Required practicals: A-level Chemistry AQA revision notes

RP1

Standard solution and titration

Make a standard solution, then titrate

Definitions

Standard solution: a solution of accurately known concentration.
End point: when the indicator changes colour. Equivalence point: when the reactants have been added in exactly the ratio of the equation. A good indicator makes the two coincide.
Concordant titres: titres that agree within 0.10 cm³. Only these are averaged.

Key results

Amount: $n = c \times \frac{V}{1000}$ with $V$ in cm³.
Burette readings are recorded to the nearest 0.05 cm³. Each reading has an uncertainty of ±0.05 cm³, so a titre (two readings) has ±0.10 cm³.
Percentage uncertainty = $\frac{absolute uncertainty}{measured value} \times 100$ .

Method

Weigh the solid by difference: weigh the weighing bottle with solid, tip the solid into a beaker, reweigh the bottle. The difference is the mass transferred, whatever stayed behind.
Dissolve the solid in a small volume of deionised water in the beaker, stirring.
Transfer the solution to a volumetric flask through a funnel, then rinse the beaker, rod and funnel with deionised water into the flask, so that no solute is left behind.
Add deionised water until the bottom of the meniscus sits on the graduation mark, read at eye level; add the last drops with a dropping pipette.
Stopper and invert the flask several times so the concentration is the same throughout.
Rinse the pipette with the solution it will deliver and the burette with the titrant, so neither is diluted by water left on the glass.
Pipette the solution into a conical flask, add a few drops of indicator, and stand the flask on a white tile.
Run in the titrant while swirling; record a rough titre, then add dropwise near the end point in accurate runs until two or more titres are concordant.

In practice0.636 g of anhydrous Na₂CO₃ ( $M_{r}$ 106.0) is made up to 250.0 cm³. 25.0 cm³ needs a mean titre of 24.00 cm³ of hydrochloric acid. Find the concentration of the acid.

$n (Na_{2} CO_{3}) = \frac{0.636}{106.0} = 6.00 \times 1 0^{- 3} mol$ in the whole 250.0 cm³
$in 25.0 cm^{3} : 6.00 \times 1 0^{- 4} mol$ one tenth of the flask
$n (HCl) = 2 \times 6.00 \times 1 0^{- 4} = 1.20 \times 1 0^{- 3} mol$ Na₂CO₃ + 2HCl
$c = \frac{1.20 \times 1 0 ^{- 3}}{0.02400} = 0.0500 mol dm^{- 3}$ titre in dm³

Notes

Indicator choice: phenolphthalein (colourless to pink in alkali) when the titration involves a strong base; methyl orange (red in acid, yellow in alkali, orange at the end point) for a strong acid with a weak base.
The conical flask may be wet with deionised water: that adds no extra moles of reagent. The burette and pipette must not be.
To reduce the percentage uncertainty of the titre, make the titre larger: a larger pipette, a more concentrated sample, or a more dilute titrant.
A funnel left in the burette, or an air bubble below the tap, changes the reading after the run starts: remove the funnel and run some liquid through the tap first.
Exam trap: averaging every titre, including the rough one. Average only the concordant ones and say which you used.

A standard solution is one whose concentration is known accurately, because it was made from an accurately weighed mass of a pure, stable solid. It is then used to find the concentration of another solution by titration. Every step is about not losing solute and not changing a volume you think you know.

RP2

Measuring an enthalpy change

Measure an enthalpy change by calorimetry

Definitions

Enthalpy change, $Δ H$ : the heat energy change at constant pressure, per mole of the reaction as written.
Specific heat capacity, $c$ : the energy needed to raise 1 g of a substance by 1 K. For dilute aqueous solutions, take the value for water, 4.18 J g⁻¹ K⁻¹.

Key results

Energy transferred: $q = m c Δ T$ , with $m$ the mass of the solution (take 1.00 g cm⁻³, so $m$ in g equals the total volume in cm³).
$Δ H = - \frac{q}{n}$ , converted to kJ mol⁻¹. The sign is negative when the temperature rose (exothermic).
Use the moles of the limiting reactant; a solid added in excess does not set $n$ .

Method

Measure a known volume of one reactant into a polystyrene cup in a beaker (the beaker keeps it upright); add a lid with a hole for the thermometer.
Record the temperature every 30 s (or every minute) for about 3 minutes to establish a steady starting temperature.
At a set time (say the 4th minute) add the second reactant (a known volume of solution, or an excess of a weighed solid) and stir. Do not take a reading at the moment of mixing.
Keep recording the temperature at regular intervals for several minutes after the maximum.
Plot temperature against time. Draw a line through the steady start, and a line through the cooling points after the maximum; extrapolate the cooling line back to the time of mixing.
The temperature change is the gap between the two lines at the time of mixing.

In practiceExcess zinc powder is added to 25.0 cm³ of 0.200 mol dm⁻³ copper(II) sulfate solution. The corrected temperature rise is 10.2 K. Find ΔH for Zn + CuSO₄ → ZnSO₄ + Cu.

$q = m c Δ T = 25.0 \times 4.18 \times 10.2 = 1066 J$ mass of solution ≈ 25.0 g
$n (CuSO_{4}) = 0.200 \times 0.0250 = 5.00 \times 1 0^{- 3} mol$ zinc is in excess, so CuSO₄ limits
$Δ H = - \frac{1.066}{5.00 \times 1 0 ^{- 3}} = - 213 kJ mol^{- 1}$ negative: the temperature rose

Notes

Why extrapolate: the solution is losing heat while the reaction is still finishing, so the highest reading is lower than the true temperature it would have reached with no heat loss.
Assumptions to state when evaluating: the solution has the density and specific heat capacity of water; no heat is absorbed by the cup or thermometer; the reaction is complete.
Enthalpy of combustion with a spirit burner and a copper can gives values far less exothermic than data-book values: heat lost to the air, incomplete combustion (soot on the can), and evaporation of the fuel from the wick.
Exam trap: using only the volume of one solution for $m$ after mixing two solutions. The mass heated is the total.

A reaction in solution heats or cools the solution it happens in. Measure the temperature change, work out the energy that change represents, and divide by the moles that reacted. The main source of error is heat lost to (or gained from) the surroundings, and the method is built around correcting for it.

RP3

Rate and temperature

Rate against temperature: the disappearing cross

Definitions

Independent variable: temperature of the reaction mixture.
Dependent variable: the time, $t$ , for the cross to disappear. Rate is taken as $1/ t$ , because the same amount of sulfur must form each time.
Control variables: the volumes and concentrations of both solutions, the total volume, the flask and the cross.

Key results

Rate is proportional to $1/ t$ .
Arrhenius: $k = A e^{- E_{a} / RT}$ , so $ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}$ . Because $1/ t \propto k$ here, a graph of $ln (1/ t)$ against $1/ T$ (T in kelvin) has gradient $- E_{a} / R$ .
$E_{a} = - gradient \times 8.31$ in J mol⁻¹; divide by 1000 for kJ mol⁻¹.

Method

Measure the thiosulfate solution into a conical flask and the acid into a test tube; warm both separately in a water bath to the chosen temperature.
Place the flask on a cross drawn on paper.
Add the acid to the flask, swirl once and start the timer at the moment of mixing.
Look down through the solution and stop the timer when the cross can no longer be seen.
Measure the temperature of the mixture, ideally at the start and end, and use the mean.
Repeat at four or five temperatures (for example 20–60 °C) and repeat each one.

In practiceThe cross disappears after 64 s at 30 °C and 31 s at 40 °C. Compare the rates and estimate the activation energy.

$rate \propto \frac{1}{t} : \frac{1}{64} = 0.0156, \frac{1}{31} = 0.0323 s^{- 1}$ the rate roughly doubles
$ln \frac{0.0323}{0.0156} = \frac{E _{a}}{R} (\frac{1}{303} - \frac{1}{313})$ Arrhenius at two temperatures, T in K
$0.725 = \frac{E _{a}}{8.31} \times 1.05 \times 1 0^{- 4}$
$E_{a} \approx 5.7 \times 1 0^{4} J mol^{- 1} = 57 kJ mol^{- 1}$ two points only: a graph of several is better

Notes

Safety: sulfur dioxide is toxic and irritates the lungs; work in a well-ventilated room and avoid high acid concentrations at high temperatures. Pour the mixture away promptly after each run.
The end point is a judgement. Use the same observer and the same lighting each time, or replace the cross with a light sensor or colorimeter.
At high temperatures the mixture cools during the run, so the recorded temperature is not constant; use the mean of start and finish.
Exam trap: plotting $ln t$ (not $ln (1/ t)$ ) and getting the sign of $E_{a}$ wrong. Check that $E_{a}$ comes out positive.

Sodium thiosulfate reacts with hydrochloric acid to form a fine precipitate of sulfur, which gradually hides a cross drawn under the flask: $Na_{2} S_{2} O_{3} + 2 HCl \to 2 NaCl + SO_{2} + S + H_{2} O$ . Timing how long the cross takes to disappear at several temperatures, with everything else kept the same, shows how rate depends on temperature.

RP4

Tests for inorganic ions

Test-tube tests for cations and anions

Key results

Group 2 ions with dilute NaOH: Mg²⁺ white precipitate; Ca²⁺ slight white precipitate; Sr²⁺ very slight; Ba²⁺ none. Solubility of the hydroxides increases down the group.
Group 2 ions with dilute H₂SO₄: Ba²⁺ dense white precipitate (BaSO₄); Sr²⁺ white precipitate; Ca²⁺ slight; Mg²⁺ none. Solubility of the sulfates decreases down the group.
NH₄⁺: warm with NaOH(aq); ammonia is given off, which turns damp red litmus blue.
OH⁻: the solution turns red litmus blue (pH above 7).
CO₃²⁻: add dilute acid; effervescence, and the gas turns limewater milky (CO₂).
SO₄²⁻: acidify with dilute HCl, then add BaCl₂(aq); white precipitate of BaSO₄.
Halides: acidify with dilute HNO₃, then add AgNO₃(aq). Cl⁻ white AgCl, dissolves in dilute NH₃; Br⁻ cream AgBr, dissolves only in concentrated NH₃; I⁻ yellow AgI, insoluble in both.

Method

Use about 1 cm³ of the test solution in a clean test tube; add reagents dropwise, then in excess, observing after each.
For an unknown, test for carbonate first (acid), then sulfate (HCl then BaCl₂), then halides (HNO₃ then AgNO₃).
For ammonium ions, warm gently and hold damp red litmus at the mouth of the tube, not touching the glass.

In practiceIdentify an unknown white solid that dissolves in water.

Add dilute nitric acid: no effervescence, so no carbonate.
Add silver nitrate solution: a cream precipitate, which dissolves in concentrated (but not dilute) ammonia, so the anion is bromide.
Warm a fresh sample with NaOH(aq): the gas turns damp red litmus blue, so ammonium ions are present.
The solid is ammonium bromide, NH₄Br.

Notes

Why acidify before AgNO₃ or BaCl₂: carbonate (and hydroxide) ions would also give precipitates (Ag₂CO₃, BaCO₃), giving a false positive. The acid removes them.
Use nitric acid for the halide test and hydrochloric acid for the sulfate test: HCl adds chloride ions (false halide result), and H₂SO₄ adds sulfate ions.
Describe observations, not identities: 'white precipitate', not 'AgCl forms'. State colours of solutions and precipitates separately.
Exam trap: 'cloudy' and 'precipitate' are different from 'colourless solution'. Clear means transparent, not colourless.

Each test turns an ion into something you can see: a precipitate, a gas, or a colour. The order of tests matters, because some ions interfere with tests for others. Carbonate in particular gives precipitates with both silver and barium ions, so it is removed by acid first.

RP5

Distilling a product

Distil a product as it forms

Key results

$CH_{3} CH_{2} OH + [O] \to CH_{3} CHO + H_{2} O$ (ethanol to ethanal).
Dichromate(VI) is reduced from orange Cr₂O₇²⁻ to green Cr³⁺.
Boiling points: ethanal 20 °C, ethanol 78 °C, ethanoic acid 118 °C.

Method

Add anti-bumping granules to the flask, so that the liquid boils smoothly rather than in violent bursts.
Warm the acidified dichromate(VI) and add the alcohol slowly (or the reverse, as directed), with the distillation apparatus already set up.
Heat with a water bath or heating mantle, never a naked flame: the organic vapours are flammable.
Run cooling water into the condenser at the lower end and out at the upper end, so the jacket stays full and the coolest water meets the vapour last.
Place the thermometer bulb level with the side arm, where it measures the temperature of the vapour that is distilling over.
Collect the distillate in a receiver cooled in ice, because the aldehyde is very volatile.

In practiceOxidising 4.60 g of ethanol ( $M_{r}$ 46.0) gives 2.20 g of ethanal ( $M_{r}$ 44.0). Find the percentage yield.

$n (ethanol) = \frac{4.60}{46.0} = 0.100 mol$
$theoretical ethanal = 0.100 \times 44.0 = 4.40 g$ 1 : 1
$yield = \frac{2.20}{4.40} \times 100 = 50.0%$ losses: evaporation, further oxidation

Notes

Never heat a sealed apparatus: the pressure would build up. The receiver end is open (or loosely covered).
Collecting only the liquid that distils within a narrow temperature range gives a purer product.
Grease or clips on the joints stop vapour leaking; keep the joints clean.
Exam trap: drawing the thermometer bulb in the liquid. In distillation it measures the vapour; in reflux there is no thermometer in the condenser at all.

Oxidising a primary alcohol with acidified potassium dichromate(VI) gives an aldehyde first and then a carboxylic acid. To stop at the aldehyde, distil it out of the mixture as soon as it forms: the aldehyde has a much lower boiling point than the alcohol and the acid (no hydrogen bonding between aldehyde molecules), so it leaves first and is no longer in contact with the oxidising agent. Refluxing the same mixture, by contrast, returns the aldehyde to the flask and gives the acid.

RP6

Tests for organic functional groups

Identifying organic functional groups in a test tube

Key results

Alkene: shake with bromine water; orange to colourless.
Aldehyde: Tollens' reagent, warmed, gives a silver mirror; or Fehling's solution, warmed, turns from blue to a brick-red precipitate. Ketones give neither.
Primary or secondary alcohol (and aldehyde): warm with acidified potassium dichromate(VI); orange to green. Tertiary alcohols stay orange.
Carboxylic acid: add sodium hydrogencarbonate (or carbonate); effervescence, and the gas turns limewater milky.

Method

Make Tollens' reagent fresh: add a drop of NaOH(aq) to AgNO₃(aq) to give a brown precipitate, then add dilute NH₃(aq) dropwise until it just dissolves, forming [Ag(NH₃)₂]⁺.
Warm Tollens', Fehling's and dichromate tests in a beaker of hot water from a kettle, not over a flame, because the organic liquids are flammable.
Use a few drops of the organic compound and a fresh portion of reagent for each test.

In practiceUnknown liquid X: identify its functional group from these results.

Sodium hydrogencarbonate: no effervescence, so not a carboxylic acid.
Bromine water: stays orange, so no C=C.
Tollens' reagent, warmed: no silver mirror, so not an aldehyde.
Acidified dichromate(VI), warmed: orange to green, so X is a primary or secondary alcohol.

Notes

Tertiary alcohols are not oxidised: the carbon bearing the −OH has no hydrogen atom to lose.
Tollens' reagent must not be stored: it can form explosive silver compounds. Rinse the tubes with dilute nitric acid afterwards.
Bromine water is harmful; use it dilute and in a well-ventilated room.
Exam trap: using acidified dichromate alone to tell an alcohol from an aldehyde. Both can go green; Tollens' or Fehling's separates them.

Each test relies on a reaction that only one functional group gives, or gives easily, under mild conditions. Because some tests overlap (aldehydes and primary or secondary alcohols are both oxidised by acidified dichromate), a set of unknowns is identified by a sequence of tests rather than one.

RP7

Initial rates and continuous monitoring

Measuring rate: initial rates and continuous monitoring

Definitions

Initial rate: the rate at $t = 0$ , found from the gradient of the tangent to a concentration–time (or volume–time) curve at the start.
Clock reaction: a reaction that produces a sudden colour change once a fixed amount of product has formed; $1/ t$ is then proportional to the initial rate.

Key results

In the clock reaction the thiosulfate removes iodine as it forms; once it is used up, iodine colours the starch. The same amount of iodine forms in each run, so rate $\propto 1/ t$ .
Order with respect to a reactant: if doubling its concentration doubles the rate, first order; quadruples, second order; no change, zero order.
Initial rate from a curve = gradient of the tangent at $t = 0$ , in the units of the $y$ -axis per second.

Method

Iodine clock (initial rates): mix hydrogen peroxide, acid, a small fixed amount of sodium thiosulfate and starch; add potassium iodide and start the timer. Stop it when the blue-black colour appears.
Change the concentration of one reactant at a time by dilution, keeping the total volume the same so the other concentrations are unchanged.
Continuous monitoring: collect the gas in a gas syringe (or measure mass loss on a balance, or absorbance with a colorimeter) at regular intervals from the moment of mixing.
Plot the measured quantity against time and draw tangents.

In practiceTripling [A] from 0.10 to 0.30 mol dm⁻³ raises the initial rate from $2.0 \times 1 0^{- 4}$ to $1.8 \times 1 0^{- 3}$ mol dm⁻³ s⁻¹, other concentrations fixed. Find the order with respect to A.

$\frac{1.8 \times 1 0 ^{- 3}}{2.0 \times 1 0 ^{- 4}} = 9$ the rate ratio
$9 = 3^{n} \Rightarrow n = 2$ concentration ratio 3
$rate \propto [A]^{2}$ second order in A

Notes

Only a small amount of thiosulfate is used, so the reactant concentrations barely change before the colour appears: the measured rate is close to the initial rate.
Draw tangents long, using two points far apart on the line, to reduce the uncertainty in the gradient.
A gas syringe can stick; a gas collected over water may dissolve (CO₂). State these as limitations.
Exam trap: changing a concentration by adding a different volume of solution without making up the total volume, which changes every other concentration as well.

Two ways to find how rate depends on concentration. In the initial rate method, many short experiments each give one rate at the start, when the concentrations are those you made up. In continuous monitoring, one experiment follows a property (gas volume, mass, colour) over time, and rates are found from tangents to the curve.

RP8

EMF of electrochemical cells

Two half-cells and a voltmeter: cell EMF

Definitions

EMF: the potential difference across a cell when no current flows.
Salt bridge: completes the circuit by allowing ions to move between the half-cells without mixing the solutions; usually filter paper soaked in saturated potassium nitrate.

Key results

$E_{cell} = E_{right (more positive)}^{⊖} - E_{left (more negative)}^{⊖}$ .
Cu²⁺/Cu (+0.34 V) with Zn²⁺/Zn (−0.76 V): $E_{cell} = 0.34 - (- 0.76) = 1.10$ V; zinc is the negative electrode.
Conventional representation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).

Method

Clean each metal electrode with emery paper to remove the oxide layer, then rinse.
Place each electrode in a 1.00 mol dm⁻³ solution of its own ions in a beaker.
Connect the two solutions with a salt bridge made fresh for each pair.
Connect the electrodes to a high-resistance voltmeter, so almost no current flows and the reading is the EMF.
Record the reading, noting which electrode is positive.

In practiceA cell is made from Ni²⁺/Ni ( $E^{⊖} = - 0.25$ V) and Cu²⁺/Cu ( $E^{⊖} = + 0.34$ V). Find the EMF and write the cell representation.

$E_{cell} = 0.34 - (- 0.25) = 0.59 V$ more positive minus more negative
$Ni (s) ∣ Ni^{2 +} (aq) ∣∣ Cu^{2 +} (aq) ∣ Cu (s)$ nickel is oxidised, so it goes on the left

Notes

Potassium nitrate is chosen because nitrate and potassium ions form no precipitates with the ions in the half-cells; KCl would precipitate AgCl with silver ions.
A measured value differs from the standard value if concentrations are not 1.00 mol dm⁻³, the temperature is not 298 K, the metal surface is oxidised, or the voltmeter draws a current.
Exam trap: subtracting the wrong way round and quoting a negative EMF. A cell EMF is positive.

Two half-cells joined by a salt bridge and a high-resistance voltmeter form a cell. The voltmeter reading is the EMF: the difference between the electrode potentials of the two half-cells. Measuring it for different pairs lets you compare how readily each metal loses electrons.

RP9

pH curves

pH curves for titrations

Key results

Weak acid–strong base: equivalence pH above 7 (the conjugate base, e.g. ethanoate, is basic). Strong acid–weak base: equivalence pH below 7 (e.g. NH₄⁺ is acidic).
At half the equivalence volume, $[HA] = [A^{-}]$ , so $pH = p K_{a}$ and $K_{a} = 1 0^{- pH}$ .
A suitable indicator changes colour entirely within the vertical section of the curve.

Method

Calibrate the pH meter with buffer solutions of known pH (for example pH 4, 7 and 10) and rinse the probe with deionised water between solutions.
Pipette the acid into a beaker or conical flask and record its pH.
Add the base from a burette in measured portions (say 2 cm³), stirring and recording the pH after each.
Add smaller portions (0.5 cm³ or less, then dropwise) near the equivalence point, where the pH changes quickly.
Continue well past the equivalence point and plot pH against volume added.

In practice25.0 cm³ of a weak acid needs 20.0 cm³ of 0.100 mol dm⁻³ NaOH; the pH at 10.0 cm³ is 3.75. Find $K_{a}$ and the acid's concentration.

$at 10.0 cm^{3} : [HA] = [A^{-}], pH = p K_{a} = 3.75$ half the equivalence volume
$K_{a} = 1 0^{- 3.75} = 1.78 \times 1 0^{- 4} mol dm^{- 3}$
$c = \frac{0.100 \times 20.0}{25.0} = 0.0800 mol dm^{- 3}$ 1 : 1 at equivalence

Notes

Phenolphthalein (pH 8.3–10) suits a weak acid with a strong base; methyl orange (3.1–4.4) suits a strong acid with a weak base. Neither suits a weak acid with a weak base, which has no steep section.
The meter reads to 0.01 pH; its accuracy depends on calibration. Temperature also changes pH readings.
Exam trap: reading $p K_{a}$ at the equivalence point. It is at half the equivalence volume.

Following the pH with a meter as base is added to acid (or acid to base) gives a pH curve. Its shape tells you the strengths of the acid and base, the equivalence point, which indicators would work, and for a weak acid its $K_{a}$ .

RP10

Preparing a pure organic solid and liquid

Prepare and purify an organic solid and an organic liquid

Key results

A pure solid melts sharply at its data-book value; impurities lower the melting point and spread it over a range.
Percentage yield = $\frac{actual mass}{theoretical mass} \times 100$ ; the theoretical mass comes from the moles of the limiting reagent.

Method

Solid (e.g. aspirin): dissolve the crude solid in the minimum volume of hot solvent, so that the solution is saturated when hot.
Filter the hot solution through fluted paper if there are insoluble impurities.
Cool slowly, then in ice: the product crystallises; soluble impurities stay in solution because they are present in small amounts.
Collect the crystals by suction filtration with a Büchner funnel; wash with a little ice-cold solvent and leave to dry.
Liquid (e.g. cyclohexene from cyclohexanol): distil the crude product from the reaction mixture.
In a separating funnel, wash with water or sodium hydrogencarbonate solution to remove acid, inverting gently and releasing the pressure through the tap often.
Run off and discard the aqueous layer, keeping the organic layer (identify the layers by density).
Dry the organic layer with anhydrous calcium chloride or magnesium sulfate until it is clear and the drying agent no longer clumps; decant or filter.
Redistil, collecting the fraction that boils within a degree or two of the product's boiling point.

In practice2.00 g of salicylic acid ( $M_{r}$ 138.0) gives 1.95 g of pure, dry aspirin ( $M_{r}$ 180.0). Find the percentage yield.

$n = \frac{2.00}{138.0} = 1.449 \times 1 0^{- 2} mol$ salicylic acid limits
$theoretical = 1.449 \times 1 0^{- 2} \times 180.0 = 2.61 g$ 1 : 1
$yield = \frac{1.95}{2.61} \times 100 = 74.7%$ losses in recrystallisation and transfer

Notes

Use the minimum of hot solvent: extra solvent keeps more product in solution when cold and lowers the yield.
Wash crystals with cold solvent, and only a little, so the product does not redissolve.
Melting point: heat slowly near the expected value in a melting-point apparatus or a Thiele tube; record the range from first liquid to all liquid.
Exam trap: saying recrystallisation removes 'all' impurities, or that a wide melting range means 'wet'. A wide range below the true value means impure.

A crude organic product contains unreacted starting materials, by-products, catalyst and solvent. Solids are purified by recrystallisation and tested by melting point; liquids are washed in a separating funnel, dried and redistilled, and tested by boiling point.

RP11

Tests for transition metal ions

Test-tube reactions of transition metal ions

Key results

Cu²⁺ (blue solution): NaOH or a little NH₃ gives a blue precipitate, [Cu(H₂O)₄(OH)₂]; excess NaOH, no change; excess NH₃, a deep blue solution of [Cu(NH₃)₄(H₂O)₂]²⁺; Na₂CO₃ gives a blue-green precipitate of CuCO₃.
Fe²⁺ (pale green solution): NaOH or NH₃ gives a green precipitate, [Fe(H₂O)₄(OH)₂], which darkens to brown in air (oxidised to iron(III)); Na₂CO₃ gives a green precipitate of FeCO₃.
Fe³⁺ (yellow-brown solution): NaOH or NH₃ gives a brown precipitate, [Fe(H₂O)₃(OH)₃]; Na₂CO₃ gives a brown precipitate and effervescence (CO₂).
Al³⁺ (colourless solution): NaOH gives a white precipitate that dissolves in excess, forming [Al(OH)₄]⁻ (amphoteric); NH₃ gives a white precipitate insoluble in excess; Na₂CO₃ gives a white precipitate and effervescence.

Method

Add the reagent dropwise to about 1 cm³ of the solution, shaking after each drop; record what forms.
Then add the reagent in excess and record any change.
Leave an iron(II) precipitate to stand and look again after a few minutes.

In practiceA pale green solution gives a green precipitate with NaOH(aq), insoluble in excess, that turns brown at the surface on standing. Identify the ion and explain the change.

The ion is iron(II), [Fe(H₂O)₆]²⁺: the green precipitate is [Fe(H₂O)₄(OH)₂].
Oxygen in the air oxidises iron(II) to iron(III), giving brown [Fe(H₂O)₃(OH)₃] where the precipitate meets the air.

Notes

Why 3+ ions give CO₂ with carbonate: their higher charge density polarises the water ligands more, so [M(H₂O)₆]³⁺ releases H⁺ readily and the solution is acidic enough to react with carbonate ions. 2+ ions are not, and simply form MCO₃.
Concentrated ammonia and sodium hydroxide are corrosive; wear eye protection.
Exam trap: writing 'Cu(OH)₂' without the water ligands when a complex formula is asked for. AQA expects the neutral complex, e.g. [Cu(H₂O)₄(OH)₂].

In water, metal ions exist as aqua complexes. Adding hydroxide ions, ammonia or carbonate removes protons from the water ligands or replaces them, giving coloured precipitates and solutions that identify the metal. The 3+ ions are more acidic than the 2+ ions, which explains their different reactions with carbonate.

RP12

Thin-layer chromatography

Separate and identify by thin-layer chromatography

Definitions

$R_{f}$ value: $\frac{distance moved by the spot}{distance moved by the solvent front}$ , both measured from the baseline.

Key results

A more polar compound is held more strongly by the silica and moves less: lower $R_{f}$ .
A spot with the same $R_{f}$ as a standard on the same plate is likely to be that compound; one spot does not prove purity.

Method

Draw a baseline in pencil about 1 cm from the bottom (ink would dissolve and run).
Spot the mixture and each standard on the baseline with a capillary tube, keeping the spots small and concentrated; let them dry.
Stand the plate in a beaker with solvent below the baseline, so the samples do not dissolve into the solvent pool; cover the beaker so the air inside is saturated with solvent vapour.
Remove the plate when the solvent is near the top and mark the solvent front at once in pencil, before it evaporates.
Show colourless spots under UV light (on a fluorescent plate) or with iodine vapour, or ninhydrin for amino acids; draw round them.
Measure distances to the centre of each spot and calculate $R_{f}$ .

In practiceOn one plate the solvent front is 6.0 cm from the baseline. A spot of the sample is 3.3 cm up; a glycine standard spot is also 3.3 cm up. Interpret.

$R_{f} = \frac{3.3}{6.0} = 0.55$ sample and standard alike
Same Rf on the same plate: the sample probably contains glycine. A second solvent would confirm it.

Notes

$R_{f}$ values depend on the solvent, the temperature and the plate, which is why standards are run alongside the sample.
Handle the plate by its edges and wear gloves: ninhydrin and many solvents are toxic or flammable; work in a fume cupboard or well-ventilated room.
Exam trap: measuring to the top of a spot, or to the top of the plate instead of the solvent front.

A TLC plate carries a thin layer of silica (or alumina), the stationary phase. A solvent, the mobile phase, rises up the plate and carries each component a distance that depends on how strongly it is adsorbed by the polar stationary phase compared with how soluble it is in the solvent. Comparing spots with standards run on the same plate identifies the components.