Chemistry 9701/34 — October/November 2018

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

Quantitative Analysis

Read through the whole method before starting any practical work. Where appropriate, prepare a table for your results in the space provided.

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

In this experiment you will determine the percentage by mass of an impure sample of sodium hydrogencarbonate, $\text{NaHCO}_3$ .

You will do this by titration with hydrochloric acid, $\text{HCl}$ . The impurity in the sample is $\mathbf{X}$ . $\mathbf{X}$ is a sodium compound which does not react with $\text{HCl}$ .

$\mathbf{FB\ 1}$ is a mixture containing sodium hydrogencarbonate and $\mathbf{X}$ .
You are supplied with approximately $6.5\text{ g}$ of $\mathbf{FB\ 1}$ . You will also use $\mathbf{FB\ 1}$ in Question 2.
$\mathbf{FB\ 2}$ is $0.105\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .
methyl orange indicator

(a)

Method

Preparing a solution of FB 1

Weigh the $100\text{ cm}^3$ beaker. Record the mass.
Add between $2.8\text{ g}$ and $3.0\text{ g}$ of $\mathbf{FB\ 1}$ to the beaker.
Reweigh the beaker with $\mathbf{FB\ 1}$ . Record the mass.
Calculate and record the mass of $\mathbf{FB\ 1}$ used.
Add approximately $50\text{ cm}^3$ of distilled water to $\mathbf{FB\ 1}$ in the beaker.
Stir the mixture with a glass rod until all the $\mathbf{FB\ 1}$ 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.
Add distilled water to the volumetric flask up to the mark.
Shake the flask thoroughly.
This solution of impure sodium hydrogencarbonate is $\mathbf{FB\ 3}$ . Label the flask $\mathbf{FB\ 3}$ .

Titration of FB 3

Fill the burette with $\mathbf{FB\ 2}$ .
Pipette $25.0\text{ cm}^3$ of $\mathbf{FB\ 3}$ into a conical flask.
Add approximately 10 drops of methyl orange indicator.
Carry out a rough titration.
Record your burette readings and the rough titre 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 $\mathbf{FB\ 2}$ added in each accurate titration.

DifficultyMedium

Worked solution

Answer

Record the two weighings and subtract to find the mass of FB 1:

mass of beaker = 45.20 g
mass of beaker + FB 1 = 48.10 g
mass of FB 1 = 48.10 − 45.20 = 2.90 g (between 2.80 and 3.00 g)

Rough titration: initial reading 0.00 cm³, final reading 27.35 cm³, rough titre = 27.35 cm³.

Accurate titrations (all burette readings to the nearest 0.05 cm³):

	1	2	3
final / cm³	26.45	52.80	26.35
initial / cm³	0.00	26.45	0.00
titre / cm³	26.45	26.35	26.35

The final accurate titre (26.35 cm³) is within 0.10 cm³ of another accurate titre (26.45 cm³).

Final answer

Mass of FB 1 = 2.90 g; concordant accurate titres within 0.10 cm³ (e.g. 26.45 and 26.35 cm³) — candidate-dependent readings.

Detailed explanation

Background Concept

This is a quantitative titration experiment. The aim is to determine the percentage by mass of sodium hydrogencarbonate in an impure sample. A known mass of the impure solid (FB 1) is dissolved and made up to exactly 250 cm³ in a volumetric flask; a 25.0 cm³ aliquot is then titrated against standard hydrochloric acid (FB 2, 0.105 mol dm⁻³) using methyl orange indicator. The reaction is a 1:1 acid–base neutralisation:

\text{NaHCO}_3(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

Because the impurity X is a sodium compound that does not react with HCl, the volume of HCl consumed is directly proportional to the amount of NaHCO₃ present. Methyl orange changes from yellow (alkaline) to pink/red (acid) at the end point.

Good titration technique demands: a rough titration to locate the end point, then accurate titrations in which the acid is added dropwise near the end point. Two titres that agree within 0.10 cm³ are called concordant and are the sign of a reliable result. Burette readings are recorded to the nearest 0.05 cm³.

Understanding the Question

Part (a) is the practical part: prepare the solution of FB 1 and carry out the titrations. The marks are not for a single numerical answer — each candidate's readings differ — but for correct technique and precise, unambiguous recording. The examiner checks: (I) the two weighings and a mass of FB 1 between 2.80 and 3.00 g; (II) that the rough and accurate readings are all recorded; (III) that the accurate titration table has correct headings and units; (IV) that all burette readings are to the nearest 0.05 cm³; and (V) that the final accurate titre is within 0.10 cm³ of another accurate titre.

Approach

Weigh the empty beaker, add 2.8–3.0 g of FB 1, reweigh, and subtract to obtain the mass of FB 1.
Dissolve in about 50 cm³ of distilled water, transfer quantitatively to the 250 cm³ volumetric flask, wash the beaker into the flask, make up to the mark, and shake.
Fill the burette with FB 2; pipette 25.0 cm³ of FB 3 into a conical flask; add about 10 drops of methyl orange.
Run a rough titration, then accurate titrations until two concordant results are obtained.
Record every reading to the nearest 0.05 cm³ in a table with headings and units.

Step-by-Step Reasoning

Weighing by difference: record the mass of the empty beaker (45.20 g), then the mass of beaker + FB 1 (48.10 g). The difference, 2.90 g, is the mass of FB 1. This must lie between 2.80 and 3.00 g, a range chosen so the titre is a sensible volume.
Preparing FB 3: the solid is dissolved and transferred quantitatively — the beaker is washed with distilled water and the washings added — so that no FB 1 is lost. The solution is made up to exactly 250 cm³ so that each 25.0 cm³ aliquot is exactly one tenth of the whole.
Rough titration: add HCl quickly until the indicator just changes colour, giving an approximate titre (27.35 cm³).
Accurate titrations: add HCl dropwise near the end point. Record initial and final burette readings to 0.05 cm³. The titre is final − initial. For example, titration 1: 26.45 − 0.00 = 26.45 cm³; titration 2: 52.80 − 26.45 = 26.35 cm³; titration 3: 26.35 − 0.00 = 26.35 cm³.
Concordance: the final accurate titre (26.35 cm³) is within 0.10 cm³ of another accurate titre (26.45 cm³), satisfying the concordance requirement.
Table headings: each column must carry a quantity and a unit — "final / cm³", "initial / cm³", "titre / cm³" (or "volume of FB 2 used / cm³"). The word "titre" or "volume of FB 2 used/added" is required; "difference" or "total volume" is not accepted.

Key Takeaways

Weighing by difference is the standard way to obtain a precise mass of a solid.
A volumetric flask gives an exact total volume; a pipette delivers an exact aliquot.
Concordant titres (within 0.10 cm³) indicate a reliable end point.
Burette readings are recorded to 0.05 cm³; the titre itself need not be.
A results table must have clear headings with units.

Common Mistakes

Using 50.00 cm³ as the initial burette reading — the mark is not awarded.
Recording readings to only 1 decimal place (e.g. 26.4 instead of 26.40 or 26.45) — must be to 0.05 cm³.
Not recording both initial and final readings for each titration.
Using the rough titre as if it were an accurate titre.
Failing to wash the beaker into the volumetric flask, losing some FB 3.
Writing "difference" or "total volume" as the titre heading — these are rejected.

Things to Be Careful About

Every burette reading, including 0.00 cm³, must be to the nearest 0.05 cm³.
No burette reading may exceed 50.00 cm³, and not more than one final reading may be 50.00.
The mass of FB 1 must be within 2.80–3.00 g.
The titre heading must be "titre / cm³" or "volume of FB 2 used / cm³".
The unit cm³ must appear with each heading or with each recorded volume.

Techniques used

weigh by differenceprepare a solution in a volumetric flaskcarry out a rough titrationperform accurate titrations to concordancerecord burette readings to 0.05 cm³construct a titration results table with headings and units

(b)

From your accurate titration results, obtain a suitable value for the volume of $\mathbf{FB\ 2}$ to be used in your calculations.
Show clearly how you obtained this value.

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

DifficultyMedium-Easy

Worked solution

Working

Mean titre = (26.45 + 26.35) ÷ 2 = 26.40 cm³

The two titres used are within 0.20 cm³ of each other.

Answer

25.0 cm³ of FB 3 required 26.40 cm³ of FB 2.

Final answer

26.40 cm³

Detailed explanation

Background Concept

The mean titre is the average of the concordant accurate titrations and is the single value carried into all subsequent calculations. The mark scheme requires the mean to be calculated from two or more titres whose total spread is no more than 0.20 cm³, and the mean must be quoted to 2 decimal places.

Understanding the Question

Part (b) asks for one suitable value of the volume of FB 2 to use in the calculations. This is the mean of the accurate titrations that agree closely. The working must be shown, or the selected readings ticked, so the examiner can see which titres were averaged.

Approach

Identify two (or more) accurate titres that agree within 0.20 cm³.
Add them and divide by the number of titres.
Round the mean to the nearest 0.01 cm³ (2 dp).

Step-by-Step Reasoning

From the example data, the accurate titres are 26.45, 26.35 and 26.35 cm³. The two closest are 26.45 and 26.35 cm³, a spread of 0.10 cm³, well within the 0.20 cm³ limit.

\text{mean titre} = \frac{26.45 + 26.35}{2} = 26.40 \text{ cm}^3

The mean is already at 2 dp. If the mean had been 26.667 cm³, it would be rounded to 26.67 cm³. A special case: a mean of 26.325 cm³ may be quoted to 3 dp (26.325) because 0.025 is a legitimate 3-dp value.

Key Takeaways

The mean titre is the average of concordant accurate titres.
The mean must be quoted to 2 decimal places.
The rough titre must never be included in the mean.

Common Mistakes

Including the rough titre in the average.
Averaging titres that are not concordant (spread greater than 0.20 cm³).
Quoting the mean to 3 dp when not required, or to 1 dp.

Things to Be Careful About

The mean must be rounded to the nearest 0.01 cm³.
Show the working or tick the selected readings so the examiner can follow your choice.

Techniques used

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

(c)

Calculations

(i)

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

DifficultyEasy

Worked solution

Answer

All answers to parts (ii)–(v) are quoted to 3 significant figures:

moles of HCl = 2.77 × 10⁻³ mol
moles of NaHCO₃ = 2.77 × 10⁻³ mol
moles of NaHCO₃ in FB 1 = 2.77 × 10⁻² mol
percentage by mass = 80.3%

Final answer

3 significant figures (e.g. 2.77 × 10⁻³ mol, 80.3%)

Detailed explanation

Background Concept

Significant figures reflect the precision of a measurement. In a calculation, the final answer should not claim more precision than the least precise input. Here, the concentration of HCl (0.105 mol dm⁻³) has 3 significant figures and the titre (e.g. 26.40 cm³) has 4, so answers to 3 or 4 significant figures are appropriate.

Understanding the Question

Part (c)(i) is an instruction, not a calculation: it tells you to give the answers to (ii)–(v) to the appropriate number of significant figures. The mark is awarded if at least three of the four answers are quoted to 3 or 4 significant figures.

Approach

After each calculation, round the result to 3 significant figures (or keep 4 if preferred). Check every answer before writing it down.

Step-by-Step Reasoning

Moles of HCl: 0.105 × 26.40/1000 = 0.002772 mol → 2.77 × 10⁻³ mol (3 sig fig).
Moles of NaHCO₃: same value, 2.77 × 10⁻³ mol.
Moles in FB 1: 10 × 2.77 × 10⁻³ = 2.77 × 10⁻² mol.
Percentage by mass: 80.3% (3 sig fig).

All four are to 3 significant figures, so the mark is secured.

Key Takeaways

Answers should be to 3 or 4 significant figures.
The mark requires at least three of the four answers to be correctly rounded.

Common Mistakes

Quoting too many significant figures (e.g. 0.0027720 mol).
Quoting too few (e.g. 0.0028 mol or 80%).
Rounding intermediate values before the final answer, introducing error.

Things to Be Careful About

The percentage should be 80.3%, not 80% or 80.29%.
Keep intermediate values in your working; round only the final answers.

Techniques used

round calculated values to 3 or 4 significant figures

(ii)

Calculate the number of moles of hydrochloric acid, $\text{HCl}$ , in the volume of $\mathbf{FB\ 2}$ calculated in (b).

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

DifficultyMedium-Easy

Worked solution

Working

\text{moles of HCl} = 0.105 \times \frac{26.40}{1000} = 2.772 \times 10^{-3} \text{ mol}

Answer

moles of HCl = 2.77 × 10⁻³ mol

Final answer

2.77 × 10⁻³ mol

Detailed explanation

Background Concept

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

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

The concentration of FB 2 is 0.105 mol dm⁻³, meaning 0.105 mol of HCl in every 1 dm³. The titre is measured in cm³, so it must be converted to dm³ by dividing by 1000.

Understanding the Question

Part (c)(ii) asks for the number of 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 NaHCO₃ → mass → percentage.

Approach

Take the mean titre from (b), convert it to dm³, and multiply by the concentration 0.105 mol dm⁻³.

Step-by-Step Reasoning

Mean titre = 26.40 cm³ = 26.40/1000 dm³ = 0.02640 dm³.

\text{moles of HCl} = 0.105 \times 0.02640 = 2.772 \times 10^{-3} \text{ mol}

Rounded to 3 significant figures: 2.77 × 10⁻³ mol.

Key Takeaways

moles = concentration × volume, with volume in dm³.
Convert cm³ to dm³ by dividing by 1000.

Common Mistakes

Forgetting to divide the titre by 1000, giving an answer 1000 times too large.
Using the rough titre instead of the mean titre from (b).
Quoting too many significant figures.

Things to Be Careful About

Use the mean titre from part (b), not the rough titre.
Quote the answer to 3 or 4 significant figures.

Techniques used

convert cm³ to dm³calculate moles from concentration and volume

(iii)

Complete and balance the equation for the reaction of sodium hydrogencarbonate with hydrochloric acid. Include state symbols.

\text{....NaHCO}_3\text{......} + \text{....HCl......} \rightarrow \text{....NaCl......} + \text{....CO}_2\text{......} + \text{.................}

Deduce the number of moles of sodium hydrogencarbonate that reacted with the number of moles of $\text{HCl}$ calculated in (ii).

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

DifficultyMedium-Easy

Worked solution

Answer

\text{NaHCO}_3(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

The reaction is 1:1, so:

\text{moles of NaHCO}_3 = \text{moles of HCl} = 2.77 \times 10^{-3} \text{ mol}

Final answer

NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l); moles of NaHCO3 = 2.77 × 10⁻³ mol

Detailed explanation

Background Concept

Sodium hydrogencarbonate reacts with hydrochloric acid in a 1:1 mole ratio:

\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}

The hydrogencarbonate ion accepts a proton from the acid, producing carbonic acid, which decomposes to CO₂ and H₂O. Because the stoichiometry is 1:1, the number of moles of NaHCO₃ that reacted equals the number of moles of HCl added.

Understanding the Question

Part (c)(iii) has two tasks: (1) complete and balance the equation with state symbols, and (2) deduce the moles of NaHCO₃ that reacted with the moles of HCl from (ii).

Approach

Balance the equation atom by atom, add the correct state symbols, then use the 1:1 ratio to state that moles of NaHCO₃ = moles of HCl.

Step-by-Step Reasoning

The reactants are NaHCO₃ and HCl; the products are NaCl, CO₂ and H₂O.
Counting atoms: 1 Na, 1 H, 1 C, 3 O on the left (NaHCO₃) plus 1 H, 1 Cl (HCl); on the right, NaCl has 1 Na and 1 Cl, CO₂ has 1 C and 2 O, H₂O has 2 H and 1 O. Total: 1 Na, 2 H, 1 C, 3 O, 1 Cl on each side — balanced with all coefficients 1.
State symbols: NaHCO₃ is dissolved in water, so (aq); HCl is the aqueous acid, (aq); NaCl is soluble, (aq); CO₂ is given off as a gas, (g); H₂O is liquid, (l).
Since the ratio is 1:1, moles of NaHCO₃ = moles of HCl = 2.77 × 10⁻³ mol.

Key Takeaways

The NaHCO₃ + HCl reaction is 1:1.
State symbols are part of the mark.

Common Mistakes

Missing state symbols.
Writing an unbalanced equation.
Using a 2:1 ratio.

Things to Be Careful About

CO₂ must be (g), H₂O must be (l).
The answer to the moles part must equal the answer to (ii) exactly.

Techniques used

balance a neutralisation equationinclude state symbolsapply the 1:1 stoichiometric ratio

(iv)

Use your answer to (iii) to calculate the number of moles of sodium hydrogencarbonate in the $\mathbf{FB\ 1}$ that you weighed out.

\text{moles of NaHCO}_3\text{ in FB 1 used} = \text{.............................. mol}

DifficultyMedium-Easy

Worked solution

Working

The 25.0 cm³ aliquot is one tenth of the 250 cm³ solution, so:

\text{moles of NaHCO}_3\text{ in FB 1} = 10 \times 2.77 \times 10^{-3} = 2.77 \times 10^{-2} \text{ mol}

Answer

moles of NaHCO₃ in FB 1 used = 2.77 × 10⁻² mol

Final answer

2.77 × 10⁻² mol

Detailed explanation

Background Concept

The titration was performed on a 25.0 cm³ aliquot of the 250 cm³ solution. The aliquot therefore contains exactly one tenth of the NaHCO₃ present in the whole of FB 3, which itself contains all of the weighed-out FB 1. To find the moles in the whole sample, multiply the moles in the aliquot by the dilution factor 250/25 = 10.

Understanding the Question

Part (c)(iv) asks for the number of moles of NaHCO₃ in the FB 1 that was weighed out, i.e. in the entire 250 cm³ solution, not just the 25.0 cm³ aliquot titrated.

Approach

Multiply the answer to (iii) by 10, because 25.0 cm³ is one tenth of 250 cm³.

Step-by-Step Reasoning

The factor is:

\frac{250 \text{ cm}^3}{25.0 \text{ cm}^3} = 10

\text{moles in FB 1} = 10 \times 2.77 \times 10^{-3} = 2.77 \times 10^{-2} \text{ mol}

Key Takeaways

The volumetric flask factor is 250/25 = 10.
Always scale the aliquot result up to the total solution volume.

Common Mistakes

Forgetting the × 10 factor.
Using the wrong factor (e.g. 25/250 = 0.1).
Confusing the aliquot volume with the total volume.

Things to Be Careful About

The factor is 10 because 25.0 cm³ is one tenth of 250 cm³.
Quote the answer to 3 or 4 significant figures.

Techniques used

scale moles from aliquot to total solutionapply the volumetric flask dilution factor

(v)

Calculate the percentage by mass of $\text{NaHCO}_3$ in $\mathbf{FB\ 1}$ .

\text{percentage by mass of NaHCO}_3\text{ in FB 1} = \text{.............................. \%}

DifficultyMedium

Worked solution

Working

$M_r(\text{NaHCO}_3) = 23 + 1 + 12 + (3 \times 16) = 84$

\text{mass of NaHCO}_3 = 2.77 \times 10^{-2} \times 84 = 2.33 \text{ g}

\text{percentage by mass} = \frac{2.33}{2.90} \times 100 = 80.3\%

Answer

percentage by mass of NaHCO₃ in FB 1 = 80.3%

Final answer

80.3%

Detailed explanation

Background Concept

The percentage by mass of a component in a mixture is:

\text{percentage by mass} = \frac{\text{mass of component}}{\text{mass of sample}} \times 100

The mass of NaHCO₃ is found from the number of moles (from part (iv)) and its molar mass:

\text{mass} = \text{moles} \times M_r

The molar mass of NaHCO₃ is the sum of the relative atomic masses: Na (23) + H (1) + C (12) + 3 × O (16) = 84 g mol⁻¹.

Understanding the Question

Part (c)(v) is the final step: convert the moles of NaHCO₃ in the whole FB 1 sample into a mass, then express that mass as a percentage of the mass of FB 1 weighed out in part (a).

Approach

Calculate $M_r$ of NaHCO₃.
Multiply moles (from (iv)) by $M_r$ to get the mass of NaHCO₃.
Divide by the mass of FB 1 (from (a)) and multiply by 100.

Step-by-Step Reasoning

$M_r(\text{NaHCO}_3) = 23 + 1 + 12 + 48 = 84$ .
Mass of NaHCO₃ = 2.77 × 10⁻² mol × 84 g mol⁻¹ = 2.33 g.
Percentage by mass = (2.33 g / 2.90 g) × 100 = 80.3%.

The answer is quoted to 3 significant figures, consistent with the precision of the data.

Key Takeaways

$M_r$ of NaHCO₃ = 84.
Percentage by mass = (mass of component / mass of sample) × 100.
The mass of sample is the mass of FB 1 from part (a).

Common Mistakes

Using the wrong $M_r$ (e.g. forgetting the water of crystallisation or miscounting oxygen atoms).
Forgetting to multiply by 100.
Using the mass of the aliquot instead of the mass of the whole FB 1 sample.
Using the moles from (iii) instead of the scaled moles from (iv).

Things to Be Careful About

Use the mass of FB 1 recorded in part (a), not the mass of the beaker or the beaker + FB 1.
Quote the percentage to 3 significant figures (80.3%, not 80% or 80.29%).
The 84 may be written as the sum of the $A_r$ values to show the working.

Techniques used

calculate molar mass of NaHCO3convert moles to masscalculate percentage by mass

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