Chemistry 9701/32 — May/June 2025

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation · Qualitative Analysis

Q1Medium-EasyManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Acids donate $\text{H}^+$ in aqueous solution. The number of moles of $\text{H}^+$ donated per mole of acid is the proticity of the acid. For example, sulfuric acid is diprotic as it donates two moles of $\text{H}^+$ per mole of acid.

In this experiment you will carry out a titration to determine the proticity of citric acid, $\text{C}_6\text{H}_8\text{O}_7$ .

FB 1 is aqueous citric acid containing $7.50\text{ g dm}^{-3}$ $\text{C}_6\text{H}_8\text{O}_7$ .
FB 2 is aqueous sodium hydroxide containing $4.50\text{ g dm}^{-3}$ $\text{NaOH}$ .
FB 3 is thymolphthalein indicator.

(a)

Method

Fill the burette with FB 2.
Pipette $25.0\text{ cm}^3$ of FB 1 into a conical flask.
Add a few drops of FB 3 to 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 in the space below, all your burette readings and the volume of FB 2 added in each accurate titration.

Results

7M Similar

DifficultyMedium-Easy

Worked solution

Answer

Rough titration: titre = 26.50 cm³

Accurate titrations (burette readings to 0.05 cm³):

	Titration 1	Titration 2	Titration 3
Final burette reading / cm³	26.40	52.75	26.35
Initial burette reading / cm³	0.00	26.40	0.00
Titre / cm³	26.40	26.35	26.35

The accurate titres 26.35 cm³ and 26.35 cm³ are concordant (within 0.10 cm³ of each other).

Final answer

See working — candidate-dependent readings; representative titres: 26.40, 26.35, 26.35 cm³

Detailed explanation

Background Concept

In a titration, a burette delivers a solution (the titrant) into a measured volume of another solution (the analyte) until the end point is reached, indicated by a colour change. The burette is read twice for each titration — the initial reading (before any titrant is added) and the final reading (at the end point) — and the titre is the difference between them. For reliable results, the titration is repeated until two or more concordant titres are obtained, i.e. titres that agree within a small tolerance (typically 0.10 cm³). Burette readings are recorded to the nearest 0.05 cm³ because the graduations on a burette are typically 0.1 cm³ apart and the eye can estimate to half a division.

Understanding the Question

This part asks you to actually perform the titration and record your results properly. You are given FB 1 (citric acid, 7.50 g dm⁻³), FB 2 (NaOH, 4.50 g dm⁻³), and FB 3 (thymolphthalein indicator). You must first do a rough titration to find the approximate end point, then repeat accurately until you have consistent results. The marks are awarded for correct recording technique: two burette readings and a titre for the rough titration, initial and final readings for at least two accurate titrations, correct headings and units, readings to 0.05 cm³, and concordant titres within 0.10 cm³.

Approach

The key is to record everything clearly and consistently. For each titration, note the initial and final burette readings to 0.05 cm³, and calculate the titre as final − initial. Repeat until two titres agree within 0.10 cm³. Use a table with proper headings and units.

Step-by-Step Reasoning

Rough titration: add FB 2 from the burette quickly until the indicator changes colour (thymolphthalein is colourless in acid, blue in base). Record the rough titre.
Accurate titrations: refill the burette, record the initial reading, add FB 2 dropwise near the end point, record the final reading. Calculate titre = final − initial.
Repeat until two titres are within 0.10 cm³ of each other.
Record all readings in a table with headings: "Initial burette reading / cm³", "Final burette reading / cm³", "Titre / cm³".

In the representative example:

Rough titre: 26.50 cm³
Accurate 1: initial 0.00, final 26.40, titre 26.40
Accurate 2: initial 26.40, final 52.75, titre 26.35
Accurate 3: initial 0.00, final 26.35, titre 26.35

The two concordant titres are 26.35 cm³ and 26.35 cm³ (identical), and 26.40 cm³ is within 0.10 cm³ of them.

Key Takeaways

Always record initial AND final burette readings, not just the titre.
Record to 0.05 cm³.
Repeat until concordant (within 0.10 cm³).
Use clear table headings with units.

Common Mistakes

Recording only the titre, not the two readings.
Recording to 0.1 cm³ instead of 0.05 cm³.
Using the rough titre in the mean.
Not repeating until concordant.

Things to Be Careful About

The mark scheme requires readings to 0.05 cm³.
Concordant titres must be within 0.10 cm³.
Include units in table headings.

Techniques used

perform a rough titrationrecord burette readings to 0.05 cm³repeat titrations to concordancetabulate titration results with units

(b)

From your accurate titration results, calculate a suitable mean value to use in your calculations. Show clearly how you obtain the mean value.

$25.0\text{ cm}^3$ of FB 1 required .............................. $\text{cm}^3$ of FB 2.

1M Similar

(c)

Calculations

(i)

Calculate the amount, in mol, of citric acid in $25.0\text{ cm}^3$ of FB 1. Show your working.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$M_r(\text{C}_6\text{H}_8\text{O}_7) = 6(12.0) + 8(1.0) + 7(16.0) = 192$

Amount of citric acid = $\frac{25.0}{1000} \times \frac{7.50}{192} = 9.77 \times 10^{-4}\text{ mol}$

Answer

$9.77 \times 10^{-4}\text{ mol}$

Final answer

9.77 × 10⁻⁴ mol

Detailed explanation

Background Concept

To find the amount (in mol) of a solute in a solution of known concentration, use: amount (mol) = concentration (mol dm⁻³) × volume (dm³). When the concentration is given in g dm⁻³, it must first be converted to mol dm⁻³ by dividing by the molar mass. Equivalently, amount = (mass concentration × volume) / molar mass, with volume in dm³.

Understanding the Question

You are given FB 1 as 7.50 g dm⁻³ citric acid, and you pipette 25.0 cm³. Find the amount of citric acid in that 25.0 cm³ sample.

Approach

Convert the volume from cm³ to dm³ (divide by 1000), then multiply by the mass concentration (7.50 g dm⁻³) to get the mass, then divide by the molar mass (192 g mol⁻¹). This can be done in one expression.

Step-by-Step Reasoning

Calculate the molar mass of citric acid, C₆H₈O₇:
$M_r = 6(12.0) + 8(1.0) + 7(16.0) = 72 + 8 + 112 = 192$
Convert volume: $25.0\text{ cm}^3 = 25.0/1000 = 0.0250\text{ dm}^3$
Amount = $0.0250 \times \frac{7.50}{192} = 9.765625 \times 10^{-4}\text{ mol}$
Round to 3 or 4 significant figures: $9.77 \times 10^{-4}\text{ mol}$ (or $9.766 \times 10^{-4}\text{ mol}$ ).

Key Takeaways

Convert cm³ to dm³ by dividing by 1000.
Convert g dm⁻³ to mol dm⁻³ by dividing by molar mass.
Quote the answer to 3 or 4 significant figures.

Common Mistakes

Forgetting to convert cm³ to dm³ (a very common error).
Using the wrong molar mass.
Not showing working.

Things to Be Careful About

The mark scheme accepts $9.766 \times 10^{-4}$ or $9.77 \times 10^{-4}\text{ mol}$ .
Include units in the answer.

Techniques used

convert cm³ to dm³calculate moles from mass concentration and molar mass

(ii)

Calculate the amount, in mol, of sodium hydroxide in the volume of FB 2 in (b).

1M Similar

(iii)

Citric acid is triprotic. Show whether your results in this experiment support this statement.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Mole ratio = $\frac{\text{amount of NaOH}}{\text{amount of citric acid}} = \frac{2.96 \times 10^{-3}}{9.77 \times 10^{-4}} = 3.03$

Answer

Mole ratio ≈ 3.0, which lies between 2.5 and 3.5, so the results support the statement that citric acid is triprotic.

Final answer

Mole ratio ≈ 3.0, supports triprotic statement

Detailed explanation

Background Concept

The proticity of an acid is the number of moles of H⁺ donated per mole of acid. In a titration, each mole of NaOH provides one mole of OH⁻, which neutralises one mole of H⁺. Therefore, at the end point, the mole ratio of NaOH to acid equals the proticity of the acid.

Understanding the Question

Citric acid is claimed to be triprotic (3 H⁺ per molecule). Compare the mole ratio of NaOH to citric acid from your results. If the ratio is about 3, the claim is supported.

Approach

Divide the amount of NaOH from (c)(ii) by the amount of citric acid from (c)(i). Compare the result to the expected value of 3. The mark scheme accepts a ratio between 2.5 and 3.5 as supporting triprotic.

Step-by-Step Reasoning

Amount of NaOH = $2.96 \times 10^{-3}\text{ mol}$ (from c(ii)).
Amount of citric acid = $9.77 \times 10^{-4}\text{ mol}$ (from c(i)).
Mole ratio = $\frac{2.96 \times 10^{-3}}{9.77 \times 10^{-4}} = 3.03$ .
Since 3.03 is between 2.5 and 3.5, the results support the statement that citric acid is triprotic.

Key Takeaways

Mole ratio = n(NaOH) / n(acid) = proticity.
A ratio near 3 supports triprotic.

Common Mistakes

Dividing the wrong way round (acid/NaOH instead of NaOH/acid).
Not stating the conclusion clearly.

Things to Be Careful About

The mark scheme accepts a stated ratio between 2.5 and 3.5 with the conclusion "supports".
If the ratio is outside this range, the conclusion must be "does not support".

Techniques used

calculate the NaOH:acid mole ratiocompare the ratio to the expected proticity

(iv)

Complete the equation for the reaction of citric acid, $\text{C}_6\text{H}_8\text{O}_7$ , with sodium hydroxide.

\text{C}_6\text{H}_8\text{O}_7 + \dots\dots \text{NaOH} \rightarrow \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots + \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots

1M Similar

(v)

Citric acid does not contain a chiral carbon atom. Draw a possible structural formula of a molecule of citric acid.

1M Similar

(d)

A student uses a pipette labelled $25.0 \pm 0.06\text{ cm}^3$ to measure FB 1. The student suggests that it is more accurate to measure the volume of FB 1 with a burette instead of the pipette.

State whether the student’s suggestion is correct. Explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The student is not correct. A burette reading has an uncertainty of ±0.05 cm³, so a titre (two readings) has an uncertainty of 2 × 0.05 = ±0.10 cm³. This is greater than the pipette's ±0.06 cm³, so the pipette is more accurate.

Final answer

Not correct — burette titre uncertainty 2 × 0.05 = 0.10 cm³ > pipette 0.06 cm³

Detailed explanation

Background Concept

Every measuring instrument has an uncertainty. A burette has graduations every 0.1 cm³, and readings are estimated to ±0.05 cm³. A titre requires two readings (initial and final), so the total uncertainty is the sum of the two individual uncertainties: 2 × 0.05 = ±0.10 cm³. A pipette delivers a fixed volume with a single stated uncertainty, here ±0.06 cm³.

Understanding the Question

The student claims that measuring the 25.0 cm³ of FB 1 with a burette is more accurate than using the pipette. You must state whether this is correct and explain why, comparing the uncertainties.

Approach

Compare the uncertainties: the burette titre uncertainty (±0.10 cm³) versus the pipette uncertainty (±0.06 cm³). The instrument with the smaller uncertainty is the more accurate one.

Step-by-Step Reasoning

Burette: each reading has an uncertainty of ±0.05 cm³.
A titre = final reading − initial reading, so the uncertainties add: 0.05 + 0.05 = ±0.10 cm³.
Pipette: stated uncertainty is ±0.06 cm³.
Since 0.10 > 0.06, the pipette has the smaller uncertainty and is therefore more accurate.
Conclusion: the student's suggestion is not correct.

Key Takeaways

Titre uncertainty = 2 × burette reading uncertainty.
Smaller uncertainty means greater accuracy.

Common Mistakes

Only considering one burette reading (±0.05) instead of two (±0.10).
Saying the student is correct without the uncertainty comparison.

Things to Be Careful About

The mark scheme requires BOTH "not correct" AND the uncertainty calculation (2 × 0.05 greater than 0.06).

Techniques used

calculate titre uncertainty from two burette readingscompare uncertainties of burette and pipette

Q2Medium-EasyPresentation of Data and Observations

You will determine the enthalpy change for the reaction of aqueous citric acid with aqueous sodium hydroxide to form aqueous sodium citrate.

The procedure involves two experiments using solid citric acid.

FB 4 is citric acid, $\text{C}_6\text{H}_8\text{O}_7$ . Use this for Experiment 1.
FB 5 is citric acid, $\text{C}_6\text{H}_8\text{O}_7$ . Use this for Experiment 2.
FB 6 is $2.00\text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$ .

(a)

Experiment 1 is the determination of the enthalpy change of solution, $\Delta H_1$ , for citric acid. This is the enthalpy change when one mole of citric acid dissolves in water.

\text{C}_6\text{H}_8\text{O}_7(\text{s}) + \text{aq} \rightarrow \text{C}_6\text{H}_8\text{O}_7(\text{aq}) \quad \Delta H_1

Method

Support one of the cups in the $250\text{ cm}^3$ beaker.
Use the $50\text{ cm}^3$ measuring cylinder to transfer $30.0\text{ cm}^3$ of distilled water into the cup.
Measure the temperature of the water in the cup. Record this temperature in the space for results.
Weigh the container with FB 4. Record the mass.
Tip all of the FB 4 into the water in the cup.
Stir the mixture until the minimum temperature is obtained. Record this temperature.
Weigh the container with any residual FB 4. Record the mass.
Calculate and record the mass of FB 4 used.
Calculate and record the temperature change.

Results

3M Similar

DifficultyEasy

Worked solution

Answer

Record the results in a table with unambiguous headings and units:

Quantity	Reading
Mass of container + FB 4 / g	...
Mass of container + residual FB 4 / g	...
Mass of FB 4 used / g	...
Initial temperature of water / °C	...
Final (minimum) temperature / °C	...
Temperature change / °C	...

Give both balance readings to the same number of decimal places (2 or 3 dp). Record all four thermometer readings to the nearest 0.5 °C. The same conventions apply to the results table in Experiment 2.

Final answer

See working (candidate-dependent results table)

Detailed explanation

Background Concept

In a calorimetry experiment, the temperature change is measured in an insulated cup. For a valid record, every measured quantity must have a heading and a unit, and readings must be recorded with appropriate precision. Balance readings should be given to the same number of decimal places because the mass used is found by difference; liquid-in-glass thermometer readings are conventionally recorded to the nearest 0.5 °C.

Understanding the Question

This part asks you to set out the results table for Experiment 1, dissolving citric acid in water. The mark scheme rewards three things: unambiguous headings with units for masses and temperatures, correct calculation of the mass used and temperature change, and consistent precision. No numerical values can be given here because they depend on your own readings.

Approach

Identify the quantities measured: mass of the container with and without residual FB 4, and the initial and final temperatures. Record them in a table with units, calculate the differences for mass and temperature, and apply the precision conventions.

Step-by-Step Reasoning

The mass of FB 4 used is found by subtracting the mass of the container plus residual solid from the mass of the container plus original solid. Both readings must have the same number of decimal places. The temperature change is final minus initial temperature. Since the temperature falls when citric acid dissolves, record it as a fall or decrease. The same style is expected in Experiment 2, where the mass of FB 5 used must lie between 4.80 and 5.00 g.

Key Takeaways

A results table is credited for clear headings with units, correct derived quantities, and consistent precision.

Common Mistakes

Missing units in headings.
Writing balance readings to different numbers of decimal places.
Recording temperatures as 21.3 °C instead of 21.0 or 21.5 °C.
Confusing “mass of container + FB 4” with “mass of FB 4”.

Things to Be Careful About

Use “temperature change / °C” as a heading, not just “temperature”. In Experiment 1 the change is a decrease; in Experiment 2 it is an increase. Use the same precision conventions in both tables.

Techniques used

label results table with quantity and unitrecord balance readings to consistent decimal placesrecord thermometer readings to nearest 0.5 °Ccalculate mass used by differencecalculate temperature change

(b)

Calculations

(i)

Calculate the energy change, in J, in your experiment.

1M Similar

DifficultyEasy

Worked solution

Working

Energy change = mass of water × specific heat capacity × temperature change

q = 30.0 \times 4.18 \times \Delta T \text{ J}

Use your measured temperature decrease for $\Delta T$ and round the answer to 2–4 significant figures.

Final answer

See working (candidate-dependent)

Detailed explanation

Background Concept

The heat energy absorbed or released by a substance is given by $q = mc\Delta T$ , where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change. For water, $c = 4.18 \text{ J g}^{-1}\text{ K}^{-1}$ . Because 30.0 cm³ of water has a mass of approximately 30.0 g, the mass can be taken as 30.0 g.

Understanding the Question

This part asks you to calculate the energy change, in J, for Experiment 1. You need to use the volume of water, the specific heat capacity, and the temperature decrease you recorded.

Approach

Substitute $m = 30.0\text{ g}$ , $c = 4.18\text{ J g}^{-1}\text{ K}^{-1}$ , and your measured $\Delta T$ into $q = mc\Delta T$ . The result is in joules.

Step-by-Step Reasoning

For example, if your temperature decrease was 5.0 °C, then $q = 30.0 \times 4.18 \times 5.0 = 627\text{ J}$ . Round to 2–4 significant figures, so 630 J to 2 sf or 627 J to 3 sf. The sign is not required here; you only need the magnitude of the energy change.

Key Takeaways

The energy change in a calorimetry experiment is found from the mass of the solution, its specific heat capacity, and the temperature change.

Common Mistakes

Using the volume in cm³ directly without treating it as a mass in grams.
Using the wrong specific heat capacity.
Forgetting to round the final answer to an appropriate number of significant figures.

Things to Be Careful About

A temperature change in °C has the same numerical value as one in K, so no conversion is needed for $\Delta T$ . Keep the unit as J for this part; conversion to kJ happens in the next part.

Techniques used

apply q = mcΔTuse mass of water from volumeround to appropriate significant figures

(ii)

Calculate the enthalpy change of solution, $\Delta H_1$ , in $\text{kJ mol}^{-1}$ , for dissolving $1.00\text{ mol}$ of solid citric acid, $\text{C}_6\text{H}_8\text{O}_7$ , in water. Show your working.

\Delta H_1 = \dots\dots \text{ sign} \quad \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \text{ value} \quad \text{kJ mol}^{-1}

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$M_r(\text{C}_6\text{H}_8\text{O}_7) = 192$

Amount of citric acid = mass of FB 4 used / 192 mol

n = \frac{m}{192} \text{ mol}

\Delta H_1 = \frac{\text{energy change from (b)(i)}}{n} \times \frac{1}{1000} \text{ kJ mol}^{-1}

= + ... kJ mol⁻¹ (positive sign, 2–4 sf)

Final answer

See working (candidate-dependent)

Detailed explanation

Background Concept

The enthalpy change of solution, $\Delta H_1$ , is the heat energy change when one mole of solute dissolves in a stated amount of solvent. To find it, you divide the measured energy change by the number of moles of solute and convert joules to kilojoules. The sign is positive if the process is endothermic, which is shown by a fall in temperature.

Understanding the Question

You must use the energy change from part (b)(i) and the mass of FB 4 used to calculate $\Delta H_1$ in kJ mol⁻¹. The mark scheme requires the working to show moles of citric acid, the division by moles, the conversion to kJ, and a positive sign.

Approach

First calculate the amount of citric acid using $n = m/M_r$ . Then divide the energy change in J by this amount, convert J to kJ by dividing by 1000, and add a positive sign because the temperature decreased.

Step-by-Step Reasoning

The molar mass of citric acid is $M_r = 6(12) + 8(1) + 7(16) = 192$ . If, for example, 1.20 g of FB 4 was used, $n = 1.20/192 = 0.00625\text{ mol}$ . If the energy change from (b)(i) was 627 J, then $\Delta H_1 = (627/0.00625) \times (1/1000) = +100.32 \text{ kJ mol}^{-1}$ , which should be quoted as about +100 kJ mol⁻¹ to 2–4 sf.

Key Takeaways

To find an enthalpy change per mole, divide the energy change by the number of moles and convert to kJ mol⁻¹. The sign is determined by whether the temperature rises or falls.

Common Mistakes

Forgetting to divide by 1000 to convert J to kJ.
Using the wrong molar mass for citric acid.
Giving a negative sign when the temperature fell.
Not showing the calculation of moles.

Things to Be Careful About

The mark scheme allows 2–4 significant figures. The sign is essential: $\Delta H_1$ is positive because dissolving citric acid in water is endothermic.

Techniques used

calculate moles from mass and Mrconvert J to kJ and divide by molesassign sign to enthalpy change

(c)

Experiment 2 is the determination of the enthalpy change, $\Delta H_2$ , for the reaction of one mole of solid citric acid with aqueous sodium hydroxide. In this experiment, aqueous sodium hydroxide, FB 6, is used in excess.

Method

Support the second cup in the beaker.
Add $4.80\text{--}5.00\text{ g}$ of FB 5 to the cup. Record your weighings.
Measure and record the temperature of FB 6 in its container.
Use the $50\text{ cm}^3$ measuring cylinder to transfer $50.0\text{ cm}^3$ of FB 6 into the cup with FB 5.
Stir the mixture until the maximum temperature is obtained. Record the maximum temperature.
Calculate and record the mass of FB 5 used.
Calculate and record the temperature change.

Results

3M Similar

DifficultyEasy

Worked solution

Answer

Record the results in a table with unambiguous headings and units:

Quantity	Reading
Mass of container + FB 5 / g	...
Mass of container + residual FB 5 / g	...
Mass of FB 5 used / g	...
Initial temperature of FB 6 / °C	...
Final (maximum) temperature / °C	...
Temperature change / °C	...

The mass of FB 5 used must be between 4.80 and 5.00 g. Give balance readings to the same number of decimal places (2 or 3 dp) and thermometer readings to the nearest 0.5 °C.

Final answer

See working (candidate-dependent results table)

Detailed explanation

Background Concept

Experiment 2 is an exothermic reaction between solid citric acid and excess aqueous sodium hydroxide. The temperature rises, so the temperature change is recorded as a rise. As in Experiment 1, the results table must have clear headings with units and consistent precision.

Understanding the Question

This part asks you to set out the results table for Experiment 2. You need to record the mass of FB 5 used and the initial and maximum temperatures of the sodium hydroxide solution. The mass of FB 5 used must lie between 4.80 and 5.00 g.

Approach

Record the container masses before and after adding FB 5, calculate the mass of FB 5 used by difference, and record the initial and maximum temperatures. Then calculate the temperature rise.

Step-by-Step Reasoning

The mass of FB 5 used is the difference between the mass of the container plus FB 5 and the mass of the container plus any residual FB 5. The temperature change is the maximum temperature minus the initial temperature of FB 6. Because the reaction is exothermic, this is a temperature rise. The same precision rules apply as in Experiment 1: balance readings to the same number of decimal places and thermometer readings to the nearest 0.5 °C.

Key Takeaways

A clear results table with units and consistent precision is essential for scoring marks in practical work.

Common Mistakes

Recording the mass of FB 5 outside the 4.80–5.00 g range.
Using inconsistent decimal places for balance readings.
Calling the temperature change a decrease when it is a rise.

Things to Be Careful About

The mark scheme treats temperature readings to 0.5 °C as acceptable. Ensure the table headings clearly distinguish between “mass of container + FB 5” and “mass of FB 5 used”.

Techniques used

record mass and temperature headings with unitscalculate mass of FB 5 used by differencecalculate temperature riseensure mass of FB 5 in 4.80–5.00 g range

(d)

Calculations

Calculate the enthalpy change of reaction, $\Delta H_2$ , in $\text{kJ mol}^{-1}$ , for the reaction of $1.00\text{ mol}$ of solid citric acid, $\text{C}_6\text{H}_8\text{O}_7$ , with aqueous sodium hydroxide. Show your working.

\Delta H_2 = \dots\dots \text{ sign} \quad \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \text{ value} \quad \text{kJ mol}^{-1}

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Energy change = 50.0 × 4.18 × temperature rise J

Amount of FB 5 = mass of FB 5 / 192 mol

n = \frac{m}{192} \text{ mol}

\Delta H_2 = \frac{\text{energy change}}{n} \times \frac{1}{1000} \text{ kJ mol}^{-1}

= − ... kJ mol⁻¹ (negative sign, 2–4 sf)

Final answer

See working (candidate-dependent)

Detailed explanation

Background Concept

In Experiment 2, solid citric acid reacts with excess aqueous sodium hydroxide. The reaction is exothermic, so the temperature rises. The energy released is calculated using $q = mc\Delta T$ , where $m$ is the mass of the sodium hydroxide solution, taken as 50.0 g because 50.0 cm³ of solution is used. The enthalpy change per mole is then found by dividing by the number of moles of citric acid and converting J to kJ.

Understanding the Question

You must calculate $\Delta H_2$ , the enthalpy change for the reaction of one mole of solid citric acid with aqueous sodium hydroxide. Use the temperature rise from Experiment 2 and the mass of FB 5 used.

Approach

Calculate the energy change using 50.0 × 4.18 × temperature rise. Then calculate the amount of citric acid using $n = m/192$ . Divide the energy change by $n$ , convert to kJ mol⁻¹, and give a negative sign because the temperature rose.

Step-by-Step Reasoning

For example, if the temperature rise was 6.0 °C, the energy change is $50.0 \times 4.18 \times 6.0 = 1254\text{ J}$ . If the mass of FB 5 was 4.90 g, $n = 4.90/192 = 0.02552\text{ mol}$ . Then $\Delta H_2 = (1254/0.02552) \times (1/1000) = -49.1 \text{ kJ mol}^{-1}$ , quoted to 2–4 sf.

Key Takeaways

The enthalpy change of an exothermic reaction is negative. The calculation follows the same pattern as for $\Delta H_1$ , but with 50.0 cm³ of solution and a temperature rise.

Common Mistakes

Using 30.0 instead of 50.0 for the volume of solution.
Forgetting the negative sign.
Not dividing by 1000 to convert J to kJ.
Using the mass of FB 5 incorrectly in the mole calculation.

Things to Be Careful About

The sign is essential: $\Delta H_2$ is negative because the reaction is exothermic. Quote the final answer to 2–4 significant figures.

Techniques used

apply q = mcΔT to 50.0 cm³ NaOHcalculate moles of citric acid from mass and Mrconvert energy to kJ per moleassign negative sign

(e)

Use your values for $\Delta H_1$ and $\Delta H_2$ to calculate the enthalpy change, $\Delta H_r$ , in $\text{kJ mol}^{-1}$ , for the reaction of $1.00\text{ mol}$ of aqueous citric acid with aqueous sodium hydroxide.

\Delta H_r = \dots\dots \text{ sign} \quad \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \text{ value} \quad \text{kJ mol}^{-1}

1M Similar

DifficultyEasy

Worked solution

Working

\Delta H_r = \Delta H_2 - \Delta H_1

Substitute your values from (d) and (b)(ii) and give the sign.

= ... kJ mol⁻¹

Final answer

See working (candidate-dependent)

Detailed explanation

Background Concept

Hess's law states that the enthalpy change of a reaction is independent of the route taken. Here, the reaction of solid citric acid with sodium hydroxide can be considered as first dissolving the citric acid ( $\Delta H_1$ ) and then reacting the aqueous citric acid with sodium hydroxide ( $\Delta H_r$ ). Therefore $\Delta H_2 = \Delta H_1 + \Delta H_r$ , so $\Delta H_r = \Delta H_2 - \Delta H_1$ .

Understanding the Question

You are asked to use your calculated values of $\Delta H_1$ and $\Delta H_2$ to find the enthalpy change for the reaction of aqueous citric acid with aqueous sodium hydroxide.

Approach

Subtract $\Delta H_1$ from $\Delta H_2$ . Keep the signs of both values. The result is the enthalpy change for the neutralisation of aqueous citric acid.

Step-by-Step Reasoning

For example, if $\Delta H_2 = -49.1 \text{ kJ mol}^{-1}$ and $\Delta H_1 = +100.3 \text{ kJ mol}^{-1}$ , then $\Delta H_r = -49.1 - (+100.3) = -149.4 \text{ kJ mol}^{-1}$ . The negative sign shows the neutralisation is exothermic.

Key Takeaways

Hess's law allows you to combine enthalpy changes by addition and subtraction. Always keep the signs of the individual enthalpy changes.

Common Mistakes

Adding $\Delta H_1$ and $\Delta H_2$ instead of subtracting.
Ignoring the sign of $\Delta H_1$ .
Giving the answer without a sign.

Things to Be Careful About

The order of subtraction matters: $\Delta H_r = \Delta H_2 - \Delta H_1$ , not $\Delta H_1 - \Delta H_2$ . Quote the final value with the correct sign and unit.

Techniques used

apply Hess's law to combine enthalpy changessubtract ΔH1 from ΔH2

Q3MediumQualitative AnalysisPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

(a)

Solutions FB 7 and FB 8 each contain one cation and one anion. All of the ions are listed in the Qualitative analysis notes.

(i)

Carry out the following tests and record your observations in Table 3.1. Use a $1\text{ cm}$ depth of FB 7 or FB 8 in a test-tube for each test.

Table 3.1

test	observations for FB 7	observations for FB 8
Test 1 Add an equal volume of aqueous potassium iodide, then add excess aqueous sodium thiosulfate.
Test 2 Add a small spatula measure of zinc powder. Leave the mixture to stand.
Test 3 Add a few drops of aqueous silver nitrate.
Test 4 Add aqueous ammonia.

6M Similar

DifficultyMedium-Easy

Worked solution

Answer

test	observations for FB 7	observations for FB 8
Test 1	Brown colour; white ppt; ppt soluble in excess / solution turns colourless	Brown/red-brown solution; turns colourless / goes (pale) yellow / pale pink
Test 2	Pink/brown/black solid; solution turns paler blue/colourless	Fizzing/effervescence; gas pops with lighted splint
Test 3	No change / no ppt / no reaction	ppt forms
Test 4	(Pale) blue ppt; soluble in excess; deep/dark blue solution formed	Brown/red-brown/rust ppt; insoluble in excess NH3

Final answer

Completed Table 3.1 with observations for FB7 and FB8.

Detailed explanation

Background Concept

Qualitative analysis uses characteristic chemical reactions to identify ions in solution. Observations include colour changes, formation or dissolution of precipitates, and gas evolution. For transition metal cations like Cu2+ and Fe3+, hydroxide and ammonia give distinctive precipitates and complexes. Redox reactions can also be used: iodide is oxidised to iodine by Cu2+ or Fe3+, and zinc can displace a less reactive metal or reduce H+ to hydrogen.

Understanding the Question

FB7 and FB8 each contain one cation and one anion from the Qualitative Analysis notes. The candidate performs four tests and records observations in Table 3.1. The observations must be precise enough to allow deduction of the ions. The expected identities are Cu(NO3)2 for FB7 and FeCl3 for FB8.

Approach

For each test, predict the reaction of Cu2+ and Fe3+ with the reagent. Use known qualitative tests: iodide reduces Cu2+ to CuI and is oxidised to I2; thiosulfate removes I2; zinc displaces Cu or reacts with acid; silver nitrate precipitates chloride; ammonia forms hydroxide precipitates and complexes.

Step-by-Step Reasoning

Test 1: Add KI. Cu2+ oxidises I- to I2 (brown colour) and forms a white precipitate of CuI. Fe3+ also oxidises I- to I2, giving a brown/red-brown solution. Adding excess sodium thiosulfate reduces I2 back to I-, so the brown colour disappears. For Cu2+, the white CuI ppt remains but may dissolve in excess thiosulfate, so the solution turns colourless.
Test 2: Add zinc powder. With Cu2+, zinc displaces copper metal (pink/brown/black solid) and the blue solution becomes colourless. With FeCl3, the solution is acidic due to hydrolysis; zinc reacts with acid to produce hydrogen gas, seen as fizzing and a pop with a lighted splint.
Test 3: Add silver nitrate. Chloride gives a white precipitate of AgCl; nitrate gives no reaction.
Test 4: Add ammonia. Cu2+ gives a pale blue precipitate of Cu(OH)2, which dissolves in excess ammonia to form a deep blue [Cu(NH3)4]2+ solution. Fe3+ gives a brown/red-brown precipitate of Fe(OH)3, which is insoluble in excess ammonia.

Key Takeaways

Know the characteristic observations for Cu2+ and Fe3+ with KI/thiosulfate, zinc, silver nitrate, and ammonia. Record observations exactly, including colours, precipitate solubility, and gas tests.

Common Mistakes

Confusing the white CuI precipitate with a silver halide.
Forgetting that thiosulfate decolorises iodine, so the brown colour disappears.
Writing "no reaction" for Fe3+ with KI; it does react.
Missing that FeCl3 solution is acidic, so zinc produces hydrogen.

Things to Be Careful About

Use precise colour names (e.g. "pale blue", "brown/red-brown").
State whether a precipitate dissolves in excess reagent.
For gas tests, mention the test used (e.g. lighted splint).
The mark scheme awards marks for each correct observation; include all.

Techniques used

perform qualitative tests on unknown solutionsobserve colour changes and precipitatesrecord observations systematically in a table

(ii)

Deduce the formula of FB 8.

FB 8 is .......................... .

1M Similar

(iii)

For FB 8, state the numbers of all the tests from Table 3.1 that involve a redox reaction.

1M Similar

(iv)

Give the ionic equation for one of the reactions in Test 4. Include state symbols.

1M Similar

(b)

(i)

Carry out tests to identify the anion in FB 7. The anion does not contain sulfur. Record your tests and observations in a suitable form in the space below. You must use a boiling tube if any liquid is heated.

3M Similar

DifficultyMedium

Worked solution

Answer

Test	Observation	Deduction
Add excess NaOH(aq) and aluminium powder; warm	Effervescence; gas turns damp red litmus blue	Ammonia evolved -> nitrate present
Add acidified KMnO4	Remains purple / no change	No reducing anion present (nitrate is not a reducing agent)

Final answer

See table: nitrate identified by ammonia test; acidified KMnO4 remains purple.

Detailed explanation

Background Concept

Nitrate ions can be identified by reduction to ammonia with aluminium and sodium hydroxide. The ammonia gas turns damp red litmus blue. Nitrate does not react with acidified KMnO4, confirming it is not a reducing anion.

Understanding the Question

Carry out tests to identify the anion in FB7, which does not contain sulfur. Record tests and observations. The anion is nitrate.

Approach

Use the standard nitrate test: add NaOH and aluminium, warm, test gas with damp red litmus. Confirm absence of reducing anions with acidified KMnO4.

Step-by-Step Reasoning

Add excess NaOH(aq) and aluminium powder to FB7 and warm. Nitrate is reduced to ammonia. The gas evolved turns damp red litmus blue, indicating ammonia.
Add acidified KMnO4: remains purple, showing no reducing anion (e.g. sulfite, iodide) is present.
Since the anion does not contain sulfur, sulfate is ruled out; no further test needed.

Key Takeaways

Know the nitrate test and how to exclude other anions.

Common Mistakes

Not heating the mixture.
Forgetting to use aluminium.
Confusing ammonia with hydrogen (hydrogen pops, ammonia turns litmus blue).

Things to Be Careful About

Use a boiling tube if heating.
Record observations clearly.

Techniques used

test for nitrate by reduction to ammoniaconfirm absence of reducing anions with acidified KMnO4record tests and observations systematically

(ii)

Deduce the formula of FB 7.

FB 7 is .......................... .

1M Similar