Chemistry 5070/32 — October/November 2022

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

Q117MObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationFree sample

P is a sample of dilute nitric acid.

P is prepared by adding $5.0\text{ cm}^3$ of concentrated nitric acid to distilled water and making the total volume of the solution up to $250\text{ cm}^3$ with distilled water.

Q is $0.316\text{ mol / dm}^3$ sodium hydroxide.

(a)

Put P into the burette.

Pipette $25.0\text{ cm}^3$ of Q into a flask and titrate with P using methyl orange indicator.

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 P used / $\text{cm}^3$
best titration results ( $\checkmark$ )

Summary

Tick ( $\checkmark$ ) the best titration results.

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

12M

DifficultyMedium-Easy

Worked solution

Answer

Complete the results table with your own readings:

initial and final readings for each titration, each recorded to 1 decimal place (e.g. 24.5), no reading above 50.0 cm³ and no initial reading equal to 50.0 cm³;
volume of P used = final reading − initial reading, calculated correctly for every titration;
tick (✓) the best (concordant) titres — those within 0.2 cm³ of each other;
average volume of P = (sum of the ticked titres) ÷ (number of ticked titres), written in the summary line.

Final answer

Complete the table with own burette readings to 1 dp; volume used = final − initial; tick concordant titres and average them.

Detailed explanation

Walkthrough

This is the practical part of the question: you carry out the titration yourself. Pipette 25.0 cm³ of Q (sodium hydroxide) into a conical flask, add a few drops of methyl orange indicator (yellow in alkali, pink/red in acid), and run P (nitric acid) in from the burette until the indicator just changes colour — the end point. Record the initial and final burette readings for each titration.

The marks fall into five groups:

Measurements — both initial and final readings present for each titration, recorded to 1 decimal place, none above 50.0 cm³, and no initial reading of 50.0 cm³.
Titres — each volume of P used = final − initial, with no subtraction errors.
Accuracy — your two best titres compared with the Supervisor's value; the closer they are, the more marks you gain.
Concordance — the ticked (best) titres should be within 0.2 cm³ of each other.
Average — the average of the ticked titres, correctly calculated.

The average titre you write in the summary is carried into part (b), so make sure it is the average of your concordant results only.

Key Takeaways

Burette readings are always recorded to 1 decimal place (e.g. 24.5 cm³), never to whole numbers or 2 decimal places.
Volume used = final reading − initial reading.
Repeat the titration until two concordant results are obtained, and average only the concordant titres.
The average titre is the key result used in all the later calculation parts.

Common Mistakes

Recording readings to 2 decimal places or to whole numbers — loses the measurements mark.
An initial reading of 50.0 cm³ (burette completely full) — not allowed.
Averaging all the titres, including those that are not concordant.
Subtraction errors when calculating the titres.

Things to Be Careful About

Read the burette at eye level to avoid parallax error.
The end point with methyl orange is the first permanent colour change (yellow → pink/red).
Rinse the burette with P and the pipette with Q before use.
Write the average titre clearly in the summary line — it is the value used in part (b).

Techniques used

record burette readings to one decimal placecalculate titre volumes from initial and final readingsselect concordant results and calculate the average

(b)

Q is $0.316\text{ mol / dm}^3$ sodium hydroxide.

The equation for the reaction is shown.

\text{NaOH} + \text{HNO}_3 \rightarrow \text{NaNO}_3 + \text{H}_2\text{O}

Use your result from (a) to calculate the concentration, in $\text{mol / dm}^3$ , of nitric acid in P.

Give your answer to three significant figures.

concentration = ______ $\text{mol / dm}^3$

DifficultyMedium

Worked solution

Working

Moles of NaOH in 25.0 cm³ of Q:

\text{moles} = \frac{25.0}{1000} \times 0.316 = 0.00790\ \text{mol}

The equation shows NaOH : HNO₃ = 1 : 1, so 0.00790 mol of HNO₃ is in the titre volume of P.

Let V = average titre of P from (a), in cm³:

\text{concentration of P} = \frac{0.00790}{V/1000} = \frac{7.90}{V}\ \text{mol / dm}^3

Answer

concentration of P = (25.0 × 0.316) / V = 7.90 / V mol / dm³, to 3 significant figures.

Final answer

(25.0 × 0.316)/V = 7.90/V mol / dm³, where V is the average titre of P from (a) in cm³

Detailed explanation

Walkthrough

Start by finding the moles of NaOH in the 25.0 cm³ pipette volume of Q. Convert cm³ to dm³ by dividing by 1000, then multiply by the concentration:

\text{moles} = \frac{25.0}{1000} \times 0.316 = 0.00790\ \text{mol}

The balanced equation shows NaOH and HNO₃ react in a 1 : 1 mole ratio, so the same number of moles of HNO₃ (0.00790 mol) must be present in the volume of P that was used — your average titre V from part (a). Concentration is moles divided by volume, so:

\text{concentration of P} = \frac{0.00790}{V/1000}\ \text{mol / dm}^3

This is the mark scheme's formula (25.0 × 0.316) / V. Give the final answer to 3 significant figures.

Key Takeaways

moles = concentration × volume, with volume in dm³.
Use the balanced equation to find the reacting mole ratio (here 1 : 1).
concentration = moles / volume.

Common Mistakes

Forgetting to convert cm³ to dm³.
Using the wrong titre (e.g. a single titre instead of the average).
Giving the answer to the wrong number of significant figures.

Things to Be Careful About

The answer must be given to 3 significant figures.
The mark scheme formula is (25.0 × 0.316) / (a): the average titre in cm³ goes straight into the denominator.

Techniques used

convert volume to dm³ and calculate moles of NaOHapply the 1:1 mole ratio from the equationdivide moles by the titre volume to find concentration

(c)

P is prepared by adding $5.0\text{ cm}^3$ of concentrated nitric acid to distilled water and making the total volume of the solution up to $250\text{ cm}^3$ with distilled water.

Use your answer from (b) to calculate the number of moles of nitric acid in $5.0\text{ cm}^3$ of concentrated nitric acid.

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

DifficultyMedium-Easy

Worked solution

Working

The 5.0 cm³ of concentrated acid was diluted to make 250 cm³ of P, so the moles in 5.0 cm³ of concentrated acid equal the moles in 250 cm³ of P:

\text{moles} = \text{concentration of P} \times \frac{250}{1000} = \frac{\text{(b)}}{4}

Answer

number of moles = (b) / 4 mol

Final answer

(b)/4 mol, where (b) is the concentration of P from part (b)

Detailed explanation

Walkthrough

The 5.0 cm³ of concentrated acid was diluted with distilled water to make 250 cm³ of P. Dilution does not change the number of moles of acid — it only spreads them through a larger volume. So the moles in 5.0 cm³ of concentrated acid equal the moles in all 250 cm³ of P:

\text{moles} = \text{concentration of P} \times \frac{250}{1000} = \frac{\text{(b)}}{4}

Key Takeaways

Dilution conserves moles: moles before = moles after.
moles = concentration × volume, with volume in dm³.

Common Mistakes

Thinking the number of moles changes on dilution.
Using the wrong volume (5.0 instead of 250 cm³).

Things to Be Careful About

Use the concentration from (b) in mol/dm³ and the volume in dm³.

Techniques used

use the dilution factor from 5.0 cm³ to 250 cm³calculate moles from concentration and volume

(d)

Use your answer from (c) to calculate the concentration, in $\text{mol / dm}^3$ , of concentrated nitric acid.

concentration = ______ $\text{mol / dm}^3$

DifficultyMedium-Easy

Worked solution

Working

\text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{\text{(c)}}{5.0/1000} = \text{(c)} \times 200

Answer

concentration = (c) × 200 mol / dm³

Final answer

Detailed explanation

Walkthrough

Now reverse the concentration calculation for the concentrated acid. You know the moles present in 5.0 cm³ of it from part (c). Convert 5.0 cm³ to dm³ and divide:

\text{concentration} = \frac{\text{(c)}}{5.0/1000} = \text{(c)} \times 200\ \text{mol / dm}^3

Dividing by 5.0/1000 is the same as multiplying by 200.

Key Takeaways

concentration = moles / volume, with volume in dm³.
Dividing by a small volume in dm³ multiplies the moles by a large factor.

Common Mistakes

Dividing by 5.0 instead of 0.005 dm³.
Unit errors in the final answer.

Things to Be Careful About

Keep the volume in dm³: 5.0 cm³ = 0.005 dm³.

Techniques used

calculate concentration from moles and volume

(e)

Use your answer from (d) to calculate the mass, in $\text{g}$ , of nitric acid, $\text{HNO}_3$ , in $1\text{ dm}^3$ of concentrated nitric acid.

[ $M_r$ : $\text{HNO}_3$ , 63]

mass = ______ $\text{g}$

DifficultyEasy

Worked solution

Working

\text{mass} = \text{moles} \times M_r = \text{(d)} \times 63

Answer

mass = (d) × 63 g

Final answer

(d) × 63 g, where (d) is the concentration from part (d)

Detailed explanation

Walkthrough

The concentration from (d) is the number of moles of HNO₃ in 1 dm³ of concentrated acid. To find the mass, multiply moles by M_r:

\text{mass} = \text{(d)} \times 63\ \text{g}

Key Takeaways

mass = moles × M_r.
A concentration in mol/dm³ is the number of moles in 1 dm³.

Common Mistakes

Using the wrong M_r value.
Forgetting the unit (the answer is in grams).

Things to Be Careful About

The unit of the answer is grams (g).

Techniques used

convert moles to mass using M_r

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