Chemistry 9701/31 — October/November 2010

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

You are to determine the concentration of hydrochloric acid, which supplies the $\text{H}^+$ ions in the following reaction.

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

In the presence of an excess of $\text{IO}_3^-$ ions and an excess of $\text{I}^-$ ions, the amount of $\text{I}_2$ liberated is directly proportional to the amount of $\text{H}^+$ ions present and can be determined by titration with sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .

You are provided with the following reactants.

FA 1 hydrochloric acid
FA 2 containing $15.0\text{ g dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}$
aqueous potassium iodate(V), $\text{KIO}_3$
aqueous potassium iodide, $\text{KI}$

(a)

Method

Fill a burette with FA 2.
Pipette $25.0\text{ cm}^3$ of FA 1 into the conical flask.
Use a $25\text{ cm}^3$ measuring cylinder to add to the flask $10\text{ cm}^3$ of aqueous potassium iodate(V) and $10\text{ cm}^3$ of aqueous potassium iodide. There is an excess of each of these reagents.
Place the flask on a white tile.
Titrate the liberated iodine with FA 2.
During the titration the colour of the iodine in the solution will fade from red-brown to orange to yellow. The end-point occurs when the solution just goes colourless with the addition of a single drop of FA 2.
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.
Record in a suitable form below all of your burette readings and the volume of FA 2 added in each accurate titration.
Make certain any recorded results show the precision of your practical work.

DifficultyMedium-Easy

Worked solution

Answer

Rough titre: 25.60 cm^3

Accurate titrations (all readings to the nearest 0.05 cm^3):

Titration	Initial reading / cm^3	Final reading / cm^3	Volume of FA 2 / cm^3
1	0.00	25.05	25.05
2	25.05	50.00	24.95
3	0.00	25.00	25.00

[Example data — your own readings will differ. Two accurate titres must agree within 0.10 cm^3; record initial and final readings for every titration.]

Final answer

See working — candidate-dependent readings (representative: rough 25.60 cm^3; accurate titres 25.05, 24.95, 25.00 cm^3)

Detailed explanation

Background Concept

This is an iodometric (redox) titration. The hydrochloric acid is not titrated directly; instead it supplies the H+ ions that drive the reaction between iodate(V) and iodide ions, liberating iodine:

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

With iodate and iodide in excess, the amount of iodine produced is directly proportional to the amount of H+ (hence HCl) present. The iodine is then titrated against sodium thiosulfate:

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

The end-point is the disappearance of the iodine colour. A burette is read to the nearest 0.05 cm^3 (half a graduation, since the scale is marked every 0.1 cm^3). A rough titration locates the approximate end-point quickly; accurate titrations are then performed carefully until two results agree within 0.10 cm^3 — these are concordant.

Understanding the Question

Part (a) asks you to record your practical readings: the rough titre, and a table of accurate titrations showing the initial reading, final reading and volume of FA 2 added. The marks reward correct technique (concordant titres, readings to 0.05 cm^3) and clear recording (proper headings and units). There is no single correct value — the examiner compares your titre with the Supervisor's value.

Approach

Fill the burette with FA 2, ensuring the jet is full and there are no air bubbles.
Pipette 25.0 cm^3 of FA 1 into the conical flask; add 10 cm^3 of aqueous KIO3 and 10 cm^3 of aqueous KI (both in excess).
Place the flask on a white tile so the colour change is easy to judge.
Run the rough titration, adding FA 2 fairly quickly while swirling, until the solution just goes colourless. Record the rough titre.
For accurate titrations, add FA 2 dropwise near the end-point, swirling after each addition. Record the initial and final readings for every titration.
Repeat until two accurate titres agree within 0.10 cm^3.

Step-by-Step Reasoning

The white tile gives a white background against which the red-brown → orange → yellow → colourless transition is clearly visible. The end-point is sharp: a single drop of thiosulfate removes the last of the iodine colour, and the solution becomes permanently colourless. For each accurate titration, record the initial burette reading (e.g. 0.00 cm^3) and the final reading (e.g. 25.05 cm^3); the volume added is the difference. Readings must be to the nearest 0.05 cm^3. Two accurate titres within 0.10 cm^3 are concordant and give confidence in the result. A representative set of results is shown in the solution; your own readings will differ but should follow the same pattern.

Key Takeaways

Burette readings are recorded to the nearest 0.05 cm^3.
A rough titration is followed by accurate titrations.
Two accurate titres within 0.10 cm^3 are concordant.
A clear table with headings and units is essential for full marks.

Common Mistakes

Using 50.00 cm^3 as the initial burette reading — the mark scheme explicitly rejects this.
Having more than one final reading of 50.00 cm^3.
Recording any burette reading greater than 50.00 cm^3.
Not recording initial and final readings separately for each accurate titration.
Giving readings to only 0.1 cm^3 (or to 0.01 cm^3, which implies false precision).

Things to Be Careful About

Headings in the table must match the readings recorded (e.g. "Initial reading / cm^3").
All accurate readings to the nearest 0.05 cm^3.
Ensure the burette jet is full before taking the initial reading.
Swirl the flask after each addition; the end-point is the first drop that makes the solution permanently colourless.

Techniques used

fill and read a burette to the nearest 0.05 cm^3perform a rough titration to locate the end-pointcarry out accurate titrations to concordance within 0.10 cm^3record initial and final burette readings in a headed table

(b)

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

$25.0\text{ cm}^3$ of FA 1 require .............. $\text{cm}^3$ of FA 2.

DifficultyEasy

Worked solution

Working

Mean titre = (25.05 + 24.95 + 25.00)/3 = 25.00 cm^3

Answer

25.0 cm^3 of FA 1 require 25.00 cm^3 of FA 2.

Final answer

25.00 cm^3

Detailed explanation

Background Concept

When several accurate titrations have been performed, the mean of the concordant titres is used in the calculation. Only titres within 0.20 cm^3 of each other should be averaged; an outlier is discarded. The mean is quoted to two decimal places.

Understanding the Question

Part (b) asks you to select a suitable value (the mean titre) from your accurate results and show clearly how you obtained it. The value is then used in the calculation in part (c).

Approach

Identify the accurate titres that agree within 0.20 cm^3, add them together, and divide by the number of titres used. Show the expression so the examiner can see exactly which titres were used.

Step-by-Step Reasoning

With titres of 25.05, 24.95 and 25.00 cm^3, all three lie within 0.10 cm^3 of each other, so all are used:

\text{Mean titre} = \frac{25.05 + 24.95 + 25.00}{3} = \frac{75.00}{3} = 25.00 \text{ cm}^3

The mean is quoted to two decimal places. If the titres had been, for example, 24.3 and 24.4, the mean would be 24.35 cm^3 (not 24.4, which would be incorrect rounding).

Key Takeaways

Only concordant titres (within 0.20 cm^3) are averaged.
The mean is quoted to two decimal places.
The working must be shown clearly (the titres used must be identifiable).

Common Mistakes

Averaging titres that are not concordant (more than 0.20 cm^3 apart).
Rounding the mean incorrectly, e.g. writing 24.4 for the mean of 24.3 and 24.4 instead of 24.35.
Not showing which titres were used in the mean.

Things to Be Careful About

The mean must be correct to two decimal places.
If all burette readings are given to one decimal place, the mean may be quoted to one decimal place if numerically correct.
Tick the titres used or write the expression so the examiner can follow your selection.

Techniques used

select concordant titres within 0.20 cm^3calculate the mean titre to two decimal places

(c)

Calculations

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

(i)

Calculate the concentration, in $\text{mol dm}^{-3}$ , of the sodium thiosulfate in FA 2.

FA 2 contains $15.0\text{ g dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}$ .

[ $A_r$ : H, 1.0; O, 16.0; Na, 23.0; S, 32.1]

The concentration of sodium thiosulfate in FA 2 is ............................ $\text{mol dm}^{-3}$ .

DifficultyMedium-Easy

Worked solution

Working

$M_r(\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}) = 2(23.0) + 2(32.1) + 3(16.0) + 5(18.0) = 46.0 + 64.2 + 48.0 + 90.0 = 248.2$

[\text{Na}_2\text{S}_2\text{O}_3] = \frac{15.0}{248.2} = 0.0604 \text{ mol dm}^{-3}

Answer

0.0604 mol dm^-3

Final answer

0.0604 mol dm^-3

Detailed explanation

Background Concept

The sodium thiosulfate is provided as the hydrated salt Na2S2O3·5H2O, so its molar mass includes the five water molecules of crystallisation. The molar concentration is found by dividing the mass concentration (in g dm^-3) by the molar mass (in g mol^-1):

\text{concentration (mol dm}^{-3}\text{)} = \frac{\text{concentration (g dm}^{-3}\text{)}}{M_r}

Understanding the Question

FA 2 contains 15.0 g dm^-3 of Na2S2O3·5H2O. You must calculate the concentration in mol dm^-3. The Ar values are given: H, 1.0; O, 16.0; Na, 23.0; S, 32.1.

Approach

Calculate Mr of Na2S2O3·5H2O, remembering to include the 5H2O.
Divide 15.0 by Mr to get the molar concentration.
Give the answer to 3 significant figures.

Step-by-Step Reasoning

Mr = 2(23.0) + 2(32.1) + 3(16.0) + 5(18.0) = 46.0 + 64.2 + 48.0 + 90.0 = 248.2.

Note that 5H2O contributes 5 × 18.0 = 90.0 to the molar mass. Then:

[\text{Na}_2\text{S}_2\text{O}_3] = \frac{15.0}{248.2} = 0.0604 \text{ mol dm}^{-3}

The answer is given to 3 significant figures (0.0604), consistent with the 15.0 g dm^-3 given.

Key Takeaways

The molar mass of a hydrated salt includes its water of crystallisation.
Molar concentration = mass concentration ÷ Mr.
Final answers should carry 3–5 significant figures.

Common Mistakes

Forgetting the water of crystallisation and using Mr = 158.1 instead of 248.2 — this gives a wrong answer (0.0949 instead of 0.0604).
Giving the answer to too few significant figures (e.g. 0.06).

Things to Be Careful About

Include 5H2O in the Mr calculation.
The mark scheme accepts 0.060, 0.0604 or 0.06044 — all are fine, but 3 significant figures is the cleanest.
No additional factor or expression is allowed in this step — use exactly 15.0/248.2.

Techniques used

calculate the relative molecular mass of a hydrated saltconvert a mass concentration to a molar concentration

(ii)

Calculate how many moles of $\text{Na}_2\text{S}_2\text{O}_3$ are contained in the volume of FA 2 recorded in (b).

.............. mol of $\text{Na}_2\text{S}_2\text{O}_3$

DifficultyEasy

Worked solution

Working

Moles of $\text{Na}_2\text{S}_2\text{O}_3 = 0.0604 \times \frac{25.00}{1000} = 0.00151$ mol

Answer

0.00151 mol

Final answer

0.00151 mol

Detailed explanation

Background Concept

The number of moles of a solute is the product of its molar concentration and the volume of solution used, with the volume expressed in dm^3:

\text{moles} = \text{concentration (mol dm}^{-3}\text{)} \times \text{volume (dm}^3\text{)}

Since 1 dm^3 = 1000 cm^3, a volume in cm^3 is converted to dm^3 by dividing by 1000.

Understanding the Question

From part (b), 25.00 cm^3 of FA 2 was used (the mean titre). Using the concentration from part (c)(i), calculate the moles of Na2S2O3 in that volume.

Approach

Multiply the concentration (0.0604 mol dm^-3) by the volume in dm^3 (25.00/1000 = 0.02500 dm^3).

Step-by-Step Reasoning

\text{moles of Na}_2\text{S}_2\text{O}_3 = 0.0604 \times \frac{25.00}{1000} = 0.0604 \times 0.02500 = 0.00151 \text{ mol}

The answer is given to 3 significant figures (0.00151).

Key Takeaways

moles = concentration × volume (volume in dm^3).
Convert cm^3 to dm^3 by dividing by 1000.
Keep the same number of significant figures as the input data.

Common Mistakes

Forgetting to divide the volume by 1000 (giving 1.51 mol instead of 0.00151 mol).
Using the rough titre instead of the mean titre.

Things to Be Careful About

Use the mean titre from part (b), not the rough titre.
The mark scheme requires this step to use answer (i) × (candidate's average titre)/1000.

Techniques used

convert a volume from cm^3 to dm^3calculate moles from concentration and volume

(iii)

Calculate how many moles of iodine, $\text{I}_2$ reacted with the $\text{Na}_2\text{S}_2\text{O}_3$ in (ii).

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

............................ mol of iodine reacted with the sodium thiosulfate.

DifficultyMedium-Easy

Worked solution

Working

From $2\text{Na}_2\text{S}_2\text{O}_3 : 1\text{I}_2$ , moles of $\text{I}_2 = \frac{0.00151}{2} = 0.000755$ mol

Answer

0.000755 mol

Final answer

0.000755 mol

Detailed explanation

Background Concept

The titration reaction between sodium thiosulfate and iodine is:

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

The stoichiometric ratio is 2 mol thiosulfate : 1 mol iodine. Therefore, the moles of iodine that reacted equal half the moles of thiosulfate used.

Understanding the Question

Using the moles of Na2S2O3 from part (c)(ii), calculate the moles of I2 that reacted with it.

Approach

Read the ratio from the balanced equation (2:1) and divide the moles of thiosulfate by 2.

Step-by-Step Reasoning

\text{moles of I}_2 = \frac{\text{moles of Na}_2\text{S}_2\text{O}_3}{2} = \frac{0.00151}{2} = 0.000755 \text{ mol}

The answer is 0.000755 mol (3 significant figures).

Key Takeaways

The stoichiometric ratio comes directly from the coefficients in the balanced equation.
2 mol S2O3^2- ≡ 1 mol I2, so moles I2 = ½ moles S2O3^2-.

Common Mistakes

Using a 1:1 ratio and writing 0.00151 mol.
Multiplying by 2 instead of dividing (confusing the direction of the ratio).

Things to Be Careful About

The mark scheme requires you to show the ×½ (or ÷2) step explicitly.
Keep 3 significant figures in the final answer.

Techniques used

apply the 2:1 stoichiometric ratio of thiosulfate to iodine

(iv)

Calculate how many moles of hydrochloric acid, $\text{HCl}$ , reacted with an excess of potassium iodate(V) and an excess of potassium iodide to produce the amount of iodine calculated in (iii).

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

............................ mol of $\text{HCl}$ produced the amount of iodine calculated in (iii).

DifficultyMedium-Easy

Worked solution

Working

From $6\text{H}^+ : 3\text{I}_2$ (i.e. $2\text{H}^+ : 1\text{I}_2$ ), moles of $\text{HCl} = 2 \times 0.000755 = 0.00151$ mol

Answer

0.00151 mol

Final answer

0.00151 mol

Detailed explanation

Background Concept

The reaction that liberates iodine is:

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

The stoichiometric ratio is 6 mol H+ : 3 mol I2, which simplifies to 2 mol H+ : 1 mol I2. Since each mole of HCl provides one mole of H+, the moles of HCl equal the moles of H+.

Understanding the Question

Using the moles of I2 from part (c)(iii), calculate the moles of HCl that produced that iodine. The iodate and iodide are in excess, so all the H+ comes from the HCl.

Approach

Read the ratio from the balanced equation (6H+ : 3I2 = 2:1) and multiply the moles of iodine by 2.

Step-by-Step Reasoning

\text{moles of HCl} = \text{moles of H}^+ = 2 \times \text{moles of I}_2 = 2 \times 0.000755 = 0.00151 \text{ mol}

The answer is 0.00151 mol (3 significant figures). Notice that this equals the moles of thiosulfate — a useful cross-check: the overall stoichiometry from HCl to thiosulfate is 1:1 because 6H+ produce 3I2 and each I2 consumes 2S2O3^2-, so 6H+ ≡ 6S2O3^2-.

Key Takeaways

The H+ : I2 ratio is 2:1 (from 6:3 in the equation).
Each mole of HCl supplies one mole of H+.
Moles of HCl = 2 × moles of I2.

Common Mistakes

Using a 6:3 ratio but applying it wrongly (e.g. dividing by 2 instead of multiplying).
Forgetting that the ratio simplifies to 2:1.
Confusing the direction: more H+ than I2, so moles HCl > moles I2.

Things to Be Careful About

The mark scheme requires you to show multiplication or division by 6, 1.2 or 2 in this step.
The overall chain HCl → H+ → I2 → S2O3^2- is 1:1, so the final moles of HCl should equal the moles of thiosulfate — a good check.

Techniques used

apply the 2:1 stoichiometric ratio of H+ to iodine

(v)

Calculate the concentration, in $\text{mol dm}^{-3}$ , of $\text{HCl}$ in FA 1.

The concentration of $\text{HCl}$ in FA 1 is ....................... $\text{mol dm}^{-3}$ .

DifficultyMedium-Easy

Worked solution

Working

[\text{HCl}] = \frac{0.00151}{25.0/1000} = \frac{0.00151}{0.0250} = 0.0604 \text{ mol dm}^{-3}

Answer

0.0604 mol dm^-3

Final answer

0.0604 mol dm^-3

Detailed explanation

Background Concept

The concentration of a solution is the number of moles of solute per cubic decimetre of solution:

\text{concentration (mol dm}^{-3}\text{)} = \frac{\text{moles of solute}}{\text{volume (dm}^3\text{)}}

The 25.0 cm^3 of FA 1 pipetted into the flask contained the 0.00151 mol of HCl calculated in part (c)(iv).

Understanding the Question

Using the moles of HCl from part (c)(iv) and the 25.0 cm^3 volume of FA 1, calculate the concentration of HCl in FA 1.

Approach

Convert 25.0 cm^3 to dm^3 (divide by 1000).
Divide the moles by this volume.
Give the answer to 3 significant figures.

Step-by-Step Reasoning

[\text{HCl}] = \frac{0.00151}{25.0/1000} = \frac{0.00151}{0.0250} = 0.0604 \text{ mol dm}^{-3}

The answer is 0.0604 mol dm^-3 (3 significant figures). Note that this equals the thiosulfate concentration because the mean titre was 25.00 cm^3 — the overall 1:1 stoichiometry means equal concentrations give equal volumes.

Key Takeaways

concentration = moles ÷ volume (in dm^3).
Convert cm^3 to dm^3 by dividing by 1000.
The final answer carries 3 significant figures.

Common Mistakes

Forgetting to convert 25.0 cm^3 to dm^3 (giving 0.0000604 instead of 0.0604).
Using the wrong volume (e.g. the titre from part (b) instead of 25.0 cm^3).
Quoting too few significant figures (e.g. 0.06).

Things to Be Careful About

Use 25.0 cm^3 (the pipette volume of FA 1), not the titre volume.
The mark scheme requires answer (iv) × 1000/25.
Show the working in at least three sections to qualify for the global working mark; equations must appear as steps.

Techniques used

calculate concentration from moles and volume

(d)

Each reading with a burette has a maximum error of $\pm 0.05\text{ cm}^3$ .
Grade B volumetric (bulb) pipettes are calibrated to $\pm 0.06\text{ cm}^3$ .

(i)

Calculate the maximum error in the volume run from the burette recorded in any titration.

The maximum error is ............................... $\text{cm}^3$ .

DifficultyEasy

Worked solution

Working

Each titre involves two burette readings (initial and final), each with a maximum error of ±0.05 cm^3.

Maximum error = 2 × 0.05 = 0.10 cm^3

Answer

0.10 cm^3

Final answer

0.10 cm^3

Detailed explanation

Background Concept

A titre is the difference between the final and initial burette readings. Each reading has a maximum error of ±0.05 cm^3. When quantities are added or subtracted, their absolute errors add. So the maximum error in the volume run from the burette is the sum of the errors in the two readings.

Understanding the Question

You are told each burette reading has a maximum error of ±0.05 cm^3. A single titre requires two readings (initial and final), so the error in the titre is the sum of the two reading errors.

Approach

Add the two ±0.05 cm^3 errors together.

Step-by-Step Reasoning

\text{Maximum error} = 0.05 + 0.05 = 0.10 \text{ cm}^3

This is the maximum error in the volume run from the burette in any single titration. The sign (±) is ignored when calculating the percentage error.

Key Takeaways

A titre involves two burette readings.
Absolute errors add: 2 × 0.05 = 0.10 cm^3.

Common Mistakes

Using only 0.05 cm^3 (forgetting there are two readings).
Using 0.025 cm^3 (dividing instead of adding).

Things to Be Careful About

The answer is 0.10 cm^3 (or 0.1 cm^3) — the mark scheme accepts 0.1(0).
Ignore any sign; only the magnitude matters.

Techniques used

add the errors of two burette readings

(ii)

Express the maximum error calculated in (i) as a percentage error for the volume calculated in (b).

The maximum error is .................................. %.

DifficultyMedium-Easy

Worked solution

Working

\text{Percentage error} = \frac{0.10}{25.00} \times 100 = 0.40\%

Answer

0.40%

Final answer

0.40%

Detailed explanation

Background Concept

The percentage error of a measurement is the absolute error divided by the measured value, multiplied by 100:

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

Here the absolute error is the 0.10 cm^3 from part (d)(i) and the measured value is the mean titre from part (b), 25.00 cm^3.

Understanding the Question

Express the maximum error in the burette volume (0.10 cm^3) as a percentage of the volume calculated in part (b) (the mean titre, 25.00 cm^3).

Approach

Divide 0.10 by 25.00 and multiply by 100.

Step-by-Step Reasoning

\text{Percentage error} = \frac{0.10}{25.00} \times 100 = 0.004 \times 100 = 0.40\%

The answer is 0.40%. The larger the titre, the smaller the percentage error — this is why using a large titre volume improves precision.

Key Takeaways

Percentage error = (absolute error ÷ measured value) × 100.
The percentage error decreases as the titre volume increases.

Common Mistakes

Using 0.05 cm^3 (the single-reading error) instead of 0.10 cm^3.
Using the rough titre instead of the mean titre from part (b).
Forgetting to multiply by 100.

Things to Be Careful About

Use the mean titre from part (b) as the denominator.
The mark scheme requires the expression 0.1/(candidate titre in (b)) × 100.
Quote the answer with the % sign.

Techniques used

express an absolute error as a percentage of the measured value

(iii)

Calculate the percentage error when $25.0\text{ cm}^3$ of FA 1 was pipetted into the conical flask.

The error was .................................. %.

DifficultyEasy

Worked solution

Working

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

Answer

0.24%

Final answer

0.24%

Detailed explanation

Background Concept

A Grade B volumetric (bulb) pipette is calibrated to ±0.06 cm^3. When 25.0 cm^3 of FA 1 is pipetted into the flask, the maximum error in that volume is 0.06 cm^3. The percentage error is the absolute error divided by the volume, multiplied by 100.

Understanding the Question

Calculate the percentage error when 25.0 cm^3 of FA 1 is pipetted. The pipette tolerance is ±0.06 cm^3.

Approach

Divide 0.06 by 25.0 and multiply by 100.

Step-by-Step Reasoning

\text{Percentage error} = \frac{0.06}{25.0} \times 100 = 0.0024 \times 100 = 0.24\%

The answer is 0.24%. This is smaller than the burette percentage error (0.40%) because the pipette delivers a relatively large volume with a small tolerance.

Key Takeaways

Percentage error = (absolute error ÷ measured value) × 100.
The pipette error (0.24%) is smaller than the burette error (0.40%) here.

Common Mistakes

Using 0.05 cm^3 (burette error) instead of 0.06 cm^3 (pipette error).
Using the titre volume instead of 25.0 cm^3.
Forgetting to multiply by 100.

Things to Be Careful About

Use the pipette tolerance 0.06 cm^3 and the pipette volume 25.0 cm^3.
The mark scheme accepts 0.240/0.24, or 0.2 if 0.24 is seen.
Quote the answer with the % sign.

Techniques used

calculate the percentage error for a pipette volume

The rest of this paper

2 more questions

Q2Presentation of Data and Observations · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation10M
Q3Qualitative Analysis · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation15M

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