1. Electrolytes & Ionization

Ionic Equilibrium: The equilibrium established between unionized molecules and ions in the solution of weak electrolytes.

Types of Substances

Non-Electrolytes: Do not conduct electricity (e.g., Urea, Glucose).
Strong Electrolytes: Ionize almost completely (α ≈ 1). (e.g., HCl, NaOH, NaCl).
Weak Electrolytes: Ionize partially (α < 1). Equilibrium exists. (e.g., CH₃COOH, NH₄OH).

Degree of Ionization (α):
Fraction of total molecules dissociated into ions.

α = (Number of molecules dissociated) / (Total number of molecules)

Ostwald's Dilution Law

For weak electrolytes, degree of ionization increases with dilution.

α = √(K_a / C)

(Valid only for weak electrolytes)

2. Acids, Bases & pH Scale

1. Ionic Product of Water (K_w):

Product of conc. of H⁺ and OH^- ions.

K_w = [H⁺][OH^-] = 10^-14 (at 25°C)

pK_w = pH + pOH = 14

2. pH Scale:

Negative logarithm of H⁺ ion concentration.

pH = -log [H⁺]

Ionization Constants

Acid Dissociation Constant (K_a): Measure of acid strength. High K_a = Strong Acid.
Base Dissociation Constant (K_b): Measure of base strength. High K_b = Strong Base.
Relation: K_a × K_b = K_w (for conjugate pair).

3. Hydrolysis of Salts

Reaction of cation or anion of salt with water to produce acidity or alkalinity.

Type	Example	Nature	pH Formula
Strong Acid + Strong Base	NaCl	Neutral	pH = 7
Weak Acid + Strong Base	CH₃COONa	Basic (Anionic Hyd.)	7 + ½(pK_a + log C)
Strong Acid + Weak Base	NH₄Cl	Acidic (Cationic Hyd.)	7 - ½(pK_b + log C)
Weak Acid + Weak Base	CH₃COONH₄	Depends	7 + ½(pK_a - pK_b)

4. Buffer Solutions

Solutions which resist change in pH upon addition of small amounts of acid or base.

Types & Henderson-Hasselbalch Equation

1. Acidic Buffer:
Weak Acid + Salt with Strong Base (e.g., CH₃COOH + CH₃COONa).

pH = pK_a + log ([Salt] / [Acid])

2. Basic Buffer:
Weak Base + Salt with Strong Acid (e.g., NH₄OH + NH₄Cl).

pOH = pK_b + log ([Salt] / [Base])

5. Solubility Product (K_sp)

Product of molar concentrations of ions in a saturated solution, raised to power of stoichiometric coefficients.

General Formula:

A_xB_y ↔ xA^y+ + yB^x-

K_sp = (xs)^x (ys)^y = x^xy^y s^(x+y)

Conditions for Precipitation

Q_sp < K_sp: Unsaturated solution (No PPT).
Q_sp = K_sp: Saturated solution (Equilibrium).
Q_sp > K_sp: Super-saturated (Precipitation occurs).

6. Common Ion Effect

Suppression of degree of dissociation of a weak electrolyte by adding a strong electrolyte containing a common ion.

Result:
1. Solubility of salt Decreases.
2. Used in purification (Salting out of soap, NaCl purification).
3. Used in salt analysis (Group II and Group IV precipitation).

Numericals & HOTS

Q1. pH Calculation

Calculate the pH of a 0.1 M solution of acetic acid (CH₃COOH) if its dissociation constant (K_a) is 1.8 × 10^-5.

Solution:
Since K_a is small, α is small.
[H⁺] = √(K_a × C)
[H⁺] = √(1.8 × 10^-5 × 0.1)
[H⁺] = √(1.8 × 10^-6)
[H⁺] = 1.34 × 10^-3 M.

pH = -log(1.34 × 10^-3)
pH = 3 - log(1.34)
pH = 3 - 0.13
pH = 2.87

Q2. Solubility Suppression

The solubility product (K_sp) of AgCl is 1.8 × 10^-10. Calculate its solubility in 0.1 M NaCl solution.

Solution:
Let solubility be 's'.
[Ag⁺] = s
[Cl^-] = s + 0.1 (from NaCl) ≈ 0.1 (since s is very small).

K_sp = [Ag⁺][Cl^-]
1.8 × 10^-10 = s × (0.1)
s = 1.8 × 10^-10 / 0.1
s = 1.8 × 10^-9 M

Q3. Henderson Equation

Calculate the pH of a buffer solution containing 0.2 M NH₄Cl and 0.1 M NH₄OH. (Given pK_b of NH₄OH = 4.75).

Solution:
This is a Basic Buffer.
pOH = pK_b + log([Salt] / [Base])
pOH = 4.75 + log(0.2 / 0.1)
pOH = 4.75 + log(2)
pOH = 4.75 + 0.30 = 5.05

pH = 14 - pOH
pH = 14 - 5.05
pH = 8.95

Q4. pH of Salt Solution

Calculate the pH of 0.1 M sodium acetate (CH₃COONa) solution. (pK_a of CH₃COOH = 4.74).

Solution:
Salt of Weak Acid + Strong Base (Anionic Hydrolysis).
Formula: pH = 7 + ½(pK_a + log C)

pH = 7 + ½(4.74 + log 0.1)
pH = 7 + ½(4.74 - 1)
pH = 7 + ½(3.74)
pH = 7 + 1.87
pH = 8.87

Q5. Will it Precipitate?

Equal volumes of 0.02 M CaCl₂ and 0.0004 M Na₂SO₄ are mixed. Will CaSO₄ precipitate? (K_sp of CaSO₄ = 2.4 × 10^-5).

Solution:
When equal volumes are mixed, concentrations are halved.
[Ca²⁺] = 0.02 / 2 = 0.01 M = 10^-2 M.
[SO₄^2-] = 0.0004 / 2 = 0.0002 M = 2 × 10^-4 M.

Ionic Product (Q_sp) = [Ca²⁺][SO₄^2-]
Q_sp = (10^-2) × (2 × 10^-4)
Q_sp = 2 × 10^-6.

Comparing: Q_sp (2 × 10^-6) < K_sp (2.4 × 10^-5).
No Precipitation.

Q6. Ion Concentration

For H₂S, K_a1 = 10^-7 and K_a2 = 10^-14. In a 0.1 M H₂S solution, what is the concentration of S^2-?

Solution:
For a diprotic acid where K_a1 >> K_a2:
1. [H⁺] comes mainly from first ionization.
[H⁺] ≈ √(K_a1C) = √(10^-7 × 0.1) = 10^-4 M.

2. Second ionization: HS^- ↔ H⁺ + S^2-
K_a2 = ([H⁺][S^2-]) / [HS^-]
Since [H⁺] ≈ [HS^-] (from step 1), they cancel out.

Therefore, [S^2-] ≈ K_a2.
[S^2-] = 10^-14 M

Q7. pH of Mixture

Calculate pH of a mixture formed by mixing 50 mL of 0.1 M HCl and 50 mL of 0.2 M H₂SO₄.

Solution:
Total Volume = 100 mL.
Moles H⁺ from HCl = 50 × 0.1 = 5 mmol.
Moles H⁺ from H₂SO₄ = 50 × 0.2 × 2 (diprotic) = 20 mmol.

Total H⁺ = 25 mmol.
[H⁺] = 25 mmol / 100 mL = 0.25 M.

pH = -log(0.25) = -log(1/4) = log 4 = 2log 2.
pH = 2(0.301)
pH = 0.602

Q8. Degree of Hydrolysis

Calculate the percentage hydrolysis of 0.1 M KCN solution. (K_a for HCN = 10^-9).

Solution:
KCN is Salt of Weak Acid + Strong Base.
Hydrolysis Constant K_h = K_w / K_a
K_h = 10^-14 / 10^-9 = 10^-5.

Degree of hydrolysis (h) = √(K_h / C)
h = √(10^-5 / 0.1) = √(10^-4)
h = 10^-2 = 0.01.

% Hydrolysis = h × 100
Answer: 1%

Q9. pH of 10^-8 M HCl

Calculate the pH of 10^-8 M HCl.

Solution:
We cannot neglect H⁺ from water (10^-7 M) as concentration is very low.
Total [H⁺] = [H⁺]_acid + [H⁺]_water.
Let [H⁺] from water = x. Then [OH^-] = x.
Total [H⁺] = 10^-8 + x.

K_w = (10^-8 + x)(x) = 10^-14.
Solving quadratic: x² + 10^-8x - 10^-14 = 0.
x ≈ 0.95 × 10^-7.
Total [H⁺] = 1.05 × 10^-7.

pH = 7 - log(1.05) ≈ 7 - 0.02
pH = 6.98 (Acidic, close to 7)

Q10. Buffer Ratio

In what ratio should sodium acetate and acetic acid be mixed to obtain a buffer of pH = 5.04? (pK_a = 4.74).

Solution:
pH = pK_a + log([Salt] / [Acid])
5.04 = 4.74 + log([Salt] / [Acid])
log([Salt] / [Acid]) = 5.04 - 4.74 = 0.30

[Salt] / [Acid] = antilog(0.30)
Since log 2 = 0.3010
Ratio [Salt] : [Acid] = 2 : 1

Important Formulae

1. Ostwald's Dilution Law

For Weak Monobasic Acid (HA):

α = √(K_a / C)

[H⁺] = Cα = √(K_a · C)

(Valid when α ≤ 0.05 or 5%)

2. pH Calculations

Basic Definitions:

pH = -log [H⁺]

pH + pOH = 14 (at 25°C)

Mixture of Strong Acids:

[H⁺]_mix = (N₁V₁ + N₂V₂) / (V₁ + V₂)

3. Salt Hydrolysis (pH)

Weak Acid + Strong Base:

pH = 7 + ½(pK_a + log C)

Strong Acid + Weak Base:

pH = 7 - ½(pK_b + log C)

Weak Acid + Weak Base:

pH = 7 + ½(pK_a - pK_b)

4. Henderson-Hasselbalch Eq.

Acidic Buffer:

pH = pK_a + log([Salt] / [Acid])

Basic Buffer:

pOH = pK_b + log([Salt] / [Base])

5. Solubility Product (K_sp)

Type	K_sp Relation
AB	s²
AB₂	4s³
A₂B₃	108s⁵

(s = Solubility in mol/L)

20 Golden Facts (NEET)

1. Temp Effect on pH: As temperature increases, K_w increases. [H⁺] increases, so pH of pure water decreases (< 7). However, water remains neutral because [H⁺] = [OH^-].
2. Conjugate Pairs: Strong Acid has a Weak Conjugate Base. Weak Acid has a Strong Conjugate Base. (e.g., HCl is strong, Cl^- is a very weak base).
3. Common Ion Effect: Adding a strong electrolyte with a common ion to a weak electrolyte suppresses the ionization of the weak electrolyte. (e.g., adding HCl to H₂S reduces S^2- conc).
4. Buffer Capacity: It is maximum when [Salt] = [Acid] or [Salt] = [Base]. At this point, pH = pK_a or pOH = pK_b.
5. Amphoteric Solvents: Water behaves as an acid with NH₃ and as a base with HCl.
6. Dilution Effect: On dilution, pH of an acidic solution increases towards 7, and pH of a basic solution decreases towards 7.
7. Very Dilute Acid: For 10^-8 M HCl, pH is NOT 8. You must add contribution from water (10^-7 M). pH ≈ 6.98.
8. Mixing Equal Volumes: Mixing pH 2 and pH 3 acids does not give pH 2.5. Since pH is log scale, resulting pH ≈ 2.26.
9. Salt of WA + WB: The pH is independent of concentration. It only depends on pK_a and pK_b.
10. Precipitation Condition: Precipitation occurs ONLY when Ionic Product (Q_sp) > Solubility Product (K_sp).
11. Group II Analysis: HCl is added before H₂S to suppress S^2- ion concentration so only Group II sulphides (low K_sp) precipitate.
12. Group IV Analysis: NH₄OH is added in excess of NH₄Cl to increase OH^- just enough to precipitate Group IV hydroxides but not Mg(OH)₂.
13. Blood Buffer: The pH of human blood is maintained at ~7.4 by the H₂CO₃ / HCO₃^- buffer system.
14. Lewis Concept: BF₃, AlCl₃ are Lewis Acids (Electron pair acceptors). NH₃, H₂O are Lewis Bases (Electron pair donors).
15. Polyprotic Acids: For H₃PO₄, K_a1 > K_a2 > K_a3. Removal of H⁺ from a negative ion is harder.
16. Solubility vs pH: Solubility of salts of weak acids (like CaF₂) increases in acidic pH because H⁺ consumes the anion (F^-).
17. Highest pH: Among 0.1 M solutions of NaCl, NH₄Cl, and NaCN, NaCN has the highest pH (Basic hydrolysis of CN^-).
18. Neutral Point: Neutral point is where [H⁺] = [OH^-]. At 90°C, neutral pH is ~6, not 7.
19. Indicator Range: Methyl Orange (3.1-4.4, Acidic range). Phenolphthalein (8.3-10, Basic range).
20. Simultaneous Solubility: If two salts share a common ion (e.g., AgCl and AgBr), the solubility of both decreases more than if they were alone.

📱 Practice MCQs for this topic inside our App