Chemistry 9701/32 — May/June 2020

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment you will determine the formula of the ion, $\text{IO}_x^-$ . To do this you will first react $\text{IO}_x^-$ ions with an excess of iodide ions, $\text{I}^-$ , to form iodine, $\text{I}_2$ .

The equation for this reaction is:

\text{IO}_x^- + y\text{I}^- + z\text{H}^+ \rightarrow \left(\frac{1 + y}{2}\right)\text{I}_2 + \frac{z}{2}\text{H}_2\text{O}

where $x$ , $y$ and $z$ are all integers.

The amount of iodine produced will then be determined by titration with thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$ .

\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}

FB 1 is a solution containing $0.0150\text{ mol dm}^{-3}$ $\text{IO}_x^-$ ions.
FB 2 is dilute sulfuric acid, $\text{H}_2\text{SO}_4$ .
FB 3 is $0.500\text{ mol dm}^{-3}$ potassium iodide, $\text{KI}$ .
FB 4 is $0.100\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
starch indicator

(a)

Method

Pipette $25.0\text{ cm}^3$ of FB 1 into a conical flask.
Use the measuring cylinder to add $25\text{ cm}^3$ of FB 2 to the conical flask.
Use the measuring cylinder to add $10\text{ cm}^3$ of FB 3 to the conical flask. The solution will turn brown as iodine is produced.
Fill the burette with FB 4.
Add FB 4 from the burette until the solution in the conical flask turns yellow.
Add 10–15 drops of starch indicator to the conical flask. The solution will turn blue-black.
Continue to add more FB 4 from the burette until the blue-black colour just disappears. This is the end-point of the titration.
Carry out a rough titration and record your burette readings in the space below.

The rough titre is .............................. $\text{cm}^3$ .

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make sure that your recorded results show the precision of your practical work.
Record in a suitable form in the space below all of your burette readings and the volume of FB 4 added in each accurate titration.

Keep FB 3 and FB 4 for use in Question 3.

DifficultyMedium-Easy

Worked solution

Answer

Rough titration:

Rough titre = 23.4 cm³

Accurate titrations (example of correctly recorded data):

Titration	Initial burette reading / cm³	Final burette reading / cm³	Volume of FB 4 used / cm³
1	0.00	23.20	23.20
2	23.20	46.35	23.15
3	0.00	23.10	23.10

Technique shown: a rough titration first, then accurate titrations with both burette readings recorded to the nearest 0.05 cm³, repeated until two accurate titres agree within 0.10 cm³. Starch indicator is added only when the solution is pale yellow, and FB 4 is added dropwise until the blue-black colour just disappears.

Final answer

Candidate-dependent: rough titre plus a table of accurate burette readings (to 0.05 cm^3) and titres, with concordant titres within 0.10 cm^3.

Detailed explanation

Background Concept

A titration determines the amount of a reagent in solution by reacting it with a standard solution of known concentration. Here the iodine generated in the flask is titrated with sodium thiosulfate. The burette can be read to 0.05 cm³ (half of the smallest 0.1 cm³ graduation), so all burette readings must be quoted to two decimal places ending in 0 or 5. Precision is demonstrated by repeating the titration until titres agree within 0.10 cm³ — these are called concordant titres.

Understanding the Question

This is the method/data-recording part of a Paper 3 titration question. You must actually carry out the titration and record: a rough titre, then as many accurate titrations as needed, showing initial reading, final reading and titre for each. The mark scheme awards marks for presentation, precision of readings and consistency, not for any particular value.

Approach

Do a rough titration to locate the end-point approximately.
Repeat accurately: run FB 4 in until the solution is pale yellow, add 10–15 drops of starch (solution turns blue-black), then add dropwise until the blue-black just disappears.
Record every burette reading to 0.05 cm³ in a table with correct headings and units.
Repeat until two accurate titres agree within 0.10 cm³.

Step-by-Step Reasoning

Rough titration: performed quickly to find the approximate end-point; its value is not used in calculations.
Accurate titrations: the flask is rinsed with distilled water between runs (the small amount of water does not change the moles of iodine present). Near the end-point the thiosulfate is added drop by drop, swirling constantly, so the end-point is not overshot.
Why starch is added late: if starch is added while the iodine concentration is high, the blue-black iodine–starch complex adsorbs thiosulfate strongly and is released slowly, causing a drifting, inaccurate end-point. Adding starch at the pale-yellow stage avoids this.
Recording: each burette reading is a single reading to 0.05 cm³ (e.g. 23.20, not 23.2); the titre is the difference. The mark scheme requires a 2×2 'box' of both readings for the accurate titrations, correct headings ('initial', 'final', 'titre/volume of FB 4 used') with units in cm³, and consistency of the final titre with another accurate titre to within 0.10 cm³.
The values shown above are representative; your own values will differ, but the format and precision requirements are identical.

Key Takeaways

Burette readings are always to 0.05 cm³; titres inherit this precision.
Concordant titres (within 0.10 cm³) demonstrate precision and are the ones used in calculations.
Starch indicator goes in near the end-point, not at the start.

Common Mistakes

Recording readings to only 1 decimal place (e.g. 23.2) — loses the precision mark.
Heading the last column 'difference' or 'total' — the mark scheme explicitly rejects these words; use 'titre' or 'volume of FB 4 used'.
Omitting units (cm³) from the table headings or entries.
Adding starch at the start of the titration, giving a sluggish, unreliable end-point.
Not repeating until titres are concordant within 0.10 cm³.

Things to Be Careful About

Every burette reading must end in 0 or 5 in the second decimal place.
The rough titre must still be recorded, even though it is not used later.
Consistent decimal places down each column of the table.
Keep FB 3 and FB 4 for Question 3 — do not discard them.

Techniques used

perform a rough titration followed by accurate titrationsrecord burette readings to 0.05 cm^3 in a correctly headed tableadd starch indicator near the end-pointrepeat titrations until concordant within 0.10 cm^3

(b)

From your accurate titration results, obtain a value for the volume of FB 4 to be used in your calculations. Show clearly how you obtained this value.

$25.0\text{ cm}^3$ of FB 1 required .............................. $\text{cm}^3$ of FB 4.

DifficultyEasy

Worked solution

Working

Titres 1, 2 and 3 are 23.20, 23.15 and 23.10 cm³. All lie within 0.20 cm³ of each other (spread = 0.10 cm³), so all three are averaged:

\text{mean titre} = \frac{23.20 + 23.15 + 23.10}{3} = 23.15 \text{ cm}^3

Answer

25.0 cm³ of FB 1 required 23.15 cm³ of FB 4. (Candidate-dependent: use your own mean of concordant titres, with the titres ticked or the averaging shown.)

Final answer

23.15 cm^3 (example; candidate-dependent mean of concordant titres)

Detailed explanation

Background Concept

The value carried into all calculations should be the best estimate of the titre. This is obtained by averaging only titres that are concordant — here the mark scheme allows averaging titres that all lie within 0.20 cm³ of each other. A rough titre is never included.

Understanding the Question

You must show how you obtained the single titre used in the calculations: either tick the titres you averaged or write out the averaging sum. The command is 'obtain a value ... show clearly how you obtained this value'.

Approach

Check the spread of your accurate titres. If they are within 0.20 cm³, average them; if one is an outlier beyond this, exclude it and average the remainder. Show the sum.

Step-by-Step Reasoning

With titres 23.20, 23.15 and 23.10 cm³, the spread is 23.20 − 23.10 = 0.10 cm³, well within 0.20 cm³, so all three qualify.
Mean = (23.20 + 23.15 + 23.10)/3 = 69.45/3 = 23.15 cm³.
The mean is quoted to 2 decimal places, matching the precision of the burette readings.

Key Takeaways

Always show which titres were averaged (ticks or an explicit sum).
Only concordant titres (within 0.20 cm³ per this mark scheme) may be averaged.
Quote the mean to the same precision as the readings (0.05 cm³).

Common Mistakes

Averaging the rough titre with the accurate titres.
Including an outlier titre that differs by more than 0.20 cm³ from the others.
Not showing any working or ticks — the mark requires the selection to be visible.

Things to Be Careful About

The mark scheme requires all averaged titres to be within 0.20 cm³, even though part (a) required the last two accurate titres to be within 0.10 cm³ — these are different criteria.
Do not round the mean prematurely; carry 23.15 (or your value) into part (c).

Techniques used

select concordant titrescalculate the mean titre

(c)

Calculations

(i)

Give your answers to (c)(ii), (c)(iii) and (c)(iv) to the appropriate number of significant figures.

DifficultyEasy

Worked solution

Answer

Answers to (c)(ii) and (c)(iii) are given to 3–4 significant figures; the answer to (c)(iv) is given as an integer.

Final answer

(ii) and (iii) to 3–4 sf; (iv) as an integer

Detailed explanation

Background Concept

In practical calculations the number of significant figures should reflect the precision of the data. Titre and concentration data support 3–4 significant figures; quoting fewer loses precision, quoting more implies false precision. A count of ions such as y must be a whole number.

Understanding the Question

This is a bookkeeping mark: the examiner checks the significant figures of your answers in (c)(ii), (c)(iii) and (c)(iv).

Approach

Give moles to 3 or 4 sf and round y to the nearest odd integer.

Step-by-Step Reasoning

(c)(ii) moles of I₂ ≈ 1.158 × 10⁻³ mol (4 sf) — acceptable.
(c)(iii) 3.75 × 10⁻⁴ mol (3 sf) — acceptable.
(c)(iv) y = 5, an integer — acceptable.

Key Takeaways

3–4 sf is the standard for titration calculations in Paper 3.
Integers (x, y, z) must be given as whole numbers, not decimals.

Common Mistakes

Quoting moles to 2 sf (e.g. 1.2 × 10⁻³) — loses this mark.
Giving y as 5.17 instead of rounding to 5.

Things to Be Careful About

Apply the sf rule to every one of (ii), (iii) and (iv); a single slip loses the mark.
Do not round intermediate values before the final step of a multi-step calculation.

Techniques used

quote calculated values to an appropriate number of significant figures

(ii)

Use your answer to (b) and the relevant equation on page 2 to calculate the number of moles of iodine that form when $25.0\text{ cm}^3$ of FB 1 react with $10\text{ cm}^3$ of FB 3.

moles of $\text{I}_2$ = .............................. mol

DifficultyMedium-Easy

Worked solution

Working

\text{moles of S}_2\text{O}_3^{2-} = \frac{23.15 \times 0.100}{1000} = 2.315 \times 10^{-3} \text{ mol}

From $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ , moles of $\text{I}_2$ = half the moles of thiosulfate:

\text{moles of I}_2 = \frac{2.315 \times 10^{-3}}{2} = 1.158 \times 10^{-3} \text{ mol}

Answer

moles of $\text{I}_2$ = $1.158 \times 10^{-3}$ mol (candidate-dependent: ½ × mean titre × 0.100/1000)

Final answer

1.158 × 10^-3 mol (example, from mean titre 23.15 cm^3)

Detailed explanation

Background Concept

The titration equation $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ shows a 1 : 2 ratio between iodine and thiosulfate. Moles of thiosulfate delivered from the burette = concentration × volume (in dm³); dividing by 2 gives the moles of iodine in the flask.

Understanding the Question

Using the mean titre from (b) and the 0.100 mol dm⁻³ concentration of FB 4, find the moles of I₂ formed from 25.0 cm³ of FB 1.

Approach

moles S₂O₃²⁻ = cV/1000, then halve. The mark scheme's formula is ½ × ((b) × 0.1/1000).

Step-by-Step Reasoning

Volume of FB 4 = 23.15 cm³ = 0.02315 dm³; concentration = 0.100 mol dm⁻³.
moles S₂O₃²⁻ = 0.100 × 0.02315 = 2.315 × 10⁻³ mol.
The 1 : 2 ratio (I₂ : S₂O₃²⁻) means moles I₂ = 2.315 × 10⁻³ / 2 = 1.158 × 10⁻³ mol (4 sf).
If your mean titre differs, substitute it: the method and the halving are the same (ecf applies).

Key Takeaways

Always convert cm³ to dm³ by dividing by 1000 before using c = n/V.
The iodine–thiosulfate ratio is always 1 : 2, so moles I₂ = ½ × moles S₂O₃²⁻.

Common Mistakes

Forgetting to halve, reporting the thiosulfate moles as iodine moles.
Using 0.500 (the KI concentration) or 0.0150 (the FB 1 concentration) instead of 0.100 mol dm⁻³.
Quoting to 2 sf instead of 3–4 sf.

Things to Be Careful About

Use your own mean titre from (b); the example value changes the answer but not the method.
Keep full precision until the final answer, then round to 3–4 sf.

Techniques used

calculate moles from a titreapply the stoichiometric ratio of the thiosulfate–iodine equation

(iii)

Calculate the number of moles of $\text{IO}_x^-$ ions in $25.0\text{ cm}^3$ of FB 1.

moles of $\text{IO}_x^-$ ions = .............................. mol

DifficultyEasy

Worked solution

Working

\text{moles of IO}_x^- = \frac{25.0 \times 0.0150}{1000} = 3.75 \times 10^{-4} \text{ mol}

Answer

moles of $\text{IO}_x^-$ ions = $3.75 \times 10^{-4}$ mol

Final answer

3.75 × 10^-4 mol

Detailed explanation

Background Concept

The amount of a solute in a solution is n = c × V, with V in dm³. FB 1 contains 0.0150 mol dm⁻³ of IOₓ⁻ ions, and a 25.0 cm³ pipette portion was used.

Understanding the Question

A direct 'calculate' command: find the moles of IOₓ⁻ in the 25.0 cm³ aliquot of FB 1. This value is independent of the titration, so it is the same for every candidate.

Approach

Multiply concentration by volume converted to dm³.

Step-by-Step Reasoning

V = 25.0 cm³ = 0.0250 dm³; c = 0.0150 mol dm⁻³.
n = 0.0150 × 0.0250 = 3.75 × 10⁻⁴ mol (3 sf).

Key Takeaways

n = cV/1000 when V is in cm³.
This value is fixed by the given data — no candidate variation.

Common Mistakes

Using 250 cm³ or forgetting the /1000 conversion.
Multiplying by the wrong concentration (e.g. 0.100 or 0.500).

Things to Be Careful About

Quote to 3 sf: 3.75 × 10⁻⁴ mol.
This moles value is used as the denominator in part (c)(iv).

Techniques used

calculate moles from concentration and volume

(iv)

Use the ratio of your answers to (c)(ii) and (c)(iii) along with the relevant equation given on page 2 to calculate the value of $y$ . (Note that $y$ is an odd integer such as 1, 3, 5, 7 etc.)
Show your working.

$y$ = ..............................

DifficultyMedium

Worked solution

Working

From the equation, the ratio moles of $\text{I}_2$ : moles of $\text{IO}_x^-$ = $(1+y)/2$ : 1.

\frac{(\text{c})(\text{ii})}{(\text{c})(\text{iii})} = \frac{1.158 \times 10^{-3}}{3.75 \times 10^{-4}} = 3.09 = \frac{1+y}{2}

1 + y = 2 \times 3.09 = 6.18 \quad \Rightarrow \quad y = 5.18 \approx 5

Answer

$y$ = 5 (candidate-dependent: round your ratio to the nearest odd integer)

Final answer

Detailed explanation

Background Concept

In the equation $\text{IO}_x^- + y\text{I}^- + z\text{H}^+ \rightarrow \left(\frac{1+y}{2}\right)\text{I}_2 + \frac{z}{2}\text{H}_2\text{O}$ , the coefficient of I₂ is (1 + y)/2. Since the experiment measures moles of I₂ produced from a known number of moles of IOₓ⁻, the experimental ratio of these moles equals (1 + y)/2. Because y must be an odd integer, the experimental ratio (which contains experimental error) is rounded to the nearest odd integer.

Understanding the Question

'Show your working' is required: display the division of (c)(ii) by (c)(iii), equate it to (1 + y)/2, and solve for y, rounding to the nearest odd integer (1, 3, 5, 7...).

Approach

Divide the moles of I₂ by the moles of IOₓ⁻; this equals (1 + y)/2. Double it and subtract 1 to get y; round to the nearest odd integer.

Step-by-Step Reasoning

Ratio = 1.158 × 10⁻³ / 3.75 × 10⁻⁴ = 3.088 ≈ 3.09.
Set 3.09 = (1 + y)/2, so 1 + y = 6.18 and y = 5.18.
The nearest odd integer to 5.18 is 5, so y = 5.
The first mark is for showing the ratio (ii)/(iii) = (1 + y)/2 explicitly; the second is for the correctly rounded odd integer. Error carried forward applies if your (c)(ii) differed.

Key Takeaways

Experimental mole ratios approximate exact stoichiometric ratios; rounding to the nearest integer (here, nearest odd integer) recovers the true coefficient.
Always display the ratio equation before solving — the working itself carries a mark.

Common Mistakes

Equating the ratio to y directly instead of (1 + y)/2, giving y ≈ 3.
Rounding 5.18 to 5.2 or to the nearest even integer instead of the nearest odd integer.
Not showing the ratio display, losing the first mark.

Things to Be Careful About

y must be odd; if your calculation gives an even number near an odd one (e.g. 6.18 → 6), choose the nearest odd integer (5).
Use unrounded values from (c)(ii) and (c)(iii) in the division.

Techniques used

deduce a stoichiometric ratio from experimental molessolve for y using the reaction equation

(v)

Use your value of $y$ to determine the formula of the $\text{IO}_x^-$ ion.

formula = ..............................

DifficultyMedium-Easy

Worked solution

Working

With $y = 5$ , the equation becomes:

\text{IO}_x^- + 5\text{I}^- + z\text{H}^+ \rightarrow 3\text{I}_2 + \frac{z}{2}\text{H}_2\text{O}

Balancing hydrogen: $z = 6$ , giving $3\text{H}_2\text{O}$ , i.e. 3 oxygen atoms on the right, so $x = 3$ .

\text{IO}_3^- + 5\text{I}^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}

Answer

formula = $\text{IO}_3^-$ (the iodate ion)

Final answer

IO3^-

Detailed explanation

Background Concept

Once y is known, the remaining unknowns x and z are fixed by balancing atoms and charge. The iodine atoms on the right are (1 + y) = 6, matching 1 + y on the left. Water carries the oxygen and hydrogen balance, so x (oxygen atoms on IOₓ⁻) equals z/2 (water molecules), and the single negative charge on IOₓ⁻ is consistent with the oxidation-state change of iodine.

Understanding the Question

'Determine the formula' means identify both x and the ion: substitute y = 5 into the general equation and balance to find x.

Approach

Write the equation with y = 5, balance H to find z, then O to find x.

Step-by-Step Reasoning

With y = 5: $\text{IO}_x^- + 5\text{I}^- + z\text{H}^+ \rightarrow 3\text{I}_2 + \frac{z}{2}\text{H}_2\text{O}$ .
Iodine balance: left has 1 + 5 = 6 I atoms; right has 3 × 2 = 6 ✓.
Charge balance: left charge = −1 − 5 + z = z − 6; right charge = 0, so z = 6.
Oxygen balance: x = z/2 = 3, so the ion is IO₃⁻.
Check with oxidation numbers: iodine in IO₃⁻ is +5; it is reduced to 0 in I₂, accepting 5 electrons, which matches the 5 I⁻ oxidised to I₂ (5 electrons released). ✓
equation: $\text{IO}_3^- + 5\text{I}^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}$ .
The mark scheme allows error carried forward: if your y was 7, 3 or 1, the corresponding ions are IO₄⁻, IO₂⁻ or IO⁻ respectively, each with its balanced equation.

Key Takeaways

Once one coefficient is known, atom and charge balance pin down the rest of the equation.
Oxidation-number checking is a fast way to confirm the formula of the oxyanion.

Common Mistakes

Confusing x (oxygen count) with the water coefficient; x = z/2, not z.
Writing IO₃⁻ but failing to check that the full equation balances.
Not applying ecf — even with a wrong y, choosing the matching ion from the list still scores.

Things to Be Careful About

The answer is the formula of the ION, IO₃⁻, including the charge.
The allowed ions are IO₄⁻, IO₃⁻, IO₂⁻ and IO⁻ for y = 7, 5, 3 and 1 respectively.

Techniques used

balance the redox equation to identify the oxyanion formula

(d)

(i)

The maximum error in the volume dispensed by the pipette is $\pm0.06\text{ cm}^3$ .

Calculate the maximum percentage error in the volume of FB 1 used.

maximum percentage error = ..............................%

DifficultyEasy

Worked solution

Working

\text{maximum percentage error} = \frac{0.06}{25.0} \times 100 = 0.24\%

Answer

maximum percentage error = 0.24%

Final answer

0.24%

Detailed explanation

Background Concept

The percentage uncertainty of a measurement is (absolute uncertainty ÷ measured value) × 100. For a pipette, the manufacturer's maximum error is applied to the nominal volume delivered.

Understanding the Question

The pipette delivers 25.0 cm³ with a maximum error of ±0.06 cm³; express this error as a percentage of the volume.

Approach

Divide the absolute error by the measured volume and multiply by 100.

Step-by-Step Reasoning

(0.06 / 25.0) × 100 = 0.24%.
The result is quoted to 2 significant figures, appropriate for an uncertainty.

Key Takeaways

Percentage error = (absolute error / measured value) × 100.
Smaller measured volumes give larger percentage errors for the same absolute uncertainty.

Common Mistakes

Dividing by 0.06 instead of by 25.0.
Forgetting to multiply by 100.
Using 0.6 instead of 0.06 cm³.

Things to Be Careful About

Use the actual volume delivered (25.0 cm³), not the titre.
Do not round to 0.2% — 0.24% is the expected answer.

Techniques used

calculate percentage uncertainty

(ii)

A student suggested that a more accurate value of $x$ could be obtained if a $10\text{ cm}^3$ pipette is used to measure FB 3 rather than the measuring cylinder.

State whether you agree with the student. Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

I do not agree. The potassium iodide (FB 3) is in excess, so its exact volume does not affect the amount of iodine produced or the titre; measuring it more precisely with a pipette would not improve the accuracy of $x$ .

Final answer

Disagree — KI is in excess, so its volume does not affect the result.

Detailed explanation

Background Concept

In a titration back-reaction like this, one reagent (the IOₓ⁻ in FB 1) is the limiting quantity measured precisely by pipette; the iodide and acid are added in excess simply to ensure the IOₓ⁻ reacts completely. The amount of iodine formed depends only on the moles of IOₓ⁻, not on the exact amount of excess iodide.

Understanding the Question

The student claims pipetting FB 3 (rather than using a measuring cylinder) would give a more accurate x. You must agree or disagree and justify it. The command 'State whether ... Explain' requires a clear verdict plus the chemical reason.

Approach

Ask: does the precision of the FB 3 volume affect the measured quantity? Since FB 3 is in excess, it does not — so the improvement is pointless.

Step-by-Step Reasoning

10 cm³ of 0.500 mol dm⁻³ KI contains 5.00 × 10⁻³ mol of I⁻, far more than the ~1.9 × 10⁻³ mol needed to reduce all the IO₃⁻ (which requires 5 × 3.75 × 10⁻⁴ = 1.875 × 10⁻³ mol). The iodide is therefore in excess.
Because it is in excess, all the IOₓ⁻ reacts completely regardless of small volume errors in FB 3; the titre — and hence x — is unchanged.
A pipette would only matter if the reagent were limiting or if its exact amount entered the calculation. It does not here.

Key Takeaways

Improvements should target quantities that actually affect the result; excess reagents need only be added roughly.
Measuring cylinders are acceptable for excess reagents; pipettes/burettes are for limiting reagents and titrations.

Common Mistakes

Agreeing with the student on the general grounds that 'pipettes are more accurate' without considering whether the volume matters.
Saying the measuring cylinder is 'not accurate enough' — vague answers score zero; the excess argument is required.

Things to Be Careful About

State the verdict explicitly (disagree) before the reason.
The key phrase is 'KI is in excess', so its volume/precision is irrelevant to the titre.

Techniques used

evaluate the effect of a proposed improvement on the measurement

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation10M
Q3Qualitative Analysis · Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation14M

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