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Import { useState } from "react"; const questions = [ // ── SOLUTIONS (1-10) ── { id: 1, chapter: "Solutions", q: "Which colligative property is used to determine the molecular weight of a polymer in solution?", options: ["Boiling point elevation", "Osmotic pressure", "Freezing point depression", "Vapor pressure lowering"], answer: 1, explanation: "Osmotic pressure is the most sensitive colligative property for high-molecular-weight solutes like polymers. Even very dilute solutions show measurable osmotic pressure, making it ideal for molecular weight determination of macromolecules." }, { id: 2, chapter: "Solutions", q: "Raoult's Law states that the partial vapor pressure of a solvent is:", options: ["Inversely proportional to solute concentration", "Equal to the mole fraction of solvent × vapor pressure of pure solvent", "Independent of temperature", "Equal to the total pressure of the system"], answer: 1, explanation: "Raoult's Law: P_A = X_A × P°_A. The partial vapor pressure of the solvent equals its mole fraction multiplied by the vapor pressure of the pure solvent. Deviation from this law indicates non-ideal behavior." }, { id: 3, chapter: "Solutions", q: "A solution of NaCl shows a greater freezing point depression than predicted by van't Hoff equation because:", options: ["NaCl is a weak electrolyte", "NaCl dissociates into two ions, increasing the number of particles", "NaCl has high molecular weight", "NaCl decreases water activity"], answer: 1, explanation: "NaCl → Na⁺ + Cl⁻ (two particles per formula unit). The van't Hoff factor 'i' = 2 for NaCl, so ΔTf = i × Kf × m = 2 × 1.86 × m. More particles = greater colligative effect." }, { id: 4, chapter: "Solutions", q: "Henry's Law applies to:", options: ["Concentrated solutions", "Dilute solutions of gases", "Saturated solutions", "Supersaturated solutions"], answer: 1, explanation: "Henry's Law (p = KH × X) applies to dilute solutions of gases, stating that the partial pressure of a gas above a solution is proportional to its mole fraction in solution. It fails at high concentrations." }, { id: 5, chapter: "Solutions", q: "The osmotic pressure of a solution at 27°C containing 6g/L of glucose (MW=180) is approximately:", options: ["0.82 atm", "8.2 atm", "0.41 atm", "4.1 atm"], answer: 0, explanation: "π = CRT = (6/180) mol/L × 0.082 L·atm/mol·K × 300 K = 0.0333 × 0.082 × 300 ≈ 0.82 atm. This demonstrates the formula π = MRT used in osmotic pressure calculations." }, { id: 6, chapter: "Solutions", q: "Which statement about ideal solutions is CORRECT?", options: ["ΔH_mix > 0", "ΔV_mix ≠ 0", "Intermolecular forces between unlike molecules equal those between like molecules", "They show positive deviation from Raoult's Law"], answer: 2, explanation: "In ideal solutions, A–B interactions = A–A = B–B interactions, so ΔH_mix = 0 and ΔV_mix = 0. Both positive and negative deviations from Raoult's Law indicate non-ideal behavior due to unequal intermolecular forces." }, { id: 7, chapter: "Solutions", q: "The term 'solubility' is best defined as:", options: ["The rate at which a solute dissolves", "The maximum amount of solute that dissolves in a given solvent at a specific temperature", "The amount of solute in 100 mL of solution", "The equilibrium constant for dissolution"], answer: 1, explanation: "Solubility is the maximum concentration of solute that can dissolve in a given solvent at a specific temperature and pressure to form a stable (saturated) solution. Beyond this, the solution becomes supersaturated." }, { id: 8, chapter: "Solutions", q: "Which expression represents the cryoscopic constant Kf for water?", options: ["1.86 K·kg/mol", "0.512 K·kg/mol", "3.4 K·kg/mol", "1.20 K·kg/mol"], answer: 0, explanation: "Kf (cryoscopic constant) for water = 1.86 K·kg/mol, while Kb (ebullioscopic constant) = 0.512 K·kg/mol. Kf > Kb for water, making freezing point depression slightly more sensitive for molecular weight determination." }, { id: 9, chapter: "Solutions", q: "A 0.9% w/v NaCl solution is isotonic with blood because:", options: ["It has the same pH as blood", "It exerts the same osmotic pressure as blood plasma (~7.4 atm)", "It contains the same ions as blood", "It has the same density as blood"], answer: 1, explanation: "0.9% w/v NaCl (normal saline) is isotonic because it exerts the same osmotic pressure (~7.4 atm or 308 mOsm/kg) as blood plasma. This prevents red blood cells from crenation (hypertonic) or hemolysis (hypotonic)." }, { id: 10, chapter: "Solutions", q: "Positive deviation from Raoult's Law occurs when:", options: ["A–B interactions are stronger than A–A and B–B", "A–B interactions are weaker than A–A and B–B", "The solution is ideal", "The solute is non-volatile"], answer: 1, explanation: "Positive deviation: actual vapor pressure > Raoult's Law prediction. This occurs when unlike (A–B) interactions are weaker than like (A–A, B–B) interactions, making molecules escape more easily. Example: ethanol–water at low ethanol concentrations." }, // ── SOLUBILIZATION (11-18) ── { id: 11, chapter: "Solubilization", q: "Solubilization by surfactants primarily occurs when the concentration exceeds:", options: ["Krafft point", "Critical Micelle Concentration (CMC)", "Cloud point", "HLB value of 10"], answer: 1, explanation: "Solubilization occurs above the CMC when micelles form. Non-polar drugs are incorporated into the hydrophobic core of micelles; polar drugs adsorb at the outer shell. Below CMC, no micelles exist to solubilize the drug." }, { id: 12, chapter: "Solubilization", q: "The Krafft point of a surfactant is defined as the temperature at which:", options: ["Surfactant becomes insoluble", "Solubility of surfactant equals its CMC", "HLB value changes", "Micelles disintegrate"], answer: 1, explanation: "Krafft point is the temperature at which the solubility of an ionic surfactant equals its CMC. Above this temperature, micelle formation increases dramatically and solubilization capacity rises sharply." }, { id: 13, chapter: "Solubilization", q: "Which method of solubilization does NOT involve micelle formation?", options: ["Use of co-solvents (cosolvency)", "Surfactant micellization", "Hydrotropic solubilization", "Both A and C"], answer: 3, explanation: "Cosolvency (using ethanol, PEG, glycerin) and hydrotropism (using sodium benzoate, nicotinamide) increase solubility without micelle formation. They work by altering solvent polarity or forming weak complexes with the drug." }, { id: 14, chapter: "Solubilization", q: "Complexation with cyclodextrins improves drug solubility by:", options: ["Increasing drug's molecular weight", "Including the drug in a hydrophobic cavity, presenting a hydrophilic exterior", "Lowering the drug's melting point", "Increasing ionization of the drug"], answer: 1, explanation: "Cyclodextrins have a hydrophobic interior cavity and hydrophilic exterior. Lipophilic drugs fit inside the cavity (inclusion complex), and the hydrophilic outer surface makes the complex water-soluble. β-cyclodextrin is most commonly used." }, { id: 15, chapter: "Solubilization", q: "The pH-solubility relationship for a weak acid drug (pKa = 5) predicts maximum solubility at:", options: ["pH < pKa", "pH = pKa", "pH > pKa", "pH = 7"], answer: 2, explanation: "For weak acids: solubility increases at pH > pKa because the ionized (conjugate base) form is water-soluble. Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). At pH > pKa, more ionized form exists → greater solubility." }, { id: 16, chapter: "Solubilization", q: "Which of the following is a hydrotrope?", options: ["Span 80", "Sodium benzoate", "Tween 20", "Lecithin"], answer: 1, explanation: "Sodium benzoate is a classic hydrotrope — a substance that increases the aqueous solubility of sparingly soluble compounds at high concentrations (typically >0.1 M) without forming true micelles. Others include urea and nicotinamide." }, { id: 17, chapter: "Solubilization", q: "Particle size reduction improves drug solubility according to:", options: ["Stokes' Law", "Noyes-Whitney equation", "Ostwald-Freundlich equation", "Henderson-Hasselbalch equation"], answer: 2, explanation: "The Ostwald-Freundlich (Kelvin) equation: log(S/S₀) = 2γM/ρRTr. As particle radius 'r' decreases, solubility S increases. This is the basis for nanosizing drugs to improve their dissolution and bioavailability." }, { id: 18, chapter: "Solubilization", q: "Salt formation of a drug base is done to improve solubility. The hydrochloride salt of a basic drug has:", options: ["Lower solubility at high pH", "Higher solubility at all pH values", "Higher solubility at low to neutral pH", "No change in solubility compared to free base"], answer: 2, explanation: "Salt forms of basic drugs (e.g., HCl salts) dissociate in water to give ionized (protonated) drug + Cl⁻. At low-to-neutral pH, the ionized form predominates → higher solubility. At high pH, the free base precipitates out." }, // ── ADSORPTION (19-24) ── { id: 19, chapter: "Adsorption", q: "The Langmuir adsorption isotherm assumes:", options: ["Multilayer adsorption on heterogeneous surface", "Monolayer adsorption on homogeneous surface with fixed adsorption sites", "Physical adsorption only", "Adsorption energy increases with surface coverage"], answer: 1, explanation: "Langmuir isotherm: θ = bC/(1+bC). Assumptions: (1) monolayer adsorption, (2) homogeneous surface with equivalent sites, (3) no interaction between adsorbed molecules, (4) dynamic equilibrium between adsorption and desorption." }, { id: 20, chapter: "Adsorption", q: "The BET (Brunauer-Emmett-Teller) theory extends Langmuir isotherm by allowing:", options: ["Chemisorption only", "Multilayer adsorption", "Negative adsorption", "Adsorption in micropores only"], answer: 1, explanation: "BET theory allows multilayer adsorption, unlike Langmuir (monolayer). It's used to determine the surface area of pharmaceutical powders by measuring N₂ gas adsorption at liquid nitrogen temperature. BET surface area is critical in micromeritics." }, { id: 21, chapter: "Adsorption", q: "Freundlich adsorption isotherm is expressed as:", options: ["x/m = KC^(1/n)", "x/m = Kp^(1/n)", "x/m = bC/(1+bC)", "x/m = KC/(1+bC)"], answer: 0, explanation: "Freundlich: x/m = KC^(1/n) where x/m = amount adsorbed per gram adsorbent, C = equilibrium concentration, K and n are constants. Linearized as: log(x/m) = log K + (1/n)log C. It applies to heterogeneous surfaces and multilayer adsorption." }, { id: 22, chapter: "Adsorption", q: "Activated charcoal is used in emergency poisoning treatment because it exhibits:", options: ["High chemical adsorption (chemisorption)", "High physical adsorption due to large surface area", "Ion exchange properties", "Emulsification of toxins"], answer: 1, explanation: "Activated charcoal has an enormous surface area (500–1500 m²/g) due to its microporous structure. It adsorbs most drugs and toxins via physical adsorption (van der Waals forces), making it useful in poisoning management." }, { id: 23, chapter: "Adsorption", q: "Negative adsorption occurs when:", options: ["Solute concentration at surface < bulk concentration", "Solute concentration at surface > bulk concentration", "Temperature is below freezing", "Pressure exceeds critical point"], answer: 0, explanation: "Negative adsorption (desorption tendency): the solute concentration at the adsorbent surface is lower than in the bulk solution. This happens when the solvent is preferentially adsorbed. Example: KCl on silica gel — water is adsorbed, not KCl." }, { id: 24, chapter: "Adsorption", q: "Gibbs adsorption equation relates surface excess concentration to:", options: ["Viscosity and temperature", "Surface tension gradient with concentration", "Osmotic pressure and molecular weight", "Particle size and density"], answer: 1, explanation: "Gibbs equation: Γ = -(1/RT)(dγ/d ln C). The surface excess Γ is related to how surface tension γ changes with concentration. Surfactants lower surface tension (dγ/dC < 0), so Γ > 0 (positive adsorption at interface)." }, // ── IONIZATION (25-29) ── { id: 25, chapter: "Ionization", q: "The Henderson-Hasselbalch equation for a weak acid is:", options: ["pH = pKa − log([HA]/[A⁻])", "pH = pKa + log([A⁻]/[HA])", "pH = pKb + log([B]/[BH⁺])", "pH = 14 − pKb"], answer: 1, explanation: "Henderson-Hasselbalch for weak acid: pH = pKa + log([A⁻]/[HA]). At pH = pKa, [A⁻]=[HA], so 50% ionization. This equation is fundamental for calculating ionization degree, buffer pH, and drug absorption across biological membranes." }, { id: 26, chapter: "Ionization", q: "The degree of ionization of aspirin (pKa = 3.5) at gastric pH 1.5 is approximately:", options: ["99% ionized", "50% ionized", "~1% ionized", "0% ionized"], answer: 2, explanation: "log([A⁻]/[HA]) = pH − pKa = 1.5 − 3.5 = −2, so [A⁻]/[HA] = 0.01 (1:100 ratio). Thus ~1% ionized, 99% unionized. Unionized aspirin is absorbed well in the stomach despite low pH — the pH-partition hypothesis." }, { id: 27, chapter: "Ionization", q: "Buffer capacity (β) is maximum when:", options: ["pH = pKa ± 2", "pH = pKa (equal concentrations of weak acid and conjugate base)", "pH = 7", "The buffer is diluted 10-fold"], answer: 1, explanation: "β = 2.303C × Ka[H⁺]/(Ka+[H⁺])². Maximum β occurs when pH = pKa, where [HA] = [A⁻]. Effective buffering range is pH = pKa ± 1. Beyond ±2 pH units from pKa, buffering capacity is negligible." }, { id: 28, chapter: "Ionization", q: "Which factor does NOT affect the degree of ionization of a drug in solution?", options: ["pH of the medium", "pKa of the drug", "Temperature", "Molecular weight of the drug"], answer:

Import { useState } from "react"; const questions = [ // ── SOLUTIONS (1-10) ── { id: 1, chapter: "Solutions", q: "Which colligative property is used to determine the molecular weight of a polymer in solution?", options: ["Boiling point elevation", "Osmotic pressure", "Freezing point depression", "Vapor pressure lowering"], answer: 1, explanation: "Osmotic pressure is the most sensitive colligative property for high-molecular-weight solutes like polymers. Even very dilute solutions show measurable osmotic pressure, making it ideal for molecular weight determination of macromolecules." }, { id: 2, chapter: "Solutions", q: "Raoult's Law states that the partial vapor pressure of a solvent is:", options: ["Inversely proportional to solute concentration", "Equal to the mole fraction of solvent × vapor pressure of pure solvent", "Independent of temperature", "Equal to the total pressure of the system"], answer: 1, explanation: "Raoult's Law: P_A = X_A × P°_A. The partial vapor pressure of the solvent equals its mole fraction multiplied by the vapor pressure of the pure solvent. Deviation from this law indicates non-ideal behavior." }, { id: 3, chapter: "Solutions", q: "A solution of NaCl shows a greater freezing point depression than predicted by van't Hoff equation because:", options: ["NaCl is a weak electrolyte", "NaCl dissociates into two ions, increasing the number of particles", "NaCl has high molecular weight", "NaCl decreases water activity"], answer: 1, explanation: "NaCl → Na⁺ + Cl⁻ (two particles per formula unit). The van't Hoff factor 'i' = 2 for NaCl, so ΔTf = i × Kf × m = 2 × 1.86 × m. More particles = greater colligative effect." }, { id: 4, chapter: "Solutions", q: "Henry's Law applies to:", options: ["Concentrated solutions", "Dilute solutions of gases", "Saturated solutions", "Supersaturated solutions"], answer: 1, explanation: "Henry's Law (p = KH × X) applies to dilute solutions of gases, stating that the partial pressure of a gas above a solution is proportional to its mole fraction in solution. It fails at high concentrations." }, { id: 5, chapter: "Solutions", q: "The osmotic pressure of a solution at 27°C containing 6g/L of glucose (MW=180) is approximately:", options: ["0.82 atm", "8.2 atm", "0.41 atm", "4.1 atm"], answer: 0, explanation: "π = CRT = (6/180) mol/L × 0.082 L·atm/mol·K × 300 K = 0.0333 × 0.082 × 300 ≈ 0.82 atm. This demonstrates the formula π = MRT used in osmotic pressure calculations." }, { id: 6, chapter: "Solutions", q: "Which statement about ideal solutions is CORRECT?", options: ["ΔH_mix > 0", "ΔV_mix ≠ 0", "Intermolecular forces between unlike molecules equal those between like molecules", "They show positive deviation from Raoult's Law"], answer: 2, explanation: "In ideal solutions, A–B interactions = A–A = B–B interactions, so ΔH_mix = 0 and ΔV_mix = 0. Both positive and negative deviations from Raoult's Law indicate non-ideal behavior due to unequal intermolecular forces." }, { id: 7, chapter: "Solutions", q: "The term 'solubility' is best defined as:", options: ["The rate at which a solute dissolves", "The maximum amount of solute that dissolves in a given solvent at a specific temperature", "The amount of solute in 100 mL of solution", "The equilibrium constant for dissolution"], answer: 1, explanation: "Solubility is the maximum concentration of solute that can dissolve in a given solvent at a specific temperature and pressure to form a stable (saturated) solution. Beyond this, the solution becomes supersaturated." }, { id: 8, chapter: "Solutions", q: "Which expression represents the cryoscopic constant Kf for water?", options: ["1.86 K·kg/mol", "0.512 K·kg/mol", "3.4 K·kg/mol", "1.20 K·kg/mol"], answer: 0, explanation: "Kf (cryoscopic constant) for water = 1.86 K·kg/mol, while Kb (ebullioscopic constant) = 0.512 K·kg/mol. Kf > Kb for water, making freezing point depression slightly more sensitive for molecular weight determination." }, { id: 9, chapter: "Solutions", q: "A 0.9% w/v NaCl solution is isotonic with blood because:", options: ["It has the same pH as blood", "It exerts the same osmotic pressure as blood plasma (~7.4 atm)", "It contains the same ions as blood", "It has the same density as blood"], answer: 1, explanation: "0.9% w/v NaCl (normal saline) is isotonic because it exerts the same osmotic pressure (~7.4 atm or 308 mOsm/kg) as blood plasma. This prevents red blood cells from crenation (hypertonic) or hemolysis (hypotonic)." }, { id: 10, chapter: "Solutions", q: "Positive deviation from Raoult's Law occurs when:", options: ["A–B interactions are stronger than A–A and B–B", "A–B interactions are weaker than A–A and B–B", "The solution is ideal", "The solute is non-volatile"], answer: 1, explanation: "Positive deviation: actual vapor pressure > Raoult's Law prediction. This occurs when unlike (A–B) interactions are weaker than like (A–A, B–B) interactions, making molecules escape more easily. Example: ethanol–water at low ethanol concentrations." }, // ── SOLUBILIZATION (11-18) ── { id: 11, chapter: "Solubilization", q: "Solubilization by surfactants primarily occurs when the concentration exceeds:", options: ["Krafft point", "Critical Micelle Concentration (CMC)", "Cloud point", "HLB value of 10"], answer: 1, explanation: "Solubilization occurs above the CMC when micelles form. Non-polar drugs are incorporated into the hydrophobic core of micelles; polar drugs adsorb at the outer shell. Below CMC, no micelles exist to solubilize the drug." }, { id: 12, chapter: "Solubilization", q: "The Krafft point of a surfactant is defined as the temperature at which:", options: ["Surfactant becomes insoluble", "Solubility of surfactant equals its CMC", "HLB value changes", "Micelles disintegrate"], answer: 1, explanation: "Krafft point is the temperature at which the solubility of an ionic surfactant equals its CMC. Above this temperature, micelle formation increases dramatically and solubilization capacity rises sharply." }, { id: 13, chapter: "Solubilization", q: "Which method of solubilization does NOT involve micelle formation?", options: ["Use of co-solvents (cosolvency)", "Surfactant micellization", "Hydrotropic solubilization", "Both A and C"], answer: 3, explanation: "Cosolvency (using ethanol, PEG, glycerin) and hydrotropism (using sodium benzoate, nicotinamide) increase solubility without micelle formation. They work by altering solvent polarity or forming weak complexes with the drug." }, { id: 14, chapter: "Solubilization", q: "Complexation with cyclodextrins improves drug solubility by:", options: ["Increasing drug's molecular weight", "Including the drug in a hydrophobic cavity, presenting a hydrophilic exterior", "Lowering the drug's melting point", "Increasing ionization of the drug"], answer: 1, explanation: "Cyclodextrins have a hydrophobic interior cavity and hydrophilic exterior. Lipophilic drugs fit inside the cavity (inclusion complex), and the hydrophilic outer surface makes the complex water-soluble. β-cyclodextrin is most commonly used." }, { id: 15, chapter: "Solubilization", q: "The pH-solubility relationship for a weak acid drug (pKa = 5) predicts maximum solubility at:", options: ["pH < pKa", "pH = pKa", "pH > pKa", "pH = 7"], answer: 2, explanation: "For weak acids: solubility increases at pH > pKa because the ionized (conjugate base) form is water-soluble. Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). At pH > pKa, more ionized form exists → greater solubility." }, { id: 16, chapter: "Solubilization", q: "Which of the following is a hydrotrope?", options: ["Span 80", "Sodium benzoate", "Tween 20", "Lecithin"], answer: 1, explanation: "Sodium benzoate is a classic hydrotrope — a substance that increases the aqueous solubility of sparingly soluble compounds at high concentrations (typically >0.1 M) without forming true micelles. Others include urea and nicotinamide." }, { id: 17, chapter: "Solubilization", q: "Particle size reduction improves drug solubility according to:", options: ["Stokes' Law", "Noyes-Whitney equation", "Ostwald-Freundlich equation", "Henderson-Hasselbalch equation"], answer: 2, explanation: "The Ostwald-Freundlich (Kelvin) equation: log(S/S₀) = 2γM/ρRTr. As particle radius 'r' decreases, solubility S increases. This is the basis for nanosizing drugs to improve their dissolution and bioavailability." }, { id: 18, chapter: "Solubilization", q: "Salt formation of a drug base is done to improve solubility. The hydrochloride salt of a basic drug has:", options: ["Lower solubility at high pH", "Higher solubility at all pH values", "Higher solubility at low to neutral pH", "No change in solubility compared to free base"], answer: 2, explanation: "Salt forms of basic drugs (e.g., HCl salts) dissociate in water to give ionized (protonated) drug + Cl⁻. At low-to-neutral pH, the ionized form predominates → higher solubility. At high pH, the free base precipitates out." }, // ── ADSORPTION (19-24) ── { id: 19, chapter: "Adsorption", q: "The Langmuir adsorption isotherm assumes:", options: ["Multilayer adsorption on heterogeneous surface", "Monolayer adsorption on homogeneous surface with fixed adsorption sites", "Physical adsorption only", "Adsorption energy increases with surface coverage"], answer: 1, explanation: "Langmuir isotherm: θ = bC/(1+bC). Assumptions: (1) monolayer adsorption, (2) homogeneous surface with equivalent sites, (3) no interaction between adsorbed molecules, (4) dynamic equilibrium between adsorption and desorption." }, { id: 20, chapter: "Adsorption", q: "The BET (Brunauer-Emmett-Teller) theory extends Langmuir isotherm by allowing:", options: ["Chemisorption only", "Multilayer adsorption", "Negative adsorption", "Adsorption in micropores only"], answer: 1, explanation: "BET theory allows multilayer adsorption, unlike Langmuir (monolayer). It's used to determine the surface area of pharmaceutical powders by measuring N₂ gas adsorption at liquid nitrogen temperature. BET surface area is critical in micromeritics." }, { id: 21, chapter: "Adsorption", q: "Freundlich adsorption isotherm is expressed as:", options: ["x/m = KC^(1/n)", "x/m = Kp^(1/n)", "x/m = bC/(1+bC)", "x/m = KC/(1+bC)"], answer: 0, explanation: "Freundlich: x/m = KC^(1/n) where x/m = amount adsorbed per gram adsorbent, C = equilibrium concentration, K and n are constants. Linearized as: log(x/m) = log K + (1/n)log C. It applies to heterogeneous surfaces and multilayer adsorption." }, { id: 22, chapter: "Adsorption", q: "Activated charcoal is used in emergency poisoning treatment because it exhibits:", options: ["High chemical adsorption (chemisorption)", "High physical adsorption due to large surface area", "Ion exchange properties", "Emulsification of toxins"], answer: 1, explanation: "Activated charcoal has an enormous surface area (500–1500 m²/g) due to its microporous structure. It adsorbs most drugs and toxins via physical adsorption (van der Waals forces), making it useful in poisoning management." }, { id: 23, chapter: "Adsorption", q: "Negative adsorption occurs when:", options: ["Solute concentration at surface < bulk concentration", "Solute concentration at surface > bulk concentration", "Temperature is below freezing", "Pressure exceeds critical point"], answer: 0, explanation: "Negative adsorption (desorption tendency): the solute concentration at the adsorbent surface is lower than in the bulk solution. This happens when the solvent is preferentially adsorbed. Example: KCl on silica gel — water is adsorbed, not KCl." }, { id: 24, chapter: "Adsorption", q: "Gibbs adsorption equation relates surface excess concentration to:", options: ["Viscosity and temperature", "Surface tension gradient with concentration", "Osmotic pressure and molecular weight", "Particle size and density"], answer: 1, explanation: "Gibbs equation: Γ = -(1/RT)(dγ/d ln C). The surface excess Γ is related to how surface tension γ changes with concentration. Surfactants lower surface tension (dγ/dC < 0), so Γ > 0 (positive adsorption at interface)." }, // ── IONIZATION (25-29) ── { id: 25, chapter: "Ionization", q: "The Henderson-Hasselbalch equation for a weak acid is:", options: ["pH = pKa − log([HA]/[A⁻])", "pH = pKa + log([A⁻]/[HA])", "pH = pKb + log([B]/[BH⁺])", "pH = 14 − pKb"], answer: 1, explanation: "Henderson-Hasselbalch for weak acid: pH = pKa + log([A⁻]/[HA]). At pH = pKa, [A⁻]=[HA], so 50% ionization. This equation is fundamental for calculating ionization degree, buffer pH, and drug absorption across biological membranes." }, { id: 26, chapter: "Ionization", q: "The degree of ionization of aspirin (pKa = 3.5) at gastric pH 1.5 is approximately:", options: ["99% ionized", "50% ionized", "~1% ionized", "0% ionized"], answer: 2, explanation: "log([A⁻]/[HA]) = pH − pKa = 1.5 − 3.5 = −2, so [A⁻]/[HA] = 0.01 (1:100 ratio). Thus ~1% ionized, 99% unionized. Unionized aspirin is absorbed well in the stomach despite low pH — the pH-partition hypothesis." }, { id: 27, chapter: "Ionization", q: "Buffer capacity (β) is maximum when:", options: ["pH = pKa ± 2", "pH = pKa (equal concentrations of weak acid and conjugate base)", "pH = 7", "The buffer is diluted 10-fold"], answer: 1, explanation: "β = 2.303C × Ka[H⁺]/(Ka+[H⁺])². Maximum β occurs when pH = pKa, where [HA] = [A⁻]. Effective buffering range is pH = pKa ± 1. Beyond ±2 pH units from pKa, buffering capacity is negligible." },