CUET UG Chemistry Physical Chemistry Ionic Equilibrium pH Buffer Solutions Solubility Product

Must‑Know

• pH is defined as pH = –log₁₀[H⁺]; for example, if [H⁺] = 10⁻³ M, then pH = 3.
• pOH = –log₁₀[OH⁻], and at 25°C, pH + pOH = 14 (derived from K_w = [H⁺][OH⁻] = 10⁻¹⁴).
• For strong acids like HCl (0.01 M), [H⁺] = 0.01 M, so pH = 2.
• For strong bases like NaOH (0.001 M), [OH⁻] = 0.001 M, so pOH = 3 and pH = 11.
• The ionization constant of water (K_w) increases with temperature; at 25°C, K_w = 1.0 × 10⁻¹⁴, but at 60°C, it is about 9.6 × 10⁻¹⁴ (verify from NCERT).
• Weak acids partially dissociate: for acetic acid, K_a = [H⁺][CH₃COO⁻]/[CH₃COOH] = 1.8 × 10⁻⁵ at 25°C.
• For a weak acid, [H⁺] ≈ √(K_a × C), where C is the initial concentration; e.g., 0.1 M CH₃COOH gives [H⁺] ≈ √(1.8×10⁻⁵ × 0.1) = 1.34×10⁻³ M.
• Henderson-Hasselbalch equation for acidic buffer: pH = pK_a + log₁₀([salt]/[acid]); e.g., for CH₃COOH/CH₃COONa buffer with equal concentrations, pH = pK_a = 4.74.
• Basic buffer uses NH₄OH and NH₄Cl; pH = 14 – pK_b – log₁₀([salt]/[base]) or pOH = pK_b + log₁₀([salt]/[base]).
• pK_a + pK_b = 14 for conjugate acid-base pairs at 25°C.
• Buffer capacity is maximum when pH = pK_a and [acid] = [salt].
• Buffer resists pH change upon addition of small amounts of acid or base; e.g., blood (pH ≈ 7.4) is buffered by H₂CO₃/HCO₃⁻ system.
• Solubility product (K_sp) is the product of ion concentrations raised to stoichiometric coefficients in a saturated solution; for AgCl, K_sp = [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰ at 25°C.
• For CaF₂, K_sp = [Ca²⁺][F⁻]²; if solubility is S mol/L, then K_sp = S × (2S)² = 4S³.
• Precipitation occurs when ionic product (IP) > K_sp; no precipitation if IP < K_sp; solution is saturated if IP = K_sp.
• Common ion effect reduces solubility; e.g., solubility of AgCl in 0.1 M NaCl is much less than in pure water due to increased [Cl⁻].
• For a salt AB₂, if solubility in water is S, then K_sp = 4S³; if K_sp = 3.2 × 10⁻¹¹, then S = ∛(K_sp/4) = ∛(8×10⁻¹²) ≈ 2×10⁻⁴ M.
• pH of salt of weak acid and strong base (e.g., CH₃COONa) is >7; hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻.
• pH of salt of strong acid and weak base (e.g., NH₄Cl) is <7; hydrolysis: NH₄⁺ ⇌ NH₃ + H⁺.
• For salt of weak acid and weak base (e.g., CH₃COONH₄), pH ≈ 7 + ½(pK_a – pK_b); if pK_a = pK_b, pH ≈ 7.

Difficulty Level

Intermediate — requires understanding of logarithmic calculations, equilibrium concepts, and application of formulas in varied contexts including buffers and solubility.

Common CUET Traps

• Trap: Assuming [H⁺] = C for weak acids.
  Avoid: Use [H⁺] = √(K_a·C) for weak monoprotic acids when approximation holds.

• Trap: Confusing K_sp expression for salts like Al₂(SO₄)₃ or Ca₃(PO₄)₂.
  Avoid: Write correct stoichiometry: for Ca₃(PO₄)₂, K_sp = [Ca²⁺]³[PO₄³⁻]².

• Trap: Thinking buffer pH changes significantly when diluted.
  Avoid: Dilution does not change [salt]/[acid] ratio, so pH remains unchanged in ideal buffers.

Practice MCQs

Q1. What is the pH of 0.001 M NaOH solution at 25°C?
A) 3
B) 11
C) 10
D) 12

Answer: B
Explanation: [OH⁻] = 0.001 M → pOH = 3 → pH = 14 – 3 = 11.
Why others fail: Option A assumes pH = –log[OH⁻], confusing pH and pOH.

Q2. Which of the following represents the K_sp expression for Fe(OH)₃?
A) [Fe³⁺][OH⁻]
B) [Fe³⁺][3OH⁻]³
C) [Fe³⁺][OH⁻]³
D) [Fe³⁺]³[OH⁻]

Answer: C
Explanation: Fe(OH)₃ ⇌ Fe³⁺ + 3OH⁻ ⇒ K_sp = [Fe³⁺][OH⁻]³.
Why others fail: Option B incorrectly includes coefficient inside concentration term.

Q3. A buffer solution contains 0.1 M CH₃COOH and 0.1 M CH₃COONa. If pK_a of acetic acid is 4.74, what is the pH of the solution?
A) 4.74
B) 5.74
C) 3.74
D) 7.00

Answer: A
Explanation: pH = pK_a + log([salt]/[acid]) = 4.74 + log(1) = 4.74.
Why others fail: Option B adds log value incorrectly, thinking salt increases pH directly.

Q4. The solubility of AgCl in pure water is S mol/L. How does it change in 0.01 M NaCl solution?
A) Remains same
B) Increases
C) Decreases due to common ion effect
D) Becomes zero

Answer: C
Explanation: Common ion (Cl⁻) suppresses dissociation of AgCl, reducing solubility.
Why others fail: Option A ignores the common ion effect, a key principle in solubility equilibria.

Q5. For which of the following salts will the value of solubility (S) be equal to √K_sp?
A) Ag₂CrO₄
B) AlCl₃
C) BaSO₄
D) Ca₃(PO₄)₂

Answer: C
Explanation: BaSO₄ ⇌ Ba²⁺ + SO₄²⁻ ⇒ K_sp = S² ⇒ S = √K_sp.
Why others fail: Option A gives K_sp = 4S³, not S²; students misapply formula based on stoichiometry.

Last‑Minute Revision

• ⚠️ pH = –log[H⁺]; never use [H⁺] directly without log.
• ⚠️ K_w = 1×10⁻¹⁴ at 25°C — use this unless temperature specified.
• ⚠️ For weak acid: [H⁺] = √(K_a·C) — approximation valid if C >> [H⁺].
• ⚠️ Strong acid: [H⁺] = concentration; strong base: [OH⁻] = concentration.
• ⚠️ Neutral solution: [H⁺] = [OH⁻] = 10⁻⁷ M at 25°C.
• ⚠️ Acidic buffer: weak acid + salt with strong base (e.g., CH₃COOH + CH₃COONa).
• ⚠️ Basic buffer: weak base + salt with strong acid (e.g., NH₄OH + NH₄Cl).
• ⚠️ Henderson-Hasselbalch: pH = pK_a + log([A⁻]/[HA]).
• ⚠️ Buffer pH unchanged on dilution — ratio stays constant.
• ⚠️ Blood buffer: H₂CO₃ / HCO₃⁻ maintains pH ~7.4.
• ⚠️ Salt of weak acid & strong base → basic (pH > 7).
• ⚠️ Salt of strong acid & weak base → acidic (pH < 7).
• ⚠️ K_sp = product of ion concentrations raised to their coefficients.
• ⚠️ For AB type salt (e.g., AgCl), S = √K_sp.
• ⚠️ For AB₂ type (e.g., CaF₂), S = ∛(K_sp/4).
• ⚠️ Common ion effect decreases solubility.
• ⚠️ Precipitation begins when ionic product > K_sp.
• ⚠️ pK_a + pK_b = 14 (at 25°C) for conjugate pairs.
• ⚠️ Mnemonic: “Low pH – High H⁺” (low numbers mean more acidic).
• ⚠️ CH₃COONH₄ solution is nearly neutral because pK_a ≈ pK_b.