In 1883, Julius Thomsen published 28 calorimetry experiments comparing strong and weak acid neutralisations — he measured ΔHn(HCl + NaOH) = −57.3 kJ/mol and ΔHn(CH₃COOH + NaOH) = −55.2 kJ/mol. The 2.1 kJ/mol difference looked trivial. Thomsen recognised it was the fingerprint of ionisation energy — proof that acetic acid was only partially ionised before mixing, consuming energy during neutralisation. This lesson explains why that small difference contains the whole story of acid strength.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A food technologist is comparing two cleaning protocols. Protocol A uses 0.5 mol/L HCl neutralised by NaOH. Protocol B uses 0.5 mol/L CH₃COOH (the active component of vinegar-based cleaners) neutralised by NaOH. Both use the same concentrations, same volumes, and the same NaOH concentration.
The technologist notices that the temperature rise during Protocol A is always larger than Protocol B — consistently, reproducibly, every time. A junior technician suggests: "Protocol B must be releasing less heat because the acetic acid reacts incompletely." The technologist shakes her head: "No — the neutralisation goes to completion in both cases. The difference in heat is caused by something that happens before the neutralisation."
Before reading on, write your prediction: what happens before the neutralisation in Protocol B that does not happen in Protocol A, and how does this reduce the net heat released?
Before comparing enthalpy values across different acid-base combinations, the reason for the −57 kJ/mol baseline must be established precisely — because every deviation from this value is explained as an energy cost subtracted from it, and the direction of that deviation is always the same.
The standard enthalpy of neutralisation for strong acid + strong base (ΔHn ≈ −57 kJ/mol) is the result of the net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l). This value is constant regardless of which strong acid and which strong base are used because:
The −57 kJ/mol is therefore the maximum enthalpy released per mole of water formed — it represents the energy released solely from forming the O–H bond in H₂O from pre-existing aqueous H⁺ and OH⁻. Any deviation toward less negative values means that some of the energy produced by the H⁺ + OH⁻ → H₂O step has been consumed elsewhere — specifically, in the ionisation of a weak acid or base.
Baseline enthalpy of neutralisation: strong + strong → ΔHn ≈ −57 kJ/mol because H⁺ and OH⁻ are pre-formed and only H⁺ + OH⁻ → H₂O occurs. Any weak species causes a more positive ΔHn (less exothermic) because ionisation/proton acceptance is endothermic. ΔHn cannot be more negative than −57 kJ/mol due to ionisation effects alone. Ranking: strong+strong > weak acid+strong base ≈ strong acid+weak base > weak+weak.
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Quick Check — Card 1
Which acid-base combination gives the most exothermic neutralisation?
We just saw that weak species cause a more positive ΔHn because of an extra energy cost. That raises a question: What is that energy cost at the molecular level — what exactly happens differently when CH₃COOH reacts with NaOH vs HCl + NaOH? This card answers it → two steps: (1) CH₃COOH → H⁺ + CH₃COO⁻ (endothermic bond breaking); (2) H⁺ + OH⁻ → H₂O (exothermic); net = 57 kJ/mol minus ionisation energy.
The energy difference between strong and weak acid neutralisations is not abstract — it is the direct, measurable consequence of having to break O–H bonds in intact weak acid molecules before those protons can react with OH⁻, and tracing this mechanism at the molecular level is what a Band 6 response requires.
When HCl meets NaOH: H⁺ ions and Cl⁻ ions are already separated in solution (complete ionisation); Na⁺ and OH⁻ are already separated. The only event on mixing is H⁺ + OH⁻ → H₂O — the formation of O–H bonds in liquid water, releasing 57 kJ/mol. No bonds are broken in the acid before this occurs.
When CH₃COOH meets NaOH: Most CH₃COOH molecules are intact (Ka = 1.8 × 10⁻⁵ → ~0.1% ionised at 0.5 mol/L). When OH⁻ is added, it rapidly reacts with the small available H⁺, driving the acetic acid equilibrium right: CH₃COOH → H⁺ + CH₃COO⁻. This ionisation is bond-breaking — the O–H bond in the carboxyl group must be broken to release the proton. Bond breaking is endothermic. This energy comes from the thermal energy of the solution, reducing the temperature rise measured.
Net heat measured = (57 kJ released from H⁺ + OH⁻ → H₂O) − (energy absorbed breaking O–H bonds in CH₃COOH). The stronger the weak acid (larger Ka), the less energy needed per mole of ionisation, and the closer ΔHn is to −57 kJ/mol.
Molecular mechanism for weak acid + NaOH: two-step process — (1) CH₃COOH → H⁺ + CH₃COO⁻ (endothermic O–H bond breaking during the reaction); (2) H⁺ + OH⁻ → H₂O (exothermic). Net heat = 57 kJ/mol minus ionisation energy. ΔH(ionisation) = ΔHn(weak+NaOH) − ΔHn(HCl+NaOH) — always a positive value.
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Quick Check — Card 2
Why does CH₃COOH + NaOH give a smaller ΔT than HCl + NaOH at equal concentration and volume?
We just saw that weak acid + strong base = two steps; ionisation energy is endothermic and reduces net heat. That raises a question: Is the explanation the same for strong acid + weak base, and can both energies be added? This card answers it → mirror image: NH₃ must accept a proton (endothermic); ΔHn(weak acid+weak base) = −57 + ΔH(acid ionisation) + ΔH(base proton acceptance).
The explanation for strong acid + weak base giving less exothermic neutralisation is exactly analogous to the weak acid case — but now it is the base that must undergo an endothermic process before the H⁺ + OH⁻ reaction can occur, and working through this mirror-image mechanism solidifies the general principle.
When HCl meets NH₃: H⁺ is fully available from HCl (complete ionisation). But NH₃ is a weak base — it has not pre-formed NH₄⁺ and OH⁻. For the H⁺ + OH⁻ → H₂O step to occur with maximum energy release, the equivalent of a strong base pathway would require OH⁻ to be freely available before mixing. For the weak base NH₃, producing that OH⁻ requires: NH₃ + H₂O → NH₄⁺ + OH⁻ — an endothermic step. This endothermic energy reduces the net heat measured.
Alternatively, the direct reaction NH₃ + H⁺ → NH₄⁺ can be written — but this does not release the full 57 kJ/mol because the proton ends up in NH₄⁺ rather than in H₂O. Either way, the thermodynamic result is the same: less heat released than −57 kJ/mol.
Approximate ΔHn values: HCl+NaOH ≈ −57; CH₃COOH+NaOH ≈ −55; HCl+NH₃ ≈ −52; CH₃COOH+NH₃ ≈ −50 kJ/mol. Energies are additive: ΔHn(weak acid+weak base) = −57 + ΔH(acid ionisation) + ΔH(base proton acceptance). More negative = more exothermic; −52 releases LESS heat than −57.
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Quick Check — Card 3
ΔHn(HCl+NaOH) = −57 kJ/mol; ΔHn(HCl+NH₃) = −52 kJ/mol. What is ΔH(ionisation) for NH₃?
We just saw the ΔHn rankings and the additive energy model. That raises a question: How do you actually measure and compare ΔHn values experimentally so the comparison is valid? This card answers it → standard design: same concentration (1.00 mol/L), same volume (50 mL each), foam cup calorimeter; ΔHn = −q/n(H₂O) in kJ/mol.
Designing an experiment to compare ΔHn for different acid-base combinations requires careful control of variables — and identifying which variables must be held constant, and why, is itself an HSC investigation design skill that frequently appears in extended response questions.
To compare ΔHn values across combinations, all non-acid/base variables must be held constant: same concentrations, same volumes, same calorimeter (polystyrene/foam cup), same initial temperature (both solutions at the same T before mixing), and the same measurement procedure (swirl gently, record T_max within 30 s).
A standard comparative design: use 50.0 mL of each acid at 1.00 mol/L mixed with 50.0 mL of each base at 1.00 mol/L for four combinations. Each gives n(H₂O) = 0.0500 mol at equivalence. Record T_initial and T_max, calculate q = mcΔT (m = 100 g), calculate ΔHn = −q/n(H₂O).
Standard ΔHn comparison design: 50.0 mL acid (1.00 mol/L) + 50.0 mL base (1.00 mol/L) in foam cup → n(H₂O) = 0.0500 mol. ΔHn = −q/n(H₂O) where q = mcΔT (m = 100.0 g, c = 4.18 J g⁻¹ °C⁻¹); divide J by 1000 for kJ/mol. Controlled variables: concentration, volume, calorimeter type, initial temperature.
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Quick Check — Card 4
50.0 mL of 1.00 mol/L HCl mixed with 50.0 mL of 1.00 mol/L NaOH gives ΔT = 6.8°C. What is ΔHn?
We just saw how to measure ΔHn using a standard foam-cup design. That raises a question: Once you have ΔHn data for an unknown acid, what can you conclude — and what can't you conclude — about its strength? This card answers it → ΔHn ≈ −57 kJ/mol = strong acid; significantly more positive = weak acid. Deviation = ΔH(ionisation). But calorimetry cannot determine Ka — use pH for that.
A technician measures ΔHn for two unknown acid + NaOH reactions: Sample X gives −57.1 kJ/mol; Sample Y gives −51.8 kJ/mol. Without any pH probe or Ka table, the thermodynamic data alone classifies Sample X as a strong acid and Sample Y as a weak acid — a 5.3 kJ/mol gap that is the fingerprint of ionisation energy consumed before the H⁺ + OH⁻ reaction can occur.
If an unknown acid is neutralised with NaOH and the measured ΔHn is close to −57 kJ/mol (within experimental error ±2–3 kJ/mol), this is evidence the acid is strong — consistent with zero ionisation energy cost. If the measured ΔHn is significantly more positive than −57 kJ/mol (e.g. −50 kJ/mol or −45 kJ/mol), this is evidence the acid is weak — the positive deviation corresponds to the endothermic ionisation energy consumed during neutralisation.
Limitation: calorimetry can identify strong vs weak, but cannot determine Ka. A larger positive deviation does not mean the acid is a stronger weak acid — in fact, a weaker acid (smaller Ka) has a more endothermic ionisation and a more positive ΔHn. Ka must be determined from pH measurements (L09 method). ΔHn only classifies and quantifies ionisation enthalpy.
| Evidence type | Strong acid indication | Weak acid indication | Limitation |
|---|---|---|---|
| ΔHn measurement | ΔHn ≈ −57 kJ/mol (within ±3 kJ/mol) | ΔHn significantly more positive than −57 | Cannot determine Ka; only classifies strong/weak |
| pH (same concentration) | pH = −log(c) exactly | pH > −log(c) | Requires calibrated pH probe and known concentration |
| Conductivity (same c) | High conductivity (~fully ionised) | Lower conductivity (fewer ions) | Requires conductivity meter; proportional only |
| Rate with Mg | Fast initial rate, slows as acid consumed | Slower initial rate; rate sustained as equilibrium shifts right | Qualitative; hard to quantify |
Using ΔHn to classify acids: ΔHn ≈ −57 kJ/mol (±3) → strong acid (zero ionisation cost); significantly more positive → weak acid (positive deviation = ΔH(ionisation)). Counter-intuitive: weaker Ka = more positive ΔHn deviation (stronger O–H bond = more endothermic). Calorimetry classifies strong/weak only — cannot determine Ka, which requires pH measurement.
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Quick Check — Card 5
An unknown acid gives ΔHn = −43 kJ/mol when neutralised with NaOH (baseline −57 kJ/mol). What can you conclude?
Worked Example 1 — Calculating and Comparing ΔHn Values
Two experiments use identical 50.0 mL + 50.0 mL, 1.00 mol/L acid + 1.00 mol/L NaOH setups. Experiment A (HCl + NaOH): T_initial = 21.0°C, T_max = 27.8°C. Experiment B (CH₃COOH + NaOH): T_initial = 21.0°C, T_max = 27.2°C. Calculate ΔHn for each and explain the difference.
GIVEN: V_total = 100.0 mL; m = 100.0 g; c = 4.18 J g⁻¹ °C⁻¹; n(acid) = 1.00 × 0.0500 = 0.0500 mol = n(H₂O). FIND: ΔHn for each experiment.
Experiment A (HCl + NaOH): ΔT = 27.8 − 21.0 = 6.8°C. q = mcΔT = 100.0 × 4.18 × 6.8 = 2842 J = 2.842 kJ. ΔHn = −q/n = −2.842/0.0500 = −56.8 kJ/mol.
Experiment B (CH₃COOH + NaOH): ΔT = 27.2 − 21.0 = 6.2°C. q = 100.0 × 4.18 × 6.2 = 2592 J = 2.592 kJ. ΔHn = −2.592/0.0500 = −51.8 kJ/mol.
Comparison and explanation: ΔHn(A) = −56.8 kJ/mol ≈ −57 kJ/mol ✓ (strong acid + strong base, consistent with theoretical). ΔHn(B) = −51.8 kJ/mol — more positive by 5.0 kJ/mol. This 5.0 kJ/mol difference is the enthalpy of ionisation of CH₃COOH. In Experiment A, HCl is fully ionised before mixing — no ionisation energy consumed; all energy is from H⁺ + OH⁻ → H₂O. In Experiment B, most CH₃COOH molecules are intact before mixing; during neutralisation, OH⁻ drives CH₃COOH ionisation (endothermic, 5.0 kJ/mol absorbed), leaving only 51.8 kJ/mol to heat the solution.
Worked Example 2 — Identifying Acid Type from ΔHn Data
Three unknown 1.00 mol/L acids (X, Y, Z) are each neutralised with 50.0 mL of 1.00 mol/L NaOH in a foam cup. Initial temperature = 20.0°C for all. Maximum temperatures: Acid X = 26.6°C; Acid Y = 26.8°C; Acid Z = 25.9°C. (a) Calculate ΔHn for each. (b) Classify each as strong or weak. (c) Calculate the enthalpy of ionisation for any weak acid identified.
GIVEN: V_total = 100.0 mL; m = 100.0 g; n(H₂O) = 0.0500 mol; T_initial = 20.0°C for all. FIND: ΔHn(X, Y, Z), classification, ΔH(ionisation).
(a) ΔHn calculations:
Acid X: ΔT = 6.6°C; q = 100.0 × 4.18 × 6.6 = 2758.8 J; ΔHn = −2758.8/0.0500/1000 = −55.2 kJ/mol
Acid Y: ΔT = 6.8°C; q = 100.0 × 4.18 × 6.8 = 2842.4 J; ΔHn = −2842.4/0.0500/1000 = −56.8 kJ/mol
Acid Z: ΔT = 5.9°C; q = 100.0 × 4.18 × 5.9 = 2466.2 J; ΔHn = −2466.2/0.0500/1000 = −49.3 kJ/mol
(b) Classification: Acid Y: ΔHn = −56.8 kJ/mol ≈ −57 kJ/mol — within experimental error of the strong acid baseline → strong acid. Acid X: ΔHn = −55.2 kJ/mol, notably more positive (+1.8 kJ/mol deviation) → weak acid. Acid Z: ΔHn = −49.3 kJ/mol, significantly more positive (+7.5 kJ/mol deviation) → weak acid, and weaker than X (larger ionisation enthalpy cost).
(c) ΔH(ionisation): Using the experimental baseline −56.8 kJ/mol (from Acid Y): Acid X: ΔH(ionisation) = −55.2 − (−56.8) = +1.6 kJ/mol. Acid Z: ΔH(ionisation) = −49.3 − (−56.8) = +7.5 kJ/mol. (Using theoretical −57.0: Acid X = +1.8 kJ/mol; Acid Z = +7.7 kJ/mol.)
Worked Example 3 — Band 6: Calorimetry, pH, and Complementary Evidence
A teacher has three unlabelled bottles: 1.00 mol/L HCl, 1.00 mol/L CH₃COOH, and 1.00 mol/L HNO₂ (Ka = 4.5 × 10⁻⁴). (a) Describe a calorimetry investigation to identify each bottle using 1.00 mol/L NaOH and a foam cup calorimeter. Include expected ΔHn values and interpretation. (b) Calculate the expected pH of each acid at 1.00 mol/L and explain how pH complements the calorimetry results. (c) Explain why neither method alone provides complete characterisation, and why combining both is more informative.
GIVEN: Three acids at 1.00 mol/L: HCl (strong), CH₃COOH (Ka = 1.8 × 10⁻⁵), HNO₂ (Ka = 4.5 × 10⁻⁴). FIND: Calorimetry design, expected ΔHn; pH values; combined interpretation.
(a) Calorimetry investigation: Mix 50.0 mL of each unknown acid (1.00 mol/L) with 50.0 mL of 1.00 mol/L NaOH in a foam cup. Record T_initial and T_max for each. Calculate q = mcΔT (m = 100.0 g), n(H₂O) = 0.0500 mol, ΔHn = −q/n.
Expected ΔHn: HCl ≈ −57 kJ/mol (no ionisation cost — strong). HNO₂: Ka = 4.5 × 10⁻⁴ (stronger weak acid) → smaller ionisation enthalpy → ΔHn ≈ −54 to −55 kJ/mol. CH₃COOH: Ka = 1.8 × 10⁻⁵ (weaker weak acid) → larger ionisation enthalpy → ΔHn ≈ −55 to −56 kJ/mol.
Ranking (most to least exothermic): HCl > CH₃COOH > HNO₂. Note: CH₃COOH and HNO₂ may be close — calorimetry may not resolve them reliably given foam cup precision (±2–3 kJ/mol).
(b) pH calculations:
HCl (strong): [H⁺] = 1.00 mol/L; pH = 0.00.
HNO₂ (Ka = 4.5 × 10⁻⁴): Check: Ka/c = 4.5 × 10⁻⁴/1.00 = 4.5 × 10⁻⁴ << 0.0025 ✓. x = √(4.5 × 10⁻⁴ × 1.00) = 2.12 × 10⁻² mol/L. Verify: 2.12% < 5% ✓. pH = −log(2.12 × 10⁻²) = 1.67.
CH₃COOH (Ka = 1.8 × 10⁻⁵): x = √(1.8 × 10⁻⁵ × 1.00) = 4.24 × 10⁻³ mol/L. Verify: 0.42% < 5% ✓. pH = −log(4.24 × 10⁻³) = 2.37.
Expected pH: HCl = 0.00; HNO₂ = 1.67; CH₃COOH = 2.37 — three clearly distinct values. The pH probe definitively separates all three (difference of 0.70 pH units between HNO₂ and CH₃COOH = a factor of 5× in [H⁺]).
(c) Complementarity: Calorimetry alone: classifies strong vs weak (ΔHn ≈ −57 or more positive) but may not resolve HNO₂ from CH₃COOH if ΔHn differences are within ±2–3 kJ/mol of foam cup precision. Also cannot determine Ka. pH measurement alone: clearly distinguishes all three by [H⁺] and can be used to calculate Ka, but does not directly reveal the thermodynamic ionisation enthalpy or confirm the mechanism of the strong/weak difference.
Together: calorimetry confirms the strong/weak classification from a thermodynamic direction; pH provides quantitative [H⁺] allowing Ka calculation for the two weak acids and unambiguous identification even when calorimetry ΔHn values are too close. The two methods access fundamentally different aspects of acid character — thermodynamic (ΔH) vs equilibrium ([H⁺], Ka). Using both eliminates ambiguity that either alone would leave unresolved.
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Strong acid + strong base gives ΔH_neut closest to ____ kJ/mol because both fully ionise. Weak acid + strong base gives a ____ value because energy is consumed to ____ the weak acid. The net ionic equation for any strong acid + strong base neutralisation is always ____.
A student performs neutralisation experiments using 50.0 mL of four different 1.00 mol/L acids mixed with 50.0 mL of 1.00 mol/L NaOH. T_initial = 20.0°C for all trials.
| Acid | T_max (°C) | ΔT (°C) | q (J) | ΔHn (kJ/mol) | Classification | ΔH(ionisation) (kJ/mol) |
|---|---|---|---|---|---|---|
| HNO₃ | 26.8 | |||||
| HF (Ka = 6.8 × 10⁻⁴) | 26.2 | |||||
| HCN (Ka = 6.2 × 10⁻¹⁰) | 24.1 | |||||
| H₂SO₄ (1st ionisation) | 26.9 |
Each student description of an experiment to compare ΔHn contains one or more errors. Identify each error and write the correction.
1. Four neutralisation experiments are conducted using equal volumes and concentrations. The measured ΔHn values are: W = −56.8 kJ/mol; X = −49.2 kJ/mol; Y = −57.1 kJ/mol; Z = −52.6 kJ/mol. Which correctly ranks them from most to least exothermic, and identifies which involve weak species?
2. A student claims: "The neutralisation of acetic acid with NaOH gives a smaller ΔT than HCl + NaOH because the acetic acid doesn't fully react." Which response correctly identifies the error?
3. The following ΔHn values are recorded for a weak acid HA neutralised with NaOH at three concentrations of HA: 1.00 mol/L → −55.2 kJ/mol; 0.100 mol/L → −55.4 kJ/mol; 0.010 mol/L → −55.8 kJ/mol. What trend is observed and what does it indicate?
4. 50.0 mL of 2.00 mol/L HCl is mixed with 50.0 mL of 2.00 mol/L NaOH in a foam cup. T_initial = 22.0°C; T_max = 35.6°C. Calculate ΔHn.
5. An unknown acid X is neutralised with NaOH using identical conditions to a HCl + NaOH reference experiment. ΔHn(HCl) = −56.9 kJ/mol; ΔHn(X) = −43.1 kJ/mol. Which statements are correct? (I) Acid X is a weak acid. (II) ΔH(ionisation of X) ≈ +13.8 kJ/mol. (III) Acid X is a weaker acid than acetic acid (Ka = 1.8 × 10⁻⁵, ΔH_ionisation ≈ +1.6 kJ/mol) because its ΔHn deviation is larger.
Question 6. A student neutralises 50.0 mL of 1.00 mol/L ammonia (NH₃, a weak base) with 50.0 mL of 1.00 mol/L HCl. T_initial = 21.5°C; T_max = 27.2°C.
(a) Calculate ΔHn for this reaction. (b) The theoretical value for strong acid + strong base is −57.0 kJ/mol. Calculate the deviation. (c) Explain at the molecular level why this deviation exists for HCl + NH₃.
Question 7. A student measures ΔHn for four combinations (all 50.0 mL + 50.0 mL, 1.00 mol/L) and obtains: HCl + NaOH = −56.5 kJ/mol; HF + NaOH = −54.8 kJ/mol; HCl + NH₃ = −51.9 kJ/mol; HF + NH₃ = −50.2 kJ/mol.
(a) Calculate ΔH(ionisation) for HF using HCl + NaOH as the baseline. (b) Calculate ΔH(ionisation) for NH₃ using HCl + NaOH as the baseline. (c) A student predicts: "HF + NH₃ should have ΔHn = −56.5 + 1.7 + 4.6 = −50.2 kJ/mol." Is this prediction correct? Explain why or why not, accounting for the sign convention carefully.
Question 8. A research team is characterising three unknown solutions (P, Q, R), each at 1.00 mol/L and believed to be acid solutions. They perform two experiments:
Experiment 1 (calorimetry): Each acid is neutralised with 50.0 mL of 1.00 mol/L NaOH (50.0 mL acid + 50.0 mL NaOH). T_initial = 20.0°C. T_max: P = 26.6°C; Q = 26.8°C; R = 24.8°C.
Experiment 2 (pH): pH of each acid at 1.00 mol/L: P = 1.78; Q = 0.00; R = 2.52.
(a) Calculate ΔHn for each acid. Classify each as strong or weak with justification. (3 marks) (b) For each weak acid, use the pH data to calculate Ka. (2 marks) (c) The team proposes that a larger ΔHn deviation from −57 kJ/mol means the acid is weaker (smaller Ka). Evaluate this claim using the data from your calculations. (2 marks)
Q1: C — More negative ΔHn = more exothermic. Ranking: Y (−57.1) > W (−56.8) > Z (−52.6) > X (−49.2). Y and W are both ≈ −57 kJ/mol — within experimental variation of strong acid + strong base; both involve only strong species. Z deviates by +4.4 kJ/mol — one weak species. X deviates by +7.8 kJ/mol — the weakest species or two weak species. Option A incorrectly says only W involves strong species. Option B ranks in the wrong direction. Option D ranks incorrectly.
Q2: B — Both neutralisations go to completion — the same moles of H₂O are formed. The smaller ΔT is thermodynamic: CH₃COOH ionisation is endothermic and consumes part of the exothermic H⁺ + OH⁻ → H₂O energy. Options A and D incorrectly claim incomplete reaction. Option C introduces molar mass irrelevantly — n(H₂O) is the same in both experiments at equal concentration and volume.
Q3: B — For a weak acid, degree of ionisation increases with dilution (Le Chatelier — lower concentration shifts HA ⇌ H⁺ + A⁻ right). At lower concentration, more HA has already ionised before mixing with NaOH — so less ionisation occurs during the neutralisation reaction itself, meaning less endothermic ionisation energy is consumed during the reaction. ΔHn becomes slightly more negative as concentration decreases. Option A incorrectly implies Ka changes with dilution. Option C has the wrong direction. Option D dismisses a systematic trend as noise.
Q4: A — n(H₂O) = 2.00 × 0.0500 = 0.100 mol (both solutions at 2.00 mol/L in 50.0 mL). m = 100.0 g. ΔT = 35.6 − 22.0 = 13.6°C. q = 100.0 × 4.18 × 13.6 = 5684.8 J = 5.685 kJ. ΔHn = −5.685/0.100 = −56.9 kJ/mol. Option B uses wrong n(H₂O). Option C uses wrong mass. Option D is logically incorrect — the concentration does not prevent calculation; ΔHn is per mole regardless of starting concentration.
Q5: C — I: ΔHn(X) = −43.1 kJ/mol is much more positive than −57 kJ/mol → acid X is weak ✓. II: ΔH(ionisation) = −43.1 − (−56.9) = +13.8 kJ/mol ✓. III: A larger positive ΔHn deviation does indicate a weaker acid — a stronger O–H bond costs more ionisation energy, giving a more positive ΔHn deviation. Acid X (deviation = +13.8 kJ/mol) is weaker than CH₃COOH (deviation ≈ +1.6 kJ/mol). All three statements are correct → C.
(a) ΔHn for HCl + NH₃:
ΔT = 27.2 − 21.5 = 5.7°C. q = 100.0 × 4.18 × 5.7 = 2382.6 J = 2.383 kJ. n(H₂O) = 1.00 × 0.0500 = 0.0500 mol. ΔHn = −2.383/0.0500 = −47.7 kJ/mol.
(b) Deviation: ΔH(ionisation of NH₃) = −47.7 − (−57.0) = +9.3 kJ/mol deviation (ΔHn is 9.3 kJ/mol less exothermic than the strong acid + strong base baseline).
(c) Molecular explanation: NH₃ is a weak base — it has not pre-formed OH⁻ before mixing (Kb = 1.8 × 10⁻⁵, so <1% has undergone NH₃ + H₂O → NH₄⁺ + OH⁻ at 1.00 mol/L). For the H⁺ + OH⁻ → H₂O reaction to release its full −57 kJ/mol, free OH⁻ must be available. With NH₃, producing OH⁻ requires the endothermic step NH₃ + H₂O → NH₄⁺ + OH⁻ (Kb process). This endothermic step absorbs energy from the thermal energy of the solution, reducing the temperature rise measured. Either way, 9.3 kJ/mol of energy per mole of neutralisation is consumed by the endothermic weak base process, reducing net heat released to 47.7 kJ/mol.
(a) ΔH(ionisation of HF):
ΔH(ionisation HF) = ΔHn(HF+NaOH) − ΔHn(HCl+NaOH) = −54.8 − (−56.5) = +1.7 kJ/mol
(b) ΔH(ionisation of NH₃):
ΔH(ionisation NH₃) = ΔHn(HCl+NH₃) − ΔHn(HCl+NaOH) = −51.9 − (−56.5) = +4.6 kJ/mol
(c) Evaluate prediction:
ΔHn(HF+NH₃) = ΔHn(strong+strong) + ΔH(ionisation HF) + ΔH(ionisation NH₃) = −56.5 + (+1.7) + (+4.6) = −56.5 + 6.3 = −50.2 kJ/mol.
This matches the measured value of −50.2 kJ/mol exactly. The prediction is correct. The energy budget is additive: each weak species independently contributes its ionisation enthalpy as an endothermic cost subtracted from the strong+strong baseline. The sign convention must be applied carefully: adding positive values (+1.7 and +4.6) to a negative baseline (−56.5) makes the result less negative (−50.2), consistent with weak+weak being the least exothermic combination.
(a) ΔHn calculations and classification:
All: m = 100.0 g; n(H₂O) = 0.0500 mol.
P: ΔT = 6.6°C; q = 2758.8 J; ΔHn = −55.2 kJ/mol. Deviation = +1.8 kJ/mol from −57 → weak acid.
Q: ΔT = 6.8°C; q = 2842.4 J; ΔHn = −56.8 kJ/mol. ≈ −57 kJ/mol → strong acid.
R: ΔT = 4.8°C; q = 2006.4 J; ΔHn = −40.1 kJ/mol. Deviation = +16.9 kJ/mol → weak acid.
(b) Ka from pH:
Ka(P): [H⁺] = 10⁻¹·⁷⁸ = 1.66 × 10⁻² mol/L. Ka = (1.66 × 10⁻²)²/(1.00 − 1.66 × 10⁻²) = (2.76 × 10⁻⁴)/0.9834 = 2.81 × 10⁻⁴.
Ka(R): [H⁺] = 10⁻²·⁵² = 3.02 × 10⁻³ mol/L. Ka = (3.02 × 10⁻³)²/(1.00 − 3.02 × 10⁻³) = (9.12 × 10⁻⁶)/0.9970 = 9.15 × 10⁻⁶.
(c) Evaluate the claim:
P: ΔHn deviation = +1.8 kJ/mol; Ka = 2.81 × 10⁻⁴ (moderately weak acid, larger Ka)
R: ΔHn deviation = +16.9 kJ/mol; Ka = 9.15 × 10⁻⁶ (more weakly acidic, smaller Ka)
The data supports the claim: P has a larger Ka (stronger weak acid) and a smaller ΔHn deviation (+1.8 kJ/mol closer to −57). R has a smaller Ka (weaker acid) and a much larger positive deviation (+16.9 kJ/mol further from −57). This is consistent with the principle: a weaker acid has a stronger, more endothermic O–H bond → larger ionisation enthalpy → larger ΔHn deviation from −57 kJ/mol. The team's claim is therefore supported by this dataset, though it should be noted that the relationship is qualitative — Ka and ΔH(ionisation) are related through the thermodynamic cycle but are not directly proportional, as they measure different aspects of acid character.
Return to the Protocol B mystery. Recall Thomsen's 1883 data: ΔHn(HCl + NaOH) = −57.3 kJ/mol versus ΔHn(CH₃COOH + NaOH) = −55.2 kJ/mol — the 2.1 kJ/mol gap is the ionisation enthalpy of acetic acid. You can now give the full molecular explanation:
A student measures the enthalpy of neutralisation for three unknown acid solutions (A, B, C), each at 1.00 mol/L, reacted with 1.00 mol/L NaOH (50.0 mL + 50.0 mL). Results: A: T_max = 26.3°C; B: T_max = 26.8°C; C: T_max = 24.5°C. T_initial = 20.0°C for all.
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