Chemistry 9701/33 — May/June 2017

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

questions

marks

120

minutes

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

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

Sodium hydrogencarbonate, $\text{NaHCO}_3$ , is used as baking soda in cooking. Baking soda may also contain small amounts of other chemicals.

In this experiment, you will determine the percentage purity by mass of an impure sample of $\text{NaHCO}_3$ by titration with sulfuric acid.

FA 1 is $0.0500\text{ mol dm}^{-3}$ sulfuric acid, $\text{H}_2\text{SO}_4$ .
FA 2 is impure $\text{NaHCO}_3$ .
methyl orange

(a)

Method

Preparing a solution of FA 2

Weigh the stoppered container of FA 2. Record the mass in the space below.
Tip all the FA 2 into the beaker.
Reweigh the container with its stopper. Record the mass.
Calculate and record the mass of FA 2 used.
Add approximately $100\text{ cm}^3$ of distilled water to the FA 2 in the beaker.
Stir the mixture with a glass rod until all the FA 2 has dissolved.
Transfer this solution into the $250\text{ cm}^3$ volumetric flask.
Wash the beaker with distilled water and transfer the washings to the volumetric flask.
Rinse the glass rod with distilled water and transfer the washings to the volumetric flask.
Make up the solution in the volumetric flask to the mark using distilled water.
Shake the flask thoroughly.
This solution of impure $\text{NaHCO}_3$ is FA 3. Label the flask FA 3.

Results

Titration

Fill the burette with FA 1.
Pipette $25.0\text{ cm}^3$ of FA 3 into a conical flask.
Add several drops of methyl orange.
Perform 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 any recorded results show the precision of your practical work.
Record in a suitable form below all of your burette readings and the volume of FA 1 added in each accurate titration.

Keep FA 1 for use in Question 2.

DifficultyMedium-Easy

Worked solution

Answer

Using the following example readings (a candidate records their own data):

Mass of FA 2

Measurement	Mass / g
container + FA 2	13.50
empty container	10.50
FA 2 used	3.00

Rough titration

Reading	Volume / $\text{cm}^3$
final	35.20
initial	0.00
rough titre	35.20

Accurate titrations

	1	2	3
final burette reading / $\text{cm}^3$	35.10	35.00	35.05
initial burette reading / $\text{cm}^3$	0.00	0.00	0.00
volume of FA 1 added / $\text{cm}^3$	35.10	35.00	35.05

Final answer

Example data: mass FA 2 = 3.00 g; accurate titres 35.10, 35.00, 35.05 cm3

Detailed explanation

Background Concept

This part tests the standard practical skills of preparing a solution of accurately known concentration and titrating it against a standard acid. Weighing by difference gives the mass of FA 2 without needing to transfer every particle: the difference between the two weighings is the mass that entered the beaker. A volumetric flask is used to make an exact total volume (250 cm $^3$ ), so the amount of FA 2 in any pipetted portion is known. In a titration, a burette delivers measured volumes to 0.05 cm $^3$ , and concordant titres (within 0.10 cm $^3$ ) show that the end point has been found reliably. Methyl orange changes from yellow in alkaline/neutral solution to red in acid, so it signals the acid–base end point.

Understanding the Question

You are asked to carry out the preparation and titration and to record your results in a clear table. The marks are for correct headings with units, readings to the correct precision, and consistent titres. The exact numbers are not known in advance, so the model below uses example data; in the exam you would record your own readings.

Approach

First weigh FA 2 by difference. Dissolve it in about 100 cm $^3$ of water, transfer quantitatively to the 250 cm $^3$ volumetric flask, washing the beaker and rod into the flask, then make up to the mark and shake. Fill the burette with FA 1, pipette 25.0 cm $^3$ of FA 3 into a conical flask, add methyl orange, and titrate. Do a rough titration first, then repeat until two or more accurate titres agree within 0.10 cm $^3$ . Record every burette reading to the nearest 0.05 cm $^3$ .

Step-by-Step Reasoning

Weigh the stoppered container with FA 2, then reweigh after tipping out the solid; the mass of FA 2 is the difference. After dissolving, every piece of apparatus that touched the solution must be rinsed into the flask so no solute is lost. The flask is filled to the graduation mark, giving exactly 250 cm $^3$ of FA 3. For the titration, the pipette delivers exactly 25.0 cm $^3$ ; methyl orange is yellow before the end point and turns red at it. The rough titre gives an approximate end point; accurate titres should be within 0.10 cm $^3$ of each other. In the example, the accurate titres are 35.10, 35.00 and 35.05 cm $^3$ , all recorded to 0.05 cm $^3$ , with headings including units.

Key Takeaways

Weighing by difference avoids transferring every grain of solid. Quantitative transfer and making up to the mark in a volumetric flask give a known total volume. Concordant titres and readings to 0.05 cm $^3$ are the evidence of reliable practical work.

Common Mistakes

Recording masses without units or without stating what was weighed.
Using 50.00 cm $^3$ as an initial burette reading, or recording any burette reading above 50.00 cm $^3$ .
Recording burette readings to 1 decimal place only.
Counting the rough titre as an accurate titre.
Only doing one accurate titration.
Not washing the beaker and glass rod into the volumetric flask.

Things to Be Careful About

The mark scheme requires each heading to include the quantity and unit, e.g. final burette reading / cm $^3$ . Burette readings must be to the nearest 0.05 cm $^3$ . Accurate titres should agree within 0.10 cm $^3$ . The mass of FA 2 must be found by subtraction, not by weighing the empty beaker only.

Techniques used

weigh by differenceprepare a solution in a volumetric flaskperform accurate titrationsrecord burette readings to the nearest 0.05 cm3construct a results table with headings and units

(b)

From your accurate titration results, obtain a suitable value for the volume of FA 1 to be used in your calculations.
Show clearly how you obtained this value.

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

DifficultyEasy

Worked solution

Working

Accurate titres: 35.10, 35.00 and 35.05 cm $^3$ .

Spread = 35.10 − 35.00 = 0.10 cm $^3$ , which is within 0.20 cm $^3$ .

Mean = (35.10 + 35.00 + 35.05) / 3 = 105.15 / 3 = 35.05 cm $^3$ .

Answer

25.0 cm $^3$ of FA 3 required 35.05 cm $^3$ of FA 1.

Final answer

35.05 cm3

Detailed explanation

Background Concept

A reliable titre is the mean of at least two accurate, concordant titrations. Concordant means the values agree closely; in this syllabus accurate titres should be within a total spread of 0.20 cm $^3$ (often 0.10 cm $^3$ of each other). Averaging reduces the effect of random reading errors.

Understanding the Question

Part (b) asks you to choose a single volume of FA 1 from your accurate titration results and show how you obtained it. The mark is for selecting two or more concordant titres and calculating their mean correctly.

Approach

Identify the accurate titres that are within a total spread of no more than 0.20 cm $^3$ . Add them and divide by the number used. Quote the mean to 2 decimal places, rounding to the nearest 0.01 cm $^3$ .

Step-by-Step Reasoning

Using the example accurate titres 35.10, 35.00 and 35.05 cm $^3$ , the spread is 35.10 − 35.00 = 0.10 cm $^3$ , which is within 0.20 cm $^3$ . The mean is (35.10 + 35.00 + 35.05) / 3 = 105.15 / 3 = 35.05 cm $^3$ . This is already to 2 dp. If the mean had been 26.665 cm $^3$ , it would be rounded to 26.67 cm $^3$ .

Key Takeaways

The mean titre is the volume used in all subsequent calculations. It must be based on concordant accurate titres only, never on the rough titre.

Common Mistakes

Including the rough titre in the mean.
Averaging only one accurate titration.
Subtracting burette readings incorrectly.
Quoting the mean to 1 dp when the readings were recorded to 2 dp.

Things to Be Careful About

The two or more titres used must have a total spread of no more than 0.20 cm $^3$ . Show your selection, for example by ticking the chosen readings or writing the calculation. Round the mean to the nearest 0.01 cm $^3$ .

Techniques used

select concordant titrescalculate the mean titreround the mean to 2 decimal places

(c)

Calculations

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

(i)

Calculate the number of moles of sulfuric acid present in the volume of FA 1 calculated in (b).

moles of $\text{H}_2\text{SO}_4$ = ............................. mol

DifficultyEasy

Worked solution

Working

$n(\text{H}_2\text{SO}_4) = c \times V = 0.0500 \times \frac{35.05}{1000} = 1.7525 \times 10^{-3} \text{ mol}$

To 3 significant figures:

$n(\text{H}_2\text{SO}_4) = 1.75 \times 10^{-3} \text{ mol}$

Answer

$1.75 \times 10^{-3}$ mol

Final answer

1.75 × 10^-3 mol

Detailed explanation

Background Concept

For a solution, amount of solute in moles is given by n = c × V, where c is concentration in mol dm $^{-3}$ and V is volume in dm $^3$ . Since burette volumes are measured in cm $^3$ , convert by dividing by 1000.

Understanding the Question

You need the amount of H2SO4 in the mean titre from (b). The concentration of FA 1 is given as 0.0500 mol dm $^{-3}$ .

Approach

Substitute the mean titre (converted to dm $^3$ ) and the concentration into n = c × V.

Step-by-Step Reasoning

Using the example mean titre 35.05 cm $^3$ :
V = 35.05 / 1000 = 0.03505 dm $^3$ .
n(H2SO4) = 0.0500 × 0.03505 = 1.7525 × 10 $^{-3}$ mol.
To 3 significant figures, this is 1.75 × 10 $^{-3}$ mol.

Key Takeaways

Always convert cm $^3$ to dm $^3$ before using n = c × V. Keep enough significant figures in intermediate steps.

Common Mistakes

Using the volume in cm $^3$ without dividing by 1000.
Rounding too early and losing accuracy.
Quoting too many or too few significant figures.

Things to Be Careful About

The mark scheme accepts answers to a minimum of 2 significant figures, but 3 or 4 significant figures are safer. Use the mean titre from (b), not the rough titre.

Techniques used

convert volume from cm3 to dm3calculate moles from concentration and volume

(ii)

Balance the equation for the reaction of sulfuric acid and sodium hydrogencarbonate. State symbols are not required.

\dots\dots\text{NaHCO}_3 + \dots\dots\text{H}_2\text{SO}_4 \rightarrow \dots\dots\text{Na}_2\text{SO}_4 + \dots\dots\text{CO}_2 + \dots\dots\text{H}_2\text{O}

DifficultyEasy

Worked solution

Answer

2\text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{CO}_2 + 2\text{H}_2\text{O}

Final answer

2NaHCO3 + H2SO4 -> Na2SO4 + 2CO2 + 2H2O

Detailed explanation

Background Concept

Sodium hydrogencarbonate is a base that reacts with sulfuric acid to give sodium sulfate, carbon dioxide and water. The hydrogencarbonate ion acts as a base, accepting a proton to form carbonic acid, which decomposes to CO2 and H2O.

Understanding the Question

You are given an unbalanced equation and asked to supply the coefficients. State symbols are not required.

Approach

Balance the equation systematically: balance metals first, then non-metals, then hydrogen and oxygen last.

Step-by-Step Reasoning

Na2SO4 contains two Na atoms, so place 2 before NaHCO3. The two NaHCO3 units provide two C atoms, so place 2 before CO2. They also provide two H atoms, and H2SO4 provides two more H atoms, so four H atoms in total; place 2 before H2O. Check oxygen: 2 NaHCO3 gives 6 O, H2SO4 gives 4 O, total 10 O; products have 4 (Na2SO4) + 4 (2CO2) + 2 (2H2O) = 10 O. The balanced equation is:

2\text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{CO}_2 + 2\text{H}_2\text{O}

Key Takeaways

The balanced equation gives the stoichiometric ratio: 2 mol NaHCO3 react with 1 mol H2SO4.

Common Mistakes

Balancing Na but forgetting to balance C, H or O.
Writing 1 before NaHCO3.
Adding state symbols incorrectly if included.

Things to Be Careful About

The mark for this part is shared with (iii): the equation and the mole ratio are marked together. Check the equation by counting atoms of each element on both sides.

Techniques used

balance a neutralisation equationdetermine stoichiometric coefficients

(iii)

Using your answers to (i) and (ii), calculate the number of moles of sodium hydrogencarbonate used in each titration.

moles of $\text{NaHCO}_3$ = ............................. mol

DifficultyEasy

Worked solution

Working

From the balanced equation, 2 mol NaHCO3 react with 1 mol H2SO4.

$n(\text{NaHCO}_3) = 2 \times n(\text{H}_2\text{SO}_4) = 2 \times 1.75 \times 10^{-3} = 3.50 \times 10^{-3} \text{ mol}$

Answer

$3.50 \times 10^{-3}$ mol

Final answer

3.50 × 10^-3 mol

Detailed explanation

Background Concept

The coefficients in a balanced equation give the mole ratio in which substances react. Here 2 mol NaHCO3 react with 1 mol H2SO4, so n(NaHCO3) = 2 × n(H2SO4).

Understanding the Question

Using the moles of H2SO4 from (i) and the balanced equation from (ii), calculate the moles of NaHCO3 in the 25.0 cm $^3$ portion titrated.

Approach

Multiply the moles of H2SO4 by the stoichiometric ratio 2/1.

Step-by-Step Reasoning

From (i), n(H2SO4) = 1.75 × 10 $^{-3}$ mol.
n(NaHCO3) = 2 × 1.75 × 10 $^{-3}$ = 3.50 × 10 $^{-3}$ mol.
This is the amount in the 25.0 cm $^3$ pipette sample of FA 3.

Key Takeaways

The mole ratio comes from the balanced equation. If the equation were wrong, this calculation would be wrong too.

Common Mistakes

Using a 1:1 ratio instead of 2:1.
Using the moles of acid from the whole flask rather than the aliquot.

Things to Be Careful About

Use the unrounded or sufficiently precise value from (i) to avoid rounding errors. The final answer should be to 3 or 4 significant figures.

Techniques used

apply mole ratio from balanced equationcalculate moles of NaHCO3

(iv)

Using your answer to (iii), calculate the mass of sodium hydrogencarbonate present in the mass of FA 2 used to prepare FA 3.

mass of $\text{NaHCO}_3$ = ............................. g

DifficultyMedium-Easy

Worked solution

Working

Moles in 25.0 cm $^3$ aliquot = $3.50 \times 10^{-3}$ mol.

Moles in 250 cm $^3$ FA 3 = $3.50 \times 10^{-3} \times 10 = 3.50 \times 10^{-2}$ mol.

$M(\text{NaHCO}_3) = 23 + 1 + 12 + 3(16) = 84$ g mol $^{-1}$ .

Mass = $3.50 \times 10^{-2} \times 84 = 2.94$ g.

Answer

2.94 g

Final answer

2.94 g

Detailed explanation

Background Concept

The titration gives the amount of NaHCO3 in only 25.0 cm $^3$ of the 250 cm $^3$ solution. Because the solution is homogeneous, the amount in the whole flask is 10 times larger. Mass is then found from amount × molar mass.

Understanding the Question

You need the mass of pure NaHCO3 in the original mass of FA 2 used to prepare FA 3. That requires scaling the aliquot moles up by the factor 250/25 = 10, then multiplying by M(NaHCO3).

Approach

Multiply the moles from (iii) by 10, then by 84 g mol $^{-1}$ .

Step-by-Step Reasoning

Moles in 25.0 cm $^3$ = 3.50 × 10 $^{-3}$ mol.
Moles in 250 cm $^3$ = 3.50 × 10 $^{-3}$ × 10 = 3.50 × 10 $^{-2}$ mol.
M(NaHCO3) = 23 + 1 + 12 + 3(16) = 84 g mol $^{-1}$ .
Mass = 3.50 × 10 $^{-2}$ × 84 = 2.94 g.

Key Takeaways

The aliquot factor is essential: the titration measures only a fraction of the prepared solution. Always check whether you need to scale up.

Common Mistakes

Forgetting to multiply by 10.
Using M(NaHCO3) = 53 or another incorrect value.
Mixing up the aliquot volume and the flask volume.

Things to Be Careful About

M(NaHCO3) is 84 g mol $^{-1}$ (Na 23, H 1, C 12, O 16 × 3). Keep significant figures consistent.

Techniques used

scale moles from aliquot to total volumecalculate mass from moles and molar mass

(v)

Calculate the percentage purity by mass of the impure sodium hydrogencarbonate sample, FA 2.

percentage purity by mass of impure $\text{NaHCO}_3$ , FA 2 = ............................. %

DifficultyMedium-Easy

Worked solution

Working

Percentage purity = (mass of NaHCO3 / mass of FA 2) × 100 = (2.94 / 3.00) × 100 = 98.0%

Answer

98.0%

Final answer

98.0%

Detailed explanation

Background Concept

Percentage purity by mass is the mass of the pure substance divided by the mass of the impure sample, multiplied by 100. It assumes the impurities do not react with the acid.

Understanding the Question

You have the mass of pure NaHCO3 in the sample from (iv) and the original mass of FA 2 used. Calculate the percentage.

Approach

Divide the mass of NaHCO3 by the mass of FA 2 and multiply by 100.

Step-by-Step Reasoning

Using the example data:
Percentage purity = (2.94 / 3.00) × 100 = 98.0%.
The answer is quoted to 3 significant figures, consistent with the data.

Key Takeaways

Percentage purity is a simple ratio. The result should be less than or equal to 100% if the sample is impure; a value above 100% usually indicates an error.

Common Mistakes

Dividing the mass of sample by the mass of pure substance.
Forgetting to multiply by 100.
Quoting the answer to too few significant figures.

Things to Be Careful About

The mark scheme requires answers in (i), (iii), (iv) and (v) to be shown to 3 or 4 significant figures. Use the mass of FA 2 used, not the mass of the container.

Techniques used

calculate percentage purity by massapply significant figures

(vi)

What did you assume about the impurities in FA 2 when you calculated the percentage purity?

DifficultyEasy

Worked solution

Answer

The impurities do not react with sulfuric acid / FA 1; they are not acidic or alkaline (they are neutral).

Final answer

The impurities do not react with sulfuric acid (are neutral).

Detailed explanation

Background Concept

Percentage purity calculated from titration assumes that only NaHCO3 in the sample reacts with the acid. If an impurity also consumed acid, the calculated amount of NaHCO3 would be too high; if an impurity were acidic or alkaline, it would affect the end point.

Understanding the Question

You are asked to state the assumption behind the calculation.

Approach

Identify what must be true about the impurities for the titration result to represent only NaHCO3.

Step-by-Step Reasoning

The calculation treats every mole of acid as reacting with NaHCO3. This is valid only if the impurities do not react with sulfuric acid, and are neither acidic nor alkaline. In other words, the impurities are neutral and inert under the titration conditions.

Key Takeaways

A titration-based purity calculation is only as valid as the assumption that the measured reaction is due entirely to the substance of interest.

Common Mistakes

Saying the impurities do not matter without specifying that they do not react.
Saying there are no impurities when the sample is known to be impure.

Things to Be Careful About

The mark scheme accepts any one of: impurities do not react with sulfuric acid/FA 1; impurities are not acidic or alkaline; impurities are neutral.

Techniques used

state the assumption about inert impurities

(vii)

A volumetric flask was labelled $250.0 \pm 0.10\text{ cm}^3$ .

Calculate the maximum percentage error when using this volumetric flask.

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

DifficultyEasy

Worked solution

Working

Maximum percentage error = (0.10 / 250.0) × 100 = 0.04%

Answer

0.04%

Final answer

0.04%

Detailed explanation

Background Concept

Percentage error (or percentage uncertainty) compares the absolute uncertainty of a measurement with the measured value: percentage error = (absolute uncertainty / measured value) × 100.

Understanding the Question

The volumetric flask is marked 250.0 ± 0.10 cm $^3$ . You need the maximum percentage error in using it to measure 250.0 cm $^3$ .

Approach

Use the uncertainty 0.10 cm $^3$ as the numerator and 250.0 cm $^3$ as the measured value.

Step-by-Step Reasoning

Percentage error = (0.10 / 250.0) × 100 = 0.04%.
The uncertainty is one part in 2500, which is 0.04%.

Key Takeaways

Percentage error allows different sources of error to be compared. Here the flask contributes a very small error.

Common Mistakes

Using 0.1 / 250 = 0.0004 and forgetting to multiply by 100.
Confusing absolute and percentage uncertainty.

Things to Be Careful About

The answer is 0.04%, not 0.4%. Use the uncertainty in the same units as the measured value.

Techniques used

calculate percentage error from absolute uncertainty and measured value

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Qualitative Analysis14M
Q3Qualitative Analysis · Analysis, Conclusions and Evaluation10M

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