Chemistry 5070/31 — October/November 2020

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

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

Q118MExperimental ContextsObservations and MeasurementsAnalysis, Conclusions and EvaluationFree sample

Citric acid is a carboxylic acid found in lemon juice.

The equation for the reaction between citric acid, $\text{H}_3\text{C}_6\text{H}_5\text{O}_7$ , and potassium hydroxide, $\text{KOH}$ , is shown.

3\text{KOH} + \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \rightarrow \text{K}_3\text{C}_6\text{H}_5\text{O}_7 + 3\text{H}_2\text{O}

The mass of citric acid dissolved in $500\text{ cm}^3$ of an aqueous solution can be determined by titration with $\text{KOH(aq)}$ .

Thymolphthalein is used to determine the end-point of the titration.

$\mathbf{P}$ is $0.100\text{ mol / dm}^3\text{ KOH(aq)}$ .

$\mathbf{Q}$ is aqueous citric acid.

(a)

Put $\mathbf{P}$ into the burette.

Pipette $25.0\text{ cm}^3$ of $\mathbf{Q}$ into a flask and titrate with $\mathbf{P}$ using three drops of thymolphthalein as the indicator.

The end-point is the first appearance of a blue colour that remains for 30 seconds.

Record your results in the table.

Repeat the titration as many times as necessary to achieve consistent results.

Results

Burette readings

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

Summary

Tick (✓) the best titration results in the table.

Using the best titration results the average volume of $\mathbf{P}$ required is ______ $\text{cm}^3$ .

12M

DifficultyMedium-Easy

Worked solution

Answer

Complete a results table showing, for each titration, a final reading and an initial reading, both to 1 decimal place (e.g. $25.4\text{ cm}^3$ , $0.0\text{ cm}^3$ ).

For each titration calculate:

$\text{volume of } \mathbf{P} \text{ used} = \text{final reading} - \text{initial reading}$

Record this titre to 1 decimal place.

Repeat the titration until at least two concordant titres are obtained. Tick the two best titres and find the average:

\text{average titre } = \frac{\text{sum of the ticked titres}}{\text{number of ticked titres}}

Write the average in the summary line.

Final answer

Completed table with readings and titres to 1 decimal place, with the average of the best concordant titres.

Detailed explanation

Walkthrough

This part is not marked for one single number; it is marked by the quality and completeness of the titration results.

Set up and perform the titration.
Put $0.100\text{ mol / dm}^3$ KOH into the burette. Pipette $25.0\text{ cm}^3$ of citric acid solution into a conical flask. Add three drops of thymolphthalein. Titrate until a blue colour first appears and remains for 30 seconds.
Record initial and final burette readings.
Both readings must be written to 1 decimal place. This includes recording $0.0\text{ cm}^3$ if needed.
Calculate each titre.
The titre is final reading minus initial reading. Subtraction errors lose the mark.
Repeat for consistency.
At least two titres need to be close together. The usual standard is that the ticked titres are all within $0.2\text{ cm}^3$ of one another.
Average only the selected best results.
The mean is taken only from the ticked titres, not from all attempted titrations.

The mark scheme also compares the two best titres with the supervisor's value. Points are awarded depending on how close they are to the supervisor's value, and separate points are given for concordance and for an average.

Key Takeaways

In a titration, every burette reading must be recorded with the same precision, usually 1 decimal place.
The titre is always final reading minus initial reading.
You repeat titrations until concordant; only then can you call the mean value reliable.
The average must be calculated from the ticked best results only.

Common Mistakes

Forgetting to record the initial reading.
Giving readings without a decimal place, e.g. writing $23$ instead of $23.0$ .
Writing an initial reading as $50.0$ , which is not realistic for a full burette.
Averaging all titrations, including the outlier, instead of ticked best results.
Subtraction mistakes in titre calculations.

Things to Be Careful About

Read the bottom of the meniscus at eye level.
If the first titration is only a trial, it may be recorded but not used in the average.
End-point is when blue remains for 30 seconds, not when it appears and then disappears.
The average should normally be given to at least 1 decimal place, often to 2.

Techniques used

fill and read a burette correctlyrecord burette readings to 1 decimal placecalculate each titre as final reading minus initial readingrepeat the titration until concordant results are obtainedchoose the best titres and calculate their average

(b)

$\mathbf{P}$ is $0.100\text{ mol / dm}^3\text{ KOH(aq)}$ .

Use your results from (a) to calculate the number of moles of $\text{KOH}$ in the average volume of $\mathbf{P}$ used.

Give your answer to three significant figures.

number of moles of $\text{KOH}$ = ______

DifficultyMedium-Easy

Worked solution

Working

Let $V$ be the average titre in $\text{cm}^3$ from part (a).

$n_{\text{KOH}} = \frac{V}{1000} \times 0.100$

Substitute the candidate's own titre and give the answer to 3 significant figures.

Answer

Use your average titre value:

number of moles of $\text{KOH}$ = $\dfrac{V}{1000} \times 0.100$ to 3 significant figures.

Final answer

average titre (cm3) / 1000 x 0.100, to 3 s.f.

Detailed explanation

Walkthrough

The standard relationship used here is:

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

Your average titre is measured in $\text{cm}^3$ , but concentration is in $\text{mol / dm}^3$ . So the first step is always:

$\text{volume in dm}^3 = \frac{\text{volume in cm}^3}{1000}$

Then multiply by the KOH concentration, $0.100\text{ mol / dm}^3$ .

For example, if your average titer was $25.0\text{ cm}^3$ , the moles would be:

$\frac{25.0}{1000} \times 0.100 = 0.00250\text{ mol}$

You must present your own final value correct to 3 significant figures.

Key Takeaways

moles = concentration x volume is the central equation for titrations.
Always change cm3 to dm3 before using this equation.
The concentration of KOH is in mol / dm3, so the volume also needs dm3.
3 significant figures means three non-zero digits in the final answer.

Common Mistakes

Forgetting to divide by 1000.
Writing concentration as 0.100 mol / cm3.
Rounding too early or giving fewer than 3 significant figures.
Confusing moles with concentration.

Things to be Careful About

The value of $V$ is the average titre from part (a), not the final burette reading.
Keep at least 3 significant figures throughout.
Units are moles (mol).

Techniques used

convert volume in cm3 to dm3 by dividing by 1000multiply volume by concentration to find molesround the answer to 3 significant figures

(c)

Use your answer from (b) to calculate the number of moles of citric acid in $25.0\text{ cm}^3$ of $\mathbf{Q}$ .

3\text{KOH} + \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \rightarrow \text{K}_3\text{C}_6\text{H}_5\text{O}_7 + 3\text{H}_2\text{O}

number of moles of citric acid in $25\text{ cm}^3$ of $\mathbf{Q}$ = ______

DifficultyMedium-Easy

Worked solution

Working

The equation shows:

$3\text{KOH} + \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \rightarrow \text{K}_3\text{C}_6\text{H}_5\text{O}_7 + 3\text{H}_2\text{O}$

So 3 moles of KOH react with 1 mole of citric acid.

$n_{\text{acid}} = \frac{n_{\text{KOH}}}{3}$

Answer

moles of citric acid in $25.0\text{ cm}^3$ = $n_{\text{KOH}}$ from (b) $\div 3$ .

Final answer

moles of citric acid in 25.0 cm3 = moles KOH from (b) / 3

Detailed explanation

Walkthrough

You already know the moles of KOH used in the titration. The balanced equation gives the mole ratio:

3 KOH react with 1 citric acid

The acid is in the flask, so every citric acid molecule is neutralised by 3 KOH molecules. Therefore:

$\text{moles citric acid} = \frac{\text{moles KOH}}{3}$

This is the only calculation in this part. It must use the value from part (b) so the mark is a follow-through mark.

Key Takeaways

The balanced equation tells you the mole ratio, not the formula masses.
Divid mO by the ratio when moving from a higher coefficient reactant to a lower coefficient reactant.
This is a stoichiometry step: 3 : 1.

Common Mistakes

Multiplying by 3 instead of dividing by 3.
Using the wrong volume (25 cm3 doesn't enter this calculation).
Forgetting to write down the formula of the acid.

Buttons

Use moles of acid in 25.0 cm3 (thread may be larger), not moles of KOH.
Keep 3 significant figures as you progress.

Techniques used

use the stoichiometric ratio from the balanced equationdivide moles of KOH by 3

(d)

Use your answer from (c) to calculate:

(i)

the concentration of citric acid in $\mathbf{Q}$ .

concentration of citric acid in $\mathbf{Q}$ = ______ $\text{mol / dm}^3$

DifficultyMedium-Easy

Worked solution

Working

Volume of $\mathbf{Q}$ in the sample = $25.0\text{ cm}^3 = \dfrac{25.0}{1000}\text{ dm}^3$ .

$c_{\text{acid}} = \frac{n_{\text{acid}}}{\frac{25.0}{1000}} = n_{\text{acid}} \times \frac{1000}{25.0}$

Here $n_{\text{acid}}$ is the value from part (c).

Answer

concentration of citric acid in $\mathbf{Q}$ = $n_{\text{acid}} \times \frac{1000}{25.0}$ $\text{mol / dm}^3$ .

Final answer

n(c) x 1000 / 25.0 mol / dm3

Detailed explanation

Walkthrough

Concentration tells us moles of solute in one dm3 of solution. In part (c) you already have the number of moles in $25.0\text{ cm}^3$ of $\mathbf{Q}$ .

Since:

$c = \frac{n}{V}$

and the volume in dm3 is $25.0/1000 = 0.025\text{ dm}^3$ , the concentration is:

$c = \frac{n_{\text{acid}}}{0.025}$

Equivalently:

$c = n_{\text{acid}} \times 40$

because $1000/25 = 40$ . This is a direct concentration calculation.

Key Takeaways

Concentration is always moles per litre (dm3).
Convert a volume in cm3 to dm3 by dividing by 1000.
moles in the 25.0 cm3 sample and concentration look similar but are not the same; concentration is per dm3.

Common Mistakes

Writing concentration = moles / volume(cm3) without converting.
Using the original concentration value from $\mathbf{P}$ instead of the acid.
Leaving the volume in cm3 instead of dm3.

Things to be Careful About

The units must be mol / dm3.
Check your arithmetic: multiplying by 40 is the same as dividing by 0.025.

Techniques used

convert cm3 to dm3 for the acid sample volumedivide moles by volume in dm3 to get concentration

(ii)

the number of moles of citric acid in $500\text{ cm}^3$ of $\mathbf{Q}$ .

number of moles of citric acid in $500\text{ cm}^3$ of $\mathbf{Q}$ = ______

DifficultyMedium-Easy

Worked solution

Working

A $500\text{ cm}^3$ portion is $500/25 = 20$ times larger than a $25\text{ cm}^3$ portion.

$n_{\text{acid in } 500\text{ cm}^3} = n_{\text{acid in }25\text{ cm}^3} \times 20$

Alternatively, use concentration from (d)(i):

$n = c \times 0.500$

Answer

moles of citric acid in $500\text{ cm}^3$ = $n$ from part (c) $\times 20$ .

Final answer

moles citric acid in 500 cm3 = n(c) x 20

Detailed explanation

Walkthrough

You have moles in a $25\text{ cm}^3$ sample. The whole solution is $500\text{ cm}^3$ , so the molar amount in the whole solution is 20 times the amount in $25\text{ cm}^3$ :

$\frac{500}{25}=20$

The answer is moles in 25 multiplied by 20.

Alternative: multiply the concentration by $0.500\text{ dm}^3$ . The mark scheme accepts either.

Key Takeaways

Moles scale with volume if the concentration is the same.
Multiplying by the ratio of volumes is a quick scaling method.
Always keep the sample volume and the total volume distinct.

Common Mistakes

Multiplying by the sample volume instead of 500/25 = 20.
Use the concentration rather than the moles.
Confusing cm3 and dm3 in the final units.

Things to be Careful About

This is the moles in the whole required solution, not in a $25\text{ cm}^3$ sample.
If you used concentration in mol / dm3, be sure to multiply by 0.500 dm3, not 500.

Techniques used

scale the moles from 25.0 cm3 to 500 cm3 by multiplying by 20

(e)

Citric acid is available in hydrated form.

The formula of hydrated citric acid is $\text{H}_3\text{C}_6\text{H}_5\text{O}_7\cdot\text{H}_2\text{O}$

Use your answer from (d)(ii) to calculate the mass of hydrated citric acid crystals needed to make $500\text{ cm}^3$ of $\mathbf{Q}$ .

[ $A_r$ : $\text{H}, 1$ ; $\text{C}, 12$ ; $\text{O}, 16$ ]

mass of hydrated citric acid in $500\text{ cm}^3$ of $\mathbf{Q}$ = ______ $\text{g}$

DifficultyMedium-Easy

Worked solution

Working

Formula of hydrated citric acid: $\text{H}_3\text{C}_6\text{H}_5\text{O}_7\cdot\text{H}_2\text{O}$

Total atoms: $\text{H}_10$ , $\text{C}_6$ , $\text{O}_8$ .

M_r = 10(1) + 6(12) + 8(16) = 10 + 72 + 128 = 210

mass = moles of hydrated citric acid $\times M_r$
moles acid in $500\text{ cm}^3$ is the value from part (d)(ii).

\text{mass} = n \times 210

Give the answer in g, correct to an appropriate number of significant figures.

Answer

mass = $210 \times$ moles of citric acid from (d)(ii) g.

Final answer

mass = 210 x n(d)(ii) g

Detailed explanation

Walkthrough

The hydrated acid contains all of the acid plus one water of crystallisation per formula unit. Its molar mass must include the water unit. Count the atoms:

H: 3 + 5 + 2 = 10
C: 6
O: 7 + 1 = 8

Then:

$M_r = 10 + 6\times 12 + 8\times16 = 210$

Once you know the moles from part (d)(ii), the mass required is:

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

Because the water is part of the crystal, the moles of hydrated citric acid is the same as the moles of anhydrous citric acid required.

Key Takeaways

Hydrated formula includes water in the mole ratio.
Calculate M_r by adding up every atom, including water
Use mass = moles × Mr to convert moles to grams.

Common Mistakes

Forgetting to include the water molecule in Mr.
Using wrong atom counts (especially O: 7 + 2 = 9? No, the hydrated water adds 1 oxygen, so total O = 7 + 2 = 9? Wait: citric acid has 7 oxygen, and H2O has 1 more, so total O = 8. A frequent slip is thinking 7 + 2 because H2O has 2 hydrogens, not 2 oxygens.)
Using Mr expressed without including the dot water.
Using mass of citric acid (anhydrous) instead of hydrated.

Things to be Careful About

The H2O in the formula contributes 2 H and 1 O.
Check your value $M_r = 210$ before multiplying.
Write final mass in g. If the value from (d)(ii) was moles in 500 cm3, you are calculating g of hydrated crystals required to make that solution.

Techniques used

calculate relative molecular mass from atomic massesmass = moles x relative molecular massuse answer from part (d)(ii)

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