Neutralization Reactions Lab Report

This report was to study Neutralization Reactions

Sample Calculations

1.
q = ms∆t – 1M HCl

q = (acid mass(g) + base mass(g))(t_f-t_i)(s)

q = -(50.2+26.1)(8.6)(4.184)

q = -2.746 kJ

2.
x mol of acid & base – 1M HCl

mol HCl = (1M HCl x (50mL))/(1000mL) = 0.05 mol HCl

mol NaOH = (2M NaOH x (25mL)/(1000mL) = 0.05 mol NaOH

3.
mol H2O – 1M HCl

mol H2O – .05mol NaOH + .05mol HCl → .05mol NaCl + .05mol H2O

mol H2O =.05 mol

4.
∆H – 1M HCl

∆H = q/mol H2O = -2.746/.05 = -54.920 kJ/mol H2O

5.
Standard Enthalpy of Formation – 1M HCl

Hf – [H2O -285.8 NaCl = -385.92 NaOH= -469.6 HCl = -92.3]

∆H = [(H2O+NaCl)-(NaOH + HCl)] = -110.27kJ/mol H2O

Analysis:

1. Compare results with the group’s results.

The enthalpy results obtained in our experiment was the highest of the class, in the 1M HCl reaction and 2M HCl reaction. There is a 36.3% deviation between our value for enthalpy and the lowest experimental value for the 1M HCl neutralization. There is a 30.1% deviation between our value for enthalpy and the other experimental value for the 2M HCl neutralization.

2. How does concentration of acid correlate with heat released?

The acid concentration has a direct correlation with the heat released, as seen in our two experiments, the 2M HCl neutralization released approximately twice the heat as the 1M HCl neutralization. After determining the experiments’ experimental H/mol value, they had discrepancy of only 1.55%, so this seems very valid.

3. Compare the molar heat of neutralization for the different acids tested and propose explanations for deviations in expectations.

The experimental results show nearly 50% error when compared to the standard enthalpy calculation. This is a very high number. Since the experiment was not capable of being conducted in a 100% closed system, there were obvious errors in calculating the change in temperature. Heat released into the air, into the Styrofoam cup, and etc. That was probably the main source of error.

Editor’s note (October 2026): The standard-enthalpy comparison in step 5 above, and the “nearly 50% error” drawn from it, are not correct. The −92.3 kJ/mol used for HCl is the standard enthalpy of formation of HCl gas, not of HCl dissolved in water, and the −110.27 kJ/mol result is not the accepted heat of neutralization. Using standard enthalpies of formation from OpenStax Chemistry 2e, Appendix G (H⁺(aq) = 0, OH⁻(aq) = −230.0 kJ/mol, H₂O(l) = −285.83 kJ/mol):

ΔH°_neut = −285.83 − (0 + (−230.0)) ≈ −55.8 kJ per mole of water

Comparing this lab’s result with that value: |−54.92 − (−55.83)| / 55.83 × 100% ≈ 1.6% error, not 50%. Published values for strong acid–strong base pairs vary a little with the data source and conditions (LibreTexts gives −56.9 kJ/mol in one text and between −57 and −58 kJ/mol in another), so against that whole range the error is roughly 2–5%. That is a good result for a Styrofoam-cup calorimeter. The student’s original measurements and analysis are left unchanged above.

Why strong acid–strong base neutralization reactions give nearly the same ΔH

Strong acids and strong bases are fully ionized in water, so in HCl + NaOH the Na⁺ and Cl⁻ ions are spectator ions that take no part in the reaction. The net ionic equation for any strong acid–strong base neutralization is the same:

H⁺(aq) + OH⁻(aq) → H₂O(l)

Because the reaction that actually happens is identical each time, the enthalpy change per mole of water formed is very similar whichever strong acid and strong base are used. Weak acids or bases (such as acetic acid or ammonia) give somewhat less exothermic values, because part of the energy goes into ionizing them. Sources: LibreTexts, Enthalpy Change of Neutralization; LibreTexts Chem1, 14.4 Thermochemistry and Calorimetry.