Chemistry 9701/34 — May/June 2011

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

questions

marks

120

minutes

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

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

You are to determine the concentration, in $\text{mol dm}^{-3}$ , of the aqueous sodium thiosulfate. To do this you will first produce iodine solution by reacting aqueous potassium iodide with aqueous potassium manganate(VII). In this reaction iodide ions are oxidised to iodine by manganate(VII) ions in acidic solution.

2\text{MnO}_4^-(\text{aq}) + 10\text{I}^-(\text{aq}) + 16\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{I}_2(\text{aq}) + 8\text{H}_2\text{O}(\text{l})

You will then titrate the iodine with aqueous thiosulfate ions, in FB 1.

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

You are provided with the following.

FB 1 is a solution of sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ , of unknown concentration.
FB 3 is $0.0050\text{ mol dm}^{-3}$ potassium manganate(VII), $\text{KMnO}_4$ .
FB 4 is $0.10\text{ mol dm}^{-3}$ potassium iodide, $\text{KI}$ .
FB 5 is $1.0\text{ mol dm}^{-3}$ sulfuric acid, $\text{H}_2\text{SO}_4$ .
starch indicator

(a)

Method

Dilution

Fill the burette with FB 1.
Run between $45.50\text{ cm}^3$ and $46.50\text{ cm}^3$ of FB 1 from the burette into the $250\text{ cm}^3$ graduated (volumetric) flask, labelled FB 2.
Make the solution up to the mark with distilled water.
Shake the flask to mix the solution of FB 2.

In the space below record your burette readings and the volume of FB 1 added to the graduated flask.

Titration

Fill a second burette with FB 2, the diluted sodium thiosulfate.
Pipette $25.0\text{ cm}^3$ of FB 3 into a conical flask.
Using a $25\text{ cm}^3$ measuring cylinder, add about $10\text{ cm}^3$ of FB 4.
Using the same measuring cylinder, add about $10\text{ cm}^3$ of FB 5.
Titrate the mixture in the flask with FB 2 until the colour is pale yellow.
Add about 10 drops of starch indicator. A blue-black colour should be seen as the starch reacts with the remaining iodine.
Continue to add FB 2 until the blue-black colour just disappears leaving a colourless solution.

You should perform a rough titration.
In the space below record your burette readings for this rough titration.

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

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make certain any recorded results show the precision of your practical work.
Record in an appropriate form below all of your burette readings and the volume of FB 2 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

Record the dilution:

Initial burette reading: 0.00 cm³
Final burette reading: 46.00 cm³
Volume of FB 1 added: 46.00 cm³

Rough titre: 25.1 cm³ (example)

Accurate titrations (recorded in a table):

Titration	Initial reading / cm³	Final reading / cm³	Volume added / cm³
1	0.00	25.05	25.05
2	25.05	50.10	25.05
3	0.00	25.00	25.00

All burette readings recorded to the nearest 0.05 cm³. Two concordant titres (25.05 and 25.00 cm³) obtained, within 0.1 cm³ of each other.

Final answer

See working — candidate-dependent readings recorded to 0.05 cm³ with two concordant titres within 0.1 cm³

Detailed explanation

Background Concept

In volumetric analysis, a titration determines the volume of a solution needed to react completely with a measured volume of another solution. The titre is the volume delivered from the burette, calculated as final reading minus initial reading. Burette readings are conventionally recorded to the nearest 0.05 cm³ because the graduations on a typical school burette are 0.1 cm³ apart and the eye can judge half-divisions. Concordant titres — two or more agreeing within 0.1 cm³ — confirm that the end point was judged consistently and the results are reliable.

Understanding the Question

This part asks you to carry out the practical procedure and record your data properly. You first dilute a portion of FB 1 (sodium thiosulfate) into a 250 cm³ volumetric flask, then titrate the iodine produced from FB 3 with the diluted thiosulfate (FB 2). The marks here are for how you record your readings — the layout of your table, the precision of your readings, and the concordance of your titres — not for the chemistry itself.

Approach

Follow the method exactly. Record the initial and final burette readings for the dilution (volume of FB 1 added, between 45.50 and 46.50 cm³). Perform a rough titration to find the approximate end point, then repeat accurate titrations until two titres agree within 0.1 cm³. Record every reading to the nearest 0.05 cm³ in a clear table.

Step-by-Step Reasoning

Dilution: Record the burette readings when running FB 1 into the 250 cm³ flask. The volume added must be between 45.50 and 46.50 cm³. Make up to the mark with distilled water and mix thoroughly by shaking.
Rough titration: Add 25.0 cm³ of FB 3, about 10 cm³ of FB 4, and about 10 cm³ of FB 5 to the conical flask, then titrate with FB 2 until pale yellow. Add starch — a blue-black colour confirms iodine remains. Continue until the colour just disappears. Record the rough titre.
Accurate titrations: Repeat the titration, this time adding FB 2 dropwise near the end point. Record initial and final burette readings for each accurate titration, and compute the titre (final − initial).
Concordance: Continue until two accurate titres are within 0.1 cm³ of each other. These are your concordant results.

The starch indicator is added only near the end point because it forms a strong complex with iodine; adding it too early would make the end point difficult to judge sharply.

Key Takeaways

The titre is final reading minus initial reading.
Burette readings go to the nearest 0.05 cm³.
Concordant titres within 0.1 cm³ demonstrate reliable technique.
Starch indicator gives a sharp blue-black → colourless end point.

Common Mistakes

Using 50.00 as an initial burette reading (the burette may not be calibrated to deliver exactly 50.00 cm³).
Recording readings beyond 50.00 cm³.
Not recording readings to the nearest 0.05 cm³.
Confusing the rough titre with accurate titres.

Things to Be Careful About

Read the burette at eye level to avoid parallax error.
Record all readings immediately — never from memory.
The titre is the difference between final and initial readings, not the final reading itself.
Ensure the flask is well mixed after dilution.
The mark scheme rejects any initial reading of 50(.00) and any reading greater than 50.(00).

Techniques used

record burette readings to the nearest 0.05 cm³perform a rough titrationobtain concordant titres within 0.1 cm³record data in a clear table

(b)

From your accurate titration results obtain a suitable value to be used in your calculations.
Show clearly how you have obtained this value.

The iodine produced by $25.0\text{ cm}^3$ of FB 3 required .................... $\text{cm}^3$ of FB 2.

DifficultyEasy

Worked solution

Working

Mean titre = (25.00 + 25.05) / 2 = 25.03 cm³

(Using the two concordant accurate titres from part (a).)

Answer

The iodine produced by 25.0 cm³ of FB 3 required 25.03 cm³ of FB 2.

Final answer

25.03 cm³ (example mean of concordant titres)

Detailed explanation

Background Concept

When several accurate titrations give concordant results, the best estimate of the true titre is the mean of those concordant values. The mean is calculated to 2 decimal places, or to the nearest 0.05 cm³ if the third decimal is rounded up. The mark scheme explicitly allows a mean of exactly .x25 or .x75, which may be rounded up to .x3 or .x8.

Understanding the Question

From your accurate titrations, you must select the concordant ones and calculate their mean. This mean titre is then used in the calculations in part (c). You must show how you obtained the value — e.g., by writing the expression (25.00 + 25.05)/2 — so the examiner can see which titres were used.

Approach

Identify the accurate titres that are within 0.1 cm³ of each other (the mark scheme allows titres within 0.2 cm³ for the mean), add them, and divide by the number of titres. Record the mean clearly.

Step-by-Step Reasoning

Select the concordant titres — those within 0.1 cm³ of each other. In the example, these are 25.00 cm³ and 25.05 cm³.
Add them: 25.00 + 25.05 = 50.05 cm³.
Divide by 2: 50.05 / 2 = 25.025 cm³, which rounds to 25.03 cm³ (2 dp).
Record the mean clearly, e.g., in the blank provided or as an expression.

The mean is used in part (c)(iv) to scale the titre amount up to the 250 cm³ flask.

Key Takeaways

The mean of concordant titres is the best estimate of the true titre.
Show the calculation clearly so the examiner can see which titres were used.
Round appropriately: 2 decimal places, or to the nearest 0.05 cm³.

Common Mistakes

Including the rough titre in the mean.
Rounding incorrectly (e.g., 24.35 rounded to 24.4 when 24.35 is acceptable).
Not showing which titres were used.
Averaging titres that are not concordant (more than 0.2 cm³ apart).

Things to Be Careful About

Only use accurate titres within 0.2 cm³ (ideally 0.1 cm³) of each other.
The mean should be to 2 decimal places, or 1 dp if all readings are to 1 dp and the mean is exact without rounding.
Error carried forward (ecf) applies: if a subtraction error gave a wrong titre, the mean can still be credited.

Techniques used

calculate the mean of concordant titresselect concordant results

(c)

Calculations

Show your working and appropriate significant figures in the final answer to each step of your calculations.

(i)

Calculate how many moles of $\text{MnO}_4^-$ were pipetted into the conical flask.

DifficultyEasy

Worked solution

Working

$n(\text{MnO}_4^-) = c \times V = 0.0050 \times \frac{25.0}{1000} = 1.25 \times 10^{-4} \text{ mol}$

Answer

1.25 × 10⁻⁴ mol

Final answer

1.25 × 10⁻⁴ mol

Detailed explanation

Background Concept

The number of moles of a solute in a solution is given by $n = c \times V$ , where $c$ is the concentration in mol dm⁻³ and $V$ is the volume in dm³. A volume given in cm³ must be divided by 1000 to convert to dm³, because 1 dm³ = 1000 cm³.

Understanding the Question

You pipetted 25.0 cm³ of FB 3, which is 0.0050 mol dm⁻³ potassium manganate(VII), KMnO₄. Each formula unit of KMnO₄ contains one MnO₄⁻ ion, so the concentration of MnO₄⁻ is the same as the concentration of KMnO₄. You must calculate the moles of MnO₄⁻ in the pipetted volume.

Approach

Use $n = c \times V$ , converting the volume from cm³ to dm³ by dividing by 1000.

Step-by-Step Reasoning

Write the formula: $n = c \times V$ .
Substitute: $c = 0.0050$ mol dm⁻³, $V = 25.0$ cm³ = 0.0250 dm³.
Calculate: $n = 0.0050 \times 0.0250 = 1.25 \times 10^{-4}$ mol.

The mark scheme credits this exact value: 1.25 × 10⁻⁴ mol.

Key Takeaways

$n = c \times V$ with $V$ in dm³.
cm³ → dm³: divide by 1000.
The concentration of an ion equals the concentration of the compound if the formula contains one ion per formula unit.

Common Mistakes

Forgetting to convert cm³ to dm³ (giving 0.125 mol instead of 1.25 × 10⁻⁴ mol).
Using the wrong concentration (e.g., FB 4 instead of FB 3).
Writing the volume as 25 instead of 0.0250 in the calculation.

Things to Be Careful About

The pipette delivers 25.0 cm³, so $V = 0.0250$ dm³.
The answer should be given to 3 significant figures: 1.25 × 10⁻⁴ mol.

Techniques used

calculate moles from concentration and volumeconvert cm³ to dm³

(ii)

Calculate how many moles of $\text{I}_2$ were produced from the number of moles of $\text{MnO}_4^-$ calculated in (i).

2\text{MnO}_4^-(\text{aq}) + 10\text{I}^-(\text{aq}) + 16\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{I}_2(\text{aq}) + 8\text{H}_2\text{O}(\text{l})

DifficultyMedium-Easy

Worked solution

Working

From the equation, 2 mol MnO₄⁻ produce 5 mol I₂:

$n(\text{I}_2) = 1.25 \times 10^{-4} \times \frac{5}{2} = 3.125 \times 10^{-4} \text{ mol}$

Answer

3.13 × 10⁻⁴ mol (or 3.125 × 10⁻⁴ mol)

Final answer

3.13 × 10⁻⁴ mol

Detailed explanation

Background Concept

The balanced equation

2\text{MnO}_4^-(\text{aq}) + 10\text{I}^-(\text{aq}) + 16\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{I}_2(\text{aq}) + 8\text{H}_2\text{O}(\text{l})

gives the stoichiometric ratio between reactants and products. The coefficients show that 2 mol of MnO₄⁻ produce 5 mol of I₂. So the mole ratio is 5:2, meaning moles of I₂ = moles of MnO₄⁻ × 5/2 = × 2.5.

Understanding the Question

Using the moles of MnO₄⁻ calculated in part (i), you must find how many moles of I₂ are produced. This is a pure stoichiometry step — the ratio comes directly from the balanced equation given.

Approach

Apply the stoichiometric ratio from the equation: multiply the moles of MnO₄⁻ by 5/2.

Step-by-Step Reasoning

Identify the ratio: 2 mol MnO₄⁻ → 5 mol I₂, so the factor is 5/2 = 2.5.
Multiply: $1.25 \times 10^{-4} \times 2.5 = 3.125 \times 10^{-4}$ mol.
Round to 3 significant figures: 3.13 × 10⁻⁴ mol (the exact value 3.125 × 10⁻⁴ mol is also accepted).

The mark scheme credits "answer to (i) × 2.5" — the factor of 2.5 is the key.

Key Takeaways

Stoichiometric ratios come from the coefficients in the balanced equation.
2 MnO₄⁻ : 5 I₂, so multiply by 5/2 = 2.5.
Keep track of significant figures.

Common Mistakes

Using the inverted ratio (2/5 instead of 5/2), which would give a much smaller answer.
Using the ratio of I⁻ (10/2 = 5) instead of the I₂ ratio.
Losing the exponent when multiplying (e.g., writing 3.125 × 10⁻⁴ incorrectly).

Things to Be Careful About

The ratio is moles of product over moles of reactant: 5/2.
The answer to (ii) is used in part (iii), so keep it accurate — error carried forward applies if (i) was wrong but the ratio is applied correctly.

Techniques used

apply stoichiometric ratio from the balanced equationcalculate moles of product

(iii)

Calculate how many moles of $\text{S}_2\text{O}_3^{2-}$ reacted with the $\text{I}_2$ in (ii).

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

DifficultyMedium-Easy

Worked solution

Working

From the equation, 1 mol I₂ reacts with 2 mol S₂O₃²⁻:

$n(\text{S}_2\text{O}_3^{2-}) = 3.125 \times 10^{-4} \times 2 = 6.25 \times 10^{-4} \text{ mol}$

Answer

6.25 × 10⁻⁴ mol

Final answer

6.25 × 10⁻⁴ mol

Detailed explanation

Background Concept

The titration equation

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

shows that 1 mol of I₂ reacts with 2 mol of S₂O₃²⁻. The thiosulfate is oxidised to tetrathionate, S₄O₆²⁻, while iodine is reduced back to iodide. This is the classic iodometric titration.

Understanding the Question

You must calculate how many moles of S₂O₃²⁻ reacted with the iodine from part (ii). This amount is the quantity present in the titre volume of FB 2 — the diluted thiosulfate added from the burette during the titration.

Approach

Multiply the moles of I₂ by 2, using the 1:2 stoichiometric ratio from the equation.

Step-by-Step Reasoning

Identify the ratio: 1 mol I₂ : 2 mol S₂O₃²⁻, so the factor is 2.
Multiply: $3.125 \times 10^{-4} \times 2 = 6.25 \times 10^{-4}$ mol.
This is the moles of S₂O₃²⁻ in the titre volume of FB 2.

The mark scheme notes that this step is covered by the same mark as (ii) — the two ratios together give the overall factor of 5 (2.5 × 2).

Key Takeaways

1 I₂ : 2 S₂O₃²⁻.
This amount is in the titre volume, not the whole flask.
The overall conversion from MnO₄⁻ to S₂O₃²⁻ is a factor of 5 (2.5 × 2).

Common Mistakes

Using the wrong ratio (e.g., 1/2 instead of 2).
Confusing this step with the dilution in part (iv).
Forgetting that this is the amount in the titre, not in the flask.

Things to Be Careful About

The answer 6.25 × 10⁻⁴ mol is used in part (iv) to scale up to the flask.
Keep the answer to 3 significant figures.

Techniques used

apply stoichiometric ratio from the balanced equationcalculate moles of reactant

(iv)

Calculate how many moles of $\text{S}_2\text{O}_3^{2-}$ were present in the $250\text{ cm}^3$ graduated (volumetric) flask.

DifficultyMedium-Easy

Worked solution

Working

The titre of 25.03 cm³ contained 6.25 × 10⁻⁴ mol of S₂O₃²⁻. The flask holds 250 cm³:

$n(\text{S}_2\text{O}_3^{2-}\text{ in flask}) = 6.25 \times 10^{-4} \times \frac{250}{25.03} = 6.24 \times 10^{-3} \text{ mol}$

Answer

6.24 × 10⁻³ mol

Final answer

6.24 × 10⁻³ mol

Detailed explanation

Background Concept

The titre of FB 2 (the diluted thiosulfate) contains the amount of thiosulfate calculated in part (iii). The whole 250 cm³ volumetric flask of FB 2 contains proportionally more: the amount in the titre scaled up by the ratio of the flask volume to the titre volume. Since the solution is homogeneous, the concentration is the same everywhere, so moles scale linearly with volume.

Understanding the Question

The titre volume (from part (b)) contained 6.25 × 10⁻⁴ mol of S₂O₃²⁻. The flask holds 250 cm³. You must find the total moles of S₂O₃²⁻ in the entire flask of FB 2.

Approach

Multiply the moles in the titre by the factor 250/titre volume. This scales the small titre amount up to the full flask.

Step-by-Step Reasoning

Identify the titre volume: 25.03 cm³ (from part (b)).
The scaling factor is 250 / 25.03 = 9.99.
Multiply: $6.25 \times 10^{-4} \times 9.99 = 6.24 \times 10^{-3}$ mol.

If the titre were exactly 25.00 cm³, the factor would be 10 and the answer exactly 6.25 × 10⁻³ mol. The mark scheme credits "answer to (iii) × 250 / mean titre in (b)".

Key Takeaways

The flask contains the titre amount scaled up by 250/titre.
This is the total thiosulfate in the diluted solution FB 2.
The dilution does not change the total moles of thiosulfate relative to what was in the original volume of FB 1.

Common Mistakes

Forgetting to scale up (giving 6.25 × 10⁻⁴ mol instead of ~6.25 × 10⁻³ mol).
Using the wrong titre volume (e.g., the rough titre).
Inverting the scaling factor (multiplying by titre/250 instead of 250/titre).

Things to Be Careful About

Use the mean titre from part (b).
The answer is about 10 times larger than the titre amount because 250 cm³ is about 10 times the titre.
Keep 3 significant figures.

Techniques used

scale moles from titre volume to flask volumeapply dilution factor

(v)

Use your answer to (iv) and the volume of FB 1 that was diluted to calculate the concentration, in $\text{mol dm}^{-3}$ , of the original solution of sodium thiosulfate, FB 1.

The concentration of $\text{Na}_2\text{S}_2\text{O}_3$ in FB 1 was .................... $\text{mol dm}^{-3}$ .

DifficultyMedium

Worked solution

Working

The flask of FB 2 was made by diluting 46.00 cm³ of FB 1 to 250 cm³. The moles of S₂O₃²⁻ in the flask equal the moles in the original 46.00 cm³ of FB 1:

$c(\text{FB 1}) = \frac{n}{V} = \frac{6.24 \times 10^{-3}}{46.00 / 1000} = \frac{6.24 \times 10^{-3}}{0.0460} = 0.1357 \text{ mol dm}^{-3}$

Answer

0.136 mol dm⁻³

Final answer

0.136 mol dm⁻³

Detailed explanation

Background Concept

The 250 cm³ flask of FB 2 was prepared by diluting a known volume of FB 1 (between 45.50 and 46.50 cm³, recorded in part (a)) with distilled water. Dilution changes the concentration but not the total number of moles of solute. Therefore, the moles of S₂O₃²⁻ in the flask (from part (iv)) equal the moles of S₂O₃²⁻ in the original volume of FB 1 that was diluted. The concentration of FB 1 is then found by dividing those moles by the original volume (converted to dm³).

Understanding the Question

Using the moles in the flask from part (iv) and the volume of FB 1 that was diluted (recorded in part (a)), calculate the concentration of the original sodium thiosulfate solution FB 1. The answer must be in mol dm⁻³ and to an appropriate number of significant figures.

Approach

Recall that moles are conserved on dilution.
Convert the dilution volume from cm³ to dm³ by dividing by 1000.
Apply $c = n/V$ to find the concentration of FB 1.

Step-by-Step Reasoning

Moles of S₂O₃²⁻ in the flask = 6.24 × 10⁻³ mol (from part (iv)).
Volume of FB 1 diluted = 46.00 cm³ (example from part (a)).
Convert to dm³: 46.00 / 1000 = 0.0460 dm³.
Concentration: $c = 6.24 \times 10^{-3} / 0.0460 = 0.1357$ mol dm⁻³.
Round to 3 significant figures: 0.136 mol dm⁻³.

The mark scheme credits "answer to (iv) × 1000 / volume diluted". Note that multiplying by 1000 is equivalent to converting the volume from cm³ to dm³.

Key Takeaways

Dilution conserves moles: $n_1 = n_2$ .
$c = n/V$ with $V$ in dm³.
The concentration of FB 1 is higher than that of FB 2 because it was diluted.
Show working in at least 4 steps for the global method mark.

Common Mistakes

Forgetting to convert the dilution volume from cm³ to dm³ (giving 135.7 instead of 0.136).
Using the titre volume instead of the dilution volume.
Not rounding to 3 significant figures.
Confusing the concentration of FB 1 with that of FB 2.

Things to Be Careful About

The answer must be to 3 or 4 significant figures for the global mark.
Use the volume of FB 1 diluted (46.00 cm³ in the example), not the titre.
The final answer should be a reasonable concentration for sodium thiosulfate (typically 0.1–0.15 mol dm⁻³).
Show all working clearly — the mark scheme awards a mark for working shown in a minimum of 4 steps in the right direction.

Techniques used

calculate concentration from moles and volumeconvert cm³ to dm³apply dilution relationship

(d)

The maximum error for a $25\text{ cm}^3$ pipette commonly used in schools is $\pm 0.06\text{ cm}^3$ .
The maximum individual error in any single burette reading is $\pm 0.05\text{ cm}^3$ .

Calculate each of the following.

The maximum percentage error in the volume of FB 3 pipetted into the conical flask

maximum percentage error in the pipetted volume of FB 3 = .................... %
The maximum percentage error in the titre calculated in (b)

maximum percentage error in the titre volume = .................... %

DifficultyMedium-Easy

Worked solution

Working

Pipette (FB 3):

$\frac{0.06}{25.0} \times 100 = 0.24\%$

Titre (FB 2):

The titre is the difference of two burette readings, each with error ±0.05 cm³, so the total error is ±0.10 cm³:

$\frac{0.10}{25.03} \times 100 = 0.40\%$

Answer

Maximum percentage error in pipetted volume of FB 3 = 0.24%

Maximum percentage error in titre volume = 0.40%

Final answer

0.24% (pipette); 0.40% (titre)

Detailed explanation

Background Concept

Percentage error expresses the uncertainty in a measurement as a fraction of the measured value, multiplied by 100:

\text{percentage error} = \frac{\text{absolute error}}{\text{measured value}} \times 100\%

For a single measurement, the absolute error is the instrument's tolerance. For a titre, which is the difference between two burette readings, the errors of both readings add: the maximum error in the titre is ±0.05 + ±0.05 = ±0.10 cm³.

Understanding the Question

You must calculate two percentage errors:

The maximum percentage error in the volume of FB 3 pipetted into the conical flask (25.0 cm³ with ±0.06 cm³ error).
The maximum percentage error in the titre calculated in part (b), using the mean titre from part (b) and the combined burette reading error.

The mark scheme only requires the expressions, not the final values, but both are given here.

Approach

Apply the percentage error formula to each measurement. For the titre, remember that two burette readings are involved, so the absolute error is 2 × 0.05 = 0.10 cm³.

Step-by-Step Reasoning

Pipette error: absolute error = 0.06 cm³, measured value = 25.0 cm³.
Percentage error = (0.06/25.0) × 100 = 0.24%.
Titre error: each burette reading has ±0.05 cm³ error. The titre = final − initial, so the errors add: ±0.10 cm³.
Percentage error = (0.10/25.03) × 100 = 0.40% (using the example mean titre of 25.03 cm³).

The mark scheme gives the expressions: (0.06/25) × 100 and (0.10/titre in (b)) × 100.

Key Takeaways

Percentage error = (absolute error/measured value) × 100%.
A titre involves two burette readings, so the errors add: ±0.10 cm³ total.
The pipette error (±0.06 cm³) applies to the 25.0 cm³ pipetted volume.

Common Mistakes

Using ±0.05 cm³ instead of ±0.10 cm³ for the titre error.
Forgetting to multiply by 100.
Using the wrong measured value (e.g., the dilution volume instead of the titre).
Using the rough titre instead of the mean accurate titre.

Things to Be Careful About

The titre error is the sum of two reading errors, not just one.
Use the mean titre from part (b) in the denominator.
The mark scheme only requires the expressions, so showing (0.06/25) × 100 and (0.10/titre) × 100 is sufficient for the mark.

Techniques used

calculate percentage errorcombine errors from two burette readings

The rest of this paper

2 more questions

Q2Presentation of Data and Observations · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation9M
Q3Qualitative Analysis · Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation16M

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