Chemistry 9701/53 — May/June 2024

Background Concept

A solution of known concentration is made by dissolving a known amount of solute in a known volume of solution. The amount in moles is $n = cV$, where $c$ is concentration in $\text{mol dm}^{-3}$ and $V$ is volume in $\text{dm}^3$. The mass of solid needed is $m = nM_r$, where $M_r$ is the molar mass. The hydrated salt $\text{Na}_2\text{S}_2\text{O}_3\cdot5\text{H}_2\text{O}$ contains water of crystallisation, so its molar mass includes five water molecules.

Understanding the Question

We need the mass of hydrated sodium thiosulfate that must be weighed out to make 500.0 cm$^3$ of 0.00200 mol dm$^{-3}$ solution. The volume is given in cm$^3$, so it must be converted to dm$^3$ before using $n = cV$.

Approach

Convert 500.0 cm$^3$ to 0.5000 dm$^3$. Calculate moles of $\text{Na}_2\text{S}_2\text{O}_3$ needed. Then multiply by the molar mass of the hydrated salt, remembering to include the five water molecules.

Step-by-Step Reasoning

1. $V = 500.0 \text{ cm}^3 = 0.5000 \text{ dm}^3$.
2. $n = 0.00200 \times 0.5000 = 1.00 \times 10^{-3} \text{ mol}$.
3. Molar mass: $M_r = 2(23.0) + 2(32.1) + 3(16.0) + 5(18.0) = 248.2 \text{ g mol}^{-1}$.
4. Mass: $m = 1.00 \times 10^{-3} \times 248.2 = 0.248 \text{ g}$.

Key Takeaways

Always use the molar mass of the actual substance weighed, including water of crystallisation. Convert volume to dm$^3$ before using $n = cV$.

Common Mistakes

• Using the molar mass of anhydrous $\text{Na}_2\text{S}_2\text{O}_3$ instead of the hydrated salt.
• Forgetting to convert cm$^3$ to dm$^3$.
• Mixing up $n = cV$ and $m = nM$.

Things to Be Careful About

• The answer should be given to 3 significant figures because the data are given to 3 significant figures.
• Include the unit g.
• If using $M_r = 248.2$, the mass is 0.248 g; using more precise atomic masses gives 0.2482 g, which is also acceptable.

Background Concept

A standard solution of an exact volume is prepared using a volumetric flask, which is calibrated to contain a precise volume at a given temperature. A measuring cylinder or beaker is not accurate enough for preparing a standard solution.

Understanding the Question

After weighing the correct mass of solid, the next step is to dissolve it and make the total volume exactly 500.0 cm$^3$. The question asks for the piece of apparatus used for this.

Approach

Recognise that making an exact volume of solution requires a volumetric flask.

Step-by-Step Reasoning

The solid is dissolved in a small volume of distilled water in a beaker, then transferred quantitatively to a 500 cm$^3$ volumetric flask. The flask is filled to the graduation mark with distilled water and shaken to mix. The volumetric flask is the only common piece of apparatus that gives a precise 500.0 cm$^3$ volume.

Key Takeaways

Volumetric flasks are used to prepare standard solutions of an exact volume; burettes and pipettes measure out volumes, not prepare fixed volumes.

Common Mistakes

• Saying 'measuring cylinder' or 'beaker' – these are not accurate enough.
• Forgetting to state the volume (500 cm$^3$).

Things to Be Careful About

Write '500 cm$^3$ volumetric flask' to score the mark.

Background Concept

Every measuring apparatus has an uncertainty. For a scale with divisions, the uncertainty of one reading is usually taken as half the smallest division. If a volume is obtained from two readings, the total uncertainty is twice this value. Percentage error is the uncertainty expressed as a percentage of the quantity measured: $\text{percentage error} = \frac{\text{uncertainty}}{\text{measured value}} \times 100$.

Understanding the Question

The syringe is graduated every 1.0 cm$^3$. We need the percentage error in measuring 5.0 cm$^3$ of potassium iodide solution. The mark scheme requires working to be shown.

Approach

Find the uncertainty of the measurement, then divide by 5.0 cm$^3$ and multiply by 100.

Step-by-Step Reasoning

1. Smallest division = 1.0 cm$^3$, so one reading has uncertainty $\pm 0.5$ cm$^3$.
2. Because the volume is measured from two readings, the total uncertainty is $2 \times 0.5 = 1.0$ cm$^3$.
3. Percentage error $= \frac{1.0}{5.0} \times 100 = 20\%$.

Key Takeaways

Percentage error decreases as the measured volume increases for a fixed apparatus uncertainty. Always show the uncertainty calculation.

Common Mistakes

• Using only $0.5$ cm$^3$ as the uncertainty, giving 10%.
• Forgetting to multiply by 100.
• Not showing working.

Things to Be Careful About

The mark scheme explicitly uses $2 \times 0.5$ because two syringe readings are involved. Give the answer as 20%.

Background Concept

Absolute uncertainty is the fixed uncertainty of the apparatus, independent of the volume measured. Percentage error is the absolute uncertainty divided by the measured volume, multiplied by 100. If the measured volume increases while the absolute uncertainty stays the same, the percentage error decreases.

Understanding the Question

We must decide whether using a larger volume of alkaline potassium iodide makes the uncertainty greater, unchanged, or smaller, and do the same for the percentage error.

Approach

Think about whether the syringe's absolute uncertainty changes when a larger volume is used. It does not. Then think about how percentage error depends on the denominator.

Step-by-Step Reasoning

• The uncertainty of the measurement depends on the graduations of the syringe, not on the volume taken, so it has no effect.
• Percentage error $= \frac{\text{uncertainty}}{\text{volume}} \times 100$. If the volume is larger and the uncertainty is unchanged, the percentage error becomes smaller.

Key Takeaways

A larger measured volume gives a smaller percentage error for the same apparatus.

Common Mistakes

• Thinking that using a larger volume makes the uncertainty larger.
• Confusing absolute uncertainty with percentage error.

Things to Be Careful About

Place the tick for 'no effect' in the uncertainty row and 'smaller effect' in the percentage error row.

Background Concept

A burette must be clean and free from any solution that would contaminate the titrant. It is rinsed with the solution that will be used, not with water, so that the concentration of the titrant is not changed. The jet below the tap must also be filled so that the initial reading corresponds to a full column of solution.

Understanding the Question

Describe how to prepare a clean burette before taking readings, using the provided sodium thiosulfate solution.

Approach

State the two essential steps: rinse with the thiosulfate solution, then fill the burette and run solution through the jet.

Step-by-Step Reasoning

1. Rinse the burette with a little sodium thiosulfate solution so any water or contaminant is removed without diluting the titrant.
2. Fill the burette with sodium thiosulfate solution.
3. Open the tap briefly to run solution out through the jet, removing any air bubble, then record the initial reading.

Key Takeaways

Rinsing with the titrant and removing air from the jet are essential for an accurate titration.

Common Mistakes

• Saying 'rinse with distilled water' – this would dilute the thiosulfate.
• Forgetting to run solution through the jet.
• Not mentioning that the burette is filled with sodium thiosulfate.

Things to Be Careful About

The mark scheme gives one mark for rinsing with thiosulfate and one for filling and running solution through the jet.

Background Concept

In a titration, the burette delivers the titrant to the analyte until the reaction is just complete. The end-point is shown by a permanent colour change. To obtain an accurate titre, the titrant is added quickly at first but dropwise near the end-point so that the exact volume is not overshot.

Understanding the Question

Describe how to carry out the one accurate titration in step 6, using the sample in the conical flask and the sodium thiosulfate in the burette.

Approach

Describe adding thiosulfate from the burette to the flask while swirling, and adding it dropwise near the end-point until a permanent colour change occurs.

Step-by-Step Reasoning

1. Run sodium thiosulfate solution from the burette into the conical flask, swirling the flask continuously to mix the reactants.
2. As the colour starts to change, slow down and add the thiosulfate dropwise.
3. Stop when a permanent colour change is seen – this is the end-point. Record the final burette reading.

Key Takeaways

Accurate titration requires swirling, slow addition near the end-point, and recognising a permanent colour change.

Common Mistakes

• Adding the thiosulfate too quickly and overshooting the end-point.
• Not swirling, so the reaction is not uniform.
• Stopping at a temporary colour change.

Things to Be Careful About

The mark scheme gives one mark for adding thiosulfate until a permanent colour change and one for adding dropwise towards the end.

Background Concept

A blank experiment uses a sample known to contain none of the analyte. Any positive result must come from the reagents or the procedure. In this case, freshly distilled water contains no dissolved oxygen, so any measured oxygen must have come from the reagents used.

Understanding the Question

The student obtained $2.26 \times 10^{-5}$ mol dm$^{-3}$ for distilled water, not zero. We need to suggest why, assuming the procedure was carried out correctly.

Approach

Identify the source of the oxygen as the reagents or solutions used in the experiment.

Step-by-Step Reasoning

The manganese(II) sulfate solution, potassium iodide solution, sulfuric acid, and even the distilled water used for rinsing can contain small amounts of dissolved oxygen. When these are added to the sample, they contribute to the measured oxygen, giving a small positive result.

Key Takeaways

A blank experiment identifies the contribution of the reagents, which can then be subtracted.

Common Mistakes

• Saying the procedure was wrong – the question says it was carried out correctly.
• Saying 'oxygen from the air' – the mark scheme specifically credits oxygen in the reagents/solutions.

Things to Be Careful About

Mention 'oxygen in the reagents/solutions' to match the mark scheme.

Background Concept

A blank correction removes the systematic contribution of the reagents. The blank value is subtracted from the measured value for the sample to give the true concentration of the analyte.

Understanding the Question

We need to suggest how the blank value of $2.26 \times 10^{-5}$ mol dm$^{-3}$ could be used to improve the answer in part (f).

Approach

Subtract the blank value from the dissolved oxygen concentration calculated in (f).

Step-by-Step Reasoning

The value from (f) includes oxygen from the reagents as well as from the river water. Subtracting the blank value removes this contribution. For example, if the value in (f) were $2.11 \times 10^{-4}$ mol dm$^{-3}$, the corrected value would be $2.11 \times 10^{-4} - 2.26 \times 10^{-5} = 1.88 \times 10^{-4}$ mol dm$^{-3}$.

Key Takeaways

A blank correction improves accuracy by removing systematic error from the reagents.

Common Mistakes

• Adding the blank value instead of subtracting.
• Not stating that the blank value is subtracted from the result in (f).

Things to Be Careful About

Use the same units for both values before subtracting.

Background Concept

The reaction between hydrochloric acid and sodium thiosulfate is exothermic. As the reaction proceeds, heat is released, causing the temperature of the reaction mixture to rise slightly above the initial water bath temperature. Because the rate of reaction is temperature-dependent, the temperature is not constant throughout the 176 seconds it takes for the cross to disappear.

Understanding the Question

The question asks what temperature measurements should be taken and when, to accurately represent the temperature at which the reaction occurred. Since the temperature changes during the reaction, a single measurement is insufficient.

Approach

To capture the temperature change, we must measure the temperature at the beginning and at the end of the reaction period. These two values will then be averaged to give a representative temperature for the reaction.

Step-by-Step Reasoning

1. Start of reaction: The temperature should be measured immediately before the reactants are mixed (step 7) or immediately after step 5 when the solutions are at bath temperature. This gives the initial temperature $T_{\text{initial}}$.
2. End of reaction: The temperature should be measured immediately after the timer is stopped (step 8), when the cross is no longer visible. This gives the final temperature $T_{\text{final}}$.
   By taking readings at both points, we account for the temperature rise due to the exothermic nature of the reaction.

Key Takeaways

For exothermic or endothermic reactions where the temperature changes significantly during the measurement period, temperature must be recorded at both the start and end of the reaction to calculate an accurate average temperature.

Common Mistakes

• Only measuring the temperature at the start or only at the end.
• Measuring the temperature of the water bath instead of the reaction mixture itself.

Things to Be Careful About

The measurement must be of the reaction mixture in flask A, not the water bath. The timing must be clearly linked to the procedure steps (before/after mixing or before/after the cross disappears).

Background Concept

When a temperature changes linearly or approximately linearly over a time period, the average temperature can be accurately represented by the arithmetic mean of the initial and final temperatures.

Understanding the Question

Having taken two temperature readings (start and end), the question asks how to combine them to get an accurate value for the temperature during the reaction.

Approach

Use the simple arithmetic mean of the two recorded temperatures.

Step-by-Step Reasoning

The temperature at the start is $T_1$ and at the end is $T_2$. The average temperature during the reaction is calculated as:
$T_{\text{avg}} = \frac{T_1 + T_2}{2}$
This mean value is then used in the Arrhenius equation analysis.

Key Takeaways

Averaging start and end temperature readings is a standard technique to account for temperature drift in kinetic experiments.

Common Mistakes

• Taking the temperature of the water bath instead of averaging the mixture readings.
• Not averaging at all and just using one of the values.

Things to Be Careful About

Ensure you explicitly state "calculate the mean" or "average". Do not overcomplicate the explanation.

Background Concept

The Arrhenius equation in its logarithmic form is: $\ln k = -\frac{E_A}{R} \left(\frac{1}{T}\right) + \ln A$. Using base-10 logarithms, this becomes $\log k = -\frac{E_A}{2.303 R} \left(\frac{1}{T}\right) + \log A$. Since $1/t$ is proportional to the rate constant $k$, $\log(1/t)$ will be linearly related to $1/T$. The gradient of this line is $-\frac{E_A}{2.303 R}$, which is approximately $-\frac{0.434 E_A}{R}$.

Understanding the Question

The question asks to plot $\log(1/t)$ against $1/T$ using the data from Table 2.2 and draw a line of best fit. This graph will be used to determine the gradient, which is then used to calculate $E_A$.

Approach

1. Set up the axes correctly based on the given grid (Fig. 2.1).
2. Plot each data point from Table 2.2 as a cross ($\times$).
3. Draw a straight line of best fit that minimises the distance to all points.

Step-by-Step Reasoning

Plotting points:

• Point 1: $x = 0.00353$, $y = -2.43$
• Point 2: $x = 0.00336$, $y = -1.99$
• Point 3: $x = 0.00325$, $y = -1.68$
• Point 4: $x = 0.00314$, $y = -1.47$
• Point 5: $x = 0.00302$, $y = -1.21$
• Point 6: $x = 0.00287$, $y = -0.82$

Line of best fit:
The points should form a roughly straight line with a positive gradient. The line should pass as close as possible to all points, with an equal number of points above and below the line. One point (e.g., $0.00287, -0.82$) might be slightly off the line, which is acceptable if the overall trend is linear.

Key Takeaways

When plotting graphs for kinetic analysis, accuracy in plotting is crucial as it directly affects the gradient calculation and subsequent determination of $E_A$.

Common Mistakes

• Plotting points incorrectly (swapping $x$ and $y$ axes, or misreading the scale).
• Drawing a line that connects the points (a "joined-up" line) instead of a line of best fit.
• Not extending the line across the entire graph.

Things to Be Careful About

• The $x$-axis is $1/T$ and the $y$-axis is $\log(1/t)$. Do not swap them.
• Use a sharp pencil for accurate plotting.
• The line of best fit must be a straight line, not a curve connecting the points.

Background Concept

The gradient of the line in a plot of $\log(1/t)$ against $1/T$ is directly related to the activation energy $E_A$ by the equation: $\text{gradient} = -\frac{0.434 E_A}{R}$, where $R = 8.31 \text{ J K}^{-1} \text{mol}^{-1}$.

Understanding the Question

The question asks to determine the gradient of the line of best fit from the graph drawn in part (d)(i). The gradient must be calculated using two points on the line of best fit, not the actual data points, to minimise plotting errors.

Approach

1. Choose two points on the line of best fit that are far apart to minimise percentage error in the gradient calculation.
2. Read the coordinates $(x_1, y_1)$ and $(x_2, y_2)$ from the graph.
3. Calculate the gradient using $m = \frac{y_2 - y_1}{x_2 - x_1}$.
4. Round the final answer to three significant figures.

Step-by-Step Reasoning

Let's assume a line of best fit passes through:

• Point 1: $x_1 = 0.00290 \text{ K}^{-1}$, $y_1 = -1.00$
• Point 2: $x_2 = 0.00350 \text{ K}^{-1}$, $y_2 = -2.30$

Expressed to three significant figures: $-2.17 \times 10^3 \text{ K}$.

The units are Kelvin (K) because the $y$-axis is dimensionless (logarithm) and the $x$-axis is in $\text{K}^{-1}$, so $\frac{\text{dimensionless}}{\text{K}^{-1}} = \text{K}$.

Key Takeaways

Always use points on the line of best fit, not the raw data points, when calculating the gradient. Use a large triangle (points far apart) to reduce percentage error.

Common Mistakes

• Using data points instead of points on the line of best fit.
• Calculating $\frac{\Delta x}{\Delta y}$ instead of $\frac{\Delta y}{\Delta x}$.
• Forgetting the negative sign (the gradient is negative, though the graph appears to have a positive slope if plotted incorrectly; remember $1/T$ increases as $T$ decreases, and $\log(1/t)$ becomes more negative as $1/T$ increases).
• Not rounding to three significant figures.

Things to Be Careful About

• The coordinates must be read accurately from the graph.
• The gradient is negative because as $1/T$ increases (temperature decreases), $\log(1/t)$ decreases (reaction is slower).
• Ensure the units are stated as K.

Background Concept

The Arrhenius equation relates the rate constant $k$ to temperature $T$ and activation energy $E_A$: $k = A e^{-E_A / RT}$. Taking the natural logarithm gives $\ln k = -\frac{E_A}{R} \left(\frac{1}{T}\right) + \ln A$. Converting to base-10 logarithms ($\log_{10} = \frac{\ln}{2.303}$):

Since $1/t$ is proportional to $k$, $\log(1/t)$ has the same form. The constant $0.434$ is approximately $1/2.303$.

Understanding the Question

The question provides the equation $\log(1/t) = -\frac{0.434 E_A}{RT} + \text{constant}$ and asks to calculate $E_A$ using the gradient from part (d)(ii).

Approach

1. Identify that the gradient of the graph equals $-\frac{0.434 E_A}{R}$.
2. Rearrange the equation to solve for $E_A$.
3. Substitute the gradient value and the gas constant $R = 8.31 \text{ J K}^{-1} \text{mol}^{-1}$.
4. Calculate $E_A$ in J mol$^{-1}$ and convert to kJ mol$^{-1}$.

Step-by-Step Reasoning

From the graph equation:

Rearranging:

Substituting the example gradient ($-2.17 \times 10^3 \text{ K}$) and $R = 8.31 \text{ J K}^{-1} \text{mol}^{-1}$:

Converting to kJ mol$^{-1}$ (divide by 1000):

Key Takeaways

The gradient of a $\log(1/t)$ vs $1/T$ graph is directly used to calculate activation energy. Always check units: $R$ is in J, so $E_A$ will be in J mol$^{-1}$, which should be converted to kJ mol$^{-1}$.

Common Mistakes

• Forgetting to rearrange the equation correctly.
• Using the wrong value for $R$ (must be $8.31 \text{ J K}^{-1} \text{mol}^{-1}$, not $0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{mol}^{-1}$).
• Not converting from J mol$^{-1}$ to kJ mol$^{-1}$.
• Using data points instead of the gradient in the calculation.

Things to Be Careful About

• The equation given uses $0.434$, which is $1/\ln(10) \approx 1/2.303$. Do not use $2.303$ in the denominator instead of $0.434$ in the numerator.
• Units: The gradient has units of K (since $y$ is dimensionless and $x$ is $\text{K}^{-1}$). $R$ is in $\text{J K}^{-1} \text{mol}^{-1}$. Therefore, $E_A = \frac{\text{K} \times \text{J K}^{-1} \text{mol}^{-1}}{\text{dimensionless}} = \text{J mol}^{-1}$.
• Always include units in the final answer as requested.

Background Concept

The reliability of experimental data in kinetics is assessed by examining how well the data points fit the expected theoretical relationship. For the Arrhenius plot, a linear relationship is expected. Deviations from linearity indicate experimental errors or anomalous results.

Understanding the Question

The question asks to use the graph to state whether the results are reliable and to justify the answer.

Approach

Examine the graph to see if the points follow the expected linear trend. Look for anomalous points (outliers) that are far from the line of best fit.

Step-by-Step Reasoning

If the results are reliable:

• State: "Yes, the results are reliable."
• Justification: "There are no (or very few) anomalous points, and most points lie on or near the line of best fit. This indicates a consistent linear relationship between $\log(1/t)$ and $1/T$, supporting the Arrhenius equation."

If the results are not fully reliable:

• State: "No, the results are not fully reliable."
• Justification: "There is an anomalous point (e.g., at $1/T = 0.00287, \log(1/t) = -0.82$), and some points do not lie on or near the line of best fit. This suggests experimental errors at certain temperatures."

Key Takeaways

Reliability is assessed by the scatter of data points around the line of best fit. A good fit indicates reliable data.

Common Mistakes

• Saying "yes" or "no" without justification.
• Using vague terms like "the graph looks good" without referencing anomalous points or the line of best fit.

Things to Be Careful About

• The justification must directly reference the graph (anomalous points, points on/near the line).
• Both "yes" and "no" are acceptable answers as long as the justification is consistent with the graph drawn.