Ionic Equilibrium

1. Arrhenius Acids

• Acids: Substances that dissociate in water to give hydrogen ions [H⁺].
• Ionization of HA:
  HA(aq) ⇌ H⁺(aq) + A⁻(aq)
  or
  HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)
• Classification:
  • Monobasic/Monoprotic (HA): Release one H⁺ per molecule. (e.g., HCl, HBr, HI, CH₃COOH, HNO₃)
  • Dibasic/Diprotic (H₂A): Release two H⁺ per molecule. (e.g., H₂SO₄, H₂SO₃, H₃PO₃, H₂C₂O₄, H₂S)
  • Tribasic/Triprotic (H₃A): Release three H⁺ per molecule. (e.g., H₃PO₄, H₃AsO₄)

2. Arrhenius Bases

• Bases: Substances that produce hydroxyl ions [OH⁻] in water.
• Ionization: BOH(aq) ⇌ B⁺(aq) + OH⁻(aq)
• Classification:
  • Monoacidic (BOH): Release one OH⁻ per molecule. (e.g., NaOH, KOH, RbOH, CsOH, NH₄OH)
  • Diacidic [B(OH)₂]: Release two OH⁻ per molecule. (e.g., Mg(OH)₂, Ca(OH)₂, Zn(OH)₂)
  • Triacidic [B(OH)₃]: Release three OH⁻ per molecule. (e.g., Al(OH)₃, Fe(OH)₃)

3. Bronsted–Lowry Concept

• Acids: Proton (H⁺) donors.
• Bases: Proton (H⁺) acceptors.

• NH₃ acts as a base by adding a proton; H₂O acts as an acid by losing a proton.

4. Conjugate Acid-Base Pair

• A pair of an acid and a base that differs only by one proton (H⁺) is a conjugate acid-base pair.

• Stronger Bronsted acids have weaker conjugate bases, and vice-versa.

5. The Lewis Concept

• Lewis Acid: Any molecule or ion that can accept an electron pair to form a coordinate bond (e.g., BF₃).
• Lewis Base: Any molecule or ion that has a lone pair of electrons to donate (e.g., NH₃).
• The reaction involves the formation of a coordinate bond:

NH₃ (Lewis base) donates its lone pair to BF₃ (Lewis acid).

6. Arrhenius Concept: Monobasic Acids

• Monobasic acids (acids providing one H⁺ ion):

H₃PO₂ has only one ionisable (P–OH) H. H₃BO₃ is monobasic because it accepts OH⁻ from water (Lewis acid): B(OH)₃ + H₂O → [B(OH)₄]⁻ + H⁺.

7. Ionic Product of Water (K_w)

• Water dissociation: H₂O ⇌ H⁺ + OH⁻
• Dissociation constant: K_a = [H⁺][OH⁻][H₂O]
• K_w = K_a[H₂O] = [H⁺][OH⁻]
• At 25°C: K_w = 10⁻¹⁴
• K_a = K_w[H₂O] = 10⁻¹⁴55.55

8. pH Scale & Neutrality

• pH is defined as pH = −log[H⁺].
• At 25°C:
  • H₂O ⇌ H⁺ + OH⁻
  • K_w = [H⁺][OH⁻] = 10⁻¹⁴
  • At neutral point, [H⁺] = [OH⁻] = 10⁻⁷ M.
  • pH = −log(10⁻⁷) = 7 (Neutral).
• At Temperature > 25°C (e.g., K_w = 10⁻¹²):
  • [H⁺] = √(K_w) = 10⁻⁶ M.
  • pH = −log(10⁻⁶) = 6 (Neutral).
  • pH < 6 is acidic; pH > 6 is basic.
  • Note: A solution with pH = 7 at this temperature is basic.
• At Temperature < 25°C (e.g., K_w = 10⁻¹⁶):
  • [H⁺] = √(K_w) = 10⁻⁸ M.
  • pH = −log(10⁻⁸) = 8 (Neutral).
  • pH < 8 is acidic; pH > 8 is basic.
  • Note: A solution with pH = 7 at this temperature is acidic.

9. pH of Strong Monobasic Acids

• For a solution of concentration C M HCl:

• Total H⁺ concentration:

10. pH of Extremely Dilute Strong Acid Solutions

• For very dilute solutions (e.g., [HCl] ≤ 10^-6 M), the contribution of H⁺ from water cannot be neglected.
• [H⁺]_total = [H⁺]_acid + [H⁺]_H2O

Case : 10^-7 M HCl

• [H⁺]_acid = 10^-7 M, [H⁺]_water = 10^-7 M
• [H⁺]_total = 10^-7 + 10^-7 = 2 × 10^-7 M
• pH = -log(2 × 10^-7) ≈ 6.7 (rough; adding 10^-7 from water over-counts because the acid suppresses water's ionisation)

Case : Precise calculation (x = [OH^-] from water)

• K_w = [H⁺][OH^-] = 10^-14
• (x + 10^-7)(x) = 10^-14
• pH = -log(x + 10^-7) ≈ 6.79

Case : 10^-8 M HCl

• (x + 10^-8)(x) = 10^-14
• pH = -log(x + 10^-8) ≈ 6.98

Case : Very extreme dilution (≤ 10^-10 M)

• When [HCl] ≤ 10^-10 M, [H⁺]_total ≈ [H⁺]_water ≈ 10^-7 M.
• Therefore, pH ≈ 7.

11. pH Calculation for Strong Bases

• [OH^-]_total = [OH^-]_base + [OH^-]_water
• If [OH^-] ≥ 10^-6 M, H₂O contribution is negligible.
• If [OH^-] < 10^-6 M, H₂O contribution must be considered.

12. Relation between pH and pOH

• pH = 14 - pOH

13. pH Calculation of Strong Acid/Base Mixtures

• For a mixture of strong acids, the resulting [H⁺] is calculated as the sum of total moles of H⁺ ions divided by the total volume of the solution.
• Molarity (M) × V_(L) = moles of solute; M × V_(mL) = mmol of solute.

14. pH Calculation of Weak Acid (Monoprotic)

• For a weak acid like CH₃COOH: CH₃COOH ⇌ CH₃COO⁻ + H⁺
• If the concentration of H⁺ from the acid (Cα) is ≥ 10⁻⁶ M, the contribution of H⁺ from water is neglected:
  pH = −log(Cα)
• If Cα < 10⁻⁶ M, the contribution of H⁺ from water (x) must be considered:
  pH = −log(x + Cα)

  where (x + Cα)x = 10⁻¹⁴

15. Method Selection for Weak Acids

16. Weak Acid and Base Equilibria

For Monobasic Weak Acid :

• Dissociation: HA ⇌ H⁺ + A^-
• Degree of dissociation (α) = √K_a√C
• Dissociation constant: K_a = Cα²1 - α ≈ Cα²
• [H⁺] = Cα = √K_aC
• pH = -log(√K_aC) = 12 [pK_a - log C]

For Monoacidic Weak Base :

• Dissociation: AOH ⇌ A⁺ + OH^-
• Degree of dissociation (α) = √K_b√C
• [OH^-] = Cα = √K_bC
• pOH = -log(√K_bC) = 12 [pK_b - log C]
• pH = 14 - pOH

17. pH Calculation of Weak Diprotic Acids

• For a weak diprotic acid H₂A (where K_a2 << K_a1):
• Dissociation steps:
• H₂A ⇌ HA⁻ + H⁺ ; K_a1
• HA⁻ ⇌ A²⁻ + H⁺ ; K_a2
• If α₁ = √(K_a1/C) ≤ 0.05 (approximation valid):

• If α₁ > 0.05, solve K_a1 = Cα₁²1−α₁ for α₁, then pH = -log(Cα₁).

18. Calculation of [A²⁻] and α₂

• For diprotic acid equilibrium:
• [H⁺]_total = Cα₁ + Cα₁α₂ ≈ Cα₁
• K_a1 ≈ Cα₁² (assuming α₁ is small).
• From the second dissociation HA⁻ ⇌ A²⁻ + H⁺:
• K_a2 = [A²⁻][H⁺][HA⁻] = (Cα₁α₂)(Cα₁)(Cα₁) = Cα₁α₂.

• Therefore, for a weak diprotic acid, the concentration of the dianion [A²⁻] is approximately equal to K_a2.

19. pH of a Mixture of Two Weak Acids

• Consider a mixture of C₁ M HA (K_a1) and C₂ M HB (K_a2).
• Equilibria:

• [H⁺]_total = C₁α₁ + C₂α₂.
• Using the law of mass action and the approximation α ≪ 1:

20. Degree of Dissociation in Mixture

• In a mixture, the dissociation of each acid is suppressed by the common ion [H⁺].
• The ratio of degrees of dissociation (d.o.d) for the two acids is:

• The individual d.o.d values in the mixture are given by:

21. Total [H⁺] Calculation

• [H⁺]_acid = C₁α₁ + C₂α₂ = √(K_a1C₁ + K_a2C₂).
• Considering the contribution of water (K_w):

22. Salt Hydrolysis Basics

• For NaCl (Salt of Strong Acid + Strong Base):

• Na⁺ comes from Base, Cl⁻ comes from Acid.
• Neither Na⁺_(aq) + H₂O nor Cl⁻_(aq) + H₂O undergo hydrolysis.

23. Relative pH of Salt Solutions

• Comparing 0.1 M solutions at 25°C:
• NaCl (SA + SB): pH = 7.
• NH₄Cl (WB + SA): NH₄⁺ + H₂O ⇌ NH₄OH + H⁺; pH < 7.
• CH₃COOH (weak acid): pH much lower than 7.
• CH₃COONH₄ (WA + WB): pH = 7 (given K_a(CH₃COOH) ≃ K_b(NH₄OH) ≃ 1.8 × 10⁻⁵).
• K₂S: K₂S → 2K⁺ + S²⁻. S²⁻ undergoes anionic hydrolysis producing OH⁻, so pH > 7.

24. Salt of Weak Base + Strong Acid (e.g., NH₄Cl)

• Dissociation: NH₄Cl_(aq) → NH₄⁺_(aq) + Cl⁻_(aq)
• Hydrolysis: NH₄⁺_(aq) + H₂O ⇌ NH₄OH + H⁺
• Let h be degree of hydrolysis (h << 1).

• h = √(K_w / (C ⋅ K_b))
• [H⁺] = Ch = √(K_w ⋅ C / K_b)
• pH = −log√(K_w ⋅ C / K_b) = 7 − 12pK_b − 12log C

Note: If h > 0.05, use K_h = K_w / K_b = Ch² / (1 − h).

25. Salt of Weak Acid and Strong Base

• Example: CH₃COONa (sodium acetate).
• Dissociation: CH₃COONa(aq) → CH₃COO⁻(aq) + Na⁺(aq).
• Anionic Hydrolysis: CH₃COO⁻(aq) + H₂O ⇌ CH₃COOH + OH⁻.

26. Salt of Weak Acid and Weak Base

• Degree of hydrolysis (h) and pH are independent of the concentration of the salt solution.
• Nature of solution at 25°C:
  • If K_a > K_b, solution is acidic (pH < 7).
  • If K_b > K_a, solution is basic (pH > 7).
  • If K_a = K_b, solution is neutral (pH = 7).

27. Neutral Salt Solutions

• A salt solution can be neutral due to:
  1. No cationic or anionic hydrolysis (e.g., NaCl).
  2. Hydrolysis of both cation and anion, where the parent weak acid and weak base have equal K_a and K_b values.
• When K_a = K_b: [H⁺] = √K_w K_aK_b = √K_w = 10⁻⁷ M ⇒ pH = 7.

28. Salt Hydrolysis Summary

29. Salt of a Polyanion (e.g., Na₃PO₄)

• Dissociation of H₃PO₄: H₃PO₄ ⇌ H₂PO₄⁻ + H⁺ (K_a1); H₂PO₄⁻ ⇌ HPO₄²⁻ + H⁺ (K_a2); HPO₄²⁻ ⇌ PO₄³⁻ + H⁺ (K_a3).
• K_a1 > K_a2 > K_a3.

30. Salt of an Amphiprotic Anion (e.g., NaHCO₃)

• Dissociation: H₂CO₃ ⇌ HCO₃⁻ + H⁺ (K_a1); HCO₃⁻ ⇌ CO₃²⁻ + H⁺ (K_a2).

31. Acidic Salts and pH Calculation

• For a salt like Na₂HPO₄, the anion is HPO₄²⁻.
• The dissociation equilibrium for the species derived from H₃PO₄:

• For the amphiprotic species HPO₄²⁻, the [H⁺] and pH are given by:

32. Properties of Na₂HPO₃

• H₃PO₃ is a dibasic acid.
• HPO₃²⁻ is not amphiprotic.
• For Na₂HPO₃, the pH is given by:

33. Titration of H₃PO₄ with NaOH

• Example: 100 mL of 0.1 M H₃PO₄ titrated with 0.1 M NaOH.

i) 0 mL NaOH :

• pH = -log√(K_a1 C)

ii) 100 mL NaOH :

• Reaction: H₃PO₄ + NaOH → NaH₂PO₄ + H₂O
• pH = pK_a1 + pK_a22

iii) 200 mL NaOH :

• Reaction: NaH₂PO₄ + NaOH → Na₂HPO₄ + H₂O
• pH = pK_a2 + pK_a32

iv) 300 mL NaOH :

• Reactions involve the complete neutralisation to Na₃PO₄.
• At the final stage, the hydrolysis of PO₄³⁻ dominates.
• pH = -log √(K_wK_a3 / C_eff), where C_eff is the concentration of the salt in the total volume.

34. Buffer Solution

• A buffer solution resists pH change on the addition of a small amount of strong acid or strong base.

35. Acidic Buffer (WA + C.B)

• Composed of a Weak Acid and its Conjugate Base.

• Effect of adding strong acid (x mol): Amount of Conjugate Base (C.B.) decreases by x, amount of Weak Acid increases by x.
• Effect of adding strong base (OH^-, y mol): Amount of Acid decreases by y, amount of C.B. increases by y.
• [CH₃COO^-]_total ≃ [CH₃COO^-]_salt.

36. Henderson-Hasselbalch Equation for Acidic Buffer

• For a buffer containing CH₃COOH + (CH₃COO)₂Ca, because of the stoichiometry, the concentration of the conjugate base effectively contains a factor of 2.
• pH = pK_a + log(2[Salt][Acid]).

37. Basic Buffer (W.B + C.A)

• Composed of a Weak Base (W.B) and its Conjugate Acid (C.A).
• NH₄OH ⇌ NH₄⁺ + OH^-
• NH₄Cl → NH₄⁺ + Cl^-
• Addition of OH^- (x mol): Amount of C.A. decreases by x, amount of Base increases by x.
• Addition of strong acid (y mol): Amount of W.B. decreases by y, amount of C.A. increases by y.

38. Basic Buffer: pOH and pH Relations

• For a basic buffer (weak base + its conjugate acid):

[CA] = [conjugate acid] (usually from salt). Note: [CA] ≠ [salt] always (e.g., (NH₄)₂S → 2NH₄⁺, concentration of cation becomes 2C).

39. Indicator Theory

• Dissociation: HIn ⇌ In⁻ + H⁺ (where HIn is undissociated, In⁻ is dissociated).

• Range of indicator: pK_a ± 1.

40. Titration Indicators

41. Buffer Solution Concepts

• A buffer solution resists change in pH upon the addition of small amounts of acid or base.
• Blood is a naturally occurring buffer solution maintained by H₂CO₃ / HCO₃⁻ concentrations.

42. Buffer Capacity

• Definition: The number of moles of strong acid or strong base added to 1 L of a buffer solution to change its pH by 1 unit.

43. Buffer Capacity Formula

• For a buffer solution of volume 1 L containing 'b' moles of salt and 'a' moles of acid:

• Condition for maximum buffer capacity: Occurs when a = b.

44. Solubility of Sparingly Soluble Salts

• Solubility: Amount of salt dissolved in 1 L of saturated solution.
• Gram solubility: In terms of grams.
• Molar solubility (S): In terms of moles.

45. Solubility Equilibrium

• For a sparingly soluble salt like AgCl:

• If molar solubility is S mol/L, then at equilibrium, [Ag⁺] = S and [Cl⁻] = S.
• Solubility Product constant (K_sp) is the product of ion concentrations:

46. General Solubility Expression

• For a salt of type M_pN_q:

• At equilibrium, concentrations are [M^+q] = pS and [N^-p] = qS.
• K_sp = [M^+q]^p [N^-p]^q = (pS)^p(qS)^q = p^pq^qS^p+q

47. Common Salt Types

AB₂ or A₂B type salt :

• Example: PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl^-(aq)

A₂B₃ type salt :

• Example: A₂B₃(s) ⇌ 2A³⁺(aq) + 3B^2-(aq)

General A_xB_y type salt :

48. Selective Precipitation

• Selective precipitation occurs based on the requirement of Ag⁺ concentration for a salt to precipitate.
• Relationship between ionic product and K_sp:

• Given: K_sp(AgCl) = 10⁻¹⁰, K_sp(AgBr) = 10⁻¹³, K_sp(Ag₂CrO₄) = 10⁻¹³.
• To find which precipitates first, calculate the required [Ag⁺] for each at 0.1 M concentration of anions:

• The salt with the lowest [Ag⁺] requirement precipitates first; thus, AgBr precipitates first.

49. Precipitation Conditions

• For a salt AY₂, precipitate forms if Ionic Product (Q) > K_sp.
• If equal volumes are mixed, final concentration of each ion is halved (concentration becomes C/2).

50. Solubility in Different Solutions

1. In pure water :

2. In acidic or basic buffer :

• Neither Ag⁺ nor Cl⁻ reacts with H⁺ / OH⁻ of the buffer, so S₂ = S₃ = √K_sp = S₁.

3. In presence of a common-ion :

• Example: AgCl in 0.1 M AgNO₃.
• AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
• Let solubility be S₄. [Ag⁺] = S₄ + 0.1, [Cl⁻] = S₄.
• K_sp = (S₄ + 0.1)S₄ ≅ 0.1S₄ (since S₄ is small).

51. Common Ion Effect and Solubility

• Solubility of a sparingly soluble salt decreases in the presence of a common ion.
• Solubility is maximum in a solvent where no common ion is present (e.g., deionised water vs. solutions containing constituent ions like KCl, AgNO₃, HCl for AgCl).

52. Simultaneous Solubility

• When two sparingly soluble salts (AgCl and AgBr) are present together:
• AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) ; AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq)
• Let [Ag⁺] = S₅ + S (where S₅ is solubility of AgCl, S is solubility of AgBr).
• K_sp(AgCl) = (S₅ + S)S₅
• K_sp(AgBr) = (S₅ + S)S
• Conditions: S₅ < √K_sp(AgCl) and S < √K_sp(AgBr).

53. Solubility with Complex Formation

• In the presence of a complexing agent (e.g., NH₃), solubility increases significantly.
• AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
• Ag⁺(aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq)
• Solubility order: S₆ (complex) > S₁ = S₂ = S₃ > S₅ > S₄.

54. Solubility of CH₃COOAg

• Salt of weak acid (CH₃COOH, K_a = 1.8 × 10^-5) and strong base.
• In pure water: CH₃COOAg(s) ⇌ CH₃COO^- + Ag⁺ (K_sp).
• Hydrolysis: CH₃COO^- + H₂O ⇌ CH₃COOH + OH^- (K_h = K_w / K_a).
• Solubility S₁ is slightly greater than √K_sp.

55. Solubility in Different Buffers

• In Acidic Buffer: CH₃COOAg(s) ⇌ CH₃COO^- + Ag⁺; CH₃COO^- + H⁺ ⇌ CH₃COOH (K_eq = 1/K_a = 10⁵).
• In Basic Buffer: The solubility S₃ ≈ √K_sp.
• Comparison: S₂ (Acidic) > S₁ (Pure Water) > S₃ (Basic).

56. Quick Formula Revision