Chemistry 9701/31 — 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 · Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Qualitative Analysis

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

In this experiment you will carry out a titration to determine the relative formula mass of a hydrated salt, $\text{FA 1}$ .

$\text{FA 1}$ is a hydrated salt.
$\text{FA 2}$ is dilute sulfuric acid.
$\text{FA 3}$ is $0.0200\text{ mol dm}^{-3}$ potassium manganate(VII).

(a)

Method

Preparing a solution of FA 1

Weigh the stoppered container of $\text{FA 1}$ . Record the mass in the space below.
Tip all the $\text{FA 1}$ into the beaker.
Reweigh the container with its stopper. Record the mass.
Calculate and record the mass of $\text{FA 1}$ used.
Add approximately $100\text{ cm}^3$ of $\text{FA 2}$ to the $\text{FA 1}$ in the beaker.
Stir the mixture until all the $\text{FA 1}$ has dissolved.
Transfer this solution into the $250\text{ cm}^3$ volumetric flask.
Rinse the beaker and 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 the hydrated salt is $\text{FA 4}$ . Label the flask $\text{FA 4}$ .

Titration

Fill the burette with $\text{FA 3}$ .
Pipette $25.0\text{ cm}^3$ of $\text{FA 4}$ into a conical flask.
Use the $25.0\text{ cm}^3$ measuring cylinder to add $10\text{ cm}^3$ of $\text{FA 2}$ to the $\text{FA 4}$ in the conical flask.
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 $\text{FA 3}$ added in each accurate titration.

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

DifficultyMedium-Easy

Worked solution

Answer

Record the following in the space provided.

Weighings (recorded to 2 decimal places):

Quantity	Mass / g
Mass of container + FA 1	27.45
Mass of empty container	20.64
Mass of FA 1 used	6.81

Rough titre: 25.10 cm³

Accurate titrations:

	Titration 1	Titration 2	Titration 3
Initial burette reading / cm³	0.00	24.45	0.00
Final burette reading / cm³	24.45	48.95	24.55
Titre / cm³	24.45	24.50	24.55

All burette readings recorded to the nearest 0.05 cm³. Accurate titres agree within 0.10 cm³ of each other.

Final answer

See working: representative weighings (6.81 g FA 1 used) and titres (24.45, 24.50, 24.55 cm³) recorded to the required precision.

Detailed explanation

Background Concept

This part of the practical assesses your ability to carry out a titration accurately and to record your data in a clear, professional manner. Three core techniques are being tested:

Weighing by difference. You weigh the container holding the solid, transfer the solid to the beaker, then reweigh the empty container. The mass of solid transferred is the difference between the two readings. This avoids errors from trying to weigh an empty beaker first and then adding solid to it.
Burette technique. A burette measures volumes to the nearest 0.05 cm³. The scale is graduated in 0.1 cm³ divisions, so you estimate to half a division. You read the bottom of the meniscus at eye level. Before starting, you must ensure the jet below the tap is filled with solution (no air bubble), because an air bubble would be expelled during the titration and would not represent solution delivered.
Concordant titres. A single titration is not reliable. You perform a rough titration to find the approximate end-point, then several accurate titrations. The accurate titres should agree within 0.10 cm³ of each other (concordant). The mean of these concordant values is used in the calculations.

The mark scheme for this part awards up to 8 marks:

B1 for recording the mass of container + FA 1, the mass of the empty container, and the correctly subtracted mass of FA 1 used, with consistent decimal places (at least 1 dp).
B1 for recording the rough titre and initial/final burette readings for two or more accurate titrations.
B1 for appropriate headings and units in the accurate titration table.
B1 for all accurate burette readings to the nearest 0.05 cm³.
B1 for the final accurate titre being within 0.10 cm³ of another accurate titre.
B3 for the spread of titres per gram of FA 1 (δ): δ ≤ 0.020 cm³ g⁻¹ earns all 3 marks; 0.020 < δ ≤ 0.040 earns 2; 0.040 < δ ≤ 0.060 earns 1.

Understanding the Question

The question gives you a step-by-step method. Your job is to carry it out and record the data in the spaces provided. The mark scheme is looking for specific things: correct headings, correct precision, and concordant results. The representative data below show what a good set of results looks like.

Approach

Follow the method exactly. Record every reading as you take it. Use a table for the accurate titrations with proper headings and units. Ensure all burette readings are to 2 decimal places (nearest 0.05 cm³). Aim for at least three accurate titrations so you have concordant values to average.

Step-by-Step Reasoning

Weighings. Record the mass of the stoppered container with FA 1 (e.g., 27.45 g). Tip all the FA 1 into the beaker. Reweigh the empty container with stopper (e.g., 20.64 g). Mass of FA 1 used = 27.45 − 20.64 = 6.81 g. Record this. Use the same number of decimal places for both weighings.
Preparing FA 4. Dissolve the FA 1 in about 100 cm³ of FA 2 (dilute sulfuric acid). Stir until dissolved. Transfer to the 250 cm³ volumetric flask. Rinse the beaker and glass rod with distilled water and add the washings (this ensures all the salt is transferred). Make up to the mark with distilled water. Shake thoroughly. Label as FA 4.
Rough titration. Fill the burette with FA 3. Pipette 25.0 cm³ of FA 4 into a conical flask. Add 10 cm³ of FA 2 from the measuring cylinder. Titrate until the end-point (a permanent pale pink colour). Record the rough titre (e.g., 25.10 cm³).
Accurate titrations. Repeat the titration, adding FA 3 dropwise near the end-point. Record initial and final burette readings for each accurate titration. Aim for concordant results (within 0.10 cm³). A typical set: 24.45, 24.50, 24.55 cm³.
Recording. Present the accurate titrations in a table with columns for initial reading, final reading, and titre, each with the unit cm³.

Key Takeaways

Weighing by difference is the standard technique for accurate mass transfer.
Burette readings must be recorded to the nearest 0.05 cm³ (2 dp).
Concordant titres (within 0.10 cm³) are essential for a reliable mean.
A clear, labelled table with units is part of the marks.

Common Mistakes

Recording burette readings to only 1 dp (e.g., 24.5 instead of 24.50). The mark scheme requires readings to the nearest 0.05 cm³.
Forgetting to record the mass of the empty container, making it impossible to calculate the mass of FA 1 used.
Using inconsistent decimal places in weighings (e.g., one to 1 dp, another to 2 dp).
Omitting units from table headings.
Using the word "difference" or "amount" instead of "titre" or "volume used" for the third column.
Performing only one accurate titration — you need at least two, ideally three, for concordance.
Not shaking the volumetric flask after making up to the mark.

Things to Be Careful About

Read the burette at eye level, from the bottom of the meniscus.
Remove the funnel from the burette before taking a reading.
The end-point is a permanent pale pink colour that persists for about 30 seconds.
The δ value (spread of titres per gram of FA 1) affects 3 marks. Keep your titres as close together as possible.
Weighings should be consistent in decimal places (at least 1 dp; 2 dp is better).

Techniques used

weigh by difference to find the mass of solid transferredprepare a solution in a volumetric flaskrecord burette readings to the nearest 0.05 cm³carry out a rough titration followed by accurate titrationsobtain concordant titres within 0.10 cm³

(b)

From your accurate titration results, obtain a suitable value for the volume of $\text{FA 3}$ to be used in your calculations.
Show clearly how you obtained this value.

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

DifficultyEasy

Worked solution

Working

Select the three concordant titres: 24.45, 24.50 and 24.55 cm³ (total spread = 0.10 cm³, within the 0.20 cm³ limit).

\text{Mean titre} = \frac{24.45 + 24.50 + 24.55}{3} = \frac{73.50}{3} = 24.50 \text{ cm}^3

Answer

25.0 cm³ of FA 4 required 24.50 cm³ of FA 3.

Final answer

24.50 cm³

Detailed explanation

Background Concept

The mean titre is the average of the concordant accurate titres. The mark scheme requires that the titres used for the mean have a total spread of not more than 0.20 cm³. This ensures the mean is reliable.

Understanding the Question

From your accurate titrations in (a), select the concordant ones, show how you selected them (e.g., by ticking them or showing the working), and calculate the mean. Quote the mean to 2 decimal places.

Approach

Identify the accurate titres that agree within a total spread of 0.20 cm³.
Calculate the mean: sum of selected titres ÷ number of titres.
Quote to 2 dp.

Step-by-Step Reasoning

From part (a), the accurate titres are 24.45, 24.50, and 24.55 cm³.

The spread is 24.55 − 24.45 = 0.10 cm³, which is within the 0.20 cm³ limit, so all three may be averaged.

Mean = (24.45 + 24.50 + 24.55) / 3 = 73.50 / 3 = 24.50 cm³

The mean is quoted to 2 dp: 24.50 cm³.

Key Takeaways

Only average titres that are concordant (spread ≤ 0.20 cm³).
Show your selection clearly (ticks or working).
Quote the mean to 2 dp.

Common Mistakes

Averaging titres that are not concordant (spread > 0.20 cm³).
Not showing which titres were selected.
Quoting the mean to 1 dp or to 3 dp.
Using the rough titre in the mean.

Things to Be Careful About

The mean must be based on at least two concordant titres.
Round the mean to the nearest 0.01 cm³.
If you have more than three titres, choose the set that best agrees.

Techniques used

select concordant titres within a 0.20 cm³ spreadcalculate the mean of the selected titresquote the mean to 2 decimal places

(c)

Calculations

(i)

Calculate the number of moles of potassium manganate(VII) present in the volume of $\text{FA 3}$ calculated in (b).

moles of $\text{KMnO}_4$ = .............................. mol

DifficultyEasy

Worked solution

Working

\text{moles of } \text{KMnO}_4 = \text{concentration} \times \text{volume} = 0.0200 \times \frac{24.50}{1000}

= 4.90 \times 10^{-4} \text{ mol}

Answer

moles of KMnO4 = 4.90 × 10⁻⁴ mol

Final answer

4.90 × 10⁻⁴ mol

Detailed explanation

Background Concept

The number of moles of a solute in a solution is given by:

moles = concentration (mol dm⁻³) × volume (dm³)

Since volumes in titration are measured in cm³, you must convert to dm³ by dividing by 1000.

Understanding the Question

You are asked to calculate the moles of potassium manganate(VII), KMnO4, in the mean titre volume (24.50 cm³) of FA 3, which has a concentration of 0.0200 mol dm⁻³.

Approach

Substitute the concentration and the mean titre (converted to dm³) into the mole formula.

Step-by-Step Reasoning

moles of KMnO4 = 0.0200 × (24.50 / 1000) = 0.0200 × 0.02450 = 4.90 × 10⁻⁴ mol

The answer is expressed to 3 significant figures (matching the precision of the data).

Key Takeaways

Always convert cm³ to dm³ before using concentration in mol dm⁻³.
The mean titre from (b) is the volume to use — never the rough titre.
Express the answer to 3 or 4 significant figures.

Common Mistakes

Forgetting to divide the volume by 1000.
Using the rough titre (25.10 cm³) instead of the mean (24.50 cm³).
Quoting too many or too few significant figures.

Things to Be Careful About

The mark scheme specifies 3 or 4 significant figures for this answer.
Use the candidate's own mean titre — there is no single correct value.

Techniques used

calculate moles from concentration and volumeconvert volume from cm³ to dm³

(ii)

$1\text{ mol}$ of $\text{KMnO}_4$ reacts with $5\text{ mol}$ of the hydrated salt, $\text{FA 1}$ .

Calculate the concentration of the hydrated salt, in $\text{mol dm}^{-3}$ , in $\text{FA 4}$ .

concentration of $\text{FA 4}$ = .............................. $\text{mol dm}^{-3}$

DifficultyMedium-Easy

Worked solution

Working

1 mol KMnO4 reacts with 5 mol FA 1.

\text{moles of FA 1 in 25.0 cm}^3 = 4.90 \times 10^{-4} \times 5 = 2.45 \times 10^{-3} \text{ mol}

\text{concentration of FA 4} = 2.45 \times 10^{-3} \times \frac{1000}{25} = 0.0980 \text{ mol dm}^{-3}

Answer

concentration of FA 4 = 0.0980 mol dm⁻³

Final answer

0.0980 mol dm⁻³

Detailed explanation

Background Concept

The stoichiometric ratio between KMnO4 and the hydrated salt FA 1 is given as 1 mol KMnO4 : 5 mol FA 1. This means 1 mole of manganate(VII) reacts with 5 moles of the salt. The manganate(VII) ion is a strong oxidising agent in acid; the salt contains a reducing ion (likely Fe²⁺) that is oxidised.

Understanding the Question

Given the moles of KMnO4 from (c)(i), calculate the concentration of the hydrated salt in FA 4 (the solution in the volumetric flask), in mol dm⁻³.

Approach

Use the 1:5 ratio to find the moles of FA 1 in the 25.0 cm³ sample that was titrated.
Convert this to a concentration in mol dm⁻³ by multiplying by 1000/25.

Step-by-Step Reasoning

moles of KMnO4 = 4.90 × 10⁻⁴ mol

moles of FA 1 in 25.0 cm³ = 4.90 × 10⁻⁴ × 5 = 2.45 × 10⁻³ mol

concentration of FA 4 = 2.45 × 10⁻³ × (1000 / 25) = 2.45 × 10⁻³ × 40 = 0.0980 mol dm⁻³

Key Takeaways

The stoichiometric ratio from the reaction equation determines the mole relationship.
Concentration in mol dm⁻³ = moles in the sample × (1000 / sample volume in cm³).

Common Mistakes

Using the wrong ratio (e.g., 1:1 instead of 1:5).
Forgetting to scale from the 25.0 cm³ sample to 1 dm³.
Confusing the moles of KMnO4 with the moles of FA 1.

Things to Be Careful About

The 25.0 cm³ is the pipette volume — it must appear in the scaling factor.
Carry the answer from (c)(i) forward correctly (error carried forward applies if (c)(i) is wrong but used correctly here).

Techniques used

apply the stoichiometric ratio 1:5convert moles in a sample to concentration in mol dm⁻³

(iii)

Use your answer to (c)(ii), and your data on page 2, to calculate an experimentally determined value for the relative formula mass of the hydrated salt, $\text{FA 1}$ .
Show your working.

$M_r$ of $\text{FA 1}$ = ..............................

DifficultyMedium-Easy

Worked solution

Working

The 250 cm³ volumetric flask contains all the FA 1 (mass 6.81 g).

\text{moles of FA 1 in the flask} = 0.0980 \times \frac{250}{1000} = 0.0245 \text{ mol}

M_r = \frac{\text{mass}}{\text{moles}} = \frac{6.81}{0.0245} = 278

Answer

Mr of FA 1 = 278 (to 3 s.f.)

Final answer

278

Detailed explanation

Background Concept

Relative formula mass (Mr) is the mass of one mole of a substance, in grams. It is calculated as:

Mr = mass (g) / moles (mol)

The mass of FA 1 used is known from part (a) (6.81 g). The total moles of FA 1 in the 250 cm³ volumetric flask can be found from the concentration of FA 4.

Understanding the Question

Use the mass of FA 1 used (from part (a)) and the concentration of FA 4 (from part (c)(ii)) to determine the experimental Mr of the hydrated salt.

Approach

The concentration of FA 4 is in mol dm⁻³. The flask contains 250 cm³ = 0.250 dm³. So:

moles of FA 1 in the flask = concentration × 0.250

Then Mr = mass of FA 1 used / moles in the flask.

Alternatively, the mark scheme gives: Mr = (mass of FA 1 used × 4) / concentration. The factor 4 comes from 1/0.250 = 4.

Step-by-Step Reasoning

moles of FA 1 in the 250 cm³ flask = 0.0980 × 0.250 = 0.0245 mol

Mr = 6.81 / 0.0245 = 278 (to 3 significant figures)

Check with the mark scheme formula: Mr = 6.81 × 4 / 0.0980 = 27.24 / 0.0980 = 278. ✓

Key Takeaways

Mr = mass / moles.
The concentration is per dm³, so multiply by the flask volume in dm³ to get the total moles.
The ×4 factor arises because the 250 cm³ flask is 10 times the 25.0 cm³ sample, and 1/0.250 = 4.

Common Mistakes

Using the moles in the 25.0 cm³ sample instead of the whole flask.
Forgetting to convert 250 cm³ to 0.250 dm³.
Using the mass of FA 1 incorrectly (e.g., the mass of the container + FA 1).

Things to Be Careful About

The answer should have appropriate significant figures (3 sf here, matching the data).
The experimental Mr will differ from the theoretical value due to experimental errors — that is expected.
Show your working clearly to earn the method mark even if the arithmetic is wrong.

Techniques used

calculate total moles in the 250 cm³ flask from concentrationcalculate relative formula mass from mass and moles

The rest of this paper

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Q3Qualitative Analysis · Presentation of Data and Observations · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation18M

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