Chemistry 9701/34 — May/June 2013

Cambridge AS Level · Advanced Practical Skills 2 · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

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

Q1Presentation of Data and ObservationsManipulation, Measurement and ObservationAnalysis, Conclusions and EvaluationFree sample

When aqueous hydrochloric acid is mixed with aqueous sodium hydroxide, the neutralisation reaction releases heat causing a rise in the temperature of the solution.

\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}

In this experiment you will mix different volumes of hydrochloric acid and sodium hydroxide but the total volume will be kept constant. For each mixture you will record the temperature rise. Since the combined volume remains the same, the temperature rise is a direct measure of the heat given out by the reaction. The maximum heat given out occurs when all the acid present is exactly neutralised by all the alkali present. By determining the volumes when this occurs you can work out the concentration of the sodium hydroxide.

FB 1 is $2.00\text{ mol dm}^{-3}$ hydrochloric acid, HCl.
FB 2 is aqueous sodium hydroxide, NaOH.

Read through the instructions carefully and prepare a table for your results in the space on page 4 before starting any practical work.

(a)

Method

Experiment 1

Support the plastic cup in the $250\text{ cm}^3$ beaker.
Fill the unlabelled burette with FB 1.
Run $26.00\text{ cm}^3$ of FB 1 from the burette into the plastic cup.
Record the temperature of FB 1, $T_1$ , in the space below.

$T_1 = \text{............. } ^\circ\text{C}$

Fill the burette labelled FB 2 with FB 2.
Run $4.00\text{ cm}^3$ of FB 2 from the burette into the plastic cup.
Stir the mixture thoroughly and record in your table the maximum temperature of the solution.
Empty the plastic cup, rinse thoroughly with water and shake dry.

Experiment 2

Support the plastic cup in the $250\text{ cm}^3$ beaker.
Run $22.00\text{ cm}^3$ of FB 1 from the burette into the plastic cup.
Run $8.00\text{ cm}^3$ of FB 2 from the burette into the plastic cup.
Stir the mixture thoroughly and record in your table the maximum temperature of the solution.
Empty the plastic cup, rinse thoroughly with water and shake dry.

Experiments 3 – 7

Repeat the experiment using 18.00, 14.00, 10.00, 6.00 and $2.00\text{ cm}^3$ of FB 1 respectively. Add sufficient FB 2 each time to make sure that the total volume remains $30.00\text{ cm}^3$ .

For each of your seven experiments, record in the space below

the volume of FB 1,
the volume of FB 2,
the maximum temperature of the solution,
the temperature rise, $\Delta T$ , where $\Delta T = \text{maximum temperature recorded} - T_1$ .

DifficultyMedium-Easy

Worked solution

Answer

Initial temperature of FB 1, $T_1 = 21.0\ ^\circ\text{C}$

Experiment	Volume of FB 1 / $\text{cm}^3$	Volume of FB 2 / $\text{cm}^3$	Maximum temperature / $^\circ\text{C}$	Temperature rise, $\Delta T$ / $^\circ\text{C}$
1	26.00	4.00	24.5	3.5
2	22.00	8.00	28.0	7.0
3	18.00	12.00	31.5	10.5
4	14.00	16.00	32.0	11.0
5	10.00	20.00	29.0	8.0
6	6.00	24.00	26.0	5.0
7	2.00	28.00	22.5	1.5

Final answer

Complete results table with 7 experiments, all temperatures to 0.5 °C, and correctly calculated Delta T

Detailed explanation

Background Concept

In a thermometric titration (continuous variation method), exothermic neutralisation produces heat:

\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}

The total volume of the reaction mixture is held constant ( $30.00\text{ cm}^3$ ), so the heat capacity of the resulting mixture remains approximately constant across all runs. Under these conditions, the temperature rise $\Delta T$ is directly proportional to the amount of heat evolved, which is determined by whichever reagent is the limiting reactant. The maximum temperature rise corresponds to the stoichiometric equivalence point where both acid and alkali are completely consumed.

Understanding the Question

The candidate must measure and record the initial temperature $T_1$ of FB 1 ( $2.00\text{ mol dm}^{-3}\text{ HCl}$ ) and then carry out seven neutralisation experiments using varying complementary volumes of FB 1 and FB 2 (aqueous $\text{NaOH}$ ) such that $V_{\text{FB 1}} + V_{\text{FB 2}} = 30.00\text{ cm}^3$ . All raw data and calculated $\Delta T$ values must be presented systematically in a single table.

Approach

Design a clear table containing columns for: Experiment number, Volume of FB 1, Volume of FB 2, Maximum temperature reached, and Temperature rise $\Delta T$ .
Ensure all column headings have correct quantities and units (e.g. $/\text{cm}^3$ and $/^\circ\text{C}$ ).
Record all temperatures using the appropriate precision of the thermometer (typically to the nearest $0.5\ ^\circ\text{C}$ , ensuring readings ending in .0 and .5 are both present).
Accurately calculate $\Delta T = T_{\text{max}} - T_1$ for each of the 7 runs.

Step-by-Step Reasoning

Table construction and headings: Every experiment (1 to 7) must have entries. Column headers must show the quantity and unit explicitly: "Volume of FB 1 / $\text{cm}^3$ ", "Volume of FB 2 / $\text{cm}^3$ ", "Maximum temperature / $^\circ\text{C}$ ", " $\Delta T$ / $^\circ\text{C}$ ".
Precision of temperature readings: Thermometers supplied in Cambridge A-Level practicals are typically graduated in $1\ ^\circ\text{C}$ divisions; thus, readings are estimated to the nearest half-degree ( $0.5\ ^\circ\text{C}$ ). Record values as $21.0$ , $24.5$ , etc.
Calculation of $\Delta T$ : Subtract the initial temperature $T_1$ consistently from each maximum temperature value.
Accuracy check: Cambridge awards marks based on comparing the candidate's $\Delta T$ value at $14.00\text{ cm}^3$ of FB 1 to the Supervisor's recorded value (typically around $11.0\ ^\circ\text{C}$ ). Closer agreement earns higher marks.

Key Takeaways

Tabulating data clearly with unambiguous headings and units is essential for practical marks.
All readings taken with a given instrument must follow a consistent precision matching the instrument resolution.

Common Mistakes

Omitting units in the table header or repeating units inside data cells.
Inconsistent precision in thermometer readings (e.g. writing 24 instead of 24.0, or giving arbitrary decimals like 24.2 °C when measuring to 0.5 °C).
Arithmetic mistakes when calculating $\Delta T$ .

Things to Be Careful About

Ensure the total volume for each experiment strictly adds up to $30.00\text{ cm}^3$ ( $26+4$ , $22+8$ , $18+12$ , $14+16$ , $10+20$ , $6+24$ , $2+28$ ).

Techniques used

construct a results table with headings and unitsrecord temperature readings to nearest 0.5 °Ccalculate temperature differences consistently

(b)

(i)

On the grid opposite, plot the temperature rise, $\Delta T$ , on the y-axis against the volume of FB 1 on the x-axis.
The scale for $\Delta T$ should extend at least $2\ ^\circ\text{C}$ above your greatest temperature rise.

Fig. 1.1

DifficultyMedium-Easy

Worked solution

Answer

Axes: Plot $\Delta T$ on the vertical ( $y$ ) axis and volume of FB 1 on the horizontal ( $x$ ) axis. Both axes are clearly labelled with their quantity and unit: $\Delta T\ /\ ^\circ\text{C}$ and $\text{volume of FB 1}\ /\ \text{cm}^3$ .
Scale:
- $x$ -axis: $0$ to $30\text{ cm}^3$ , using a sensible linear scale (e.g. 2 large squares = $5\text{ cm}^3$ or 1 large square = $2\text{ cm}^3$ ) occupying at least 5 large grid squares.
- $y$ -axis: Linear scale from $0$ to at least $14\ ^\circ\text{C}$ (extending at least $2\ ^\circ\text{C}$ above the maximum $\Delta T$ of $11.0\ ^\circ\text{C}$ ), occupying at least 6 large grid squares vertically.
Plotting: All seven data points from the results table plotted accurately to within half a small square using small, sharp crosses ( $imes$ ) or circled dots $(\odot)$ .

Final answer

Graph plotted with Delta T on y-axis against volume of FB 1 on x-axis

Detailed explanation

Background Concept

In graphical data presentation, plotting the dependent variable on the $y$ -axis and the independent variable on the $x$ -axis allows the visual identification of trends, relationships, and intersections that represent key chemical points (such as the stoichiometric endpoint).

Understanding the Question

The question asks to plot $\Delta T$ (dependent variable) on the $y$ -axis against the volume of FB 1 (independent variable) on the $x$ -axis on Fig. 1.1. The $y$ -axis scale must extend at least $2\ ^\circ\text{C}$ beyond the maximum recorded $\Delta T$ .

Approach

Check the data range: Volume of FB 1 ranges from $2.00$ to $26.00\text{ cm}^3$ . $\Delta T$ ranges from $1.5$ to $11.0\ ^\circ\text{C}$ .
Set up the scales so that the plotted points occupy more than half of the grid in both dimensions, avoiding awkward ratios (such as multiples of 3, 7, etc.).
Ensure the $y$ -axis extends to at least $11.0 + 2.0 = 13.0\ ^\circ\text{C}$ (e.g. up to $14.0\ ^\circ\text{C}$ or $15.0\ ^\circ\text{C}$ ).
Label both axes clearly.
Plot each pair of coordinates precisely.

Step-by-Step Reasoning

Scale selection: A standard grid has 8–10 large squares horizontally and 12–14 vertically. For FB 1 volume ( $0$ to $30\text{ cm}^3$ ), choosing 1 large square ( $2\text{ cm}$ ) = $2.5\text{ cm}^3$ or 2 large squares = $5\text{ cm}^3$ gives an easy-to-use linear scale. For $\Delta T$ ( $0$ to $14\ ^\circ\text{C}$ ), choosing 1 large square = $1\ ^\circ\text{C}$ or 2 large squares = $2\ ^\circ\text{C}$ spans across more than 6 large vertical squares.
Plotting precision: Every point must be within $\pm 0.5$ small square of its true position. Clean, sharp marks are essential.

Key Takeaways

Graph scales must be linear, easy to interpolate (multiples of 1, 2, 5, or 10), and span over half of the available grid.

Common Mistakes

Inverting the axes (plotting volume on the $y$ -axis).
Failing to extend the $y$ -axis scale by the required $2\ ^\circ\text{C}$ above the highest reading.
Using non-linear scales or awkward increments that make plotting and reading off difficult.

Things to Be Careful About

Ensure points are marked clearly with fine pencil crosses rather than large, smudged blobs.

Techniques used

choose appropriate graph scaleslabel axes with quantities and unitsplot data points accurately on a grid

(ii)

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

DifficultyMedium-Easy

Worked solution

Answer

Draw a straight line of best fit through the points where $\Delta T$ increases (the lower volumes of FB 1 where FB 1 is the limiting reactant).
Draw a second straight line of best fit through the points where $\Delta T$ decreases (the higher volumes of FB 1 where FB 2 is the limiting reactant).
Both lines are drawn cleanly using a ruler and extended so that they intersect clearly.

Final answer

Two intersecting straight lines of best fit drawn

Detailed explanation

Background Concept

In the method of continuous variation (Job's method) for an exothermic neutralisation:

When acid (FB 1) is limiting (low volume of FB 1), increasing FB 1 produces more water and releases more heat, so $\Delta T$ increases linearly with the volume of FB 1.
Beyond the equivalence point, alkali (FB 2) becomes limiting. Increasing FB 1 further means less FB 2 is present, so less neutralisation occurs and $\Delta T$ decreases linearly with increasing volume of FB 1.
Thus, the theoretical curve consists of two straight lines that intersect at the exact stoichiometric point.

Understanding the Question

The question instructs the candidate to draw two distinct lines of best fit: one through the rising points and one through the falling points.

Approach

Identify which points form the rising portion (typically volumes 2.00, 6.00, 10.00, and 14.00 $\text{cm}^3$ ).
Use a clear ruler to draw a straight line of best fit through these points, balancing points evenly on either side if they do not lie perfectly on the line.
Identify which points form the falling portion (typically volumes 18.00, 22.00, and 26.00 $\text{cm}^3$ , with 14.00 near the peak).
Draw the second straight line of best fit.
Extend both lines until they intersect cleanly at a single sharp point.

Step-by-Step Reasoning

A continuous curved line must not be drawn; the underlying chemistry is piece-wise linear, governed by two different limiting reactants.
The intersection marks the exact stoichiometric equivalence point, which usually falls between two tested points (e.g., around $13.5$ to $14.5\text{ cm}^3$ ).

Key Takeaways

Continuous variation thermometric titrations produce two intersecting straight lines whose intersection defines the equivalence point.

Common Mistakes

Drawing a single smooth curved line over the data points instead of two distinct intersecting straight lines.
Drawing lines that stop short of intersecting.

Things to Be Careful About

Ensure the lines are drawn with a sharp pencil and a ruler; lines must not be 'tramlined' (doubled) or kinked.

Techniques used

draw straight lines of best fitidentify intersecting linear trends

(iii)

From your graph, determine the value of the volume of FB 1 where the two lines of best fit intersect.

$\text{volume of FB 1} = \text{................................................ } \text{cm}^3$

DifficultyMedium-Easy

Worked solution

Answer

$\text{volume of FB 1} = 13.50\text{ cm}^3$

Final answer

13.50 cm^3

Detailed explanation

Background Concept

The point of intersection of the two best-fit lines corresponds to the exact equivalence point of the titration, where the ratio of FB 1 to FB 2 provides stoichiometric amounts of $\text{H}^+$ and $\text{OH}^-$ with neither reagent in excess.

Understanding the Question

The candidate must read the value on the horizontal ( $x$ ) axis (volume of FB 1) directly below the point where the two straight lines of best fit intersect.

Approach

Locate the exact intersection of the two lines of best fit drawn in (b)(ii).
Project vertically down to the $x$ -axis.
Read the volume of FB 1 to within half a small square according to the scale used.

Step-by-Step Reasoning

For typical experimental data where the concentration of FB 2 is approximately $2.4\text{ mol dm}^{-3}$ to $2.5\text{ mol dm}^{-3}$ , the equivalence point volume of FB 1 ( $2.00\text{ mol dm}^{-3}$ ) lies near $13.50\text{ cm}^3$ .
For this representative dataset, the intersection occurs at $13.50\text{ cm}^3$ .

Key Takeaways

Accurately reading intersection coordinates allows determination of stoichiometric volumes without needing an experiment that directly lands on that exact volume.

Common Mistakes

Misreading the graph scale (e.g. reading 13.25 as 13.5 if each small square represents 0.5 or 0.25).
Giving the maximum experimental data point volume rather than the coordinate of the line intersection.

Things to Be Careful About

Include appropriate precision and ensure the value is consistent with the graph drawn.

Techniques used

read coordinate from graph intersection

(c)

Calculations

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

(i)

Calculate how many moles of hydrochloric acid are contained in the volume recorded in (b)(iii).

$\text{moles of HCl} = \text{............... mol}$

DifficultyMedium-Easy

Worked solution

Working

\begin{aligned} \text{moles of HCl} &= c \times V \ &= 2.00\text{ mol dm}^{-3} \times \frac{13.50}{1000}\text{ dm}^3 \ &= 0.0270\text{ mol} \end{aligned}

Answer

$\text{moles of HCl} = 0.0270\text{ mol}$

Final answer

0.0270 mol

Detailed explanation

Background Concept

The number of moles of solute in a solution is given by:

n = c \times V

where $c$ is the concentration in $\text{mol dm}^{-3}$ and $V$ is the volume in $\text{dm}^3$ (with $V\text{ (dm}^3\text{)} = \frac{V\text{ (cm}^3\text{)}}{1000}$ ).

Understanding the Question

Calculate the amount (in moles) of $\text{HCl}$ present in the equivalence volume of FB 1 determined in (b)(iii) ( $13.50\text{ cm}^3$ ), given that FB 1 has a concentration of $2.00\text{ mol dm}^{-3}$ .

Approach

Convert the volume from $\text{cm}^3$ to $\text{dm}^3$ by dividing by $1000$ .
Multiply by the concentration of FB 1 ( $2.00\text{ mol dm}^{-3}$ ).
Express the final answer to an appropriate number of significant figures (3 or 4 significant figures).

Step-by-Step Reasoning

\text{moles of HCl} = \frac{2.00 \times 13.50}{1000} = 0.0270\text{ mol}

This answer has 3 significant figures, matching the precision of the concentration given ( $2.00\text{ mol dm}^{-3}$ ) and the volume measurement.

Key Takeaways

Always convert volumes in $\text{cm}^3$ to $\text{dm}^3$ before multiplying by concentration in $\text{mol dm}^{-3}$ .

Common Mistakes

Forgetting to divide by $1000$ , resulting in a value 1000 times too large.
Expressing the answer to only 1 significant figure (e.g. $0.03\text{ mol}$ ).

Things to Be Careful About

Ensure the value used for the volume of FB 1 is carried directly from (b)(iii).

Techniques used

calculate moles in a solution volumeapply correct significant figures

(ii)

Calculate how many moles of sodium hydroxide would react completely with the number of moles of hydrochloric acid in (c)(i).

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

DifficultyEasy

Worked solution

Answer

$\text{moles of NaOH} = 0.0270\text{ mol}$

Final answer

0.0270 mol

Detailed explanation

Background Concept

From the balanced chemical equation:

\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}

the stoichiometric mole ratio between $\text{HCl}$ and $\text{NaOH}$ is $1:1$ .

Understanding the Question

The question asks for the number of moles of sodium hydroxide that reacts completely with the moles of hydrochloric acid calculated in (c)(i).

Approach

Apply the $1:1$ stoichiometric molar ratio from the equation: $\text{moles of NaOH} = \text{moles of HCl}$ .

Step-by-Step Reasoning

Since $1\text{ mole}$ of $\text{HCl}$ reacts with $1\text{ mole}$ of $\text{NaOH}$ :

\text{moles of NaOH} = \text{moles of HCl} = 0.0270\text{ mol}

Key Takeaways

In equimolar reactions, the reacting amounts in moles are identical.

Common Mistakes

Attempting unnecessary calculations or changing the number of significant figures.

Things to Be Careful About

Transfer the exact value from (c)(i) without rounding errors.

Techniques used

deduce stoichiometric reacting moles

(iii)

Calculate the concentration of FB 2. Remember that the combined volume of FB 1 and FB 2 in each experiment was $30.00\text{ cm}^3$ .

$\text{concentration of FB 2} = \text{.................. } \text{mol dm}^{-3}$

DifficultyMedium-Easy

Worked solution

Working

\begin{aligned} \text{Volume of FB 2 at neutralisation} &= 30.00 - 13.50 = 16.50\text{ cm}^3 \ \text{concentration of FB 2} &= \frac{\text{moles of NaOH}}{\text{volume of FB 2 in dm}^3} \ &= \frac{0.0270}{\frac{16.50}{1000}} \ &= \frac{0.0270 \times 1000}{16.50} \ &= 1.64\text{ mol dm}^{-3} \end{aligned}

Answer

$\text{concentration of FB 2} = 1.64\text{ mol dm}^{-3}$

Final answer

1.64 mol dm^-3

Detailed explanation

Background Concept

The concentration of a solution is determined by:

c = \frac{n}{V}

where $n$ is the amount in moles and $V$ is the volume in $\text{dm}^3$ .
In this experiment, the combined volume of acid and alkali is always $30.00\text{ cm}^3$ . Therefore, at the equivalence point:

V_{\text{FB 2}} = 30.00 - V_{\text{FB 1}}

Understanding the Question

Calculate the concentration of FB 2 (aqueous $\text{NaOH}$ ) using the moles of $\text{NaOH}$ from (c)(ii) and the volume of FB 2 present at the point of neutralisation.

Approach

Find the volume of FB 2 at equivalence: $V_{\text{FB 2}} = 30.00 - V_{\text{FB 1 (b)(iii)}}$ .
Divide the moles of $\text{NaOH}$ from (c)(ii) by this volume (in $\text{dm}^3$ ).
Express the final value to 3 or 4 significant figures.

Step-by-Step Reasoning

Volume of FB 2 used: $V_{\text{FB 2}} = 30.00 - 13.50 = 16.50\text{ cm}^3$
Concentration calculation: $c(\text{NaOH}) = \frac{1000 \times 0.0270}{16.50} = 1.6363...\text{ mol dm}^{-3}$
Rounding to 3 significant figures gives $1.64\text{ mol dm}^{-3}$ . (Or $1.636\text{ mol dm}^{-3}$ to 4 significant figures).

Key Takeaways

In continuous variation experiments, remember to subtract the deduced volume from the constant total volume to find the volume of the second reagent.

Common Mistakes

Dividing by the total volume ( $30.00\text{ cm}^3$ ) instead of the volume of FB 2 ( $16.50\text{ cm}^3$ ).
Incorrect rounding or giving answers to an inappropriate number of significant figures (such as 1 or 2 sig figs).

Things to Be Careful About

Ensure all values in parts (c)(i), (c)(ii), and (c)(iii) are displayed consistently to 3 or 4 significant figures, as required by the mark scheme.

Techniques used

calculate concentration from moles and volumeapply correct significant figures

(d)

A student decided to modify the experiment. The total volume of the solution was increased to $50\text{ cm}^3$ and temperature rises were recorded for 5, 10, 15, 20, 25, 30, 35, 40 and $45\text{ cm}^3$ of FB 2. The volumes were measured using a $50\text{ cm}^3$ measuring cylinder. Discuss how these changes would affect the accuracy with which the concentration of FB 2 could be determined.

DifficultyMedium

Worked solution

Answer

Any two of:

Apparatus precision: Using a measuring cylinder instead of a burette decreases accuracy because a measuring cylinder has a larger uncertainty / calibration error (less precision).
Number of data points: Having 9 experiments instead of 7 increases accuracy because more points allow more reliable lines of best fit to be drawn and a more accurate intercept to be determined.
Total volume change: Increasing the total volume to $50\text{ cm}^3$ makes no difference to the accuracy because the concentration of solutions remains the same, so the maximum temperature rise remains essentially unchanged.

Final answer

Measuring cylinder decreases accuracy due to greater volume uncertainty; greater number of data points increases accuracy by providing a more reliable best-fit line and intercept.

Detailed explanation

Background Concept

Experimental accuracy depends on the magnitude of systematic and random uncertainties in measurements. Modifying practical procedures affects accuracy in several ways:

Choice of apparatus: A burette has a typical reading uncertainty of $\pm 0.05\text{ cm}^3$ , whereas a $50\text{ cm}^3$ measuring cylinder typically has an uncertainty of $\pm 0.5\text{ cm}^3$ or $\pm 1\text{ cm}^3$ . The larger uncertainty leads to greater percentage error in volume measurements.
Sampling frequency / number of data points: Having a greater number of points distributed along the rising and falling limbs reduces random error and improves the certainty of the lines of best fit and their intersection.
Scale of reaction: When the total volume is scaled up proportionally while maintaining solution concentrations, the heat released scales up proportionally with the mass of solution heated ( $q \propto V$ and $m \propto V$ ). Since $\Delta T = \frac{q}{m c}$ , the theoretical temperature rise is independent of total volume.

Understanding the Question

The candidate must discuss how three modifications affect the accuracy of the determined concentration of FB 2:

Increasing the total volume from $30\text{ cm}^3$ to $50\text{ cm}^3$ .
Increasing the number of experiments from 7 to 9.
Using a $50\text{ cm}^3$ measuring cylinder instead of burettes to measure volumes.

Approach

Select two distinct factors and state clearly whether each increases, decreases, or does not affect accuracy, providing the chemical and physical reasoning behind each.

Step-by-Step Reasoning

Factor 1 (Measuring cylinder): A $50\text{ cm}^3$ measuring cylinder is significantly less precise than a burette. The percentage error in measuring each volume increases, which scatters data points further from the true best-fit lines, thereby decreasing accuracy.
Factor 2 (Number of readings): Recording 9 readings instead of 7 gives more data points across the curve. This makes anomalies easier to spot and allows the straight lines of best fit to be positioned with greater confidence, thereby increasing accuracy in determining the intersection.
Factor 3 (Total volume): The temperature rise $\Delta T = \frac{q}{m c}$ . If concentration is unchanged, doubling volume doubles $q$ and doubles $m$ , so $\Delta T$ remains identical. Therefore, changing total volume has no effect on accuracy.

Key Takeaways

Swapping high-precision apparatus (burette) for lower-precision apparatus (measuring cylinder) increases measurement uncertainty.
Collecting more data points improves the reliability of best-fit lines and graphical intersections.

Common Mistakes

Vaguely stating that a measuring cylinder is "human error" without referencing precision, graduation divisions, or percentage error.
Assuming that a larger volume must cause a larger temperature rise.

Things to Be Careful About

Ensure comparative language is used when discussing apparatus (e.g. "measuring cylinder is less accurate / less precise than a burette").

Techniques used

evaluate experimental modificationscompare apparatus precision and apparatus percentage error

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2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation9M
Q3Qualitative Analysis · Presentation of Data and Observations15M

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