The Hook: Why Do Some Solutions Conduct Electricity?
Drop your phone in pure water—it might survive. Drop it in seawater—it’s toast! Why? Ions make the difference.
Tap water, salt water, and battery acid all conduct electricity, but pure water and sugar solution don’t. The secret lies in ionic equilibrium—the presence of free-moving ions in solution.
Real-world application: Your body’s nervous system works on ionic signals (Na⁺, K⁺). Sports drinks contain electrolytes to restore ionic balance. Even your car battery relies on ionic equilibrium!
JEE Reality: Ionic equilibrium is the bridge between physical chemistry and inorganic chemistry. Expect 3-4 questions combining concepts from acids, bases, salts, and equilibrium.
The Core Concept
What is Ionic Equilibrium?
Ionic equilibrium is the equilibrium established between ions and unionized molecules in a solution.
$$AB \rightleftharpoons A^+ + B^-$$This equilibrium exists for:
- Weak acids (CH₃COOH ⇌ H⁺ + CH₃COO⁻)
- Weak bases (NH₄OH ⇌ NH₄⁺ + OH⁻)
- Sparingly soluble salts (AgCl ⇌ Ag⁺ + Cl⁻)
In simple terms: Not all molecules break into ions. Some stay intact, some break apart, and an equilibrium exists between the two.
Electrolytes: Strong vs Weak
Strong Electrolytes
Definition: Substances that completely ionize in aqueous solution.
$$NaCl \xrightarrow{H_2O} Na^+ + Cl^-$$(100% ionization)
Examples:
- Strong acids: HCl, HNO₃, H₂SO₄, HClO₄
- Strong bases: NaOH, KOH, Ba(OH)₂
- Most salts: NaCl, KBr, MgSO₄
Characteristics:
- Degree of ionization (α) = 1 (or 100%)
- Good conductors of electricity
- No equilibrium exists (complete dissociation)
Weak Electrolytes
Definition: Substances that partially ionize in aqueous solution.
$$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$(only ~1% ionizes)
Examples:
- Weak acids: CH₃COOH, HF, H₃PO₄, HCN
- Weak bases: NH₄OH, organic amines
- Some salts: HgCl₂, CdCl₂
Characteristics:
- Degree of ionization (α) < 1 (typically 1-10%)
- Poor conductors compared to strong electrolytes
- Equilibrium exists between ionized and unionized forms
Strong electrolytes: Ionic bonds easily break in polar solvents like water (high lattice energy overcome by hydration energy)
Weak electrolytes: Covalent bonds partially break; most molecules stay intact
Degree of Ionization (α)
$$\alpha = \frac{\text{Number of moles dissociated}}{\text{Total number of moles taken}}$$Range: $0 \leq \alpha \leq 1$ (or 0% to 100%)
For Strong Electrolytes
- α = 1 (complete ionization)
- Example: HCl, NaCl, NaOH
For Weak Electrolytes
- α « 1 (partial ionization)
- Depends on:
- Nature of electrolyte (stronger acid → higher α)
- Concentration (dilution increases α)
- Temperature (usually increases α)
Ostwald’s Dilution Law
For weak electrolyte $AB \rightleftharpoons A^+ + B^-$ with initial concentration C:
$$K = \frac{C\alpha \cdot C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$$If α « 1 (very weak electrolyte):
$$\boxed{\alpha = \sqrt{\frac{K}{C}}}$$Key Insight: Dilution increases degree of ionization!
At infinite dilution, even weak acids are fully ionized (α → 1).
Memory Tricks & Patterns
Mnemonic for Strong Acids
“Heaven’s Clouds Shower Nitric Perchlorates Briskly”
- HCl - Hydrochloric acid
- HClO₄ - Perchloric acid
- H₂SO₄ - Sulfuric acid
- HNO₃ - Nitric acid
- HBr - Hydrobromic acid
- HI - Hydroiodic acid (bonus)
All others (like CH₃COOH, HF, H₃PO₄) are weak acids!
Mnemonic for Strong Bases
“Soluble Hydroxides: Group 1 + Ba, Sr, Ca (barely)”
- Group 1 (Li, Na, K, Rb, Cs): All strong bases
- Group 2: Ba(OH)₂ and Sr(OH)₂ strong, Ca(OH)₂ moderately strong, Mg(OH)₂ weak
All others (like NH₄OH, Al(OH)₃) are weak bases!
Pattern Recognition: Dilution Effect
| Property | Effect of Dilution (adding water) |
|---|---|
| Degree of ionization (α) | Increases |
| Number of ions | Increases (more molecules ionize) |
| Concentration of ions | Decreases (dilution effect) |
| Conductivity | Initially increases, then decreases |
Memory: “Dilution encourages ionization but spreads ions apart”
Ionization of Weak Acids
For weak acid HA:
$$HA \rightleftharpoons H^+ + A^-$$Ionization constant (Ka):
$$\boxed{K_a = \frac{[H^+][A^-]}{[HA]}}$$Calculating [H⁺] from Ka
For initial concentration C and degree of ionization α:
| HA | H⁺ | A⁻ | |
|---|---|---|---|
| I | C | 0 | 0 |
| C | -Cα | +Cα | +Cα |
| E | C(1-α) | Cα | Cα |
If α « 1:
$$K_a \approx C\alpha^2$$ $$\boxed{[H^+] = C\alpha = \sqrt{K_a \cdot C}}$$Examples of Ka Values
| Acid | Formula | Ka at 25°C | Strength |
|---|---|---|---|
| Hydrochloric | HCl | ~10⁷ | Very strong |
| Sulfuric (first) | H₂SO₄ | ~10² | Strong |
| Phosphoric (first) | H₃PO₄ | 7.5×10⁻³ | Moderately weak |
| Acetic | CH₃COOH | 1.8×10⁻⁵ | Weak |
| Carbonic (first) | H₂CO₃ | 4.3×10⁻⁷ | Very weak |
| Hydrocyanic | HCN | 6.2×10⁻¹⁰ | Extremely weak |
Pattern: Larger Ka → Stronger acid → More ionization
Ionization of Weak Bases
For weak base BOH:
$$BOH \rightleftharpoons B^+ + OH^-$$Ionization constant (Kb):
$$\boxed{K_b = \frac{[B^+][OH^-]}{[BOH]}}$$Calculating [OH⁻] from Kb
Similar to weak acids, if α « 1:
$$\boxed{[OH^-] = \sqrt{K_b \cdot C}}$$Examples of Kb Values
| Base | Formula | Kb at 25°C | Strength |
|---|---|---|---|
| Sodium hydroxide | NaOH | ~10⁶ | Very strong |
| Methylamine | CH₃NH₂ | 4.4×10⁻⁴ | Weak |
| Ammonia | NH₃ | 1.8×10⁻⁵ | Weak |
| Pyridine | C₅H₅N | 1.7×10⁻⁹ | Very weak |
Relationship Between Ka and Kb
For conjugate acid-base pair:
$$\boxed{K_a \times K_b = K_w}$$Where $K_w = 1.0 \times 10^{-14}$ at 25°C (ionic product of water)
Alternative form:
$$\boxed{pK_a + pK_b = 14}$$Why This Relationship?
For conjugate pair HA and A⁻:
$$HA \rightleftharpoons H^+ + A^-$$($K_a$)
$$A^- + H_2O \rightleftharpoons HA + OH^-$$($K_b$)
Adding these reactions:
$$H_2O \rightleftharpoons H^+ + OH^-$$($K_w$)
Therefore: $K_a \times K_b = K_w$
Given: $K_a$ of CH₃COOH = 1.8×10⁻⁵
Find: $K_b$ of CH₃COO⁻
Solution:
$$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$$This is frequently asked in JEE!
Common Ion Effect
Definition: The suppression of ionization of a weak electrolyte by adding a strong electrolyte containing a common ion.
Example: CH₃COOH + CH₃COONa
$$CH_3COOH \rightleftharpoons H^+ + CH_3COO^-$$(weak acid, partially ionized)
Add CH₃COONa (strong electrolyte):
$$CH_3COONa \xrightarrow{100\%} CH_3COO^- + Na^+$$Effect:
- High [CH₃COO⁻] from salt
- By Le Chatelier’s Principle, equilibrium shifts left
- Ionization of CH₃COOH suppressed
- [H⁺] decreases
Result: Solution becomes less acidic (pH increases)
General Rule
Adding common ion:
- Suppresses ionization of weak electrolyte
- Shifts equilibrium toward unionized form
- Decreases degree of ionization (α)
Isohydric Solutions
Definition: Solutions of different weak acids/bases having the same H⁺ or OH⁻ concentration.
Example: Mixing two weak acids HA₁ and HA₂
For HA₁: $[H^+] = \sqrt{K_{a1}C_1}$
For HA₂: $[H^+] = \sqrt{K_{a2}C_2}$
In mixture, both contribute to same [H⁺]:
Isohydric condition:
$$\sqrt{K_{a1}C_1} = \sqrt{K_{a2}C_2}$$Common Mistakes to Avoid
Wrong: Treating all acids as completely ionized
Correct:
- Only 7 common strong acids (HCl, HNO₃, H₂SO₄, HClO₄, HBr, HI, HClO₃)
- Rest are weak—use Ka and equilibrium expressions
JEE Trap: Given “0.1M HF solution”, students directly write [H⁺] = 0.1M
Correct approach: HF is weak, use $[H^+] = \sqrt{K_a \cdot C}$
Strong/Weak = Degree of ionization (α) Concentrated/Dilute = Amount of solute in solution
Examples:
- Dilute HCl (0.01M HCl): Strong acid, dilute solution
- Concentrated CH₃COOH (10M acetic acid): Weak acid, concentrated solution
Don’t confuse: A weak acid can be concentrated, and a strong acid can be dilute!
Common Error: “Adding CH₃COONa to CH₃COOH increases acidity”
Correct: Common ion (CH₃COO⁻) shifts equilibrium left, decreases [H⁺], increases pH (less acidic)
Memory: Common ion suppresses ionization of weak electrolyte
Interactive Demo: Visualize Equilibrium Dynamics
See how ionic equilibrium and common ion effect work in real-time.
Wrong: “Diluting a solution always increases its conductivity”
Correct: Conductivity depends on number of ions × concentration
- Initially, dilution increases α (more ions form) → conductivity increases
- Further dilution spreads ions apart → conductivity decreases
- Maximum conductivity at some intermediate dilution
Graph: Conductivity vs dilution shows a maximum!
Practice Problems
Level 1: Foundation (NCERT)
Classify as strong or weak electrolytes:
- HCl
- CH₃COOH
- NaOH
- NH₄OH
- KCl
Solution:
- HCl - Strong acid
- CH₃COOH - Weak acid
- NaOH - Strong base
- NH₄OH - Weak base
- KCl - Strong electrolyte (salt)
A weak acid with Ka = 1.0×10⁻⁵ is dissolved to make 0.1M solution. Calculate degree of ionization.
Solution: Using $\alpha = \sqrt{\frac{K_a}{C}}$:
$$\alpha = \sqrt{\frac{1.0 \times 10^{-5}}{0.1}} = \sqrt{1.0 \times 10^{-4}} = 0.01$$Answer: α = 0.01 or 1% ionization
Check: $\frac{K_a}{C} = 10^{-4} < 0.0025$, so approximation is valid ✓
Explain what happens when solid CH₃COONa is added to CH₃COOH solution.
Solution:
$$CH_3COOH \rightleftharpoons H^+ + CH_3COO^-$$CH₃COONa provides CH₃COO⁻ ions (common ion)
Effect:
- Equilibrium shifts left (Le Chatelier)
- Ionization of CH₃COOH suppressed
- [H⁺] decreases
- pH increases (less acidic)
This is the principle behind buffer solutions!
Level 2: JEE Main
A 0.1M weak acid solution is 2% ionized. Calculate Ka.
Solution: Given: C = 0.1M, α = 0.02
$$K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.1 \times (0.02)^2}{1-0.02}$$ $$K_a = \frac{0.1 \times 0.0004}{0.98} = \frac{4 \times 10^{-5}}{0.98} = 4.08 \times 10^{-5}$$Alternatively, if we use approximation (α « 1):
$$K_a \approx C\alpha^2 = 0.1 \times (0.02)^2 = 4 \times 10^{-5}$$(Very close! Approximation valid when α < 0.05)
A weak acid has degree of ionization 0.01 at 0.1M concentration. What is the degree of ionization when diluted to 0.01M?
Solution: Using $\alpha = \sqrt{\frac{K_a}{C}}$:
At C₁ = 0.1M: $\alpha_1 = 0.01$
$$K_a = C_1\alpha_1^2 = 0.1 \times (0.01)^2 = 1 \times 10^{-5}$$At C₂ = 0.01M:
$$\alpha_2 = \sqrt{\frac{K_a}{C_2}} = \sqrt{\frac{1 \times 10^{-5}}{0.01}} = \sqrt{1 \times 10^{-3}} = 0.0316$$Answer: α₂ = 0.0316 or 3.16%
Insight: 10× dilution → ~3× increase in degree of ionization!
Ka of HCN is 6.2×10⁻¹⁰. Calculate Kb of CN⁻.
Solution: For conjugate acid-base pair:
$$K_a \times K_b = K_w = 1.0 \times 10^{-14}$$ $$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-10}} = 1.61 \times 10^{-5}$$Interpretation: CN⁻ is a much stronger base (Kb = 1.61×10⁻⁵) than HCN is an acid (Ka = 6.2×10⁻¹⁰)
This is why NaCN solutions are basic!
Level 3: JEE Advanced
0.1M CH₃COOH (Ka = 1.8×10⁻⁵) solution has what [H⁺]?
If 0.1M CH₃COONa is added, what is new [H⁺]?
Solution:
Part 1: Pure CH₃COOH
$$[H^+] = \sqrt{K_a \cdot C} = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}}$$ $$[H^+] = 1.34 \times 10^{-3} \text{ M}$$Part 2: With 0.1M CH₃COONa
CH₃COONa completely ionizes: [CH₃COO⁻] = 0.1M (from salt)
Let additional [H⁺] from acid = x (very small)
| CH₃COOH | H⁺ | CH₃COO⁻ | |
|---|---|---|---|
| I | 0.1 | 0 | 0.1 |
| C | -x | +x | +x |
| E | 0.1-x | x | 0.1+x |
Since Ka is small and common ion present, x « 0.1:
$$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x \times 0.1}{0.1} = x$$ $$x = K_a = 1.8 \times 10^{-5} \text{ M}$$New [H⁺] = 1.8×10⁻⁵ M
Comparison:
- Without salt: [H⁺] = 1.34×10⁻³ M
- With salt: [H⁺] = 1.8×10⁻⁵ M
Reduction factor: ~75 times less acidic!
100 mL of 0.2M CH₃COOH (Ka₁ = 1.8×10⁻⁵) is mixed with 100 mL of 0.2M HCN (Ka₂ = 6.2×10⁻¹⁰). Calculate [H⁺] in the mixture.
Solution:
After mixing:
- [CH₃COOH] = 0.1M
- [HCN] = 0.1M
Since Ka₁ » Ka₂, CH₃COOH is much stronger and dominates H⁺ production.
HCN’s contribution is negligible. Common ion effect (H⁺ from CH₃COOH) suppresses HCN ionization even more.
Approximate calculation: Treat as pure CH₃COOH solution
$$[H^+] \approx \sqrt{K_{a1} \cdot C_1} = \sqrt{1.8 \times 10^{-5} \times 0.1} = 1.34 \times 10^{-3} \text{ M}$$More accurate calculation:
Total [H⁺] = x
$$K_{a1} = \frac{x \cdot x}{0.1 - x} \approx \frac{x^2}{0.1}$$(dominant)
$$x^2 = 1.8 \times 10^{-6}$$ $$x = 1.34 \times 10^{-3} \text{ M}$$HCN contribution check:
$$K_{a2} = \frac{[H^+][CN^-]}{[HCN]} = \frac{1.34 \times 10^{-3} \times [CN^-]}{0.1}$$ $$[CN^-] = \frac{6.2 \times 10^{-10} \times 0.1}{1.34 \times 10^{-3}} = 4.6 \times 10^{-8}$$M
Negligible compared to 1.34×10⁻³ M ✓
Answer: [H⁺] = 1.34×10⁻³ M (dominated by CH₃COOH)
For H₂S: Ka₁ = 1×10⁻⁷, Ka₂ = 1.3×10⁻¹⁴
Calculate [H⁺], [HS⁻], and [S²⁻] in 0.1M H₂S solution.
Solution:
First ionization:
$$H_2S \rightleftharpoons H^+ + HS^-$$(Ka₁)
$$[H^+] = \sqrt{K_{a1} \cdot C} = \sqrt{1 \times 10^{-7} \times 0.1} = \sqrt{1 \times 10^{-8}}$$ $$[H^+] = 1 \times 10^{-4} \text{ M}$$Since ionization is small, [HS⁻] ≈ [H⁺]:
$$[HS^-] = 1 \times 10^{-4} \text{ M}$$Second ionization:
$$HS^- \rightleftharpoons H^+ + S^{2-}$$(Ka₂)
High [H⁺] from first ionization suppresses second ionization heavily.
$$K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]} = \frac{1 \times 10^{-4} \times [S^{2-}]}{1 \times 10^{-4}}$$ $$[S^{2-}] = K_{a2} = 1.3 \times 10^{-14} \text{ M}$$Summary:
- [H⁺] = 1×10⁻⁴ M
- [HS⁻] = 1×10⁻⁴ M
- [S²⁻] = 1.3×10⁻¹⁴ M (extremely small!)
Key Insight: For polyprotic weak acids, first ionization dominates. Second ionization contributes negligibly to [H⁺].
Quick Revision Box
| Concept | Formula/Rule |
|---|---|
| Strong electrolyte | α = 1 (100% ionization) |
| Weak electrolyte | α « 1, use equilibrium expressions |
| Degree of ionization (weak) | $\alpha = \sqrt{K/C}$ (when α « 1) |
| Weak acid ionization | $K_a = \frac{[H^+][A^-]}{[HA]}$ |
| Weak base ionization | $K_b = \frac{[B^+][OH^-]}{[BOH]}$ |
| Conjugate pair relation | $K_a \times K_b = K_w = 10^{-14}$ |
| [H⁺] from weak acid | $[H^+] = \sqrt{K_a \cdot C}$ |
| [OH⁻] from weak base | $[OH^-] = \sqrt{K_b \cdot C}$ |
| Common ion effect | Suppresses ionization of weak electrolyte |
| Dilution effect | Increases degree of ionization (α increases) |
When to Use This
Is it a strong electrolyte?
- Strong acid: HCl, HNO₃, H₂SO₄, HClO₄, HBr, HI
- Strong base: Group 1 hydroxides, Ba(OH)₂, Sr(OH)₂
- Most salts
YES → Direct calculation
- [H⁺] = Cacid for strong acids
- [OH⁻] = Cbase for strong bases
NO → Use equilibrium approach
- Write equilibrium expression
- Use Ka or Kb
- Apply ICE table if needed
- Consider common ion effect if salt present
Connection to Other Topics
Within Equilibrium Chapter
- Chemical Equilibrium - Uses same K, Q, ICE table concepts
- Le Chatelier’s Principle - Common ion effect is Le Chatelier applied
- Acids and Bases - Foundation for acid-base theories
- pH and Buffer Solutions - Quantitative applications of Ka, Kb
- Solubility Product - Ionic equilibrium for sparingly soluble salts
Electrochemistry Links
- Electrode Potentials - Ion concentrations affect E
- Nernst Equation - EMF depends on ion concentration
- Electrochemical Cells - Electrolyte solutions in cells
- Electrolysis - Ionic conduction in solutions
Solutions and Thermodynamics
- Solutions - Concentration Methods - Molarity and concentration calculations
- Colligative Properties - van’t Hoff factor for electrolytes
- Thermodynamics - Gibbs Energy - ΔG and spontaneity of ionization
Practical Applications
- Qualitative Analysis - Testing for ions
- Volumetric Analysis - Acid-base titrations
Teacher’s Summary
- Strong vs weak = degree of ionization - Not concentration!
- Only 7 common strong acids - Rest are weak, use Ka
- Ostwald’s law: α = √(K/C) - Dilution increases ionization
- Ka × Kb = Kw for conjugate pairs - High-yield JEE formula
- Common ion suppresses ionization - Decreases α, shifts equilibrium
“Ionic equilibrium is all about the battle between staying together and breaking apart. Strong electrolytes surrender completely; weak ones keep fighting!”
JEE Strategy:
- Memorize the 7 strong acids (everything else is weak!)
- Master Ka-Kb conversion using Kw
- Common ion effect appears in every ionic equilibrium chapter
- Polyprotic acids: First ionization dominates (Ka₁ » Ka₂)
- Link to buffer problems—this is the foundation!
What’s Next?
Now that you understand ionic equilibrium, dive deeper into acid-base theories:
- Acids and Bases - Arrhenius, Bronsted-Lowry, and Lewis concepts
- pH and Buffer Solutions - Quantitative pH calculations and buffer design
- Solubility Product - Ionic equilibrium of sparingly soluble salts