Qualitative analysis For each test you should record all your observations in the spaces provided. Examples of observations include: colour changes seen the formation of any precipitate and its solubility (where appropriate) in an excess of the reagent added the formation of any gas and its identification (where appropriate) by a suitable test. You should record clearly at what stage in a test an observation is made. Where no change is observed, you should write ‘no change’.…

(a)(i) FA7 = AgNO3(aq), FA8 = CH3COOH(aq), FA9 = HCl(aq) · (a)(ii) Mg(s) + 2H+(aq) -> Mg2+(aq) + H2(g) · (b)(i) See table. Underlying salts: FA10 = MgSO4(aq) and FA11 = Zn(NO3)2(aq). · (b)(ii) FA10: Mg2+ and SO4^2-; FA11: Zn2+ and unknown anion

Chemistry 9701/31 — May/June 2024

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

questions

marks

120

minutes

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

Q1MediumPresentation of Data and ObservationsManipulation, Measurement and ObservationAnalysis, Conclusions and Evaluation

In this experiment you will determine the enthalpy change, $\Delta H_r$ , for the reaction shown.

\text{NaOH(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{NaHCO}_3\text{(aq)} \quad \text{enthalpy change} = \Delta H_r

You will react each of sodium hydroxide and sodium hydrogencarbonate with excess dilute sulfuric acid. You will determine the enthalpy change for each reaction, then use Hess’s law to calculate $\Delta H_r$ .

(a)

Reaction of sodium hydroxide with sulfuric acid

2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)} \quad \text{enthalpy change} = \Delta H_1

FA 1 is $2.00\text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$ .
FA 2 is $2.00\text{ mol dm}^{-3}$ sulfuric acid, $\text{H}_2\text{SO}_4$ .

Method

Support a cup in the $250\text{ cm}^3$ beaker.
Use the $50\text{ cm}^3$ measuring cylinder to transfer $30.0\text{ cm}^3$ of FA 1 into the cup.
Place the thermometer in FA 1 and tilt the cup, if necessary, so that the bulb of the thermometer is fully covered. Record the temperature of FA 1.
Use the $25\text{ cm}^3$ measuring cylinder to add $20.0\text{ cm}^3$ of FA 2 to the FA 1 in the cup.
Stir the mixture.
Measure and record the maximum temperature reached.
Calculate and record the change in temperature.

Results

Prepare a table for your results in the space provided.

2M Similar

DifficultyEasy

Worked solution

Answer

Record the following three values, each with the unit °C:

initial temperature of FA 1: 20.0 °C
maximum temperature reached: 30.0 °C
temperature rise: 10.0 °C

(Representative values — the candidate records their own readings; the accuracy mark compares the temperature rise with the supervisor's value.)

Final answer

Table of initial temperature, maximum temperature and temperature rise with units (representative: 20.0 °C, 30.0 °C, 10.0 °C)

Detailed explanation

Background Concept

This is a simple calorimetry experiment. The neutralisation of sodium hydroxide solution with dilute sulfuric acid is exothermic — heat is released, so the temperature of the mixture rises. The key measured quantities are the initial temperature of the NaOH solution (before the acid is added) and the maximum temperature reached after mixing and stirring. The temperature rise is found by subtraction.

Understanding the Question

Part (a) asks you to prepare a results table in the space provided and record three pieces of data: the initial temperature of FA 1, the maximum temperature reached, and the calculated temperature rise. The mark scheme awards one mark for showing these three values clearly with units, and a second accuracy mark for the temperature rise being close to the supervisor's value.

Approach

The method gives the sequence: record the temperature of FA 1, add 20.0 cm³ of FA 2, stir, and record the maximum temperature. The temperature rise is maximum minus initial. Present the three values in a small table with clear headings and the unit °C on every value.

Step-by-Step Reasoning

The first mark (I) requires three pieces of data: two thermometer readings and the correctly calculated temperature rise, all listed with units. Formal headings are not required, but it must be clear what each figure refers to — a simple table with headings such as "initial temperature / °C", "maximum temperature / °C" and "temperature rise / °C" is the safest format. The accuracy mark (II) compares your temperature rise with the supervisor's: if the supervisor's value is ≥ 10 °C, your value must be within 1.5 °C; if between 5 and 10 °C, within 1.0 °C; if below 5 °C, within 0.5 °C. A representative set of readings would be: initial temperature 20.0 °C, maximum temperature 30.0 °C, temperature rise 10.0 °C. The thermometer bulb must be fully covered by the solution to obtain a reliable reading.

Key Takeaways

Always record raw readings with their units.
Show the derived quantity (temperature rise) calculated from the raw readings.
Keep consistent precision (e.g. to 0.5 °C) across all thermometer readings.

Common Mistakes

Forgetting to record the temperature rise (the third piece of data).
Recording readings without units, or with units on only some values.
Recording a thermometer reading below 10 °C — the mark scheme explicitly does not award the mark for this.
Confusing which reading is initial and which is maximum.

Things to Be Careful About

The mark scheme ignores the precision of the readings for mark I, but the accuracy mark depends on the temperature rise matching the supervisor's value. Ensure the thermometer bulb is fully covered by the solution when reading the initial temperature, and stir thoroughly before reading the maximum.

Techniques used

record thermometer readingscalculate temperature risepresent data in a table with units

(b)

Calculations

(i)

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

1M Similar

(ii)

Calculate the amount, in mol, of sulfuric acid that reacted with FA 1 in your experiment.

1M Similar

DifficultyEasy

Worked solution

Working

Moles of NaOH:

n(\text{NaOH}) = 0.0300 \times 2.00 = 0.0600\ \text{mol}

From $2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ , 2 mol NaOH reacts with 1 mol H₂SO₄:

n(\text{H}_2\text{SO}_4) = \frac{0.0600}{2} = 0.0300\ \text{mol}

Answer

0.0300 mol

Final answer

0.0300 mol

Detailed explanation

Background Concept

The amount of H₂SO₄ that reacts is determined by the stoichiometry of the neutralisation equation: $2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$ . Two moles of NaOH react with one mole of H₂SO₄. The NaOH is the limiting reagent here because the acid is added in excess.

Understanding the Question

Calculate the amount, in mol, of sulfuric acid that reacted with FA 1 in your experiment.

Approach

First find the moles of NaOH from its concentration and volume, then use the 2:1 stoichiometric ratio to find the moles of H₂SO₄.

Step-by-Step Reasoning

Volume of FA 1 = 30.0 cm³ = 0.0300 dm³. Moles NaOH = $0.0300 \times 2.00 = 0.0600$ mol. From the equation, 2 mol NaOH reacts with 1 mol H₂SO₄, so moles H₂SO₄ = $0.0600/2 = 0.0300$ mol. The mark scheme writes this compactly as 0.030 × 2 × 0.5 = 0.03 mol.

Key Takeaways

Convert cm³ to dm³ (divide by 1000) before using concentration in mol dm⁻³.
Apply the stoichiometric ratio from the balanced equation.

Common Mistakes

Using the volume of acid (20.0 cm³) instead of the volume of NaOH (30.0 cm³).
Forgetting to halve the moles of NaOH for the 2:1 ratio.

Things to Be Careful About

The answer is the amount of acid that actually reacts, which is set by the limiting reagent (NaOH), not the total acid added.

Techniques used

calculate moles from concentration and volumeapply stoichiometric ratio

(iii)

Calculate the enthalpy change of reaction, $\Delta H_1$ , in $\text{kJ mol}^{-1}$ of sulfuric acid, for the neutralisation of $\text{NaOH(aq)}$ with $\text{H}_2\text{SO}_4\text{(aq)}$ .

Show your working.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\Delta H_1 = \frac{2090}{1000 \times 0.0300} = -69.7\ \text{kJ mol}^{-1}

Answer

ΔH₁ = −69.7 kJ mol⁻¹ (negative sign, 2–4 s.f.)

Final answer

-69.7 kJ mol^-1

Detailed explanation

Background Concept

The enthalpy change of reaction per mole of a specified substance is the energy change divided by the number of moles of that substance. Here it is per mole of H₂SO₄, so $\Delta H_1 = \text{energy}/(1000 \times \text{moles H}_2\text{SO}_4)$ , with the division by 1000 converting J to kJ. Because neutralisation is exothermic (heat is released), $\Delta H_1$ is negative.

Understanding the Question

Calculate $\Delta H_1$ in kJ mol⁻¹ of sulfuric acid, showing your working. You combine the energy from part (b)(i) and the moles from part (b)(ii).

Approach

Divide the energy (in J) by the moles of H₂SO₄, convert to kJ by dividing by 1000, and attach a negative sign.

Step-by-Step Reasoning

$\Delta H_1 = 2090/(1000 \times 0.0300) = 2090/30.0 = 69.7$ kJ mol⁻¹, so $\Delta H_1 = -69.7$ kJ mol⁻¹. The mark scheme requires the negative sign and the answer to 2–4 significant figures.

Key Takeaways

Exothermic reactions have negative ΔH.
The question specifies per mole of H₂SO₄, so use moles of acid, not moles of NaOH.

Common Mistakes

Missing the negative sign — the mark scheme explicitly requires it.
Using moles of NaOH instead of moles of H₂SO₄.
Forgetting to divide by 1000 to convert J to kJ.

Things to Be Careful About

The mark scheme requires BOTH the negative sign AND 2–4 significant figures. Show the working so the method mark can be awarded even if the arithmetic is slightly off.

Techniques used

divide energy by moles to find enthalpy changeapply sign convention for exothermic reaction

(c)

Reaction of sodium hydrogencarbonate with sulfuric acid

2\text{NaHCO}_3\text{(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)} + 2\text{CO}_2\text{(g)} \quad \text{enthalpy change} = \Delta H_2

FA 2 is $2.00\text{ mol dm}^{-3}$ sulfuric acid, $\text{H}_2\text{SO}_4$ .
FA 3 is sodium hydrogencarbonate, $\text{NaHCO}_3$ .

Method

Support the other cup in the $250\text{ cm}^3$ beaker.
Use the $25\text{ cm}^3$ measuring cylinder to transfer $25.0\text{ cm}^3$ of FA 2 into the cup.
Place the thermometer in FA 2 and tilt the cup, if necessary, so that the bulb of the thermometer is fully covered. Record the temperature of FA 2.
Weigh the container with FA 3. Record the mass.
Adding small quantities at a time, tip all the FA 3 from the container into the FA 2 in the cup.
Stir the mixture.
Measure and record the minimum temperature reached.
Calculate and record the change in temperature.
Weigh the container with any residual FA 3. Record the mass.
Calculate and record the mass of FA 3 added.

Results

Prepare a table for your results in the space provided.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Record the following six values in a table, each with the correct unit:

mass of container + FA 3 / g: 12.34
mass of container (with residual FA 3) / g: 6.50
mass of FA 3 added / g: 5.84
initial temperature of FA 2 / °C: 20.0
minimum temperature / °C: 10.5
temperature fall / °C: 9.5

(Representative values — the candidate records their own readings. Masses to 2 or 3 d.p., temperatures to .0 or .5.)

Final answer

Six-heading table with units (representative: masses 12.34 g, 6.50 g, 5.84 g; temperatures 20.0 °C, 10.5 °C, fall 9.5 °C)

Detailed explanation

Background Concept

The reaction of sodium hydrogencarbonate with sulfuric acid is endothermic — it absorbs heat from the surroundings, so the temperature of the mixture falls. The experiment measures the minimum temperature reached. The mass of NaHCO₃ actually added is found by difference: weigh the container with FA 3 before adding, then reweigh the container with any residual solid after, and subtract.

Understanding the Question

Prepare a results table recording six pieces of data: mass of container + FA 3, mass of container (with residual FA 3), mass of FA 3 added, initial temperature of FA 2, minimum temperature, and temperature fall.

Approach

Follow the method: weigh the container with FA 3, record the initial temperature of the acid, add the solid in small portions with stirring, record the minimum temperature, reweigh the container, then calculate the mass of solid added and the temperature fall. Present all six values in a table with units.

Step-by-Step Reasoning

The mark scheme awards mark I for six correct headings with units: (mass of) container + FA 3; (mass of) container; (mass of) FA 3 added; initial temperature; minimum/final temperature; temperature change. Mark II requires all four balance/thermometer readings shown to appropriate precision — temperatures to .0 or .5, masses to 2 or 3 decimal places consistently. Mark III requires the subtractions to be correct: mass of solid between 5.00–7.00 g and temperature fall between 8.0–12.0 °C. Representative values: container + FA 3 = 12.34 g, container residual = 6.50 g, so mass FA 3 = 5.84 g; initial temperature 20.0 °C, minimum 10.5 °C, so fall 9.5 °C.

Key Takeaways

Mass by difference is the standard technique for finding how much solid was added.
Endothermic reactions show a temperature fall (minimum temperature).
Consistency of precision across readings is part of the mark.

Common Mistakes

Missing one of the six headings or their units.
Inconsistent decimal places in the masses.
Recording the temperature fall as a rise (wrong sign).

Things to Be Careful About

All four thermometer readings across parts (a) and (c) must be to .0 or .5. Both masses must be to the same number of decimal places (either both 2 or both 3).

Techniques used

record masses and temperaturescalculate mass by differencepresent data with headings and units

(d)

Use your data to calculate the enthalpy change, $\Delta H_2$ , in $\text{kJ mol}^{-1}$ of sulfuric acid, for the reaction of $\text{NaHCO}_3\text{(s)}$ with $\text{H}_2\text{SO}_4\text{(aq)}$ .

Show your working.

3M Similar

DifficultyMedium

Worked solution

Working

Energy absorbed:

q = 25 \times 4.18 \times 9.5 = 993\ \text{J}

Moles of NaHCO₃ added:

n(\text{NaHCO}_3) = \frac{5.84}{84} = 0.0695\ \text{mol}

From $2\text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + 2\text{CO}_2$ , moles H₂SO₄ = 0.5 × 0.0695 = 0.0348 mol:

\Delta H_2 = \frac{993}{1000 \times 0.0348} = +28.6\ \text{kJ mol}^{-1}

Answer

ΔH₂ = +28.6 kJ mol⁻¹ (positive sign, ≥ 2 s.f.)

Final answer

+28.6 kJ mol^-1

Detailed explanation

Background Concept

The same calorimetry principles apply as in part (b), but now the reaction is endothermic, so the temperature falls and $\Delta H_2$ is positive. The energy absorbed is $q = mc\Delta T$ with $m = 25$ g (25.0 cm³ of acid solution). The amount of H₂SO₄ that reacts is found from the mass of NaHCO₃ added: the equation $2\text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + 2\text{CO}_2$ shows 2 mol NaHCO₃ per 1 mol H₂SO₄, and $M_r(\text{NaHCO}_3) = 84$ .

Understanding the Question

Calculate $\Delta H_2$ in kJ mol⁻¹ of H₂SO₄, showing your working. Three method marks are available: energy, moles of acid, and the enthalpy division.

Approach

M1: energy = 25 × 4.18 × temperature fall. M2: moles H₂SO₄ = 0.5 × (mass NaHCO₃/84). M3: $\Delta H_2$ = energy/(1000 × moles H₂SO₄), with a positive sign.

Step-by-Step Reasoning

Using the representative values (mass NaHCO₃ = 5.84 g, temperature fall = 9.5 °C):
Energy = $25 \times 4.18 \times 9.5 = 993$ J.
Moles NaHCO₃ = $5.84/84 = 0.0695$ mol; moles H₂SO₄ = $0.5 \times 0.0695 = 0.0348$ mol.
$\Delta H_2 = 993/(1000 \times 0.0348) = +28.6$ kJ mol⁻¹.
The mark scheme allows the expressions for M1 and M2 to be unevaluated, but M3 requires the positive sign and the answer to two or more significant figures.

Key Takeaways

The mass term is 25 g because only 25.0 cm³ of solution is present.
The positive sign reflects the endothermic reaction.
The stoichiometric ratio halves the moles of NaHCO₃ to get moles of acid.

Common Mistakes

Using 50 g instead of 25 g for the mass.
Using moles of NaHCO₃ instead of moles of H₂SO₄ in the final division.
Forgetting the positive sign.

Things to Be Careful About

The mark scheme requires the positive sign and ≥ 2 s.f. for the final answer. Show all three steps so the method marks are secure.

Techniques used

apply q = mcΔTcalculate moles from mass and molar massapply stoichiometric ratiocompute enthalpy change with sign

(e)

The enthalpy change when one mole of sodium hydrogencarbonate dissolves in water is $\Delta H_3$ .

\text{NaHCO}_3\text{(s)} + \text{aq} \rightarrow \text{NaHCO}_3\text{(aq)} \quad \text{enthalpy change} = \Delta H_3

Using the symbols $\Delta H_1$ , $\Delta H_2$ and $\Delta H_3$ in your answer, use Hess’s law to deduce an expression for $\Delta H_r$ .

\text{NaOH(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{NaHCO}_3\text{(aq)} \quad \text{enthalpy change} = \Delta H_r

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\Delta H_r = 0.5\Delta H_1 - 0.5\Delta H_2 + \Delta H_3

Final answer

ΔH_r = 0.5ΔH_1 − 0.5ΔH_2 + ΔH_3

Detailed explanation

Background Concept

Hess's law states that the enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same. We can combine the three given reactions (with scaling and reversal as needed) to construct the target reaction, and sum the corresponding enthalpy changes.

Understanding the Question

Express $\Delta H_r$ in terms of $\Delta H_1$ , $\Delta H_2$ and $\Delta H_3$ . The target reaction is $\text{NaOH(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{NaHCO}_3\text{(aq)}$ .

Approach

Build the target reaction from the given ones: take 0.5 × reaction (1), reverse reaction (2) and take 0.5 × it, and add reaction (3). Then sum the enthalpy changes with the appropriate signs.

Step-by-Step Reasoning

0.5 × (1): $0.5 \times [2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}]$ gives $\text{NaOH} + 0.5\text{H}_2\text{SO}_4 \rightarrow 0.5\text{Na}_2\text{SO}_4 + \text{H}_2\text{O}$ , with $0.5\Delta H_1$ .
−0.5 × (2): reversing and halving gives $0.5\text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \rightarrow \text{NaHCO}_3 + 0.5\text{H}_2\text{SO}_4$ , with $-0.5\Delta H_2$ .

(3): $\text{NaHCO}_3\text{(s)} + \text{aq} \rightarrow \text{NaHCO}_3\text{(aq)}$ , with $\Delta H_3$ .
Adding these three, the intermediate species (H₂SO₄, Na₂SO₄, H₂O, NaHCO₃(s)) cancel, leaving $\text{NaOH(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{NaHCO}_3\text{(aq)}$ . Therefore $\Delta H_r = 0.5\Delta H_1 - 0.5\Delta H_2 + \Delta H_3$ .

Key Takeaways

Reversing a reaction flips the sign of its ΔH.
Multiplying a reaction by a factor multiplies its ΔH by the same factor.
Intermediate species must cancel for the combination to be valid.

Common Mistakes

Forgetting to halve $\Delta H_1$ and $\Delta H_2$ (the given equations are for 2 mol of NaOH/NaHCO₃).
Using a plus sign before $0.5\Delta H_2$ instead of a minus (because reaction (2) is reversed).
Omitting $\Delta H_3$ .

Things to Be Careful About

The answer must use exactly the symbols $\Delta H_1$ , $\Delta H_2$ , $\Delta H_3$ as instructed.

Techniques used

apply Hess's lawcombine enthalpy changes with scaling and reversal

(f)

A student suggested that the experiment in (c) would be more accurate if $25.0\text{ cm}^3$ of $3.00\text{ mol dm}^{-3}$ sulfuric acid was used instead of $25.0\text{ cm}^3$ of $2.00\text{ mol dm}^{-3}$ sulfuric acid.

State whether the student’s suggestion is correct.

Explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The student is not correct. In experiment (c) the sulfuric acid is already in excess, so using 3.00 mol dm⁻³ acid would not change the amount of acid that reacts; the temperature fall (and hence ΔH₂) would be unchanged.

Final answer

Not correct — acid is already in excess, so temperature fall is unchanged

Detailed explanation

Background Concept

In a limiting-reagent situation, the extent of reaction is set by the limiting reagent. In experiment (c), the acid is added in excess, so the amount of H₂SO₄ that reacts is determined by the amount of NaHCO₃, not by the acid concentration.

Understanding the Question

Evaluate whether using 25.0 cm³ of 3.00 mol dm⁻³ H₂SO₄ instead of 25.0 cm³ of 2.00 mol dm⁻³ would make the experiment more accurate.

Approach

Check whether the acid is already in excess in the original experiment. If it is, increasing its concentration changes nothing about the amount that reacts or the temperature fall.

Step-by-Step Reasoning

In experiment (c): 25.0 cm³ of 2.00 mol dm⁻³ acid contains $0.0250 \times 2.00 = 0.050$ mol H₂SO₄. The ~5.84 g of NaHCO₃ is $5.84/84 = 0.0695$ mol, which needs $0.0348$ mol H₂SO₄ (half, from the 2:1 ratio). Since 0.050 > 0.0348, the acid is already in excess. Using 3.00 mol dm⁻³ would give 0.075 mol, still in excess; the same amount of acid reacts, so the temperature fall is unchanged. The student's suggestion is therefore not correct.

Key Takeaways

Adding more of an already-excess reagent does not change the extent of reaction.
The temperature change depends on the amount of limiting reagent, not on the excess.

Common Mistakes

Saying the suggestion is correct.
Arguing that more concentrated acid releases more heat, without recognising the acid is already in excess.

Things to Be Careful About

The mark scheme's key point is that the acid is used in excess in experiment (c), so the temperature fall is unchanged.

Techniques used

evaluate experimental modificationidentify limiting reagent and excess

Q2Medium-EasyManipulation, Measurement and ObservationPresentation of Data and Observations

Sodium sulfite is oxidised when it reacts with excess iodine.

\text{Na}_2\text{SO}_3\text{(aq)} + \text{I}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{HI(aq)}

The remaining iodine is then titrated using aqueous sodium thiosulfate.

\text{I}_2\text{(aq)} + 2\text{Na}_2\text{S}_2\text{O}_3\text{(aq)} \rightarrow \text{Na}_2\text{S}_4\text{O}_6\text{(aq)} + 2\text{NaI(aq)}

You will determine the integer value of $x$ in the formula of hydrated sodium sulfite, $\text{Na}_2\text{SO}_3\cdot x\text{H}_2\text{O}$ , by titration.

FA 4 is aqueous sodium thiosulfate containing $14.24\text{ g}$ of $\text{Na}_2\text{S}_2\text{O}_3$ in $1.00\text{ dm}^3$ .

FA 5 is aqueous iodine, prepared as shown.

$5.00\text{ g}$ of hydrated sodium sulfite is added to $600\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ aqueous iodine.
The mixture is allowed to stand to ensure that all the sodium sulfite has been oxidised.
The mixture containing the remaining iodine is made up to $1.00\text{ dm}^3$ with distilled water.

FA 6 is starch indicator.

(a)

Method

Fill the burette with FA 4.
Pipette $25.0\text{ cm}^3$ of FA 5 into a conical flask.
Add FA 4 from the burette into the conical flask until the colour of the solution changes to yellow.
Add 10 drops of FA 6 to the conical flask. Continue titrating until the blue-black colour just disappears.
Perform a rough titration and record your burette readings in the space below.

Record, in a suitable form in the space below, all your burette readings and the volume of FA 4 added in each accurate titration.

7M Similar

DifficultyMedium-Easy

Worked solution

Answer

Example table using your own readings:

Titration	rough	1	2
final burette reading / cm3	22.55	22.30	44.65
initial burette reading / cm3	0.00	0.00	22.30
titre / cm3	22.55	22.30	22.35

All burette readings are recorded to 0.05 cm3. The accurate titres 22.30 cm3 and 22.35 cm3 are concordant (within 0.10 cm3).

Final answer

Candidate-dependent; example table with rough and two accurate titres (22.30 and 22.35 cm3).

Detailed explanation

Background Concept

In an iodine-thiosulfate titration, iodine is titrated with sodium thiosulfate. Starch is added near the end point because it forms an intense blue-black complex with iodine; the end point is when the blue-black colour just disappears. Reliable results require readings to 0.05 cm3, at least two accurate titrations agreeing within 0.10 cm3, and a clear record with headings and units.

Understanding the Question

This part asks you to perform the titration and record all burette readings in a suitable table. The marks are for the quality of the recorded data: rough readings, initial and final readings for accurate titrations, the titre values, correct headings with units, readings to 0.05 cm3, and concordant accurate titres.

Approach

Fill the burette with FA 4, pipette 25.0 cm3 of FA 5 into a conical flask, and titrate until the solution turns yellow. Add starch and continue until the blue-black colour just disappears. Do a rough titration first, then repeat accurately. Record all readings in a table with columns for rough, 1 and 2, and rows for final reading, initial reading and titre.

Step-by-Step Reasoning

Read the burette to the nearest 0.05 cm3.
The rough titre gives an approximate volume, so the accurate titrations can be done quickly at first and dropwise near the end.
Starch is added only when the solution is yellow; adding it earlier can make the end point less sharp.
A suitable example set is: rough titre 22.55 cm3; accurate titres 22.30 cm3 and 22.35 cm3. These agree within 0.10 cm3, so they are concordant.
The table must show initial and final readings separately, not just the titre, and every heading must include the unit / cm3.

Key Takeaways

A good titration record is complete, tidy and unambiguous: all readings, units, precision to 0.05 cm3, and concordant accurate titres.

Common Mistakes

Omitting the rough titration readings.
Writing headings without units, e.g. "volume" instead of "volume / cm3".
Recording readings to 0.1 cm3 instead of 0.05 cm3.
Using accurate titres that differ by more than 0.10 cm3.
Recording only the titre and not the initial and final readings.

Things to Be Careful About

Read the burette at eye level, from the bottom of the meniscus.
Record all volumes in cm3.
Do not include the rough titre when calculating the mean in part (b).

Techniques used

fill and read a burette accuratelyperform a rough titration and accurate titrationsrecord readings in a table with headings and unitsselect concordant accurate titres

(b)

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

1M Similar

(c)

Calculations

(i)

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

1M Similar

(ii)

Calculate the amount, in mol, of sodium thiosulfate present in the volume of FA 4 in (b).

1M Similar

(iii)

Calculate the amount, in mol, of iodine in $1.00\text{ dm}^3$ of FA 5.

1M Similar

(iv)

Use the information given and your answer to (c)(iii) to calculate the amount, in mol, of iodine that reacted with sodium sulfite when solution FA 5 was prepared.

1M Similar

(v)

Use your answer to (c)(iv) to calculate the relative formula mass, $M_r$ , of hydrated sodium sulfite.

1M Similar

(vi)

Calculate the value of $x$ .

Show your working.

1M Similar

Q3MediumQualitative AnalysisManipulation, Measurement and ObservationAnalysis, Conclusions and Evaluation

Qualitative analysis

For each test you should record all your observations in the spaces provided.

Examples of observations include:

colour changes seen
the formation of any precipitate and its solubility (where appropriate) in an excess of the reagent added
the formation of any gas and its identification (where appropriate) by a suitable test.

You should record clearly at what stage in a test an observation is made.

Where no change is observed, you should write ‘no change’.

Where reagents are selected for use in a test, the name or correct formula of the element or compound must be given.

If any solution is warmed, a boiling tube must be used. If a solid is heated, a hard-glass test-tube must be used.

Rinse and reuse test-tubes and boiling tubes where possible.

No additional tests should be attempted.

(a)

FA 7, FA 8 and FA 9 are dilute ethanoic acid, dilute hydrochloric acid and aqueous silver nitrate but not necessarily in that order. The solutions of acids have equal concentrations.

(i)

You are supplied with strips of magnesium ribbon. You must not use any other reagents in this part of the question.

Carry out tests to identify each of the three solutions, FA 7, FA 8 and FA 9.
Obtain as much evidence as you can for your identifications.
Use a $1\text{ cm}$ depth of solution in a test-tube for each test you carry out.

Record all your observations.

5M Similar

DifficultyMedium

Worked solution

Answer

Use a $1\text{ cm}$ depth of each solution and a strip of magnesium ribbon.

Solution	Observation with magnesium ribbon
FA7	Dark grey/black solid (precipitate/deposit/coating/layer) forms on the magnesium; no effervescence.
FA8	Effervescence/fizzing; reaction is slower/less vigorous than with FA9. Gas pops with a lighted splint; magnesium dissolves, a colourless solution forms, and heat is produced.
FA9	Effervescence/fizzing; reaction is faster/more vigorous than with FA8. Gas pops with a lighted splint; magnesium dissolves, a colourless solution forms, and heat is produced.

Mixing FA7 with FA9 gives a white precipitate.

Identities: FA7 = $\text{AgNO}_3(\text{aq})$ , FA8 = $\text{CH}_3\text{COOH}(\text{aq})$ , FA9 = $\text{HCl}(\text{aq})$ .

Final answer

FA7 = AgNO3(aq), FA8 = CH3COOH(aq), FA9 = HCl(aq)

Detailed explanation

Background Concept

Magnesium is a reactive metal. In acid it is oxidised to $\text{Mg}^{2+}$ and hydrogen ions are reduced to hydrogen gas:

\text{Mg}(\text{s}) + 2\text{H}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2(\text{g})

With a salt of a less reactive metal, such as silver nitrate, magnesium instead reduces the metal ion, depositing metallic silver. Because the two acids have the same concentration, the only difference between them is their degree of ionisation: hydrochloric acid is a strong acid (fully ionised, so high $[\text{H}^+]$ ) whereas ethanoic acid is a weak acid (partially ionised, so lower $[\text{H}^+]$ ). Thus the acid with the higher $[\text{H}^+]$ reacts faster with magnesium.

Understanding the Question

Three unlabelled solutions have been provided; they are silver nitrate, ethanoic acid and hydrochloric acid, all of the same concentration. You may not use any reagent other than strips of magnesium. The task is to record observations that allow each solution to be identified. The question expects observations in the order in which they happen: the appearance of a solid deposit, any effervescence, and the relative vigour of the reaction.

Approach

Test each solution separately with a small strip of magnesium. Look for three things:

Is gas evolved? If not, and a dark solid forms, the solution is silver nitrate.
Is gas evolved in both of the remaining solutions? Then both are acids.
Which acid reacts faster? The faster one is hydrochloric acid; the slower is ethanoic acid.
Finally, test any evolved gas with a lighted splint and, if convenient, mix the silver nitrate solution with the hydrochloric acid solution to confirm the white silver chloride precipitate. Do not use any other reagent.

Step-by-Step Reasoning

Add magnesium to FA7. With silver nitrate the reducing action of magnesium converts $\text{Ag}^+$ to metallic silver, seen as a dark grey/black solid on the ribbon. No hydrogen is produced, so there is no effervescence. This identifies FA7 as silver nitrate.
Add magnesium to FA8 and FA9. Both are acids, so both fizz as hydrogen is produced. The gas can be confirmed by a lighted splint: a small pop. The magnesium also dissolves and a colourless solution forms, with heat produced.
Compare the two acid reactions. Equal concentrations mean the strong acid, HCl (FA9), provides more $\text{H}^+$ ions and reacts much faster/more vigorously than the weak acid, ethanoic acid (FA8). This is the key comparison.
Optional confirmation: mixing FA7 with FA9 gives a white precipitate of silver chloride, $\text{AgCl}$ , confirming both the silver nitrate and the hydrochloric acid.

Every observation in the mark scheme is explicitly awarded: the dark solid for FA7, effervescence in both acids, the rate comparison, and at least one additional confirmation such as the splint pop, dissolution/heat, or white precipitate on mixing.

Key Takeaways

Reactive metals can displace less reactive metals from their salts.
Equimolar concentrations of a strong acid and a weak acid have different $[\text{H}^+]$ and therefore different reaction rates with a metal.
A single reagent can be enough if you plan the sequence of observations: gas, solid, and rate comparison.

Common Mistakes

Writing only 'black' for FA7 without specifying that a solid/precipitate/deposit/coating forms; the mark scheme requires the physical form to be stated.
Recording 'effervescence in both acids' without comparing the rates of FA8 and FA9.
Using extra reagents such as universal indicator; the question forbids additional tests.
Forgetting to record the splint pop, which is the direct test for hydrogen.

Things to Be Careful About

Record 'no effervescence' for FA7; omitting the negative observation loses the contrast.
In the rate comparison, phrase it clearly as FA9 faster/more vigorous than FA8 (or the reverse, ORA).
State at what stage the observation is made, for example 'on adding magnesium' rather than a vague later description.
If a solution is warmed, use a boiling tube; if a solid is heated, use a hard-glass test-tube. Here no heating is needed.

Techniques used

compare rates of effervescencetest gas with a lighted splintidentify solutions solely from observationsconfirm silver ions by mixing solutions

(ii)

Give the ionic equation for the reaction of magnesium with FA 8. Include state symbols.

1M Similar

DifficultyEasy

Worked solution

Answer

\text{Mg}(\text{s}) + 2\text{H}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2(\text{g})

Final answer

Mg(s) + 2H+(aq) -> Mg2+(aq) + H2(g)

Detailed explanation

Background Concept

An ionic equation shows only the species actually taking part in a reaction, excluding spectator ions. For magnesium and an acid, the acid provides hydrogen ions that are reduced to hydrogen gas, while magnesium is oxidised to magnesium ions. In ethanoic acid the acetate ion does not take part in the reaction, so it is omitted.

Understanding the Question

FA8 is ethanoic acid. The question asks for the net ionic equation for the reaction of magnesium with this acid and explicitly requires state symbols. You need to show the metal, the hydrogen ion, the magnesium ion and the hydrogen gas.

Approach

Start from the full balanced equation with ethanoic acid, cancel the spectator acetate ions, and make sure atoms and charge balance.

Step-by-Step Reasoning

The full reaction is:

\text{Mg}(\text{s}) + 2\text{CH}_3\text{COOH}(\text{aq}) \rightarrow (\text{CH}_3\text{COO})_2\text{Mg}(\text{aq}) + \text{H}_2(\text{g})

The acetate ions, $\text{CH}_3\text{COO}^-$ , appear on both sides and are spectators; remove them.
The remaining species give:

\text{Mg}(\text{s}) + 2\text{H}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2(\text{g})

Both atoms and charge balance: left has 2+ charge from two hydrogen ions; right has 2+ charge from one magnesium ion.

Key Takeaways

Ionic equations omit spectator ions; only species that change are shown.
The reacting species in every acid are hydrogen ions, so the same net ionic equation applies to hydrochloric acid, sulfuric acid and ethanoic acid.
State symbols are required and carry marks.

Common Mistakes

Including $\text{CH}_3\text{COO}^-$ in the ionic equation.
Forgetting the coefficient 2 in front of $\text{H}^+$ .
Omitting state symbols, particularly (s) for magnesium and (g) for hydrogen.
Writing $\text{Mg}^{2+}$ without the 2+ charge.

Things to Be Careful About

The mark scheme allows only the correct ionic equation; an unbalanced equation does not score.
Make sure the number of electrons is conserved: Mg loses 2 electrons, but they are not written in the net ionic equation because they are transferred to 2H+.
Use correct notation: $\text{H}_2$ , not H or H2 without subscript.

Techniques used

write a balanced ionic equationinclude state symbolsomit spectator ions

(b)

FA 10 and FA 11 are both aqueous solutions of salts, each of which contains one cation and one anion listed in the Qualitative analysis notes.

(i)

Carry out the following tests and record your observations in Table 3.1.

For Tests 1 and 2, use a $1\text{ cm}$ depth of FA 10 or FA 11 in a test-tube.
For Test 3, use a $1\text{ cm}$ depth of FA 10 or FA 11 in a boiling tube.

Table 3.1

test	observations for FA 10	observations for FA 11
Test 1 Add aqueous ammonia.
Test 2 Add a few drops of aqueous barium chloride or aqueous barium nitrate, then
add dilute hydrochloric acid.
Test 3 Add aqueous sodium hydroxide, then
warm the mixture carefully, then
add one piece of aluminium foil.

5M Similar

DifficultyMedium

Worked solution

Answer

Test	FA10	FA11
Test 1 – add aqueous ammonia	White precipitate; precipitate is insoluble in excess ammonia.	White precipitate; precipitate dissolves in excess ammonia to give a colourless solution.
Test 2 – add a few drops of aqueous barium chloride (or barium nitrate), then dilute hydrochloric acid	White precipitate forms; precipitate is insoluble / no change after adding dilute HCl.	No reaction / no change / no precipitate; no reaction / no change / no precipitate after adding dilute HCl.
Test 3 – add aqueous sodium hydroxide, warm, then add aluminium foil	NaOH: white precipitate, insoluble in excess. On warming: no change. After adding Al: effervescence; gas pops with a lighted splint (or gas does not turn red litmus blue).	NaOH: white precipitate, soluble in excess to give a colourless solution. On warming: no change. After adding Al: effervescence; gas turns red litmus blue.

Final answer

See table. Underlying salts: FA10 = MgSO4(aq) and FA11 = Zn(NO3)2(aq).

Detailed explanation

Background Concept

Aqueous ammonia and sodium hydroxide both precipitate insoluble metal hydroxides. The important difference is solubility in excess: $\text{Mg(OH)}_2$ is not amphoteric and stays as a white precipitate, while $\text{Zn(OH)}_2$ dissolves in excess ammonia (forming the complex $[\text{Zn(NH}_3)_4]^{2+}$ ) and in excess hydroxide (forming the zincate ion). This distinguishes $\text{Mg}^{2+}$ from $\text{Zn}^{2+}$ .
Barium ions precipitate sulfate as white, acid-insoluble barium sulfate:

\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})

Nitrate gives no such precipitate with barium ions.
Finally, aluminium reduces nitrate ions in hot alkaline solution to ammonia, which turns red litmus blue. Aluminium also reacts with sodium hydroxide to give hydrogen, so both solutions may effervesce when aluminium is added; the gas test tells you whether the gas is hydrogen (pops, does not affect litmus) or ammonia (turns red litmus blue).

Understanding the Question

Two salt solutions, FA10 and FA11, each contain one cation and one anion. You are asked to carry out three tests and record observations in a table. Each test has two parts and each mark generally requires two observations: e.g. 'white precipitate' and 'insoluble in excess'.

Approach

Use the ammonia and sodium hydroxide tests to identify the cation by the solubility of the hydroxide in excess. Use barium chloride and dilute hydrochloric acid to test specifically for sulfate. Use the sodium hydroxide/aluminium test to detect nitrate through ammonia gas, while remembering that hydrogen is also evolved from aluminium with alkali.

Step-by-Step Reasoning

Test 1 (ammonia): Both FA10 and FA11 give a white precipitate because both $\text{Mg}^{2+}$ and $\text{Zn}^{2+}$ form insoluble hydroxides. In excess ammonia, FA10's precipitate remains, so FA10 contains $\text{Mg}^{2+}$ . FA11's precipitate dissolves, showing $\text{Zn}^{2+}$ .
Test 2 (barium chloride then acid): FA10 gives a white precipitate that does not dissolve in dilute HCl, confirming sulfate as the anion. FA11 gives no precipitate with barium ions, so sulfate is absent.
Test 3 (sodium hydroxide): FA10 gives a white precipitate insoluble in excess NaOH, consistent with $\text{Mg}^{2+}$ . FA11 gives a white precipitate that dissolves in excess NaOH, consistent with $\text{Zn}^{2+}$ . On warming, nothing is observed. When aluminium foil is added, both solutions effervesce because aluminium reacts with NaOH to produce hydrogen. For FA10 the gas pops with a lighted splint and does not turn red litmus blue; for FA11 the gas also makes the mixture fizz but turns red litmus blue, showing ammonia is present, which is the classic nitrate test.

Record every observation clearly and at the correct stage.

Key Takeaways

Solubility of hydroxide in excess ammonia or sodium hydroxide is a reliable cation test for amphoteric hydroxides.
Barium chloride/nitrate followed by dilute acid is the standard confirmatory test for sulfate.
Aluminium + sodium hydroxide is the test for nitrate: ammonia is evolved and turns red litmus blue. Hydrogen from Al + NaOH causes effervescence in all cases, so test the gas.

Common Mistakes

Writing only 'white precipitate' without stating whether it is soluble in excess; both observations are needed.
Stating that FA11 gives no reaction with barium chloride but omitting 'no change/no precipitate' after adding the acid.
Confusing the two gas observations: both may fizz, but only nitrate produces ammonia; the gas test is the discriminator.
Treating 'heat' as a separate observation row when no positive gas test occurs; the question includes it, but no ammonia is given before aluminium is added.

Things to Be Careful About

When stating excess reagent, indicate at which stage the observation is made.
For the barium test, the acid is added after the precipitate forms; record the result both before and after acid.
Red litmus turns blue only in the presence of ammonia; a splint pop confirms hydrogen.
In this paper, FA10 is magnesium sulfate and FA11 is zinc nitrate, but the candidate must derive this solely from the observations recorded.

Techniques used

test cations with aqueous ammoniatest for sulfate with barium chloride/nitrate and dilute acidtest for nitrate with aluminium and sodium hydroxidetest gases with a lighted splint and red litmus

(ii)

Use your observations in (b)(i) to complete Table 3.2 by identifying the formulae of the ions present in FA 10 and FA 11.

If you cannot identify an ion write ‘unknown’.

Table 3.2

	cation	anion
FA 10
FA 11

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

	cation	anion
FA10	$\text{Mg}^{2+}$	$\text{SO}_4^{2-}$
FA11	$\text{Zn}^{2+}$	unknown

Final answer

FA10: Mg2+ and SO4^2-; FA11: Zn2+ and unknown anion

Detailed explanation

Background Concept

The identity of an ion is deduced from the observations made with specific reagents. A cation is identified by the colour and solubility of the hydroxide in excess ammonia/sodium hydroxide. An anion is identified by a specific precipitate test, such as the white acid-insoluble barium sulfate precipitate.

Understanding the Question

Using the observations recorded in part (b)(i), write the formulae of the cation and anion in each solution. If an ion cannot be identified, write 'unknown'. The mark scheme requires the formula with its charge, not just the name.

Approach

Cation: compare the results with ammonia and excess sodium hydroxide.
Anion: apply the barium chloride/acid result; a white precipitate insoluble in acid means sulfate.
For FA11, the anion is not confirmed by the given tests, so it is 'unknown' in this mark scheme.

Step-by-Step Reasoning

FA10 gave a white precipitate with both ammonia and NaOH, and the precipitate was insoluble in excess of both reagents. This matches magnesium hydroxide, $\text{Mg(OH)}_2$ , so the cation is $\text{Mg}^{2+}$ .
FA10 gave a white precipitate with barium chloride that remained after adding HCl. This is barium sulfate, so the anion is $\text{SO}_4^{2-}$ .
FA11 gave a white precipitate with ammonia and NaOH that dissolved in excess of either reagent. This is characteristic of zinc hydroxide, so the cation is $\text{Zn}^{2+}$ .
FA11 gave no precipitate with barium chloride, so sulfate is absent. Although the sodium hydroxide/aluminium gas test may suggest nitrate, this mark scheme asks for 'unknown' for the anion of FA11.

Key Takeaways

Magnesium hydroxide is insoluble in excess; zinc hydroxide is amphoteric and dissolves in excess alkali/ammonia.
A white precipitate insoluble in dilute acid with barium ions indicates sulfate.
Ions must be written as charged formulae, not neutral symbols.

Common Mistakes

Writing 'Mg' or 'Zn' without the 2+ charge.
Writing 'SO4' without the 2- charge.
Giving $\text{NO}_3^-$ for FA11 if the observations did not confirm it; the mark scheme records 'unknown'.
Swapping cation and anion in the table.

Things to Be Careful About

Charge balance and correct subscript/superscript position: $\text{SO}_4^{2-}$ , not SO4-.
The mark at this part depends on the observations recorded earlier; if an earlier observation is missing, the deduction cannot be justified.
If you cannot identify an ion, it is better to write 'unknown' than to guess incorrectly, but use all the positive evidence you have.

Techniques used

deduce cation from hydroxide solubilitydeduce anion from barium chloride resultwrite formula of ions with correct charges