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) Test 1: no change. Test 2: white ppt, soluble in excess NaOH; fizzing with Al; gas turns damp red litmus blue. Test 3: white ppt, soluble in excess NH3. · (a)(ii) Zn2+; NO2- or NO3- · (a)(iii) NO3- (nitrate) · (b)(i) See table of observations. · (b)(ii) FB 8: I-, FB 9: CO3^2- · (b)(iii) 2H+ + CO3^2- -> H2O + CO2

Chemistry 9701/32 — May/June 2024

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

questions

marks

120

minutes

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

Q1MediumAnalysis, Conclusions and EvaluationPresentation of Data and ObservationsManipulation, Measurement and Observation

Students are told to plan and carry out an experiment to determine the enthalpy change, $\Delta H_1$ , when one mole of anhydrous sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ , is hydrated.

\text{Na}_2\text{S}_2\text{O}_3(\text{s}) + 5\text{H}_2\text{O}(\text{l}) \rightarrow \text{Na}_2\text{S}_2\text{O}_3\cdot 5\text{H}_2\text{O}(\text{s})

One student suggested adding five moles of water to one mole of anhydrous sodium thiosulfate and measuring the temperature change.

Their teacher said that method would not work and suggested another method using Hess’s law.

You will carry out the teacher’s method to determine the enthalpy change, $\Delta H_2$ , when one mole of hydrated sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3\cdot 5\text{H}_2\text{O}$ , is dissolved in water.

\text{Na}_2\text{S}_2\text{O}_3\cdot 5\text{H}_2\text{O}(\text{s}) + \text{water} \rightarrow \text{Na}_2\text{S}_2\text{O}_3(\text{aq})

You will do this by adding a known mass of hydrated sodium thiosulfate to a known volume of water and measuring the temperature change when the solid dissolves.

(a)

Explain why the student’s suggestion to add five moles of water to one mole of anhydrous sodium thiosulfate would not be a suitable method to determine $\Delta H_1$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The temperature change of a solid cannot be accurately measured (a thermometer cannot be placed in a solid to record a reliable temperature change).

Final answer

Cannot accurately measure the temperature change of a solid.

Detailed explanation

Background Concept

In calorimetry experiments, the temperature change of a solution is measured using a thermometer immersed in the liquid. The thermometer bulb must be fully surrounded by the substance whose temperature is being measured for an accurate reading. When a solid reacts with a small amount of liquid (as in the hydration of anhydrous sodium thiosulfate with five moles of water), the resulting mixture is predominantly solid or a thick paste, making it impossible to obtain a reliable, uniform temperature reading.

Understanding the Question

The question asks why the student's proposed method — adding exactly five moles of water to one mole of anhydrous sodium thiosulfate and measuring the temperature change — would not work. The command word is "explain," so a reason must be given, not just a statement.

Approach

Consider the practical difficulties: the stoichiometric amount of water (five moles per mole of solid) is a very small volume relative to the mass of solid. Think about what the mixture would look like and whether a thermometer could give a meaningful reading.

Step-by-Step Reasoning

Five moles of water is 90 g (90 cm³), and one mole of anhydrous Na₂S₂O₃ is 158 g. The water-to-solid ratio is very low, so the mixture would be a thick paste or largely solid. A thermometer cannot be properly immersed in such a mixture to give an accurate, uniform temperature reading. Additionally, the reaction may not go to completion (not all the solid would form the pentahydrate), and some of the solid might simply dissolve rather than hydrate, introducing competing processes.

The mark scheme accepts any one of three points: (1) not all would form pentahydrate, (2) some would dissolve, or (3) cannot accurately measure the temperature change of a solid. The third is the most fundamental practical reason.

Key Takeaways

Direct calorimetry requires the substance to be in a state (usually liquid/solution) where temperature can be uniformly and accurately measured. When a reaction produces a solid or paste, indirect methods such as Hess's law must be used instead.

Common Mistakes

Saying "the reaction is too slow" — this is not the issue; hydration is rapid.
Saying "the temperature change is too small" — it is actually large, but unmeasurable in a solid.
Failing to give a reason (just stating it "wouldn't work" without explaining why).

Things to Be Careful About

The question asks for one reason only (1 mark). Any of the three accepted points will score. Do not write multiple reasons hoping one is right — just give one clear, correct explanation.

Techniques used

identify practical limitations of a proposed methodexplain why a direct measurement is not feasible

(b)

Teacher’s method

FB 1 is hydrated sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3\cdot 5\text{H}_2\text{O}$ .

Support the cup in the $250\text{ cm}^3$ beaker.
Use the $25\text{ cm}^3$ measuring cylinder to transfer $25.0\text{ cm}^3$ of distilled water into the cup.
Place the thermometer in the water and tilt the cup, if necessary, so that the bulb of the thermometer is fully covered. Record the temperature of the water in the space for results.
Weigh the container with FB 1. Record the mass.
Add all of the FB 1 to the water in the cup.
Stir the mixture. Measure and record the minimum temperature reached.
Reweigh the container and any residual FB 1. Record the mass.
Calculate and record the mass of FB 1 added.
Calculate and record the change in temperature.

Results

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

A representative results table (candidate's own readings will differ):

Measurement	Value / unit
Mass of container + FB 1	29.50 g
Mass of container + residue	24.50 g
Mass of FB 1 added	5.00 g
Initial temperature of water	21.0 °C
Final (minimum) temperature of solution	15.0 °C
Change in temperature, ΔT	6.0 °C

All balance readings recorded to 2 decimal places; thermometer readings to 0.1 °C (or nearest 0.5 °C as appropriate for the thermometer used). All subtractions correct.

Final answer

Table with correct headings and units, balance readings to 2 dp, thermometer readings consistent, all subtractions correct.

Detailed explanation

Background Concept

In Paper 3 practical examinations, candidates must record their own experimental data in a clearly presented table. The mark scheme rewards: (I) unambiguous headings with units for every entry, (II) consistent decimal places for balance readings (typically 2 dp, e.g. 24.50 g) and thermometer readings (to 0.1 °C or nearest 0.5 °C depending on the thermometer precision), all written in the table, and all subtractions performed correctly. Accuracy marks (III and IV) compare the candidate's ΔT/mass ratio to the supervisor's value.

Understanding the Question

Part (b) requires the candidate to carry out the teacher's method and record all measurements in a table. The key measurements are: mass of container + solid, mass of container + residue, mass of solid added (by subtraction), initial temperature of water, final/minimum temperature of solution, and change in temperature (by subtraction).

Approach

Set up a table with clear headings that include both the quantity and the unit. Record each reading as taken (not calculated), then show the derived values. Ensure balance readings are to 2 dp and thermometer readings are to the nearest 0.1 °C (or 0.5 °C if that is the thermometer's precision).

Step-by-Step Reasoning

Weigh container + FB 1: e.g. 29.50 g (2 dp balance reading).
After adding FB 1 to water, weigh container + residue: e.g. 24.50 g.
Mass of FB 1 added = 29.50 − 24.50 = 5.00 g.
Record initial temperature of water before adding solid: e.g. 21.0 °C.
After stirring, record minimum temperature reached: e.g. 15.0 °C.
ΔT = 21.0 − 15.0 = 6.0 °C.

The table must have headings that are unambiguous — "Mass of container + FB 1 / g" is better than just "mass." Units can be in the heading (e.g. "/g" or "(g)") or stated after each value.

For the accuracy marks, the supervisor calculates ΔT/mass and compares to the candidate's value. A ratio within 30% earns mark III; within 15% earns mark IV.

Key Takeaways

Always record raw readings in a table with clear headings including units.
Balance readings: 2 decimal places (e.g. 24.50 g).
Thermometer readings: to the precision of the thermometer (typically 0.1 °C or 0.5 °C).
Derived quantities (mass added, ΔT) should also appear in the table.

Common Mistakes

Writing mass to 1 dp (e.g. 5.0 g instead of 5.00 g) — inconsistent with balance precision.
Omitting units from column headings.
Recording ΔT without showing the subtraction in the table.
Using a thermometer reading like 21.37 °C when the thermometer only reads to 0.5 °C.

Things to Be Careful About

The heading must be unambiguous: "Mass of FB 1" could be confused with initial or final mass. Use "Mass of container + FB 1" and "Mass of container + residue."
All subtractions must be arithmetically correct — a simple error here loses mark II.
The temperature change is the initial minus the final (since the final is lower for an endothermic dissolution), giving a positive ΔT value.

Techniques used

record measurements with correct units and decimal placesconstruct a results table with unambiguous headingsperform subtraction to find mass added and temperature change

(c)

Calculations

(i)

Calculate the energy change, in J, when FB 1 is added to water.

1M Similar

DifficultyEasy

Worked solution

Working

q = 4.18 \times \Delta T \times 25

Using representative values from part (b): $\Delta T = 6.0\text{ °C}$

q = 4.18 \times 6.0 \times 25 = 627\text{ J}

Answer

$627\text{ J}$ (to 3 significant figures)

Final answer

627 J

Detailed explanation

Background Concept

In a simple calorimetry experiment, the heat energy absorbed or released by the reaction is calculated from the temperature change of the surrounding water using $q = mc\Delta T$ , where $m$ is the mass of water (in g), $c$ is the specific heat capacity of water ( $4.18\text{ J g}^{-1}\text{°C}^{-1}$ ), and $\Delta T$ is the temperature change (in °C). The volume of water is assumed to have a mass numerically equal to its volume in cm³ (since density of water ≈ 1 g cm⁻³).

Understanding the Question

The question asks for the energy change when FB 1 is added to 25.0 cm³ of water. The mass of water is 25 g (from 25.0 cm³), the specific heat capacity is 4.18 J g⁻¹ °C⁻¹, and the temperature change is taken from the candidate's own results in part (b).

Approach

Substitute directly into $q = mc\Delta T$ . The sign is ignored at this stage (the mark scheme says "ignore sign"), as the sign is dealt with in part (c)(ii).

Step-by-Step Reasoning

Mass of water = 25 g (from 25.0 cm³, assuming density = 1 g cm⁻³)
$c = 4.18\text{ J g}^{-1}\text{°C}^{-1}$
$\Delta T$ = temperature change from part (b) = 6.0 °C (representative)
$q = 4.18 \times 25 \times 6.0 = 627\text{ J}$

The answer must be given to 2–4 significant figures. 627 J (3 s.f.) is appropriate.

Key Takeaways

Always use the mass of water (not the mass of solute) in $q = mc\Delta T$ for solution calorimetry.
25.0 cm³ of water = 25 g.
The energy change is positive here (energy absorbed by the dissolution process, since temperature fell).

Common Mistakes

Using the mass of FB 1 instead of the mass of water.
Forgetting to multiply by 25 (using only 4.18 × ΔT).
Giving the answer to only 1 or 2 significant figures (e.g. 600 J or 630 J when 627 J is more appropriate).
Including a negative sign (the mark scheme says ignore sign at this stage).

Things to Be Careful About

The mark scheme specifically states the formula as $4.18 \times \text{temperature change} \times 25$ . The order of multiplication does not matter, but all three factors must be present.
Answer to 2–4 significant figures.

Techniques used

apply q = mcΔT for aqueous solution calorimetryuse specific heat capacity of water and mass of water

(ii)

Calculate the enthalpy change, $\Delta H_2$ , in $\text{kJ mol}^{-1}$ , when one mole of hydrated sodium thiosulfate, FB 1, dissolves in water.

\Delta H_2 = \text{sign} \quad \text{value}

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\text{moles of } \text{Na}_2\text{S}_2\text{O}_3\cdot 5\text{H}_2\text{O} = \frac{5.00}{248.2} = 0.02015\text{ mol}

\Delta H_2 = +\frac{627}{0.02015 \times 1000} = +31.1\text{ kJ mol}^{-1}

Answer

$\Delta H_2 = +31.1\text{ kJ mol}^{-1}$

Final answer

+31.1 kJ mol⁻¹

Detailed explanation

Background Concept

Enthalpy change of solution is the heat energy change when one mole of a substance dissolves in water under standard conditions. It is calculated by dividing the total energy change (from calorimetry) by the number of moles of solute dissolved, and converting from J to kJ. The sign is determined by whether the temperature increased (exothermic, negative ΔH) or decreased (endothermic, positive ΔH).

Understanding the Question

Using the energy change calculated in (c)(i) and the mass of FB 1 recorded in (b), calculate ΔH₂ — the enthalpy change when one mole of hydrated sodium thiosulfate dissolves. The molar mass of Na₂S₂O₃·5H₂O is 248.2 g mol⁻¹.

Approach

Calculate moles of FB 1 from mass ÷ molar mass.
Divide the energy change (in J) by the number of moles to get J mol⁻¹.
Convert to kJ mol⁻¹ by dividing by 1000.
Assign the correct sign: since the temperature decreased (endothermic process), ΔH₂ is positive.

Step-by-Step Reasoning

Mass of FB 1 = 5.00 g (representative from part b)
$M_r$ of Na₂S₂O₃·5H₂O = (2×23) + (2×32.1) + (3×16) + 5×(2×1 + 16) = 46 + 64.2 + 48 + 90 = 248.2
Moles = 5.00/248.2 = 0.02015 mol
Energy change from (c)(i) = 627 J
ΔH₂ = +627/(0.02015 × 1000) = +627/20.15 = +31.1 kJ mol⁻¹

The positive sign indicates an endothermic process (temperature fell during dissolution).

Key Takeaways

Always divide by the number of moles of the substance whose enthalpy change is being determined (here, moles of FB 1, not moles of water).
Convert J to kJ by dividing by 1000.
The sign of ΔH depends on the direction of temperature change: temperature decrease → endothermic → positive ΔH.

Common Mistakes

Forgetting to convert from J to kJ (giving 31 100 instead of 31.1).
Using the wrong molar mass (e.g. 158 for anhydrous instead of 248.2 for pentahydrate).
Getting the sign wrong — writing a negative value when the process is endothermic.
Dividing by mass instead of moles.

Things to Be Careful About

The mark scheme requires the positive sign to be explicitly stated ("+31.1 kJ mol⁻¹").
Answer must be to 2–4 significant figures.
This answer is carried forward into part (c)(iii), so an error here propagates (ecf applies).

Techniques used

calculate moles from mass and molar massconvert energy change to enthalpy change per moledetermine sign of enthalpy change from temperature change direction

(iii)

The enthalpy change, $\Delta H_3$ , when one mole of anhydrous sodium thiosulfate is dissolved in water is $-8.1\text{ kJ mol}^{-1}$ .

\text{Na}_2\text{S}_2\text{O}_3(\text{s}) + \text{water} \rightarrow \text{Na}_2\text{S}_2\text{O}_3(\text{aq}) \quad \Delta H_3 = -8.1\text{ kJ mol}^{-1}

Use your answer to (c)(ii) and the information given to construct a Hess’s cycle to calculate $\Delta H_1$ in $\text{kJ mol}^{-1}$ . Show clearly how you used the data.

(If you were unable to calculate an answer in (c)(ii), assume a value of $31.6\text{ kJ mol}^{-1}$ . Note this may not be the correct value and the sign has been omitted.)

\text{Na}_2\text{S}_2\text{O}_3(\text{s}) + 5\text{H}_2\text{O}(\text{l}) \rightarrow \text{Na}_2\text{S}_2\text{O}_3\cdot 5\text{H}_2\text{O}(\text{s}) \quad \Delta H_1

\Delta H_1 = \text{sign} \quad \text{value}

2M Similar

DifficultyMedium

Worked solution

Working

Hess's cycle:

From the cycle:

\Delta H_3 = \Delta H_1 + \Delta H_2

\Delta H_1 = \Delta H_3 - \Delta H_2 = -8.1 - (+31.1) = -39.2\text{ kJ mol}^{-1}

Answer

$\Delta H_1 = -39.2\text{ kJ mol}^{-1}$

Final answer

-39.2 kJ mol⁻¹

Detailed explanation

Background Concept

Hess's law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. This allows us to calculate an enthalpy change that cannot be measured directly by combining known enthalpy changes of related reactions. The key is to construct a cycle where the desired reaction is linked to measurable reactions through a common intermediate or product.

Understanding the Question

We need to find ΔH₁ (hydration of anhydrous Na₂S₂O₃ to pentahydrate) which cannot be measured directly (as established in part a). We are given:

ΔH₂ (from our experiment): Na₂S₂O₃·5H₂O(s) + water → Na₂S₂O₃(aq), ΔH₂ = +31.1 kJ mol⁻¹
ΔH₃ (given): Na₂S₂O₃(s) + water → Na₂S₂O₃(aq), ΔH₃ = −8.1 kJ mol⁻¹

The target: Na₂S₂O₃(s) + 5H₂O(l) → Na₂S₂O₃·5H₂O(s), ΔH₁ = ?

Approach

Construct a Hess's cycle with the reactants (Na₂S₂O₃(s) + 5H₂O(l)) at the top, the product (Na₂S₂O₃·5H₂O(s)) also at the top (connected by ΔH₁), and the common aqueous product (Na₂S₂O₃(aq)) at the bottom. Two downward arrows connect the top species to the bottom, labelled ΔH₃ and ΔH₂ respectively.

Step-by-Step Reasoning

The cycle has:

Top left: Na₂S₂O₃(s) + 5H₂O(l)
Top right: Na₂S₂O₃·5H₂O(s)
Bottom: Na₂S₂O₃(aq)

Arrows:

Left downward arrow: ΔH₃ = −8.1 kJ mol⁻¹ (anhydrous dissolving directly)
Right downward arrow: ΔH₂ = +31.1 kJ mol⁻¹ (pentahydrate dissolving)
Top horizontal arrow (left to right): ΔH₁ (hydration, the unknown)

By Hess's law, going from top-left to bottom via two routes:

Route 1 (direct): ΔH₃ = −8.1
Route 2 (via top-right): ΔH₁ + ΔH₂

Therefore: ΔH₃ = ΔH₁ + ΔH₂

Rearranging: ΔH₁ = ΔH₃ − ΔH₂ = −8.1 − (+31.1) = −39.2 kJ mol⁻¹

The negative sign indicates the hydration is exothermic, which is chemically sensible (forming crystal lattice bonds releases energy).

Key Takeaways

Hess's law allows calculation of enthalpy changes that cannot be measured directly.
The cycle must have a common starting point and a common ending point for the two routes.
The algebraic relationship ΔH₃ = ΔH₁ + ΔH₂ comes from equating the two routes from the same start to the same end.

Common Mistakes

Drawing arrows in the wrong direction (both should point downward to the common aqueous product).
Writing ΔH₁ = ΔH₂ − ΔH₃ instead of ΔH₁ = ΔH₃ − ΔH₂ (sign error in rearrangement).
Forgetting the positive sign on ΔH₂ when substituting (writing −8.1 − 31.1 instead of −8.1 − (+31.1)).
Omitting the sign in the final answer.

Things to Be Careful About

The mark scheme requires TWO downward arrows in the diagram (not one up and one down).
Labels must be correct: ΔH₃ (or −8.1) on the left arrow, ΔH₂ (or the candidate's answer) on the right arrow.
The final answer must include the sign (negative).
If the candidate could not calculate (c)(ii), they should use +31.6 kJ mol⁻¹ as stated, giving ΔH₁ = −8.1 − 31.6 = −39.7 kJ mol⁻¹.

Techniques used

construct a Hess's law cycleapply Hess's law to find an unknown enthalpy change from known values

(d)

A sample of FB 1 was contaminated with anhydrous sodium thiosulfate.

State what effect this would have on the temperature change in (b). Explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The temperature change would be less (smaller decrease / the minimum temperature would be higher) because the dissolution of anhydrous sodium thiosulfate is exothermic (ΔH₃ = −8.1 kJ mol⁻¹), which partially offsets the endothermic dissolution of the pentahydrate.

Final answer

Temperature change is less (smaller) because anhydrous sodium thiosulfate dissolves exothermically, offsetting the endothermic effect.

Detailed explanation

Background Concept

When a sample intended to be purely hydrated sodium thiosulfate (Na₂S₂O₃·5H₂O) is contaminated with some anhydrous sodium thiosulfate (Na₂S₂O₃), the dissolution process involves two competing thermal effects: the pentahydrate dissolves endothermically (temperature decreases, ΔH₂ > 0), while the anhydrous form dissolves exothermically (temperature increases, ΔH₃ < 0). The net temperature change observed will be the sum of these opposing effects, resulting in a smaller overall temperature decrease than if the sample were pure pentahydrate.

Understanding the Question

The question asks for the effect of contamination on the temperature change measured in part (b), with an explanation. The command word is "state" (the effect) and "explain" (the reason). One mark is available, so a concise correct statement with reason is needed.

Approach

Consider what happens when the contaminated sample is added to water: both the pentahydrate and the anhydrous form dissolve. The anhydrous form's dissolution is exothermic (we know ΔH₃ = −8.1 kJ mol⁻¹ from the question stem). This exothermic process releases heat, partially compensating for the endothermic dissolution of the pentahydrate.

Step-by-Step Reasoning

Pure pentahydrate dissolving: endothermic → temperature decreases by ΔT.
Anhydrous dissolving: exothermic → temperature increases.
Contaminated sample: both processes occur simultaneously → the temperature decrease is less than it would be for pure pentahydrate.
Therefore, the measured ΔT is smaller (the minimum temperature reached is higher than expected for a pure sample).

This means the calculated ΔH₂ would be too small (less positive), which in turn makes the calculated ΔH₁ less negative than the true value.

Key Takeaways

Contamination with a species that undergoes an opposing thermal process reduces the magnitude of the observed temperature change.
Understanding the sign of each enthalpy change is essential for predicting the direction of the error.

Common Mistakes

Saying the temperature change would be greater (confusing the direction of the exothermic contribution).
Saying the anhydrous form is endothermic (it is not — ΔH₃ is negative, exothermic).
Stating the effect without giving the reason (the question asks to explain).

Things to Be Careful About

The mark scheme accepts "less" or "decreased" for the effect, and requires the explanation to mention that the anhydrous form's reaction is exothermic or causes a temperature increase.
Be precise: the temperature change (ΔT) is smaller, not that the temperature itself is higher (though the minimum temperature reached is indeed higher).

Techniques used

predict the effect of an impurity on a measured temperature changeidentify the thermal nature of a competing process

Q2MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Iodate ions contain iodine and oxygen. They have the formula $\text{IO}_x^-$ where $x$ is an integer.

In this experiment you will determine the value of $x$ in an iodate. You will first react $\text{IO}_x^-$ with an excess of iodide ions, $\text{I}^-$ , to form iodine, $\text{I}_2$ .

The amount of iodine produced is then determined by titration with thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$ .

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

FB 2 is $0.100\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
FB 3 is a solution containing $0.0140\text{ mol dm}^{-3}$ $\text{IO}_x^-$ ions.
FB 4 is dilute sulfuric acid, $\text{H}_2\text{SO}_4$ .
FB 5 is $0.500\text{ mol dm}^{-3}$ potassium iodide, $\text{KI}$ .
FB 6 is starch indicator.

(a)

Method

Fill the burette with FB 2.
Pipette $25.0\text{ cm}^3$ of FB 3 into a conical flask.
Use the $25\text{ cm}^3$ measuring cylinder to add $10\text{ cm}^3$ of FB 4 to the conical flask.
Use the same measuring cylinder to add $10\text{ cm}^3$ of FB 5 to the conical flask.
Add FB 2 from the burette until the solution turns yellow.
Add 10–15 drops of FB 6 to the solution in the conical flask.
Continue to add more FB 2 from the burette until the blue-black colour just disappears.
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.

Keep the remaining FB 2 for use in Question 3.

7M Similar

DifficultyMedium-Easy

Worked solution

Answer

Complete the rough and accurate titration record. All burette readings must be recorded to the nearest $0.05\text{ cm}^3$ . Include headings with units for every column. Example completed record (replace with your own readings):

Titration	rough	1	2
initial burette reading / $\text{cm}^3$	0.00	0.00	0.00
final burette reading / $\text{cm}^3$	28.50	28.00	28.00
titre / $\text{cm}^3$	28.50	28.00	28.00

Keep the remaining FB 2 for Question 3.

Final answer

Candidate-dependent: complete rough and accurate titration table, readings to nearest 0.05 cm3 (example shown: accurate titres 28.00 cm3).

Detailed explanation

Background Concept

The experiment is an iodometric titration. Iodate ions $\text{IO}_x^-$ react with excess iodide in acid to produce iodine, $\text{I}_2$ . The amount of $\text{I}_2$ formed is then measured by titrating with thiosulfate. The reaction is

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

The endpoint is shown by starch indicator, which forms a blue-black complex with iodine; when all iodine has reacted, the blue-black colour disappears. Burette readings are made at the bottom of the meniscus and recorded to the nearest $0.05\text{ cm}^3$ .

Understanding the Question

Part (a) asks you to carry out the titration and record both a rough titre and accurate titres. The marking points reward the completeness and quality of the recorded data: two burette readings and a titre for the rough run; initial and final readings and a titre for at least two accurate runs; clear headings with units; readings to the nearest $0.05\text{ cm}^3$ ; and accurate titres that are close together.

Approach

First do a rough titration to find the approximate endpoint. Then repeat the titration carefully, adding starch only near the endpoint, until you have at least two concordant accurate titres. Present all readings in a table with columns for initial reading, final reading and titre, each with its unit.

Step-by-Step Reasoning

Before starting, ensure the burette is clean and rinsed with FB 2, fill it, and record the initial reading.
Pipette $25.0\text{ cm}^3$ of FB 3 into the conical flask, then add $10\text{ cm}^3$ of FB 4 and $10\text{ cm}^3$ of FB 5 using the same measuring cylinder.
Run in FB 2 until the solution turns yellow; this yellow is iodine in solution. Add 10-15 drops of starch, which turns the mixture blue-black, then continue the titration dropwise until the blue-black colour just disappears.
Record the final burette reading and calculate the titre.
Repeat until two or more accurate titres agree within $0.10\text{ cm}^3$ .
Record rough and accurate readings in a table with units, as in the sample.

Key Takeaways

A well-presented titration record contains all raw readings, clear headings with units, readings to the precision of the burette, and concordant accurate titres.

Common Mistakes

Forgetting to show units in every column heading.
Recording readings to only one decimal place; the burette can be read to the nearest $0.05\text{ cm}^3$ .
Omitting the rough titre or its two burette readings.
Averaging titres that are too far apart.
Adding starch at the start instead of near the endpoint, which makes the endpoint harder to judge.

Things to Be Careful About

The exact readings you record are candidate-dependent and must be used in all later calculations. Read the burette at eye level at the bottom of the meniscus and keep the remaining FB 2 for Question 3.

Techniques used

perform an iodometric titration using a buretterecord burette readings to the nearest 0.05 cm3record a results 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.

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Select the accurate titres that agree closely. Tick or underline the two titres used.

For example, using the accurate titres $28.00\text{ cm}^3$ and $28.00\text{ cm}^3$ :

\text{mean titre} = \frac{28.00 + 28.00}{2} = 28.00\text{ cm}^3

Answer

$28.00\text{ cm}^3$ (candidate-dependent; choose your own two concordant accurate titres and show your selection).

Final answer

28.00 cm3 (example; candidate-dependent)

Detailed explanation

Background Concept

The mean titre is used in all subsequent calculations. Not every titre should be used: only accurate titres that are close together are averaged. The rough titre is never included in the mean because it was obtained quickly and is not reliable.

Understanding the Question

Part (b) asks you to calculate a suitable mean titre from your accurate titration results. The mark scheme requires two (or more) accurate titres whose total spread is not more than $0.20\text{ cm}^3$ , a clear indication of which titres were used, and a mean quoted to two decimal places.

Approach

Look at your accurate titres and choose two or more that agree closely. Add them together and divide by the number of values. Show the working or tick the selected readings so that the examiner can see which titres were used.

Step-by-Step Reasoning

Identify the accurate titres, ignoring the rough value.
Check that the chosen titres are within $0.20\text{ cm}^3$ of each other; preferably within $0.10\text{ cm}^3$ .
Add the selected values and divide by how many there are.
Round the result correctly to two decimal places, i.e. to the nearest $0.01\text{ cm}^3$ .

With two identical accurate titres of $28.00\text{ cm}^3$ , the mean is exactly $28.00\text{ cm}^3$ .

Key Takeaways

The mean is only as good as the titres used. Always choose concordant accurate readingshjälper and show which values were averaged.

Common Mistakes

Including the rough titre in the mean.
Averaging titres that are spread by more than $0.20\text{ cm}^3$ .
Not showing which titres were used, so the examiner cannot award the mark.
Rounding to one decimal place instead of two.

Things to Be Careful About

The mean is expressed to two decimal places, even if it appears to be a whole number of cubic centimetres. Use the value you obtain here in parts (c)(ii) and (c)(iv).

Techniques used

select concordant accurate titrescalculate the mean titreround the mean to two decimal places

(c)

Calculations

(i)

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

1M Similar

(ii)

Use your answer to (b) and the information given to calculate the amount, in mol, of iodine formed when $25.0\text{ cm}^3$ of FB 3 reacts with $10\text{ cm}^3$ of FB 5.

amount of $\text{I}_2$ = .............................. mol

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using the mean titre from (b), for example $28.00\text{ cm}^3$ :

n(\text{S}_2\text{O}_3^{2-}) = \frac{28.00}{1000} \times 0.100 = 2.80 \times 10^{-3}\text{ mol}

From

\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}

n(\text{I}_2) = \frac{1}{2} n(\text{S}_2\text{O}_3^{2-}) = \frac{1}{2} \times 2.80 \times 10^{-3} = 1.40 \times 10^{-3}\text{ mol}

Answer

$1.40 \times 10^{-3}\text{ mol}$ (candidate-dependent, using the mean titre from part (b)).

Final answer

1.40 × 10^-3 mol (using mean titre 28.00 cm3; candidate-dependent)

Detailed explanation

Background Concept

The titration reaction is

\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}

One mole of iodine reacts with two moles of thiosulfate. Therefore, to find moles of iodine from the titration, first find the moles of thiosulfate used, then halve that value. The amount of thiosulfate is calculated from $n = cV$ , where volume is in $\text{dm}^3$ .

Understanding the Question

Part (c)(ii) asks for the amount of iodine formed when $25.0\text{ cm}^3$ of FB 3 reacts with $10\text{ cm}^3$ of FB 5. The iodine is then titrated by FB 2, so the titre tells us the moles of iodine directly through the 1:2 stoichiometry.

Approach

Convert the mean titre from $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000. Multiply by the concentration of FB 2, $0.100\text{ mol dm}^{-3}$ , to obtain moles of thiosulfate. Then halve to obtain moles of iodine.

Step-by-Step Reasoning

Mean titre: $28.00\text{ cm}^3$ .
Volume in $\text{dm}^3$ : $28.00/1000 = 0.02800\text{ dm}^3$ .
Moles of thiosulfate: $0.02800 \times 0.100 = 2.80 \times 10^{-3}\text{ mol}$ .
Mole ratio from the balanced equation: $\text{I}_2 : \text{S}_2\text{O}_3^{2-} = 1 : 2$ .
Moles of iodine: $\frac{1}{2} \times 2.80 \times 10^{-3} = 1.40 \times 10^{-3}\text{ mol}$ .

Key Takeaways

Always balance the titration equation before using the mole ratio. The most common error is forgetting the factor of $\frac{1}{2}$ for iodine.

Common Mistakes

Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ .
Using a 1:1 ratio instead of 1:2 between iodine and thiosulfate.
Quoting the answer to too many significant figures.

Things to Be Careful About

Use the mean titre exactly as calculated in part (b), and carry the value of iodine moles into part (c)(iv) without rounding prematurely.

Techniques used

convert mean titre from cm3 to dm3calculate moles of thiosulfate from concentration and volumeuse the 1:2 stoichiometric ratio between I2 and S2O3^2-

(iii)

Calculate the amount, in mol, of $\text{IO}_x^-$ ions in $25.0\text{ cm}^3$ of FB 3.

amount of $\text{IO}_x^-$ ions = .............................. mol

1M Similar

DifficultyEasy

Worked solution

Working

n(\text{IO}_x^-) = cV = 0.0140 \times \frac{25.0}{1000} = 3.50 \times 10^{-4}\text{ mol}

Answer

$3.50 \times 10^{-4}\text{ mol}$

Final answer

3.50 × 10^-4 mol

Detailed explanation

Background Concept

The amount of a solute in solution is given by $n = cV$ , where $c$ is the concentration in $\text{mol dm}^{-3}$ and $V$ is the volume in $\text{dm}^3$ . Since the volume is given in $\text{cm}^3$ , divide by 1000 to convert it to $\text{dm}^3$ .

Understanding the Question

This part asks for the moles of iodate ions in the $25.0\text{ cm}^3$ sample of FB 3. The concentration of FB 3 is given as $0.0140\text{ mol dm}^{-3}$ , so no titration data are needed.

Approach

Substitute the given volume and concentration directly into $n = cV$ .

Step-by-Step Reasoning

Volume: $25.0\text{ cm}^3 = 25.0/1000 = 0.0250\text{ dm}^3$ .
Concentration: $0.0140\text{ mol dm}^{-3}$ .
Moles: $0.0140 \times 0.0250 = 3.50 \times 10^{-4}\text{ mol}$ .

The answer is given to 3 significant figures, matching the data.

Key Takeaways

$n = cV$ is the standard way to find moles from a solution. Always check that volume is in $\text{dm}^3$ before multiplying.

Common Mistakes

Using the volume in $\text{cm}^3$ without converting, giving an answer 1000 times too large.
Quoting too many or too few significant figures.

Things to Be Careful About

This result does not depend on the titre; it is fixed by the stated concentration and volume. Use $3.50 \times 10^{-4}\text{ mol}$ in part (c)(iv).

Techniques used

calculate moles from concentration and volumeconvert cm3 to dm3

(iv)

An unbalanced equation for the reaction of $\text{IO}_x^-$ ions with iodide ions, $\text{I}^-$ , and hydrogen ions, $\text{H}^+$ , is shown.

\text{IO}_x^- + \dots \text{I}^- + \dots \text{H}^+ \rightarrow \dots \text{I}_2 + \dots \text{H}_2\text{O}

Use the ratio of your answers to (c)(ii) and (c)(iii) to balance this equation and determine the value of $x$ .
Show your working.

ratio $\text{IO}_x^- : \text{I}_2 = 1 : \dots$

$x = \dots$

2M Similar

DifficultyMedium

Worked solution

Working

Using the example values from (c)(ii) and (c)(iii):

\frac{n(\text{I}_2)}{n(\text{IO}_x^-)} = \frac{1.40 \times 10^{-3}}{3.50 \times 10^{-4}} = 4.00

So the ratio $\text{IO}_x^- : \text{I}_2 = 1 : 4$ ; each iodate ion forms four molecules of $\text{I}_2$ , hence $x = 4$ .

Balanced equation:

\text{IO}_4^- + 7\text{I}^- + 8\text{H}^+ \rightarrow 4\text{I}_2 + 4\text{H}_2\text{O}

Answer

$x = 4$

Final answer

x = 4; balanced equation: IO4^- + 7I^- + 8H+ -> 4I2 + 4H2O

Detailed explanation

Background Concept

Each iodate ion $\text{IO}_x^-$ is reduced by iodide to give a number of iodine molecules. The balanced general equation has the form

\text{IO}_x^- + (2x-1)\text{I}^- + 2x\text{H}^+ \rightarrow x\text{I}_2 + x\text{H}_2\text{O}

Thus the ratio $\text{IO}_x^- : \text{I}_2$ is $1 : x$ . The value of $x$ can be found by comparing the moles of iodine formed with the moles of iodate ions used.

Understanding the Question

Part (c)(iv) asks you to use the mole ratio from parts (c)(ii) and (c)(iii) to find the integer $x$ , then balance the full equation for the reaction. The marking scheme awards one mark for obtaining an integer $x$ from the ratio and another for balancing the equation with that $x$ .

Approach

Divide the moles of iodine by the moles of iodate ions. The result should be a whole number, which is the value of $x$ . Then balance the equation for that value of $x$ .

Step-by-Step Reasoning

Moles of iodine: $1.40 \times 10^{-3}\text{ mol}$ .
Moles of iodate: $3.50 \times 10^{-4}\text{ mol}$ .
Ratio: $1.40 \times 10^{-3} / 3.50 \times 10^{-4} = 4.00$ .
Therefore each $\text{IO}_x^-$ produces 4 $\text{I}_2$ , so $x = 4$ .
The balanced equation for $x = 4$ is

\text{IO}_4^- + 7\text{I}^- + 8\text{H}^+ \rightarrow 4\text{I}_2 + 4\text{H}_2\text{O}

Check atoms and charges: left has $\text{IO}_4^-$ + $7\text{I}^-$ + $8\text{H}^+$ , total charge $(-1) + 7(-1) + 8(+1) = 0$ ; right has 8 iodine atoms and 8 hydrogen atoms, total charge 0.

Key Takeaways

The mole ratio directly gives $x$ because the stoichiometry of the balanced equation fixes the number of iodine molecules produced per iodate ion. Balancing redox equations requires balancing atoms and charge together.

Common Mistakes

Not dividing the two mole amounts, or dividing in the wrong order.
Rounding the ratio incorrectly; the ratio must give an integer $x$ .
Balancing the equation without checking that both atoms and charges balance.

Things to Be Careful About

The final value of $x$ must be an integer. In this worked example $x = 4$ , but if your mean titre differs, your ratio may lead to a different integer value such as $2$ , $3$ , or $5$ ; use your own values.

Techniques used

calculate the mole ratio of IO_x^- to I2deduce the integer x from the ratiobalance the redox equation for the iodate-iodide reaction

(d)

A student carries out the same experiment as in (a) but uses $0.0140\text{ mol dm}^{-3}$ $\text{IO}_2^-$ ions in place of FB 3.

Tick the correct box in Table 2.1. Explain your answer.

(If you were unable to determine the value of $x$ in (c)(iv) or you determined the value of $x$ to be 2, assume $x = 4$ . Note that this may not be the correct value).

Table 2.1


Volume of FB 2 will be smaller.
Volume of FB 2 will be unchanged.
Volume of FB 2 will be larger.

2M Similar

DifficultyMedium

Worked solution

Answer

Tick: Volume of FB 2 will be smaller.

Because $\text{IO}_2^-$ contains iodine in a lower oxidation state than $\text{IO}_4^-$ , less iodine is produced per mole of iodate, so a smaller volume of thiosulfate is needed.

Balanced equation for $\text{IO}_2^-$ :

\text{IO}_2^- + 3\text{I}^- + 4\text{H}^+ \rightarrow 2\text{I}_2 + 2\text{H}_2\text{O}

Each $\text{IO}_2^-$ gives only 2 molecules of $\text{I}_2$ , whereas each $\text{IO}_4^-$ gives 4 molecules of $\text{I}_2$ .

Final answer

Smaller

Detailed explanation

Background Concept

The volume of thiosulfate needed is directly proportional to the amount of iodine produced. The amount of iodine produced per iodate ion depends on the oxidation state of iodine in the iodate. In $\text{IO}_4^-$ , iodine has oxidation state $+7$ , while in $\text{IO}_2^-$ , iodine has oxidation state $+3$ ; both are reduced to iodine, $\text{I}_2$ , in which iodine is in oxidation state 0. A larger decrease in oxidation state corresponds to more iodine produced per ion.

Understanding the Question

Part (d) asks what happens to the titre if the same concentration of iodate is used, but the ion is changed from the original iodate to $\text{IO}_2^-$ . You must tick one of three options and explain. The marking scheme awards a mark for identifying that the volume is smaller and a second mark for the reasoning, either through oxidation state or through a balanced equation.

Approach

Write or consider the balanced equation for the reaction of $\text{IO}_2^-$ with iodide. It produces fewer iodine molecules per iodate ion than $\text{IO}_4^-$ doesasi, so less iodine is formed and hence less thiosulfate is needed.

Step-by-Step Reasoning

For the original ion, using $x = 4$ , each $\text{IO}_4^-$ produces 4 $\text{I}_2$ :

\text{IO}_4^- + 7\text{I}^- + 8\text{H}^+ \rightarrow 4\text{I}_2 + 4\text{H}_2\text{O}

For $\text{IO}_2^-$ , balance the equation:

\text{IO}_2^- + 3\text{I}^- + 4\text{H}^+ \rightarrow 2\text{I}_2 + 2\text{H}_2\text{O}

Each $\text{IO}_2^-$ produces only 2 moles of $\text{I}_2$ instead of 4 moles.
Since thiosulfate reacts 1:2 with iodine, the titre is smaller when less iodine is produced.

Alternatively, compare oxidation states: iodine in $\text{IO}_4^-$ is $+7$ ; in $\text{IO}_2^-$ it is $+3$ . The change to $0$ is smaller for $\text{IO}_2^-$ , so less iodine is produced.

Key Takeaways

The titre volume is a direct measure of the iodine produced. A change in the oxidation state of iodine in the iodate changes the stoichiometry and therefore the titre.

Common Mistakes

Ticking "larger" by thinking a smaller oxidation state means more iodine.
Giving the observation without any explanation; the second mark requires a reason.
Writing an unbalanced equation for $\text{IO}_2^-$ .

Things to Be Careful About

If you were unable to find $x$ in part (c)(iv), the question tells you to assume $x = 4$ . The comparison between $\text{IO}_4^-$ and $\text{IO}_2^-$ still leads to "smaller". Ensure the ticked box and the explanation are consistent.

Techniques used

compare oxidation states of iodine in IO4^- and IO2^-predict the relative amount of iodine produceduse a balanced equation for IO2^- to support the conclusion

Q3MediumQualitative AnalysisAnalysis, Conclusions and EvaluationPresentation of Data and Observations

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.

FB 7 is an aqueous solution of a salt containing one cation and one anion, both of which are listed in the Qualitative analysis notes.

(a)

(i)

Carry out the following tests using a $1\text{ cm}^3$ depth of FB 7 in a test-tube for each test. Record your observations in Table 3.1.

Table 3.1

test	observations
Test 1 Add aqueous barium chloride or aqueous barium nitrate.
Test 2 Add aqueous sodium hydroxide, then transfer the mixture to a boiling tube and warm gently, then add a small piece of aluminium foil.
Test 3 Add aqueous ammonia.

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Test 1 - Add aqueous barium chloride/nitrate:

No visible change / no precipitate.

Test 2 - Add aqueous sodium hydroxide:

White precipitate forms.
Precipitate dissolves in excess NaOH to give a colourless solution.
On warming with aluminium foil: effervescence/fizzing.
Gas evolved turns damp red litmus paper blue (ammonia).

Test 3 - Add aqueous ammonia:

White precipitate forms.
Precipitate dissolves in excess aqueous ammonia to give a colourless solution.

Final answer

Test 1: no change. Test 2: white ppt, soluble in excess NaOH; fizzing with Al; gas turns damp red litmus blue. Test 3: white ppt, soluble in excess NH3.

Detailed explanation

Background Concept

Qualitative analysis identifies ions by characteristic reactions. $\text{Zn}^{2+}$ forms an amphoteric hydroxide: with NaOH or NH3 it gives white $\text{Zn(OH)}_2$ , which dissolves in excess reagent because the hydroxide/ammine complex is soluble. Nitrate and nitrite ions are reduced by aluminium in hot alkaline solution to ammonia; ammonia is an alkaline gas that turns damp red litmus paper blue. Barium chloride/nitrate is used to test for sulfate, carbonate and sulfite; with nitrate alone no precipitate forms.

Understanding the Question

You are given FB 7, a solution of a salt with one cation and one anion. Three tests are specified: barium chloride/nitrate, NaOH with warming and Al foil, and aqueous ammonia. You must record observations at each stage. The tests are chosen to reveal the cation (via hydroxide/ammine behaviour) and the anion (via ammonia gas from nitrate/nitrite). The mark scheme rewards precise observations, not just final identities.

Approach

Perform each test on a fresh 1 cm3 sample. For each, note the initial effect, then what happens when excess reagent is added, and any gas produced. For Test 2, the sequence matters: first NaOH, then warming, then Al foil. Test any gas with damp red litmus.

Step-by-Step Reasoning

Test 1: $\text{Ba}^{2+}$ would precipitate sulfate, carbonate or sulfite. With nitrate only, no precipitate forms; record 'no visible change' or 'no precipitate'.

Test 2: NaOH first gives white $\text{Zn(OH)}_2$ . On adding excess NaOH, the precipitate dissolves to a colourless solution because $[\text{Zn(OH)}_4]^{2-}$ forms. Warming alone produces no gas. Adding Al foil in alkaline solution reduces nitrate/nitrite to NH3; you see effervescence/fizzing (H2 also forms) and the gas turns damp red litmus blue. Record both the fizzing and the litmus result.

Test 3: Aqueous ammonia also gives white $\text{Zn(OH)}_2$ . In excess ammonia, the precipitate dissolves to a colourless solution because the zinc ammine complex forms. This solubility in excess NH3 is a key distinction from $\text{Al}^{3+}$ , whose hydroxide is insoluble in excess NH3.

Key Takeaways

$\text{Zn}^{2+}$ is amphoteric: its hydroxide dissolves in excess NaOH and excess NH3.
Nitrate/nitrite are detected by reduction to NH3 with Al and hot NaOH.
Always record the stage at which a change occurs.

Common Mistakes

Writing 'white precipitate' but omitting 'soluble in excess' – the solubility is the identifying point.
Forgetting to test the gas with damp red litmus.
Writing 'no reaction' instead of 'no visible change' for Test 1.
Using a test-tube for warming; the instructions require a boiling tube.

Things to Be Careful About

Use fresh samples for each test.
Record observations in the table exactly as they occur.
If no change, write 'no change' as instructed.
Do not attempt additional tests.

Techniques used

test for Zn2+ with NaOH and excesstest for Zn2+ with aqueous ammonia and excesstest for nitrate/nitrite by warming with NaOH and aluminium foiltest for ammonia gas with damp red litmus

(ii)

The results of the tests in Table 3.1 allow you to deduce one cation and two possible anions in FB 7.

Deduce which ions may be present in FB 7. Give the formulae of the ions.

cation ....................

anion ..................... or .....................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Cation: $\text{Zn}^{2+}$

Anion: $\text{NO}_2^-$ or $\text{NO}_3^-$

Final answer

Zn2+; NO2- or NO3-

Detailed explanation

Background Concept

In qualitative analysis, the identity of a cation is deduced from the behaviour of its hydroxide with NaOH and NH3. $\text{Zn}^{2+}$ is one of the few cations whose hydroxide is soluble in both excess NaOH and excess NH3. For anions, nitrate and nitrite are both reduced by aluminium in hot alkali to ammonia gas. Therefore the observation of ammonia gas in Test 2 limits the anion to $\text{NO}_2^-$ or $\text{NO}_3^-$ .

Understanding the Question

The observations from Table 3.1 must be used to deduce one cation and two possible anions. You are not asked to identify the exact anion yet; that is done in part (iii). The phrase 'may be present' signals that the tests give a shortlist, not a unique answer.

Approach

First identify the cation from the hydroxide solubility pattern, especially Test 3 with NH3. Then identify the anion family from the gas produced in Test 2. Write the formulae with charges.

Step-by-Step Reasoning

Cation: Test 3 gives a white precipitate that dissolves in excess aqueous ammonia. This is characteristic of $\text{Zn}^{2+}$ . ( $\text{Al}^{3+}$ gives a white ppt with NH3 but it does not dissolve in excess NH3; $\text{Pb}^{2+}$ gives a white ppt soluble in excess NaOH but not NH3.) So cation = $\text{Zn}^{2+}$ .

Anion: Test 2 produces ammonia gas after adding Al foil to the hot alkaline mixture. Both nitrate and nitrite are reduced to NH3 under these conditions. So the anion is either $\text{NO}_2^-$ or $\text{NO}_3^-$ . The barium chloride test gave no precipitate, which rules out sulfate, carbonate and sulfite, consistent with nitrate/nitrite.

Key Takeaways

Amphoteric hydroxides identify $\text{Zn}^{2+}$ .
Al/NaOH reduction identifies nitrate and nitrite together; further tests distinguish them.

Common Mistakes

Giving only $\text{NO}_3^-$ and forgetting $\text{NO}_2^-$ .
Writing $\text{Zn}^+$ instead of $\text{Zn}^{2+}$ .
Identifying the cation as $\text{Al}^{3+}$ because both give white hydroxides; the solubility in excess NH3 is the discriminator.

Things to Be Careful About

Include charges on ion formulae.
Use 'or' between the two possible anions, not 'and'.
The deduction depends on correct observations in part (i); if an observation is wrong, the deduction may be wrong.

Techniques used

deduce cation from amphoteric hydroxide behaviourdeduce possible anions from ammonia gas testlink observations to ion identities

(iii)

Describe a test to identify which of the possible anions is present in FB 7.

Carry out your test. Record your observations and conclusion.

test ....................................................................................................................................

observations ......................................................................................................................

conclusion .........................................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Test: Add a few drops of acidified aqueous potassium manganate(VII) to a fresh sample of FB 7.

Observations: The purple colour remains / no decolourisation.

Conclusion: The anion is $\text{NO}_3^-$ (nitrate); nitrite would decolourise the manganate(VII).

Final answer

NO3- (nitrate)

Detailed explanation

Background Concept

Nitrate and nitrite can be distinguished by redox behaviour. Nitrite, $\text{NO}_2^-$ , is a reducing agent: it decolourises acidified potassium manganate(VII), reducing $\text{MnO}_4^-$ (purple) to $\text{Mn}^{2+}$ (colourless/pale pink). Nitrate, $\text{NO}_3^-$ , does not reduce $\text{MnO}_4^-$ under these conditions, so the purple colour remains. Alternatively, adding a mineral acid to a nitrite produces brown nitrogen dioxide gas, whereas nitrate gives no such reaction.

Understanding the Question

Part (ii) gave two possible anions. This part asks for a test that will decide which one is actually present in FB 7. You must name the reagent, carry out the test, record the observation and state the conclusion. The mark scheme accepts either the $\text{KMnO}_4$ test or the acid test.

Approach

Choose a reagent that reacts differently with $\text{NO}_2^-$ and $\text{NO}_3^-$ . Acidified $\text{KMnO}_4$ is the cleanest: it is decolourised by nitrite but not by nitrate. Add a few drops to a fresh sample and observe the colour.

Step-by-Step Reasoning

Add a few drops of acidified aqueous $\text{KMnO}_4$ to a fresh sample of FB 7.
Observe the colour. If the purple colour remains, nitrite is absent, so the anion is $\text{NO}_3^-$ . If the purple colour were discharged, nitrite would be present.
Since FB 7 is $\text{Zn(NO}_3)_2$ , the observation is 'purple colour remains' and the conclusion is $\text{NO}_3^-$ .

Alternative acid test: add dilute sulfuric acid. With nitrate there is no effervescence and no brown gas; with nitrite, brown $\text{NO}_2$ fumes would appear. The conclusion is the same.

Key Takeaways

Nitrite reduces acidified $\text{KMnO}_4$ ; nitrate does not.
A negative test (no change) can be just as diagnostic as a positive one.

Common Mistakes

Using unacidified $\text{KMnO}_4$ ; the redox reaction requires acid.
Saying the $\text{KMnO}_4$ is decolourised for nitrate.
Forgetting to state the conclusion (which anion is present).
Using 'no reaction' without linking it to $\text{NO}_3^-$ .

Things to Be Careful About

Use a fresh sample; do not reuse the mixture from part (i).
Name the reagent precisely: acidified aqueous potassium manganate(VII).
Record the observation as 'purple colour remains' or 'no decolourisation'.
If using the acid test, name the acid and look for effervescence/brown gas.

Techniques used

distinguish nitrite from nitrate using acidified KMnO4record observation and draw conclusion

(b)

(i)

FB 8 and FB 9 are solutions of Group 1 salts. Each contains one anion, both of which are listed in the Qualitative analysis notes. One of the anions contains oxygen but not nitrogen. The other anion is a halide.

Carry out tests to identify the two anions. Record your tests and observations in a suitable form in the space below.

4M Similar

DifficultyMedium

Worked solution

Answer

Test	FB 8	FB 9
Add aqueous silver nitrate	Pale yellow precipitate	White precipitate (may darken)
Add excess aqueous ammonia	Precipitate remains (insoluble)	–
Add dilute hydrochloric acid	No visible change	Effervescence; gas turns limewater milky

Final answer

See table of observations.

Detailed explanation

Background Concept

Halide ions are identified by precipitation with aqueous silver nitrate: AgCl is white, AgBr is cream, AgI is pale yellow. The precipitate's solubility in aqueous ammonia distinguishes them: AgCl dissolves in dilute NH3, AgBr dissolves in concentrated NH3, AgI is insoluble in both. Carbonate is identified by adding an acid: CO2 gas is evolved, which turns limewater milky. Group 1 salts are chosen so the cation does not interfere.

Understanding the Question

You have two Group 1 salt solutions, FB 8 and FB 9. One contains an oxygen-containing anion that is not nitrogen (carbonate), the other a halide. You must carry out tests to identify both anions and record the tests and observations in a suitable form. The mark scheme requires a clear table with test and observations for each salt, and the correct reagents: silver nitrate and a named acid.

Approach

Use silver nitrate to test for the halide; observe the precipitate colour and its solubility in aqueous ammonia. Use a named acid to test for carbonate; test any gas with limewater. Record results in a table with tests as rows and FB 8/FB 9 as columns.

Step-by-Step Reasoning

Add aqueous silver nitrate to samples of FB 8 and FB 9. FB 8 gives a pale yellow precipitate, characteristic of AgI. FB 9 gives a white precipitate (silver carbonate, which may darken).
Add excess aqueous ammonia to the FB 8 precipitate: it remains, confirming AgI (not AgCl or AgBr). This is the key observation that identifies iodide.
Add dilute hydrochloric acid to FB 9: effervescence occurs. Collect the gas and bubble it through limewater; a white precipitate/milky appearance confirms CO2, so the anion is carbonate.
Present these observations in a table.

Key Takeaways

AgI is pale yellow and insoluble in aqueous ammonia.
Carbonate + acid gives CO2, which turns limewater milky.
A results table should have clear row/column headers with test and salt.

Common Mistakes

Using only silver nitrate and not testing solubility in ammonia – you cannot distinguish iodide from bromide.
Using an unnamed acid; the mark scheme requires a named acid.
Forgetting to test the gas with limewater.
Writing observations without identifying the salt.

Things to Be Careful About

Use 'aqueous' for silver nitrate.
Record 'pale yellow' not just 'yellow'.
For the carbonate test, the gas must be identified by limewater.
The table is worth a mark: include test and observations for each salt.

Techniques used

test for halides with silver nitrate and ammoniatest for carbonate with acid and limewaterconstruct a results tablerecord observations systematically

(ii)

Use your observations in (b)(i) to complete Table 3.2 by identifying the formulae of the anions in FB 8 and FB 9.

Table 3.2

	FB 8	FB 9
anion

1M Similar

(iii)

Write an ionic equation for a reaction observed in (b)(i) for one of the anions tested. Include state symbols.

1M Similar

DifficultyEasy

Worked solution

Answer

2\text{H}^+(\text{aq}) + \text{CO}_3^{2-}(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

Final answer

2H+ + CO3^2- -> H2O + CO2

Detailed explanation

Background Concept

An ionic equation shows only the species that actually change, omitting spectator ions. For carbonate + acid, the carbonate ion reacts with H+ to form water and carbon dioxide. For silver nitrate + iodide, Ag+ and I- combine to form insoluble AgI.

Understanding the Question

Write an ionic equation for one reaction observed in (b)(i), with state symbols. You may choose either the iodide precipitation or the carbonate/acid reaction, but the equation must match an observation you actually recorded.

Approach

Choose the carbonate reaction because it is straightforward: H+ + CO3^2- -> H2O + CO2. Balance atoms and charge, add state symbols.

Step-by-Step Reasoning

For the carbonate test with dilute HCl:

The acid provides H+(aq); the carbonate provides CO3^2-(aq).
The products are H2O(l) and CO2(g).
Balance: two H+ are needed to balance the two negative charges and to form one water and one CO2.
Ionic equation: $2\text{H}^+(\text{aq}) + \text{CO}_3^{2-}(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$ .
Alternatively, for the iodide test: $\text{Ag}^+(\text{aq}) + \text{I}^-(\text{aq}) \rightarrow \text{AgI}(\text{s})$ .

Key Takeaways

Ionic equations omit spectator ions.
State symbols are essential: (aq), (l), (g), (s).
Charges must balance.

Common Mistakes

Including Na+ and Cl- as spectator ions.
Missing state symbols.
Writing H+ + CO3^2- -> H2CO3 without decomposition.
For AgI, forgetting (s).

Things to Be Careful About

The equation must match the observation you recorded in (b)(i).
Use correct charges and subscripts.
If you choose the carbonate equation, include the (g) on CO2.

Techniques used

write ionic equation for silver iodide precipitationwrite ionic equation for carbonate with acid