Chemistry 9701/32 — May/June 2021

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

Washing soda consists of hydrated sodium carbonate, $\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}$ . When it is stored it loses some of its water of crystallisation to leave $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ . Since water has been lost $x$ is no longer an integer.

You will carry out a titration to determine the value of $x$ . You will titrate a solution of the sodium carbonate with hydrochloric acid.

The equation for the reaction is shown.

\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{CO}_2\text{(aq)} + (x+1)\text{H}_2\text{O(l)}

FB 1 is an aqueous solution containing $11.30\text{ g dm}^{-3}$ of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .
FB 2 is $0.100\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .
bromophenol blue indicator

(a)

Method

Fill the burette with FB 2.
Pipette $25.0\text{ cm}^3$ of FB 1 into a conical flask.
Add a few drops of bromophenol blue indicator.
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 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 2 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

Carry out a rough titration first, then repeat accurate titrations until at least two concordant results (agreeing within 0.10 cm³) are obtained. Record every burette reading to the nearest 0.05 cm³ in a table with correct headings and units.

Representative results (example only — your own readings will differ):

	Rough	1	2	3
Initial burette reading / cm³	0.00	0.00	0.00	0.00
Final burette reading / cm³	19.80	19.75	19.70	19.75
Volume of FB 2 added / cm³	19.80	19.75	19.70	19.75

Final answer

See working — representative titre table (candidate-dependent)

Detailed explanation

Background Concept

A titration is a quantitative technique for finding the concentration of a solution by reacting a measured volume of it with a solution of known concentration (the standard solution). Here, hydrochloric acid of known concentration (0.100 mol dm⁻³) is run from a burette into a measured 25.0 cm³ portion of sodium carbonate solution in a conical flask. An indicator signals the end point. Bromophenol blue is chosen because the carbonate–acid reaction produces carbon dioxide, which makes the solution slightly acidic; bromophenol blue changes from blue (alkaline) to yellow (acidic) at around pH 3–4.6, matching the end point of this reaction.

The marks in this part reward technique and correct recording: reading the burette to the nearest 0.05 cm³, carrying out a rough titration first, repeating accurate titrations until concordant, and presenting the results in a properly headed table with units.

Understanding the Question

You are asked to perform the titration and record your results. The examiner cannot watch you work, so the recorded table is the evidence of your practical skill. Marks are given for: (i) a table with initial/final readings and volume added for the rough and accurate titrations; (ii) correct headings and units; (iii) readings to the nearest 0.05 cm³; (iv) concordant accurate titres; and (v) accuracy compared with the supervisor's value.

Approach

Fill the burette with FB 2 (the acid) and check for leaks.
Use a pipette to deliver exactly 25.0 cm³ of FB 1 (the carbonate) into a conical flask.
Add a few drops of bromophenol blue indicator.
Perform a rough titration to find the approximate end point.
Repeat with accurate titrations, adding acid dropwise near the end point, until at least two concordant results (within 0.10 cm³) are obtained.
Record all readings in a table with headings and units.

Step-by-Step Reasoning

The burette holds the acid because we need to measure the volume of acid delivered; the pipette delivers a fixed, accurate 25.0 cm³ of carbonate. The rough titration gives the approximate titre so accurate titrations can be done quickly. Near the end point, add acid dropwise while swirling until the indicator just changes colour permanently — this is the end point.

Burette readings must be to the nearest 0.05 cm³ because the scale is graduated every 0.1 cm³ and can be estimated to half a division. A reading such as 19.75 cm³ shows proper precision; 19.7 cm³ (0.1 cm³) or 20 cm³ (1 cm³) loses the precision mark.

The table needs three columns: initial burette reading, final burette reading, and volume of FB 2 added (final − initial), each with the unit cm³. The rough titre is kept separate from the accurate ones.

Concordance: at least two accurate titres must agree within 0.10 cm³. The final accurate titre must be within 0.1 cm³ of another accurate titre to earn the concordance mark.

Key Takeaways

Read a burette to the nearest 0.05 cm³.
Record results in a table with headings and units.
Do a rough titration, then accurate titrations to concordance (within 0.10 cm³).
The recorded table is the evidence of technique and accuracy.

Common Mistakes

Recording readings to 0.01 cm³ (a burette cannot be read that precisely) or to whole cm³ (too imprecise).
Using 50.00 as the initial burette reading (explicitly disallowed by the mark scheme).
Omitting units in the table headings.
Using headings such as "difference" or "change" instead of "volume of FB 2 added".
Stopping after only one accurate titration — at least two concordant titres are required.

Things to Be Careful About

The mark scheme disallows 50.00 as an initial reading, more than one final reading of 50.00, and any reading above 50.00.
Do not give accurate burette readings to 0 decimal places (except 0.00).
The volume of FB 2 added must be clearly recorded for each accurate titration.

Techniques used

fill and use a burette correctlycarry out a rough titration then accurate titrationsread a burette to the nearest 0.05 cm³record titration results in a table with headings and units

(b)

From your accurate titration results, obtain a value for the volume of FB 2 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 2.

DifficultyEasy

Worked solution

Answer

Select the accurate titres that agree within 0.20 cm³: 19.75, 19.70 and 19.75 cm³.

\text{mean titre} = \frac{19.75 + 19.70 + 19.75}{3} = 19.73\text{ cm}^3

$25.0\text{ cm}^3$ of FB 1 required $19.73\text{ cm}^3$ of FB 2.

Final answer

19.73 cm³ (example mean of concordant titres)

Detailed explanation

Background Concept

After carrying out several accurate titrations, you must decide which results to use. The mark scheme requires that the value used in calculations is the mean of two or more titres that all lie within 0.20 cm³ of each other. This ensures the chosen value is reliable and reproducible.

Understanding the Question

This part asks you to select the concordant titres from your accurate results and average them, showing your working. The mark is awarded for averaging two or more titres that are all within 0.20 cm³, with the working or selection clearly shown (e.g. ticks next to the chosen titres).

Approach

Look at your accurate titres (ignoring the rough titre).
Identify the group of two or more titres all within 0.20 cm³ of each other.
Average them and express the mean to the nearest 0.01 cm³.

Step-by-Step Reasoning

Using the representative results from part (a): accurate titres are 19.75, 19.70 and 19.75 cm³. These all lie within 0.20 cm³ of each other (19.70 to 19.75 is a spread of only 0.05 cm³), so all three are used.

\text{mean} = \frac{19.75 + 19.70 + 19.75}{3} = \frac{59.20}{3} = 19.733\ldots = 19.73\text{ cm}^3

The mean is expressed to the nearest 0.01 cm³, consistent with the precision of the individual readings.

Key Takeaways

Use only concordant titres (within 0.20 cm³) for the mean.
Ignore the rough titre.
Show your selection (ticks) and the averaging working.

Common Mistakes

Averaging all titres including a non-concordant one.
Including the rough titre in the mean.
Not showing which titres were selected.
Giving the mean to too many decimal places (e.g. 19.733333).

Things to Be Careful About

The mark scheme requires the selected titres to be within 0.20 cm³ of each other.
Express the mean to the nearest 0.01 cm³.

Techniques used

select concordant accurate titrescalculate a mean titre

(c)

Calculations

(i)

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

DifficultyEasy

Worked solution

Answer

Answers to (c)(ii) and (c)(iii) are given to 3–4 significant figures.

Final answer

3–4 significant figures

Detailed explanation

Background Concept

Significant figures reflect the precision of a measurement. The titre (19.73 cm³) and the concentration (0.100 mol dm⁻³) are known to 3–4 significant figures, so the calculated moles and concentration should be quoted to 3–4 significant figures to avoid implying false precision.

Understanding the Question

This is a directive part: it tells you the precision expected in the answers to (c)(ii) and (c)(iii). The mark is awarded simply for giving those answers to 3–4 significant figures.

Approach

When you calculate moles of HCl and the concentration of the carbonate, round the final answers to 3–4 significant figures (not to 1 or 2, and not to 5+).

Step-by-Step Reasoning

The input data are: concentration of HCl = 0.100 mol dm⁻³ (3 sf) and titre = 19.73 cm³ (4 sf). The limiting precision is 3 sf, so answers to 3–4 sf are acceptable. For example, moles of HCl = 1.973 × 10⁻³ mol (4 sf) and concentration = 0.0395 mol dm⁻³ (3 sf) both satisfy this.

Key Takeaways

Match the significant figures of your answer to the precision of the data.
3–4 sf is the accepted convention here.

Common Mistakes

Giving answers to 1 or 2 significant figures (too imprecise).
Giving answers to 5 or more significant figures (false precision).

Things to Be Careful About

The final value of x in (c)(iv) is also expected to 2–4 sf by its own mark scheme.

Techniques used

apply significant-figure conventions to calculated answers

(ii)

Calculate the number of moles of hydrochloric acid present in the volume of FB 2 you calculated in (b).

\text{moles of HCl} = \text{.............................. mol}

DifficultyMedium-Easy

Worked solution

Working

\text{moles of HCl} = \frac{0.100 \times 19.73}{1000} = 1.973 \times 10^{-3}\text{ mol}

Answer

$1.97 \times 10^{-3}\text{ mol}$ (to 3 sf)

Final answer

1.97 × 10⁻³ mol

Detailed explanation

Background Concept

The number of moles of a solute in a solution is found from its concentration and volume:

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

Since the titre is in cm³, it must be converted to dm³ by dividing by 1000.

Understanding the Question

You are asked to calculate the moles of HCl in the volume of FB 2 found in part (b). This is the first step of the calculation chain: titre → moles of HCl → moles of carbonate → concentration of carbonate → Mr → x.

Approach

Substitute the concentration of FB 2 (0.100 mol dm⁻³) and the mean titre (19.73 cm³, converted to dm³) into the moles formula.

Step-by-Step Reasoning

\text{moles of HCl} = 0.100 \times \frac{19.73}{1000} = 0.100 \times 0.01973 = 1.973 \times 10^{-3}\text{ mol}

The answer is quoted to 3–4 sf as required by (c)(i): 1.97 × 10⁻³ mol (3 sf) or 1.973 × 10⁻³ mol (4 sf).

Key Takeaways

Always convert cm³ to dm³ (÷1000) before using the moles formula.
moles = concentration × volume is the fundamental relationship.

Common Mistakes

Forgetting to divide the titre by 1000.
Quoting the answer to the wrong number of significant figures.
Mixing up cm³ and dm³.

Things to Be Careful About

Use the titre from part (b), not the rough titre.
The mark scheme formula is exactly $\frac{0.1 \times \text{vol in (b)}}{1000}$ .

Techniques used

calculate moles from concentration and volume

(iii)

Use the equation on page 2, and your answer to (c)(ii), to calculate the concentration, in $\text{mol dm}^{-3}$ , of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ present in FB 1.

\text{concentration of Na}_2\text{CO}_3\cdot x\text{H}_2\text{O} = \text{.............................. mol dm}^{-3}

DifficultyMedium-Easy

Worked solution

Working

From the equation, 1 mol of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ reacts with 2 mol of HCl.

\text{moles of } \text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O in } 25.0\text{ cm}^3 = \frac{1.973 \times 10^{-3}}{2} = 9.865 \times 10^{-4}\text{ mol}

\text{concentration} = 9.865 \times 10^{-4} \times \frac{1000}{25.0} = 0.0395\text{ mol dm}^{-3}

Answer

$0.0395\text{ mol dm}^{-3}$

Final answer

0.0395 mol dm⁻³

Detailed explanation

Background Concept

The balanced equation shows the stoichiometric ratio:

\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{CO}_2\text{(aq)} + (x+1)\text{H}_2\text{O(l)}

1 mole of carbonate reacts with 2 moles of HCl. So the moles of carbonate in the 25.0 cm³ sample are half the moles of HCl.

Understanding the Question

You must use the moles of HCl from (c)(ii) and the stoichiometry of the equation to find the moles of carbonate in the 25.0 cm³ sample, then scale up to a concentration in mol dm⁻³ (per 1000 cm³).

Approach

Divide moles of HCl by 2 to get moles of carbonate in 25.0 cm³.
Multiply by 1000/25.0 = 40 to express per dm³.

Step-by-Step Reasoning

\text{moles of carbonate in } 25.0\text{ cm}^3 = \frac{1.973 \times 10^{-3}}{2} = 9.865 \times 10^{-4}\text{ mol}

Since 25.0 cm³ is 1/40 of a dm³, the concentration is:

\text{concentration} = 9.865 \times 10^{-4} \times 40 = 0.03946 \approx 0.0395\text{ mol dm}^{-3}

The answer is given to 3 sf to satisfy (c)(i).

Key Takeaways

Use the stoichiometric ratio from the balanced equation (here 1:2).
To convert moles in a portion to concentration, multiply by 1000/volume(cm³).

Common Mistakes

Forgetting to divide by 2 (using the HCl moles directly as carbonate moles).
Confusing the scaling factor (using 25.0/1000 instead of 1000/25.0).
Omitting the unit mol dm⁻³.

Things to Be Careful About

The mark scheme credits: (c)(ii) ÷ 2 and × 40.
Keep the working visible so method marks can be awarded even if the arithmetic slips.

Techniques used

apply the stoichiometric ratio from the balanced equationconvert moles in 25.0 cm³ to concentration in mol dm⁻³

(iv)

Calculate the value of $x$ in this sample of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

Show your working.

x = \text{..............................}

DifficultyMedium

Worked solution

Working

M_r\text{ of } \text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O} = \frac{11.30}{0.03946} = 286.4

M_r\text{ of } \text{Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 106

\text{mass of } x\text{H}_2\text{O} = 286.4 - 106 = 180.4

x = \frac{180.4}{18} = 10.0

Answer

$x = 10.0$

Final answer

x = 10.0

Detailed explanation

Background Concept

The concentration of the hydrated carbonate in FB 1 (0.0395 mol dm⁻³) and its mass concentration (11.30 g dm⁻³) are related by:

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

so $M_r = \frac{\text{mass concentration}}{\text{molar concentration}}$ . The molar mass of the hydrated salt is $106 + 18x$ (anhydrous Na₂CO₃ = 106, each H₂O = 18). Once $M_r$ is known, x is found by subtracting 106 and dividing by 18.

Understanding the Question

You are given the mass concentration of FB 1 (11.30 g dm⁻³) and have calculated its molar concentration in (c)(iii). The task is to use these to find the molar mass, then back-calculate the number of water molecules of crystallisation, x.

Approach

Divide the mass concentration by the molar concentration to get $M_r$ .
Subtract the molar mass of anhydrous Na₂CO₃ (106).
Divide the remainder by 18 (molar mass of H₂O) to get x.

Step-by-Step Reasoning

Using the value from (c)(iii), 0.03946 mol dm⁻³:

M_r = \frac{11.30}{0.03946} = 286.4

This is the molar mass of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

The anhydrous salt: $M_r = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106$ .

The mass of water of crystallisation per formula unit is therefore:

286.4 - 106 = 180.4

Each water molecule has $M_r = 18$ , so:

x = \frac{180.4}{18} = 10.02 \approx 10.0

The value of x is 10.0, consistent with the original washing soda formula $\text{Na}_2\text{CO}_3\cdot 10\text{H}_2\text{O}$ (the sample here has lost essentially no water). The answer is given to 2–4 sf as required.

Key Takeaways

$M_r = \frac{\text{mass concentration}}{\text{molar concentration}}$ .
Molar mass of a hydrate = molar mass of anhydrous salt + x × 18.
Back-calculate x by subtracting the anhydrous mass and dividing by 18.

Common Mistakes

Using the mass of the whole solution instead of the mass concentration.
Forgetting to subtract the anhydrous molar mass (106) before dividing by 18.
Using 18.0 but not accounting for the number of waters correctly.
Arithmetic slip in $M_r$ of Na₂CO₃ (e.g. forgetting the two sodium atoms contribute 46).

Things to Be Careful About

The mark scheme awards M1 for $M_r = 11.30/\text{moles from (c)(iii)}$ , M2 for $x\text{H}_2\text{O} = M_r - 106$ , and M3 for dividing by 18 with the answer to 2–4 sf.
An alternative valid route uses the ratio of moles of anhydrous salt to moles of water; either method is credited.

Techniques used

calculate molar mass from concentration and mass concentrationdeduce the number of water molecules of crystallisationsubtract the anhydrous molar mass and divide by 18

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation11M
Q3Qualitative Analysis · Analysis, Conclusions and Evaluation15M

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