Calorimeter: q = (m·c·ΔT) + C_cal·ΔT where C_cal is calorimeter constant. ΔH_neut for strong-acid + strong-base ≈ −57.1 kJ/mol (constant, due to common reaction H⁺ + OH⁻ → H2O). ΔH_soln depends on the solute. Account for radiation loss by extrapolating cooling curve.
-- NCERT Chemistry Lab Manual Class 12, Unit 3, p. 24Enthalpy Solution Neutralization
8 MCQs3 revision cards9-step worked example
Source: NCERT Practical and Analytical ChemistryOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Enthalpy Solution Neutralization, explained for NEET
The trap that costs marks here: you pick the wrong indicator for a titration because you matched it to the acid's pH rather than to the equivalence-point pH. NEET exploits this by giving you a weak-acid/strong-base pair and offering methyl orange as a plausible distractor.
Core concept — enthalpy of neutralization. When a strong acid neutralizes a strong base in dilute aqueous solution, the net ionic reaction is always H⁺(aq) + OH⁻(aq) → H₂O(l), and the enthalpy change is approximately −57.1 kJ mol⁻¹ (NCERT Chemistry Lab Manual Class 12, Unit 3, page 24). This value holds regardless of which strong acid or strong base is used — because the spectator ions contribute no enthalpy change.
For weak acid + strong base (or vice versa), the measured heat of neutralization is less exothermic than −57.1 kJ mol⁻¹. The "missing" energy is consumed in dissociating the weak electrolyte. This difference equals the enthalpy of ionization of the weak species.
Enthalpy of solution is the enthalpy change when one mole of solute dissolves in a large excess of solvent to form a solution of infinite dilution. It can be endothermic (e.g., NH₄Cl in water) or exothermic (e.g., NaOH in water). NEET practical-chemistry questions may ask you to identify which dissolution is exothermic from a calorimetry setup.
Titration bridge: in the practical, you measure heat of neutralization using a calorimeter and confirm stoichiometry via titration. The normality equation (N₁V₁ = N₂V₂) gives the equivalence point; the indicator's colour-change pH must bracket that equivalence-point pH — not the initial pH of either solution.
Watch-out: if the equivalence point lies above pH 7 (weak acid + strong base), phenolphthalein (range 8.2–10) is correct; methyl orange (range 3.1–4.4) will signal the end-point too early, giving a systematic error.
Can you answer these Enthalpy Solution Neutralization MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
The enthalpy of neutralization of HCl with NaOH is −57.1 kJ mol⁻¹. The enthalpy of neutralization of CH₃COOH with NaOH is −55.2 kJ mol⁻¹. What is the enthalpy of ionization of acetic acid?
Show answer and why every option is right or wrong
Answer: C. C is correct. Neutralising a weak acid releases less heat than a strong acid because some of the energy is spent ionising the weak acid first. So ΔH_ionisation = ΔH_neut(weak) − ΔH_neut(strong) = (−55.2) − (−57.1) = +1.9 kJ mol⁻¹, positive as an endothermic step must be. Subtract in that order and the sign comes out right on its own — reversing the two terms gives −1.9 and then needs a sign to be talked away, which is a sign that the subtraction was set up backwards. (NCERT Chemistry Lab Manual Class 12, Unit 3, page 24.)
Why A is wrong: A is wrong because adding the two values (−57.1 + −55.2) confuses enthalpy of ionization with a sum of neutralization enthalpies — these are not additive in that way.
Why B is wrong: B is wrong because −1.9 kJ mol⁻¹ gives the correct magnitude but wrong sign; ionization of a weak acid is endothermic (positive ΔH). (Trap: sign-flip error.)
Why D is wrong: D is wrong because 112.3 is the arithmetic sum of the two magnitudes, which has no physical meaning in this context.
MCQ 2Easy RecallPractice
The enthalpy of neutralization for any strong acid–strong base pair in dilute solution is approximately the same because:
Show answer and why every option is right or wrong
Answer: B. Strong acids and strong bases are fully dissociated in dilute solution; the only enthalpy-producing step is the combination of H⁺ and OH⁻ to form water. Spectator ions do not contribute. (NCERT Chemistry Lab Manual Class 12, Unit 3, page 24.)
Why A is wrong: A is wrong because different strong acids (HCl, HNO₃, H₂SO₄) have very different bond energies; the constancy arises from net ionic reaction, not molecular bond energy.
Why C is wrong: C is wrong because spectator ions remain fully solvated and undergo no net enthalpy change — they neither release nor absorb heat during neutralization.
Why D is wrong: D is wrong because specific heat capacity affects temperature measurement, not the enthalpy of the reaction itself.
MCQ 3Direct ApplicationPractice
In a titration of 0.1 N NaOH against 0.1 N HCl, 25.0 mL of NaOH is used. What volume of HCl is required to reach the equivalence point?
Show answer and why every option is right or wrong
Answer: D. By N₁V₁ = N₂V₂: 0.1 × V_HCl = 0.1 × 25.0, so V_HCl = 25.0 mL. Equal normalities require equal volumes.
Why A is wrong: A is wrong because halving the volume implies a 2:1 stoichiometry, but HCl–NaOH is 1:1 in equivalent terms.
Why B is wrong: B is wrong because 5.0 mL corresponds to dividing by 5, which has no stoichiometric or dilution basis here.
Why C is wrong: C is wrong because doubling the volume would only apply if the HCl normality were half the NaOH normality (0.05 N), which is not stated.
MCQ 4Direct ApplicationPractice
Which indicator is suitable for a titration of acetic acid (weak acid) against NaOH (strong base)?
Show answer and why every option is right or wrong
Answer: A. Weak acid + strong base gives an equivalence-point pH > 7 (due to hydrolysis of the acetate ion). Only phenolphthalein's transition range (8.2–10) brackets that region. (Trap: indicator selection — matching range to equivalence-point pH, not to initial pH.)
Why B is wrong: B is wrong because methyl orange transitions at pH 3.1–4.4, which is far below the equivalence-point pH (~8.7) for this combination; end-point would be detected prematurely. (Trap: trap: indicator selection.)
Why C is wrong: C is wrong because methyl red transitions at pH 4.4–6.2, still below the equivalence-point pH; gives early end-point detection.
Why D is wrong: D is wrong because indicator choice depends on the equivalence-point pH; 'any indicator' is valid only for strong acid + strong base where the pH jump is steep across all ranges.
MCQ 5Concept TrapPractice
During a titration, the solution turned pink on adding one drop of phenolphthalein-indicated NaOH, and the colour persisted through a full minute of swirling. The student then added two more drops "to be sure". What error has been committed?
Show answer and why every option is right or wrong
Answer: B. B is correct. The end-point is the FIRST persistent colour change — a pink that survives ≥30 s of swirling. That drop had already delivered it. Every drop added afterwards is titrant past the end-point, which inflates the burette reading and therefore the calculated concentration. Note the contrast with a pink that appears and then FADES: that one signals the end-point has not been reached, and there continuing drop-wise is the correct procedure, not an error. (Trap: end-point overshoot.)
Why A is wrong: A is wrong because the colour persisted for a full minute, which is the end-point criterion met. Under-titration is the opposite case — a pink that appears and then fades on swirling, where more titrant genuinely is needed.
Why C is wrong: C is wrong because it names over-titration but misstates where the end-point lies. The end-point IS the first persistent colour, the drop that was already added — not some point before it.
Why D is wrong: D is wrong because extra drops do not confirm an end-point, they pass it. Confirmation comes from repeating the whole titration until concordant readings agree within 0.1 mL, not from adding more titrant to a finished one.
MCQ 6Easy RecallPractice
What is the unit of normality?
Show answer and why every option is right or wrong
Answer: C. Normality is defined as the number of gram equivalents of solute per litre of solution, expressed as eq/L.
Why A is wrong: A is wrong because mol/L is the unit of molarity, not normality. Normality uses equivalents, which account for the n-factor.
Why B is wrong: B is wrong because g/L is mass concentration (or 'strength'), not normality.
Why D is wrong: D is wrong because mol/kg is the unit of molality, not normality.
MCQ 7CalculationPractice
The enthalpy of neutralization of HF (a weak acid) with KOH is measured as −68.6 kJ mol⁻¹. Given that ΔH_neutralization for strong acid + strong base = −57.1 kJ mol⁻¹, what does this imply about HF ionization?
Show answer and why every option is right or wrong
Answer: D. ΔH_neutralization(measured) = ΔH_neutralization(H⁺+OH⁻) + ΔH_ionization. Here −68.6 = −57.1 + ΔH_ionization gives ΔH_ionization = −11.5 kJ mol⁻¹. This means HF's ionization is exothermic — an unusual feature of HF among weak acids, attributed to the very high hydration enthalpy of F⁻.
Why A is wrong: A is wrong because +11.5 kJ mol⁻¹ would apply if the measured enthalpy were LESS exothermic than −57.1 (as for CH₃COOH). Here it is MORE exothermic, so the ionization contributes exothermically.
Why B is wrong: B is wrong because there is no thermodynamic prohibition against measured neutralization enthalpy exceeding −57.1 kJ mol⁻¹ — it occurs when the ionization step itself is exothermic.
Why C is wrong: C is wrong because HF is definitively a weak acid (Ka = 6.8 × 10⁻⁴); its anomalous enthalpy does not reclassify it as strong.
MCQ 8Direct ApplicationPractice
25 mL of 0.1 M H₂SO₄ is titrated against 0.1 M NaOH. Using the molarity-stoichiometry equation (MₐVₐ/nₐ = M_bV_b/n_b), calculate the volume of NaOH required.
Show answer and why every option is right or wrong
Answer: A. Balanced equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Here nₐ = 1 (H₂SO₄), n_b = 2 (NaOH). Applying MₐVₐ/nₐ = M_bV_b/n_b: (0.1 × 25)/1 = (0.1 × V_b)/2. Solving: V_b = 50 mL.
Why B is wrong: B is wrong because 25 mL assumes 1:1 stoichiometry (nₐ = n_b = 1), ignoring that H₂SO₄ is diprotic and requires 2 mol NaOH per mol acid.
Why C is wrong: C is wrong because 12.5 mL results from inverting the stoichiometric ratio (dividing by 2 instead of multiplying), confusing which species has the higher coefficient.
Why D is wrong: D is wrong because 100 mL results from squaring the stoichiometric factor or doubling twice, which has no chemical basis.
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Enthalpy Solution Neutralization: quick recall before you leave
How do you solve a Enthalpy Solution Neutralization question? A worked example
- 1
Given
• V_NaOH = 50 mL, N_NaOH = 0.1 N• V_HCl = 20 mL, N_HCl = ?• Temperature rise ΔT = 0.86°C• Total solution mass = 100 g (density 1 g/mL, volume 100 mL)• Specific heat c = 4.18 J g⁻¹ K⁻¹
- 2
Required
(a) Normality of HCl
(b) Enthalpy of neutralization (kJ mol⁻¹) - 3
Concept
At the equivalence point, equivalents of acid = equivalents of base (N₁V₁ = N₂V₂). The heat released equals mcΔT, and dividing by moles of water formed gives ΔH per mole.
- 4
Formula
(a) N_HCl × V_HCl = N_NaOH × V_NaOH
(b) q = mcΔT; ΔH = −q / moles of H₂O formed - 5
Substitution
(a) N_HCl × 20 = 0.1 × 50
(b) q = 100 × 4.18 × 0.86; moles H₂O = N_NaOH × V_NaOH / 1000 = 0.1 × 50 / 1000 - 6
Calculation
(a) N_HCl = 5.0 / 20 = 0.25 N
(b) q = 100 × 4.18 × 0.86 = 359.48 J = 0.35948 kJ
Moles of water = 0.005 mol
ΔH = −0.35948 / 0.005 = −71.9 kJ mol⁻¹
Note: the specific heat (4.18 J g⁻¹ K⁻¹) and density (1 g/mL) are exact problem-defined constants and do not limit significant figures. - 7
Final answer
(a) N_HCl = 0.25 N
(b) ΔH_neutralization ≈ −71.9 kJ mol⁻¹
(The theoretical value for strong acid + strong base is −57.1 kJ mol⁻¹; the measured value is MORE exothermic, and heat loss cannot explain that — heat escaping the cup lowers ΔT and would make the result LESS exothermic, not more. An over-large result points the other way: ΔT read too high, or the 100 g taken as the whole heat-absorbing mass when the calorimeter itself also warmed. Work out which direction an error pushes the answer before naming it; that is the discussion these NEET practical questions are really after.) - 8
Common trap
Overshooting the end-point (adding excess NaOH past the first persistent colour change) inflates V_NaOH, causing the calculated N_HCl to appear lower than actual. NEET may present a scenario where the student identifies why a calculated normality is anomalously low.
- 9
Similar NEET-style question
"In a titration, 0.1 N H₂SO₄ required 30 mL of NaOH to reach the phenolphthalein end-point. If the burette had an unnoticed +0.1 mL zero error, what is the percentage error in the calculated normality of NaOH?" (Tests: zero-error correction + N₁V₁ = N₂V₂.)
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What to remember before solving Enthalpy Solution Neutralization questions
More in Practical and Analytical Chemistry: 6 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
Enthalpy Solution Neutralization questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
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Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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