Chemistry 9701/33 — February/March 2022

Background Concept

In a thermometric titration, the reaction is exothermic, so the temperature rises as acid is added until the alkali is completely neutralised (the end-point). After this point, adding more acid does not produce heat, but the cooler added acid dilutes the mixture and absorbs heat from the solution, causing the temperature to fall. Plotting temperature against volume of titrant yields a graph with two distinct linear regions: a rising line (before the end-point) and a falling line (after the end-point). The true end-point temperature is found by extrapolating both lines to their intersection, which corrects for heat loss that occurred during the reaction and between additions.

Understanding the Question

Part (b)(i) asks the candidate to plot the temperature vs. volume data on the provided grid. Specific requirements include axis labels, scale selection (occupying > half the grid), accurate plotting, and drawing two separate lines of best fit that are extrapolated to intersect.

Approach

Describe the exact construction of the graph: axis assignment, scale rules, plotting rules, and the specific method for drawing and extrapolating the two lines of best fit to find the corrected maximum temperature.

Step-by-Step Reasoning

1. Axis assignment: Temperature is the dependent variable, so it goes on the $y$-axis. Volume of FA 2 (sulfuric acid) is the independent variable, so it goes on the $x$-axis. Both must have clear labels and units (°C and cm³).
2. Scale selection: The scales must be linear and chosen so that the data points occupy more than half the length of each axis. This ensures the graph is large enough to read accurately. The $y$-axis must extend to at least $4.0 \text{ °C}$ above the highest recorded temperature to allow room for the extrapolated intersection.
3. Plotting: All 10 data points must be plotted accurately. If any point is clearly an outlier (e.g., due to a missed temperature reading or poor stirring), it should be marked as anomalous and excluded from the line of best fit.
4. Lines of best fit: Draw one straight line through the rising points (before the end-point) and another through the falling points (after the end-point). These lines should not necessarily pass through all points but should balance the scatter around them.
5. Extrapolation: Extend both lines beyond the data range until they cross. The intersection represents the theoretical maximum temperature if no heat were lost. This intersection temperature must be at or above the highest actual recorded temperature.

Key Takeaways

• Thermometric graphs require two lines of best fit (rising and falling) extrapolated to an intersection.
• The intersection gives the corrected maximum temperature, accounting for heat loss.
• Scales must be chosen to maximise the use of the graph paper for accuracy.

Common Mistakes

• Drawing a single smooth curve through all points instead of two distinct straight lines.
• Forgetting to extrapolate the lines; reading the maximum temperature directly from the highest plotted point.
• Using a non-linear scale or choosing scales that make the graph too small.

Things to Be Careful About

• The intersection point must be at a temperature equal to or higher than the highest recorded temperature. If the intersection is lower, the lines are drawn incorrectly.
• Ensure the $y$-axis scale includes the required $4.0 \text{ °C}$ buffer above the max recorded temperature.

Background Concept

The intersection of the two extrapolated lines of best fit in a thermometric titration graph represents the theoretical end-point. The $x$-coordinate of this intersection is the exact volume of titrant (acid) required to completely neutralise the analyte (alkali), corrected for any heat loss that shifted the actual maximum temperature to a lower volume.

Understanding the Question

Part (b)(ii) asks the candidate to determine the volume of sulfuric acid (FA 2) needed to neutralise the 25.0 cm³ of sodium hydroxide (FA 1) using the graph constructed in part (b)(i).

Approach

Explain that the volume is read directly from the $x$-axis at the point where the two extrapolated lines of best fit intersect. Provide a representative example value.

Step-by-Step Reasoning

1. Locate the intersection point of the two lines of best fit on the graph.
2. Draw a vertical line down from this intersection to the $x$-axis (volume of FA 2).
3. Read the value on the $x$-axis. Since the $x$-axis has major markings every 5.00 cm³ and minor markings every 0.50 or 1.00 cm³, the value can be read to 1 or 2 decimal places (e.g., 24.50 cm³).
4. This volume is the corrected volume of acid required for neutralisation.

Key Takeaways

• The $x$-coordinate of the intersection gives the corrected volume of titrant at the end-point.
• Readings from graphs should be estimated to one decimal place beyond the smallest grid division.

Common Mistakes

• Reading the volume at the highest recorded temperature point instead of the intersection of the extrapolated lines.
• Not reading to the correct number of decimal places (the mark scheme allows 1 or 2 dp).

Things to Be Careful About

• Ensure the value read is from the $x$-axis, not the $y$-axis.
• The value must be physically plausible (e.g., between 0 and 45.00 cm³ based on the data range).

Background Concept

The concentration of an unknown solution can be determined from a titration using the balanced chemical equation and the mole ratio between the reactants. For the reaction between sodium hydroxide and sulfuric acid:

2 moles of NaOH react with 1 mole of H₂SO₄. This 2:1 ratio must be used when calculating moles of acid from moles of alkali.

Understanding the Question

Part (b)(iii) asks for the concentration of sulfuric acid in FA 2. The candidate must use the volume of acid determined from the graph in part (b)(ii) and the known concentration and volume of sodium hydroxide in FA 1.

Approach

Calculate the moles of NaOH, use the 2:1 stoichiometric ratio to find moles of H₂SO₄, then divide by the volume of H₂SO₄ (in dm³) to find its concentration. Show the formula and substitute the value from (b)(ii).

Step-by-Step Reasoning

1. Moles of NaOH: $n = c \times V = 1.90 \text{ mol dm}^{-3} \times 0.0250 \text{ dm}^3 = 0.0475 \text{ mol}$.
2. Mole ratio: From the equation, $n(\text{H}_2\text{SO}_4) = \frac{1}{2} n(\text{NaOH}) = \frac{0.0475}{2} = 0.02375 \text{ mol}$.
3. Concentration of H₂SO₄: $c = \frac{n}{V} = \frac{0.02375 \text{ mol}}{V_{\text{acid}} \text{ in dm}^3}$.
4. Using the mark scheme formula: The mark scheme provides a direct formula: $\text{Concentration} = \frac{25 \times 1.9}{2 \times \text{volume from (b)(ii)}}$. This is derived by combining the steps above and keeping volumes in cm³ (the 1000 factors cancel out).
5. Substitution: If the volume from (b)(ii) is 24.50 cm³, then $c = \frac{25 \times 1.9}{2 \times 24.50} = \frac{47.5}{49.0} = 0.9694 \approx 0.970 \text{ mol dm}^{-3}$.
6. Significant figures: The answer should be given to 2-4 significant figures. 0.970 has 3 sf, which is appropriate given the data (1.90 has 3 sf, 25.0 has 3 sf).

Key Takeaways

• Always use the correct stoichiometric ratio from the balanced equation.
• The mark scheme formula $\frac{c_1 V_1}{n_1} = \frac{c_2 V_2}{n_2}$ can be rearranged to find any unknown concentration.
• Keep volumes in consistent units; if both are in cm³, the 1000 conversion factors cancel.

Common Mistakes

• Forgetting the 2:1 mole ratio and calculating the acid concentration as if it were a 1:1 reaction (which would give double the correct value).
• Not converting the volume from cm³ to dm³ before dividing (though the mark scheme formula bypasses this by keeping both volumes in cm³).
• Rounding intermediate values too early, leading to a final answer with incorrect significant figures.

Things to Be Careful About

• The volume used must be the one read from the intersection of the lines in part (b)(ii), not the volume at the highest recorded temperature.
• Ensure the final answer has the correct unit: mol dm⁻³.

Background Concept

The enthalpy change of reaction is calculated using the equation $q = mc\Delta T$, where $\Delta T$ is the temperature change. In a thermometric titration, heat is lost to the surroundings, so the maximum recorded temperature is lower than the true temperature that would have been reached in an adiabatic (perfectly insulated) system. To find the true $\Delta T$, the graph is used to extrapolate back to the theoretical end-point temperature.

Understanding the Question

Part (c)(i) asks how the student used their graph to determine the value of $\Delta T$ to use in the equation $q = mc\Delta T$.

Approach

Explain that $\Delta T$ is the difference between the theoretical maximum temperature (found at the intersection of the extrapolated lines) and the initial temperature of the alkali before any acid was added.

Step-by-Step Reasoning

1. The initial temperature is the temperature of the 25.0 cm³ of NaOH solution recorded in the results table at 0.00 cm³ of acid added.
2. The theoretical maximum temperature is the $y$-coordinate of the intersection of the two extrapolated lines of best fit on the graph.
3. $\Delta T$ is calculated as: $\Delta T = T_{\text{final (theoretical)}} - T_{\text{initial}}$.
4. An alternative (and acceptable) method is to use the $y$-intercept of the rising line of best fit as the initial temperature (correcting for any heat loss that occurred before the first reading was taken), so $\Delta T = T_{\text{intersection}} - T_{y\text{-intercept}}$.

Key Takeaways

• $\Delta T$ for enthalpy calculations in thermometric titrations must be based on the extrapolated end-point temperature, not the highest recorded temperature.
• The initial temperature is typically the reading at 0.00 cm³ of titrant added.

Common Mistakes

• Using the highest recorded temperature as $T_{\text{final}}$ instead of the extrapolated intersection temperature.
• Forgetting to subtract the initial temperature (calculating $\Delta T$ as just the final temperature).

Things to Be Careful About

• Ensure the initial temperature used is the one recorded at 0.00 cm³, not an estimated value from the $y$-axis intercept unless explicitly using that method.
• $\Delta T$ must be positive for an exothermic reaction (final T > initial T).

Background Concept

Percentage error is a measure of accuracy, comparing an experimental value to a known theoretical or accepted value. It is calculated as the absolute difference between the two values, divided by the theoretical value, multiplied by 100 to give a percentage. For enthalpy changes, which are negative for exothermic reactions, the absolute value is used in the denominator to ensure the percentage error is positive.

Understanding the Question

Part (c)(ii) provides the student's calculated enthalpy of neutralisation ($-55.2 \text{ kJ mol}^{-1}$) and the theoretical value ($-57.6 \text{ kJ mol}^{-1}$). The candidate must calculate the percentage error.

Approach

Apply the percentage error formula: $\frac{\text{difference}}{\text{theoretical value}} \times 100$. Use absolute values to ensure a positive percentage.

Step-by-Step Reasoning

1. Difference: $|-55.2 - (-57.6)| = |-55.2 + 57.6| = 2.4 \text{ kJ mol}^{-1}$.
2. Theoretical value: $|-57.6| = 57.6 \text{ kJ mol}^{-1}$.
3. Calculation: $\frac{2.4}{57.6} \times 100 = 4.1666...\%$.
4. Rounding: The mark scheme accepts 4.167, 4.17, or 4.2%. Typically, 3 significant figures (4.17%) is appropriate.

Key Takeaways

• Percentage error is always positive; use absolute values.
• The denominator is the theoretical (accepted) value, not the experimental value.

Common Mistakes

• Forgetting to use absolute values, resulting in a negative percentage error.
• Dividing by the experimental value instead of the theoretical value.
• Not multiplying by 100 (giving 0.0417 instead of 4.17%).

Things to Be Careful About

• Enthalpy values are negative; ensure the signs are handled correctly in the subtraction: $-55.2 - (-57.6) = +2.4$.
• The mark scheme accepts a range of rounded values (4.167, 4.17, 4.2).

Background Concept

The theoretical enthalpy of neutralisation for a strong acid and strong base is $-57.6 \text{ kJ mol}^{-1}$. Experimental values are often less negative (e.g., $-55.2 \text{ kJ mol}^{-1}$) due to systematic errors that cause the measured temperature rise ($\Delta T$) to be smaller than it should be. Since $q = mc\Delta T$ and $\Delta H = -q/n$, a smaller $\Delta T$ leads to a smaller calculated $q$ and thus a less negative $\Delta H$.

Understanding the Question

Part (c)(iii) asks why the student's result ($-55.2 \text{ kJ mol}^{-1}$) was less negative than the theoretical value ($-57.6 \text{ kJ mol}^{-1}$). The candidate must suggest a reason and explain how it affects the result.

Approach

Identify a source of error in the experimental method that would cause the measured temperature rise to be lower than the true value. Explain the chain of reasoning: error -> lower $\Delta T$ -> lower $q$ -> less negative $\Delta H$.

Step-by-Step Reasoning

1. Heat loss (most common): The apparatus (cup, beaker, thermometer) is not perfectly insulated. Heat is lost to the surroundings during the reaction and, crucially, between the 5.00 cm³ additions of acid. This means the temperature recorded after each addition is lower than it would be in an adiabatic system. The extrapolated intersection temperature is therefore lower than the true maximum. A lower $\Delta T$ gives a lower $q$, and thus a less negative $\Delta H$.
2. Insufficient readings: The method adds acid in 5.00 cm³ increments. If the end-point is between two additions (e.g., at 24.5 cm³), the temperature may have already started to fall before the true maximum was recorded. If there are not enough readings near the end-point, the falling line of best fit may be inaccurate, lowering the intersection temperature.
3. Volume measurement error: The 25.0 cm³ of NaOH was measured with a measuring cylinder, which has a lower accuracy (±0.5 cm³) than a pipette (±0.06 cm³). If the actual volume of NaOH was less than 25.0 cm³ (e.g., 24.5 cm³), there would be fewer moles of NaOH to react, releasing less total heat. This would result in a lower temperature rise and a less negative $\Delta H$.

Key Takeaways

• Experimental enthalpy values are often less negative than theoretical values due to heat loss.
• Errors that lower the measured $\Delta T$ will always result in a less negative calculated $\Delta H$.
• Measuring instruments matter: a measuring cylinder introduces more error than a pipette.

Common Mistakes

• Saying "heat was lost" without explaining how this leads to a less negative $\Delta H$ (must link lower $\Delta T$ to lower $q$ to less negative $\Delta H$).
• Suggesting heat was gained from the surroundings (which would make the result more negative).
• Saying "human error" or "inaccurate equipment" without specifying the chemical/physical reason.

Things to Be Careful About

• The explanation must be a cause-and-effect chain: Error -> effect on $\Delta T$ -> effect on calculated $\Delta H$.
• Ensure the direction of the error is correct: a less negative result means the magnitude of $\Delta H$ is too small, so $\Delta T$ must have been too small.