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. 198Weak Strong Electrolytes
8 MCQs3 revision cards9-step worked example
Source: NCERT EquilibriumOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Weak Strong Electrolytes, explained for NEET
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 6, page 195).
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).
Can you answer these Weak Strong Electrolytes 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?
Show answer and why every option is right or wrong
Answer: D. NaOH is a strong base that dissociates completely in water: NaOH → Na⁺ + OH⁻ (NCERT Class 11 Chemistry Chapter 6, page 195).
Why A is wrong: A is wrong because acetic acid (CH₃COOH) is a weak acid — it partially dissociates in water (Ka ≈ 1.8 × 10⁻⁵).
Why B is wrong: B is wrong because ammonia (NH₃) is a weak base — it partially ionises in water (Kb ≈ 1.8 × 10⁻⁵).
Why C is wrong: C is wrong because carbonic acid (H₂CO₃) is a weak diprotic acid — both dissociation steps are partial.
MCQ 2Easy RecallPractice
The degree of dissociation of a weak electrolyte increases when:
Show answer and why every option is right or wrong
Answer: A. By Ostwald's dilution law, α = √(Ka/C). As concentration C decreases (dilution), α increases (NCERT Class 11 Chemistry Chapter 7).
Why B is wrong: B is wrong because increasing concentration C makes α = √(Ka/C) smaller — more molecules present but a smaller fraction dissociates.
Why C is wrong: C is wrong because adding a common ion suppresses dissociation (Le Chatelier's principle shifts the equilibrium back toward the undissociated form).
Why D is wrong: D is wrong because the dissociation of a weak electrolyte is usually endothermic, so lowering the temperature shifts the equilibrium back toward the undissociated form and decreases α.
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⁻?
Show answer and why every option is right or wrong
Answer: B. Ka × Kb = Kw = 1.0 × 10⁻¹⁴. So Kb = 1.0 × 10⁻¹⁴ / 4.0 × 10⁻⁶ = 2.5 × 10⁻⁹ (NCERT Class 11 Chemistry Chapter 6, page 195).
Why A is wrong: A is wrong — this results from dividing 10⁻¹⁴ by 4.0 × 10⁻⁷ instead of 4.0 × 10⁻⁶ (exponent error in the denominator).
Why C is wrong: C is wrong — this comes from dividing 10⁻¹⁴ by 2.5 × 10⁻⁷, a mis-manipulation of the division.
Why D is wrong: D is wrong — this results from multiplying Ka × Kw instead of dividing Kw by Ka.
MCQ 4Direct ApplicationPractice
The pH of a 0.01 M solution of a strong acid HX (monoprotic, fully dissociated) is:
Show answer and why every option is right or wrong
Answer: D. Strong acid fully dissociates: [H⁺] = 0.01 M = 10⁻² M. pH = −log(10⁻²) = 2.
Why A is wrong: A is wrong — pH = 1 corresponds to [H⁺] = 0.1 M, not 0.01 M. This confuses the concentration by a factor of 10.
Why B is wrong: B is wrong — pH = 7 is for pure water. A 0.01 M strong acid solution is decidedly acidic.
Why C is wrong: C is wrong — pH = 3 corresponds to [H⁺] = 10⁻³ M, which would apply if only a tenth of the acid dissociated (but HX is a strong acid).
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:
Show answer and why every option is right or wrong
Answer: A. [H⁺] ≈ √(Ka × C) = √(1.0 × 10⁻⁶ × 0.1) = √(10⁻⁷) = 10⁻³·⁵ ≈ 3.16 × 10⁻⁴ M. This is the standard weak-acid approximation (NCERT Class 11 Chemistry Chapter 7).
Why B is wrong: B is wrong — 10⁻³ M results from taking √(Ka) alone and ignoring the concentration factor, or from an incorrect exponent manipulation.
Why C is wrong: C is wrong — [H⁺] = 0.1 M treats the weak acid as fully dissociated (strong acid error). This is the classic trap: ignoring that HA is weak.
Why D is wrong: D is wrong — 10⁻⁶ M confuses Ka itself with [H⁺]. Ka is the equilibrium constant, not the hydrogen ion concentration.
MCQ 6Easy RecallPractice
For a conjugate acid-base pair at 25 °C, pKa + pKb equals:
Show answer and why every option is right or wrong
Answer: B. Ka × Kb = Kw = 10⁻¹⁴ at 25 °C. Taking −log of both sides: pKa + pKb = pKw = 14 (NCERT Class 11 Chemistry Chapter 6, page 195).
Why A is wrong: A is wrong — 7 is the pH of pure water, not the sum pKa + pKb. This confuses pH with pKw.
Why C is wrong: C is wrong — pKa + pKb = 1 has no chemical basis. This likely confuses Ka + Kb = some small number with the logarithmic relationship.
Why D is wrong: D is wrong — the sum pKa + pKb = 14 is a fixed relationship at 25 °C for ANY conjugate pair, regardless of acid or base strength. The individual values vary, but the sum is constant.
MCQ 7Concept TrapPractice
A weak acid HA at concentration C₁ has degree of dissociation α₁ = 0.03. A second solution of the same acid is prepared at concentration C₂ = C₁/9 (nine times more dilute). Assuming α ≪ 1 in both cases, what is the degree of dissociation α₂ of the second solution?
Show answer and why every option is right or wrong
Answer: C. By Ostwald's dilution law, α = √(Ka/C), so for a fixed Ka, α is inversely proportional to √C. Diluting nine-fold (C₂ = C₁/9) increases α by a factor of √9 = 3, giving α₂ = 3 × 0.03 = 0.09 — degree of ionization varying with concentration is described in NCERT Class 11 Chemistry, Chapter 6, page 199.
Why A is wrong: A is wrong — 0.27 comes from scaling α linearly by the concentration factor (0.03 × 9), instead of by its square root (√9 = 3). The concentration ratio, not α, scales linearly.
Why B is wrong: B is wrong — 0.01 assumes dilution DECREASES α (0.03 ÷ 3), the opposite of Ostwald's dilution law: dissociation increases, not decreases, on dilution.
Why D is wrong: D is wrong — 0.03 assumes α is unaffected by dilution, ignoring that α depends on C through α = √(Ka/C).
MCQ 8CalculationPractice
The pH of a 0.04 M weak monoprotic acid (Ka = 1.0 × 10⁻⁴) is:
Show answer and why every option is right or wrong
Answer: C. [H⁺] = √(Ka × C) = √(1.0 × 10⁻⁴ × 0.04) = √(4.0 × 10⁻⁶) = 2.0 × 10⁻³ M. pH = −log(2.0 × 10⁻³) = 3 − log 2 = 3 − 0.30 = 2.7.
Why A is wrong: A is wrong — pH = 4.0 comes from using −log(Ka) = −log(10⁻⁴) = 4, confusing Ka with [H⁺]. Ka is NOT the hydrogen ion concentration for a weak acid.
Why B is wrong: B is wrong — pH = 2.0 results from treating the weak acid as a strong acid: [H⁺] = 0.04 M → pH = −log(0.04) ≈ 1.4, and then this doesn't even give 2.0. Alternatively, it comes from √(10⁻⁴) = 10⁻² (pH = 2) — ignoring the concentration factor C = 0.04 in the square root.
Why D is wrong: D is wrong — pH = 1.4 results from −log(0.04) = 1.4, treating 0.04 M as [H⁺] directly (the strong-acid-assumption error). Weak acid: use [H⁺] = √(Ka·C), not [H⁺] = C.
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Weak Strong Electrolytes: quick recall before you leave
How do you solve a Weak Strong Electrolytes question? A 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.)
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What to remember before solving Weak Strong Electrolytes questions
Which Weak Strong Electrolytes formulas do you need for NEET?
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
More in Equilibrium: 8 exam traps and mistakes · 4 formulas · 3 question patterns from its other lessons.
Weak Strong Electrolytes questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Weak Strong Electrolytes?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
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
Sources
NCERT refs: Class 11 Chemistry Chapter 6, p.195
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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