Formula
pH and pOH
pH = -log[H⁺]; pOH = -log[OH⁻]. At 25°C: pH + pOH = 14. Strong acid: pH = -log(C); strong base: pOH = -log(C).
-- NCERT Class 11 Chemistry, Ch. 6, p. 191pH Scale
8 MCQs3 revision cards9-step worked example
Source: NCERT EquilibriumPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 25 Sep 2026
pH Scale, explained for NEET
The high-frequency trap in pH problems: treating a weak acid as if it fully dissociates. A 0.1 M acetic acid solution does NOT have pH = 1. Only a fraction ionises, and ignoring that costs you the question outright.
pH and pOH — the definitions. pH = −log₁₀[H⁺]. pOH = −log₁₀[OH⁻]. At 25 °C, pH + pOH = 14 because Kw = [H⁺][OH⁻] = 10⁻¹⁴ (NCERT Class 11 Chemistry Chapter 6, page 194). Pure water: [H⁺] = [OH⁻] = 10⁻⁷ mol/L, so pH = pOH = 7.
Strong vs weak — the fork that decides your method. For a strong acid like HCl at concentration C, assume complete dissociation: [H⁺] = C, pH = −log C. For a weak acid (acetic acid, Ka ≈ 1.8 × 10⁻⁵), only a small fraction α ionises. The working formula is [H⁺] ≈ √(Ka · C), valid when α is small (C ≫ Ka). This is where most NEET errors originate — students skip the √(Ka · C) step and plug C directly into pH = −log C.
The Ka–Kb–Kw bridge. For a conjugate acid-base pair, Ka × Kb = Kw. Equivalently, pKa + pKb = 14. If a question gives Kb for ammonia and asks pH of ammonium chloride solution, you need Ka = Kw/Kb first.
Logarithm arithmetic. NEET pH questions reduce to log manipulation. Know these cold: log 2 ≈ 0.301, log 3 ≈ 0.477, log 5 ≈ 0.699. A one-unit pH change is a tenfold change in [H⁺].
Watch-out: when the problem says "weak acid" but gives a neat concentration like 0.01 M, the temptation is to write pH = 2 directly. That only works for strong acids. For weak acids, always reach for [H⁺] ≈ √(Ka · C).
Can you answer these pH Scale 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
The pH of a 1.0 × 10⁻³ M HCl solution at 25 °C is:
Show answer and why every option is right or wrong
Answer: A. HCl is a strong acid and fully dissociates. [H⁺] = 1.0 × 10⁻³ M. pH = −log(10⁻³) = 3. (NCERT Class 11 Chemistry Chapter 6, page 191.)
Why B is wrong: B (pH = 11) would be the pOH, not the pH. pOH = 14 − 3 = 11. Confusing pH with pOH is a common slip.
Why C is wrong: C (pH = 7) applies to pure water, not to an acid solution of known concentration.
Why D is wrong: D (pH = −3) results from incorrectly taking −log of the exponent's sign. pH = −log(10⁻³) = −(−3) = 3, not −3.
MCQ 2Easy RecallPractice
At 25 °C, if the pH of a solution is 4.5, its pOH is:
Show answer and why every option is right or wrong
Answer: D. At 25 °C, pH + pOH = 14. pOH = 14 − 4.5 = 9.5. (NCERT Class 11 Chemistry Chapter 6, page 191.)
Why A is wrong: A assumes pH = pOH. That equality holds only for pure water (pH = pOH = 7), not for any arbitrary solution.
Why B is wrong: B adds pH and 14 instead of subtracting. The relation is pH + pOH = 14, so pOH = 14 − pH.
Why C is wrong: C applies a sign error. pOH is always positive for aqueous solutions between pH 0 and 14.
MCQ 3Easy RecallPractice
Which of the following correctly represents the ion product of water at 25 °C?
Show answer and why every option is right or wrong
Answer: C. Kw is the product (not sum) of [H⁺] and [OH⁻], equal to 10⁻¹⁴ at 25 °C. (NCERT Class 11 Chemistry Chapter 6, page 191.)
Why A is wrong: A uses addition instead of multiplication. Kw is defined as the product [H⁺] × [OH⁻], not the sum.
Why B is wrong: B has the correct product form but wrong numerical value. 10⁻⁷ is the concentration of each ion in pure water, not Kw itself.
Why D is wrong: D squares each ion concentration, which has no basis in the autoionisation equilibrium H₂O ⇌ H⁺ + OH⁻.
MCQ 4Direct ApplicationPractice
The pH of a 0.1 M weak monoprotic acid HA with Ka = 1.0 × 10⁻⁵ is approximately (log 2 = 0.301):
Show answer and why every option is right or wrong
Answer: A. For a weak acid, [H⁺] ≈ √(Ka × C) = √(10⁻⁵ × 0.1) = √(10⁻⁶) = 10⁻³ M. pH = −log(10⁻³) = 3.0. The key step is recognising that the acid does NOT fully dissociate.
Why B is wrong: B (pH = 5) confuses pKa with pH. pKa = −log(10⁻⁵) = 5, but pH ≠ pKa unless [salt] = [acid] in a buffer.
Why C is wrong: C (pH = 1) results from treating the weak acid as fully dissociated: pH = −log(0.1) = 1. This is the trap addressed by mistake mistake: ph weak acid fully dissoc — weak acids only partially ionise.
Why D is wrong: D (pH = 2.5) may arise from halving the pKa (5/2 = 2.5) without correctly applying the square-root formula.
MCQ 5Direct ApplicationPractice
The Kb of ammonia is 1.8 × 10⁻⁵ at 25 °C. The Ka of its conjugate acid NH₄⁺ is approximately:
Show answer and why every option is right or wrong
Answer: C. Ka × Kb = Kw = 1.0 × 10⁻¹⁴. Ka = Kw / Kb = 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰ ≈ 5.6 × 10⁻¹⁰.
Why A is wrong: A assumes Ka = Kb. That would only be true for a symmetric amphoteric species, not for a conjugate pair in general.
Why B is wrong: B likely arises from subtracting exponents incorrectly: 10⁻¹⁴ / 10⁻⁵ = 10⁻⁹, but the coefficient 1/1.8 ≈ 0.56 shifts the answer to the 10⁻¹⁰ range.
Why D is wrong: D reverses the calculation — multiplying Kb by a factor instead of dividing Kw by Kb.
MCQ 6Direct ApplicationPractice
A 0.01 M NaOH solution at 25 °C has a pH of:
Show answer and why every option is right or wrong
Answer: B. NaOH is a strong base: [OH⁻] = 0.01 = 10⁻² M. pOH = 2. pH = 14 − 2 = 12.
Why A is wrong: A (pH = 2) gives −log(0.01) = 2, which is the pOH, not the pH. This swap is a common error with strong base solutions.
Why C is wrong: C (pH = 7) is the pH of pure water. A 0.01 M NaOH solution is strongly basic.
Why D is wrong: D (pH = 14) would require [OH⁻] = 1 M (pOH = 0), not 0.01 M.
MCQ 7CalculationPractice
A weak monoprotic acid HA has Ka = 4.0 × 10⁻⁶. What is the pH of a 0.1 M solution? (Given: log 2 = 0.301)
Show answer and why every option is right or wrong
Answer: B. [H⁺] = √(Ka × C) = √(4.0 × 10⁻⁶ × 0.1) = √(4.0 × 10⁻⁷) = 2.0 × 10⁻³·⁵. More precisely: √(4 × 10⁻⁷) = 2 × 10⁻³·⁵ = 2 × 10⁻³·⁵. pH = −log(2 × 10⁻³·⁵) = 3.5 − log 2 = 3.5 − 0.301 = 3.199 ≈ 3.2.
Why A is wrong: A (2.7) may come from an arithmetic slip — e.g., using √(4 × 10⁻⁶) = 2 × 10⁻³ and pH = 3 − 0.301 = 2.7. This misses dividing Ka by 10 to account for the concentration factor.
Why C is wrong: C (5.4) equals pKa (−log(4 × 10⁻⁶) = 6 − 0.602 = 5.4). pKa is NOT the pH of the acid solution.
Why D is wrong: D (1.0) treats the weak acid as fully dissociated (pH = −log 0.1 = 1). Weak acids partially ionise.
MCQ 8Concept TrapPractice
If [H⁺] in a solution increases by a factor of 100, the pH:
Show answer and why every option is right or wrong
Answer: D. pH = −log[H⁺]. If [H⁺] increases 100-fold, the new pH = −log(100 × [H⁺]) = −log[H⁺] − log 100 = original pH − 2. pH decreases by 2.
Why A is wrong: A reverses the direction. Increasing [H⁺] makes the solution more acidic, so pH goes down, not up.
Why B is wrong: B assumes pH scales linearly with [H⁺]. pH is logarithmic: doubling [H⁺] changes pH by only log 2 ≈ 0.3, not by a factor of 2.
Why C is wrong: C confuses the multiplicative change in [H⁺] with an arithmetic change in pH. The log scale converts a 100× change to a 2-unit shift.
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pH Scale: quick recall before you leave
How do you solve a pH Scale question? A worked example
Pattern: NEET pattern: ph weak acid base (pH of weak acid/base — observed in NEET 2021, 2022, 2024, 2025)
- 1
Given
A 0.025 M solution of a weak monoprotic acid HA at 25 °C. Ka = 1.0 × 10⁻⁵. Given: log 5 = 0.699, log 2 = 0.301.
- 2
Required
Find the pH of the solution.
- 3
Concept
HA is a weak acid — it does NOT fully dissociate. We use the approximation [H⁺] ≈ √(Ka × C), valid because C (0.025) ≫ Ka (10⁻⁵).
- 4
Formula
[H⁺] = √(Ka × C)
pH = −log₁₀[H⁺] - 5
Substitution
[H⁺] = √(1.0 × 10⁻⁵ × 2.5 × 10⁻²)
[H⁺] = √(2.5 × 10⁻⁷) - 6
Calculation
√(2.5 × 10⁻⁷) = √2.5 × √(10⁻⁷)
√2.5 = √(25/10) = 5/√10 = 5/3.162 ≈ 1.581
√(10⁻⁷) = 10⁻³·⁵
[H⁺] = 1.581 × 10⁻³·⁵ = 1.581 × 3.162 × 10⁻⁴ = 5.0 × 10⁻⁴ M
pH = −log(5.0 × 10⁻⁴) = −log 5 − log(10⁻⁴) = −0.699 + 4 = 3.301
Note: the coefficient 1.0 in Ka is exact (defined), so it does not limit the precision. The given log values (log 5, log 2) are the precision-limiting data here. - 7
Final answer
pH ≈ 3.3
- 8
Common trap
The temptation is to write [H⁺] = 0.025 M → pH = −log(0.025) = 1.6. That treats the weak acid as fully dissociated — the single most tested error in NEET pH questions. The correct [H⁺] via √(Ka·C) is 5.0 × 10⁻⁴, giving pH ≈ 3.3, nearly double the wrong answer.
- 9
Similar NEET-style question
"Calculate the pH of a 0.04 M solution of a weak acid with Ka = 2.5 × 10⁻⁶ at 25 °C." (Same method: [H⁺] = √(2.5 × 10⁻⁶ × 0.04) = √(10⁻⁷) = 10⁻³·⁵; pH ≈ 3.5.)
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What to remember before solving pH Scale questions
Which pH Scale formulas do you need for NEET?
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
Where do students lose marks on pH Scale?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
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).
More in Equilibrium: 7 exam traps and mistakes · 4 formulas · 3 question patterns from its other lessons.
pH Scale questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys.
NEET 2025
1A, B and C only
2A and B only
3A and C only
4B, C and D only
NTA Answer: Option 1(final)
How does NEET ask about pH Scale?
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.194
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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