How to Calculate Enthalpy of Neutralisation When Volumes Differ

Mix 25.0 mL of acid with 40.0 mL of base in a calorimeter. The temperature rises. Which volume should you use to calculate the heat released?

Both volumes. The entire mixture absorbs the heat, so both volumes contribute to its mass.

But which volume determines how many moles actually react? Neither volume on its own. You must calculate the moles of acid and base separately, identify the limiting reactant, and use the stoichiometry to determine how many moles of water form.

This is the key to calculating the enthalpy of neutralisation when acid and base volumes differ: use the total volume to calculate heat, but use the amount that actually reacts to calculate molar enthalpy.

01Why the two volumes have different jobs

Imagine mixing hydrochloric acid and sodium hydroxide in an insulated cup.

The reaction releases energy. That energy warms the solution, and a thermometer records the resulting temperature increase.

There are two different quantities to keep track of:

• The mass of solution that warms up.
• The amount of acid and base that actually react.

These are not necessarily proportional to each other.

For example, suppose you mix 25.0 mL of hydrochloric acid with 40.0 mL of sodium hydroxide. Even if the acid is completely consumed, the excess sodium hydroxide remains in the cup. It still absorbs heat and contributes to the mass of the solution.

System and surroundings: where does the heat go?

Before calculating anything, identify the system and surroundings.

• System: The neutralisation reaction itself, where the reactants are converted into products.
• Surroundings: The aqueous solution and the calorimeter, including the cup and other equipment that absorbs heat.

During an exothermic neutralisation, energy transfers from the reacting system to its surroundings.

The solution gets warmer because it gains energy.

This gives us two important signs:

The energy conservation relationship is:

\[ q_{\text{reaction}}=-q_{\text{solution}} \]

The thermometer measures the temperature change of the surroundings, not the heat released directly by the reaction.

At approximately constant atmospheric pressure, the heat transferred by the reaction corresponds to its enthalpy change.

What should the calorimeter show?

Predict what happens if the same amount of heat is released into two different volumes of solution.

The larger volume should experience a smaller temperature increase because the released energy is distributed across more solution.

This is why you cannot ignore the base just because it is in excess. Even if it doesn’t react, it still contributes to the energy absorbed by the surroundings.

02The calculation method: five steps

For most HSC-style neutralisation calorimetry questions, follow this order.

Step 1: Calculate the total solution mass

When the acid and base have different volumes, add them to find the total volume of the mixture.

\[ V_{\text{total}}=V_{\text{acid}}+V_{\text{base}} \]

In a typical school calorimetry question, you are asked to assume:

• The density of each solution is \(1.00\ \mathrm{g\,mL^{-1}}\).
• The final volume is approximately the sum of the starting volumes.

Therefore:

\[ m_{\text{solution}}\approx V_{\text{total}} \]

Here, the numerical value of the mass in grams equals the total volume in millilitres.

For example, mixing 25.0 mL and 40.0 mL gives:

\[ m_{\text{solution}}=65.0\ \mathrm{g} \]

Notice that we use both volumes, even if one reactant is in excess.

If the question provides actual solution masses or different densities, use those instead of this approximation.

Step 2: Calculate the moles of each reactant

Now calculate the amount of acid and base separately.

Use:

\[ n=CV \]

where:

• \(n\) is the amount in moles.
• \(C\) is the molar concentration in \(\mathrm{mol\,L^{-1}}\).
• \(V\) is the volume in litres.

Convert each volume from millilitres to litres before substituting.

For example:

\[ 25.0\ \mathrm{mL}=0.0250\ \mathrm{L} \]

Do not add the acid and base volumes before calculating their moles. Their concentrations may be different.

Step 3: Identify the limiting reactant

The limiting reactant is the reactant that is completely consumed first.

For a simple reaction between a monoprotic acid and a base containing one hydroxide ion per formula unit:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

The mole ratio is 1:1.

Therefore:

\[ n(\ce{H2O})= \min\left(n(\ce{HCl}),n(\ce{NaOH})\right) \]

This gives the amount of water formed.

For reactions with different mole ratios, use the balanced equation rather than automatically choosing the smaller number of moles.

For example:

\[ \ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O} \]

One mole of sulfuric acid requires two moles of sodium hydroxide for complete neutralisation.

If you have 2.00 mmol of sulfuric acid and 3.00 mmol of sodium hydroxide, sodium hydroxide is limiting. The amount of water formed is 3.00 mmol, not 2.00 mmol.

The balanced equation always determines the correct amount.

Step 4: Calculate the heat absorbed by the solution

Use the calorimetry equation:

\[ q_{\text{solution}}=mc\Delta T \]

where:

• \(q_{\text{solution}}\) is heat absorbed in joules (J).
• \(m\) is the total solution mass in grams (g).
• \(c\) is the specific heat capacity of the solution.
• \(\Delta T\) is the temperature change in degrees Celsius.

For dilute aqueous solutions, assume:

\[ c=4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}} \]

Calculate the temperature change using:

\[ \Delta T=T_{\text{final}}-T_{\text{initial}} \]

For an exothermic reaction, the temperature increases, so \(\Delta T\) is positive.

Step 5: Calculate the enthalpy of neutralisation

Remember that the solution gains heat while the reaction releases it.

Therefore:

\[ q_{\text{reaction}}=-mc\Delta T \]

The molar enthalpy of neutralisation is the enthalpy change per mole of water formed.

\[ \boxed{ \Delta H_{\text{neut}}= \frac{q_{\text{reaction}}}{n(\ce{H2O})} } \]

If your heat is in joules, divide the result by 1000 to express the enthalpy in \(\mathrm{kJ\,mol^{-1}}\).

Your final answer should normally be negative for an exothermic neutralisation.

04The common mistake: using the limiting reactant’s volume for everything

A tempting approach is to think that because hydrochloric acid is the limiting reactant, its 25.0 mL volume should be used throughout the calculation.

That would be incorrect.

The acid determines how much neutralisation can occur, but it does not determine the total mass of solution absorbing the heat.

Compare the two calculations:

There is another mistake worth avoiding: dividing the reaction heat by the sum of the initial acid and base moles.

That would give you an energy per mole of combined starting reactants, not the enthalpy of neutralisation per mole of water formed.

Keep the two calculations separate.

06What assumptions does the calorimetry method make?

The calculations above use a simplified calorimeter model. It is useful, but it is not a perfect representation of a real experiment.

1. The solution has the same properties as water

We assume:

\[ \rho=1.00\ \mathrm{g\,mL^{-1}} \]

and:

\[ c=4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}} \]

These are approximations for dilute aqueous solutions.

If a question supplies a different density, specific heat capacity, or actual solution mass, use the supplied values.

2. The calorimeter does not absorb significant heat

A polystyrene cup reduces heat transfer to the surroundings, but it does not eliminate it.

Some energy may warm the cup, lid, thermometer, and other equipment.

If the calorimeter absorbs heat, the energy balance becomes:

\[ q_{\text{reaction}}= -(q_{\text{solution}}+q_{\text{calorimeter}}) \]

If the calorimeter constant is supplied, calculate:

\[ q_{\text{calorimeter}}=C_{\text{cal}}\Delta T \]

where \(C_{\text{cal}}\) is the calorimeter heat capacity in \(\mathrm{J\,^{\circ}C^{-1}}\).

The corrected calculation therefore includes both the solution and the calorimeter.

If you ignore the calorimeter’s heat capacity when it is significant, you underestimate the heat released by an exothermic reaction. Your calculated enthalpy will be less negative than the actual value.

3. The initial solutions have the same temperature

The usual calculation:

\[ \Delta T=T_{\text{final}}-T_{\text{initial}} \]

assumes both solutions begin at the same temperature.

If the acid and base start at different temperatures, you cannot simply use the acid’s initial temperature.

You must account for the thermal energy already present in each solution before mixing.

We will examine this more carefully in Question 5.

4. The reaction proceeds to completion

The simple limiting-reactant method assumes the neutralisation proceeds according to the balanced equation.

For a strong acid reacting with a strong base, this is generally an appropriate approximation.

Weak acids and bases require additional consideration. Their ionisation behaviour contributes to the overall energy change, so their measured enthalpy of neutralisation may differ from that of a strong acid and strong base.

For dilute strong acid-strong base neutralisation, a value near \(-57\ \mathrm{kJ\,mol^{-1}}\) is a useful approximate reference. It is a comparison point, not a value you should force your experimental calculation to produce.

07Questions and solutions

The following questions use illustrative data. They progress from a straightforward unequal-volume calculation to problems involving experimental assumptions and competing sources of heat.

Question 1

A student mixes 20.0 mL of 0.100 mol L\(^{-1}\) nitric acid with 30.0 mL of 0.100 mol L\(^{-1}\) potassium hydroxide.

The initial temperature of both solutions is 24.0°C, and the maximum temperature after mixing is 24.55°C.

Assume the density of the solution is \(1.00\ \mathrm{g\,mL^{-1}}\), its specific heat capacity is \(4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}\), and heat absorbed by the calorimeter is negligible.

Calculate the molar enthalpy of neutralisation.

Solution 1

The molar enthalpy of neutralisation is \(-57.5\ \mathrm{kJ\,mol^{-1}}\). The key is to use the total 50.0 g of solution for the heat calculation and the 0.00200 mol of water formed for the enthalpy calculation.

Step 1: Calculate the total solution mass.

\[ m=(20.0+30.0)\times1.00 \]

\[ m=50.0\ \mathrm{g} \]

Step 2: Calculate the moles of each reactant.

For nitric acid:

\[ n(\ce{HNO3})=0.100\times0.0200 \]

\[ n(\ce{HNO3})=0.00200\ \mathrm{mol} \]

For potassium hydroxide:

\[ n(\ce{KOH})=0.100\times0.0300 \]

\[ n(\ce{KOH})=0.00300\ \mathrm{mol} \]

Step 3: Identify the limiting reactant.

The reaction is:

\[ \ce{HNO3 + KOH -> KNO3 + H2O} \]

Nitric acid is limiting because it has fewer moles.

Therefore:

\[ n(\ce{H2O})=0.00200\ \mathrm{mol} \]

Step 4: Calculate the temperature change.

\[ \Delta T=24.55-24.0 \]

\[ \Delta T=0.55\,^{\circ}\mathrm{C} \]

Step 5: Calculate the heat absorbed by the solution.

\[ \begin{aligned} q_{\text{solution}} &=50.0\times4.18\times0.55\\ &=114.95\ \mathrm{J} \end{aligned} \]

Step 6: Calculate the enthalpy change.

\[ q_{\text{reaction}}=-114.95\ \mathrm{J} \]

\[ \begin{aligned} \Delta H_{\text{neut}} &=\frac{-114.95}{0.00200}\\ &=-57\,475\ \mathrm{J\,mol^{-1}}\\ &=\boxed{-57.5\ \mathrm{kJ\,mol^{-1}}} \end{aligned} \]

The negative result indicates that heat was released during neutralisation.

Question 2

A student reacts 25.0 mL of 0.120 mol L\(^{-1}\) hydrochloric acid with 35.0 mL of 0.0800 mol L\(^{-1}\) sodium hydroxide.

Both solutions initially have a temperature of 20.00°C. The maximum temperature after mixing is 20.63°C.

Assume the solution has a density of \(1.00\ \mathrm{g\,mL^{-1}}\), a specific heat capacity of \(4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}\), and negligible heat absorption by the calorimeter.

Calculate the molar enthalpy of neutralisation. Explain why using the larger volume to identify the limiting reactant would give an incorrect result.

Solution 2

The molar enthalpy of neutralisation is \(-56.4\ \mathrm{kJ\,mol^{-1}}\). Sodium hydroxide is the limiting reactant, even though it has the larger volume.

Step 1: Calculate the total solution mass.

\[ m=(25.0+35.0)\times1.00 \]

\[ m=60.0\ \mathrm{g} \]

Step 2: Calculate the moles of each reactant.

For hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=0.120\times0.0250\\ &=0.00300\ \mathrm{mol} \end{aligned} \]

For sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=0.0800\times0.0350\\ &=0.00280\ \mathrm{mol} \end{aligned} \]

Step 3: Identify the limiting reactant.

The reaction is:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

The ratio is 1:1.

Sodium hydroxide has fewer moles, so it is limiting.

Therefore:

\[ n(\ce{H2O})=0.00280\ \mathrm{mol} \]

The larger volume does not necessarily contain more moles. Concentration must also be considered.

Step 4: Calculate the temperature change.

\[ \Delta T=20.63-20.00 \]

\[ \Delta T=0.63\,^{\circ}\mathrm{C} \]

Step 5: Calculate the heat absorbed.

\[ \begin{aligned} q_{\text{solution}} &=60.0\times4.18\times0.63\\ &=157.99\ \mathrm{J} \end{aligned} \]

Step 6: Calculate the molar enthalpy.

\[ q_{\text{reaction}}=-157.99\ \mathrm{J} \]

\[ \begin{aligned} \Delta H_{\text{neut}} &=\frac{-157.99}{0.00280}\\ &=-56\,425\ \mathrm{J\,mol^{-1}}\\ &=\boxed{-56.4\ \mathrm{kJ\,mol^{-1}}} \end{aligned} \]

The important distinction is that volume and concentration together determine the number of moles. Comparing volumes alone is not a valid way to identify the limiting reactant.

Question 3

A student mixes 15.0 mL of 0.200 mol L\(^{-1}\) hydrochloric acid with 50.0 mL of 0.0800 mol L\(^{-1}\) sodium hydroxide.

The temperature increases from 19.50°C to 20.10°C.

Assume the density is \(1.00\ \mathrm{g\,mL^{-1}}\), the specific heat capacity is \(4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}\), and calorimeter heat absorption is negligible.

Two students suggest different calculation methods:

• Student A uses 15.0 g as the solution mass because the hydrochloric acid is the limiting reactant and the sodium hydroxide is in excess.
• Student B uses 65.0 g as the solution mass but divides the reaction heat by the sum of the initial moles of acid and base.

Identify the error in each approach and calculate the correct molar enthalpy of neutralisation.

Solution 3

Neither student has the complete correct method. Student A excludes solution that still absorbs heat, while Student B uses an incorrect denominator. The correct molar enthalpy is \(-54.3\ \mathrm{kJ\,mol^{-1}}\).

Step 1: Identify the correct solution mass.

Both solutions are present in the calorimeter:

\[ m=(15.0+50.0)\times1.00 \]

\[ m=65.0\ \mathrm{g} \]

Student A’s mistake is excluding the excess sodium hydroxide from the solution mass.

Even though the excess base does not react, it still warms up and absorbs energy.

Step 2: Calculate the moles of each reactant.

For hydrochloric acid:

\[ n(\ce{HCl})=0.200\times0.0150 \]

\[ n(\ce{HCl})=0.00300\ \mathrm{mol} \]

For sodium hydroxide:

\[ n(\ce{NaOH})=0.0800\times0.0500 \]

\[ n(\ce{NaOH})=0.00400\ \mathrm{mol} \]

Step 3: Determine the amount of water formed.

The balanced equation is:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

Hydrochloric acid is limiting.

Therefore:

\[ n(\ce{H2O})=0.00300\ \mathrm{mol} \]

Student B’s mistake is using the combined starting moles:

\[ 0.00300+0.00400=0.00700\ \mathrm{mol} \]

The enthalpy of neutralisation is calculated per mole of water formed, not per mole of combined starting reactants.

Step 4: Calculate the temperature change.

\[ \Delta T=20.10-19.50 \]

\[ \Delta T=0.60\,^{\circ}\mathrm{C} \]

Step 5: Calculate the heat absorbed.

\[ \begin{aligned} q_{\text{solution}} &=65.0\times4.18\times0.60\\ &=163.02\ \mathrm{J} \end{aligned} \]

Step 6: Calculate the molar enthalpy.

\[ q_{\text{reaction}}=-163.02\ \mathrm{J} \]

\[ \begin{aligned} \Delta H_{\text{neut}} &=\frac{-163.02}{0.00300}\\ &=-54\,340\ \mathrm{J\,mol^{-1}}\\ &=\boxed{-54.3\ \mathrm{kJ\,mol^{-1}} } \end{aligned} \]

The two mistakes affect different parts of the calculation.

• Student A underestimates the heat absorbed because the mass is too small.
• Student B calculates energy per mole of initial reactants rather than per mole of water formed.

These errors come from treating the heat calculation and the stoichiometric calculation as if they measure the same thing.

Question 4

A student mixes 30.0 mL of 0.100 mol L\(^{-1}\) hydrochloric acid with 20.0 mL of 0.150 mol L\(^{-1}\) sodium hydroxide.

Both solutions initially have a temperature of 20.00°C. The final temperature is 20.75°C.

The total mass of solution is 50.0 g, its specific heat capacity is \(4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}\), and the calorimeter has a heat capacity of \(18.0\ \mathrm{J\,^{\circ}C^{-1}}\).

Calculate the molar enthalpy of neutralisation, accounting for heat absorbed by both the solution and the calorimeter.

Explain why ignoring the calorimeter heat capacity changes the calculated result.

Solution 4

The corrected molar enthalpy of neutralisation is \(-56.8\ \mathrm{kJ\,mol^{-1}}\). The calorimeter must be included because it absorbs some of the energy released by the reaction.

Step 1: Calculate the moles of each reactant.

For hydrochloric acid:

\[ n(\ce{HCl})=0.100\times0.0300 \]

\[ n(\ce{HCl})=0.00300\ \mathrm{mol} \]

For sodium hydroxide:

\[ n(\ce{NaOH})=0.150\times0.0200 \]

\[ n(\ce{NaOH})=0.00300\ \mathrm{mol} \]

The reactants are present in exactly the stoichiometric ratio.

Therefore:

\[ n(\ce{H2O})=0.00300\ \mathrm{mol} \]

Step 2: Calculate the temperature change.

\[ \Delta T=20.75-20.00 \]

\[ \Delta T=0.75\,^{\circ}\mathrm{C} \]

Step 3: Calculate the heat absorbed by the solution.

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=50.0\times4.18\times0.75\\ &=156.75\ \mathrm{J} \end{aligned} \]

Step 4: Calculate the heat absorbed by the calorimeter.

Use:

\[ q_{\text{cal}}=C_{\text{cal}}\Delta T \]

Substitute:

\[ \begin{aligned} q_{\text{cal}} &=18.0\times0.75\\ &=13.50\ \mathrm{J} \end{aligned} \]

Step 5: Calculate the total heat released by the reaction.

The reaction supplies energy to both the solution and the calorimeter.

\[ \begin{aligned} q_{\text{reaction}} &=-(q_{\text{solution}}+q_{\text{cal}})\\ &=-(156.75+13.50)\\ &=-170.25\ \mathrm{J} \end{aligned} \]

Step 6: Calculate the molar enthalpy.

\[ \begin{aligned} \Delta H_{\text{neut}} &=\frac{-170.25}{0.00300}\\ &=-56\,750\ \mathrm{J\,mol^{-1}}\\ &=\boxed{-56.8\ \mathrm{kJ\,mol^{-1}}} \end{aligned} \]

If you ignored the calorimeter, you would calculate:

\[ \Delta H_{\text{neut}}= \frac{-156.75}{0.00300} \]

\[ =-52.3\ \mathrm{kJ\,mol^{-1}} \]

This is less negative than the corrected result.

The reason is simple: the solution does not absorb all the energy released. Some energy also warms the calorimeter.

Question 5

This final question introduces a complication that changes how you calculate the temperature change.

A student mixes two solutions in an insulated calorimeter:

• 40.0 mL of 0.100 mol L\(^{-1}\) hydrochloric acid initially at 20.0°C.
• 20.0 mL of 0.200 mol L\(^{-1}\) sodium hydroxide initially at 26.0°C.

The final mixture reaches a temperature of 22.9°C.

Assume:

• Both solutions have a density of \(1.00\ \mathrm{g\,mL^{-1}}\).
• Both solutions have a specific heat capacity of \(4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}\).
• The calorimeter has negligible heat capacity.
• No heat is lost to the surroundings.
• The two solutions would reach thermal equilibrium without reacting if simply mixed.

Calculate the molar enthalpy of neutralisation.

Explain why using \(22.9-20.0\) as the reaction’s temperature change produces an incorrect result.

Solution 5

The correct molar enthalpy of neutralisation is \(-56.4\ \mathrm{kJ\,mol^{-1}}\).

The crucial point is that the two solutions start at different temperatures. Some of the observed temperature increase occurs because the warmer sodium hydroxide transfers energy to the cooler acid, even without a chemical reaction.

We must separate this ordinary thermal mixing from the heat released by neutralisation.

Step 1: Calculate the moles of each reactant.

For hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=0.100\times0.0400\\ &=0.00400\ \mathrm{mol} \end{aligned} \]

For sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=0.200\times0.0200\\ &=0.00400\ \mathrm{mol} \end{aligned} \]

Step 2: Determine the amount of water formed.

The reaction is:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

The reactants have equal moles and react in a 1:1 ratio.

Therefore:

\[ n(\ce{H2O})=0.00400\ \mathrm{mol} \]

Step 3: Calculate the mass of each solution.

For the acid:

\[ m_{\text{acid}}=40.0\ \mathrm{g} \]

For the base:

\[ m_{\text{base}}=20.0\ \mathrm{g} \]

The total mass is:

\[ m_{\text{total}}=60.0\ \mathrm{g} \]

Step 4: Calculate the temperature the mixture would reach without reacting.

This is the important step.

Imagine mixing the two solutions without allowing any chemical reaction to occur.

The warmer sodium hydroxide would transfer thermal energy to the cooler hydrochloric acid until both reached the same temperature.

Since both solutions have the same specific heat capacity, the equilibrium temperature is the mass-weighted average:

\[ T_{\text{mix}}= \frac{m_aT_a+m_bT_b}{m_a+m_b} \]

Substitute the values:

\[ T_{\text{mix}}= \frac{40.0(20.0)+20.0(26.0)} {60.0} \]

\[ T_{\text{mix}}= \frac{800+520}{60.0} \]

\[ T_{\text{mix}}=22.0\,^{\circ}\mathrm{C} \]

This is the temperature expected from thermal mixing alone.

Step 5: Calculate the additional temperature increase caused by the reaction.

The actual final temperature is 22.9°C.

The temperature increase attributable to neutralisation is therefore:

\[ \Delta T_{\text{reaction}}=22.9-22.0 \]

\[ \Delta T_{\text{reaction}}=0.9\,^{\circ}\mathrm{C} \]

Step 6: Calculate the heat absorbed by the solution.

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=60.0\times4.18\times0.9\\ &=225.72\ \mathrm{J} \end{aligned} \]

Step 7: Calculate the molar enthalpy of neutralisation.

The reaction releases the energy absorbed by the solution:

\[ q_{\text{reaction}}=-225.72\ \mathrm{J} \]

Therefore:

\[ \begin{aligned} \Delta H_{\text{neut}} &=\frac{-225.72}{0.00400}\\ &=-56\,430\ \mathrm{J\,mol^{-1}}\\ &=\boxed{-56.4\ \mathrm{kJ\,mol^{-1}}} \end{aligned} \]

Why the tempting method fails

A student might calculate:

\[ \Delta T=22.9-20.0=2.9\,^{\circ}\mathrm{C} \]

This appears reasonable because 20.0°C is the initial temperature of the acid.

However, it ignores the fact that the sodium hydroxide already started at 26.0°C.

Even without neutralisation, the warmer base would have increased the temperature of the acid during mixing.

Using 2.9°C would incorrectly attribute this pre-existing thermal energy to the chemical reaction, producing an enthalpy of approximately:

\[ -181.8\ \mathrm{kJ\,mol^{-1}} \]

That is not the heat released by neutralisation alone.

The correct calculation separates the energy transferred during ordinary mixing from the additional energy released by the chemical reaction.

This is why the starting conditions matter just as much as the final thermometer reading.

08What to remember when solving these questions

When acid and base volumes differ, keep three separate calculations in mind:

Also remember:

• A reactant in excess still contributes to the solution’s mass.
• The limiting reactant determines how much neutralisation occurs.
• The solution gains heat, so its \(q\) is positive when it warms.
• The reaction loses heat, so its \(q\) is negative when it is exothermic.
• If the calorimeter absorbs heat, include its contribution.
• If the solutions start at different temperatures, account for thermal mixing before calculating the reaction’s temperature increase.

Once you can separate these calculations, the next useful step is understanding why the strength of the acid and base can also change the measured enthalpy. A weak acid, for example, involves an additional ionisation energy contribution.

Continue with Neutralisation Models for Strong and Weak Acids and Bases to explore that difference, or revisit Acid-Base Reactions: Proton Transfer and Neutralisation if you need to refresh the underlying reaction chemistry.