Formula
Ka, Kb, and ionization
Weak acid HA: Ka = [H⁺][A⁻]/[HA]. Weak base B: Kb = [BH⁺][OH⁻]/[B]. Conjugate pair: Ka × Kb = Kw = 10⁻¹⁴ at 25°C. pKa = -log(Ka).
-- NCERT Class 11 Chemistry, Ch. 6, p. 28Weak Strong Electrolytes
8 MCQs3 revision cards9-step worked example
Source: NCERT EquilibriumPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The trap: You see "0.1 M acetic acid" in a NEET stem and compute pH = −log(0.1) = 1. That answer is wrong by two pH units — because acetic acid is a weak electrolyte and does not fully dissociate.
Strong vs. weak electrolytes. A strong electrolyte dissociates completely in water. HCl → H⁺ + Cl⁻: every molecule ionises, so [H⁺] equals the initial acid concentration. NaOH, KNO₃, H₂SO₄ (first dissociation) — all strong. A weak electrolyte dissociates only partially. Acetic acid (CH₃COOH), ammonia (NH₃), carbonic acid (H₂CO₃) — only a small fraction of molecules produce ions at equilibrium (NCERT Class 11 Chemistry Chapter 7, page 28).
Degree of dissociation (α). For a weak acid HA at concentration C: HA ⇌ H⁺ + A⁻. At equilibrium, [H⁺] = Cα and [A⁻] = Cα, with undissociated [HA] = C(1 − α). The acid dissociation constant is Ka = Cα²/(1 − α). When α ≪ 1 (the standard NEET approximation): Ka ≈ Cα², so α ≈ √(Ka/C) and [H⁺] ≈ √(Ka · C).
The conjugate link. For a conjugate acid-base pair: Ka × Kb = Kw = 10⁻¹⁴ at 25 °C, or equivalently pKa + pKb = 14. A stronger acid (larger Ka) has a weaker conjugate base (smaller Kb). This relationship is how NEET questions bridge between weak-acid and weak-base calculations.
Degree of dissociation depends on dilution. Ostwald's dilution law: α = √(Ka/C). As you dilute (C decreases), α increases. More dilute → more dissociation — but [H⁺] = √(Ka · C) still decreases because the concentration drop dominates.
Watch-out for NEET: When a stem says "weak acid" or gives Ka ≪ 1, you must use [H⁺] ≈ √(Ka · C). Treating it as a strong acid (using [H⁺] = C directly) is a common trap that costs 5 marks (4 lost + 1 penalty).
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 1Easy RecallPractice
Which of the following is a strong electrolyte in aqueous solution?
MCQ 2Easy RecallPractice
The degree of dissociation of a weak electrolyte increases when:
MCQ 3Direct ApplicationPractice
If Ka for a weak acid HA is 4.0 × 10⁻⁶ at 25 °C, what is Kb for its conjugate base A⁻?
MCQ 4Direct ApplicationPractice
The pH of a 0.01 M solution of a strong acid HX (monoprotic, fully dissociated) is:
MCQ 5Direct ApplicationPractice
A 0.1 M solution of a weak acid HA has Ka = 1.0 × 10⁻⁶. The approximate [H⁺] in the solution is:
MCQ 6Easy RecallPractice
For a conjugate acid-base pair at 25 °C, pKa + pKb equals:
MCQ 7Direct ApplicationPractice
0.1 M CH₃COOH has Ka = 1.8 × 10⁻⁵. Its degree of dissociation (α) is approximately:
MCQ 8CalculationPractice
The pH of a 0.04 M weak monoprotic acid (Ka = 1.0 × 10⁻⁴) is:
Quick recall before you leave
Worked Example
Pattern: NEET pattern: ph weak acid base — Compute pH of a weak acid solution using Ka.
- 1
Given
A 0.02 M solution of a weak monoprotic acid HA at 25 °C. Ka = 2.0 × 10⁻⁵.
- 2
Required
Find the pH of the solution.
- 3
Concept
HA is a weak acid, so it partially dissociates: HA ⇌ H⁺ + A⁻. We use the weak-acid approximation [H⁺] ≈ √(Ka × C), valid when α ≪ 1.
- 4
Formula
[H⁺] = √(Ka × C) pH = −log₁₀[H⁺]
- 5
Substitution
[H⁺] = √(2.0 × 10⁻⁵ × 0.02) [H⁺] = √(2.0 × 10⁻⁵ × 2.0 × 10⁻²) [H⁺] = √(4.0 × 10⁻⁷)
- 6
Calculation
[H⁺] = 2.0 × 10⁻³·⁵ = √4 × 10⁻³·⁵ = 2.0 × 10⁻³·⁵ More precisely: √(4.0 × 10⁻⁷) = 2.0 × 10⁻³·⁵ = 6.32 × 10⁻⁴ M. Check approximation: α = [H⁺]/C = 6.32 × 10⁻⁴ / 0.02 = 0.032 (3.2% < 5%). Approximation valid.
- 7
Final answer
pH = −log(6.32 × 10⁻⁴) pH = −log(6.32) − log(10⁻⁴) pH = −0.80 + 4 = 3.2 **Note on exact values:** The multiplier 2.0 in Ka = 2.0 × 10⁻⁵ and C = 0.02 M are problem-defined values taken as exact for the purpose of significant-figure analysis. The final answer (pH = 3.2) is reported to 2 significant figures consistent with the given data.
- 8
Common trap
Treating the weak acid as strong: [H⁺] = 0.02 M → pH = −log(0.02) = 1.7. This is off by 1.5 pH units. The giveaway is that Ka is provided — if it were a strong acid, Ka would be irrelevant to the calculation.
- 9
Similar NEET-style question
Calculate the pH of 0.05 M NH₄OH given Kb = 1.8 × 10⁻⁵. (Hint: find pOH first using [OH⁻] = √(Kb × C), then pH = 14 − pOH.) ---
Before solving, remember these
Formulas
Henderson-Hasselbalch (buffer)
pH of acidic buffer in terms of conjugate base/acid concentrations. For basic buffer: pOH = pKb + log10([salt]/[base]).
| Symbol | Quantity | SI Unit |
|---|---|---|
| pKa | -log Ka | - |
| [salt] | conjugate base conc | mol/L |
| [acid] | weak acid conc | mol/L |
Valid when
- Buffer (weak acid + conjugate base)
- Concentrations not too dilute
- Approximate (assumes negligible dissociation)
Ka, Kb, Kw relationship
Stronger acid → weaker conjugate base, and vice versa. pKa + pKb = 14.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Ka | acid dissociation | - |
| Kb | base dissociation | - |
| Kw | water 10^-14 | - |
Valid when
- Conjugate acid-base pair
- 25°C
Equilibrium constant K_p and K_c
Convert between pressure-based and concentration-based equilibrium constants. T in K; R = 0.0821 L atm/mol/K (when K_p in atm).
| Symbol | Quantity | SI Unit |
|---|---|---|
| K_p | pressure constant | - |
| K_c | concentration constant | - |
| Δn | mole change | - |
Valid when
- Gas-phase equilibrium
- Same temperature
Solubility product
Equilibrium constant for sparingly soluble salt. Q < K_sp: dissolves; Q > K_sp: precipitates.
| Symbol | Quantity | SI Unit |
|---|---|---|
| K_sp | solubility product | - |
Valid when
- Sparingly soluble salt
- Saturated solution
pH and pOH
Logarithmic acidity scale. Pure water at 25°C: pH = 7 = pOH.
| Symbol | Quantity | SI Unit |
|---|---|---|
| [H+] | hydrogen ion conc | mol/L |
| [OH-] | hydroxide conc | mol/L |
Valid when
- Aqueous solution
- Use Kw = 10^-14 at 25°C
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: Overthinking
Student claims catalyst shifts equilibrium toward products. Catalyst speeds up forward AND reverse equally; equilibrium position unchanged.
When it triggers
Question lists catalyst addition among options for shifting equilibrium.
How to avoid
Catalyst lowers activation energy of BOTH forward and reverse reactions equally. Time to reach equilibrium decreases; equilibrium position is unchanged.
Category: Sign Convention
Δn = (mol gas product) - (mol gas reactant). Sign matters; K_p = K_c (RT)^Δn.
When it triggers
Convert K_p ↔ K_c for gas-phase reaction.
How to avoid
Count moles of gas only (ignore solids/liquids). Δn = product - reactant. If Δn = 0, K_p = K_c. If Δn = +1, K_p = K_c × RT.
Category: Similar Terms
For salt M_aX_b: K_sp = [M⁺]^a [X⁻]^b. Student uses [M⁺][X⁻] regardless of stoichiometry.
When it triggers
K_sp problem with non-1:1 salt (e.g. CaF₂, Mg(OH)₂, Ag₂CrO₄).
How to avoid
Write dissolution: M_aX_b → aM + bX. Then K_sp = [M]^a · [X]^b. For CaF₂ ↔ Ca + 2F: K_sp = s · (2s)² = 4s³.
Root cause: concept gap
Correction
Catalyst speeds up forward AND reverse equally. Equilibrium position is unchanged. Time to reach equilibrium decreases.
Root cause: concept gap
Correction
Δn = (moles GAS product) - (moles GAS reactant). Solids and liquids excluded.
Root cause: formula misuse
Correction
For M_aX_b: K_sp = [M]^a · [X]^b. CaF₂: K_sp = s · (2s)² = 4s³.
Root cause: concept gap
Correction
Increase P → shift toward FEWER moles of gas. Decrease P → shift toward MORE moles. Reactions with same number of moles of gas: P doesn't shift equilibrium.
Root cause: concept gap
Correction
Weak acid: only fraction ionises. [H+] ≈ √(Ka · C). For 0.1 M acetic acid (Ka ~10⁻⁵): [H+] ~10⁻³ (pH ~3), not 1 (pH=1).
Past Year Questions
15 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
1B, D, and E only
2A, B, and C only
3A, B, and D only
4B, C, and D only
NTA Answer: Option 3(final)
NEET 2025
1Sucrose
2D-Glucose
3D-Fructose
4Maltose
NTA Answer: Option 3(final)
NEET 2025
10.021
283.1
32.077 × 105
40.033
NTA Answer: Option 4(final)
NEET 2024Revised key
C [A] = [B] = [C] = 2 × 10–3 M. Then, which of the following is correct?
1Reaction is at equilibrium.
2Reaction has a tendency to go in forward direction.
3Reaction has a tendency to go in backward direction.
4Reaction has gone to completion in forward direction.
NTA Answer: Option 3(revised_final)
NEET 2023
1l = 2n + 1
2n = 2l2 + 1 m m n –1
3n = l + 2
4l m m 2
NTA Answer: Option 4(final)
NEET 2023
Amongst the given options which of the following molecules/ ion acts as a Lewis acid?
1H O
2BF 2 3
3OH–
4NH 3
NTA Answer: Option 2(final)
NEET 2023
Which complex compound is most stable?
1CoNH NO 3 3 3 3
2CoCl en NO 2 2 3
3CoNH SO 3 6 2 4 3
4CoNH H OBrNO 3 4 2 3 2
NTA Answer: Option 2(final)
NEET 2022
1Both diamond and graphite are used as dry lubricants.
2Diamond and graphite have two dimensional network.
3Diamond is covalent and graphite is ionic.
4Diamond is sp3 hybridised and graphite is sp2 hybridized.
NTA Answer: Option 4(final)
NEET 2022
1p =p, where = mole fraction of ith gas in gaseous mixture i i i i p = pressure of ith gas in pure state i
2p = p + p + p 1 2 3 RT RT RT
3p=n +n +n 1 V 2 V 3 V
4p = p, where p = partial pressure of ith gas i i i = mole fraction of ith gas in gaseous mixture i
NTA Answer: Option 1(final)
NEET 2022
14.9
22.5
3498.6
449.8
NTA Answer: Option 1(final)
NEET 2022
11.2 × 1021
24.38 × 10–32
31.9 × 10–63
42.4 × 1031
NTA Answer: Option 2(final)
NEET 2022
11.05 V
21.0385 V
31.385 V
40.9615 V Answer (NA)
NTA Answer: Option 5(final)
NEET 2021
1and 8
2and 4 sp3 sp2
3and 6
4and 6
NTA Answer: Option 4(final)
NEET 2021
1Beryllium chloride
2Calcium chloride
3Strontium chloride
4Magnesium chloride
NTA Answer: Option 1(final)
NEET 2021
126.02
22.518
32.602
425.18
NTA Answer: Option 4(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Convert between K_p and K_c using K_p = K_c (RT)^Δn. Identify Δn carefully.
Multi StepMedium
Common distractors
wrong delta n sign
Counts moles incorrectly
Predict equilibrium shift on changing concentration, P, T, or adding catalyst.
InterpretationMedium
Common distractors
treats catalyst as shifting
Believes catalyst shifts equilibrium
Compute pH of weak acid/base solution using Ka or Kb. Approximation: [H+] ≈ √(Ka·C).
Multi StepMedium
Common distractors
ignores weak vs strong
Treats weak acid as fully dissociated
Compute solubility from K_sp or vice versa. Apply common-ion effect when relevant.
Multi StepMedium
Common distractors
forgets stoichiometric coefficient power
Uses [M+]^1 [X-]^1 for M(X)_2 salt
Sources
NCERT refs: Class 11 Chemistry Chapter 7, p.28
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