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 Class 11 Chemistry, Ch. 5, p. 184Enthalpy Solution Neutralization
8 MCQs3 revision cards9-step worked example
Source: NCERT Unit 20PYQ coverage: NEET 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
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 Class 11 Chemistry, Chapter 5, page 184). 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.
Practice 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?
MCQ 2Easy RecallPractice
The enthalpy of neutralization for any strong acid–strong base pair in dilute solution is approximately the same because:
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?
MCQ 4Direct ApplicationPractice
Which indicator is suitable for a titration of acetic acid (weak acid) against NaOH (strong base)?
MCQ 5Concept TrapPractice
During a titration, the solution turned pink on adding one drop of phenolphthalein-indicated NaOH but the colour disappeared after 10 seconds of swirling. The student added two more drops. What error has been committed?
MCQ 6Easy RecallPractice
What is the SI unit of 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?
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.
Quick recall before you leave
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 higher experimental value suggests heat loss was not fully accounted for, which is typical of a simple calorimeter — a common discussion point in NEET practical questions.)
- 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.) ---
Before solving, remember these
Formulas
Molarity-stoichiometry titration
Use when normality is awkward (e.g., diprotic acids). Stoichiometric coefficients from balanced equation.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | molarity | mol/L |
| V | volume | L |
| n | coefficient | - |
Valid when
- Balanced equation known
- Same end-point
Normality equation in titration
Equivalents of acid = equivalents of base at end-point. Or for redox: equivalents of oxidant = equivalents of reductant.
| Symbol | Quantity | SI Unit |
|---|---|---|
| N | normality | eq/L |
| V | volume | mL or L |
Valid when
- Same titration end-point
- Equivalent factors known
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Cations like Pb²⁺ precipitate in BOTH Group I (with HCl) and Group II (with H2S) — assigning to only one group misses the redundancy.
When it triggers
Cation that appears in two analytical groups, e.g. Pb²⁺ (Group I + Group II) or Hg²⁺ vs Hg2²⁺.
How to avoid
Apply confirmatory tests for each candidate group; do not assume mutual exclusivity.
Category: Overthinking
Continuing to add titrant past the first persistent colour change because the colour seemed to fade after a swirl.
When it triggers
Question describes 'colour faded after swirling' or 'persistent colour' — distinguishes transient vs end-point.
How to avoid
End-point = first PERSISTENT colour change (lasts ≥30 s). Transient fades back to original on swirling.
Category: Similar Terms
Phenolphthalein (pH 8.2–10) and methyl orange (pH 3.1–4.4) only mark equivalence when the eq-pt pH falls within their range; using the wrong indicator gives an end-point that disagrees with the actual equivalence point.
When it triggers
Titration prompt mentions a specific weak/strong combination but asks which indicator is suitable.
How to avoid
Match the indicator's pH-change range to the equivalence-point pH: phenolphthalein for eq-pt > 7, methyl orange for eq-pt < 7.
Root cause: concept gap
Correction
Pb²⁺ appears in both Group I (PbCl2 white ppt, soluble in hot water) and Group II (PbS black ppt). Confirm with hot-water solubility test.
Root cause: concept gap
Correction
Strong-acid + strong-base: any indicator (eq pt = 7). Weak-acid + strong-base: phenolphthalein (eq pt > 7). Strong-acid + weak-base: methyl orange (eq pt < 7).
Root cause: rushed under time pressure
Correction
Slow drop-wise addition near the end-point; first persistent colour change is the end-point. Re-do if overshot.
Past Year Questions
3 questions from NEET 2024, 2025. Answers verified against NTA official keys.
NEET 2025
1A-III, B-II, C-I, D-IV
2A-III, B-IV, C-II, D-I
3A-III, B-IV, C-I, D-II
4A-III, B-II, C-IV, D-I
NTA Answer: Option 3(final)
NEET 2024Revised key
1dilute hydrochloric acid
2concentrated sulphuric acid
3dilute nitric acid
4dilute sulphuric acid
NTA Answer: Option 4(revised_final)
NEET 2024Revised key
1B, A, D, C, E
2B, C, A, D, E
3E, C, D, B, A
4E, A, B, C, D
NTA Answer: Option 1(revised_final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Practical chemistry — titrimetric end-point selection, indicator–pH matching, qualitative-analysis cation/anion identification, organic functional-group tests.
Direct ApplicationMedium
Common distractors
wrong zero error direction
Sign convention is easily flipped.
Sources
NCERT refs: Class 11 Chemistry Chapter 5, p.184
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