Measuring the rate of a reaction
Full theory, orders & rate–concentration graphs: Rate equations · Kinetics: measuring rate
Aim: measure the rate of a reaction by two different methods: an initial-rate method (the iodine clock) and a continuous-monitoring method (collecting a gas as it is produced), and use each to find an order of reaction. AQA require both; you will have done one of each.
They answer the same question from opposite ends. An initial-rate method measures one number per run (how long a fixed, tiny amount of product takes to appear), then you change one concentration and run it again. A continuous-monitoring method follows one run all the way through, so a single experiment gives you a whole curve to take gradients from.
Method A: the iodine clock (an initial-rate method)
Hydrogen peroxide oxidises iodide ions to iodine. A small, fixed amount of thiosulfate is also present, and it mops up the iodine as fast as it forms:
H2O2(aq) + 2H+(aq) + 2I−(aq) → I2(aq) + 2H2O(l)
2S2O32−(aq) + I2(aq) → 2I−(aq) + S4O62−(aq)
While any thiosulfate is left there is no free iodine, so the mixture stays colourless. The moment the thiosulfate runs out, iodine builds up and the starch turns the whole beaker blue-black: that is your clock stopping. Because the same fixed amount of iodine is made every time, 1/t is proportional to the initial rate.
- Rinse a burette with potassium iodide solution and fill it.
- Transfer 10.0 cm3 of hydrogen peroxide solution to a clean, dry 100 cm3 beaker (you add this last).
- Into a clean, dry 250 cm3 beaker put 25 cm3 of dilute sulfuric acid, then 20 cm3 of water, then about 1 cm3 of starch solution.
- Run 5.0 cm3 of potassium iodide solution in from the burette.
- Add 5.0 cm3 of sodium thiosulfate solution last, and stir.
- Tip in the hydrogen peroxide, start the timer immediately, and keep stirring.
- Stop the timer the instant the mixture turns blue-black. Record the time in a table of your own design.
- Rinse and dry the beaker, then repeat with 10.0, 15.0, 20.0 and 25.0 cm3 of potassium iodide, reducing the water to 15, 10, 5 and 0 cm3 so the total volume never changes.
Method B: collecting a gas (a continuous-monitoring method)
- Measure 50 cm3 of 0.8 mol dm−3 hydrochloric acid into a 100 cm3 conical flask.
- Clamp a 100 cm3 gas syringe and connect it to a bung and delivery tube (or collect the gas over water in an inverted measuring cylinder).
- Add one 6 cm strip of magnesium ribbon, push the bung firmly home and start the timer at the same moment.
- Record the volume of hydrogen every 15 s for 2.5 minutes.
- Repeat with 0.4 mol dm−3 acid, made by mixing 25 cm3 of the 0.8 mol dm−3 acid with 25 cm3 of water.
Apparatus & techniques
- measuring rate by two different methods
- burette & measuring cylinder to a stated precision
- collecting a gas in a syringe or over water
- stopwatch, and judging a colour end point
- tangents, gradients & rate–concentration graphs
Safety. Wear eye protection throughout. The 0.8 and 0.4 mol dm−3 hydrochloric acid is below the 2 mol dm−3 at which it becomes an irritant, so it is classed as low hazard, but a splash can still harm the eyes. Whether the dilute sulfuric acid is an irritant depends on its concentration, so treat it with the same care; the hydrogen peroxide solution is an irritant that bleaches skin and clothing. Hydrogen is flammable: keep the gas syringe away from flames, and never seal the apparatus so tightly that pressure can build up. Wash any splashes off at once.
What you should see. In the clock, nothing at all happens for most of the run, then the whole beaker goes blue-black in about a second. In the gas-collection run, the plunger moves fastest at the very start and slows steadily as the magnesium, the limiting reactant, is used up, finally stopping when it has all gone.
Data analysis: getting an order out of each method
The two methods need different processing, and both are examined.
| Method | What you measure | How it becomes a rate | What you plot |
|---|---|---|---|
| Initial rate (clock) | time t for the colour to appear | rate ∝ 1/t, because the same amount of I2 is made every run | 1/t against the concentration you changed |
| Continuous (gas) | volume of gas at each time | gradient of a tangent to the curve | volume against time, then rate against concentration |
Reading a straight-line rate–concentration graph is the part that carries the marks: a horizontal line means zero order, a straight line through the origin means first order, and an upward curve means second order.
- Keep the total volume constant. The water is not padding: it is there so the only concentration that changes is the one you are investigating. Say this if asked why water is added.
- Keep the temperature the same in every run, with the beaker and the hydrogen peroxide in a water bath: a warmer run is faster for a reason that has nothing to do with concentration.
- Add the thiosulfate last, and start the clock as the peroxide goes in. Any delay between mixing and starting the timer is a systematic error that makes every time too short.
- The clock measures the initial rate only because so little reactant is used before the colour appears: the concentrations are still effectively their starting values.
- Gas escapes before the bung is in. In the continuous method the gas made first, when the reaction is fastest, is never collected, so the initial rate comes out too low. Add the magnesium and bung the flask in one movement.
- Draw the tangent at t = 0 to the curve itself, touching it at the origin, and take the gradient over a large triangle. A tangent drawn as a chord, or over two small squares, is the commonest lost mark on this practical.
- A loose cotton-wool plug, not a bung, if you are following mass loss on a balance: it stops acid spraying out but still lets the gas escape.
Exam questions: RP7
- Place a conical flask on a balance and add approximately 20 g of large marble chips.
- Add 50 cm3 of 0.4 mol dm−3 hydrochloric acid.
- Place a loose cotton wool plug in the neck of the flask.
- Zero the mass reading on the balance.
- Start a timer.
- Record the loss in mass (mt) every 30 seconds for 4 minutes.
- Wait for the reaction to finish and record the total mass loss (mtotal).
- Plot a graph of (mtotal − mt) against time.
Suggest why a loose cotton wool plug is placed in the neck of the flask, instead of leaving the flask open or inserting a bung.
Show answer
Instead of leaving the flask open: to avoid acid / solution / liquid escaping, or to avoid it splashing / spraying / spitting out. 1 mark
Instead of inserting a bung: to allow the gas / CO2 to escape. 1 mark
Ignore evaporation, spilling, impurities getting in, and “pressure would build up”. A wrongly named gas scores nothing.
20 g of large marble chips is a large excess of calcium carbonate. Suggest why using a large excess of calcium carbonate means that the rate is only affected by the changing concentration of the hydrochloric acid.
Show answer
So that the surface area / mass / amount of calcium carbonate stays approximately constant. 1 mark
Ignore “concentration stays constant”, “volume”, “so HCl is the limiting factor”, and simply restating that the rate is only affected by [HCl].
The mass of carbon dioxide produced in time t is equal to mt. The total mass of CO2 produced during the reaction is equal to mtotal. Explain why (mtotal − mt) is proportional to the concentration of hydrochloric acid remaining in the flask at time t.
Show answer
mt, the mass of CO2 produced in time t, is proportional to the amount / concentration of HCl that has reacted at time t. 1 mark
mtotal, the total mass of CO2 produced, is proportional to the total amount / concentration of HCl that reacted / was present initially, so (mtotal − mt) is proportional to the HCl present at time t. 1 mark
Equally accepted: (mtotal − mt) is the mass of CO2 still to be produced, and that is proportional to the HCl still to react. “Equal to” or “represents” are allowed for “proportional to”.
The rate of reaction, calculated from the gradient of the curve at five different times, is shown below, together with the corresponding value of (mtotal − mt).
| Rate of reaction / g s−1 | (mtotal − mt) / g |
|---|---|
| 23.0 × 10−4 | 0.340 |
| 19.0 × 10−4 | 0.280 |
| 15.7 × 10−4 | 0.225 |
| 11.5 × 10−4 | 0.170 |
| 6.67 × 10−4 | 0.100 |
Plot the rate of reaction (y-axis) against (mtotal − mt) (x-axis) and draw a line of best fit. Then state how your graph confirms that the rate equation for this reaction is rate = k[HCl].
Show answer
Scales chosen so the plotted points (and the origin, if shown) occupy more than half of each axis, with both axes labelled including units. 1 mark
Points plotted to within half a small square. 1 mark
A suitable straight line, within one square of every point. 1 mark
How it confirms the order: the graph is a straight line and it passes through the origin. 1 mark
“Constant gradient” is allowed for “straight line”, but the line must also be stated to go through the origin: “directly proportional” on its own is ignored. A reversed scale loses the first mark but the rest carry through by error.
In this experiment the variable measured is mass loss. The rate of this reaction at a constant temperature can be investigated in other ways. Suggest two other variables that can be measured instead of mass loss.
Show answer
Any two from: volume of gas / CO2 · pH · concentration of HCl / acid / H+ · conductivity. 2 marks
Temperature scores nothing. “Volume” or “concentration” unqualified scores nothing either. Say of what.
Explain why the reaction is fastest at the start.
Show answer
The concentration of hydrogen peroxide / the number of H2O2 molecules is highest at the start. 1 mark
So there are more frequent successful collisions. 1 mark
The table below shows data from a similar experiment. Plot the data and draw a line of best fit, then use your graph to determine the order of reaction with respect to H2O2, stating how the graph shows this order.
| [H2O2] / mol dm−3 | Rate / mol dm−3 s−1 |
|---|---|
| 0.02 | 0.000 49 |
| 0.03 | 0.000 73 |
| 0.05 | 0.001 24 |
| 0.07 | 0.001 68 |
| 0.09 | 0.002 19 |
Show answer
Order = 1 (first order with respect to H2O2). 1 mark
How the graph shows it: the rate–concentration graph is a straight line through the origin, so the rate is directly proportional to [H2O2]. 1 mark
Check it yourself without plotting: 0.001 24 / 0.05 = 0.0248, and 0.002 19 / 0.09 = 0.0243: rate / concentration is constant, which is exactly what first order means.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
Measuring the EMF of an electrochemical cell
Full theory, cell notation & feasibility: Electrode potentials & cells · Redox
Aim: build a standard cell from two half-cells, measure its EMF with a high-resistance voltmeter, and then use one metal as a reference to rank several metals by their electrode potentials.
- Clean a piece of copper and a piece of zinc foil with emery paper or fine sandpaper.
- Degrease each metal with cotton wool and propanone.
- Stand the copper in a 100 cm3 beaker holding about 50 cm3 of 1.00 mol dm−3 CuSO4.
- Stand the zinc in a second beaker holding about 50 cm3 of 1.00 mol dm−3 ZnSO4.
- Plug one end of a U-tube with cotton wool, fill it with 2.0 mol dm−3 sodium chloride solution, then plug the free end with cotton wool soaked in the same solution.
- Invert the U-tube so a plugged end dips into each beaker: that is your salt bridge.
- Clip a lead to each metal and connect them to a digital (high-resistance) voltmeter. Read the voltage.
- Keep the copper half-cell from the first method (copper foil in 1.00 mol dm−3 CuSO4) as your fixed reference, clipped to the positive terminal of the voltmeter.
- Clean and degrease the next metal exactly as before, and stand it in a beaker of 1.00 mol dm−3 solution of that metal’s own ions: zinc in ZnSO4, iron in FeSO4, magnesium in MgSO4, silver in AgNO3.
- Join the new half-cell to the copper reference with a fresh salt bridge: a U-tube, or a strip of filter paper, soaked in saturated potassium nitrate.
- Clip the second lead to the new metal, then note the voltage and its sign and record it in a table.
- Repeat with each metal provided, remaking the salt bridge each time, then write the conventional representation for every cell you have made.
- Use your readings to work out which pair would give the largest EMF.
Magnesium can be used as the reference instead of copper. Whichever you choose, every reading is measured against that one metal rather than against a standard hydrogen electrode, so the raw readings are cell EMFs relative to that reference. Under standard conditions, its known standard electrode potential lets you calculate the other standard electrode potentials.
Why each metal needs its own solution. An electrode potential is the position of the equilibrium Mn+(aq) + ne− ⇋ M(s). Stand the metal in anything else (sodium chloride solution, damp filter paper) and there are no Mn+ ions for that equilibrium to sit in, so there is nothing defined to measure and no cell you could write down. Use potassium nitrate, not sodium chloride, in the salt bridge if silver is one of your metals: chloride ions would precipitate AgCl on the electrode and kill the reading.
Apparatus & techniques
- preparing and using an electrochemical cell
- cleaning & degreasing electrodes
- making and placing a salt bridge
- reading a high-resistance voltmeter, with sign
- safe handling of solutions of transition-metal salts
Safety. Wear eye protection. Propanone is highly flammable: keep it away from flames and use it in a well-ventilated room. Copper(II) sulfate solution is harmful if swallowed and an irritant, and is dangerous to the environment, so it must not go down the sink; zinc sulfate solution is an irritant too. Emery paper leaves sharp metal edges. Handle the cleaned foil by its flat faces.
What you should see. The standard zinc–copper cell reads about 1.10 V, with the copper positive and the zinc negative. If the leads are the other way round the meter simply shows −1.10 V, which tells you the same thing. A reading that drifts, or is far too low, almost always means the electrodes were not cleaned or the salt bridge is not making contact.
Data analysis: from two readings to a cell
The number the voltmeter gives you is the EMF. Turning it into chemistry means deciding which half-cell is the positive electrode, which is the negative, and writing the cell down in the conventional way:
E⦵cell = E⦵(more positive half-cell) − E⦵(more negative half-cell)
| Half-cell | Electrode reaction | E⦵ / V |
|---|---|---|
| Ca2+/Ca | Ca2+(aq) + 2e− → Ca(s) | −2.87 |
| Mg2+/Mg | Mg2+(aq) + 2e− → Mg(s) | −2.38 |
| Ti2+/Ti | Ti2+(aq) + 2e− → Ti(s) | −1.63 |
| Zn2+/Zn | Zn2+(aq) + 2e− → Zn(s) | −0.76 |
| Fe2+/Fe | Fe2+(aq) + 2e− → Fe(s) | −0.44 |
| Cu2+/Cu | Cu2+(aq) + 2e− → Cu(s) | +0.34 |
| Ag+/Ag | Ag+(aq) + e− → Ag(s) | +0.80 |
Zinc has the more negative electrode potential, so zinc is the negative electrode and copper the positive:
E⦵cell = (+0.34) − (−0.76) = +1.10 V
The zinc is oxidised (it is the better reducing agent, so it gives up electrons) and the copper(II) ions are reduced. Electrons flow through the wire from zinc to copper, and the conventional representation is written with the more negative half-cell on the left:
Zn(s) | Zn2+(aq) ‖ Cu2+(aq) | Cu(s)
Read it outwards from the double line: the oxidised species of each pair sits next to the salt bridge, the reduced species at the outside.
Build a cell from any two half-cells. The widget puts the more negative one on the left, marks the electrodes, works out the EMF and writes the conventional representation. Check yourself against it before you look.
first half-cell
second half-cell
- Clean and degrease the electrodes. Emery paper takes off the oxide layer; propanone takes off grease. Either one left on gives a low, drifting reading.
- The voltmeter must be high-resistance (a digital one is). Almost no current then flows, so the concentrations stay at 1.00 mol dm−3 and the reading is the maximum potential difference, which is what EMF means.
- The salt bridge completes the circuit by letting ions move between the half-cells and keeping each solution electrically neutral. It must not react with either solution: potassium nitrate is the usual choice, and you must not use a chloride bridge with silver ions because silver chloride would precipitate.
- Standard conditions or it is not a standard EMF: all solutions 1.00 mol dm−3, 298 K, and 100 kPa for any gas. Adding water to one half-cell shifts its equilibrium and changes the reading.
- The size of the electrode makes no difference. A bigger piece of metal, or a bigger platinum electrode, changes how fast charge could flow, not the potential difference, a favourite one-mark question.
- Record the sign. A negative reading is a real result, not a mistake: with the reference on the positive terminal, it says the test metal has the more positive electrode potential, so the test metal is the positive electrode of that cell.
Exam questions: RP8
A cell is set up to measure the standard electrode potential for the half-cell Mg2+(aq) + 2e− → Mg(s), using a magnesium electrode in magnesium chloride solution joined by a salt bridge to a standard hydrogen electrode. State the purpose of the salt bridge, and identify an ionic compound that could be used in it.
Show answer
Purpose: to complete the circuit, or to allow ions to move between the half-cells, or to maintain electrical neutrality. 1 mark
Identity: potassium nitrate or sodium nitrate, or any soluble ionic compound that does not react with H+, magnesium ions or chloride ions. 1 mark
“Allows electrons to flow” scores nothing: electrons travel through the wire, ions through the bridge.
State how, if at all, the EMF of this cell will change if the surface area of the platinum electrode is increased.
Show answer
No change. 1 mark
The standard electrode potential for the half-cell is E⦵ = −2.38 V. Water is added to the beaker containing the magnesium chloride solution. What is the effect on the magnitude of the EMF of the cell? Tick one box.
The voltmeter in this cell is replaced by a bulb. Give an equation for the overall reaction that occurs when the cell is operating.
Show answer
Mg + 2HCl → MgCl2 + H2
1 mark: Mg + 2H+ → Mg2+ + H2 is equally accepted, state symbols are ignored, and multiples are allowed.
Standard electrode potentials are measured by comparison with the standard hydrogen electrode. State the substances and conditions needed in a standard hydrogen electrode.
Show answer
H2(g) and 100 kPa. 1 mark
1 mol dm−3 and HCl / HNO3 / H+, or 0.5 mol dm−3 H2SO4. 1 mark
Pt electrode and a temperature of 298 K. 1 mark
1 bar is allowed for 100 kPa; 1 atm and 101 kPa are not. Each mark needs both halves of the pair.
Describe an experiment a student does to show that the standard electrode potential for the TiO2+(aq) / Ti(s) electrode is −0.88 V. The student is provided with the Cu2+(aq) / Cu(s) electrode set up ready to use, solid titanium(IV) oxysulfate (Mr = 159.9), 0.50 mol dm−3 sulfuric acid, a strip of titanium, and laboratory apparatus and chemicals. Your answer should include how to prepare the solution, how to connect the electrodes, the measurements taken, and how they are used to calculate the electrode potential.
Show answer
Marked by levels of response across three stages. The indicative chemistry is:
Stage 1: preparing the solution. Weigh 7.995 / 8.00 g of TiOSO4; dissolve it in the 0.50 mol dm−3 sulfuric acid; transfer to a volumetric flask and make up to the mark.
Stage 2: setting up the cell. A piece of titanium immersed in the acidified TiO2+(aq); the two solutions connected by a salt bridge; the two metals connected through a high-resistance voltmeter. A labelled diagram earns these marks.
Stage 3: measurement and calculation. Record the EMF of the cell; use Ecell = ERHS − ELHS, i.e. Ecell = Ecopper − Etitanium; the reading should be +1.22 V with copper on the right (or −1.22 V with copper on the left).
Level 3 — 5–6 marks all three stages covered, correct and virtually complete, with clear practical detail (weighing into a beaker or by difference plus washings, not straight into the volumetric flask; a suitable, saturated salt-bridge solution).
Level 2 — 3–4 marks all three stages covered but incomplete or containing inaccuracies, or two stages covered well.
Level 1 — 1–2 marks two stages covered but incomplete, or one stage covered well.
Where 7.995 g comes from: 50 cm3 of a 1.00 mol dm−3 solution needs 0.0500 mol, and 0.0500 × 159.9 = 7.995 g.
Give the half-equation for the electrode reaction in the TiO2+(aq) / Ti(s) electrode in acidic conditions.
Show answer
TiO2+ + 2H+ + 4e− → Ti + H2O
1 mark: the reverse reaction is allowed, state symbols are ignored, and multiples, fractions or an equilibrium arrow are all accepted.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.
How pH changes in an acid–base titration
Full theory, Ka, buffers & indicators: Acids and bases: pH curves
Aim: follow the pH with a meter as alkali is run into acid, and plot the titration curve. AQA name two cases: a weak acid with a strong base, and a strong acid with a weak base. The curve you get tells you where the equivalence point is, which indicator would work, and (for a weak acid) its Ka.
- Rinse the probe thoroughly with deionised water and shake off the excess.
- Stand it in the pH 7.00 buffer with the bulb fully immersed, and record the reading.
- Rinse again, then repeat with the pH 4.00 and pH 9.20 buffers.
- Plot your reading (x) against the true pH of the buffer (y). The line may be straight or slightly curved. It is what you use to correct every reading in Part 2.
- Rinse a burette with 0.100 mol dm−3 ethanoic acid, fill it, and run exactly 20.0 cm3 into a clean 100 cm3 beaker. Label the burette so you do not mix the two up.
- Rinse and fill a second burette with 0.100 mol dm−3 sodium hydroxide solution.
- Rinse the probe, clamp it with the bulb fully immersed in the acid, stir gently and record the starting pH.
- Add the alkali in 2.0 cm3 portions up to 18 cm3, stirring and recording the pH after each addition.
- Switch to 0.20 cm3 portions from 18 cm3 to 22.0 cm3: that is where the pH is changing fastest.
- Go back to 2.0 cm3 portions until 40 cm3 have been added, then rinse the probe.
Apparatus & techniques
- using a burette to a stated precision
- calibrating & using a pH meter
- sampling more often where the change is fastest
- plotting and interpreting a titration curve
- safe handling of acids & alkalis
Safety. Wear eye protection. Sodium hydroxide solution is corrosive: it attacks skin and, in particular, eyes; wash any splash off immediately with plenty of water. Ethanoic acid is an irritant with a strong smell, so keep the room ventilated. Fill burettes below eye level using a funnel, and clamp the pH probe rather than holding it.
What you should see. The pH climbs slowly at first, then barely moves at all through the middle of the titration: that flat stretch is a buffer. Around 20 cm3 it leaps by several pH units for a fraction of a cubic centimetre, which is why you switch to 0.20 cm3 portions there. After that it flattens out again as the excess alkali takes over.
Data analysis: reading the curve
Correct every reading with your calibration line first, then plot corrected pH (y) against volume of alkali (x) and join the points with a smooth curve. Four things come straight off it:
| Feature | Where to look | What it gives you |
|---|---|---|
| Starting pH | the y-intercept | High for a weak acid (about 2.8 here): a strong acid of the same concentration would start near pH 1 |
| Buffer region | the flat stretch before the rise | A mixture of the weak acid and its salt resists pH change |
| Half-neutralisation | half the equivalence volume | pH = pKa, so Ka = 10−pH |
| Equivalence point | the middle of the vertical section | The volume needed, and a pH above 7 for weak acid + strong base |
| Vertical section | its top and bottom pH | An indicator only works if its range lies inside it |
- Why calibrate at all? A meter drifts, so its raw reading is not the true pH. Three buffers spanning the range you will measure let you correct every reading, and using three rather than one checks the response is linear.
- Rinse the probe with deionised water between every solution, and never wipe the bulb dry: carry-over from the last solution is a systematic error, and the glass bulb is fragile.
- Add dropwise near the equivalence point. The reason to give is that the pH changes very rapidly there for a very small addition of alkali: 2.0 cm3 portions would jump straight over the vertical section.
- Stir after every addition before reading, or the probe measures a local pocket rather than the bulk solution.
- Join the points with a smooth curve, not straight lines, and never force it through every point.
- The equivalence point of a weak acid with a strong base is above pH 7, because the salt formed (the conjugate base) is itself slightly basic. For a strong acid with a weak base it is below 7.
Exam questions: RP9
Propanoic acid is a weak acid. State the meaning of weak in this context.
Show answer
Only a small fraction of the molecules dissociate / ionise when added to water: it partially dissociates. 1 mark
“The reaction is reversible” scores nothing, and neither does “hydrogen ions do not fully dissociate”. Naming the wrong ions loses the mark.
Suggest why a student doing this experiment would add the sodium hydroxide solution dropwise around the equivalence point.
Show answer
Because there is a large pH change for a small addition of alkali there. 1 mark
The mark needs the idea of a rapid or large change; “the gradient is steep” is allowed. “So they do not miss the equivalence point” on its own is ignored.
Give an expression for Ka for propanoic acid (CH3CH2COOH). Use this expression to show that pH = pKa when half of the propanoic acid has reacted with sodium hydroxide.
Show answer
Ka = [H+][CH3CH2COO−] ÷ [CH3CH2COOH]
1 mark: [H3O+] is allowed for [H+]; round brackets for concentration are not.
At half-neutralisation, half the acid has been converted to its salt, so [CH3CH2COOH] = [CH3CH2COO−]. 1 mark
Those two terms cancel, leaving Ka = [H+], and taking −log of both sides gives pKa = pH. 1 mark
[HA] = [A−], or the ratio [HA]/[A−] = 1, earns the second mark just as well; log 1 = 0 earns the third.
Use the pH from the curve above, when half of the propanoic acid has reacted, to calculate Ka at 298 K.
Show answer
The equivalence volume is 20.0 cm3, so read the pH at 10.0 cm3: pH ≈ 4.55. 1 mark: anything from 4.4 to 4.7 is accepted.
Ka = 10−4.55 = 2.8 × 10−5 mol dm−3
1 mark: to at least 2 s.f. A pH in the accepted band gives Ka between 2.0 × 10−5 and 4.0 × 10−5.
Using the pH at the start (2.8) scores zero: that is the pH of the pure acid, not the half-neutralised mixture.
When sodium hydroxide solution is added to aqueous propanoic acid, the solution formed acts as a buffer when between 5 cm3 and 15 cm3 have been added. Explain why the pH stays approximately constant during this part of the experiment.
Show answer
The OH− added reacts with the propanoic acid (or with the H+ in solution). 1 mark
The ratio of [CH3CH2COOH] to [CH3CH2COO−] stays almost constant: equivalently, the dissociation equilibrium shifts right to replace the H+ removed. 1 mark
Methyl orange and universal indicator are not suitable indicators for the titration of propanoic acid with sodium hydroxide. State the reason why each is unsuitable.
Show answer
Methyl orange: it would not change colour at the equivalence point: its pH range does not lie inside the rapid pH change (it changes below the equivalence point, outside pH 6–11). 1 mark
Universal indicator: it gives a range of colours through the titration rather than one distinct colour change at the end point. 1 mark
Saying only that this is a weak acid–strong base titration is ignored: the mark is for what the indicator does.
Source: AQA A-Level Chemistry past papers.
Source: AQA A-Level Chemistry past papers.