Ionic Equilibrium – Comprehensive Notes

Ionic equilibrium is a crucial chapter in physical chemistry, dealing with the equilibrium established between ions and undissociated molecules in aqueous solutions, particularly involving weak acids, weak bases, and sparingly soluble salts. A strong grasp of this topic is essential for quantitative analysis and forms the basis for many other concepts in chemistry for NEET and JEE Main.

1. Introduction to Electrolytes

Electrolytes: Substances that dissociate or ionize in a solvent (usually water) to produce ions, thereby conducting electricity.
- Strong Electrolytes: Ionize almost completely (close to 100%) in aqueous solutions. The equilibrium lies far to the right, mostly existing as ions.
  - Examples: Strong acids (HCl, H2SO4, HNO3), Strong bases (NaOH, KOH, Ca(OH)2), Most salts (NaCl, KNO3, CuSO4).
- Weak Electrolytes: Ionize only partially in aqueous solutions, establishing an equilibrium between undissociated molecules and ions.
  - Examples: Weak acids (CH3COOH, H2CO3, HCN), Weak bases (NH4OH, Mg(OH)2), Water (H2O).
Non-electrolytes: Substances that do not ionize in solution and hence do not conduct electricity.
- Examples: Glucose (C6H12O6), Urea (NH2CONH2), Sucrose (C12H22O11).

2. Concepts of Acids and Bases

Several theories define acids and bases:

2.1. Arrhenius Concept (1884):
- Acid: A substance that dissociates in water to give hydrogen ions (H+). (Later understood as H3O+ or hydronium ions).
  - Example: HCl(aq) -> H+(aq) + Cl-(aq) or HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)
- Base: A substance that dissociates in water to give hydroxide ions (OH-).
  - Example: NaOH(aq) -> Na+(aq) + OH-(aq)
- Limitations:
  - Applicable only to aqueous solutions.
  - Could not explain the basicity of substances like NH3, which do not contain OH- groups.
  - Could not explain the acidic nature of CO2, SO2, which do not contain H+ ions.
2.2. Bronsted-Lowry Concept (1923):
- Acid: A substance (molecule or ion) that can donate a proton (H+). (Proton Donor).
- Base: A substance (molecule or ion) that can accept a proton (H+). (Proton Acceptor).
- Conjugate Acid-Base Pairs: When an acid donates a proton, it forms its conjugate base. When a base accepts a proton, it forms its conjugate acid. These differ by only one proton.
  - Acid – H+ = Conjugate Base
  - Base + H+ = Conjugate Acid
  - Example: CH3COOH (Acid 1) + H2O (Base 2) <=> CH3COO- (Conjugate Base 1) + H3O+ (Conjugate Acid 2)
  - Example: NH3 (Base 1) + H2O (Acid 2) <=> NH4+ (Conjugate Acid 1) + OH- (Conjugate Base 2)
- Amphoteric/Amphiprotic Substances: Substances that can act as both Bronsted acids and Bronsted bases (e.g., H2O, HCO3-, HSO4-).
- Strength of Conjugate Pairs: Strong acids have weak conjugate bases, and weak acids have strong conjugate bases. (Similarly for bases).
2.3. Lewis Concept (1923):
- Acid: An electron-pair acceptor. (Often contains an empty orbital).
  - Examples: H+, BF3, AlCl3, CO2 (can accept lone pair from O).
- Base: An electron-pair donor. (Often contains a lone pair of electrons).
  - Examples: NH3, H2O, OH-, Cl-.
- Formation of Adduct: The bond formed is a coordinate (dative) bond.
  - Example: NH3 + BF3 -> H3N:BF3 (Ammonia is Lewis base, BF3 is Lewis acid).
- Advantages: Broader concept, explains reactions that do not involve proton transfer (e.g., anhydrous reactions). Explains the acidity of electron-deficient molecules and basicity of species with lone pairs.

3. Ionization of Weak Acids and Bases (Ionic Equilibria)

3.1. Degree of Ionization (α):
- The fraction of the total number of molecules of an electrolyte that ionizes in solution.
- For strong electrolytes, α≈1.
- For weak electrolytes, 0<α<1.
- α increases with dilution and with increasing temperature.
3.2. Ostwald’s Dilution Law:
- Applicable to weak electrolytes. Relates the degree of ionization (α) to the concentration (C) and ionization constant (Ka or Kb).
- For a weak acid (HA): HA <=> H+ + A- Initial Conc: C 0 0 At Equilibrium: C(1-α) C$\alpha$ C$\alpha$ Ka=[H+][A−]/[HA]=(Cα×Cα)/C(1−α)=Cα2/(1−α) If α is very small (for very weak acids, α<<1), then (1−α)≈1. So, Ka≈Cα2⇒α=Ka/C
- For a weak base (BOH): BOH <=> B+ + OH- Kb=[B+][OH−]/[BOH]=Cα2/(1−α) If α is very small, α=Kb/C
- Implication: As concentration (C) decreases (i.e., dilution increases), α increases.
3.3. Ionization Constants:
- Acid Dissociation Constant (Ka): For a weak acid, HA. Ka=[H3O+][A−]/[HA] (or [H+][A−]/[HA]) Higher Ka implies stronger acid.
- Base Dissociation Constant (Kb): For a weak base, BOH. Kb=[B+][OH−]/[BOH] Higher Kb implies stronger base.
- Relationship between Ka and Kb for Conjugate Acid-Base Pairs: For a conjugate acid-base pair (e.g., HA and A-): Ka×Kb=Kw Where Kw is the ionic product of water.

4. Ionic Product of Water (Kw) and pH Scale

4.1. Ionic Product of Water (Kw):
- Water undergoes auto-ionization: H2O(l) + H2O(l) <=> H3O+(aq) + OH-(aq)
- Kw=[H3O+][OH−] (or [H+][OH−])
- At 25°C, Kw=1.0×10−14
- Kw increases with temperature.
4.2. pH and pOH Scale:
- pH: Negative logarithm (base 10) of the hydrogen ion concentration (in mol/L). pH=−log[H+] or pH=−log[H3O+]
- pOH: Negative logarithm (base 10) of the hydroxide ion concentration (in mol/L). pOH=−log[OH−]
- Relationship: pH+pOH=pKw At 25°C, pH+pOH=14
- Nature of Solution based on pH (at 25°C):
  - Acidic solution: pH<7
  - Basic (Alkaline) solution: pH>7
  - Neutral solution: pH=7
4.3. pH Calculations:
- Strong Acids: [H+]=Concentration of acid. (e.g., for 0.01 M HCl, [H+] = 0.01 M, pH = 2).
- Strong Bases: [OH−]=Concentration of base. (e.g., for 0.01 M NaOH, [OH-] = 0.01 M, pOH = 2, pH = 12).
- Weak Acids: [H+]=KaC (approximation, if α is small). Or using quadratic if α is not small.
  - α=[H+]/C or α=Ka/C
- Weak Bases: [OH−]=KbC (approximation, if α is small). Or using quadratic if α is not small.
  - α=[OH−]/C or α=Kb/C

5. Hydrolysis of Salts

Definition: The reaction of an ion (cation or anion, or both) derived from a salt with water, resulting in the production of H3O+ or OH- ions, which changes the pH of the solution.
Hydrolysis Constant (Kh): The equilibrium constant for a hydrolysis reaction.
Degree of Hydrolysis (h): The fraction of the salt that undergoes hydrolysis at equilibrium.
5.1. Types of Salts and pH of their Solutions:
- 1. Salt of Strong Acid and Strong Base (e.g., NaCl, KNO3):
  - Ions (Na+, Cl-) do not react with water.
  - Solution is neutral (pH=7).
  - No hydrolysis. Kh is not defined.
- 2. Salt of Weak Acid and Strong Base (e.g., CH3COONa, KCN):
  - Anion (e.g., CH3COO-) undergoes hydrolysis: CH3COO- + H2O <=> CH3COOH + OH-
  - Cation (Na+) does not hydrolyze.
  - Solution is basic (pH>7).
  - Kh=Kw/Ka
  - Degree of hydrolysis, h=Kh/C=Kw/(KaC)
  - pH=7+(1/2)pKa+(1/2)logC
- 3. Salt of Strong Acid and Weak Base (e.g., NH4Cl, FeCl3):
  - Cation (e.g., NH4+) undergoes hydrolysis: NH4+ + H2O <=> NH4OH + H+
  - Anion (Cl-) does not hydrolyze.
  - Solution is acidic (pH<7).
  - Kh=Kw/Kb
  - Degree of hydrolysis, h=Kh/C=Kw/(KbC)
  - pH=7−(1/2)pKb−(1/2)logC
- 4. Salt of Weak Acid and Weak Base (e.g., CH3COONH4, NH4CN):
  - Both cation (NH4+) and anion (CH3COO-) undergo hydrolysis.
  - The nature of the solution (acidic, basic, or neutral) depends on the relative strengths of Ka and Kb of the parent weak acid and weak base.
  - If Ka=Kb, solution is neutral.
  - If Ka>Kb, solution is acidic.
  - If Kb>Ka, solution is basic.
  - Kh=Kw/(Ka×Kb)
  - Degree of hydrolysis, h=Kh (independent of concentration, for monovalent ions).
  - pH=7+(1/2)pKa−(1/2)pKb

6. Buffer Solutions

Definition: A solution that resists change in its pH upon the addition of small amounts of strong acid or strong base.
Buffer Action: The ability of a buffer to maintain a relatively constant pH due to the presence of both an acid and a base component that can neutralize added H+ or OH- ions.
6.1. Types of Buffer Solutions:
- 1. Acidic Buffer: A mixture of a weak acid and its salt with a strong base.
  - Examples: CH3COOH + CH3COONa; H2CO3 + NaHCO3.
  - Mechanism:
    - When strong acid (H+) is added: H+ + A- (from salt) -> HA (weak acid)
    - When strong base (OH-) is added: OH- + HA (weak acid) -> A- (from salt) + H2O
  - Henderson-Hasselbalch Equation: For an acidic buffer: pH=pKa+log([Salt]/[Acid]) Where pKa=−logKa.
- 2. Basic Buffer: A mixture of a weak base and its salt with a strong acid.
  - Examples: NH4OH + NH4Cl; Pyridine + Pyridinium chloride.
  - Mechanism:
    - When strong acid (H+) is added: H+ + BOH (weak base) -> B+ (from salt) + H2O
    - When strong base (OH-) is added: OH- + B+ (from salt) -> BOH (weak base)
  - Henderson-Hasselbalch Equation: For a basic buffer: pOH=pKb+log([Salt]/[Base]) Where pKb=−logKb. Then calculate pH using pH=14−pOH.
6.2. Buffer Capacity (β):
- The ability of a buffer solution to resist changes in pH. Quantitatively, it is defined as the number of moles of acid or base required to change the pH of 1 liter of buffer solution by one unit.
- Maximum buffer capacity is achieved when [Salt]=[Acid] (or [Salt]=[Base]), i.e., when pH=pKa (or pOH=pKb).
- A buffer is generally effective within a pH range of pKa±1.

7. Solubility Product (Ksp)

Definition: For a sparingly soluble salt, the solubility product constant (Ksp) is the product of the molar concentrations of its constituent ions, each raised to the power of its stoichiometric coefficient in the balanced dissociation equilibrium.
General form for a salt AxBy: AxBy(s)<=>xAy+(aq)+yBx−(aq) Ksp=[Ay+]x[Bx−]y
- Only for sparingly soluble salts in equilibrium with their solid phase.
- Pure solids are not included in the expression.
7.1. Solubility (s):
- The molar concentration of the metal cation (or anion, depending on stoichiometry) in a saturated solution of the salt.
- Relationship between Solubility (s) and Ksp:
  - For AB(s)<=>A+(aq)+B−(aq) Ksp=[A+][B−]=s×s=s2⇒s=Ksp
  - For AB2(s)<=>A2+(aq)+2B−(aq) Ksp=[A2+][B−]2=s×(2s)2=4s3⇒s=3Ksp/4
  - For A2B(s)<=>2A+(aq)+B2−(aq) Ksp=[A+]2[B2−]=(2s)2×s=4s3⇒s=3Ksp/4
  - For AxBy(s)<=>xAy+(aq)+yBx−(aq) Ksp=(xs)x(ys)y=xxyysx+y⇒s=(x+y)Ksp/(xxyy)
7.2. Common Ion Effect on Solubility:
- The solubility of a sparingly soluble salt is significantly decreased by the presence of a common ion (an ion already present in the solution) from another source (e.g., from a soluble salt or acid/base).
- Explanation (Le Chatelier’s Principle): Adding a common ion shifts the dissolution equilibrium of the sparingly soluble salt to the left, favoring precipitation and reducing its solubility.
  - Example: Solubility of AgCl (s) in water vs. in NaCl solution. AgCl (s) <=> Ag+(aq) + Cl-(aq) Adding NaCl provides common Cl- ions, shifting the equilibrium to the left, decreasing [Ag+] (solubility of AgCl).
7.3. Conditions for Precipitation:
- Ionic Product (Q): Similar in form to Ksp but calculated using instantaneous (non-equilibrium) concentrations of ions in solution.
- Criteria for Precipitation:
  - If Q>Ksp: The solution is supersaturated, and precipitation will occur until equilibrium is reached.
  - If Q<Ksp: The solution is unsaturated, and no precipitation will occur. More salt can dissolve.
  - If Q=Ksp: The solution is saturated, and the system is at equilibrium. No net precipitation or dissolution.

This comprehensive set of notes on Ionic Equilibrium covers all the major concepts and calculation types typically tested in NEET and JEE Main. Pay close attention to the definitions, relationships between constants (Ka, Kb, Kw, Kh, Ksp), and the application of principles like Le Chatelier’s Principle to solve problems on buffers and solubility. Practice a wide variety of numerical problems to master this chapter.