NCERT Class 11 Chemistry • Equilibrium

Ionic Equilibrium

📄 Source PDF: kech106.pdf

A focused NEET Faculty Exam Pack — exact concepts, a complete formula & approximation bank, trap-avoidance shortcuts, and 10 exam-level MCQs with the fastest solving methods. No long derivations.

1 The NEET Cheat Sheet — Core Concepts & Exceptions

The simple idea

Acids and bases are two opposite kinds of chemicals — acids are sour (think lemon), bases are slippery (think soap). Scientists use three rules to decide what counts as an acid or base, each one a little broader than the one before.

Acid–Base concepts

Arrhenius

💡 The oldest, simplest rule, and it only works in water: an acid is something that lets go of a tiny positive bit (H⁺), and a base lets go of OH⁻. That's it.

Acid → furnishes H⁺ (H₃O⁺) in water · Base → furnishes OH⁻ in water.

📌 Acids: HCl, HNO₃, H₂SO₄ · Bases: NaOH, KOH, LiOH.

⚠ Limited to aqueous solutions only; cannot explain basicity of NH₃ (no OH⁻) or acidity of CO₂.

Brønsted–Lowry

💡 Picture H⁺ as a ball. The acid is the player who throws the ball; the base is the one who catches it. They always come in throw-and-catch pairs (called conjugate pairs).

Acid = proton (H⁺) donor · Base = proton acceptor. Always in conjugate pairs differing by one H⁺.

📌 HCl + H₂O → H₃O⁺ + Cl⁻ (Cl⁻ = conjugate base) · NH₃ + H₂O → NH₄⁺ + OH⁻. Water is amphoteric.

⚠ Strong acid ⇒ very weak conjugate base (Cl⁻, NO₃⁻, HSO₄⁻); strong base ⇒ weak conjugate acid. Cannot explain acids/bases without transferable H⁺.

Lewis

💡 The widest rule of all — it's about pairs of electrons (tiny shared dots), not H⁺. The acid grabs a spare pair of electrons; the base offers one. So even chemicals with no H at all can be acids.

Acid = electron-pair acceptor · Base = electron-pair donor.

📌 Lewis acids: BF₃, AlCl₃, BCl₃, H⁺, Mg²⁺, CO₂ · Lewis bases: NH₃, OH⁻, F⁻, H₂O.

⚠ Most general, but does not account for relative acid/base strength and labels even non-protonic species as acids.

How a conjugate pair works (Brønsted)

HA+H₂OH₃O⁺+A⁻
red = acid & its conjugate base (HA / A⁻)blue = base & its conjugate acid (H₂O / H₃O⁺)

Lose a proton → become the conjugate base · gain a proton → become the conjugate acid.

Exceptions, limitations & temperature effects

The pH scale (with NCERT Table 6.5 values)

01234567891011121314
Gastric juice 1.2Lemon juice 2.2Soft drink/vinegar 3.0Tomato juice 4.2Black coffee 5.0Milk 6.8Human blood 7.4Egg white/sea water 7.8Milk of magnesia 10Lime water 10.50.1 M NaOH 13

Lower than 7 = acidic · 7 = neutral (at 25 °C) · higher than 7 = basic. Each whole step is a ×10 change in [H⁺]. Data: NCERT Table 6.5.

2 The Formula & Approximation Bank

Every formula NEET tests, grouped by topic — with the time-saving approximation beside each. A one-line plain summary sits above each group.

pH & Water Equilibrium

💡 pH is just a short way of saying how acidic something is. Instead of writing a tiny number like 0.0000001, we write pH = 7. Smaller pH means more acidic, bigger means more basic.

FormulaApproximation / Notes
Kw = [H⁺][OH⁻] = 10⁻¹⁴ (at 298 K)pKw = pH + pOH = 14 at 298 K only.
pH = −log[H⁺] ; pOH = −log[OH⁻][H⁺] = 10⁻ᵖᴴ. Strong acid: [H⁺] = normality.
Strong base: [OH⁻] = C, pH = 14 − pOHFor C ≥ 10⁻⁶ M ignore water; below that, add water's 10⁻⁷.

Weak Acid / Base & Ostwald's Law

💡 A weak acid only lets go of a few of its H⁺ bits. The fraction that actually splits up is called the degree of dissociation (α) — like counting how many kids in a class raised their hands.

FormulaApproximation / Notes
Ka = Cα²/(1−α)If α < 0.05 ⇒ 1−α ≈ 1 ⇒ Ka ≈ Cα².
α = √(Ka / C) (Ostwald)Valid for weak electrolytes, α < 5%.
[H⁺] = √(Ka·C) ⇒ pH = ½(pKa − log C)Monoprotic weak acid; assumes α small.
[OH⁻] = √(Kb·C) ⇒ pOH = ½(pKb − log C)Weak base, α small.
pKa + pKb = pKw = 14 (conjugate pair, 298 K)Use to convert between Ka and Kb.

Salt Hydrolysis

💡 When a salt dissolves, its pieces sometimes react with water and quietly make the solution a little acidic or a little basic. That reaction with water is called hydrolysis.

FormulaApproximation / Notes
Weak acid + strong base (e.g. CH₃COONa): pH = 7 + ½(pKa + log C)Solution basic (pH > 7). Kh = Kw/Ka.
Weak base + strong acid (e.g. NH₄Cl): pH = 7 − ½(pKb + log C)Solution acidic (pH < 7). Kh = Kw/Kb.
Weak acid + weak base (e.g. CH₃COONH₄): pH = 7 + ½(pKa − pKb)Independent of concentration! Kh = Kw/(Ka·Kb).
Degree of hydrolysis h = √(Kh / C)Valid when h small.

Buffer Solutions (Henderson–Hasselbalch)

💡 A buffer is a shock-absorber for pH. Add a little acid or base and the pH barely moves. It's made of a weak acid together with its own salt working as a team.

FormulaApproximation / Notes
Acidic buffer: pH = pKa + log([salt]/[acid])Best buffering when [salt]≈[acid] ⇒ pH = pKa.
Basic buffer: pOH = pKb + log([salt]/[base])pH = 14 − pOH (298 K).
At half-neutralisation point: pH = pKaHalf the weak acid neutralised by strong base.

Solubility Product (Ksp)

💡 Some salts barely dissolve in water. Ksp is a number that tells you exactly how much can dissolve before the rest just stays as a solid at the bottom.

FormulaApproximation / Notes
AB type (AgCl): Ksp = s² ⇒ s = √Ksps = molar solubility (mol L⁻¹).
AB₂ / A₂B (CaF₂, Ag₂CrO₄): Ksp = 4s³ ⇒ s = (Ksp/4)^⅓Watch the stoichiometric coefficients.
AₓBᵧ: Ksp = xˣ·yʸ·s^(x+y)General formula.
Common ion (conc C): s = Ksp / CⁿSolubility ↓ sharply (common ion effect).
Precipitation occurs when Qsp > KspQsp = Ksp ⇒ saturated; < ⇒ unsaturated.

📈 Interactive Titration Curve

The simple idea

Slowly adding a base to an acid changes its pH. If you plot pH against the amount of base added, you get an S-shaped curve with a sudden jump at the “equivalence point” (where acid and base exactly cancel). Drag the sliders and watch the curve and the pH change.

Current pH: 1.00
Equivalence point: 25.0 mL
Region: acid

Acid volume fixed at 25 mL. Strong-acid curves jump through pH 7; weak-acid curves have a buffer plateau (flat region) around pH = pKa and an equivalence point above 7.

3 The “Trap Avoidance” & Shortcut Guide

The simple idea

These are the exact spots where students lose easy marks. Each card shows the trap, then the quick fix.

1

Buffer vs Salt-Hydrolysis (the mixing trap)

After mixing, first work out millimoles (M × mL). Then read off this table:

After mixing you are left with…It is aUse this
Weak acid in excess over strong base (salt + leftover acid)BUFFERpH = pKa + log([salt]/[acid])
Equal moles — complete neutralisation (only salt left)SALT HYDROLYSISpH = 7 + ½(pKa + log C)
Strong acid / base in excessSTRONGpH from the leftover strong reagent

⚡ Shortcut: at the half-neutralisation point (half the weak acid used up) → pH = pKa instantly.

2

Common-Ion Effect in Ksp problems

Adding a common ion suppresses solubility — the salt dissolves far less.
  • Set the common-ion concentration ≈ C (ignore the tiny extra from the dissolving salt).
  • For an AB salt: s = Ksp / C.  For AB₂: s = √(Ksp / 4C²)-type — but in NEET just substitute C and solve.
  • Result: solubility drops by orders of magnitude vs pure water.
3

The 10⁻⁸ M strong-acid trap

Wrong: pH of 10⁻⁸ M HCl = 8  → that would mean a strong acid is basic! Impossible.
Right: water’s own H⁺ matters here → total [H⁺] ≈ 1.05×10⁻⁷ → pH ≈ 6.98 (just below 7).

Rule: whenever the acid/base concentration is ≤ 10⁻⁶ M, you must add water’s contribution.

4

α-approximation & log shortcuts

  • Use α = √(Ka/C) only when α < 5% (otherwise solve the full quadratic).
  • Salt of a weak acid + weak base ⇒ pH is independent of concentration (pure ½(pKa − pKb)).

Memorise these log values to fly through calculations:

log 2 = 0.30log 3 = 0.48log 4 = 0.60log 5 = 0.70log 6 = 0.78log 7 = 0.85log 8 = 0.90

4 High-Yield Practice MCQs

The simple idea

Try each question first, then open the solution to see the quickest route — using the formulas and log shortcuts from above, not long textbook working.

10 NEET-level questions with numerically close options — 2 theory, 3 pH/Ostwald, 3 buffers/hydrolysis, 2 Ksp. Tap “Show fastest solution”.

TheoryQ1. Which of the following can act as a Lewis acid but NOT as a Brønsted acid?

  1. HCl
  2. BF₃
  3. CH₃COOH
  4. NH₄⁺
Show fastest solutionAnswer: (b) BF₃
BF₃ has an electron-deficient boron (empty orbital) → accepts an electron pair (Lewis acid), but it has no H⁺ to donate, so it is not a Brønsted acid. HCl, CH₃COOH and NH₄⁺ all donate protons (Brønsted acids).

TheoryQ2. The conjugate base of the dihydrogen phosphate ion (H₂PO₄⁻) is:

  1. H₃PO₄
  2. HPO₄²⁻
  3. PO₄³⁻
  4. H₂PO₄⁻
Show fastest solutionAnswer: (b) HPO₄²⁻
A conjugate base is formed by removing one H⁺: H₂PO₄⁻ − H⁺ → HPO₄²⁻. (H₃PO₄ is its conjugate acid; PO₄³⁻ is two protons away.)

pH / OstwaldQ3. The pH of a 0.001 M HCl solution is:

  1. 2.0
  2. 3.0
  3. 3.5
  4. 11.0
Show fastest solutionAnswer: (b) 3.0
Strong acid fully ionises: [H⁺] = 10⁻³ ⇒ pH = −log10⁻³ = 3.0. (11.0 is the pOH — a classic distractor.)

pH / OstwaldQ4. A 0.1 M weak monoprotic acid has Ka = 1.0×10⁻⁵. Its pH is approximately:

  1. 2.0
  2. 3.0
  3. 4.0
  4. 5.0
Show fastest solutionAnswer: (b) 3.0
Fast method: pH = ½(pKa − log C) = ½(5 − (−1)) = ½(6) = 3.0. Or [H⁺] = √(Ka·C) = √(10⁻⁵·10⁻¹) = √10⁻⁶ = 10⁻³.

pH / OstwaldQ5. The degree of dissociation of 0.01 M acetic acid (Ka = 1.8×10⁻⁵) is about:

  1. 0.42 %
  2. 1.8 %
  3. 4.2 %
  4. 13 %
Show fastest solutionAnswer: (c) 4.2 %
α = √(Ka/C) = √(1.8×10⁻⁵ / 10⁻²) = √(1.8×10⁻³) = √0.0018 ≈ 0.042 = 4.2 %. (α < 5%, so the approximation is valid.)

Buffer / HydrolysisQ6. Equal volumes of 0.2 M CH₃COOH and 0.2 M CH₃COONa are mixed (pKa = 4.74). The pH is:

  1. 3.74
  2. 4.74
  3. 5.74
  4. 7.00
Show fastest solutionAnswer: (b) 4.74
This is a buffer with equal salt and acid (ratio = 1). pH = pKa + log(1) = pKa + 0 = 4.74. Mixing equal volumes keeps the ratio 1, so concentration cancels.

Buffer / HydrolysisQ7. For an acetic-acid/acetate buffer (pKa = 4.74), what is the pH when [salt]/[acid] = 10?

  1. 3.74
  2. 4.74
  3. 5.74
  4. 6.74
Show fastest solutionAnswer: (c) 5.74
pH = pKa + log([salt]/[acid]) = 4.74 + log10 = 4.74 + 1 = 5.74. (Tripling distractors come from sign errors in log.)

Buffer / HydrolysisQ8. The pH of a 0.1 M sodium acetate solution (Ka of acetic acid = 1.0×10⁻⁵) is:

  1. 5.0
  2. 7.0
  3. 8.5
  4. 9.0
Show fastest solutionAnswer: (d) 9.0
Salt of weak acid + strong base ⇒ pH = 7 + ½(pKa + log C) = 7 + ½(5 + (−1)) = 7 + 2 = 9.0 (basic). Choosing 5.0 forgets it's the SALT (basic), not the acid.

KspQ9. The solubility product of AgCl is 1.8×10⁻¹⁰. Its molar solubility in pure water is:

  1. 1.8×10⁻¹⁰
  2. 1.34×10⁻⁵
  3. 1.8×10⁻⁵
  4. 9.0×10⁻¹¹
Show fastest solutionAnswer: (b) 1.34×10⁻⁵
AB type: Ksp = s² ⇒ s = √(1.8×10⁻¹⁰) = √1.8 ×10⁻⁵ ≈ 1.34×10⁻⁵ mol L⁻¹. (1.8×10⁻¹⁰ wrongly skips the square root.)

KspQ10. The molar solubility of AgCl (Ksp = 1.8×10⁻¹⁰) in 0.1 M NaCl is:

  1. 1.34×10⁻⁵
  2. 1.8×10⁻⁹
  3. 1.8×10⁻¹⁰
  4. 1.8×10⁻¹¹
Show fastest solutionAnswer: (b) 1.8×10⁻⁹
Common-ion effect: [Cl⁻] ≈ 0.1 M (from NaCl). s = Ksp/[Cl⁻] = 1.8×10⁻¹⁰ / 0.1 = 1.8×10⁻⁹ mol L⁻¹ — far lower than in pure water (1.34×10⁻⁵), exactly as the common-ion effect predicts.