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. 24The 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).
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The pH of a 1.0 × 10⁻³ M HCl solution at 25 °C is:
At 25 °C, if the pH of a solution is 4.5, its pOH is:
Which of the following correctly represents the ion product of water at 25 °C?
The pH of a 0.1 M weak monoprotic acid HA with Ka = 1.0 × 10⁻⁵ is approximately (log 2 = 0.301):
The Kb of ammonia is 1.8 × 10⁻⁵ at 25 °C. The Ka of its conjugate acid NH₄⁺ is approximately:
A 0.01 M NaOH solution at 25 °C has a pH of:
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)
If [H⁺] in a solution increases by a factor of 100, the pH:
Pattern: NEET pattern: ph weak acid base (pH of weak acid/base — observed in NEET 2021, 2022, 2024, 2025)
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.
Required
Find the pH of the solution.
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).
Formula
[H⁺] = √(Ka × C) pH = −log₁₀[H⁺]
Substitution
[H⁺] = √(1.0 × 10⁻⁵ × 2.5 × 10⁻²) [H⁺] = √(2.5 × 10⁻⁷)
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.
Final answer
pH ≈ 3.3
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.
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.) ---
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. 24pH 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 |
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 | - |
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 | - |
Equilibrium constant for sparingly soluble salt. Q < K_sp: dissolves; Q > K_sp: precipitates.
| Symbol | Quantity | SI Unit |
|---|---|---|
| K_sp | solubility product | - |
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 |
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.
Question lists catalyst addition among options for shifting equilibrium.
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.
Convert K_p ↔ K_c for gas-phase reaction.
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.
K_sp problem with non-1:1 salt (e.g. CaF₂, Mg(OH)₂, Ag₂CrO₄).
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
Catalyst speeds up forward AND reverse equally. Equilibrium position is unchanged. Time to reach equilibrium decreases.
Root cause: concept gap
Δn = (moles GAS product) - (moles GAS reactant). Solids and liquids excluded.
Root cause: formula misuse
For M_aX_b: K_sp = [M]^a · [X]^b. CaF₂: K_sp = s · (2s)² = 4s³.
Root cause: concept gap
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
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).
15 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
C [A] = [B] = [C] = 2 × 10–3 M. Then, which of the following is correct?
Amongst the given options which of the following molecules/ ion acts as a Lewis acid?
Which complex compound is most stable?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
wrong delta n sign
Counts moles incorrectly
treats catalyst as shifting
Believes catalyst shifts equilibrium
ignores weak vs strong
Treats weak acid as fully dissociated
forgets stoichiometric coefficient power
Uses [M+]^1 [X-]^1 for M(X)_2 salt
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