Chemistry 9701/51 — May/June 2010

Background Concept

In an exothermic neutralisation reaction, the temperature of the mixture increases as the acid and base react. Once one of the reactants is completely used up (the equivalence point or neutralisation point), no more heat is produced. Any further addition of the other reactant (which is typically at room temperature and cooler than the hot mixture) will cause the overall temperature of the solution to decrease. Plotting temperature against the volume of titrant added yields a characteristic graph with a distinct peak.

Understanding the Question

The question asks you to design the quantitative basis for a calorimetry experiment. You must choose a starting volume of NaOH, calculate the exact volume of H₂SO₄ needed to neutralise it using stoichiometry, and sketch the expected temperature-volume graph, clearly marking the neutralisation point.

Approach

1. Choose a volume of NaOH between 10 and 80 cm³. A round number like 50 cm³ is practical and easy to measure.
2. Use the given concentrations and the balanced equation to find the moles of NaOH, then the moles of H₂SO₄ required, and finally the volume of H₂SO₄.
3. Sketch the graph: rising temperature up to the calculated volume, then falling temperature afterwards. Mark the peak.

Step-by-Step Reasoning

• Selecting the volume: The mark scheme allows any volume between 10 and 80 cm³. Let's select $x = 50 \text{ cm}^3$ of $3 \text{ mol dm}^{-3}$ NaOH.
• Calculating the acid volume:
  • Moles of NaOH = concentration × volume = $3 \times \frac{50}{1000} = 0.15 \text{ mol}$.
  • The balanced equation is $2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$. The stoichiometric ratio of NaOH to H₂SO₄ is 2:1.
  • Moles of H₂SO₄ needed = $\frac{0.15}{2} = 0.075 \text{ mol}$.
  • Volume of H₂SO₄ = $\frac{\text{moles}}{\text{concentration}} = \frac{0.075}{2} = 0.0375 \text{ dm}^3 = 37.5 \text{ cm}^3$.
  • Alternatively, using the ratio directly: Volume of H₂SO₄ = $\frac{1}{2} \times \frac{3}{2} \times 50 = 0.75 \times 50 = 37.5 \text{ cm}^3$.
• Sketching the graph:
  • Axes: y-axis is 'temperature of the solution / °C', x-axis is 'volume of acid added / cm³'.
  • Shape: Start at an initial temperature (e.g., 20°C). Draw a line going up to the point (37.5, max temp). Then draw a line going down from that peak. Straight lines or curves are both acceptable.
  • Label: Mark the peak of the graph and label it 'neutralisation point' or indicate that it occurs at 37.5 cm³.

Key Takeaways

• Always choose a volume that is easy to measure and leaves room for the titrant volume to be reasonable.
• The peak of a temperature vs. volume graph in a neutralisation experiment indicates the exact stoichiometric equivalence point.
• After the equivalence point, the temperature drops because excess reactant is added at a lower temperature, cooling the mixture.

Common Mistakes

• Choosing a volume of NaOH outside the 10-80 cm³ range.
• Forgetting to use the 2:1 mole ratio from the balanced equation and assuming a 1:1 ratio.
• Sketching a graph that plateaus instead of peaking, or failing to mark the neutralisation point.
• Drawing the temperature continuing to rise after the neutralisation point.

Things to Be Careful About

• Ensure the volume units are consistent (convert cm³ to dm³ when using $n = c \times V$).
• The graph must clearly show the maximum temperature and link it to the calculated volume of acid.

Background Concept

In quantitative experiments, different apparatus is used depending on the required precision. A measuring cylinder is for rough measurements, while a burette or pipette is used for accurate volumetric measurements.

Understanding the Question

The independent variable is the volume of acid added. We need an apparatus that can measure this volume accurately and allow for gradual addition.

Approach

Identify the standard laboratory equipment used for accurate, variable volume additions in titrations and calorimetry.

Step-by-Step Reasoning

• A burette allows for precise measurement of the volume added (to 0.05 cm³) and controlled, gradual addition.
• A pipette could also be used if adding in fixed, accurate portions (e.g., a 10 cm³ pipette used multiple times).
• A measuring cylinder is not accurate enough for this type of quantitative experiment.

Background Concept

To accurately find the maximum temperature (neutralisation point), you cannot just add the acid all at once, nor can you add it dropwise (which would take too long and allow too much heat loss). You need to add it in controlled increments.

Understanding the Question

How do you use the burette/pipette to control the independent variable effectively to capture the peak temperature?

Approach

Describe the practical technique of adding the titrant in portions and recording the temperature after each portion.

Step-by-Step Reasoning

• If you add dropwise, heat loss to the surroundings will be significant, and the peak will be missed or flattened.
• If you add it all at once, you won't know the exact volume at which neutralisation occurred.
• Therefore, you add the acid in successive portions (e.g., 5 cm³ or 10 cm³ increments using the burette).
• After each addition, you stir and record the highest temperature reached before it starts to fall.
• This allows you to plot the graph and interpolate the exact volume at the peak.

Background Concept

A good data table organises raw measurements and calculated values logically. It must include column headings with quantities and units, consistent decimal places, and provisions for repeats to allow calculation of a mean.

Understanding the Question

Draw a table to record all measurements needed to determine the enthalpy change of neutralisation.

Approach

Identify the raw data (volumes, temperatures) and the calculated data (temperature change) needed. Structure it for multiple repeats.

Step-by-Step Reasoning

• Raw data: We need to record the volume of acid added (in portions) and the temperature of the solution after each addition.
• Initial temperature: We need the starting temperature of the NaOH before any acid is added (at 0 cm³ acid).
• Calculated data: The temperature change, $\Delta T = T_{\text{final}} - T_{\text{initial}}$ (or max temp - initial temp), is needed for the heat calculation $q = mc\Delta T$.
• Repeats: The experiment should be repeated at least 3 times to identify anomalies and calculate a mean $\Delta T$.
• Units: Every column heading must include the unit (e.g., 'Volume of H₂SO₄ added / cm³').
• Alternative layout: A separate column for 'Initial temperature' if it's assumed constant, or recording it for each repeat.

Key Takeaways

• Always include units in column headings.
• Include columns for raw data and derived data.
• Structure the table to clearly show repeats.

Background Concept

In simple calorimetry, the heat energy absorbed or released by the solution is calculated using the formula:

$q = mc\Delta T$

where:

• $m$ is the mass of the solution (in grams or cm³, assuming density $\approx 1 \text{ g cm}^{-3}$).
• $c$ is the specific heat capacity (in J g⁻¹ °C⁻¹ or J cm⁻³ °C⁻¹).
• $\Delta T$ is the temperature change (in °C).

The question simplifies this by giving a combined constant: $4.3 \text{ J}$ raises the temperature of $1 \text{ cm}^3$ of solution by $1 \text{ °C}$. This effectively gives $m \times c = 4.3 \text{ J cm}^{-3} \text{ °C}^{-1}$.

Understanding the Question

Show how to calculate the total heat energy produced up to the neutralisation point, using $\Delta T$ and the values from part (a).

Approach

1. Calculate the total volume of the mixture at neutralisation.
2. Multiply by the given constant (4.3) and the temperature change ($\Delta T$).

Step-by-Step Reasoning

• Total volume: At the neutralisation point, all the NaOH and the calculated volume of H₂SO₄ have been mixed. Total volume = $V_{\text{NaOH}} + V_{\text{H}_2\text{SO}_4} = 50 + 37.5 = 87.5 \text{ cm}^3$.
• Heat energy: $q = \text{Total volume} \times 4.3 \times \Delta T$.
• Substitute the values: $q = 87.5 \times 4.3 \times \Delta T = 376.25 \Delta T \text{ J}$.
• The mark scheme accepts the expression with the numerical values added: $(50 + 37.5) \times 4.3 \times \Delta T$.
• Units are not strictly required in the expression, but the result is in Joules (J).

Background Concept

The enthalpy change of neutralisation, $\Delta H_{\text{neutralisation}}$, is the heat energy change when 1 mole of water is formed from the reaction of an acid and a base (or equivalently, 1 mole of H⁺ reacts with 1 mole of OH⁻).

where $q$ is the heat energy in Joules (or kJ) and $n$ is the number of moles of water formed (or moles of limiting reactant that determines the moles of water).

The sign of $\Delta H$ is negative for exothermic reactions (heat is released).

Understanding the Question

Convert the energy calculated in part (h) (which is in Joules) into $\Delta H$ in kJ mol⁻¹ for the given equation:

Show the sign and the expression.

Approach

1. Determine the number of moles relevant to the equation (moles of NaOH or moles of H₂O).
2. Convert the energy from J to kJ.
3. Divide energy by moles.
4. Add the negative sign because neutralisation is exothermic.

Step-by-Step Reasoning

• Moles: The target equation shows 1 mole of NaOH producing 1 mole of H₂O. In our experiment, we used $0.15 \text{ mol}$ of NaOH, which produces $0.15 \text{ mol}$ of H₂O. (Alternatively, moles of H₂SO₄ = $0.075 \text{ mol}$. The equation uses $1/2$ mole of H₂SO₄, so moles of 'reactions' = $0.075 / 0.5 = 0.15 \text{ mol}$. Same result).
• Energy conversion: The energy from (h) is in Joules. To get kJ, divide by 1000: $q_{\text{kJ}} = \frac{\text{answer to (h)}}{1000}$.
• Calculation: $\Delta H = -\frac{q_{\text{kJ}}}{n} = -\frac{\text{answer to (h)} / 1000}{0.15}$.
• Sign: Must be explicitly stated as negative (–) because the reaction is exothermic.

Key Takeaways

• Always check the definition of the enthalpy change (per mole of what? Here, per mole of H₂O or per mole of NaOH).
• Remember to convert J to kJ by dividing by 1000.
• Exothermic reactions have a negative $\Delta H$.

Common Mistakes

• Forgetting to convert Joules to kJ (resulting in a value 1000 times too large).
• Using the wrong number of moles (e.g., using moles of H₂SO₄ without multiplying by 2, or using total moles of ions).
• Forgetting the negative sign for an exothermic reaction.
• Using the total volume in the denominator instead of moles.

Things to Be Careful About

• The expression must clearly show the division by moles and the conversion to kJ.
• The mark scheme allows 'moles of NaOH' or 'moles of H₂O' in the denominator. Using 'moles of H₂SO₄' is only acceptable if multiplied by 2.