Chemistry 5070/32 — October/November 2018

Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme

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Topics Observations and Measurements · Experimental Contexts · Analysis, Conclusions and Evaluation · Qualitative Analysis

Q118MObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationFree sample

You are to determine the amount of water of crystallisation present in a sample of hydrated sodium carbonate by titrating with hydrochloric acid.

\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2

P is a solution containing $12.27\text{ g / dm}^3$ of the sample of hydrated sodium carbonate.

Q is $0.110\text{ mol / dm}^3$ hydrochloric acid.

(a)

Put Q into the burette.

Pipette a $25.0\text{ cm}^3$ (or $20.0\text{ cm}^3$ ) portion of P into a flask and titrate with Q, using the indicator provided.

Record your results in the table, repeating the titration as many times as you consider necessary to achieve consistent results.

Results

Burette readings

titration number	1	2
final reading / $\text{cm}^3$
initial reading / $\text{cm}^3$
volume of Q used / $\text{cm}^3$
best titration results (✓)

Summary

Tick (✓) the best titration results.

Using the best titration results, the average volume of Q required was ______ $\text{cm}^3$ .

Volume of solution P used was ______ $\text{cm}^3$ .

12M

DifficultyMedium-Easy

Worked solution

Answer

Complete the results table as follows:

Record an initial and a final burette reading for each titration, each to 1 decimal place (e.g. 0.0, 25.3)
No reading above 50.0 cm³; no initial reading of 50.0 cm³
Calculate each titre: volume of Q used = final reading − initial reading
Tick (✓) the most consistent (concordant) titres
Calculate the average volume of Q from the ticked titres only
Record the volume of P used (25.0 cm³ or 20.0 cm³ as pipetted)

Final answer

See working — candidate-dependent table of burette readings, titres, ticked concordant results and average titre

Detailed explanation

Walkthrough

This part is the practical titration itself. You:

Rinse and fill the burette with Q (hydrochloric acid). Record the initial reading.
Use a pipette to transfer exactly 25.0 cm³ (or 20.0 cm³) of P (the sodium carbonate solution) into a conical flask.
Add a few drops of the indicator provided (likely methyl orange, since this is a strong acid–weak base titration — the end point is when the solution just turns from yellow to pink/orange).
Run Q from the burette into the flask, swirling continuously, until the indicator changes colour. Record the final reading.
Repeat until you have at least two concordant results (within 0.2 cm³ of each other).
Calculate each titre as final − initial.
Tick the best (most concordant) results.
Calculate the average of the ticked titres.

The mark scheme awards marks for:

Measurements (1): both readings present for each titration, recorded to 1 dp, no reading > 50.0, no initial reading of 50.0
Titres (1): all titres calculated correctly (no subtraction errors)
Accuracy (6): how close your titres are to the supervisor's value
Concordance (3): how close your ticked values are to each other
Average (1): correct average of selected titres

Key Takeaways

Burette readings are always recorded to 1 decimal place (e.g. 0.0, 25.3) — never 0 or 25.
Titre = final reading − initial reading.
Concordant results are those within 0.2 cm³ of each other.
The average is calculated from the ticked (concordant) results only.

Common Mistakes

Recording burette readings as whole numbers (e.g. 25 instead of 25.0) — loses the Measurements mark.
Subtraction errors when calculating titres.
Averaging all titres instead of only the ticked (concordant) ones.
Ticking results that are not actually concordant.

Things to Be Careful About

The burette scale reads from 0 at the top to 50.0 at the bottom.
Read the burette at eye level, from the bottom of the meniscus.
No initial reading should be 50.0 cm³ (the burette is full) — you need room to add acid.
The indicator colour change must be permanent (one drop past the end point).

Techniques used

record burette readings to one decimal placecalculate titre volumes from final minus initial readingsidentify concordant resultscalculate the average titre

(b)

Q is $0.110\text{ mol / dm}^3$ hydrochloric acid.

Calculate the number of moles of hydrochloric acid present in the average volume of Q.

moles of hydrochloric acid in the average volume of Q = ______

DifficultyMedium-Easy

Worked solution

Working

\text{moles of HCl} = \frac{\text{average volume of Q in cm}^3}{1000} \times 0.110

For example, with an average titre of 25.3 cm³:

\text{moles of HCl} = \frac{25.3}{1000} \times 0.110 = 0.00278\text{ mol}

Answer

moles of HCl = 0.00278 mol (for the example average titre of 25.3 cm³; substitute your own average titre from part (a))

Final answer

See working — depends on the candidate's average titre; example: 0.00278 mol

Detailed explanation

Walkthrough

To find the number of moles of HCl used in the titration, use the relationship:

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

The concentration of Q is given as 0.110 mol/dm³. The volume is the average titre you recorded in part (a), in cm³. Since concentration is in mol/dm³, you must first convert the volume from cm³ to dm³ by dividing by 1000.

For the mark scheme's example average titre of 25.3 cm³:

\text{moles of HCl} = \frac{25.3}{1000} \times 0.110 = 0.00278\text{ mol}

Key Takeaways

The mole is the amount of substance; moles = concentration × volume (with volume in dm³).
Always convert cm³ to dm³ by dividing by 1000 before using concentration in mol/dm³.

Common Mistakes

Forgetting to divide the volume by 1000 before multiplying by the concentration.
Using the wrong volume (e.g. the pipette volume of P instead of the average titre of Q).

Things to Be Careful About

The average titre is in cm³; the concentration is in mol/dm³ — the units must match.
The answer should be given to 3 significant figures (e.g. 0.00278, not 0.0028).

Techniques used

convert volume from cm³ to dm³calculate moles from concentration and volume

(c)

Using your answer from (b) and the equation shown, calculate the number of moles of sodium carbonate in the volume of P used in the titration.

\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2

moles of sodium carbonate in the volume of P used = ______

DifficultyMedium-Easy

Worked solution

Working

From the equation:

\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2

1 mol Na₂CO₃ reacts with 2 mol HCl, so:

\text{moles of Na}_2\text{CO}_3 = \frac{\text{moles of HCl}}{2}

Using the example answer from (b), 0.00278 mol:

\text{moles of Na}_2\text{CO}_3 = \frac{0.00278}{2} = 0.00139\text{ mol}

Answer

moles of sodium carbonate = 0.00139 mol (using the example answer from (b); substitute your own answer from (b))

Final answer

See working — depends on the answer to (b); example: 0.00139 mol

Detailed explanation

Walkthrough

The balanced equation shows that 1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid:

\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2

The coefficients tell us the mole ratio: for every 1 mol of Na₂CO₃, 2 mol of HCl are needed. So the moles of sodium carbonate in the pipette volume of P is exactly half the moles of HCl you calculated in part (b).

Key Takeaways

The coefficients in a balanced equation give the mole ratio of reactants.
Here the ratio Na₂CO₃ : HCl is 1 : 2, so moles of Na₂CO₃ = moles of HCl ÷ 2.

Common Mistakes

Using the ratio the wrong way round (multiplying by 2 instead of dividing by 2).
Forgetting that the answer must be carried forward from (b) — if (b) is wrong, this will be wrong too (error carried forward).

Things to Be Careful About

The equation shows 2HCl, so the mole ratio is 1:2 — check the coefficients carefully.
The answer should be given to 3 significant figures.

Techniques used

apply the 1:2 mole ratio from the balanced equation

(d)

Using your answer from (c), calculate the concentration, in $\text{mol / dm}^3$ , of sodium carbonate in P.

concentration of sodium carbonate in P = ______ $\text{mol / dm}^3$

DifficultyMedium-Easy

Worked solution

Working

\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3}

The volume of P used is 25.0 cm³ = 0.0250 dm³ (or 20.0 cm³ = 0.0200 dm³ if that pipette was used).

Using the example answer from (c), 0.00139 mol, and a 25.0 cm³ pipette:

\text{concentration} = \frac{0.00139}{0.0250} = 0.0557\text{ mol / dm}^3

Answer

concentration of sodium carbonate in P = 0.0557 mol/dm³ (using the example answer from (c) and a 25.0 cm³ pipette; substitute your own values)

Final answer

See working — depends on the answers to (c) and the pipette volume; example: 0.0557 mol/dm³

Detailed explanation

Walkthrough

Concentration is the amount of solute (in moles) dissolved in 1 dm³ of solution:

\text{concentration} = \frac{\text{moles of solute}}{\text{volume of solution in dm}^3}

The moles of sodium carbonate (from part (c)) are in the pipette volume of P — either 25.0 cm³ or 20.0 cm³. Convert this volume to dm³ by dividing by 1000, then divide the moles by this volume.

For the example values (0.00139 mol in 25.0 cm³ = 0.0250 dm³):

\text{concentration} = \frac{0.00139}{0.0250} = 0.0557\text{ mol / dm}^3

Key Takeaways

Concentration in mol/dm³ = moles ÷ volume in dm³.
The pipette volume must be converted from cm³ to dm³ before dividing.

Common Mistakes

Forgetting to convert the pipette volume from cm³ to dm³.
Using the average titre volume instead of the pipette volume of P.

Things to Be Careful About

Use the correct pipette volume — check whether the question says 25.0 cm³ or 20.0 cm³.
The answer should be given to 3 significant figures.

Techniques used

calculate concentration from moles and volume in dm³

(e)

Using your answer from (d), calculate the mass, in $\text{g}$ , of sodium carbonate in $1.00\text{ dm}^3$ of P.
[ $M_r$ : $\text{Na}_2\text{CO}_3$ , 106]

mass of sodium carbonate in $1.00\text{ dm}^3$ of P = ______ $\text{g}$

DifficultyMedium-Easy

Worked solution

Working

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

Using the example concentration from (d), 0.0557 mol/dm³, in 1.00 dm³ there are 0.0557 mol of Na₂CO₃:

\text{mass} = 0.0557 \times 106 = 5.90\text{ g}

Answer

mass of sodium carbonate in 1.00 dm³ of P = 5.90 g (using the example answer from (d); substitute your own answer from (d))

Final answer

See working — depends on the answer to (d); example: 5.90 g

Detailed explanation

Walkthrough

The concentration from part (d) tells us how many moles of sodium carbonate are in 1 dm³ of solution. To find the mass, multiply the number of moles by the relative molecular mass, M_r:

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

The question gives M_r of Na₂CO₃ = 106. So for the example concentration of 0.0557 mol/dm³:

\text{mass} = 0.0557 \times 106 = 5.90\text{ g}

This is the mass of anhydrous sodium carbonate in 1.00 dm³ of P.

Key Takeaways

mass = moles × M_r.
The concentration in mol/dm³ is numerically equal to the number of moles in 1 dm³.

Common Mistakes

Using the wrong M_r value.
Forgetting to multiply by the volume (here it is exactly 1.00 dm³, so no extra step).

Things to Be Careful About

The answer should be given to 3 significant figures (5.90 g).

Techniques used

convert moles to mass using relative molecular mass

(f)

Using your answer from (e), calculate the mass of water of crystallisation present in the $12.27\text{ g}$ of the hydrated sodium carbonate used to make P.

mass of water of crystallisation present = ______ $\text{g}$

DifficultyMedium-Easy

Worked solution

Working

The 12.27 g of hydrated sodium carbonate is made up of anhydrous Na₂CO₃ plus water of crystallisation.

\text{mass of water} = 12.27 - \text{mass of Na}_2\text{CO}_3

Using the example answer from (e), 5.90 g:

\text{mass of water} = 12.27 - 5.90 = 6.37\text{ g}

Answer

mass of water of crystallisation = 6.37 g (using the example answer from (e); substitute your own answer from (e))

Final answer

See working — depends on the answer to (e); example: 6.37 g

Detailed explanation

Walkthrough

The sample of hydrated sodium carbonate (12.27 g) is the total mass of the anhydrous salt plus the water of crystallisation. The mass of anhydrous Na₂CO₃ in 1.00 dm³ of P (from part (e)) is the same as the mass of anhydrous salt that was originally dissolved to make that 1 dm³ of solution.

So:

\text{mass of water of crystallisation} = \text{total mass of hydrated salt} - \text{mass of anhydrous salt}

For the example values:

\text{mass of water} = 12.27 - 5.90 = 6.37\text{ g}

Key Takeaways

A hydrated salt = anhydrous salt + water of crystallisation.
The mass of water is found by subtraction.

Common Mistakes

Adding the masses instead of subtracting.
Using the wrong total mass (12.27 g is the mass of the hydrated salt used to make 1 dm³ of P).

Things to Be Careful About

The answer should be given to 3 significant figures (6.37 g).

Techniques used

subtract the mass of anhydrous carbonate from the hydrated salt

(g)

Calculate the percentage by mass of water of crystallisation present in the hydrated sodium carbonate used to make P.

percentage by mass of water of crystallisation = ______

DifficultyMedium-Easy

Worked solution

Working

\text{percentage by mass of water} = \frac{\text{mass of water}}{\text{total mass of hydrated salt}} \times 100

Using the example answer from (f), 6.37 g:

\text{percentage} = \frac{6.37}{12.27} \times 100 = 51.9\%

Answer

percentage by mass of water of crystallisation = 51.9% (using the example answer from (f); substitute your own answer from (f))

Final answer

See working — depends on the answer to (f); example: 51.9%

Detailed explanation

Walkthrough

Percentage by mass is the mass of the component divided by the total mass, multiplied by 100:

\text{percentage} = \frac{\text{mass of water of crystallisation}}{\text{total mass of hydrated salt}} \times 100

The total mass is 12.27 g (the hydrated salt used to make P). For the example mass of water of 6.37 g:

\text{percentage} = \frac{6.37}{12.27} \times 100 = 51.9\%

Key Takeaways

Percentage by mass = (mass of component ÷ total mass) × 100.
The total mass here is the mass of the hydrated salt, 12.27 g.

Common Mistakes

Using the mass of anhydrous salt instead of the total mass in the denominator.
Forgetting to multiply by 100.

Things to Be Careful About

The answer should be given to 3 significant figures (51.9%).
The % sign must be included in the answer.

Techniques used

calculate percentage by mass

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