Ph Scale

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 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 7, page 24). 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).


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

The pH of a 1.0 × 10⁻³ M HCl solution at 25 °C is:

MCQ 2Easy RecallPractice

At 25 °C, if the pH of a solution is 4.5, its pOH is:

MCQ 3Easy RecallPractice

Which of the following correctly represents the ion product of water at 25 °C?

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):

MCQ 5Direct ApplicationPractice

The Kb of ammonia is 1.8 × 10⁻⁵ at 25 °C. The Ka of its conjugate acid NH₄⁺ is approximately:

MCQ 6Direct ApplicationPractice

A 0.01 M NaOH solution at 25 °C has a pH of:

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)

MCQ 8Concept TrapPractice

If [H⁺] in a solution increases by a factor of 100, the pH:

Quick recall before you leave

Worked Example

Pattern: NEET pattern: ph weak acid base (pH of weak acid/base — observed in NEET 2021, 2022, 2024, 2025)

  1. 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. 2

    Required

    Find the pH of the solution.

  3. 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. 4

    Formula

    [H⁺] = √(Ka × C) pH = −log₁₀[H⁺]

  5. 5

    Substitution

    [H⁺] = √(1.0 × 10⁻⁵ × 2.5 × 10⁻²) [H⁺] = √(2.5 × 10⁻⁷)

  6. 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. 7

    Final answer

    pH ≈ 3.3

  8. 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. 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.) ---

Before solving, remember these

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. 24

Formulas

Henderson-Hasselbalch (buffer)

pH of acidic buffer in terms of conjugate base/acid concentrations. For basic buffer: pOH = pKb + log10([salt]/[base]).

SymbolQuantitySI Unit
pKa-log Ka-
[salt]conjugate base concmol/L
[acid]weak acid concmol/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.

SymbolQuantitySI Unit
Kaacid dissociation-
Kbbase dissociation-
Kwwater 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).

SymbolQuantitySI Unit
K_ppressure constant-
K_cconcentration constant-
Δnmole change-

Valid when

  • Gas-phase equilibrium
  • Same temperature

Solubility product

Equilibrium constant for sparingly soluble salt. Q < K_sp: dissolves; Q > K_sp: precipitates.

SymbolQuantitySI Unit
K_spsolubility product-

Valid when

  • Sparingly soluble salt
  • Saturated solution

pH and pOH

Logarithmic acidity scale. Pure water at 25°C: pH = 7 = pOH.

SymbolQuantitySI Unit
[H+]hydrogen ion concmol/L
[OH-]hydroxide concmol/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.

Past Year Questions

15 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2023

Which complex compound is most stable?

1CoNH  NO    3 3 3 3
2CoCl en NO  2 2 3
3CoNH   SO   3 6 2 4 3
4CoNH  H OBrNO   3 4 2  3 2
NTA Answer: Option 2(final)
NEET 2022

Choose the correct statement:

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)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 11 Chemistry Chapter 7, p.24

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