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ChemistryMediumClass 11

Ionic Equilibrium — pH + Buffer + Ksp

Ionic Equilibrium

48

JEE Qs

9%

Hard

120

min

Master the art of identifying the type of ionic equilibrium problem and choosing the correct set of approximations and formulas to solve it efficiently.

🧮 Key Formulas

pH = -log₁₀[H⁺]
pOH = -log₁₀[OH⁻]
pH + pOH = 14 (at 298K)
K_w = [H⁺][OH⁻] = 10⁻¹⁴ (at 298K)
For weak acid HA: K_a = [H⁺][A⁻]/[HA]
[H⁺] = sqrt(K_a * C_HA) (for weak acid, if dissociation is negligible)
For weak base BOH: K_b = [B⁺][OH⁻]/[BOH]
[OH⁻] = sqrt(K_b * C_BOH) (for weak base, if dissociation is negligible)
K_a * K_b = K_w (for a conjugate acid-base pair)
Henderson-Hasselbalch equation (acidic buffer): pH = pK_a + log₁₀([Salt]/[Acid]) or pH = pK_a + log₁₀([Conjugate Base]/[Weak Acid])
Henderson-Hasselbalch equation (basic buffer): pOH = pK_b + log₁₀([Salt]/[Base]) or pOH = pK_b + log₁₀([Conjugate Acid]/[Weak Base])
For a sparingly soluble salt A_x B_y: K_sp = [Aʸ⁺]ˣ [Bˣ⁻]ʸ
For salt AB, K_sp = s²
For salt AB₂, K_sp = 4s³
For salt A₂B, K_sp = 4s³
For salt AₓBᵧ, K_sp = xˣ yʸ s^(x+y)
Ionic Product Q_sp

✅ Key Points for JEE

  • 1Always identify the type of system (strong acid/base, weak acid/base, buffer, salt hydrolysis, solubility product) to apply the correct formulas and approximations.
  • 2For weak acid/base calculations, use the approximation C - x ≈ C only when the dissociation constant (Ka or Kb) is small and initial concentration is reasonably high (verify with 5% rule if x < 5% of C).
  • 3Buffer solutions resist pH changes upon addition of small amounts of acid or base; their pH depends on the ratio of conjugate acid-base pair concentrations, as described by Henderson-Hasselbalch equation.
  • 4The common ion effect reduces the solubility of a sparingly soluble salt when a soluble salt containing a common ion is added to the solution.
  • 5Precipitation occurs if the ionic product (Q_sp) exceeds the solubility product constant (K_sp); if Q_sp < K_sp, no precipitation occurs, and if Q_sp = K_sp, the solution is saturated.

⚠️ Common Mistakes

  • Incorrectly applying formulas for strong acids/bases to weak ones, or vice versa, leading to calculation errors.
  • Ignoring dilution effects when mixing solutions, which changes initial concentrations before equilibrium calculations.
  • Stoichiometric errors in K_sp expressions (e.g., K_sp for CaF₂ is [Ca²⁺][F⁻]², not [Ca²⁺][F⁻]).
  • Not considering the salt hydrolysis (reaction of ions with water) which can affect the pH of solutions of salts of weak acids/bases.

📝 Practice Questions

See all

Q74.If 1 mM solution of ethylamine produces pH = 9, then the ionization constant (Kb) of ethylamine is 10−x . The value of x is ______ (nearest integer). [The degree of ionization of ethylamine can be neglected with respect to unity.]

2025·NumericalMedium

Q61. Ksp for Cr(OH)3 is 1.6 × 10−30 . What is the molar solubility of this salt in water? (1) 1.8×10−30 (2) 5√1.8 × 10−30 27 × 10−30 (3) 4√1.6×10−3027 (4) 2√1.6

2025·MCQEasy

Q58.The molar solubility(s) of zirconium phosphate with molecular formula (Zr4+)3(PO3−4 )4 is given by relation : (1) Ksp 13 (2) Ksp 17 ( 9612 ) ( 6912 ) (3) Ksp 17 (4) Ksp 16 ( 8435 ) ( 5348 )

2025·MCQMedium

Q62.Which of the following happens when NH4OH is added gradually to the solution containing 1 M A2+ and 1MB3+ ions? Given : Ksp [A(OH)2] = 9 × 10−10 and Ksp [B(OH)3] = 27 × 10−18 at 298 K . (1) Both A(OH)2 and B(OH)3 do not show (2) A(OH)2 will precipitate before B(OH)3 precipitation with NH4OH (3) B(OH)3 will precipitate before A(OH)2 (4) A(OH)2 and B(OH)3 will precipitate together excess HCHO alkali−

2025·MCQMedium

Q68.pH of water is 7 at 25∘C. If water is heated to 80∘C., it's pH will : (1) Decrease (2) H+ concentration increases, OH− concentration decreases (3) Remains the same (4) Increase 2025 (23 Jan Shift 2) JEE Main Previous Year Paper

2025·MCQEasy

Q58.A weak acid HA has degree of dissociation x . Which option gives the correct expression of ( pH pKa )? (1) 0 (2) log(1 + 2x) (3) log ( 1−xx ) (4) log ( 1−xx )

2025·MCQMedium

NCERT Chapters

  • Class 11 Chemistry Part 1 Ch 7: Equilibrium