Calorimetry Virtual Lab: Enthalpy of Neutralization

Updated 2026-10-06

This calorimetry virtual lab matches the AQA GCSE required practical on temperature changes, and it supports the calorimetry work in AP Chemistry. Students mix hydrochloric acid and sodium hydroxide in a polystyrene cup and record the highest temperature. They repeat the run with different volumes of alkali, plot a graph and find the volume that exactly neutralizes the acid. Then they calculate the enthalpy change of neutralization from Q = m·c·ΔT. One class period gives a full data set, a two-line graph and a ΔH value with a clear source of error.

Calorimetry – Enthalpy of Neutralization of HCl + NaOH

Curriculum links

AQA GCSE Chemistry 4.5.1.1 (exothermic and endothermic reactions) and its required practical on temperature changes. In Combined Science: Trilogy the same content is 5.5.1.1.
AP Chemistry Unit 6, topic 6.4 Heat Capacity and Calorimetry (ENE-2.D).
NGSS HS-PS1-4: a chemical reaction releases or absorbs energy.
Other courses: this is the standard enthalpy of neutralization practical in IB Chemistry, A-level and upper-secondary chemistry in Germany and elsewhere. Volumes are in mL and concentrations in mol/L, so the method transfers unchanged.

Simulic is not affiliated with or endorsed by the College Board or AQA.

Before the lab (5 min)

Ask students to commit to a prediction, on paper or as question 1 of the class link:

"25 mL of 1.0 M HCl is mixed with 1.0 M NaOH. You repeat the mixing with 10, 20, 30, 40 and 50 mL of NaOH. How does the highest temperature change as the NaOH volume goes up?"

Most students say "it keeps rising". Don't correct them yet.

Method in the simulation

Every run is a fresh mixture: both solutions start at 25.0 °C and go into a clean cup with a lid. In a school practical, students often add the alkali in portions to one cup instead. The data have the same shape.

Set V(HCl) to 25 mL, C(HCl) to 1.0 M, C(NaOH) to 1.0 M and Heat loss to 5%.
Set V(NaOH) to 10 mL. Press Pour and Stir.
Watch the thermometer. It rises for a few seconds, then falls slowly. Record the highest reading. The run is added to the results table once the peak has passed.
Press New Run, change only V(NaOH), and repeat for 15, 20, 25, 30, 35, 40 and 50 mL.
Plot the highest temperature against the volume of NaOH. Draw one best-fit line through the rising points and one through the falling points.
For the 25 mL run, calculate Q = m·c·ΔT and ΔH = −Q ÷ n(H₂O). Take m = 50 g and c = 4.18 J/g·°C.

V(NaOH) (mL)	Highest temperature (°C)	ΔT (°C)
10
15
20
25
30
35
40
50

Expected results

These readings come from the simulation with the settings above (25 mL of 1.0 M HCl, 1.0 M NaOH, 5% heat loss):

V(NaOH) (mL)	10	15	20	25	30	35	40	50
Highest T (°C)	28.6	29.8	30.7	31.4	30.8	30.3	29.9	29.3
ΔT (°C)	3.6	4.8	5.7	6.4	5.8	5.3	4.9	4.3

The 10 mL reading can come out 0.1 °C higher on some devices. The peak is at 25 mL, where the moles of acid and alkali are equal (0.025 mol each). Before that, extra NaOH neutralizes more acid and releases more heat. After it, the acid is used up: extra NaOH releases nothing and only adds mass to warm.

For the 25 mL run: Q = 50 × 4.18 × 6.4 = 1,338 J, so ΔH = −1,338 ÷ 0.025 = −53.5 kJ/mol. The theoretical value in the simulation is −57.3 kJ/mol. The difference comes from the 5% heat loss and from the cup cooling while the temperature rises.

Questions for students

(Prediction, asked again after the lab) How does the highest temperature change as the NaOH volume increases from 10 to 50 mL?
Which variables must stay the same in every run?
With 25 mL of each solution and 5% heat loss, what is the temperature rise?
Use that rise to calculate the enthalpy change of neutralization in kJ/mol.
Why is your value smaller in size than −57.3 kJ/mol, and how could the method be improved?

Answers for teachers: (1) It rises to a peak at 25 mL, then falls. (2) The volume and concentration of HCl, the concentration of NaOH and the heat-loss setting. (3) 6.4 °C (accept 6.3–6.5). (4) −53.5 kJ/mol (accept −52.5 to −54.5). (5) Heat escapes to the cup, the lid and the air, so ΔT is too small. Better insulation helps, or recording the temperature over time and extrapolating the cooling line back to the moment of mixing.

Common misconceptions

"More reactant always means a bigger temperature rise." Only the limiting reactant releases heat. Excess solution just adds mass.
"Double the volumes, double the ΔT." Twice the heat warms twice the mass, so ΔT stays at 6.4 °C. The question-writing guide turns this into a prediction question.
"ΔH is positive because the temperature went up." The surroundings gained the energy, so the reaction lost it: ΔH is negative.

Extension

Heat loss as a variable: repeat the 25 mL + 25 mL run with heat loss at 0%, 15% and 30%. The simulation gives ΔT = 6.9, 5.5 and 4.3 °C, and ΔH = −57.7, −46.0 and −35.9 kJ/mol. Students see how a systematic error pushes the result in one direction. At 0% the temperature never falls, so the run is only added to the table when the stopwatch reaches 3:00 (about 25 seconds). The −57.7 differs from −57.3 only because the thermometer rounds to 0.1 °C.
AP-style calorimetry: change both concentrations to 2.0 M and predict ΔT before running it. Ask why ΔH per mole stays about the same.

FAQ

Can students do the whole practical without a lab?

Yes. They can collect every reading, draw the graph and calculate ΔH in the simulation. Use it before a wet lab to teach the method, or instead of one when time or equipment is short.

Why does the simulation start every run at 25.0 °C?

It removes one source of scatter so students can focus on the pattern. In a real practical, measure the starting temperature of each solution and use the mean.

How do I give students a different practical setup?

Change the starting values on the class link, for example 30 mL of HCl. The peak then moves to 30 mL of NaOH. Write the new settings into question 3 and recalculate the answers before class.