₂ and SO₂ are all denser than air — downward delivery.

Question 4

Collection of a gas over water is only suitable if the gas is

A denser than air

B coloured

C insoluble in water

D less dense than air

The gas bubbles up through water into the collecting vessel, so it must not dissolve on the way. Insoluble (or nearly insoluble) gases like H₂ and O₂ qualify; ammonia emphatically does not.

Question 5

Which method should be used to collect chlorine gas?

A Over water

B Upward delivery

C Downward delivery

D Through concentrated sulfuric acid

Chlorine is soluble in water (rules out A) and denser than air (M_r = 71), so it sinks into an upright vessel: downward delivery. D describes drying, not collecting.

Question 6

To follow the rate of a reaction by measuring the volume of gas produced against time, the best collection apparatus is a

A gas syringe

B gas jar by downward delivery

C inverted test tube by upward delivery

D balloon

A gas syringe has a graduated scale, so the volume can be read at intervals — exactly what a rate experiment needs. Delivery methods collect gas but do not measure it.

Question 7

In the mixture formed when sugar dissolves in tea, the sugar is the

A solvent

B solute

C solution

D residue

The substance that dissolves is the solute; the liquid doing the dissolving (the water) is the solvent; together they make the solution.

Question 8

A saturated solution is one that

A contains a large amount of solute

B cannot dissolve any solute at any temperature

C contains the maximum amount of dissolved solute at that temperature

D has equal masses of solute and solvent

The definition needs both parts: maximum dissolved solute and at that temperature. Warm the solution and it can usually dissolve more — which is why B is wrong.

Question 9

Which is the most appropriate apparatus for adding approximately 50 cm³ of dilute acid, in excess, to a reaction mixture?

A Burette

B 25 cm³ pipette used twice

C Measuring cylinder

D Teat pipette (dropper)

The acid is in excess — its exact volume does not matter — so the approximate measuring cylinder is the appropriate choice. Precision instruments are unnecessary here.

Question 10

Why must ammonia NOT be dried by bubbling it through concentrated sulfuric acid?

A The acid is not a drying agent

B The acid would react with the alkaline ammonia

C Ammonia is insoluble in the acid

D The acid would dilute the ammonia

Ammonia is an alkaline gas; concentrated sulfuric acid would neutralise (react with) it, absorbing the gas itself. Use the basic drying agent calcium oxide instead.

Question 11

Which is a correctly recorded burette reading?

A 24.3 cm³

B 24.33 cm³

C 24.35 cm³

D 24 cm³

Burettes are read to the nearest 0.05 cm³, so readings have two decimal places ending in 0 or 5. 24.33 pretends to a precision the scale does not have; 24.3 and 24 are under-recorded.

Question 12

Hydrogen can be collected over water because it is

A less dense than air

B insoluble in water

C flammable

D colourless

The property that permits over-water collection is insolubility. Hydrogen's low density is true but irrelevant to this method (it explains why upward delivery would also work).

Question 13

Air behaves as though its relative molecular mass is about 29. Which gas would sink in air and so be collected by downward delivery?

A Methane (Mᵣ = 16)

B Ammonia (Mᵣ = 17)

C Hydrogen (Mᵣ = 2)

D Sulfur dioxide (Mᵣ = 64)

A gas denser than air has M_r greater than ~29. Only SO₂ (64) qualifies — it sinks and fills an upright vessel. The others are all lighter than air.

Question 14

When reading the volume of a liquid in a burette, you should read

A the bottom of the meniscus, at eye level

B the top of the meniscus, at eye level

C the bottom of the meniscus, from above

D either edge of the liquid surface

Standard technique: eye level with the liquid, read the bottom of the curved meniscus. Reading from above or below introduces parallax error.

Question 15

A solution is saturated at 60 °C. What is most likely to happen when it cools to 20 °C?

A More solute dissolves

B Crystals of solute form

C The solvent evaporates

D Nothing changes

Solubility of most solids falls as temperature falls. The cooled solution cannot hold all its solute, so the excess crystallises out — the basis of crystallisation as a purification method.

Question 16

Which combination is correct for measuring the quantities in an experiment?

A Time — thermometer; temperature — stopwatch

B Mass — balance; temperature — thermometer; time — stopwatch

C Volume — balance; mass — pipette

D Temperature — gas syringe; time — burette

Balance for mass, thermometer for temperature, stopwatch for time — the basic trio. The others scramble instruments and quantities.

Question 17

Anhydrous calcium chloride in a U-tube is placed between a gas generator and the collection vessel. Its purpose is to

A speed up the reaction

B remove acidic impurities from the gas

C dry the gas by absorbing water vapour

D measure the volume of gas

Anhydrous calcium chloride is a drying agent — it absorbs the water vapour that the gas carries from the aqueous reaction mixture, and reacts with the gas itself no further.

Question 18

100 cm³ of water at 25 °C dissolves a maximum of 36 g of sodium chloride. A student stirs 40 g of NaCl into 100 cm³ of water at 25 °C. What is observed?

A All the salt dissolves to give an unsaturated solution

B 36 g dissolves; about 4 g remains as undissolved solid in a saturated solution

C Nothing dissolves because too much was added

D The solution boils

The water dissolves its maximum (36 g) and becomes saturated; the excess 4 g stays as solid at the bottom. Adding extra solute past saturation changes nothing but the pile of undissolved solid.

Question 19

Why is a gas syringe better than collection over water for measuring the volume of carbon dioxide produced in an experiment?

A CO₂ is denser than air

B CO₂ is slightly soluble in water, so some would dissolve and be lost

C CO₂ would react with the glass

D Gas syringes are cheaper

CO₂ dissolves slightly in water (forming carbonic acid), so over-water collection under-records the volume. The syringe keeps the gas away from water and reads the volume directly.

Question 20

Ethanol is used to dissolve a plant pigment that does not dissolve in water. In this experiment, ethanol is acting as

A a solute

B a solvent

C a saturated solution

D an indicator

The liquid doing the dissolving is the solvent — and this example shows solvents other than water matter: a substance insoluble in water may dissolve well in ethanol or propanone.

12.2 Acid–Base Titrations ▼

The Big Idea: Finding the Exact Moment of Neutralisation

A titration answers one question with beautiful precision: exactly how much acid reacts with a measured volume of alkali (or vice versa)? Because both acid and alkali are usually colourless, an indicator stands guard in the flask and announces the end-point — the moment neutralisation is complete — with a colour change. The whole procedure is a fixed ritual, and examiners ask for its steps, its apparatus and its readings year after year.

The Procedure — Learn It as Numbered Steps

Step	Action	Why
1	Use a volumetric pipette (with safety filler) to transfer 25.0 cm³ of alkali into a conical flask	Fixed, accurately known volume; conical shape allows swirling without splashing
2	Add a few drops of indicator; place the flask on a white tile	The indicator signals the end-point; the white tile makes the colour change easy to see
3	Fill the burette with acid (rinse it with the acid first); record the initial reading to 0.05 cm³	Rinsing prevents dilution by leftover water; the titre is final minus initial reading
4	Add acid from the burette, swirling the flask; slow to drop by drop near the end-point	Swirling mixes reactants; dropwise addition avoids overshooting the end-point
5	Stop at the first permanent colour change (the end-point); record the final reading	The end-point marks exact neutralisation
6	Repeat until you have concordant titres — within 0.10 cm³ of each other; average the concordant values only	Repeats catch mistakes; a first "rough" titration finds the approximate end-point quickly

Indicators and Their Colour Changes

Indicator	Colour in acid	Colour in alkali	Notes
Methyl orange	Red	Yellow	Passes through orange at neutral; sharp, easily seen change
Thymolphthalein	Colourless	Blue	Acid → alkali: colourless to blue; alkali → acid: blue to colourless
Litmus	Red	Blue	Fine for identifying acid/alkali but its change is too gradual for a sharp titration end-point

State the Direction of the Change

An end-point answer must fit the experiment's direction. Adding acid to alkali with methyl orange: yellow → red (accept orange as the end-point). With thymolphthalein: blue → colourless. Adding alkali to acid reverses both. Writing only "it changes colour" or the wrong direction costs the mark. And never say the solution "turns clear" — colourless is the word; clear means transparent, and a blue solution is also clear.

Reading the Burette and Getting Concordant Results

A burette's scale runs from 0 at the top downward — because it measures liquid delivered, not liquid contained. The titre is final reading − initial reading. Results are concordant when titres are within 0.10 cm³ of each other; only concordant titres are averaged (the rough titration is never included).

titre = final burette reading − initial burette reading

Example: initial 1.20 cm³, final 25.65 cm³ → titre = 24.45 cm³. Titres of 24.45, 24.50 and 25.10 cm³: the first two are concordant (differ by 0.05); 25.10 is discarded. Average = (24.45 + 24.50) ÷ 2 = 24.48 cm³ (24.475 rounds to 24.48).

Supplement

Titration calculations — the four-step machine

Every titration calculation runs the same course. Step 1: moles of the known solution = concentration × volume in dm³. Step 2: use the balanced equation's mole ratio to get moles of the unknown. Step 3: concentration of unknown = moles ÷ its volume in dm³. Step 4: sanity-check units and significant figures. Example: 25.0 cm³ of NaOH needs 20.0 cm³ of 0.100 mol/dm³ HCl. Moles HCl = 0.100 × 0.0200 = 0.00200. Ratio 1:1, so moles NaOH = 0.00200. Concentration NaOH = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³. Watch ratios: H₂SO₄ + 2NaOH means moles of NaOH = 2 × moles of H₂SO₄.

Worked Example 1 Describe, step by step, how to carry out a titration to find the volume of dilute sulfuric acid needed to neutralise 25.0 cm³ of aqueous sodium hydroxide, using thymolphthalein as the indicator. Include the colour change at the end-point. [6]

Step 1: Set up the flask

Use a volumetric pipette with a safety filler to transfer exactly 25.0 cm³ of sodium hydroxide into a conical flask. Add a few drops of thymolphthalein — the solution turns blue (alkaline). Stand the flask on a white tile.

Step 2: Set up the burette

Rinse the burette with the acid, fill it, remove the air bubble from the jet, and take the initial reading at eye level, bottom of the meniscus, to the nearest 0.05 cm³.

Step 3: Titrate

Run acid into the flask, swirling constantly. As blue flashes of colour last longer, slow the addition to drop by drop.

Step 4: The end-point and the repeats

Stop at the exact drop where the solution turns from blue to colourless and stays colourless on swirling — the end-point. Record the final reading. Repeat the whole titration until two titres agree within 0.10 cm³ (concordant), and average those.

Pipette 25.0 cm³ NaOH into a conical flask [1]; add a few drops of thymolphthalein, flask on a white tile [1]; fill burette with acid, record initial reading to 0.05 cm³ [1]; add acid with swirling, dropwise near the end-point [1]; stop at the first permanent colour change, blue → colourless [1]; repeat to obtain concordant titres (within 0.10 cm³) and average them [1].

Worked Example 2 A student's four titres are: rough 26.50 cm³, then 25.85, 26.30 and 25.90 cm³. (a) Which titres should be used to calculate the average, and why? (b) Calculate the average titre. [3]

Step 1: Discard the rough

The rough (first) titration is deliberately quick and approximate — it only locates the end-point region. It is never included in the average.

Step 2: Find the concordant pair

Concordant = within 0.10 cm³ of each other. 25.85 and 25.90 differ by 0.05 cm³ — concordant. 26.30 is 0.40–0.45 cm³ away from both — an outlier, discarded.

Step 3: Average the concordant titres only

(25.85 + 25.90) ÷ 2 = 51.75 ÷ 2 = 25.875 → 25.88 cm³. Quote it to two decimal places like any burette-derived value.

(a) Use 25.85 and 25.90 cm³: they are concordant (within 0.10 cm³); the rough is approximate and 26.30 is not concordant [2]. (b) Average = 25.88 cm³ [1].

Worked Example 3 (Supplement) 25.0 cm³ of aqueous sodium hydroxide is exactly neutralised by 22.50 cm³ of 0.0500 mol/dm³ sulfuric acid: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the sodium hydroxide in mol/dm³. [3]

Step 1: Moles of the known (the acid)

Convert cm³ to dm³: 22.50 cm³ = 0.02250 dm³. Moles H₂SO₄ = 0.0500 × 0.02250 = 0.001125 mol.

Step 2: Mole ratio from the equation

1 H₂SO₄ : 2 NaOH — each acid molecule provides two H⁺. Moles NaOH = 2 × 0.001125 = 0.00225 mol. This doubling is where most marks are lost.

Step 3: Concentration of the unknown

Concentration = moles ÷ volume = 0.00225 ÷ 0.0250 = 0.0900 mol/dm³.

Moles H₂SO₄ = 0.0500 × 0.02250 = 0.001125 mol [1]; moles NaOH = 2 × 0.001125 = 0.00225 mol [1]; concentration = 0.00225 ÷ 0.0250 = 0.0900 mol/dm³ [1].

Exam Tips for 12.2

1. The apparatus quartet: burette (acid), volumetric pipette (fixed 25.0 cm³ of alkali), conical flask, white tile. Name all four when describing the method.

2. Methyl orange: red in acid, yellow in alkali. Thymolphthalein: colourless in acid, blue in alkali. Give the change in the correct direction for the experiment described.

3. "Colourless", never "clear". Clear describes transparency; colourless describes colour. Examiners penalise "the solution went clear".

4. End-point = first permanent colour change. Add dropwise near it, swirling; a single drop can carry you across.

5. Burette readings: two decimal places, ending 0 or 5. The scale runs downward from the top, so the final reading is the larger number, and titre = final − initial.

6. Concordant = within 0.10 cm³. Average concordant titres only; exclude the rough titration and outliers.

7. Why a conical flask and not a beaker? Swirling without spillage. Why a white tile? To see the colour change sharply.

8. It does not matter that indicator "contaminates" the flask — a few drops are negligible — but rinsing matters: burette rinsed with acid, pipette with alkali, flask with distilled water only.

9. (Supplement) Calculation ratio check: diprotic acids (H₂SO₄) neutralise two moles of NaOH per mole of acid. Read the balanced equation before multiplying.

⚖️ Apply It: Real-World Chemistry

Titration is the quality-control workhorse of the chemical world — from vinegar factories in Modena to pharmaceutical labs in Basel, someone is watching for a colour change right now.

A quality-control chemist at a pickle factory in Tamil Nadu titrates each batch of vinegar against standard sodium hydroxide to check the ethanoic acid concentration. Her results one morning: rough 22.90, then 22.35, 22.40, 23.05 cm³. Company rules say the average must come from concordant titres only.

Which titres does she average, and why is the 23.05 result excluded even though it is not obviously "wrong"?

▼

Concordance Is a Rule, Not a Feeling

Concordant titres agree within 0.10 cm³. 22.35 and 22.40 differ by 0.05 — concordant. 23.05 is 0.65 away: excluded. The rough is always excluded. Average = (22.35 + 22.40) ÷ 2 = 22.38 cm³.

Why the Outlier Cannot Stay

An outlier usually means a procedural slip — overshooting the end-point, a misread scale, an air bubble in the jet. Including it would drag the average away from the true value; the concordance rule filters mistakes out automatically.

Chemistry Connection

"Which titres would you use, and why?" is a guaranteed Paper 6 question. The full answer names the rule (within 0.10 cm³), excludes the rough, and averages only the concordant pair.

Two students in a London school titrate the same alkali with the same acid. Amaya uses methyl orange and stops when her flask turns from yellow to the first trace of orange-red. Ben uses no indicator at all, saying he will stop "when it looks neutral". Their teacher predicts Ben's result will be useless.

Why is an indicator essential here, and what exactly should Amaya report as her end-point colour change?

▼

Neutral Looks Like Everything Else

Both solutions are colourless; the mixture looks identical before, at and after neutralisation. There is nothing to see without an indicator — Ben literally cannot detect the end-point.

Amaya's Report

Adding acid to alkali with methyl orange: the flask starts yellow (alkaline) and the end-point is the first permanent change towards orange/red — reported as yellow to red (orange accepted at the exact end-point). She should add the final portion dropwise so one drop marks the change.

Chemistry Connection

The indicator is a tiny spy in the flask reporting pH as colour. Note the direction matters: alkali-into-acid with methyl orange would be red to yellow — same indicator, opposite report.

A trainee lab technician in Nairobi sets up a titration but makes three small mistakes: she rinses the burette with distilled water just before filling it with acid; she measures the alkali into the flask with a measuring cylinder; and she reads the burette from below, looking up at the meniscus.

Explain the effect of each mistake on the titration results.

▼

Water in the Burette Dilutes the Acid

Residual water makes the acid slightly more dilute, so a larger titre is needed — a systematic error. The burette must be rinsed with the acid it will contain.

The Cylinder and the Parallax

A measuring cylinder delivers only an approximate 25 cm³, so the moles of alkali differ from the assumed value — use a pipette. Reading the meniscus from below makes the reading appear too high or low (parallax error); the eye must be level with the meniscus.

Chemistry Connection

Error-analysis questions want cause and consequence: name the fault, then say which way it shifts the titre. "Rinsed with water → acid diluted → titre too large" is a complete chain worth full marks.

A pharmaceutical company in Basel checks every batch of antacid tablets by dissolving a tablet and titrating the alkaline mixture with standard hydrochloric acid using thymolphthalein. The batch record must state the colour change observed. One analyst writes "the solution went clear"; the reviewing chemist rejects the record and makes her rewrite it.

What should the record say, and why was "went clear" rejected?

▼

The Correct Description

Acid is being added to an alkaline solution containing thymolphthalein, so the flask starts blue and the end-point is blue → colourless. That exact phrase belongs in the record.

Clear vs Colourless

"Clear" means transparent — not cloudy. The blue solution was already clear! The change at the end-point is a change of colour, so the only acceptable word is colourless. In analytical records (and exams), this distinction is enforced strictly.

Chemistry Connection

Three commonly confused words: clear (transparent), colourless (no colour), decolourised (colour removed from something that had it — e.g. KMnO₄ in the sulfite test). Choosing the right one is worth marks in at least three different Topic 12 tests.

A water-quality lab in Kolkata measures the acidity of monsoon rain samples. 25.0 cm³ of each sample is titrated against 0.00500 mol/dm³ sodium hydroxide. For one sample the average titre is 12.50 cm³. The lab reports the rain's acid content assuming the acid behaves as a monoprotic acid HA (ratio 1:1).

(Supplement) Calculate the concentration of acid in the rain sample.

▼

Moles of the Known

Moles NaOH = concentration × volume in dm³ = 0.00500 × 0.01250 = 6.25 × 10⁻⁵ mol.

Ratio, Then Concentration

Ratio 1:1, so moles HA = 6.25 × 10⁻⁵ mol in 25.0 cm³ = 0.0250 dm³. Concentration = 6.25 × 10⁻⁵ ÷ 0.0250 = 2.5 × 10⁻³ mol/dm³.

Chemistry Connection

Same machine as every titration calculation: moles of known → ratio → concentration of unknown. Real labs run it daily; the exam version simply plugs in kinder numbers. Notice the dilute NaOH — chosen so the tiny amount of acid in rain still gives a measurable titre.

Practice Questions: 12.2

20 multiple choice questions. Click an option to check your answer.

Your Score 0 / 20

Question 1

In a typical school titration, the burette contains the

A acid, added gradually to the alkali in the flask

B indicator, added drop by drop

C alkali and the indicator together

D distilled water for rinsing

The standard arrangement: acid in the burette, a pipetted volume of alkali plus a few drops of indicator in the conical flask. (The roles can be reversed, but this is the classic setup the exam describes.)

Question 2

The 25.0 cm³ of alkali is transferred to the conical flask using a

A measuring cylinder

B volumetric pipette with a safety filler

C second burette

D beaker with a spout

A volumetric pipette delivers exactly 25.0 cm³ every time — the fixed, repeatable volume the titration depends on. The safety filler is how liquid is drawn up; mouths are never used.

Question 3

The conical flask stands on a white tile so that

A the flask does not scratch the bench

B the colour change at the end-point is seen clearly

C heat is reflected back into the flask

D spilled acid is neutralised by the tile

The white background makes the indicator's colour change easy to see, so the titration can be stopped at exactly the right drop. Nothing to do with heat or safety.

Question 4

Methyl orange in dilute hydrochloric acid is

A yellow

B red

C colourless

D blue

Methyl orange: red in acid, yellow in alkali. Blue and colourless belong to thymolphthalein — never mix up the two indicators' colour pairs.

Question 5

Thymolphthalein in aqueous sodium hydroxide is

A red

B yellow

C blue

D colourless

Thymolphthalein: colourless in acid, blue in alkali. NaOH is an alkali, so the solution is blue. Adding acid until the blue just disappears marks the end-point.

Question 6

Acid is run from a burette into alkali containing methyl orange. The colour change at the end-point is

A red to yellow

B yellow to red

C blue to colourless

D colourless to blue

The flask starts alkaline (yellow) and becomes acidic (red) as acid is added — so yellow → red. Always state the direction: the mark is for "from… to…", not just the two colours.

Question 7

The end-point of a titration is the point at which

A the burette is empty

B the flask begins to feel warm

C the indicator just changes colour, showing neutralisation is complete

D bubbles of gas stop forming

The end-point is the moment the indicator just changes colour permanently — the acid has exactly neutralised the alkali. One extra drop overshoots it, which is why the final additions are made drop by drop.

Question 8

Which of these could be a correctly recorded burette reading?

A 23.4 cm³

B 23.42 cm³

C 23.45 cm³

D 23 cm³

Burettes are read to the nearest 0.05 cm³, so every reading has two decimal places ending in 0 or 5: 23.40, 23.45, 23.50… 23.42 pretends to a precision the scale does not have; 23.4 has too few places.

Question 9

Two titres are concordant if they are

A within 1.0 cm³ of each other

B within 0.10 cm³ of each other

C exactly equal

D both larger than the rough titre

Concordant = within 0.10 cm³ of each other. They do not need to be identical — that would be an unreasonable demand — but they must agree closely enough to show the experiment is reproducible.

Question 10

A student's titres are 26.90, 25.45 and 25.50 cm³. The best average titre is

A 25.95 cm³, the mean of all three

B 25.48 cm³, the mean of the two concordant titres

C 26.90 cm³, the first careful result

D 25.45 cm³, the smallest value

25.45 and 25.50 agree within 0.10 cm³ — concordant. 26.90 (the rough titre) is discarded. Mean = (25.45 + 25.50) ÷ 2 = 25.475 ≈ 25.48 cm³. Averaging in a rough titre is one of the most common titration errors.

Question 11

The first titration of a series is usually done quickly because it is

A the most accurate one

B a rough (trial) titration to find the approximate end-point

C needed to rinse the burette

D used to test the indicator works

The rough titration locates the end-point approximately. In later runs the student can add acid rapidly to within 1–2 cm³ of that value, then go drop by drop — fast overall, precise where it matters. The rough value is never used in the mean.

Question 12

Near the expected end-point, the acid should be added

A quickly, to save time

B drop by drop, with swirling after each drop

C in 5 cm³ portions

D only after removing the indicator

One drop is about 0.05 cm³ — the same as the burette's precision. Adding dropwise with constant swirling means the end-point is not overshot, so the titre is trustworthy.

Question 13

The conical flask is swirled throughout the titration to

A mix the solutions so they react completely

B speed up evaporation of the water

C keep the indicator dissolved

D cool the mixture down

Swirling mixes each addition of acid into the whole solution, so the colour seen reflects the true state of the reaction. Without mixing, a local red flash can be mistaken for the end-point.

Question 14

Only a few drops of indicator are used because

A indicators are dangerously corrosive

B indicators are themselves weak acids or bases and too much would affect the result

C the colour would be too pale otherwise

D indicator is expensive

Indicators are weak acids/bases, so large amounts would consume a measurable amount of titrant and shift the end-point. A few drops give a clear colour while leaving the chemistry essentially untouched.

Question 15

Before filling, the burette should be rinsed with

A the alkali it will measure

B a little of the acid it will contain

C tap water only

D indicator solution

Rinsing with the solution it will hold means any droplets left clinging to the glass are that same solution — not water that would dilute the acid and make the titre too big.

Question 16

Rinsing the conical flask with distilled water just before the titration

A ruins the result by diluting the alkali

B is acceptable, because the number of moles of alkali pipetted in is unchanged

C must be done with alkali instead

D changes the indicator colour

The titration counts moles, not concentration. Extra water changes the alkali's concentration in the flask but not the amount of it, so the same volume of acid is still needed. Rinsing the flask with alkali would add extra moles — that is the wrong rinse.

Question 17

25.0 cm³ of 0.100 mol/dm³ NaOH is exactly neutralised by 20.0 cm³ of HCl. The concentration of the HCl is

A 0.080 mol/dm³

B 0.125 mol/dm³

C 0.100 mol/dm³

D 0.250 mol/dm³

Moles NaOH = 0.100 × 0.0250 = 2.50 × 10⁻³. Ratio 1:1, so moles HCl = 2.50 × 10⁻³ in 0.0200 dm³. Concentration = 2.50 × 10⁻³ ÷ 0.0200 = 0.125 mol/dm³. Smaller volume, same moles → more concentrated.

Question 18

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. If 25.0 cm³ of NaOH contains 2.0 × 10⁻³ mol, the moles of H₂SO₄ needed are

A 4.0 × 10⁻³ mol

B 2.0 × 10⁻³ mol

C 1.0 × 10⁻³ mol

D 0.5 × 10⁻³ mol

The ratio H₂SO₄ : NaOH is 1 : 2 — each H₂SO₄ provides two H⁺. So moles of acid = 2.0 × 10⁻³ ÷ 2 = 1.0 × 10⁻³ mol. Forgetting the ratio (answer B) is the classic slip.

Question 19

All burette and pipette readings are taken

A from the top of the meniscus, looking down

B from the bottom of the meniscus, at eye level

C from wherever the liquid touches the glass

D after removing the funnel and shaking the burette

Read the bottom of the meniscus at eye level to avoid parallax error. (Do remove the filling funnel before reading — a drop falling from it mid-titration would change the reading — but never shake a burette.)

Question 20

Titrations are repeated until concordant results are obtained because this

A uses up the leftover acid safely

B shows the titre is reliable and lets random error be averaged out

C makes the end-point colour brighter

D is required to clean the pipette

Two titres within 0.10 cm³ show the measurement is reproducible; averaging them reduces random error. One result alone could hide an overshoot or a misread scale.

12.3 Chromatography ▼

The Big Idea: A Race That Unmasks Mixtures

Paper chromatography separates the substances in a mixture by making them race up a piece of paper, carried by a solvent. Each substance travels at its own characteristic speed, because each has its own balance between two competing pulls: how soluble it is in the solvent (which drags it up the paper) and how strongly it is attracted to the paper (which holds it back). Very soluble, weakly held substances finish near the top; poorly soluble, strongly held ones barely leave the start. One mixture goes in; a ladder of separated spots comes out — and every spot is a substance unmasked.

Setting Up the Race — and the Two Rules That Save It

The method: draw a baseline in pencil near the bottom of the paper, place a small concentrated spot of the mixture on the line, and stand the paper in a beaker containing a shallow layer of solvent with the solvent level below the baseline. The solvent rises through the paper by capillary action, carrying the substances with it. When the solvent has nearly reached the top, take the paper out and immediately mark the solvent front — the furthest line the solvent reached — because it becomes invisible as the paper dries.

The Two Rules Examiners Test Every Year

Rule 1: the baseline must be drawn in pencil. Graphite is insoluble — it stays put. An ink line is itself a mixture of dyes: the solvent would dissolve it and carry its dyes up the paper, contaminating the chromatogram and moving the reference line the R_f measurements depend on.

Rule 2: the solvent level must start below the baseline. If the spots start underwater, they simply dissolve off the paper into the solvent in the beaker — the race is over before it begins, and the chromatogram comes out blank.

Reading a Chromatogram

Three questions, three answers straight off the paper. Is the sample pure? A pure substance produces a single spot; two or more spots mean a mixture. What does the mixture contain? Run known reference substances alongside the unknown on the same paper: a spot in the unknown at the same height as a known spot is (almost certainly) the same substance, because identical substances travel identically under identical conditions. What if nothing is visible? Colourless substances — sugars, amino acids — still separate, but their spots are invisible. Spraying the dried paper with a locating agent reacts with the substances to form coloured spots you can see (ninhydrin, for example, turns amino acids purple).

Supplement

The R_f Value

Heights on one particular chromatogram depend on how long the solvent ran. What is constant for a given substance is the ratio:

R_f = distance moved by substance ÷ distance moved by solvent

Both distances are measured from the baseline — the substance's distance to the centre of its spot, the solvent's to the solvent front. Because no substance can outrun the solvent that carries it, R_f is always less than 1. It has no units (centimetres divided by centimetres cancel) and is usually quoted to two decimal places. Under the same conditions — same solvent, same paper, same temperature — a substance always gives the same R_f, so measured values can be matched against published tables to identify unknowns. An R_f greater than 1 on your calculator means one thing: you divided the wrong way up.

Why Rₖ Values Differ — the One-Sentence Explanation

A substance with a high R_f is more soluble in the solvent and less strongly attracted to the paper; a low-R_f substance is the reverse. Any "explain why dye A moved further than dye B" question wants exactly this comparison — solubility in the solvent versus attraction to the paper.

Worked Example 1 On a chromatogram, the solvent front is 8.0 cm from the baseline. A dye spot has its centre 4.4 cm from the baseline. (a) Calculate the R_f value of the dye. (b) The same dye is run again on a larger paper where the solvent travels 12.0 cm. How far will the dye move? [3]

Step 1: Apply the definition

R_f = distance moved by substance ÷ distance moved by solvent = 4.4 ÷ 8.0 = 0.55. No units. Sanity check: less than 1. ✓

Step 2: Use the fact that Rₖ is constant

Same dye, same solvent, same paper → same R_f. So distance moved = R_f × distance moved by solvent = 0.55 × 12.0 = 6.6 cm.

(a) R_f = 4.4 ÷ 8.0 = 0.55 [1, with working shown]. (b) 0.55 × 12.0 = 6.6 cm from the baseline [2]. The spot's height changes with the paper; its ratio never does.

Worked Example 2 A food-testing laboratory runs a chromatogram of a street-food colouring X alongside three permitted dyes P, Q and R. X gives three spots: two at the same heights as P and R, and one that matches nothing. Q gives one spot matching nothing in X. What can be concluded about X? [3]

Step 1: Count the spots

X gives three spots, so X is a mixture of (at least) three substances. One spot would have meant pure.

Step 2: Match the heights

Spots at the same height as P and R (same R_f, same conditions) identify two of the components: X contains P and R. Nothing in X lines up with Q, so X does not contain Q.

Step 3: Handle the stranger

The third spot matches none of the references. Chromatography can show it is present and different, but cannot name it — the lab would need to run more reference dyes (or measure its R_f against published values) to identify it.

X is a mixture [1] containing dyes P and R but not Q [1]; it also contains one unidentified substance whose spot matches no reference, which needs further comparison to identify [1].

Worked Example 3 A biochemist separates the amino acids in a protein sample by paper chromatography. When the run finishes, the paper appears completely blank. (a) Explain why nothing is visible. (b) Describe what the biochemist should do, and state what would then be seen. [3]

Step 1: Blank does not mean failed

Amino acids are colourless. They have separated perfectly well — each has travelled its own distance up the paper — but colourless spots on white paper are invisible.

Step 2: The locating agent

Dry the paper, then spray it with (or dip it in) a locating agent — a chemical that reacts with the amino acids to form coloured products. Ninhydrin is the classic choice for amino acids, giving purple spots (gentle warming develops the colour).

(a) The amino acids are colourless, so their separated spots cannot be seen [1]. (b) Spray the dried chromatogram with a locating agent [1]; it reacts with the amino acids to give visible (coloured) spots, whose positions and R_f values can then be measured [1].

Exam Tips for 12.3

1. Pencil, because graphite is insoluble. The reason is the mark: ink would dissolve in the solvent and run up the paper with the sample.

2. Solvent below the baseline, or the spots dissolve into the solvent and are washed off the paper. Another reason-is-the-mark favourite.

3. Pure = one spot. Mixture = more than one. The number of spots is the minimum number of substances (two substances with identical Rₖ would overlap).

4. Identify by matching heights with a known substance run on the same paper — same height under the same conditions means same Rₖ, which means (almost certainly) the same substance.

5. Rₖ = substance ÷ solvent, both measured from the baseline, substance to the centre of the spot. Two decimal places, no units.

6. Rₖ is ALWAYS less than 1. Nothing travels further than the solvent carrying it. If your value exceeds 1, invert your fraction.

7. Mark the solvent front immediately — it disappears as the paper dries, and without it no Rₖ can be calculated.

8. Locating agent = makes colourless substances visible by reacting with them to form coloured spots. Name the purpose, not just the word.

🔬 Apply It: Real-World Chemistry

From food-safety labs in Jaipur to forensic document examiners in London, chromatography is the cheap, fast unmasker of mixtures.

A food-safety inspector in Jaipur suspects that the brilliant yellow colour of sweets from one shop comes from metanil yellow — a banned industrial dye — rather than permitted food colouring. The lab has a genuine sample of metanil yellow.

Describe how chromatography settles the question.

▼

The Method

Extract the colouring from the sweet and spot it on a pencil baseline alongside a spot of genuine metanil yellow. Run the paper with the solvent level below the baseline. If the sweet extract produces a spot at the same height as the metanil yellow reference (same R_f), the banned dye is present.

The Verdict

Matching heights under identical conditions is strong evidence of identity; a lab confirms by repeating in a second, different solvent — a coincidence in one solvent almost never survives two.

Chemistry Connection

This is exactly the syllabus skill "identify substances by comparison with known substances" — wearing a food-safety uniform. FSSAI labs across India run this comparison routinely on sweets, spices and street-food colourings.

A forensic document examiner in London is asked whether the signature on a disputed will was written with the same pen as the rest of the document. She cuts tiny discs of ink from both, and runs them side by side in the same solvent.

The signature ink gives four spots; the document ink gives three, all at heights matching three of the signature's four. What can she conclude?

▼

Reading the Spots

Both inks are mixtures of dyes. The signature ink contains four dyes; the document ink only three. Since one dye in the signature has no partner in the document ink, the two inks have different compositions — different pens.

The Limits

Chromatography shows the inks differ; it cannot say who held the pen or when. Careful scientists claim exactly what the chromatogram shows and no more — a good habit for exam answers too.

Chemistry Connection

Ink analysis is the classic demonstration that a single colour can hide a mixture — and it is why the baseline is drawn in pencil. An ink baseline would add its own dyes to every lane of the evidence.

A student in Chennai sets up her first chromatography experiment: she draws the baseline with her blue gel pen, pours in solvent until it covers the spots "so they get a good soaking", and returns later to find a smeared blue mess and no separate spots at all.

Explain both mistakes and their consequences.

▼

Mistake 1: The Ink Baseline

Gel-pen ink is a mixture of soluble dyes. The solvent dissolved the baseline and carried its dyes up the paper — the blue smear — contaminating every lane and destroying the reference line from which distances are measured. Pencil graphite is insoluble and would have stayed put.

Mistake 2: Solvent Above the Baseline

With the spots below the liquid surface, the samples dissolved off the paper directly into the solvent in the beaker instead of travelling up the paper. That is why no separated spots appeared — the mixture is now uselessly diluted in the beaker.

Chemistry Connection

These two errors are the most asked "what did the student do wrong?" questions in Topic 12. Each one links a rule to its reason: pencil because insoluble; solvent below baseline because otherwise the sample washes away.

A quality-control chemist at a Hyderabad pharmaceutical plant checks each batch of a drug by chromatography. A pure batch should give one spot with R_f 0.62 in the standard solvent. Today's batch gives the 0.62 spot plus a faint second spot at R_f 0.38.

What has the chromatogram revealed, and why does it matter?

▼

Two Spots = Not Pure

A pure substance gives a single spot. The extra spot at R_f 0.38 is a second substance — an impurity, perhaps an unreacted starting material or a by-product of the synthesis. Its faintness suggests a small amount, but chromatography has caught it.

Why It Matters

In medicines, impurities can be harmful or change the dose's effect, so pharmaceutical standards demand demonstrated purity. The batch fails QC and is purified or discarded — and the impurity's R_f can be compared with known substances to work out where the process went wrong.

Chemistry Connection

This scenario stitches 12.3 to 12.4: chromatography detects impurity; melting-point measurement confirms it (impurities lower and broaden the melting point); recrystallisation removes it. One quality-control story, three syllabus points.

A sports-science lab screens an athlete's drink for caffeine. On the chromatogram the solvent front is 12.0 cm from the baseline; the suspect spot's centre is 7.2 cm up. Published data: caffeine's R_f in this solvent is 0.60.

(Supplement) Does the drink contain caffeine? Show the calculation.

▼

Calculate the Rₖ

R_f = distance moved by substance ÷ distance moved by solvent = 7.2 ÷ 12.0 = 0.60.

Compare With the Reference

The measured value matches caffeine's published R_f of 0.60 in the same solvent — strong evidence the spot is caffeine. A rigorous lab confirms by running a genuine caffeine reference alongside, or repeating in a different solvent.

Chemistry Connection

Notice the double check available: match against a published R_f value or against a known substance on the same paper. Same logic, two costumes — and both appear in exam questions.

Practice Questions: 12.3

20 multiple choice questions. Click an option to check your answer.

Your Score 0 / 20

Question 1

The baseline of a chromatogram is drawn in pencil because graphite

A is cheaper than ink

B is insoluble in the solvent, so the line does not move or interfere

C conducts electricity along the paper

D shows up better on white paper

Graphite is insoluble — the reference line stays exactly where it was drawn. Ink is a mixture of soluble dyes that would travel up the paper with the samples and contaminate the chromatogram.

Question 2

At the start of a run, the solvent level must be below the baseline. Otherwise the

A solvent evaporates too quickly

B paper becomes too heavy and falls into the beaker

C spots dissolve off the paper into the solvent in the beaker

D solvent front cannot be marked

Submerged spots simply wash off into the solvent before the run begins — the chromatogram comes out blank and the sample is lost into the beaker.

Question 3

The solvent front is

A the line where the samples are spotted

B the furthest position reached by the solvent up the paper

C the level of solvent in the beaker

D the fastest-moving coloured spot

The solvent front is the highest point the rising solvent reaches. It must be marked in pencil immediately the paper is removed, because it vanishes as the paper dries — and without it no Rₖ can be calculated.

Question 4

Substances in a mixture separate during chromatography because they differ in

A colour and brightness

B boiling point

C solubility in the solvent and attraction to the paper

D density

Each substance strikes its own balance between being carried by the solvent (solubility) and being held by the paper (attraction/adsorption). Boiling point is distillation's business, not chromatography's.

Question 5

On a chromatogram, a pure substance produces

A a single spot

B no spots at all

C a spot exactly at the solvent front

D a continuous streak from baseline to front

One substance, one spot. Two or more spots reveal a mixture — this is the quickest purity check in chemistry. (A colourless pure substance also gives one spot, visible after a locating agent.)

Question 6

A mixture gives four spots on a chromatogram. The mixture contains

A exactly four substances

B at least four substances

C at most four substances

D four elements

At least four: two different substances with the same Rₖ in this solvent would sit on top of each other as one spot. Running the mixture again in a different solvent could split such a hidden pair.

Question 7

To show that a dye in a mixture is the permitted colouring E102, the best method is to

A compare the colours of the spots by eye

B run a known sample of E102 alongside the mixture on the same paper and compare spot heights

C measure how quickly the spot fades in sunlight

D weigh the spot after cutting it out

Identification is by comparison with a known substance under identical conditions: same paper, same solvent, same run. Matching heights (matching Rₖ) indicate the same substance; colour alone can deceive.

Question 8

A locating agent is used in chromatography to

A speed up the movement of the solvent

B stick the paper to the beaker

C make colourless substances visible as coloured spots

D dissolve the baseline

Sugars and amino acids separate invisibly. A locating agent reacts with them to form coloured products, revealing where each spot sits so Rₖ values can be measured.

Question 9

Which substances would most need a locating agent after chromatography?

A The dyes in black ink

B The pigments in leaf extract

C The amino acids from a protein

D The colourings in a fizzy orange drink

Amino acids are colourless — invisible on paper without a locating agent (ninhydrin turns them purple). Inks, leaf pigments and drink colourings announce themselves in colour.

Question 10

(Supplement) Rₖ is defined as

A distance moved by solvent ÷ distance moved by substance

B distance moved by substance ÷ distance moved by solvent

C distance moved by substance × distance moved by solvent

D time taken by substance ÷ time taken by solvent

Substance over solvent — the slower runner divided by the pacesetter, giving a fraction below 1. Inverting the ratio (option A) is the error that produces impossible values above 1.

Question 11

(Supplement) A spot's centre is 3.6 cm from the baseline; the solvent front is 9.0 cm from the baseline. Rₖ =

A 2.5

B 0.40

C 0.36

D 5.4

Rₖ = 3.6 ÷ 9.0 = 0.40. Option A (2.5) is the upside-down division — instantly recognisable as wrong because it exceeds 1.

Question 12

(Supplement) Rₖ values are always less than 1 because

A the paper is never long enough

B a substance cannot travel further than the solvent that carries it

C examiners round them down

D the baseline is drawn above the solvent

The solvent is the carrier — substances ride with it, always some distance behind the front. Even a substance that never grips the paper can at best match the solvent, never overtake it.

Question 13

(Supplement) When measuring the distance moved by a substance, you measure from the baseline to the

A top edge of the spot

B bottom edge of the spot

C centre of the spot

D solvent front

Spots spread slightly as they travel, so the centre is taken as the substance's true position. Measuring to an edge biases the Rₖ high or low.

Question 14

(Supplement) A dye has Rₖ 0.75 in a solvent. If the solvent front moves 8.0 cm, the dye's spot will be found

A 7.5 cm from the baseline

B 6.0 cm from the baseline

C 0.75 cm from the baseline

D 8.0 cm from the baseline

Distance = Rₖ × solvent distance = 0.75 × 8.0 = 6.0 cm. The ratio is the fixed property; the distance scales with however far the solvent runs.

Question 15

Dye P travels further up the paper than dye Q in the same run. Compared with Q, P is

A more soluble in the solvent and less strongly attracted to the paper

B less soluble in the solvent and more strongly attracted to the paper

C darker in colour

D made of smaller atoms

High travel = strong pull from the solvent (solubility) and weak grip from the paper. This two-part comparison is the standard mark-scheme sentence for "explain why one dye moved further".

Question 16

A spot that stays on the baseline after the run is

A the purest substance in the mixture

B insoluble in this solvent (or held extremely strongly by the paper)

C evidence the experiment failed completely

D always water

No movement means the solvent could not carry it — it is insoluble in that solvent (Rₖ = 0). Trying a different solvent may set it moving; the rest of the chromatogram is still valid.

Question 17

Two different substances could appear as one spot on a chromatogram if they

A have the same colour

B have the same Rₖ in that solvent

C were spotted at the same time

D are both solids at room temperature

Same Rₖ in one solvent = same finishing position = overlapping spots. The remedy is to rerun the mixture in a different solvent, where the coincidence almost never repeats.

Question 18

(Supplement) Rₖ values have

A units of cm

B units of cm/s

C no units, because they are a ratio of two distances

D units of cm²

Centimetres divided by centimetres cancel: Rₖ is a dimensionless ratio. Writing "0.62 cm" is a real mark-loser in Paper 6.

Question 19

(Supplement) A measured Rₖ can be compared with published values to identify a substance only if the

A paper was cut to the same length

B same solvent (and conditions) were used as for the published value

C spots are the same size

D experiment was done at night

Rₖ is constant only for a given solvent and conditions — change the solvent and every Rₖ changes. Paper length does not matter, because Rₖ is a ratio.

Question 20

Why must the solvent front be marked as soon as the paper is removed?

A The spots keep moving after removal

B The solvent evaporates as the paper dries, making the front invisible

C The paper shrinks when dry

D The baseline fades in the light

The wet edge disappears on drying. Lose it and you lose the denominator of every Rₖ on the paper — the whole quantitative half of the experiment.

12.4 Separation, Purity & Identification ▼

The Big Idea: Purify, Prove It Pure, Then Name It

This section is Topic 12's engine room, and the exam's favourite hunting ground. Three linked skills: separate a mixture with the right technique, prove the product is pure using melting and boiling points, and identify an unknown substance with the classic test-tube toolkit — flame tests, precipitate tests and gas tests. Learn the tables here until they are reflexes: Paper 6 (and plenty of Paper 4) is built directly on them.

Purity: The Sharp-Point Rule

A pure substance — a single element or compound, nothing else mixed in — melts and boils at sharp, exact, fixed temperatures: ice at exactly 0 °C, water boiling at exactly 100 °C. Impurities wreck this precision in a beautifully consistent way: they lower the melting point and raise the boiling point, and they smear both over a range of temperatures instead of a sharp point. So melting- and boiling-point data are both an identity check (compare with known values) and a purity check (sharp = pure; shifted and spread = impure). This is why salt scattered on icy roads melts the ice, and why sea water boils above 100 °C.

Purity matters best where chemistry meets the body: foodstuffs and drugs. An impurity in a medicine could be toxic, could change the dose that reaches the patient, or could cause side effects — so pharmaceutical companies must demonstrate purity batch after batch. The same logic protects food additives and colourings. "Pure" on a juice carton means "nothing added"; in chemistry it means a single substance — the exam expects the chemical meaning.

The Direction Matters — Memorise the Asymmetry

Impurities push the two points in opposite directions: melting point DOWN, boiling point UP (and both become ranges). Students who guess "impurities raise both" hand back marks every session. Anchor it with the two everyday examples: salted roads (ice melts below 0 °C) and salty pasta water (boils above 100 °C).

Choosing a Separation Method

Every separation exploits a difference in properties between the things being separated. Ask what kind of mixture you have, and the method chooses itself:

Mixture	Method	Property exploited & how it works
Insoluble solid + liquid (sand in water)	Filtration	Particle size: the liquid (filtrate) passes through the filter paper's pores; the insoluble solid (residue) cannot
Dissolved solid, solid wanted (copper(II) sulfate solution)	Crystallisation	Solubility falls as temperature falls: heat the solution to evaporate some solvent until saturated (crystallisation point), cool slowly so crystals form, filter them off, dry between filter papers
Dissolved solid, liquid wanted (pure water from salt water)	Simple distillation	Large boiling-point difference: the solvent boils off, is condensed in a (Liebig) condenser and collected; the solute stays behind in the flask
Two (or more) miscible liquids (ethanol + water; crude oil)	Fractional distillation	Different boiling points: with a fractionating column, the vapour reaching the top is repeatedly condensed and re-evaporated until only the lowest-boiling liquid passes over; the column stays at that liquid's boiling point until it has all distilled
Two solids, one soluble (salt + sand)	Dissolve → filter → crystallise	Solubility difference: dissolve the salt in water, filter off the sand, then crystallise the salt from the filtrate

Crystallise — Do NOT Evaporate to Dryness

To recover a salt such as hydrated copper(II) sulfate, heat only until a saturated solution forms (test: a drop on a cold glass slide grows crystals), then cool slowly, filter off the crystals and dry them between filter papers. Boiling the solution to dryness is the trap answer: the fierce heat drives off the water of crystallisation (leaving white anhydrous powder, not blue crystals) and can decompose the salt entirely. Gentle crystallisation keeps the compound intact.

Identification 1: Flame Tests for Metal Cations

Dip a clean wire (cleaned in concentrated hydrochloric acid) into the solid or solution, hold it in a hot non-luminous Bunsen flame, and read the colour:

Cation	Flame colour
Lithium, Li⁺	Red
Sodium, Na⁺	Yellow
Potassium, K⁺	Lilac
Calcium, Ca²⁺	Orange-red
Copper(II), Cu²⁺	Blue-green

Identification 2: Cations With NaOH(aq) and NH₃(aq)

Add the reagent a little at a time, then in excess, and watch for a precipitate of the metal hydroxide — then watch what excess does. The excess behaviour is what separates the look-alikes. This is the single most important table in Topic 12:

Cation	With NaOH(aq)	With NH₃(aq)
Ammonium, NH₄⁺	No precipitate; ammonia gas given off on warming (turns damp red litmus blue)	— (no reaction)
Aluminium, Al³⁺	White precipitate, soluble in excess giving a colourless solution	White precipitate, insoluble in excess
Calcium, Ca²⁺	White precipitate, insoluble in excess	No precipitate (or only a very slight one)
Chromium(III), Cr³⁺	Grey-green precipitate, soluble in excess	Grey-green precipitate, insoluble in excess
Copper(II), Cu²⁺	Light blue precipitate, insoluble in excess	Light blue precipitate, soluble in excess giving a dark blue solution
Iron(II), Fe²⁺	Green precipitate, insoluble in excess	Green precipitate, insoluble in excess
Iron(III), Fe³⁺	Red-brown precipitate, insoluble in excess	Red-brown precipitate, insoluble in excess
Zinc, Zn²⁺	White precipitate, soluble in excess giving a colourless solution	White precipitate, soluble in excess giving a colourless solution

The Three White Precipitates — and How Excess Untangles Them

Al³⁺, Ca²⁺ and Zn²⁺ all give white precipitates with NaOH. Ca²⁺: insoluble in excess NaOH — identified at once. Al³⁺ and Zn²⁺ both dissolve in excess NaOH, so switch reagent: with excess ammonia, zinc's precipitate dissolves but aluminium's does not. One sentence to memorise: zinc dissolves in excess of both; aluminium only in excess NaOH; calcium in neither.

Identification 3: Anion Tests

Anion	Test	Positive result
Carbonate, CO₃²⁻	Add dilute acid	Effervescence; the gas turns limewater milky (CO₂)
Chloride, Cl⁻	Acidify with dilute nitric acid, add aqueous silver nitrate	White precipitate (AgCl)
Bromide, Br⁻	Acidify with dilute nitric acid, add aqueous silver nitrate	Cream precipitate (AgBr)
Iodide, I⁻	Acidify with dilute nitric acid, add aqueous silver nitrate	Yellow precipitate (AgI)
Nitrate, NO₃⁻	Add aqueous NaOH and aluminium foil; warm carefully	Ammonia given off — turns damp red litmus paper blue
Sulfate, SO₄²⁻	Acidify with dilute nitric acid, add aqueous barium nitrate (or acidified barium chloride)	White precipitate (BaSO₄)
Sulfite, SO₃²⁻	Add dilute acid and warm; pass the gas into acidified aqueous potassium manganate(VII)	SO₂ released, which turns the purple KMnO₄ colourless (decolourises it)

Why Acidify First?

The nitric acid added before silver nitrate (halides) or barium nitrate (sulfate) destroys carbonate ions, which would otherwise give white precipitates of Ag₂CO₃ or BaCO₃ and fake a positive result. The acid guarantees that any precipitate that survives is the real thing. Halide colour ladder, in mass order: chloride white → bromide cream → iodide yellow — the colour deepens down the group.

Identification 4: Gas Tests

Gas	Test	Positive result
Ammonia, NH₃	Hold damp red litmus paper in the gas	Turns blue
Carbon dioxide, CO₂	Bubble through limewater	Limewater turns milky (cloudy white)
Chlorine, Cl₂	Hold damp litmus paper in the gas	Litmus is bleached (damp blue litmus turns red, then white)
Hydrogen, H₂	Hold a lighted splint at the mouth of the tube	Burns with a squeaky 'pop'
Oxygen, O₂	Insert a glowing splint into the tube	Splint relights
Sulfur dioxide, SO₂	Pass into acidified aqueous potassium manganate(VII)	Purple solution turns colourless

Lighted vs Glowing — the Splint Swap That Costs Thousands of Marks

Hydrogen: LIGHTED splint → pop. Oxygen: GLOWING splint → relights. Swap the splints and both tests fail (a glowing splint in hydrogen does little; a lighted splint in oxygen just keeps burning). Say the full pair each time: the apparatus and the observation. And for ammonia and chlorine the litmus must be damp — the gases must dissolve to act.

Worked Example 1 A white solid dissolves in water. A flame test gives a lilac colour. A solution of the solid, acidified with dilute nitric acid, gives a cream precipitate with aqueous silver nitrate. Identify the solid. [3]

Step 1: The cation from the flame

Lilac flame → potassium, K⁺. (Red would be lithium, yellow sodium, orange-red calcium, blue-green copper(II) — and copper would not give a white solid.)

Step 2: The anion from the precipitate

Acidified silver nitrate is the halide test. The colour ladder: white = chloride, cream = bromide, yellow = iodide. So the anion is Br⁻.

Step 3: Assemble the name

K⁺ + Br⁻ → potassium bromide, KBr. Check the story: white, soluble, group 1 salt — all consistent.

Cation: potassium (lilac flame) [1]. Anion: bromide (cream precipitate with acidified AgNO₃) [1]. The solid is potassium bromide, KBr [1].

Worked Example 2 Two colourless solutions, one containing aluminium ions and one containing zinc ions, have lost their labels. Aqueous sodium hydroxide alone cannot tell them apart. Explain why, and describe a test that can. [4]

Step 1: Why NaOH fails

With NaOH, both ions give a white precipitate which dissolves in excess to a colourless solution — identical observations at every stage. No distinction possible.

Step 2: Switch to ammonia

With aqueous ammonia, both again give white precipitates — but in excess ammonia the zinc hydroxide dissolves (colourless solution) while the aluminium hydroxide does not.

Step 3: Write the expected observations

Add NH₃(aq) slowly to each, then in excess. Precipitate dissolves in excess → Zn²⁺. Precipitate remains → Al³⁺.

Both ions give white precipitates soluble in excess NaOH, so the observations are identical [1]. Test: add aqueous ammonia until in excess [1]. Zinc: white precipitate dissolves in excess [1]. Aluminium: white precipitate insoluble in excess [1].

Worked Example 3 Describe how to obtain (a) pure dry sand and pure salt crystals from a mixture of sand and salt, and (b) pure water from sea water. Name each technique used. [6]

Step 1: Exploit the solubility difference

Add water and stir: the salt dissolves, the sand does not. This is the deliberate creation of a separable mixture — insoluble solid plus solution.

Step 2: Filtration, then crystallisation

Filter: sand stays on the paper as the residue (wash it with distilled water and dry it); salt solution passes through as the filtrate. Then crystallise the filtrate: heat until saturated (crystallisation point), cool slowly, filter off the crystals, dry between filter papers. Do not boil to dryness.

Step 3: Sea water → pure water needs the liquid, not the solid

Use simple distillation: boil the sea water; the water evaporates, leaving the dissolved salts behind; the vapour is condensed in a Liebig condenser and collected as pure (distilled) water.

(a) Add water and stir to dissolve the salt [1]; filter — sand is the residue, dried [1]; crystallise the filtrate: evaporate to saturation, cool, filter and dry the salt crystals [2]. (b) Simple distillation [1]: water boils off and is condensed and collected; the salt remains in the flask [1].

Exam Tips for 12.4

1. Observations, not conclusions. Write "white precipitate forms, insoluble in excess" — not "calcium is present". If the question says describe what you would see, the marks are for colours, precipitates, gases and changes.

2. Impurities: melting point DOWN, boiling point UP, both over a range. A drug melting 5 °C low and over a spread of temperatures is impure — standard 2-marker.

3. Crystallise gently; never evaporate to dryness when the salt is hydrated — you would lose the water of crystallisation or decompose the compound.

4. Filtration vocabulary: residue (stays on the paper) and filtrate (passes through). Use both words; they are often a mark each.

5. Fractional distillation needs both ideas: the liquids have different boiling points, and the column lets the lower-boiling one through first while returning the other as liquid.

6. The excess sentence: zinc's white precipitate dissolves in excess of both NaOH and NH₃; aluminium's only in excess NaOH; calcium's in neither. Fe(II) green, Fe(III) red-brown, Cu(II) light blue — none dissolve in excess NaOH.

7. Acidify before AgNO₃ or Ba(NO₃)₂ — the acid removes carbonate, which would give a false white precipitate.

8. Gas tests are apparatus + observation pairs: lighted splint/pop (H₂), glowing splint/relights (O₂), limewater/milky (CO₂), damp litmus/bleached (Cl₂), damp red litmus/blue (NH₃), acidified KMnO₄/purple to colourless (SO₂). Both halves or no mark.

🔬 Apply It: Real-World Chemistry

Fireworks over the Hooghly, salt pans in Gujarat, drug factories in Basel — section 12.4 is working everywhere.

During Diwali, a fireworks maker in Sivakasi builds a display: one shell bursts crimson-red, one brilliant yellow, one lilac-violet, and one blue-green. Each colour comes from a metal salt packed into the shell.

Suggest which metal ions produce each colour.

▼

Fireworks Are Flame Tests at Scale

Crimson-red → lithium salts (strontium in real fireworks gives a similar red, but lithium is the IGCSE answer). Yellow → sodium. Lilac → potassium. Blue-green → copper(II). Calcium salts would give orange-red.

The Chemistry Behind the Colour

The heat of the burning shell excites electrons in the metal ions; as they fall back they emit light of characteristic energy — hence characteristic colour. Same physics in the Bunsen flame, just quieter.

Chemistry Connection

A flame test question wearing festival clothes. Notice how sodium's yellow dominates mixtures — exactly why the test wire must be cleaned in concentrated HCl until no yellow shows before testing anything else.

A quality-control chemist at a pharmaceutical plant in Basel receives a batch of paracetamol. Pure paracetamol melts sharply at 169 °C. The batch melts between 158 °C and 165 °C.

What do the numbers reveal, and why must the batch be rejected?

▼

Read the Melting Data

The batch melts below 169 °C and over a 7-degree range instead of sharply. Both signatures point the same way: the sample is impure — impurities lower the melting point and spread it out.

Why Purity Is Non-Negotiable in Drugs

An unknown impurity could be toxic, alter the effective dose, or cause side effects. Medicines must be demonstrably pure, batch after batch — melting-point analysis is one of the fastest checks. The batch goes back for purification (recrystallisation) and retesting.

Chemistry Connection

This is the syllabus line "explain the importance of purity in substances in everyday life, e.g. foodstuffs and drugs" with real numbers attached. Sharp point = pure; low and broad = impure — the same rule Tara can quote for full marks.

In the salt pans of Gujarat, sea water is trapped in shallow beds and left in the sun. Weeks later, workers rake up white salt crystals. Inland, a school student tries to copy the process quickly by boiling copper(II) sulfate solution to dryness over a roaring Bunsen — and gets a dull white-grey powder instead of blue crystals.

Why did the salt pans work and the student's shortcut fail?

▼

The Salt Pans: Slow Crystallisation

Gentle solar evaporation concentrates the brine to saturation; from then on, salt crystallises steadily as more water leaves. Slow crystal growth at modest temperature — textbook crystallisation.

The Shortcut: Evaporation to Dryness

Fierce boiling to dryness drove off copper(II) sulfate's water of crystallisation: the blue hydrated crystals CuSO₄·5H₂O became white anhydrous CuSO₄ (with some decomposition possible on further heating). The compound's crystal form was destroyed, not revealed.

Chemistry Connection

"Heat to the point of crystallisation, then cool slowly" is the mark-scheme phrase. Evaporating to dryness is the named wrong answer — and the white anhydrous powder links straight back to Topic 10's test for water.

A water-testing lab in Chennai checks a borewell sample for sulfate contamination from industrial effluent. The technician acidifies a portion with dilute nitric acid, adds aqueous barium nitrate, and a dense white precipitate appears immediately.

What does the result show, and why was the acid added first?

▼

The Positive Result

White precipitate with acidified barium nitrate = sulfate ions present: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s), insoluble and white.

The Acid's Job

Groundwater often contains carbonate/hydrogencarbonate ions, and BaCO₃ is also a white precipitate. The nitric acid destroys the carbonate first (fizzing off CO₂), so any precipitate that still forms must be barium sulfate. No acid, no certainty.

Chemistry Connection

"Why acidify?" is the supplement-grade mark hiding inside every sulfate and halide test. The answer is always the same: to remove carbonate (and other interfering) ions that would give a false positive.

A farm supplier in Punjab suspects a bag sold as ammonium nitrate fertiliser has been adulterated. A lab warms a sample with aqueous sodium hydroxide: a gas turns damp red litmus blue. A fresh sample is then warmed with NaOH and aluminium foil: the same result, but noticeably stronger.

Explain what each test shows and why both were needed.

▼

Test 1: NaOH Alone

Warming with NaOH releases ammonia from ammonium ions: NH₄⁺ + OH⁻ → NH₃ + H₂O. Damp red litmus turning blue confirms NH₃ — so the sample contains NH₄⁺.

Test 2: NaOH + Aluminium Foil

Aluminium reduces nitrate ions to ammonia in alkaline solution. The extra ammonia beyond test 1's tells the lab NO₃⁻ is also present. Without test 1 first, the ammonia from NH₄⁺ alone could masquerade as a positive nitrate result.

Chemistry Connection

The nitrate test's hidden trap: on an ammonium salt it is only valid as a comparison — NaOH alone first, then NaOH + Al. Ammonium nitrate, NH₄NO₃, triggers both, which is exactly what an unadulterated bag should do. Topic 10's fertiliser chemistry meets Topic 12's detective kit.

Practice Questions: 12.4

20 multiple choice questions. Click an option to check your answer.

Your Score 0 / 20

Question 1

Compared with the pure substance, an impure sample

A melts at a higher temperature and boils at a lower one

B melts at a lower temperature and boils at a higher one, both over a range

C melts and boils at exactly the same temperatures

D cannot melt at all

Impurities push the two points in opposite directions — melting point down, boiling point up — and blur both into ranges. Sharp, exact values are the fingerprint of purity.

Question 2

Purity is critically important in drugs because impurities could

A make the tablets a different colour

B be harmful or change the effect of the dose

C make the medicine cheaper to produce

D raise the melting point of the drug

The safety argument earns the mark: impurities may be toxic or alter the amount of active drug the patient receives. (D is also chemically backwards — impurities lower melting points.)

Question 3

Sand is separated from a sand–water mixture by

A crystallisation

B simple distillation

C filtration

D chromatography

Sand is an insoluble solid: it stays on the filter paper as the residue while water passes through as the filtrate. Filtration only works because the sand never dissolved.

Question 4

To obtain blue crystals of hydrated copper(II) sulfate from its solution, you should

A boil the solution rapidly until completely dry

B evaporate to the point of crystallisation, cool slowly, filter and dry the crystals

C freeze the solution overnight

D filter the solution and keep the residue

Gentle crystallisation preserves the water of crystallisation. Boiling to dryness (A) gives white anhydrous powder or decomposition products. In D there is no residue — a dissolved salt passes straight through the paper.

Question 5

Pure water is obtained from salt solution by

A filtration

B crystallisation

C simple distillation

D adding a locating agent

You want the solvent, so boil it off and condense the vapour — simple distillation. Filtration cannot touch dissolved salt; crystallisation recovers the salt but lets the water escape.

Question 6

Fractional distillation separates ethanol (b.p. 78 °C) from water (b.p. 100 °C) because

A ethanol is insoluble in water

B the liquids have different boiling points, and the column lets the lower-boiling vapour pass over first

C ethanol is denser than water

D water evaporates first because it is more common

Miscible liquids need the fractionating column: rising vapour is repeatedly condensed and re-evaporated until only ethanol (78 °C) reaches the top. The thermometer holds at 78 °C until the ethanol has gone, then climbs.

Question 7

In a flame test, lithium and potassium give, respectively,

A yellow and red

B red and lilac

C lilac and orange-red

D blue-green and yellow

Li⁺ red, Na⁺ yellow, K⁺ lilac, Ca²⁺ orange-red, Cu²⁺ blue-green. Five colours, five gift marks — but only if the pairings are exact.

Question 8

A salt colours the flame blue-green. The cation is

A Ca²⁺

B Na⁺

C Cu²⁺

D K⁺

Blue-green = copper(II). Cross-check available: Cu²⁺ would also give a light blue precipitate with NaOH, dissolving in excess ammonia to deep blue.

Question 9

Aqueous sodium hydroxide gives a green precipitate with a solution containing

A iron(III) ions

B iron(II) ions

C copper(II) ions

D zinc ions

Fe²⁺ green; Fe³⁺ red-brown; Cu²⁺ light blue; Zn²⁺ white. The iron pair is the most-swapped duo in the syllabus — green goes with the lower oxidation state.

Question 10

A solution gives a white precipitate with NaOH(aq) that dissolves in excess NaOH, and a white precipitate with NH₃(aq) that does NOT dissolve in excess NH₃. The cation is

A Al³⁺

B Zn²⁺

C Ca²⁺

D Cu²⁺

Dissolves in excess NaOH but not excess ammonia = aluminium. Zinc dissolves in excess of both; calcium in neither; copper's precipitates are blue, not white.

Question 11

A white precipitate with NaOH(aq) that is insoluble in excess NaOH indicates

A Zn²⁺

B Al³⁺

C Ca²⁺

D NH₄⁺

Of the three white precipitates, only calcium hydroxide stays undissolved in excess NaOH. Ammonium gives no precipitate at all — it gives off ammonia gas on warming.

Question 12

Warming a solid with aqueous sodium hydroxide releases a gas that turns damp red litmus paper blue. The solid contains

A carbonate ions

B ammonium ions

C sulfate ions

D chloride ions

NH₄⁺ + OH⁻ →(warm) NH₃ + H₂O. Ammonia is the only common alkaline gas — damp red litmus turning blue identifies it uniquely, and hence the ammonium ion.

Question 13

Dilute acid is added to a white solid and the gas produced turns limewater milky. The solid contains

A carbonate ions

B sulfate ions

C nitrate ions

D bromide ions

Acid + carbonate → salt + water + CO₂, and CO₂ is the gas that turns limewater milky. Effervescence with acid is the carbonate's signature entrance.

Question 14

Acidified silver nitrate gives a yellow precipitate with a solution containing

A chloride ions

B bromide ions

C iodide ions

D sulfate ions

The halide ladder with AgNO₃: Cl⁻ white, Br⁻ cream, I⁻ yellow — deeper colour as you descend the group. Sulfate's white precipitate needs barium, not silver.

Question 15

The test for nitrate ions in solution uses

A acidified barium nitrate

B acidified silver nitrate

C aqueous sodium hydroxide and aluminium foil, warmed

D limewater

Aluminium reduces NO₃⁻ to ammonia in warm alkaline solution; the NH₃ turns damp red litmus blue. (If ammonium ions might be present, test with NaOH alone first — they release ammonia without any aluminium.)

Question 16

Before adding barium nitrate to test for sulfate ions, the solution is acidified to

A make the precipitate whiter

B remove carbonate ions that would give a false white precipitate

C dissolve the barium nitrate faster

D neutralise the sulfate ions

BaCO₃ is white too. The acid destroys any carbonate (fizzing off CO₂) so a surviving white precipitate can only be BaSO₄. The same logic protects the silver nitrate halide tests.

Question 17

(Supplement) Warm dilute acid is added to a solid and the gas evolved turns acidified potassium manganate(VII) from purple to colourless. The solid contains

A carbonate ions

B sulfite ions

C sulfate ions

D iodide ions

Sulfites release SO₂ with warm acid, and SO₂ decolourises purple KMnO₄. Sulfates do not release any gas with dilute acid — the sulfite/sulfate pair is separated by exactly this test.

Question 18

A lighted splint held at the mouth of a test tube gives a squeaky pop. The gas is

A oxygen

B hydrogen

C carbon dioxide

D ammonia

Lighted splint + pop = hydrogen (the pop is tiny explosive combustion: 2H₂ + O₂ → 2H₂O). Oxygen's test uses a glowing splint, which relights. Never swap the splints.

Question 19

Chlorine gas is identified because it turns damp litmus paper

A blue

B permanently red

C white — the litmus is bleached

D milky

Chlorine bleaches damp litmus (blue litmus flashes red first, then whitens). Turning damp red litmus blue is ammonia; "milky" belongs to limewater, not litmus.

Question 20

Anhydrous cobalt(II) chloride paper can be used to show that a liquid contains water because it turns

A from blue to pink

B from pink to blue

C from white to blue

D from purple to colourless

Cobalt(II) chloride: blue → pink with water. (Anhydrous copper(II) sulfate: white → blue.) Both show water is present; only a sharp boiling point at exactly 100 °C shows it is pure.

Topic 12: Experimental Techniques and Chemical Analysis

The Big Idea: Every Measurement Needs the Right Instrument

The Three Volume Instruments — Precision Compared

Collecting Gases: Three Questions Decide the Method

Judging density from relative molecular mass

Drying a Gas

The Language of Solutions

The Big Idea: Finding the Exact Moment of Neutralisation

The Procedure — Learn It as Numbered Steps

Indicators and Their Colour Changes

Reading the Burette and Getting Concordant Results

Titration calculations — the four-step machine

The Big Idea: A Race That Unmasks Mixtures

Setting Up the Race — and the Two Rules That Save It

Reading a Chromatogram

The Rf Value

The Big Idea: Purify, Prove It Pure, Then Name It

Purity: The Sharp-Point Rule

Choosing a Separation Method

Identification 1: Flame Tests for Metal Cations

Identification 2: Cations With NaOH(aq) and NH₃(aq)

Identification 3: Anion Tests

Identification 4: Gas Tests

The R_f Value