Chapters

Everything NCERT Class 11 Unit 6 (Equilibrium) says about ionic equilibrium (sections 6.9–6.13: electrolytes, acids and bases, ionisation constants, pH, buffers, hydrolysis of salts, solubility product) that can be asked, plus what JEE Main adds beyond the book (marked Beyond NCERT). The first part of Unit 6 (sections 6.1–6.8, K, Q and Le Chatelier) is in Chemical Equilibrium. The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula in the chapter on one screen, for the last pass. Details and examples are in the sections below. At 25 °C.

Quantity	Formula	Watch
Water	Kw = [H⁺][OH⁻] = 10⁻¹⁴	Rises with T
pH	pH = −log[H⁺]; pH + pOH = pKw = 14	14 only at 25 °C
Weak acid	Ka = cα²/(1 − α) ≈ cα²	α = √(Ka/c) when α ≪ 1
Weak acid pH	[H⁺] = √(Ka c); pH = ½(pKa − log c)	Check α < 5%
Weak base	[OH⁻] = √(Kb c); pOH = ½(pKb − log c)
Conjugate pair	Ka × Kb = Kw; pKa + pKb = 14
Polyprotic	Ka1 ≫ Ka2; [H⁺] from step 1; [A²⁻] ≈ Ka2	Overall K = Ka1 × Ka2
Common ion	α = Ka / [H⁺ already there]
Acidic buffer	pH = pKa + log([salt]/[acid])	Henderson–Hasselbalch
Basic buffer	pOH = pKb + log([salt]/[base])
Salt, weak acid + strong base	pH = 7 + ½pKa + ½ log c	Beyond NCERT
Salt, strong acid + weak base	pH = 7 − ½pKb − ½ log c	Beyond NCERT
Salt, weak acid + weak base	pH = 7 + ½(pKa − pKb)	No c
Solubility product	MₓXᵧ: Ksp = xˣ yʸ S^(x+y)	AB: S²; AB₂: 4S³; AB₃: 27S⁴; A₂B₃: 108S⁵
Precipitation	Q(ionic product) > Ksp	Use concentrations after mixing
Common ion on S	AB in c M of B⁻: S = Ksp/c	Section 12 for other types

0 · What JEE Main asks from this chapter

104 questions in JEE Main 2019–2026, about 1 per paper, mostly easy and medium. About 40% are numerical-answer questions. Grouped by type (counts are approximate):

Type	Share	Typical question
Ksp: solubility, common ion, precipitation, which precipitates first	~28%	Solubility of PbI₂ in 0.1 M Pb(NO₃)₂, Ksp 8 × 10⁻⁹
pH of acids, bases, mixtures and dilutions	~25%	600 mL 0.01 M HCl + 400 mL 0.01 M H₂SO₄: pH
Buffers, half-neutralisation	~20%	20 mL 0.1 M NaOH + 50 mL 0.1 M CH₃COOH: pH
Acid–base concepts, indicators, titration curves	~12%	Indicator for strong acid vs weak base
Salt hydrolysis	~8%	pH of 0.005 M calcium lactate, pKa 5
Polyprotic acids, Kw and temperature	~7%	[X²⁻] in 0.1 M H₂X with Ka2 = 10⁻¹³

Most numerical answers ask for pH to two decimals as "x × 10⁻²". Learn log 2 = 0.30, log 3 = 0.48, log 5 = 0.70, log 7 = 0.85.

1 · Electrolytes and ionic equilibrium

• Electrolytes conduct electricity in aqueous solution; non-electrolytes (sugar) don't (Faraday's classification).
• Strong electrolytes ionise almost completely (NaCl). Weak electrolytes ionise partly (acetic acid, under 5%), and an equilibrium sets up between the ions and the un-ionised molecules: ionic equilibrium.
• Acids, bases and salts are electrolytes, strong or weak.
• Water's high dielectric constant (80) cuts the attraction between Na⁺ and Cl⁻ by a factor of 80, and hydration keeps the ions apart, so NaCl dissolves and its ions move freely.
• How far a substance ionises depends on the strength of its bond and on how well its ions are solvated. NCERT uses dissociation (ions already in the solid separate) and ionisation (a neutral molecule splits into ions) interchangeably.

2 · Acids, bases and salts: three definitions

Everyday properties: acids taste sour (Latin acidus), turn blue litmus red and release H₂ with some metals. Bases taste bitter, feel soapy and turn red litmus blue. Acid + base in the right proportion → salt. Examples: HCl in gastric juice; acetic acid in vinegar; citric and ascorbic acid in lemon; tartaric acid in tamarind; washing soda is a base.

Theory	Acid; base	Limits
Arrhenius	gives H⁺(aq) in water; gives OH⁻(aq) in water	water only; can't explain NH₃ (no OH)
Brønsted–Lowry	proton (H⁺) donor; proton acceptor	needs a proton transfer
Lewis (1923)	electron-pair acceptor; electron-pair donor	widest

Hydronium ion: a bare proton (radius ~10⁻¹⁵ m, intense field) can't exist free in water. It binds a lone pair of water to give trigonal pyramidal H₃O⁺, found even in solids (H₃O⁺Cl⁻). In solution it is further hydrated: H₅O₂⁺, H₇O₃⁺, H₉O₄⁺. OH⁻ is hydrated too (H₃O₂⁻, H₅O₃⁻, H₇O₄⁻). H⁺(aq) and H₃O⁺ mean the same here.

Conjugate pairs (Brønsted–Lowry): an acid and a base that differ by one proton. The conjugate acid has one H⁺ more; the conjugate base one H⁺ less.

• The equilibrium lies towards the weaker acid and weaker base: the stronger acid gives its proton to the stronger base.
• Strong acid → very weak conjugate base, and the reverse. HClO₄, HCl, HBr, HI, HNO₃, H₂SO₄ give ClO₄⁻, Cl⁻, Br⁻, I⁻, NO₃⁻, HSO₄⁻, far weaker bases than water. Weak acids (HNO₂, HF, CH₃COOH) have strong conjugate bases; NH₂⁻, O²⁻ and H⁻ are much stronger bases than water.
• Water is amphoteric: a base with HCl (HCl + H₂O → H₃O⁺ + Cl⁻), an acid with NH₃ (NH₃ + H₂O ⇌ NH₄⁺ + OH⁻). This is a Brønsted explanation (2022: the assertion that Lewis theory explains it was false).

Example (NCERT Problems 6.12–6.14) · Conjugate bases of HF, H₂SO₄, HCO₃⁻: F⁻, HSO₄⁻, CO₃²⁻. Conjugate acids of NH₂⁻, NH₃, HCOO⁻: NH₃, NH₄⁺, HCOOH.

Amphiprotic species	Conjugate acid	Conjugate base
H₂O	H₃O⁺	OH⁻
HCO₃⁻	H₂CO₃	CO₃²⁻
HSO₄⁻	H₂SO₄	SO₄²⁻
NH₃	NH₄⁺	NH₂⁻
H₂PO₄⁻ (2023)	H₃PO₄	HPO₄²⁻

Lewis acids and bases: for bases, Lewis and Brønsted agree (both use a lone pair). Many Lewis acids have no proton.

• BF₃ + :NH₃ → BF₃:NH₃ (BF₃ is electron-deficient).
• Lewis acids: electron-deficient species, cations: BF₃, BCl₃, AlCl₃, Co³⁺, Mg²⁺, H⁺.
• Lewis bases: lone-pair donors: H₂O, NH₃, OH⁻, F⁻.
• NCERT exercise 6.36: of H₂O, BF₃, H⁺, NH₄⁺, the Lewis acids are BF₃ and H⁺ (NH₄⁺ has neither a vacant orbital nor a lone pair).

3 · Ionisation of water, Kw and pH

Strong acids and bases (almost completely ionised): HClO₄, HCl, HBr, HI, HNO₃, H₂SO₄; LiOH, NaOH, KOH, CsOH, Ba(OH)₂.

Self-ionisation of water: H₂O + H₂O ⇌ H₃O⁺ + OH⁻. [H₂O] is constant (pure liquid) and goes into the constant:

• Pure water is 55.55 M, so only about 2 in 10⁹ molecules are ionised (10⁻⁷/55.55 = 1.8 × 10⁻⁹).
• Kw depends on temperature. Ionisation is endothermic, so Kw rises with T. Hot pure water has pH below 7 but is still neutral ([H⁺] = [OH⁻]).
• Acidic: [H₃O⁺] > [OH⁻]; neutral: equal; basic: [H₃O⁺] < [OH⁻].

pH = −log of the hydrogen-ion activity; in dilute solutions (< 0.01 M) the activity equals [H⁺]/(mol L⁻¹), so pH is unitless.

• 10⁻² M HCl: pH 2. NaOH with [OH⁻] = 10⁻⁴: [H⁺] = 10⁻¹⁰, pH 10. Pure water at 25 °C: 7.
• Logarithmic: one pH unit = a factor of 10 in [H⁺]; a factor of 1000 = 3 units (2023).
• pH paper (four strips) reads to about ±0.5 in 1–14; a pH meter to 0.001.

pH of common substances (NCERT Table 6.5):

Fluid	pH
Conc. HCl	~−1.0
1 M HCl	~0
Gastric juice	~1.2
Lemon juice	~2.2
Soft drinks, vinegar	~3.0
Tomato juice	~4.2
Black coffee	5.0
Human saliva	6.4
Milk	6.8
Human blood	7.4
Egg white, sea water	7.8
Milk of magnesia	10
Lime water	10.5
0.1 M NaOH	13
Saturated NaOH	~15

Example (NCERT Problem 6.16) · Soft drink with [H⁺] = 3.8 × 10⁻³ M: pH = 3 − log 3.8 = 3 − 0.58 = 2.42.

Example (NCERT Problem 6.17) · 1.0 × 10⁻⁸ M HCl. Don't answer 8: an acid can't be basic. Water's own H⁺ matters: (10⁻⁸ + x)x = 10⁻¹⁴, so x = 9.5 × 10⁻⁸ and [H⁺] = 1.05 × 10⁻⁷. pH = 6.98. (2023: "pH of 10⁻⁸ M HCl is 8" was the false statement.)

Example (2019) · 2H₂O ⇌ H₃O⁺ + OH⁻, K = 10⁻¹⁴: ΔG° = 2.303 × 8.314 × 298 × 14 = +80 kJ mol⁻¹.

Example (NCERT exercise 6.65) · Kw = 2.7 × 10⁻¹⁴ at 310 K: [H⁺] = √(2.7 × 10⁻¹⁴) = 1.64 × 10⁻⁷, pH of neutral water = 6.78.

Strong acids and bases, mixtures and dilution · Beyond NCERT as a method (NCERT exercises 6.48, 6.49, 6.66)

1. Count mmol of H⁺ (M × mL × number of H⁺) and of OH⁻.
2. Subtract; the excess decides.
3. Divide the excess by the total volume.

Example (2023) · 600 mL 0.01 M HCl (6 mmol H⁺) + 400 mL 0.01 M H₂SO₄ (8 mmol H⁺): 14 mmol in 1000 mL, [H⁺] = 0.014. pH = 2 − log 1.4 = 1.85 (186 × 10⁻²).

Example (2021) · 50 mL 1 M HCl + 30 mL 1 M NaOH: 20 mmol H⁺ left in 80 mL, [H⁺] = 0.25. pH = 0.602 (6021 × 10⁻⁴).

Example (2020) · 4 g NaOH in 100 L (0.001 M) and 9.8 g H₂SO₄ in 100 L (0.001 M, 0.002 M H⁺). 40 L of the first (0.04 mol OH⁻) + 10 L of the second (0.02 mol H⁺): 0.02 mol OH⁻ in 50 L = 4 × 10⁻⁴ M. pOH = 3.4, pH = 10.6.

Example (2023, 2025) · pH 1 → 2 needs [H⁺] ÷ 10: 1 L becomes 10 L, add 9000 mL water. pH 1 diluted with an equal volume: 0.05 M, pH 1.3.

Example (2023) · Ca(OH)₂ for 100 mL of pH 12: [OH⁻] = 0.01, Ca(OH)₂ = 0.005 M = 0.5 mmol.

4 · Weak acids: Ka

HX + H₂O ⇌ H₃O⁺ + X⁻. From concentration c with degree of ionisation α:

• Larger Ka = stronger acid. Ka is dimensionless (standard state 1 M). pKa = −log Ka: smaller pKa, stronger acid.
• α rises on dilution (α ∝ 1/√c): Ostwald's dilution law (the name is Beyond NCERT).
• Percent dissociation = [HA]dissociated / [HA]initial × 100.
• The approximation 1 − α ≈ 1 is fine when α < 5%; otherwise solve the quadratic.

NCERT's step method: (1) list the species as Brønsted acids and bases; (2) write every proton transfer; (3) the reaction with the largest K is primary (HF, Ka 3.2 × 10⁻⁴ ≫ Kw); (4) make the initial–change–equilibrium table in α; (5) solve for α; (6) find the concentrations; (7) pH = −log[H₃O⁺].

Ka at 298 K (NCERT Table 6.6):

Acid	Ka
HF	3.5 × 10⁻⁴
HNO₂	4.5 × 10⁻⁴
HCOOH	1.8 × 10⁻⁴
C₆H₅COOH (benzoic)	6.5 × 10⁻⁵
CH₃COOH	1.74 × 10⁻⁵
Niacin	1.5 × 10⁻⁵
HClO	3.0 × 10⁻⁸
HCN	4.9 × 10⁻¹⁰
C₆H₅OH (phenol)	1.3 × 10⁻¹⁰

Example (NCERT Problem 6.18) · 0.02 M HF, Ka = 3.2 × 10⁻⁴: α is not small. α² + 0.016α − 0.016 = 0, α = 0.12. [H₃O⁺] = [F⁻] = 2.4 × 10⁻³ M, [HF] = 17.6 × 10⁻³ M, pH = 2.62.

Example (NCERT Problem 6.19) · 0.1 M monobasic acid, pH 4.50: [H⁺] = [A⁻] = 3.16 × 10⁻⁵. Ka = (3.16 × 10⁻⁵)²/0.1 = 1.0 × 10⁻⁸, pKa = 8.

Example (NCERT Problem 6.20) · 0.08 M HOCl, Ka = 2.5 × 10⁻⁵: x² = 2.0 × 10⁻⁶, [H⁺] = 1.41 × 10⁻³. pH = 2.85, dissociation 1.76%.

Example (2022, 2026) · 0.2 M butyric acid, Ka = 2 × 10⁻⁵: [H⁺] = √(4 × 10⁻⁶) = 2 × 10⁻³, pH = 2.7. 10 mM HA, pKa 4: pH = ½(4 + 2) = 3.

Example (2025) · For any weak acid at degree of dissociation x: pH − pKa = log(x/(1 − x)), since [A⁻]/[HA] = x/(1 − x).

General weak electrolyte AₓBᵧ (2026): K = c^(x+y−1) xˣ yʸ α^(x+y) when α ≪ 1, so α = [K / (c^(x+y−1) xˣ yʸ)]^(1/(x+y)).

5 · Weak bases: Kb, and Ka × Kb = Kw

MOH ⇌ M⁺ + OH⁻, or B + H₂O ⇌ BH⁺ + OH⁻:

Amines are derivatives of NH₃ and weak bases; methylamine, codeine, quinine and nicotine are very weak.

Kb at 298 K (NCERT Table 6.7):

Base	Kb
(CH₃)₂NH, dimethylamine	5.4 × 10⁻⁴
(C₂H₅)₃N, triethylamine	6.45 × 10⁻⁵
NH₃	1.77 × 10⁻⁵
Quinine	1.10 × 10⁻⁶
Pyridine	1.77 × 10⁻⁹
Aniline	4.27 × 10⁻¹⁰
Urea	1.3 × 10⁻¹⁴

Conjugate pair: adding the two ionisations gives 2H₂O ⇌ H₃O⁺ + OH⁻, and K for a sum of reactions is the product of the Ks:

NH₄⁺: Ka = 5.6 × 10⁻¹⁰; NH₃: Kb = 1.8 × 10⁻⁵; product 1.0 × 10⁻¹⁴. pKa of NH₄⁺ = 9.25.

Example (NCERT Problem 6.23) · 0.05 M NH₃, Kb = 1.77 × 10⁻⁵: α = √(1.77 × 10⁻⁵/0.05) = 0.0188; [OH⁻] = 9.4 × 10⁻⁴; pH = 10.97. Ka of NH₄⁺ = 10⁻¹⁴/1.77 × 10⁻⁵ = 5.64 × 10⁻¹⁰.

Example (NCERT Problem 6.21) · 0.004 M hydrazine, pH 9.7: pOH = 4.3, [OH⁻] = 5.0 × 10⁻⁵. Kb = (5.0 × 10⁻⁵)²/0.004 = 6.3 × 10⁻⁷, pKb = 6.2. NCERT prints [H⁺] = 1.67 × 10⁻¹⁰ (it is 2.0 × 10⁻¹⁰), so its Kb = 8.96 × 10⁻⁷ and pKb 6.04 are off; the method is right.

Example (2025) · 1 mM ethylamine, pH 9: [OH⁻] = 10⁻⁵, Kb = 10⁻¹⁰/10⁻³ = 10⁻⁷.

6 · Polybasic acids and what makes an acid strong

Polyprotic acids (oxalic, sulphuric, phosphoric) lose protons in steps:

• Ka1 > Ka2 > Ka3: removing H⁺ from a negative ion is harder (H₂CO₃ vs HCO₃⁻; H₂PO₄⁻ vs HPO₄²⁻).
• H₃O⁺ comes mainly from the first step.
• For H₂X ⇌ 2H⁺ + X²⁻, K = Ka1 × Ka2 (2022).
• Beyond NCERT shortcuts: in a solution of H₂X alone, [HX⁻] ≈ [H⁺] and [X²⁻] ≈ Ka2. If a strong acid fixes [H⁺], then [HX⁻] = Ka1[H₂X]/[H⁺] and [X²⁻] = Ka1Ka2[H₂X]/[H⁺]².

Example (2026) · 0.1 M H₂X, Ka1 = 2.5 × 10⁻⁸, Ka2 = 1.0 × 10⁻¹³: [X²⁻] ≈ Ka2 = 100 × 10⁻¹⁵ M. With 0.1 M HCl present and Ka1 = 8.1 × 10⁻⁸: [HA⁻] = 8.1 × 10⁻⁸ × 0.1/0.1 = 8.1 × 10⁻⁸ M.

Example (2021) · 0.588 M H₂SO₃, Ka1 = 1.7 × 10⁻²: Ka1 is large, so solve x²/(0.588 − x) = 0.017: x ≈ 0.09, pH ≈ 1.

Factors affecting acid strength (H–A):

• Weaker H–A bond → stronger acid. More polar H–A bond → stronger acid.
• Down a group, bond strength decides: A gets bigger, H–A gets weaker: HF ≪ HCl ≪ HBr ≪ HI; H₂S is stronger than H₂O.
• Across a period, polarity decides: A more electronegative → stronger acid: CH₄ < NH₃ < H₂O < HF.

7 · Common ion effect

Adding an ion already in the equilibrium shifts it back (Le Chatelier). Acetate added to acetic acid lowers [H⁺]; H⁺ added lowers the ionisation.

Example (NCERT) · 0.05 M acetic acid + 0.05 M acetate: x(0.05 + x)/(0.05 − x) = 1.8 × 10⁻⁵, so x = [H⁺] = 1.8 × 10⁻⁵, pH = 4.74 (= pKa: equal acid and salt).

Degree of ionisation with a common ion: α = Ka / [H⁺ from the strong acid] (or Kb / [OH⁻]).

Example (2021) · 0.01 mol HA (Ka = 2 × 10⁻⁶) in 1 L of 0.1 M HCl: α = 2 × 10⁻⁶/0.1 = 2 × 10⁻⁵.

Example (NCERT Problem 6.24) · 0.10 M NH₃: [OH⁻] = √(1.77 × 10⁻⁶) = 1.33 × 10⁻³, pH 11.12. Add 25 mL 0.10 M HCl to 50 mL: 2.5 mmol NH₃ neutralised, leaving 2.5 mmol NH₃ and 2.5 mmol NH₄⁺ in 75 mL. Then [OH⁻] = Kb = 1.77 × 10⁻⁵, pH 9.24.

8 · Hydrolysis of salts

Hydrolysis: the cation, anion or both of a salt react with water to re-form the parent acid or base, changing the pH.

• Cations of strong bases (Na⁺, K⁺, Ca²⁺, Ba²⁺) and anions of strong acids (Cl⁻, Br⁻, NO₃⁻, ClO₄⁻) are only hydrated, not hydrolysed. Strong acid + strong base salt: pH 7 (NaCl, KBr).
• Weak acid + strong base (CH₃COONa): CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Basic, pH > 7. Also NaCN, NaNO₂, KF.
• Strong acid + weak base (NH₄Cl): NH₄⁺ + H₂O ⇌ NH₄OH + H⁺. Acidic, pH < 7. Also NH₄NO₃.
• Weak acid + weak base (CH₃COONH₄): both ions hydrolyse. The degree of hydrolysis doesn't depend on concentration, and

pH > 7 if pKa > pKb (the base is the stronger), < 7 if the acid is the stronger. Ammonium carbonate is basic (2024).

Example (NCERT Problem 6.25) · Ammonium acetate, pKa 4.76, pKb 4.75: pH = 7 + ½(0.01) = 7.005. Ka = Kb = 1.8 × 10⁻⁵ (2024): 7.

pH of a hydrolysing salt · Beyond NCERT (c = salt concentration)

Example (2019) · 0.02 M NH₄Cl, Kb = 10⁻⁵: pH = 7 − 2.5 − ½ log 0.02 = 4.5 + 0.85 = 5.35.

Example (2023) · 0.005 M calcium lactate gives 0.01 M lactate (two per formula). pKa 5: pH = 7 + 2.5 + ½ log 0.01 = 8.5.

Example (NCERT exercise 6.61) · 0.04 M NaNO₂, Ka(HNO₂) = 4.5 × 10⁻⁴: Kh = 2.22 × 10⁻¹¹, [OH⁻] = √(Kh c) = 9.4 × 10⁻⁷. pH = 7.97, h = 2.36 × 10⁻⁵.

Example (2022) · 20 mL 0.1 M NH₄OH + 40 mL 0.05 M HCl: both 2 mmol, exactly neutralised. 2 mmol NH₄Cl in 60 mL = 0.033 M. pH = 7 − 2.5 + ½(1.48) = 5.2.

9 · Buffer solutions

A buffer resists a change in pH on dilution or on adding a little acid or alkali. Blood, urine, many medicines and cosmetics need a fixed pH. Blood's buffer is H₂CO₃/HCO₃⁻ (2024).

• Acidic buffer: weak acid + its salt with a strong base. CH₃COOH + CH₃COONa buffers near pH 4.75.
• Basic buffer: weak base + its salt with a strong acid. NH₄OH + NH₄Cl buffers near pH 9.25.

Henderson–Hasselbalch equation: from Ka = [H₃O⁺][A⁻]/[HA], take logs. The weak acid barely ionises, so [HA] ≈ acid taken; nearly all A⁻ comes from the salt, so [A⁻] ≈ salt taken:

• Equal acid and salt: pH = pKa. So choose an acid whose pKa is close to the pH wanted.
• Dilution doesn't change a buffer's pH: the ratio stays the same.
• Use mmol in the ratio: the volume cancels.
• A buffer is not "salt and acid in any quantities" (2024): the ratio must be near 1 (roughly 0.1 to 10) for it to work.

Example (NCERT Problem 6.22) · 0.2 M NH₄Cl + 0.1 M NH₃, pKb 4.75: [OH⁻] = 1.77 × 10⁻⁵ × 0.1/0.2 = 0.88 × 10⁻⁵. pH = 8.95.

Buffers made by part-neutralisation: a weak acid with less than an equivalent of strong base leaves acid + salt.

Example (2023) · 20 mL 0.1 M NaOH + 50 mL 0.1 M CH₃COOH: 2 mmol salt, 3 mmol acid left. pH = 4.76 + log(2/3) = 4.58.

Example (2022, 2026) · Half-neutralisation: half the acid turned to salt, ratio 1, pH = pKa. 50 mL 0.1 M CH₃COOH + 25 mL 0.1 M NaOH: 4.76. 20 mL 0.2 M HX (Ka 5 × 10⁻⁴) + 10 mL 0.2 M NaOH: 3.3; before any NaOH, [H⁺] = √(5 × 10⁻⁴ × 0.2) = 0.01, pH 2.0.

Example (2019) · 20 mL 0.1 M H₂SO₄ (4 mmol H⁺) + 30 mL 0.2 M NH₄OH (6 mmol): 4 mmol NH₄⁺, 2 mmol NH₃ left. pOH = 4.7 + log 2 = 5.0, pH 9.0.

Example (2023) · 1 L with 0.1 mol each NH₃ and NH₄Cl; add 0.02 mol HCl: NH₃ 0.08, NH₄⁺ 0.12. pOH = 4.745 + log 1.5 = 4.921, pH = 9.079.

Example (2022) · Buffer of pH 8.26 from 0.2 M NH₃ (pKb 4.74): pOH 5.74 = 4.74 + log(salt/0.2), salt = 2 M = 107 g NH₄Cl per litre.

Example (2021) · pH 5.74 from 1.0 M acetic acid (pKa 4.74): log(salt/1) = 1, salt = 10 M.

10 · Indicators and titration curves

An indicator is a weak organic acid (HIn) whose acid form and conjugate base form (In⁻) have different colours: HIn + H₂O ⇌ H₃O⁺ + In⁻. Phenolphthalein and bromothymol blue are examples.

Indicators in titrations · Beyond NCERT (only the HIn idea is in NCERT)

Indicator	Range (pH)	Colour change
Methyl orange	3.1–4.4	red (quinonoid, acid) → yellow (benzenoid, base)
Phenolphthalein	8.2–10	colourless (acid) → pink (base, ionised form)

Titration	Equivalence pH	Indicator
Strong acid + strong base	7; jump ~3–11	either
Weak acid + strong base	> 7	phenolphthalein
Strong acid + weak base	< 7	methyl orange
Weak acid + weak base	no sharp jump	none suitable

The weak acid starts higher, rises slowly through its buffer region (pH = pKa half-way), and its jump starts above 7: only phenolphthalein changes colour inside it. The strong acid's jump covers both ranges.

Example (2023) · Incorrect: "methyl orange may be used for a weak acid vs weak base titration." Correct: methyl orange for strong acid–weak base; phenolphthalein for weak acid–strong base and for strong–strong.

Example (2022) · Phenolphthalein is a weak acid that is colourless in acid and pink in base: in base it does ionise (the reason given, "it doesn't dissociate in basic medium", was false).

Conductometric titration · Beyond NCERT (2025) · Titrating a mixture of HCl and CH₃COOH with NaOH: HCl is neutralised first (conductance falls as fast H⁺ is replaced by Na⁺), then CH₃COOH (conductance rises slowly), then excess NaOH (rises steeply). The two breaks give the two volumes.

11 · Solubility equilibria and Ksp

Solubility depends on lattice enthalpy (must be overcome) and solvation enthalpy (always negative, released). A salt dissolves in a solvent only if its solvation enthalpy outweighs its lattice enthalpy; non-polar solvents solvate too weakly, so salts don't dissolve in them. CaCl₂ is so soluble it is hygroscopic; LiF is nearly insoluble.

Category	Solubility
Soluble	> 0.1 M
Slightly soluble	0.01–0.1 M
Sparingly soluble	< 0.01 M

Solubility product: for a sparingly soluble salt in contact with its saturated solution, [solid] is constant and goes into the constant:

For MₓXᵧ ⇌ xM^(p+) + yX^(q−), with molar solubility S:

Type	Example	Ksp
AB	AgCl, BaSO₄	S²
AB₂ / A₂B	PbI₂, Ag₂CrO₄, Ca(OH)₂	4S³
AB₃ / A₃B	Al(OH)₃, Cr(OH)₃	27S⁴
A₂B₃ / A₃B₂	Bi₂S₃, Ca₃(PO₄)₂	108S⁵
A₃B₄	Zr₃(PO₄)₄	6912S⁷

• Compare solubilities through S, not Ksp, unless the salts are the same type.
• Converting g L⁻¹: S (mol L⁻¹) = solubility (g L⁻¹) / molar mass. "W g per 100 mL" is 10W g L⁻¹.

Example (NCERT Problem 6.26) · A₂X₃, Ksp = 1.1 × 10⁻²³: 108S⁵ = 1.1 × 10⁻²³, S⁵ ≈ 1 × 10⁻²⁵, S = 1.0 × 10⁻⁵ M.

Example (NCERT Problem 6.27) · AgCN (Ksp 6 × 10⁻¹⁷): S = 7.8 × 10⁻⁹ M. Ni(OH)₂ (Ksp 2.0 × 10⁻¹⁵): 4S³ = 2 × 10⁻¹⁵, S = 7.9 × 10⁻⁶ M (NCERT prints 0.58 × 10⁻⁴, a slip). Ni(OH)₂ is more soluble despite the similar Ksp.

Example (2021) · A₂X and MX, both Ksp = 4.0 × 10⁻¹²: S(A₂X) = (10⁻¹²)^(1/3) = 10⁻⁴; S(MX) = 2 × 10⁻⁶. Ratio 50.

Example (2023) · AgCl, 1.434 × 10⁻³ g L⁻¹ ÷ 143.4 = 1.0 × 10⁻⁵ M. Ksp = 10⁻¹⁰, −log Ksp = 10.

Ionic product Q (Qsp) has the same form with the actual concentrations:

• Q > Ksp: precipitates until Q = Ksp.
• Q = Ksp: saturated. Q < Ksp: more dissolves.
• After mixing two solutions, dilute both before computing Q (equal volumes halve each).

Example (2023) · 25 mL 0.050 M Ba(NO₃)₂ + 25 mL 0.020 M NaF: [Ba²⁺] = 0.025, [F⁻] = 0.010. Q = 0.025 × 10⁻⁴ = 2.5 × 10⁻⁶; Q/Ksp = 2.5 × 10⁻⁶/0.5 × 10⁻⁶ = 5, so BaF₂ precipitates.

Example (NCERT exercise 6.69) · Equal volumes of 0.002 M NaIO₃ and Cu(ClO₃)₂: each becomes 0.001 M. Q = 10⁻³ × (10⁻³)² = 10⁻⁹ < 7.4 × 10⁻⁸: no precipitate.

Which precipitates first · Beyond NCERT · Work out the concentration of the added ion each salt needs; the smaller one precipitates first.

Example (2021) · 0.1 M Cl⁻ and 0.001 M CrO₄²⁻, AgNO₃ added. AgCl needs Ag⁺ = 1.7 × 10⁻¹⁰/0.1 = 1.7 × 10⁻⁹; Ag₂CrO₄ needs √(1.9 × 10⁻¹²/0.001) = 4.4 × 10⁻⁵. AgCl first, though its Ksp is larger.

Example (2024) · Mg(OH)₂ (Ksp 10⁻¹¹) from 0.10 M Mg²⁺: [OH⁻]² = 10⁻¹⁰, [OH⁻] = 10⁻⁵, precipitates from pH 9.

Example (2026) · 0.01 M X²⁺ and Y²⁺ saturated with H₂S (0.1 M, Ka1Ka2 = 10⁻²¹): [S²⁻] = 10⁻²²/[H⁺]². YS (Ksp 4 × 10⁻¹⁶) needs [S²⁻] = 4 × 10⁻¹⁴, so [H⁺]² = 2.5 × 10⁻⁹, [H⁺] = 5 × 10⁻⁵, pH ≈ 4. This is how group II and IV cations are separated in salt analysis: low pH keeps [S²⁻] small.

12 · Common ion effect on solubility

Adding one of the salt's ions makes Q > Ksp, so the salt precipitates until Q = Ksp again: solubility falls. Removing an ion makes more dissolve.

• NaCl precipitates, very pure, when HCl gas is passed into its saturated solution (sulphate impurities stay in solution). (2024: statement II true.)
• Used to precipitate an ion almost completely for gravimetric estimation: Ag⁺ as AgCl, Fe³⁺ as its hydroxide, Ba²⁺ as BaSO₄.

Method: the common ion's concentration c is far larger than what the salt adds, so put c for that ion and S (times its coefficient) for the other.

Example (NCERT Problem 6.28) · Ni(OH)₂ (Ksp 2.0 × 10⁻¹⁵) in 0.10 M NaOH: S(0.10)² = 2.0 × 10⁻¹⁵, S = 2.0 × 10⁻¹³ M, against 7.9 × 10⁻⁶ M in water.

Example (2021) · PbI₂ (Ksp 8.0 × 10⁻⁹) in 0.1 M Pb(NO₃)₂: 0.1 × (2S)² = 8 × 10⁻⁹, S² = 2 × 10⁻⁸, S = 1.41 × 10⁻⁴ M.

Example (2019) · Ag₂CO₃ (Ksp 8 × 10⁻¹²) in 0.1 M AgNO₃: S = 8 × 10⁻¹²/0.01 = 8 × 10⁻¹⁰ M. Al(OH)₃ (Ksp 2.4 × 10⁻²⁴) in 0.2 M NaOH: S = 2.4 × 10⁻²⁴/0.008 = 3 × 10⁻²² M.

Example (2019) · Cd(OH)₂ with S = 1.84 × 10⁻⁵ M in water: Ksp = 4S³ = 2.49 × 10⁻¹⁴. In a pH 12 buffer ([OH⁻] = 0.01): S = 2.49 × 10⁻¹⁴/10⁻⁴ = 2.49 × 10⁻¹⁰ M.

Salts of weak acids dissolve more at low pH. The anion (phosphate, cyanide, carbonate, sulphide) is protonated, which lowers [X⁻] and lets more salt dissolve. For MX with HX of constant Ka (NCERT eq. 6.46):

Example (2021) · AgCN (Ksp 2.2 × 10⁻¹⁶, Ka(HCN) 6.2 × 10⁻¹⁰) in a pH 3 buffer: S = √(2.2 × 10⁻¹⁶ × 10⁻³ / 6.2 × 10⁻¹⁰) = 1.9 × 10⁻⁵ M, about 1000 times more than in water.

Complex formation raises solubility · Beyond NCERT · AgCl dissolves in NH₃ because Ag⁺ is taken up as [Ag(NH₃)₂]⁺ (2021 asked how much NH₃ brings [Ag⁺] down to 5 × 10⁻⁸ M using its formation constant).

13 · Traps

Trap	What goes wrong	Do this
Very dilute acid	pH of 10⁻⁸ M HCl = 8	Add water's 10⁻⁷: pH 6.98
Total volume	Dividing by one solution's volume	Excess mmol ÷ total mL
H₂SO₄, Ba(OH)₂	Counting one H⁺ or OH⁻	Two each: 0.01 M H₂SO₄ is 0.02 M H⁺
pH + pOH = 14	Using it at 80 °C	Only at 25 °C; hot neutral water has pH < 7
Hot water	Calling it acidic	Still neutral: [H⁺] = [OH⁻]
Ksp for AB₂	Ksp = S³	S × (2S)² = 4S³
Comparing solubility	Smaller Ksp = less soluble, across types	Convert each to S
Precipitation	Using concentrations before mixing	Halve both for equal volumes
Which precipitates first	Lowest Ksp first	Lowest ion concentration needed first
Common ion	Dropping the coefficient	PbI₂ in Pb²⁺: c(2S)²; in I⁻: S·c²
Buffer ratio	Using concentrations before neutralising	mmol after the reaction; volume cancels
Salt of weak + strong	Using c of the salt for a 2:1 salt	Calcium lactate 0.005 M gives 0.01 M lactate
Equivalence of weak acid	pH 7 at equivalence	Salt hydrolyses: > 7 for weak acid, < 7 for weak base
Indicator	Phenolphthalein for strong acid–weak base	Methyl orange there; nothing for weak–weak
Kb from pH	Using [H⁺] directly	[OH⁻] = 10^−(14 − pH)

14 · Numbers worth knowing

• Kw = 10⁻¹⁴ at 25 °C; 2.7 × 10⁻¹⁴ at 310 K (pH 6.78).
• Ka(CH₃COOH) = 1.8 × 10⁻⁵, pKa 4.74–4.76. Kb(NH₃) = 1.8 × 10⁻⁵, pKb 4.74–4.75; pKa(NH₄⁺) = 9.25.
• Buffers: acetate ~4.75, ammonia ~9.25. Blood 7.4 (H₂CO₃/HCO₃⁻).
• Ksp(BaSO₄) = 1.1 × 10⁻¹⁰; Ksp(AgCl) ≈ 1.8 × 10⁻¹⁰.
• log 2 = 0.301, log 3 = 0.477, log 5 = 0.699, log 7 = 0.845; ln 10 = 2.303.
• Water: 55.55 M; dielectric constant 80.

15 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• HCl in gastric juice is secreted at 1.2–1.5 L a day and is essential for digestion.
• Michael Faraday (1791–1867): bookbinder's apprentice, Davy's assistant, Director of the Royal Institution laboratories (1825), first Fullerian Professor of Chemistry (1833). Laws of electrolysis (1834); Christmas lectures, The Chemical History of a Candle; about 450 papers; declined all honours.
• Svante Arrhenius (1859–1927): thesis on the conductivity of electrolyte solutions (Uppsala, 1884); the Arrhenius equation (1899); first to discuss the "greenhouse effect" by that name; Nobel Prize in Chemistry 1903 for electrolytic dissociation.
• NCERT's suggested activities: pH paper on juices, soft drinks, body fluids and salt solutions; acetate buffers; indicators; acid–base titrations; common ion effect on solubility; pH meter vs pH paper.