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) Purple crystals → black residue, crackling, oxygen gas (relights glowing splint) · (a)(ii) Test 1: purple → yellow; Test 2: purple → colourless, effervescence, oxygen (relights glowing splint) · (a)(iii) Dark green solution · (b) Test 1: no change; Test 2: white ppt insoluble in acid (sulfate); Test 3: off-white ppt darkening, then black solid + O2 · (c) Test 1: no change; Test 2: chlorine (bleaches litmus); Test 3: oxygen (relights splint) · (d)(i) Manganese (Mn) ·…

9701/31 Oct/Nov 2024 — Chemistry Solved Past Paper

Q1MediumManipulation, Measurement and ObservationPresentation of Data and Observations

Quantitative analysis

Read through the whole method before starting any practical work. Where appropriate, prepare a table for your results in the space provided.

Show the precision of the apparatus you used in the data you record.

Show your working and appropriate significant figures in the answer to each step of your calculations.

The neutralisation of an acid by an alkali is an exothermic reaction. The concentration of an acid can be found by measuring the temperature change when the acid reacts with an alkali.

You will determine the concentration of sulfuric acid by adding aqueous sodium hydroxide of known concentration to the sulfuric acid and measuring the temperature change.

FA 1 is $2.01\text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$ .
FA 2 is sulfuric acid, $\text{H}_2\text{SO}_4$ .

(a)

Method

Support the cup in the $250\text{ cm}^3$ beaker.
Pipette $25.0\text{ cm}^3$ of FA 1 into the cup.
Place the thermometer into the FA 1 in the cup. Tilt the cup if necessary to ensure the bulb of the thermometer is fully covered. Record the temperature of FA 1 in Table 1.1. This is the temperature when the volume of FA 2 is $0.00\text{ cm}^3$ .
Fill the burette with FA 2.
Run $5.00\text{ cm}^3$ of FA 2 into the cup containing FA 1.
Stir the mixture and record the maximum temperature in Table 1.1.
Run further $5.00\text{ cm}^3$ portions of FA 2 into the same cup.
After each addition of FA 2 stir the contents of the cup. Record the maximum temperature for each addition.

Table 1.1

total volume of FA 2 / $\text{cm}^3$	0.00	5.00	10.00	15.00	20.00	25.00	30.00	35.00	40.00
temperature / $^\circ\text{C}$

Keep the rest of FA 2 for use in Question 2.

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Table 1.1 (Example Data)

total volume of FA 2 / $\text{cm}^3$	0.00	5.00	10.00	15.00	20.00	25.00	30.00	35.00	40.00
temperature / $^\circ\text{C}$	20.0	22.5	25.0	27.5	30.0	33.5	32.0	30.5	29.0

Note: The temperatures above are an example set. In a real exam, you must record your own readings from the thermometer. Ensure all readings are recorded to the precision of your apparatus (e.g., ending in $.0$ or $.5$ for a $0.5\text{ }^\circ\text{C}$ scale, or $.0$ , $.1$ , $.2$ etc. for a $0.1\text{ }^\circ\text{C}$ scale).

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

In this practical, the neutralisation reaction between an acid and an alkali is exothermic. As the reaction proceeds, heat is released, causing the temperature of the solution to rise. Once the limiting reactant is completely used up (the equivalence point), no more heat is generated. Further additions of the reactant will simply cool the mixture down because the added room-temperature liquid absorbs heat from the already warmed solution. By plotting temperature against the volume of titrant added, the peak of the curve (or the intersection of the rising and falling lines) identifies the exact volume needed for complete neutralisation.

Understanding the Question

Part (a) asks you to carry out the practical method and record the temperature at each $5.00\text{ cm}^3$ addition of FA 2 ( $\text{H}_2\text{SO}_4$ ) into the $25.0\text{ cm}^3$ of FA 1 ( $\text{NaOH}$ ). The mark scheme rewards correct technique: recording all 9 volumes, using the correct precision (readings ending in $.0$ or $.5$ if using a $0.5\text{ }^\circ\text{C}$ thermometer, or appropriate decimal places for a $0.1\text{ }^\circ\text{C}$ thermometer), and ensuring the data is plausible (within $1.0\text{ }^\circ\text{C}$ or $2.0\text{ }^\circ\text{C}$ of the supervisor's true values).

Approach

Since this is a practical question, the exact numerical answers depend on the actual reagents provided by the supervisor. The strategy is to:

Follow the method precisely, ensuring the thermometer bulb is fully immersed.
Record the initial temperature at $0.00\text{ cm}^3$ .
Add $5.00\text{ cm}^3$ portions of FA 2, stirring well after each addition, and record the maximum temperature reached before it starts to drop.
Record all data to the correct number of decimal places matching the apparatus precision.

Step-by-Step Reasoning

Initial reading: At $0.00\text{ cm}^3$ of FA 2, the temperature is just the initial temperature of the $\text{NaOH}$ solution (e.g., $20.0\text{ }^\circ\text{C}$ ).
Additions: As FA 2 is added, the neutralisation reaction $2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ releases heat. The temperature rises. For example, at $5.00\text{ cm}^3$ it might be $22.5\text{ }^\circ\text{C}$ , at $10.00\text{ cm}^3$ it is $25.0\text{ }^\circ\text{C}$ , and so on.
Maximum temperature: The temperature peaks when the $\text{NaOH}$ is exactly neutralised. In this example, the peak is at $25.00\text{ cm}^3$ ( $33.5\text{ }^\circ\text{C}$ ).
Cooling phase: After the equivalence point, adding more FA 2 does not produce heat. The excess cold liquid cools the mixture, so the temperature drops (e.g., $32.0\text{ }^\circ\text{C}$ at $30.00\text{ cm}^3$ , $30.5\text{ }^\circ\text{C}$ at $35.00\text{ cm}^3$ ).
Precision: Thermometers in these practicals typically have $0.1\text{ }^\circ\text{C}$ or $0.5\text{ }^\circ\text{C}$ graduations. If reading to $0.5\text{ }^\circ\text{C}$ , all values must end in $.0$ or $.5$ . The mark scheme checks that your readings are consistent with this precision and do not deviate wildly from the supervisor's expected values (which accounts for heat loss to the surroundings).

Key Takeaways

Always record the maximum temperature reached after each addition, not the temperature at a random time.
Ensure the thermometer bulb is fully submerged in the liquid to get an accurate reading of the solution's temperature.
Record data to the precision of the measuring instrument (e.g., $0.1\text{ }^\circ\text{C}$ or $0.5\text{ }^\circ\text{C}$ ).

Common Mistakes

Recording the wrong temperature: Taking the temperature too early before the reaction is complete, or taking it after heat loss has significantly cooled the mixture. Always record the highest temperature reached.
Incorrect precision: Writing $22.37\text{ }^\circ\text{C}$ when the thermometer only has $0.5\text{ }^\circ\text{C}$ markings. The mark scheme explicitly rejects readings that do not match the apparatus precision (e.g., not ending in $.0$ or $.5$ for a $0.5$ scale).
Forgetting the initial reading: Not recording the temperature at $0.00\text{ cm}^3$ .

Things to Be Careful About

Stirring: Stir thoroughly after each addition to ensure the heat is evenly distributed and the maximum temperature is reached quickly.
Heat loss: Some heat will always be lost to the cup and surroundings. This is why the maximum temperature is slightly lower than the theoretical value, and why the cooling curve slopes downwards. The intersection of the extrapolated lines corrects for this heat loss.
Supervisor comparison: The mark scheme awards marks based on how close your maximum temperature is to the supervisor's true maximum. If your max temp is $>2.0\text{ }^\circ\text{C}$ lower, you lose marks for technique (III/IV). This usually happens if the student fails to stir properly, reads the temperature too late, or has poor insulation.

Techniques used

record temperature changes during neutralisationuse appropriate significant figures for thermometer readings

(b)

Plot a graph of temperature of solution ( $y$ -axis) against total volume of FA 2 added ( $x$ -axis) on the grid. Select a scale for the $y$ -axis to include a value $3\text{ }^\circ\text{C}$ above your maximum temperature reading. Label any points you consider to be anomalous.

Draw two lines of best fit through the points on your graph. Draw the first line for the increase in temperature and the second line after the maximum temperature was reached. Extrapolate the lines so they intersect. This intersection corresponds to the volume of FA 2 needed to neutralise the FA 1 in your experiment in (a).

Graph Grid 1

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

5M Similar

DifficultyMedium

Worked solution

Working

Graph Description:

x-axis: Total volume of FA 2 / $\text{cm}^3$ (scale: $0$ to $45\text{ cm}^3$ , e.g., $1\text{ cm}^3$ per 2 small squares).
y-axis: Temperature / $^\circ\text{C}$ (scale: e.g., $18\text{ }^\circ\text{C}$ to $36\text{ }^\circ\text{C}$ , ensuring it extends at least $3\text{ }^\circ\text{C}$ above the maximum recorded temperature).
Points: Plot all 9 data points from Table 1.1 accurately.
Lines of best fit:
1. A smooth curve (or straight line) through the rising temperature points (from $0.00$ to the maximum).
2. A straight line (or smooth curve) through the falling temperature points (after the maximum).
Intersection: Extrapolate both lines so they cross. The x-coordinate of this intersection is the volume of FA 2 required for neutralisation.

Using the example data above, the lines intersect at a volume of $25.0\text{ cm}^3$ .

Answer

$25.0\text{ cm}^3$ of FA 2

Final answer

25.0

Detailed explanation

Background Concept

When an exothermic reaction occurs in a calorimeter (like a polystyrene cup), the temperature rises until the limiting reactant is consumed. After this point, adding more reactant at room temperature cools the mixture. Plotting temperature against volume of titrant added yields a curve that rises to a peak and then falls. Because heat is lost to the surroundings during the experiment, the observed peak is lower than the theoretical maximum. To find the true equivalence point (the volume needed for exact neutralisation), we draw two lines of best fit: one for the heating phase and one for the cooling phase, and extrapolate them to intersect. The x-value at this intersection gives the corrected volume of titrant.

Understanding the Question

Part (b) requires you to plot the data collected in part (a) on the provided grid. You must draw two lines of best fit (one for the temperature increase, one for the decrease after the maximum) and extrapolate them to find the volume of FA 2 that exactly neutralised the FA 1. The final answer is this volume, read to 1 decimal place.

Approach

Set up axes: Choose a suitable scale. The x-axis must cover $0.00$ to at least $40.00\text{ cm}^3$ . The y-axis must start below the initial temperature and go at least $3\text{ }^\circ\text{C}$ above your maximum temperature.
Plot points: Carefully plot all 9 (volume, temperature) pairs from your table.
Draw lines: Draw a smooth curve through the rising points. Draw a straight line (or smooth curve) through the falling points. Do not force a single curve through all points; the cooling phase is typically linear or slightly curved due to varying heat loss rates.
Extrapolate: Extend both lines past your data points until they cross.
Read intersection: Read the x-value (volume) at the intersection point to 1 decimal place.

Step-by-Step Reasoning

Axis labels and scales: The x-axis is "total volume of FA 2 / $\text{cm}^3$ ". A good scale is $10\text{ cm}^3$ per $20$ small squares (i.e., $0.5\text{ cm}^3$ per small square). The y-axis is "temperature / $^\circ\text{C}$ ". If your max temp is $33.5\text{ }^\circ\text{C}$ , the axis should go up to at least $36.5\text{ }^\circ\text{C}$ . A scale of $2\text{ }^\circ\text{C}$ per $20$ small squares ( $0.1\text{ }^\circ\text{C}$ per small square) is suitable.
Plotting: Plot $(0.00, 20.0)$ , $(5.00, 22.5)$ , $(10.00, 25.0)$ , $(15.00, 27.5)$ , $(20.00, 30.0)$ , $(25.00, 33.5)$ , $(30.00, 32.0)$ , $(35.00, 30.5)$ , $(40.00, 29.0)$ .
Lines of best fit:
- The heating line passes through $(0.00, 20.0)$ to $(25.00, 33.5)$ . It may be slightly curved (concave down) due to increasing heat loss as the temperature difference with the surroundings grows.
- The cooling line passes through $(25.00, 33.5)$ , $(30.00, 32.0)$ , $(35.00, 30.5)$ , $(40.00, 29.0)$ . This is roughly linear: every $5\text{ cm}^3$ adds $1.5\text{ }^\circ\text{C}$ of cooling.
Intersection: Extrapolating the heating line backwards and the cooling line backwards, they intersect at approximately $25.0\text{ cm}^3$ and $33.5\text{ }^\circ\text{C}$ (in this idealised example).
Reading the value: The volume at the intersection is $25.0\text{ cm}^3$ . This must be read to 1 decimal place as per the mark scheme.

Key Takeaways

The intersection of the two extrapolated lines corrects for heat loss to the surroundings, giving a more accurate equivalence volume than simply reading the peak temperature's volume.
The line for the temperature increase should be a smooth curve; the line for cooling can be straight or curved.
Both lines must be extrapolated to intersect at or above the highest recorded temperature.

Common Mistakes

Drawing a single curve through all points: The mark scheme explicitly requires two lines of best fit. A single smooth curve through the peak will not extrapolate correctly to find the true equivalence point.
Incorrect axis scales: If the y-axis does not extend $3\text{ }^\circ\text{C}$ above the maximum temperature, or if the scales are not linear and suitable (e.g., not based on 1, 2, or 5 per 20 small squares), marks are lost.
Reading the intersection incorrectly: The volume must be read to 1 decimal place. Reading to 0 decimal places or misreading the grid will lose the final accuracy mark.
Plotting points inaccurately: Ensure points are plotted to within half a small square.

Things to Be Careful About

Labelled axes: Both axes must have clear labels with names and units (e.g., "Volume / $\text{cm}^3$ " and "Temperature / $^\circ\text{C}$ ").
Anomalous points: If any point is clearly off the line (e.g., due to a spill or poor stirring), label it as anomalous and do not include it in the line of best fit. However, in a well-run experiment, all points should follow the trend.
Extrapolation: The lines must be extended past the last data points. The intersection should logically occur at or above the highest recorded temperature, representing the theoretical maximum temperature if no heat was lost.

Techniques used

plot temperature against volumedraw lines of best fit for heating and cooling phasesextrapolate to find equivalence point

(c)

3M

(i)

Calculate the amount, in mol, of sodium hydroxide, FA 1, pipetted into the cup.

amount of $\text{NaOH}$ = .............................. mol

1M Similar

DifficultyEasy

Worked solution

Working

\text{amount of NaOH} = \frac{25.0 \times 2.01}{1000} = 5.025 \times 10^{-2} \text{ mol}

Answer

$5.025 \times 10^{-2}$ mol

Final answer

5.025e-2

Detailed explanation

Background Concept

The amount (number of moles) of a solute in a solution is calculated using the equation $n = c \times V$ , where $n$ is the amount in moles, $c$ is the concentration in $\text{mol dm}^{-3}$ , and $V$ is the volume in $\text{dm}^3$ . Since volumes are often given in $\text{cm}^3$ , the formula is adapted to $n = \frac{c \times V_{\text{cm}^3}}{1000}$ .

Understanding the Question

Part (c)(i) asks for the amount, in moles, of sodium hydroxide (FA 1) that was pipetted into the cup. The volume pipetted is $25.0\text{ cm}^3$ and the concentration is $2.01\text{ mol dm}^{-3}$ .

Approach

Use the formula $n = \frac{c \times V}{1000}$ , substituting the given values and calculating the result to 3 or 4 significant figures.

Step-by-Step Reasoning

Concentration of FA 1, $c = 2.01\text{ mol dm}^{-3}$ .
Volume of FA 1, $V = 25.0\text{ cm}^3$ .
Amount of $\text{NaOH} = \frac{2.01 \times 25.0}{1000} = 0.05025\text{ mol}$ .
In scientific notation, this is $5.025 \times 10^{-2}\text{ mol}$ . The mark scheme accepts this or $5.03 \times 10^{-2}\text{ mol}$ (to 3 sf).

Key Takeaways

Always convert volume from $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000 when using concentration in $\text{mol dm}^{-3}$ .
Maintain appropriate significant figures (3 or 4 sf is standard for these calculations).

Common Mistakes

Forgetting to divide by 1000, resulting in an answer of $50.25$ instead of $0.05025$ .
Using the wrong volume (e.g., using the volume of FA 2 from part (b) instead of the $25.0\text{ cm}^3$ of FA 1).

Things to Be Careful About

The mark scheme requires the answer to 3 or 4 significant figures. $5.025 \times 10^{-2}$ is 4 sf, which is acceptable. $5.03 \times 10^{-2}$ is 3 sf, also acceptable. Avoid 2 sf ( $5.0 \times 10^{-2}$ ) as it may lose an accuracy mark.

Techniques used

calculate moles from volume and concentration

(ii)

The equation for this neutralisation reaction is shown.

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)}

Calculate the concentration, in $\text{mol dm}^{-3}$ , of sulfuric acid in FA 2.

Show your working.

concentration of $\text{H}_2\text{SO}_4$ = .............................. $\text{mol dm}^{-3}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Step 1: Calculate moles of $\text{H}_2\text{SO}_4$
From the balanced equation:
$2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$

The stoichiometric ratio of $\text{NaOH}$ to $\text{H}_2\text{SO}_4$ is $2 : 1$ .

\text{amount of H}_2\text{SO}_4 = \frac{\text{amount of NaOH}}{2} = \frac{5.025 \times 10^{-2}}{2} = 2.5125 \times 10^{-2} \text{ mol}

Step 2: Calculate concentration of $\text{H}_2\text{SO}_4$
From part (b), the volume of FA 2 required for neutralisation is $25.0\text{ cm}^3$ .

\text{concentration of H}_2\text{SO}_4 = \frac{\text{amount of H}_2\text{SO}_4 \times 1000}{\text{volume of H}_2\text{SO}_4 \text{ in cm}^3}

\text{concentration of H}_2\text{SO}_4 = \frac{2.5125 \times 10^{-2} \times 1000}{25.0} = 1.005 \text{ mol dm}^{-3}

Answer

$1.005\text{ mol dm}^{-3}$

Final answer

1.005

Detailed explanation

Background Concept

Stoichiometry allows us to relate the amounts of reactants and products in a chemical reaction using the balanced chemical equation. For the neutralisation of sulfuric acid by sodium hydroxide:
$2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$
2 moles of $\text{NaOH}$ react with 1 mole of $\text{H}_2\text{SO}_4$ . Therefore, the amount of $\text{H}_2\text{SO}_4$ is half the amount of $\text{NaOH}$ .
Once the moles of $\text{H}_2\text{SO}_4$ are known, its concentration can be found using $c = \frac{n \times 1000}{V}$ , where $V$ is in $\text{cm}^3$ .

Understanding the Question

Part (c)(ii) asks for the concentration of sulfuric acid (FA 2). You must use the moles of $\text{NaOH}$ calculated in part (c)(i), apply the $2:1$ stoichiometric ratio to find the moles of $\text{H}_2\text{SO}_4$ , and then use the volume of FA 2 obtained from the graph in part (b) to calculate the concentration.

Approach

Use the moles of $\text{NaOH}$ from (c)(i) and divide by 2 to get moles of $\text{H}_2\text{SO}_4$ .
Use the volume of FA 2 from part (b) (e.g., $25.0\text{ cm}^3$ ).
Calculate concentration: $c = \frac{n}{V_{\text{dm}^3}} = \frac{n \times 1000}{V_{\text{cm}^3}}$ .

Step-by-Step Reasoning

Moles of $\text{NaOH}$ : $5.025 \times 10^{-2}\text{ mol}$ (from part c(i)).
Moles of $\text{H}_2\text{SO}_4$ : From the $2:1$ ratio, $n(\text{H}_2\text{SO}_4) = \frac{5.025 \times 10^{-2}}{2} = 2.5125 \times 10^{-2}\text{ mol}$ . (The mark scheme awards a method mark for this step, accepting $2.51 \times 10^{-2}$ or $2.513 \times 10^{-2}$ if using rounded values from (c)(i)).
Volume of $\text{H}_2\text{SO}_4$ : From the graph intersection in part (b), let's assume $25.0\text{ cm}^3$ .
Concentration: $c = \frac{2.5125 \times 10^{-2} \times 1000}{25.0} = \frac{25.125}{25.0} = 1.005\text{ mol dm}^{-3}$ .
Significant figures: The mark scheme requires the final answer to 3 or 4 significant figures. $1.005$ is 4 sf, which is correct. $1.01$ (3 sf) would also be acceptable.

Key Takeaways

Always use the stoichiometric ratio from the balanced equation to find the moles of the unknown reactant.
For diprotic acids like $\text{H}_2\text{SO}_4$ , the mole ratio with a monoprotic base like $\text{NaOH}$ is $1:2$ (acid:base).
The volume used in the concentration calculation must be the corrected volume from the graph intersection, not just the volume at the maximum temperature peak (though in this idealised example they are the same).

Common Mistakes

Wrong stoichiometric ratio: Using a $1:1$ ratio instead of $2:1$ , which would double the calculated concentration of $\text{H}_2\text{SO}_4$ to $2.01\text{ mol dm}^{-3}$ .
Using the wrong volume: Using $25.0\text{ cm}^3$ (the volume of FA 1) instead of the volume of FA 2 read from the graph intersection.
Forgetting to multiply by 1000: Calculating $c = \frac{2.5125 \times 10^{-2}}{25.0} = 0.001005$ , forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ .

Things to Be Careful About

Error carried forward (ecf): If you made a mistake in part (c)(i) but carried it forward correctly to part (c)(ii), you can still earn the method marks for the stoichiometry and concentration calculation, provided your working is consistent.
Significant figures: The final answer must be to 3 or 4 sf. $1.0\text{ mol dm}^{-3}$ (2 sf) will lose an accuracy mark.
Units: Ensure the final answer includes the correct units, $\text{mol dm}^{-3}$ .

Techniques used

use stoichiometric ratio from balanced equationcalculate concentration from moles and volume

Q2MediumPresentation of Data and ObservationsManipulation, Measurement and ObservationAnalysis, Conclusions and Evaluation

Acids react with carbonates to produce carbon dioxide gas.

\text{Na}_2\text{CO}_3\text{(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}

This reaction can be used to determine the concentration of acid, using the mass of carbon dioxide released.

You will determine the concentration of sulfuric acid in FA 2.

FA 2 is the solution used in Question 1.
FA 3 is sodium carbonate, $\text{Na}_2\text{CO}_3$ .

(a)

Method

Use the $25\text{ cm}^3$ measuring cylinder to transfer $25.0\text{ cm}^3$ of FA 2 into the conical flask.
Weigh the flask with the acid. Record the mass.
Weigh the container with FA 3. Record the mass.
Carefully tip all of FA 3 into the acid in the conical flask. Swirl the contents of the flask and leave the flask to stand with occasional swirling until the fizzing stops.
Weigh the container with any residual FA 3. Record the mass.
Calculate and record the mass of FA 3 added to the flask.
Calculate and record the total mass of flask + acid + FA 3.
Weigh the flask and contents when the fizzing has stopped. Record the mass.
Calculate and record the mass of carbon dioxide given off during the experiment.

Results

5M Similar

DifficultyMedium-Easy

Worked solution

Answer

Record the readings in a table with headings and units (all masses in g), e.g.:

Reading	Mass / g
flask + acid (FA 2)	130.00
container + FA 3	12.50
container + residual FA 3	10.00
FA 3 added	2.50
flask + acid + FA 3	132.50
flask and contents after reaction	131.60
carbon dioxide given off	0.90

(Example values — candidate records own readings to 2 or 3 dp.)

Calculations:

mass of FA 3 added = 12.50 − 10.00 = 2.50 g
total mass = 130.00 + 2.50 = 132.50 g
mass of CO2 = 132.50 − 131.60 = 0.90 g

Final answer

See table in working — candidate-dependent readings; example: mass CO2 = 0.90 g

Detailed explanation

Background Concept

This experiment uses the mass-loss (gas evolution) method to determine the concentration of an acid. A known mass of solid sodium carbonate is added to a known volume of sulfuric acid. The reaction produces carbon dioxide gas, which escapes from the open flask. By weighing the flask and contents before and after the reaction, the mass of CO2 lost can be found. From the stoichiometry of the balanced equation, the amount of CO2 equals the amount of H2SO4 that reacted, allowing the acid concentration to be calculated.

The reaction is:

\text{Na}_2\text{CO}_3\text{(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}

The key measurements are four direct balance readings:

Mass of flask + acid (before reaction)
Mass of container + FA 3 (before tipping in)
Mass of container + residual FA 3 (after tipping)
Mass of flask and contents after reaction

From these, three derived values are calculated:

Mass of FA 3 added = reading 2 − reading 3
Total initial mass = reading 1 + (mass of FA 3 added)
Mass of CO2 given off = total initial mass − reading 4

Understanding the Question

Part (a) asks you to record your experimental results in a table. The marks are awarded for: correct headings with units; all four balance readings shown to a consistent 2 or 3 decimal places; and correct calculation of the derived masses (FA 3 added, total mass, and CO2 mass).

Approach

Set up a results table with all the required headings, each with the unit g. Record the four direct balance readings to the same number of decimal places, then calculate the three derived values by simple subtraction.

Step-by-Step Reasoning

Create a table with the required headings. The mark scheme lists seven headings (or equivalent wording), each with unit g:
- (Mass of) flask + acid / FA 2
- (Mass of) container + FA 3
- (Mass of) container (empty/with residue)
- (Mass of) FA 3 added
- (Mass of) flask + acid + FA 3 (added)
- (Mass of) flask and contents after reaction
- (Mass of) carbon dioxide / mass loss
Record the four direct balance readings to 2 or 3 dp consistently.
Calculate mass of FA 3 added = (container + FA 3) − (container + residue).
Calculate total initial mass = (flask + acid) + (mass FA 3 added).
Calculate mass of CO2 = total initial mass − (flask after reaction).

Using the example values in the solution: 12.50 − 10.00 = 2.50 g; 130.00 + 2.50 = 132.50 g; 132.50 − 131.60 = 0.90 g.

Key Takeaways

In a mass-loss experiment, careful weighing before and after is essential.
Derived masses are found by subtraction; the CO2 mass is the difference between the total initial mass and the final mass.
Consistent decimal places (2 or 3 dp) are required for all balance readings.

Common Mistakes

Missing units in table headings.
Mixing 2 dp and 3 dp readings (inconsistent precision).
Recording only some readings instead of all four.
Using the wrong subtraction to find the mass of FA 3 added.
Forgetting that the CO2 mass is the total initial minus final mass.

Things to Be Careful About

The mass of FA 3 added is the difference between the container + FA 3 and the container + residue — NOT the difference between the container + FA 3 and the empty container.
The mass of CO2 is the difference between the total initial mass (flask + acid + FA 3) and the final flask mass.

Techniques used

record balance readings to consistent precisioncalculate derived masses by differenceconstruct a results table with headings and units

(b)

4M

(i)

Calculate the amount, in mol, of carbon dioxide given off in the reaction.

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

1M Similar

(ii)

The sodium carbonate, FA 3, was in excess in the reaction with sulfuric acid. Show by calculation that the sodium carbonate was in excess. Use your answer to (b)(i).

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$M_r(\text{Na}_2\text{CO}_3) = 2(23) + 12 + 3(16) = 106$

$\text{moles Na}_2\text{CO}_3 \text{ added} = \frac{2.50}{106} = 0.0236 \text{ mol}$

From the equation, 1 mol $\text{Na}_2\text{CO}_3$ reacts with 1 mol $\text{H}_2\text{SO}_4$ to give 1 mol $\text{CO}_2$ , so moles $\text{Na}_2\text{CO}_3$ required = moles $\text{CO}_2$ = 0.0205 mol.

Answer

moles $\text{Na}_2\text{CO}_3$ available (0.0236 mol) > moles required (0.0205 mol), so $\text{Na}_2\text{CO}_3$ is in excess.

Final answer

Na2CO3 is in excess (0.0236 mol > 0.0205 mol)

Detailed explanation

Background Concept

To show that a reagent is in excess, compare the amount available with the amount required by the stoichiometry. From the balanced equation:

\text{Na}_2\text{CO}_3\text{(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}

1 mol Na2CO3 reacts with 1 mol H2SO4 to give 1 mol CO2. Therefore, the amount of Na2CO3 required equals the amount of CO2 produced.

The molar mass of Na2CO3 is 2(23) + 12 + 3(16) = 106 g/mol.

Understanding the Question

Show by calculation that the Na2CO3 (FA 3) was in excess. Use the mass of FA 3 from part (a) and the moles of CO2 from part (b)(i).

Approach

Calculate moles of Na2CO3 = mass of FA 3 / 106.
Compare with moles of CO2 (which equals moles of Na2CO3 required). If moles Na2CO3 > moles CO2, then Na2CO3 is in excess.

Step-by-Step Reasoning

Using the example values:

Mass of FA 3 = 2.50 g
Moles Na2CO3 = 2.50 / 106 = 0.0236 mol
Moles CO2 = 0.0205 mol (from b(i))
Since 0.0236 > 0.0205, Na2CO3 is in excess.

Key Takeaways

To identify the excess reagent, compare moles available with moles required.
The stoichiometric ratio (1:1 here) determines the moles required.

Common Mistakes

Using the wrong molar mass for Na2CO3.
Comparing masses instead of moles (which is acceptable only if the molar masses are accounted for).
Not stating the comparison clearly.

Things to Be Careful About

The mark scheme accepts either a comparison of moles or of masses, but the comparison must be explicit.

Techniques used

calculate moles of a solid from masscompare moles to identify the excess reagent

(iii)

Calculate the concentration, in $\text{mol dm}^{-3}$ , of sulfuric acid in FA 2.

concentration of $\text{H}_2\text{SO}_4$ = .............................. $\text{mol dm}^{-3}$

1M Similar

(c)

4M

(i)

A student does not have a conical flask and uses a small beaker for the reaction. Explain why a conical flask is better.

1M Similar

(ii)

Two students made suggestions of how they thought the experiment in (a) could be adapted to determine the concentration of sulfuric acid, FA 2, by using other reactions. In each case their teacher told them that this method was not suitable.

Explain, in each case, why the method is not suitable. Do not consider factors based on quantities of any reagent.

Student 1 suggested using magnesium in place of sodium carbonate.

...........................................................................................................................................

Student 2 suggested using calcium carbonate in place of sodium carbonate.

...........................................................................................................................................

3M Similar

DifficultyMedium

Worked solution

Answer

Student 1 (magnesium): Hydrogen has a very low density, so the mass of gas lost is very small; the error in measuring this small mass loss gives a large percentage error.

Student 2 (calcium carbonate): Calcium sulfate is insoluble (sparingly soluble); it coats the calcium carbonate solid, preventing further reaction, so the reaction does not go to completion.

Final answer

Mg: H2 low density → large % error; CaCO3: CaSO4 insoluble coats solid

Detailed explanation

Background Concept

The mass-loss method relies on measuring the mass of gas produced. For it to work well, the gas must have a significant mass (so the mass loss is measurable with acceptable percentage error), and the reaction must go to completion without any solid product coating the reactant.

Understanding the Question

Explain why magnesium and calcium carbonate are not suitable alternatives to sodium carbonate. Do not consider quantities of reagents.

Approach

For Student 1 (Mg): consider the density of hydrogen and the resulting percentage error in the mass loss.
For Student 2 (CaCO3): consider the solubility of the calcium sulfate product and its effect on the reaction.

Step-by-Step Reasoning

Student 1 (magnesium):
Mg + H2SO4 → MgSO4 + H2. Hydrogen has a very low density, so the mass of H2 produced is very small. The error in measuring this small mass loss gives a large percentage error, making the method unreliable.

Student 2 (calcium carbonate):
CaCO3 + H2SO4 → CaSO4 + CO2 + H2O. Calcium sulfate is insoluble (sparingly soluble) in water. It coats the surface of the calcium carbonate solid, preventing further reaction. As a result, the reaction does not go to completion, and the mass of CO2 measured would be too low.

Key Takeaways

The gas produced must have a significant mass for the mass-loss method to be accurate.
An insoluble product can coat the reactant and stop the reaction prematurely.

Common Mistakes

For Student 1: saying Mg is "too reactive" without mentioning the low density of H2 and the resulting percentage error.
For Student 2: not mentioning that CaSO4 coats the solid / prevents further reaction.

Things to Be Careful About

The instruction says not to consider factors based on quantities of any reagent.
The mark scheme requires the coating effect for Student 2.

Techniques used

evaluate suitability of alternative reagentsrelate gas density to measurement errorrecognise insoluble product coating the reactant

(d)

State the uncertainty in a single reading of your balance.

uncertainty = $\pm$ .............................. g

Calculate the maximum percentage error in the mass of FA 3 that you weighed out in (a).

maximum percentage error = .............................. %

1M Similar

DifficultyMedium-Easy

Worked solution

Working

uncertainty in one balance reading = ±0.01 g (2 dp balance).
Mass of FA 3 is found from two readings, so total uncertainty = 2 × 0.01 = 0.02 g.

$\text{maximum percentage error} = \frac{2 \times 0.01}{2.50} \times 100 = 0.8\%$

Answer

uncertainty = ±0.01 g; maximum percentage error = 0.8% (using example mass 2.50 g)

Final answer

±0.01 g; 0.8%

Detailed explanation

Background Concept

The uncertainty of a measurement depends on the precision of the instrument. A balance reading to 2 decimal places has an uncertainty of ±0.01 g (some balances quote ±0.005 g). The mass of FA 3 is found by difference: mass FA 3 = (container + FA 3) − (container + residue). This involves two balance readings, so the total uncertainty is the sum of the two uncertainties: 2 × U.

The maximum percentage error is:

\text{percentage error} = \frac{\text{total uncertainty}}{\text{mass FA 3}} \times 100

Understanding the Question

State the uncertainty in a single balance reading, then calculate the maximum percentage error in the mass of FA 3 weighed out in part (a).

Approach

State U based on the balance's precision (e.g., ±0.01 g for a 2 dp balance).
Multiply by 2 because two readings are involved.
Divide by the mass of FA 3 and multiply by 100.

Step-by-Step Reasoning

Using the example:

U = ±0.01 g (2 dp balance)
Total uncertainty = 2 × 0.01 = 0.02 g
Mass FA 3 = 2.50 g
Percentage error = (0.02 / 2.50) × 100 = 0.8%

Key Takeaways

Uncertainty in a difference is the sum of the uncertainties of the two readings.
Percentage error = (uncertainty / measured value) × 100.

Common Mistakes

Using U instead of 2U (forgetting that two readings are involved).
Not converting the fraction to a percentage.

Things to Be Careful About

The factor of 2 is essential because the mass of FA 3 is a difference of two balance readings.
The mark scheme accepts U = 0.01 g or 0.005 g for a 2 dp balance, and 0.001 g or 0.0005 g for a 3 dp balance.

Techniques used

state instrument uncertaintypropagate uncertainty through a differencecalculate percentage error

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.

FA 4, FA 5 and FA 6 are compounds of the same metal in different oxidation states.

(a)

5M

(i)

Place a small spatula measure of FA 4 in a hard-glass test-tube. Heat the tube gently at first and then more strongly.

Record all your observations.

Leave the tube and contents to cool and keep for use in (a)(iii).

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Dark purple crystals initially
On heating: solid crackles / pops and jumps about
Black powder / residue forms
Gas evolved relights a glowing splint

Final answer

Purple crystals → black residue, crackling, oxygen gas (relights glowing splint)

Detailed explanation

Background Concept

Potassium manganate(VII), $\text{KMnO}_4$ , is a strong oxidising agent that is thermally unstable. On strong heating it decomposes according to:

2\text{KMnO}_4(\text{s}) \rightarrow \text{K}_2\text{MnO}_4(\text{s}) + \text{MnO}_2(\text{s}) + \text{O}_2(\text{g})

The purple crystals ( $\text{MnO}_4^-$ is intensely purple) decompose to a black solid ( $\text{MnO}_2$ , mixed with some dark green $\text{K}_2\text{MnO}_4$ ) and evolve oxygen gas. Oxygen relights a glowing splint — the standard test for oxygen.

Understanding the Question

You are asked to heat a small spatula measure of FA 4 (which is $\text{KMnO}_4$ ) in a hard-glass test-tube, gently then strongly, and record all observations. The instruction to use a hard-glass test-tube signals that a solid is being heated strongly — the tube must withstand the temperature. You must record colour changes, any gas evolved and its test, and any physical behaviour of the solid (crackling, jumping).

Approach

Heat gently first (to drive off any moisture without violent decomposition), then more strongly to decompose the permanganate. Watch the solid: its colour, whether it moves or crackles, and whether a gas is given off. Test any gas with a glowing splint — oxygen relights it. Record the colour of the residue.

Step-by-Step Reasoning

Initially the solid is dark purple — the colour of $\text{KMnO}_4$ crystals.
As it is heated strongly, the crystals decompose vigorously: they may 'jump about' or 'crackle / pop' as oxygen is released rapidly.
The residue turns black — this is $\text{MnO}_2$ (manganese(IV) oxide), mixed with some dark green $\text{K}_2\text{MnO}_4$ (manganate(VI)), but the overall appearance is black.
A gas is evolved. Test with a glowing splint: it relights, confirming oxygen. The decomposition is:

2\text{KMnO}_4(\text{s}) \rightarrow \text{K}_2\text{MnO}_4(\text{s}) + \text{MnO}_2(\text{s}) + \text{O}_2(\text{g})

Key Takeaways

$\text{KMnO}_4$ decomposes on heating to give $\text{O}_2$ and a black residue ( $\text{MnO}_2$ ).
The glowing splint test confirms oxygen.
Record observations at each stage: initial colour, behaviour on heating, colour of residue, gas test.

Common Mistakes

Writing 'brown' instead of 'black' for the residue — the mark scheme accepts black powder/residue/solid.
Forgetting to test the gas with a glowing splint, or writing 'limewater turns milky' (that is $\text{CO}_2$ , not $\text{O}_2$ ).
Not recording the crackling/jumping of the solid — this is a creditable observation.
Using a boiling tube instead of a hard-glass test-tube for heating a solid.

Things to Be Careful About

Record the colour change (purple → black) and the physical behaviour (crackling).
The gas test must be stated: 'relights a glowing splint' — the identifying test for oxygen.
The mark scheme rounds down: with 4 observations worth 2 marks, you need all of them for full marks.

Techniques used

heat a solid in a hard-glass test-tubeobserve colour changes and physical behaviour on heatingtest for oxygen with a glowing splint

(ii)

Dissolve a small spatula measure of FA 4 in approximately $5\text{ cm}$ depth of distilled water in a boiling tube and add approximately $1\text{ cm}$ depth of dilute sulfuric acid.

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

Table 3.1

test	observations
Test 1 To a $1\text{ cm}$ depth of aqueous FA 4 in a test-tube add aqueous iron(II) sulfate.
Test 2 To a $1\text{ cm}$ depth of aqueous FA 4 in a test-tube add hydrogen peroxide.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Test 1 (+ $\text{FeSO}_4(\text{aq})$ ): Purple solution; becomes yellow solution.

Test 2 (+ $\text{H}_2\text{O}_2(\text{aq})$ ): Purple solution decolourises / becomes colourless; effervescence / fizzing; gas relights a glowing splint.

Final answer

Test 1: purple → yellow; Test 2: purple → colourless, effervescence, oxygen (relights glowing splint)

Detailed explanation

Background Concept

$\text{KMnO}_4$ (FA 4) is a powerful oxidising agent; the $\text{MnO}_4^-$ ion is intensely purple. In acidic solution it is reduced to colourless $\text{Mn}^{2+}$ . $\text{Fe}^{2+}$ (from iron(II) sulfate) is oxidised to $\text{Fe}^{3+}$ , which is yellow in aqueous solution. Hydrogen peroxide, $\text{H}_2\text{O}_2$ , also acts as a reducing agent here: it reduces $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ (decolourising the purple), and is itself oxidised to oxygen gas, which gives effervescence and relights a glowing splint.

Understanding the Question

FA 4 is dissolved in water with dilute sulfuric acid (to provide the acidic medium for the redox reactions). Two tests are done: adding iron(II) sulfate, and adding hydrogen peroxide. For each, record the observations — colour changes, effervescence, and any gas test.

Approach

Recognise that both tests are redox reactions involving the purple $\text{MnO}_4^-$ ion. In each case, watch for the disappearance of the purple colour (reduction of Mn(VII) to Mn(II)) and for any gas produced. For the $\text{H}_2\text{O}_2$ test, the gas is oxygen — confirm with a glowing splint.

Step-by-Step Reasoning

Test 1: The solution is purple ( $\text{MnO}_4^-$ ). Adding $\text{FeSO}_4$ introduces $\text{Fe}^{2+}$ ions. $\text{Fe}^{2+}$ reduces $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ (colourless/pale pink), while $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ , which is yellow in solution. So the observation is: purple solution becomes yellow.
Test 2: Adding $\text{H}_2\text{O}_2$ to the purple solution. $\text{H}_2\text{O}_2$ reduces $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ (purple decolourises), and is oxidised to $\text{O}_2$ :

\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-

The gas bubbles (effervescence) and relights a glowing splint — oxygen.

Key Takeaways

$\text{MnO}_4^-$ is purple; its reduction to $\text{Mn}^{2+}$ decolourises the solution.
$\text{Fe}^{3+}$ is yellow in aqueous solution.
$\text{H}_2\text{O}_2$ can act as a reducing agent, being oxidised to $\text{O}_2$ .
The glowing splint test identifies oxygen.

Common Mistakes

Writing 'becomes brown' instead of 'yellow' for the $\text{Fe}^{3+}$ — the mark scheme specifically wants yellow.
Forgetting the effervescence and gas test in Test 2.
Not specifying that the purple solution decolourises.

Things to Be Careful About

The acid (dilute $\text{H}_2\text{SO}_4$ ) is added to keep the medium acidic — needed for the redox reactions to proceed as described.
Record the colour change precisely: 'purple → yellow' for Test 1, 'purple → colourless' for Test 2.
The mark scheme rounds down: 5 observations worth 2 marks — you need them all for full marks.

Techniques used

observe redox colour changes on adding reagentstest for oxygen with a glowing splintrecord observations at each stage

(iii)

To the cooled test-tube in (a)(i) add a $5\text{ cm}$ depth of distilled water. Observe and record the colour formed.

1M Similar

(b)

Dissolve a small spatula measure of FA 5 in a boiling tube half-filled with distilled water. Warming may be needed to dissolve the FA 5.

Carry out the following tests and record your observations in Table 3.2.

For each of the tests use a $1\text{ cm}$ depth of this FA 5 solution in a test-tube.

Table 3.2

test	observations
Test 1 Add dilute nitric acid, then add aqueous silver nitrate.
Test 2 Add aqueous barium chloride or barium nitrate, then add dilute nitric acid.
Test 3 Add aqueous sodium hydroxide, then add hydrogen peroxide.

3M Similar

DifficultyMedium

Worked solution

Answer

Test 1 (+ $\text{HNO}_3$ , then $\text{AgNO}_3$ ): No change with nitric acid; no change with silver nitrate.

Test 2 (+ $\text{BaCl}_2$ / $\text{Ba(NO}_3)_2$ , then $\text{HNO}_3$ ): White precipitate with barium ions; precipitate does not dissolve on adding dilute nitric acid.

Test 3 (+ $\text{NaOH}$ , then $\text{H}_2\text{O}_2$ ): Off-white / cream / fawn / pale brown precipitate with NaOH; precipitate darkens and is insoluble in excess NaOH. On adding $\text{H}_2\text{O}_2$ : black / dark brown solid forms; effervescence; gas relights a glowing splint.

Final answer

Test 1: no change; Test 2: white ppt insoluble in acid (sulfate); Test 3: off-white ppt darkening, then black solid + O2

Detailed explanation

Background Concept

FA 5 is $\text{MnSO}_4$ , containing $\text{Mn}^{2+}$ and sulfate ions. The tests probe the anion and cation:

$\text{AgNO}_3$ (after acidifying with $\text{HNO}_3$ ) tests for halides — sulfate gives no precipitate, so no change.
$\text{BaCl}_2$ / $\text{Ba(NO}_3)_2$ tests for sulfate: $\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4$ , a white precipitate insoluble in dilute acid.
$\text{NaOH}$ tests for the cation: $\text{Mn}^{2+}$ gives $\text{Mn(OH)}_2$ , an off-white/cream precipitate that darkens in air (oxidation to $\text{MnO}_2$ ) and is insoluble in excess $\text{NaOH}$ .
$\text{H}_2\text{O}_2$ oxidises $\text{Mn(OH)}_2$ to $\text{MnO}_2$ (black/dark brown) and decomposes to $\text{O}_2$ (effervescence, relights splint).

Understanding the Question

Three tests on the FA 5 solution ( $\text{MnSO}_4$ ). Record observations for each. The tests identify the sulfate anion (Test 2) and the $\text{Mn}^{2+}$ cation (Test 3), and confirm absence of halides (Test 1).

Approach

Work through each test: what reagent is added, what ion it reacts with, and what is observed. For Test 1, $\text{AgNO}_3$ gives no reaction with sulfate. For Test 2, $\text{Ba}^{2+}$ precipitates sulfate as white $\text{BaSO}_4$ , insoluble in acid. For Test 3, $\text{NaOH}$ precipitates $\text{Mn(OH)}_2$ (off-white), which darkens; $\text{H}_2\text{O}_2$ then oxidises it to black $\text{MnO}_2$ and gives off $\text{O}_2$ .

Step-by-Step Reasoning

Test 1: $\text{HNO}_3$ acidifies the solution (removes carbonate interference). $\text{AgNO}_3$ would precipitate $\text{AgCl}$ / $\text{AgBr}$ / $\text{AgI}$ if halides were present — but sulfate gives no precipitate, so no change.
Test 2: $\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4(\text{s})$ , white precipitate. Adding dilute $\text{HNO}_3$ : $\text{BaSO}_4$ is insoluble in acid, so the precipitate does not dissolve. (This distinguishes sulfate from carbonate, which would dissolve with effervescence.)
Test 3: $\text{Mn}^{2+} + 2\text{OH}^- \rightarrow \text{Mn(OH)}_2(\text{s})$ , off-white/cream precipitate. It darkens on standing (air oxidation to $\text{MnO}_2$ ) and is insoluble in excess $\text{NaOH}$ . Adding $\text{H}_2\text{O}_2$ : $\text{Mn(OH)}_2$ is oxidised to black/dark brown $\text{MnO}_2$ ; $\text{H}_2\text{O}_2$ also decomposes to $\text{O}_2$ , giving effervescence and a splint that relights.

Key Takeaways

$\text{BaSO}_4$ is a white precipitate insoluble in dilute acid — the test for sulfate.
$\text{AgNO}_3$ tests for halides; sulfate gives no precipitate.
$\text{Mn(OH)}_2$ is off-white/cream, darkens in air, insoluble in excess $\text{NaOH}$ .
$\text{H}_2\text{O}_2$ oxidises Mn(II) to Mn(IV) ( $\text{MnO}_2$ , black) and releases $\text{O}_2$ .

Common Mistakes

Writing 'white precipitate that dissolves in acid' for $\text{BaSO}_4$ — it does NOT dissolve in dilute acid.
Confusing the $\text{Mn(OH)}_2$ colour (off-white/cream) with white.
Forgetting that the ppt darkens or is insoluble in excess $\text{NaOH}$ .
Not testing the gas from $\text{H}_2\text{O}_2$ with a glowing splint.

Things to Be Careful About

The order matters: acidify with $\text{HNO}_3$ before $\text{AgNO}_3$ (to remove carbonates), and add acid after $\text{BaCl}_2$ to confirm the precipitate is insoluble sulfate.
The mark scheme requires the specific colour words: off-white/cream/fawn/buff/pale brown for $\text{Mn(OH)}_2$ , black/dark brown for $\text{MnO}_2$ .
The mark scheme rounds down: 7 observations worth 3 marks — you need them all for full marks.

Techniques used

test for sulfate with barium ions and dilute acidtest for halides with silver nitrate and nitric acidtest for Mn2+ with sodium hydroxidetest for oxygen with a glowing splint

(c)

Carry out the following tests and record your observations in Table 3.3. Identify any gases produced.

For each of the tests use a small spatula measure of FA 6 in a test-tube.

Table 3.3

test	observations and gases produced
Test 1 Add a $1\text{ cm}$ depth of dilute nitric acid.
Test 2 Add a few drops of concentrated hydrochloric acid. CARE Hydrochloric acid is corrosive. Fill the test-tube with water as soon as you have made your observation.
Test 3 Add a $1\text{ cm}$ depth of hydrogen peroxide.

3M Similar

DifficultyMedium

Worked solution

Answer

Test 1 (+ dilute $\text{HNO}_3$ ): No change; solid does not dissolve.

Test 2 (+ conc $\text{HCl}$ ): Bubbles / effervescence; gas bleaches damp litmus paper; chlorine gas produced.

Test 3 (+ $\text{H}_2\text{O}_2$ ): Effervescence; gas relights a glowing splint; oxygen gas produced.

Final answer

Test 1: no change; Test 2: chlorine (bleaches litmus); Test 3: oxygen (relights splint)

Detailed explanation

Background Concept

FA 6 is $\text{MnO}_2$ , manganese(IV) oxide — a black, insoluble solid. It is unreactive towards dilute acids (no change with $\text{HNO}_3$ ). With concentrated $\text{HCl}$ it acts as an oxidising agent:

\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}

producing chlorine gas (green-yellow, bleaches damp litmus). $\text{MnO}_2$ also catalyses the decomposition of $\text{H}_2\text{O}_2$ :

2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2

producing oxygen (relights a glowing splint).

Understanding the Question

Three tests on solid FA 6 ( $\text{MnO}_2$ ): dilute $\text{HNO}_3$ , concentrated $\text{HCl}$ , and $\text{H}_2\text{O}_2$ . Record observations and identify any gases.

Approach

Recognise $\text{MnO}_2$ is insoluble in dilute acids and unreactive towards them. With concentrated $\text{HCl}$ , the Mn(IV) oxidises $\text{Cl}^-$ to $\text{Cl}_2$ . With $\text{H}_2\text{O}_2$ , $\text{MnO}_2$ is a catalyst for its decomposition to $\text{O}_2$ . Identify each gas by its test (chlorine bleaches litmus; oxygen relights a glowing splint).

Step-by-Step Reasoning

Test 1: $\text{MnO}_2$ is insoluble in dilute nitric acid — no change, solid does not dissolve.
Test 2: Concentrated $\text{HCl}$ . $\text{MnO}_2$ oxidises $\text{Cl}^-$ to $\text{Cl}_2$ :

\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}

Chlorine is a green-yellow gas that bleaches damp litmus paper. Bubbles of gas are seen.

Test 3: $\text{H}_2\text{O}_2$ . $\text{MnO}_2$ catalyses $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$ . Oxygen is evolved — effervescence, and a glowing splint relights.

Key Takeaways

$\text{MnO}_2$ is insoluble in dilute acids.
$\text{MnO}_2$ + conc $\text{HCl}$ → $\text{Cl}_2$ (chlorine bleaches litmus).
$\text{MnO}_2$ catalyses $\text{H}_2\text{O}_2$ decomposition to $\text{O}_2$ .
Gas tests: $\text{Cl}_2$ bleaches damp litmus; $\text{O}_2$ relights a glowing splint.

Common Mistakes

Writing 'dissolves' for $\text{MnO}_2$ in $\text{HNO}_3$ — it does not.
Forgetting to identify chlorine by its bleaching of litmus.
Writing 'limewater turns milky' for the gas from $\text{H}_2\text{O}_2$ — it is oxygen, not $\text{CO}_2$ .
Not stating that $\text{MnO}_2$ is a catalyst for $\text{H}_2\text{O}_2$ decomposition.

Things to Be Careful About

The concentrated $\text{HCl}$ must be handled with care (corrosive) — the question warns to fill the tube with water after observing.
Chlorine is identified by bleaching damp litmus; oxygen by relighting a glowing splint.
The mark scheme rounds down: 6 observations worth 3 marks — you need them all for full marks.

Techniques used

test solubility in dilute acidtest for chlorine with damp litmustest for oxygen with a glowing splintidentify gases from observations

(d)

3M

(i)

Use your observations from the tests on FA 4, FA 5 and FA 6 to suggest the identity of the metal present in all 3 compounds.

Metal identity ..............................

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Manganese (Mn)

Final answer

Manganese (Mn)

Detailed explanation

Background Concept

Manganese forms compounds in many oxidation states, each with characteristic colours: $\text{MnO}_4^-$ (+7) is purple, $\text{Mn}^{2+}$ (+2) is pale pink/almost colourless, $\text{MnO}_2$ (+4) is black, $\text{MnO}_4^{2-}$ (+6) is dark green. The observations across FA 4, FA 5 and FA 6 — purple, off-white precipitates darkening to black, black solid, dark green solution — all point to manganese.

Understanding the Question

Using the observations from all the tests, suggest the identity of the metal present in all three compounds. The answer is manganese.

Approach

Correlate the characteristic colours and reactions: purple permanganate (FA 4), $\text{Mn}^{2+}$ giving off-white $\text{Mn(OH)}_2$ darkening to black $\text{MnO}_2$ (FA 5), black $\text{MnO}_2$ reacting with conc $\text{HCl}$ to give $\text{Cl}_2$ (FA 6). These are all manganese chemistry.

Step-by-Step Reasoning

FA 4 is purple — $\text{MnO}_4^-$ (permanganate), Mn in +7.
FA 5 gives an off-white precipitate with $\text{NaOH}$ that darkens — $\text{Mn(OH)}_2$ , typical of $\text{Mn}^{2+}$ .
FA 6 is a black solid that produces $\text{Cl}_2$ with conc $\text{HCl}$ and catalyses $\text{H}_2\text{O}_2$ decomposition — $\text{MnO}_2$ .
The green solution in (a)(iii) ( $\text{MnO}_4^{2-}$ ) is also characteristic of manganese.

All three are compounds of manganese.

Key Takeaways

Manganese has distinctive colours across oxidation states: purple (+7), green (+6), black (+4), pale pink/colourless (+2).
Recognising these colours allows identification of the metal.

Common Mistakes

Suggesting a different transition metal (e.g. iron) — the purple permanganate and green manganate colours are unique to manganese.
Not linking the colours to the metal.

Things to Be Careful About

The answer must be 'manganese' or 'Mn'.

Techniques used

deduce metal from characteristic colours and reactionscorrelate observations across three compounds

(ii)

Identify the oxidation state of the metal in each compound.

FA 4 ..............................

FA 5 ..............................

FA 6 ..............................

2M Similar

Chemistry 9701/31 — October/November 2024

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer