Chemistry 9701/31 — May/June 2013

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

questions

marks

120

minutes

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

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

The reaction between sulfuric acid and sodium hydroxide is exothermic.

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

By measuring the temperature changes that occur when different volumes of the acid are added to a fixed volume of the alkali, it is possible to determine the neutralisation point. This is the point at which just enough acid has been added to react with all the alkali present.

The aim of the investigation is to determine the concentration of the sulfuric acid.

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

Read through the instructions carefully and prepare a table for your results before starting any practical work.

(a)

Method

Support a plastic cup in a $250\text{ cm}^3$ beaker.
Use a pipette to transfer $25.0\text{ cm}^3$ of FA 1 into the plastic cup.
Record the temperature of FA 1, $T_1$ , in the space below.

T_1 = \dots\dots\dots\dots\dots ^\circ\text{C}

Fill the burette labelled FA 2 with FA 2.
Add $5.00\text{ cm}^3$ of FA 2 from the burette to the plastic cup.
Stir the mixture thoroughly and record the temperature of the solution.
Add a further $5.00\text{ cm}^3$ of FA 2 to the plastic cup and again record the temperature.
Repeat the addition of $5.00\text{ cm}^3$ portions of FA 2 until you have added a total of $50.00\text{ cm}^3$ of FA 2 to the plastic cup. Measure the temperature after each addition.
Record in your table below the total volume of FA 2 added and the temperature of the solution after each addition.

DifficultyMedium-Easy

Worked solution

Answer

A results table is constructed with the following columns and 10 rows of data (volumes from 5.00 to 50.00 cm³ in 5.00 cm³ increments):

Volume of FA 2 added /cm³	Temperature of solution /°C
0.00	$T_1$ (e.g. 21.0)
5.00	(e.g. 23.5)
10.00	(e.g. 25.0)
15.00	(e.g. 26.5)
20.00	(e.g. 27.5)
25.00	(e.g. 27.0)
30.00	(e.g. 26.0)
35.00	(e.g. 25.0)
40.00	(e.g. 24.0)
45.00	(e.g. 23.5)
50.00	(e.g. 22.5)

Requirements:

All volumes recorded to 2 decimal places (e.g. 5.00, 10.00).
All temperatures recorded to the nearest 0.5 °C (at least one ending in .0 and one ending in .5).
$T_1$ recorded separately before any acid is added.

Final answer

Candidate-dependent data table with 10 volumes (5.00–50.00 cm³) and corresponding temperatures to nearest 0.5 °C

Detailed explanation

Background Concept

In a thermochemistry practical, the temperature change of a reaction is measured to determine when a reaction is complete. The neutralisation of a strong acid with a strong base is exothermic, so the temperature rises as acid is added until all the base has reacted. Beyond the neutralisation point, adding more (cooler) acid dilutes the solution and the temperature falls. The maximum temperature (or the intersection of two lines of best fit on a processed graph) identifies the neutralisation volume.

Understanding the Question

Part (a) asks the candidate to carry out the experiment and record raw data. The mark scheme rewards: (1) a table with space for 10 volume entries, (2) correct headings with units and consistent decimal places for volumes, (3) temperatures recorded to the nearest 0.5 °C with at least one reading ending .0 and one ending .5, and (4–5) quality marks based on how close the temperature rise at 25 cm³ is to the supervisor's value.

Approach

Set up a table before starting. Record $T_1$ first. Then add 5.00 cm³ portions of FA 2 from the burette, stir, and record the temperature after each addition up to 50.00 cm³ total. Use a thermometer that reads to 0.5 °C (or estimate to the nearest 0.5 °C from a 0–50 °C thermometer).

Step-by-Step Reasoning

Table layout (B1): The table must have 10 rows for the 10 additions (5, 10, 15, 20, 25, 30, 35, 40, 45, 50 cm³). A row for $T_1$ at volume 0 is also appropriate.
Headings and units (B1): Column headings must include the quantity and unit, e.g. "Volume of FA 2 / cm³" and "Temperature / °C". All volume entries must be to the same number of decimal places (2 d.p. since the burette reads to 0.05 cm³).
Temperature recording (B1): The thermometer scale allows readings to the nearest 0.5 °C. The mark scheme explicitly requires at least one reading ending in .0 and one ending in .5 to demonstrate the candidate is reading to 0.5 °C precision rather than rounding to whole degrees.
Quality (B2): The temperature rise at 25 cm³ of FA 2 added is compared to the supervisor's value. Within ±1 °C earns 2 marks; within ±2 °C earns 1 mark. This rewards careful technique: thorough stirring, waiting for the maximum temperature, and reading the thermometer at eye level.

Key Takeaways

Always prepare the table before starting the experiment.
Record all readings to the precision the apparatus allows.
Consistency in decimal places across a column is essential for PDO marks.
Stirring and reading the thermometer correctly directly affect the quality marks.

Common Mistakes

Recording temperatures to whole degrees only (e.g. 24, 25, 26) — the mark scheme requires readings to 0.5 °C.
Inconsistent decimal places in the volume column (e.g. 5, 10, 15 instead of 5.00, 10.00, 15.00).
Omitting units from column headings.
Not recording $T_1$ separately before the first addition.

Things to Be Careful About

The burette reads to 0.05 cm³, so volumes should be recorded to 2 decimal places.
The thermometer must be read to the nearest 0.5 °C — do not interpolate to 0.1 °C as this is beyond the scale's precision.
Stir thoroughly after each addition and record the maximum temperature reached.

Techniques used

construct a results table with appropriate headings and unitsrecord temperature readings to the nearest 0.5 °Cadd measured volumes from a burette in fixed increments

(b)

After each addition of acid, the temperature rise, $\Delta T$ , is given by,

\Delta T = \text{temperature recorded} - T_1

The total volume of solution in the plastic cup, $V_T$ is given by,

V_T = \text{volume of FA 2} + \text{volume of FA 1}

The heat given out by the reaction is proportional to the temperature rise, $\Delta T$ , multiplied by the total volume of solution in the plastic cup, $V_T$ .

Use your experimental results to complete the following table.

You should include:

the volume of FA 2
the total volume in the plastic cup, $V_T$
the temperature of the solution
the temperature rise, $\Delta T$
the total volume $\times$ the temperature rise, $(V_T \times \Delta T)$

Q1_table_b

DifficultyMedium-Easy

Worked solution

Answer

Using the representative data from part (a), with $T_1 = 21.0\ ^\circ\text{C}$ :

Volume of FA 2 /cm³	$V_T$ /cm³	Temperature /°C	$\Delta T$ /°C	$V_T \times \Delta T$ /cm³·°C
5.00	30.00	23.5	2.5	75.0
10.00	35.00	25.0	4.0	140.0
15.00	40.00	26.5	5.5	220.0
20.00	45.00	27.5	6.5	292.5
25.00	50.00	27.0	6.0	300.0
30.00	55.00	26.0	5.0	275.0
35.00	60.00	25.0	4.0	240.0
40.00	65.00	24.0	3.0	195.0
45.00	70.00	23.5	2.5	175.0
50.00	75.00	22.5	1.5	112.5

Where:

$V_T = V_{\text{FA 2}} + 25.0\text{ cm}^3$
$\Delta T = T - T_1$
Product = $V_T \times \Delta T$

Final answer

Candidate-dependent table with correctly calculated $\Delta T$ , $V_T$ , and $V_T \times \Delta T$ columns

Detailed explanation

Background Concept

The quantity $V_T \times \Delta T$ is proportional to the heat evolved by the reaction. The reasoning is: $q = mc\Delta T$ , and since the specific heat capacity and density of the dilute aqueous solution are approximately constant, the mass is proportional to $V_T$ . Therefore $q \propto V_T \times \Delta T$ . Plotting this against the volume of acid added gives a graph that rises to a maximum and then falls, with the peak (or intersection of the two linear portions) indicating the neutralisation point.

Understanding the Question

Part (b) asks the candidate to extend the raw data table by adding three calculated columns: $V_T$ , $\Delta T$ , and $V_T \times \Delta T$ . The mark scheme requires at least 8 correctly calculated results.

Approach

For each row: add 25.0 to the volume of FA 2 to get $V_T$ ; subtract $T_1$ from the recorded temperature to get $\Delta T$ ; multiply these two values to get the product.

Step-by-Step Reasoning

$V_T$ (total volume): The fixed volume of FA 1 is 25.0 cm³. So $V_T = V_{\text{FA 2}} + 25.0$ . For example, at 15.00 cm³ of FA 2, $V_T = 40.00$ cm³.
$\Delta T$ : Subtract the initial temperature $T_1$ from the recorded temperature. If $T_1 = 21.0$ °C and the reading is 26.5 °C, then $\Delta T = 5.5$ °C.
$V_T \times \Delta T$ : Multiply the two calculated values. $40.00 \times 5.5 = 220.0$ cm³·°C.
The mark scheme says "assume correct data from (a)" — so if the raw data is consistent, the calculations just need to be arithmetically correct. A minimum of 8 results must be correct for the B1.

Key Takeaways

$V_T$ accounts for the increasing total volume as acid is added — this correction is what makes the graph peak at the true neutralisation point rather than being skewed by dilution.
The product $V_T \times \Delta T$ corrects for the fact that later additions are heating a larger volume of solution.
All three calculated columns must be internally consistent with the raw data.

Common Mistakes

Forgetting to add 25.0 cm³ when calculating $V_T$ (using only the volume of FA 2).
Using $T_1$ incorrectly or forgetting to subtract it.
Arithmetic errors in the multiplication, especially with decimal places.
Recording $\Delta T$ as negative for readings below $T_1$ (should not happen if data is good, but can occur at large volumes due to heat loss).

Things to Be Careful About

Significant figures: $V_T$ should be to 2 d.p. (matching the burette readings), $\Delta T$ to 1 d.p. (matching the thermometer precision), and the product to a reasonable number of figures.
The mark scheme requires a minimum of 8 correct results — ensure all rows are completed.

Techniques used

calculate temperature rise from initial temperaturecalculate total volume of solutionmultiply total volume by temperature rise

(c)

(i)

On the grid below, plot the values of $(V_T \times \Delta T)$ on the $y$ -axis against the volume of FA 2 on the $x$ -axis.

Q1_grid_c

DifficultyMedium-Easy

Worked solution

Answer

Plot $(V_T \times \Delta T)$ on the $y$ -axis against volume of FA 2 on the $x$ -axis.

$y$ -axis: $V_T \times \Delta T$ (units: cm³·°C)
$x$ -axis: Volume of FA 2 (units: cm³)
Choose scales so that plotted points occupy more than half of each axis.
Plot all points accurately (within half a small square), using at least 8 points.
Points should show a rise to a maximum and then a fall (approximately V-shaped or inverted-V profile).

Final answer

Graph of $V_T \times \Delta T$ against volume of FA 2 showing rise and fall, with all points plotted accurately

Detailed explanation

Background Concept

Graphical analysis of experimental data is a core practical skill. In this experiment, the graph of $V_T \times \Delta T$ versus volume of acid added should show a linear increase (before neutralisation, all added acid reacts and heat is proportional to moles reacted) followed by a linear decrease (after neutralisation, no more reaction occurs and the temperature falls as cooler acid is added). The intersection of these two linear portions gives the neutralisation volume.

Understanding the Question

Part (c)(i) is worth 3 marks: one for correct axis labels, one for appropriate scale choice, and one for accurate plotting. The graph grid provided has fine squares (1 mm) with larger squares (5 mm), so the examiner checks that points fall within half a small square of their correct position.

Approach

Decide on axis ranges: $x$ -axis from 0 to 50 cm³ (or 5 to 50), $y$ -axis from 0 to slightly above the maximum $V_T \times \Delta T$ value.
Choose a scale that spreads points over more than half the grid in each direction.
Plot each point carefully using a sharp pencil.

Step-by-Step Reasoning

Axis labels (B1): The $y$ -axis must be labelled $V_T \times \Delta T$ and the $x$ -axis must be labelled "Volume of FA 2" (or equivalent). Units may be included but are not required by the mark scheme ("ignore units").
Scale (B1): The points must span more than half the available grid in both directions. With 10 points from 5 to 50 cm³ on the $x$ -axis, a scale of 1 cm³ per small square or 2 cm³ per small square works. On the $y$ -axis, the range of $V_T \times \Delta T$ values (e.g. 75 to 300) should fill more than half the grid height.
Plotting (B1): All points must be within half a small square of the correct position. The examiner specifically checks points at $V$ = 5, 10, 15, 20, and 25 cm³. A minimum of 8 points must be correctly plotted.

Key Takeaways

Always label axes with the quantity being plotted (not just "volume" or "temperature").
Choose scales that maximise the spread of points across the grid.
Use a sharp pencil and mark points as small crosses or dots.

Common Mistakes

Plotting $\Delta T$ instead of $V_T \times \Delta T$ on the $y$ -axis.
Choosing a scale that clusters all points in one corner of the grid.
Plotting points inaccurately (more than half a small square from the correct position).
Forgetting to plot all 10 points.

Things to Be Careful About

The mark scheme says "Only include 0 if a point is plotted there" — do not start the axis at 0 unless a data point falls there.
Points at the extremes (5 cm³ and 50 cm³) are often where errors occur due to small $\Delta T$ values.
The graph should clearly show a peak, confirming the neutralisation point.

Techniques used

plot points on a graph with labelled axeschoose appropriate scale to use more than half of each axis

(ii)

Draw a straight line of best fit through the points where the values of $(V_T \times \Delta T)$ are increasing. Draw a second straight line of best fit through the points where the values of $(V_T \times \Delta T)$ are decreasing.

DifficultyMedium-Easy

Worked solution

Answer

Draw two straight lines of best fit:

Line 1: Through the points where $V_T \times \Delta T$ is increasing (the ascending branch, from 5 cm³ up to approximately the peak region).
Line 2: Through the points where $V_T \times \Delta T$ is decreasing (the descending branch, from approximately the peak region to 50 cm³).

Both lines must be straight (not curved) and should pass as close as possible to the points on their respective branches, with an approximately equal number of points on each side of the line.

Final answer

Two straight lines of best fit — one through the increasing points and one through the decreasing points

Detailed explanation

Background Concept

Before the neutralisation point, each additional 5 cm³ of acid reacts with excess NaOH, releasing heat proportional to the moles of acid added. After the neutralisation point, no more reaction occurs and the temperature falls as the excess cool acid dilutes the warm solution. This produces two approximately linear regions on the $V_T \times \Delta T$ graph, and the intersection gives the exact neutralisation volume.

Understanding the Question

The candidate must recognise the peak of the graph and draw one straight line through the ascending points and another through the descending points. The mark scheme awards 1 mark (B1) for drawing both lines correctly.

Approach

Identify the highest point (or the region where the trend changes from increasing to decreasing). Draw a ruler through the points on the left side of the peak, extending it slightly past the peak. Then draw a second ruler through the points on the right side, extending it slightly past the peak. The intersection of the two extended lines gives the neutralisation volume.

Step-by-Step Reasoning

Ascending line: This should pass through or near the points at 5, 10, 15, 20 cm³ (and possibly 25 cm³ if it is still on the rising side). The line should be straight — do not join points or draw a curve.
Descending line: This should pass through or near the points at 30, 35, 40, 45, 50 cm³ (and possibly 25 cm³ if it is on the falling side). Again, straight line only.
Key point: The lines are extrapolated to meet — the intersection may not coincide with any plotted point. This is the whole purpose of using two lines rather than simply reading the peak.
Common issue: The point at the peak (e.g. 25 cm³) may lie on either side of the intersection or near it. The candidate must decide which branch it belongs to based on the overall trend.

Key Takeaways

Lines of best fit must be straight, not curves.
The purpose of two lines is to find the intersection, which is more accurate than reading the maximum of a curved peak.
Each line should have roughly equal scatter of points above and below it.

Common Mistakes

Drawing a single smooth curve through all points instead of two straight lines.
Drawing lines that do not extend far enough to intersect.
Including the peak point in both lines (it should belong to one branch or the other).
Drawing the lines freehand rather than using a ruler.

Things to Be Careful About

The mark scheme specifically requires "straight lines" — a curve or a series of joined points earns no mark.
Both lines must be drawn; one line alone does not earn the B1.
The lines should be extended sufficiently to clearly show the intersection point.

Techniques used

draw a straight line of best fit through increasing data pointsdraw a second straight line of best fit through decreasing data points

(iii)

From your graph, determine the volume of FA 2 where the two lines of best fit intersect.

\text{volume of FA 2} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ cm}^3

DifficultyMedium-Easy

Worked solution

Answer

Read the $x$ -coordinate (volume of FA 2) at the point where the two lines of best fit intersect.

\text{Volume of FA 2 at intersection} = \text{(candidate's reading, e.g. 12.5 cm}^3\text{)}

The answer should be within 0.5 cm³ of the supervisor's value.

Final answer

Candidate-dependent: volume of FA 2 at intersection of the two lines (within 0.5 cm³ of supervisor value)

Detailed explanation

Background Concept

The intersection of the two lines of best fit represents the volume of acid at which the reaction is exactly complete — the neutralisation point. At this volume, all the NaOH has just been consumed. The two linear regions meet at this point because the mechanism of heat generation changes: before it, heat is produced by reaction; after it, no reaction occurs and the temperature simply drops due to dilution by cooler acid.

Understanding the Question

This is a graph-reading task. The candidate must identify the intersection point of the two lines drawn in part (c)(ii) and read off the corresponding $x$ -axis value (volume of FA 2). The mark scheme allows a tolerance of ±0.5 cm³ from the supervisor's value.

Approach

Locate the point where the two straight lines cross. Draw a vertical line down from this intersection to the $x$ -axis and read the volume. Alternatively, draw a horizontal line from the intersection to the $y$ -axis and then use the line equations to solve for $x$ .

Step-by-Step Reasoning

The intersection point should be clearly visible on the graph where the ascending line meets the descending line.
Read the volume of FA 2 at this point. For example, if the supervisor's value is 12.5 cm³, any reading between 12.0 and 13.0 cm³ earns the mark.
The mark scheme says "Ignore sf" — so 12.5, 12.50, or 12.5 cm³ are all acceptable.
This value is then used in part (d)(iii) to calculate the concentration.

Key Takeaways

The intersection method is more precise than reading the maximum because it uses all the data on both sides of the peak.
Always read from the lines, not from the plotted points themselves.

Common Mistakes

Reading the $y$ -coordinate instead of the $x$ -coordinate at the intersection.
Reading the volume at the highest plotted point rather than at the intersection of the extrapolated lines.
Being imprecise in reading from the graph (the ±0.5 cm³ tolerance is generous, but careless reading can exceed it).

Things to Be Careful About

The answer must be a volume of FA 2 (the $x$ -axis value), not a temperature or a $V_T \times \Delta T$ value.
The value will be carried forward into part (d)(iii), so an error here affects the final concentration calculation (though ecf is allowed).

Techniques used

read a value from the intersection of two lines on a graph

(d)

The value you recorded in (c)(iii) is the volume of FA 2 which is needed to neutralise $25.0\text{ cm}^3$ of FA 1. In the following calculations you will determine the concentration of FA 2.

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

(i)

Calculate how many moles of sodium hydroxide are contained in $25.0\text{ cm}^3$ of FA 1.

\text{moles of NaOH} = \dots\dots\dots\dots\dots\dots\dots\dots\text{ mol}

DifficultyEasy

Worked solution

Working

n(\text{NaOH}) = c \times V = 2.00 \times \frac{25.0}{1000} = 0.0500\text{ mol}

Answer

\text{moles of NaOH} = 0.0500\text{ mol}

Final answer

0.0500 mol

Detailed explanation

Background Concept

The number of moles of a solute in solution is calculated from $n = c \times V$ , where $c$ is the concentration in mol dm⁻³ and $V$ is the volume in dm³. Since the volume is given in cm³, it must be divided by 1000 to convert to dm³.

Understanding the Question

FA 1 is 2.00 mol dm⁻³ NaOH, and 25.0 cm³ was pipetted into the cup. The task is simply to find how many moles of NaOH this represents.

Approach

Apply $n = c \times V$ directly, converting cm³ to dm³.

Step-by-Step Reasoning

$c = 2.00$ mol dm⁻³
$V = 25.0$ cm³ $= 25.0/1000 = 0.0250$ dm³
$n = 2.00 \times 0.0250 = 0.0500$ mol

The mark scheme accepts 0.050 as well (allowing 2 or 3 significant figures).

Key Takeaways

Always convert cm³ to dm³ before using $n = cV$ .
The number of significant figures in the answer should reflect the precision of the data (25.0 cm³ has 3 sf, 2.00 mol dm⁻³ has 3 sf, so 0.0500 mol is appropriate).

Common Mistakes

Forgetting to divide by 1000 (giving 50.0 mol, which is absurd).
Using the volume of FA 2 instead of FA 1.
Incorrectly using the stoichiometric ratio at this stage (the ratio is applied in part (d)(ii), not here).

Things to Be Careful About

The question asks for moles of NaOH specifically, not moles of H₂SO₄.
Report to 3 significant figures (0.0500) or at minimum 2 (0.050) as the mark scheme allows.

Techniques used

calculate moles from concentration and volume

(ii)

Calculate how many moles of sulfuric acid would react with the number of moles of NaOH in (i).

\text{moles of H}_2\text{SO}_4 = \dots\dots\dots\dots\dots\dots\dots\dots\text{ mol}

DifficultyMedium-Easy

Worked solution

Working

From the balanced equation:

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

The mole ratio is $\text{H}_2\text{SO}_4 : \text{NaOH} = 1 : 2$

n(\text{H}_2\text{SO}_4) = \frac{n(\text{NaOH})}{2} = \frac{0.0500}{2} = 0.0250\text{ mol}

Answer

\text{moles of H}_2\text{SO}_4 = 0.0250\text{ mol}

Final answer

0.0250 mol

Detailed explanation

Background Concept

Stoichiometric calculations use the mole ratio from a balanced chemical equation to convert between the amounts of reactants and products. The equation $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ shows that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.

Understanding the Question

Given the moles of NaOH calculated in part (d)(i), determine how many moles of H₂SO₄ are needed to react with it completely (i.e. at the neutralisation point).

Approach

Use the 1:2 ratio: moles of H₂SO₄ = moles of NaOH ÷ 2.

Step-by-Step Reasoning

The equation shows 1 mol H₂SO₄ reacts with 2 mol NaOH.
Therefore, moles of H₂SO₄ = moles of NaOH / 2 = 0.0500 / 2 = 0.0250 mol.
The mark scheme allows ecf from (i): if the candidate made an error in (d)(i) but correctly divides by 2 here, the mark is still awarded.

Key Takeaways

Always read the ratio from the balanced equation, not from memory or assumption.
The species with the smaller coefficient in the denominator gets divided — here NaOH has coefficient 2, so divide by 2.

Common Mistakes

Multiplying by 2 instead of dividing (a very common error when the ratio direction is confused).
Using the ratio from the wrong equation or misreading the coefficients.
Forgetting that the ratio is H₂SO₄:NaOH = 1:2, not 2:1.

Things to Be Careful About

The mark scheme explicitly allows ecf from (d)(i), so even if the moles of NaOH are wrong, dividing by 2 correctly earns this mark.
Report to 3 significant figures (0.0250) to match the precision of the given data.

Techniques used

apply stoichiometric ratio from balanced equation

(iii)

Calculate the concentration of FA 2.

\text{concentration of FA 2} = \dots\dots\dots\dots\dots\dots\dots\dots\text{ mol dm}^{-3}

DifficultyMedium-Easy

Worked solution

Working

c(\text{H}_2\text{SO}_4) = \frac{n}{V} = \frac{0.0250}{V_{\text{FA 2}}/1000} = \frac{1000 \times 0.0250}{V_{\text{FA 2}}}

Using a representative volume of FA 2 = 12.5 cm³ from part (c)(iii):

c = \frac{1000 \times 0.0250}{12.5} = 2.00\text{ mol dm}^{-3}

Answer

\text{concentration of FA 2} = 2.00\text{ mol dm}^{-3}

(Note: the actual answer depends on the volume read from the graph in part (c)(iii).)

Final answer

Candidate-dependent: 1000 × 0.0250 / (volume from c(iii)) mol dm⁻³

Detailed explanation

Background Concept

Concentration is defined as moles of solute per unit volume of solution: $c = n/V$ . When the volume is given in cm³, the formula becomes $c = 1000n/V$ (with $V$ in cm³) to account for the conversion to dm³.

Understanding the Question

The volume of FA 2 needed to neutralise 25.0 cm³ of FA 1 was determined experimentally in part (c)(iii). Using the moles of H₂SO₄ from (d)(ii) and this volume, calculate the concentration of the sulfuric acid.

Approach

Apply $c = 1000n/V$ where $n$ is from (d)(ii) and $V$ is from (c)(iii).

Step-by-Step Reasoning

$n(\text{H}_2\text{SO}_4) = 0.0250$ mol (from d(ii))
$V_{\text{FA 2}}$ = volume read from graph (from c(iii)), e.g. 12.5 cm³
$c = \frac{1000 \times 0.0250}{12.5} = 2.00$ mol dm⁻³
The mark scheme formula is: $1000 \times \text{(d)(ii)} / \text{(c)(iii)}$
Ecf is allowed from (d)(ii): if the moles are wrong but the division is correct, the mark can still be awarded.
Significant figures: the answer should be given to 2–4 sf. The mark scheme penalises sf errors only once.

Key Takeaways

The formula $c = 1000n/V$ (with $V$ in cm³) is a convenient shortcut that avoids explicitly converting to dm³.
This is the final answer of the investigation — the concentration of the unknown acid.
Ecf means that an error in reading the graph does not necessarily lose the final mark if the method is correct.

Common Mistakes

Forgetting to multiply by 1000 (giving an answer 1000 times too small).
Dividing by the volume of FA 1 (25.0 cm³) instead of the volume of FA 2.
Using the wrong number of moles (e.g. using moles of NaOH instead of H₂SO₄).
Incorrect significant figures in the final answer.

Things to Be Careful About

The mark scheme specifies 2–4 significant figures for the final answer.
The volume used must be the one from (c)(iii) — the neutralisation volume of FA 2, not 25.0 cm³.
SF errors are penalised only once across the whole of part (d), so do not over-worry about sf if the method is correct.

Techniques used

calculate concentration from moles and volumeconvert cm³ to dm³ in concentration calculation

(e)

Other than heat losses from the plastic cup to the surroundings, suggest an additional source of error in this experiment and how this error could be reduced.

DifficultyMedium-Easy

Worked solution

Answer

Source of error: The thermometer can only be read to the nearest 0.5 °C, limiting the precision of temperature measurements.

Improvement: Use a thermometer with smaller scale divisions (e.g. a 0–50 °C thermometer graduated in 0.1 °C divisions) to allow more precise temperature readings.

(Alternative acceptable answers: uncertainty in locating the intersection of the two lines — take more readings of FA 2 in the region near the maximum; initial temperatures of acid and alkali may differ — measure both before starting.)

Final answer

Limited precision of thermometer (0.5 °C divisions) — use a thermometer with smaller scale divisions (e.g. 0.1 °C graduations)

Detailed explanation

Background Concept

In any practical investigation, identifying sources of error and proposing improvements is a key evaluation skill. The mark scheme explicitly excludes "heat losses from the plastic cup to the surroundings" as this was already stated in the question. Acceptable answers must be specific — vague responses like "human error" or "be more careful" are rejected.

Understanding the Question

The question asks for an additional source of error (not heat loss) and a method to reduce it. The mark scheme lists several creditable answers, all of which must be specific and paired with a matching improvement.

Approach

Consider what could go wrong in this experiment other than heat loss:

The thermometer's precision (only readable to 0.5 °C).
The uncertainty in reading the intersection point from the graph.
The initial temperatures of the acid and alkali may not be identical.
The burette reading itself has uncertainty.

Step-by-Step Reasoning

Thermometer precision (B1): The thermometer used reads to 0.5 °C. This means each temperature measurement has an uncertainty of ±0.25 °C, which propagates into $\Delta T$ and then into $V_T \times \Delta T$ . The improvement is to use a thermometer with finer graduations (e.g. 0–50 °C with 0.1 °C divisions). Note: the mark scheme rejects "more accurate thermometer" or "electronic thermometer" or "parallax error" as answers — the response must specifically mention smaller scale divisions.
Intersection uncertainty (B1): The two lines of best fit may not intersect at a well-defined point, especially if data near the peak is sparse. The improvement is to take additional readings of FA 2 in the region around the expected maximum (e.g. add 1 cm³ portions near the peak) to get more data points close to the intersection.
Different initial temperatures (B1): If the acid and alkali start at different temperatures, the measured $\Delta T$ is not purely due to the reaction. The improvement is to measure the initial temperature of both FA 1 and FA 2 and ensure they are the same (or account for the difference).

Key Takeaways

Error and improvement answers must be paired and specific.
"Heat loss" is already excluded — think of other systematic or random errors.
The mark scheme rejects generic answers: "more accurate" is not sufficient; you must say what specifically would be more accurate and how.

Common Mistakes

Writing "human error" or "parallax error" — the mark scheme explicitly rejects these.
Writing "use a more accurate thermometer" without specifying what makes it more accurate (the scale divisions).
Writing "use an electronic thermometer" — this is also rejected by the mark scheme.
Suggesting an improvement that does not address the stated error.
Repeating "heat loss to surroundings" as the error (it is excluded by the question).

Things to Be Careful About

The answer must be specific: name the quantity affected and explain why.
The improvement must directly address the named error.
The mark scheme provides several acceptable answers — any one specific, well-paired error/improvement earns the B1.

Techniques used

identify a specific source of experimental errorpropose a targeted improvement to reduce that error

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation10M
Q3Manipulation, Measurement and Observation · Qualitative Analysis15M

Loading the full paper…