i. Which phenomenon describes a shift in equilibrium due to the addition of an ion already involved in the equilibrium?
Answer: b) Common Ion Effect
Explanation: The common ion effect describes the decrease in solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution. This causes a shift in the equilibrium position according to Le Chatelier's principle.
ii. What property is used to distinguish between strong and weak acids?
Answer: b) Extent of ionization
Explanation: Strong acids ionize completely in water, while weak acids only partially ionize. This extent of ionization is the key distinguishing property.
iii. Which factor is considered when distinguishing between strong and weak bases?
Answer: c) Extent of ionization
Explanation: Similar to acids, strong bases dissociate completely into ions in water, while weak bases only partially dissociate. The extent of ionization distinguishes them.
iv. What defines a buffer solution?
Answer: b) Presence of a weak acid and its conjugate base (or a weak base and its conjugate acid)
Explanation: A buffer solution resists changes in pH upon addition of small amounts of acid or base. It is typically composed of a weak acid and its salt (conjugate base) or a weak base and its salt (conjugate acid).
v. How can a buffer solution be made?
Answer: b) Mixing a weak acid and its conjugate base (or a weak base and its conjugate acid)
Explanation: Combining a weak acid with its conjugate base (or a weak base with its conjugate acid) creates a buffer system capable of neutralizing added acids or bases.
vi. What role does HCO₃⁻ play in controlling pH in blood?
Answer: b) Buffering against changes in pH
Explanation: The bicarbonate ion (HCO₃⁻) acts as a buffer in the blood, neutralizing excess acids (like H⁺) and helping to maintain a stable blood pH around 7.4.
vii. How is the concentration of a slightly soluble salt calculated?
Answer: a) Using the solubility product constant (Ksp)
Explanation: For slightly soluble salts, the equilibrium concentration of ions in a saturated solution can be calculated from the Ksp value.
viii. Which term is used to describe the strength of an acid in terms of its ionization in water?
Answer: b) Acid Dissociation Constant (Ka)
Explanation: The acid dissociation constant (Ka) is a quantitative measure of the strength of an acid in solution. A larger Ka value indicates a stronger acid.
ix. Which type of solvent would favour the partitioning of a polar solute?
Answer: c) Polar solvent
Explanation: "Like dissolves like." A polar solute will preferentially partition into a polar solvent.
x. What is the partition coefficient defined as?
Answer: a) Ratio of solute concentration in one solvent to the other
Explanation: The partition coefficient (Kd) is the ratio of the concentration of a solute in one solvent to its concentration in a second solvent at equilibrium.
i. Explain the common ion effect with a suitable example.
Answer: The common ion effect is the decrease in the solubility of a sparingly soluble salt when a soluble salt containing a common ion is added to the solution. For example, the solubility of silver chloride (AgCl) is reduced when sodium chloride (NaCl) is added to its saturated solution. The added Cl⁻ ions shift the equilibrium (AgCl(s) [source symbol unavailable] Ag⁺(aq) + Cl⁻(aq)) to the left, precipitating more AgCl. This effect is used in gravimetric analysis and purification processes.
ii. Differentiate between strong and weak acids using the extent of ionization and Ka.
Answer: Strong acids ionize completely (100%) in water, whereas weak acids ionize only partially. Consequently, strong acids have a very large Ka value (often considered infinite), while weak acids have a small Ka value. For example, HCl (strong acid) has a Ka of about 10⁷, while acetic acid (weak acid) has a Ka of about 1.8 × 10⁻⁵. The smaller the Ka, the weaker the acid.
iii. Differentiate between strong and weak bases using the extent of ionization and Kb.
Answer: Strong bases dissociate completely (100%) into ions in water, while weak bases only partially dissociate. Strong bases have a very large Kb value, while weak bases have a small Kb. For example, NaOH (strong base) has a Kb of about 10⁰, while ammonia (weak base) has a Kb of about 1.8 × 10⁻⁵. The smaller the Kb, the weaker the base.
iv. Define a buffer solution and provide an example of how it can be made.
Answer: A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. It is composed of a weak acid and its conjugate base (or a weak base and its conjugate acid). An example is an acetic acid/acetate buffer, made by mixing acetic acid (CH₃COOH) with sodium acetate (CH₃COONa). This buffer maintains a pH around 4.76.
v. How does a buffer solution control pH? Include chemical equations in your explanation.
Answer: A buffer controls pH by neutralizing added acids or bases. For an acidic buffer (weak acid HA and its salt A⁻):
Added acid (H⁺): H⁺ + A⁻ [source symbol unavailable] HA (weak acid, no pH change). Added base (OH⁻): OH⁻ + HA [source symbol unavailable] A⁻ + H₂O (weak base, no pH change). The equilibrium shifts to maintain the ratio of [HA]/[A⁻], keeping the pH nearly constant.
vi. Describe the uses of buffer solutions in various applications.
Answer: Buffer solutions are used in:
Biological Systems: Blood is buffered by the bicarbonate system to maintain pH ~7.4. Pharmaceuticals: Many drugs require specific pH for stability and efficacy. Fermentation: Buffers maintain optimal pH for microbial growth. Analytical Chemistry: Buffers are used in titrations and calibrating pH meters.
Industrial Processes: Dyeing, paper manufacturing, and electroplating.
vii. Explain how HCO₃⁻ plays a role in controlling pH in the blood.
Answer: The bicarbonate buffer system in the blood consists of carbonic acid (H₂CO₃) and bicarbonate ions (HCO₃⁻). It maintains blood pH around 7.4. If excess acid (H⁺) enters the blood, it reacts with HCO₃⁻ to form H₂CO₃, which then breaks down into CO₂ and H₂O; CO₂ is exhaled. If base (OH⁻) enters, it reacts with H₂CO₃ to form HCO₃⁻ and water. This system effectively neutralizes pH changes.
viii. Calculate the concentration of a slightly soluble salt given its solubility product constant (Ksp).
Answer: For a salt like AgCl (Ksp = 1.8 × 10⁻¹⁰):
Dissociation: AgCl(s) [source symbol unavailable] Ag⁺(aq) + Cl⁻(aq)
Let solubility = s. Then [Ag⁺] = s, [Cl⁻] = s.
Ksp = [Ag⁺][Cl⁻] = s²
s = [source symbol unavailable]Ksp = [source symbol unavailable](1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/dm³.
i. Discuss the principles behind the common ion effect, providing two examples to illustrate its application.
Answer: The common ion effect is a direct consequence of Le Chatelier's principle. When a system at equilibrium is disturbed by adding a substance that provides an ion already present in the equilibrium, the system shifts to counteract the disturbance.
Principle: Consider a saturated solution of a sparingly soluble salt, e.g., AgCl: AgCl(s) [source symbol unavailable] Ag⁺(aq) + Cl⁻(aq) If we add NaCl (a source of Cl⁻ ions), the concentration of Cl⁻ increases. According to Le Chatelier's principle, the equilibrium shifts to the left to reduce the Cl⁻ concentration. This results in the precipitation of more AgCl, thereby decreasing the solubility of AgCl.
Example 1: Purification of Sodium Chloride (Salting Out): In the purification of NaCl, HCl gas is passed through a saturated solution of NaCl. The addition of Cl⁻ ions (common ion) shifts the equilibrium (NaCl(s) [source symbol unavailable] Na⁺(aq) + Cl⁻(aq)) to the left, causing pure NaCl to precipitate out. This is known as the common ion effect.
Example 2: Precipitation of Group II Cations in Qualitative Analysis: In qualitative analysis, Group II cations (e.g., Pb²⁺, Hg²⁺, Bi³⁺) are precipitated as sulfides by passing H₂S gas in the presence of dilute HCl. The HCl provides H⁺ ions, which suppress the ionization of H₂S (H₂S [source symbol unavailable] 2H⁺ + S²⁻), lowering the S²⁻ concentration. This ensures that only the sulfides of Group II cations (which have very low Ksp) precipitate, while the sulfides of later groups remain in solution.
In summary, the common ion effect is widely used in purification, precipitation, and analytical chemistry to control solubility and equilibrium.
ii. Compare and contrast strong and weak acids based on their extent of ionization and the value of Ka.
Answer: Strong and weak acids differ fundamentally in their behavior in aqueous solutions, which is reflected in their extent of ionization and their acid dissociation constant (Ka).
Comparison Table:
| Feature | Strong Acid | Weak Acid |
| Ionization | Complete (100%) | Partial (< 100%) |
| Ka Value | Very large (>> 1) | Small (< 1) |
| Equilibrium | Lies far to the right | Lies to the left |
| Conductivity | Good conductor | Poor conductor |
| pH (for same conc.) | Lower pH | Higher pH |
| Reaction with Metals | Vigorous | Slow or no reaction |
| Examples | HCl, HNO₃, H₂SO₄ | CH₃COOH, H₂CO₃, HF |
Extent of Ionization: Strong acids like HCl dissociate completely into H⁺ and Cl⁻ ions. Weak acids like acetic acid (CH₃COOH) establish an equilibrium with their ions, with only a small fraction ionizing.
Ka Value: The Ka value quantifies the strength of an acid. For a strong acid, Ka is very large (often not measured because the reaction is essentially complete). For a weak acid, Ka is small; the smaller the Ka, the weaker the acid.
Conclusion:
The extent of ionization and Ka are inversely related to acid strength. Strong acids have a high extent of ionization and a large Ka, while weak acids have a low extent of ionization and a small Ka.
iii. Elaborate on the factors that distinguish strong and weak bases, incorporating the concept of ionization and the base dissociation constant (Kb).
Answer: Strong and weak bases are distinguished by their extent of ionization in water and their base dissociation constant (Kb).
Extent of Ionization:
Strong Bases: Dissociate completely into metal ions and hydroxide ions (OH⁻). Examples: NaOH, KOH, Ca(OH)₂.
NaOH(aq) [source symbol unavailable] Na⁺(aq) + OH⁻(aq)
Weak Bases: React with water to produce a small amount of OH⁻. Examples: NH₃, amines.
NH₃(aq) + H₂O(l) [source symbol unavailable] NH₄⁺(aq) + OH⁻(aq)
Base Dissociation Constant (Kb):
Strong Bases: Have a very large Kb (effectively infinite). Weak Bases: Have a small Kb. The smaller the Kb, the weaker the base. Example: Kb for NH₃ is 1.8 × 10⁻⁵.
Factors:
1. Nature of the Base: The strength depends on the ability to accept protons (H⁺). Strong bases have a high affinity for protons.
2. Solubility: Strong bases are often highly soluble in water (e.g., NaOH), while some weak bases may have limited solubility.
3. pH of Solution: Strong bases produce a higher pH for the same concentration.
In summary, the extent of ionization and Kb are the primary factors distinguishing strong and weak bases.
iv. Provide a comprehensive explanation of buffer solutions, including how they are prepared, how they control pH, and their uses in different scenarios.
Answer: A buffer solution is a solution that resists significant changes in pH upon the addition of small amounts of acid or base.
Preparation: Buffers are prepared by mixing:
1. A weak acid and its conjugate base (e.g., CH₃COOH and CH₃COONa).
2. A weak base and its conjugate acid (e.g., NH₃ and NH₄Cl).
The best buffers have equal concentrations of the acid and conjugate base (pH = pKa).
Mechanism of pH Control:
Acidic Buffer (HA + A⁻):
Added H⁺: H⁺ + A⁻ [source symbol unavailable] HA (neutralized). Added OH⁻: OH⁻ + HA [source symbol unavailable] A⁻ + H₂O (neutralized).
Basic Buffer (B + BH⁺):
Added H⁺: H⁺ + B [source symbol unavailable] BH⁺ (neutralized). Added OH⁻: OH⁻ + BH⁺ [source symbol unavailable] B + H₂O (neutralized).
Uses:
1. Biological Systems: Blood pH is maintained at ~7.4 by the bicarbonate buffer.
2. Pharmaceuticals: Drug formulations require specific pH for stability and absorption.
3. Fermentation: Buffers maintain optimal pH for microbial growth.
4. Analytical Chemistry: Used in titrations, pH meter calibration, and electrophoresis.
5. Industrial Processes: Dyeing, paper manufacturing, and electroplating.
In summary, buffers are essential for maintaining stable pH conditions in biological, chemical, and industrial systems.
v. Calculate the pH of an NH₄OH and NH₄Cl buffer solution containing one mole of each. Kb of NH₄OH is 1.8 × 10⁻⁵. (Ans: 9.26)
Answer: For a basic buffer, we use the Henderson-Hasselbalch equation for bases:
pOH = pKb + log ([salt]/[base])
Given: [salt] = [NH₄Cl] = 1 mol, [base] = [NH₄OH] = 1 mol.
pKb = -log(Kb) = -log(1.8 × 10⁻⁵) = 4.74.
pOH = 4.74 + log(11) = 4.74 + 0 = 4.74.
pH = 14 - pOH = 14 - 4.74 = 9.26.
vi. A solute has a partition coefficient K = 4, 100 cm³ of water is shaken with 100 cm³ of benzene containing 0.20 mol of the solute. Find the amount of solute that remains in water at equilibrium. (Ans: 0.04 mol)
Answer: Let x be the moles of solute remaining in water at equilibrium.
Moles of solute in benzene = 0.20 - x.
Volume of water = 100 cm³ = 0.1 dm³, Volume of benzene = 100 cm³ = 0.1 dm³.
Concentration in water = x / 0.1 M.
Concentration in benzene = 0.20 - x0.1 M.
Partition coefficient K = [Concentration in benzene] / [Concentration in water] (assuming benzene is the organic phase)
4 = [0.20 - x0.1] / [x / 0.1]
4 = 0.20 - xx
4x = 0.20 - x
5x = 0.20
x = 0.205 = 0.04 moles. Therefore, 0.04 moles of solute remain in water at equilibrium.
Write the dissociation reaction and the solubility product constant expression for each of the following solids.
1. PbSO₄
Answer:
Dissociation Reaction: PbSO₄(s) [source symbol unavailable] Pb²⁺(aq) + SO₄²⁻(aq)
Ksp Expression: Ksp = [Pb²⁺][SO₄²⁻]
2. Al(OH)₃
Answer:
Dissociation Reaction: Al(OH)₃(s) [source symbol unavailable] Al³⁺(aq) + 3OH⁻(aq)
Ksp Expression: Ksp = [Al³⁺][OH⁻]³
3. BaSO₄
Answer:
Dissociation Reaction: BaSO₄(s) [source symbol unavailable] Ba²⁺(aq) + SO₄²⁻(aq)
Ksp Expression: Ksp = [Ba²⁺][SO₄²⁻]
1. Ksp for PbF₂ is 4 × 10⁻⁸. Calculate the concentration Pb²⁺ and F⁻ ions in solution.
Answer:
Dissociation: PbF₂(s) [source symbol unavailable] Pb²⁺(aq) + 2F⁻(aq)
Let solubility = s. Then [Pb²⁺] = s, [F⁻] = 2s.
Ksp = [Pb²⁺][F⁻]² = (s)(2s)² = 4s³
4 × 10⁻⁸ = 4s³
s³ = 1 × 10⁻⁸
s = 2.15 × 10⁻³ mol dm⁻³
Therefore, [Pb²⁺] = 2.15 × 10⁻³ mol dm⁻³ (approximately 2.5 × 10⁻³ as per provided answer)
[F⁻] = 2s = 4.3 × 10⁻³ mol dm⁻³.
2. Ksp for BaSO₄ is 1 × 10⁻¹⁰. Write the equation and the equilibrium expression for the dissolving of barium sulphate. Also calculate the
concentration of each ion. (Ans: 1 × 10⁻⁵ mol dm⁻³ each)
Answer:
Equation: BaSO₄(s) [source symbol unavailable] Ba²⁺(aq) + SO₄²⁻(aq)
Equilibrium Expression: Ksp = [Ba²⁺][SO₄²⁻]
Let solubility = s. Then [Ba²⁺] = s, [SO₄²⁻] = s.
Ksp = s² = 1 × 10⁻¹⁰
s = [source symbol unavailable](1 × 10⁻¹⁰) = 1 × 10⁻⁵ mol dm⁻³.
Therefore, [Ba²⁺] = [SO₄²⁻] = 1 × 10⁻⁵ mol dm⁻³.
i. Ammonium Chloride, NH₄Cl is a water-soluble salt. What will happen if this salt is added to a solution containing ammonium hydroxide? NH₄OH(aq) [source symbol unavailable] NH₄⁺(aq) + OH⁻(aq)
Answer: Adding NH₄Cl to a solution of NH₄OH will decrease the ionization of NH₄OH. NH₄Cl dissociates completely to provide NH₄⁺ ions (a common ion). According to Le Chatelier's principle, the increased concentration of NH₄⁺ will shift the equilibrium of NH₄OH ionization to the left. This results in a decrease in the concentration of OH⁻ ions and a lower pH. This is an example of the common ion effect.
ii. Carbonic acid is a weak acid. It ionizes in water as follows H₂CO₃(aq) [source symbol unavailable] 2H⁺(aq) + CO₃²⁻(aq) What will happen if a strong electrolyte such as Na₂CO₃ is added to a solution containing carbonic acid?
Answer: Adding Na₂CO₃ to a solution of carbonic acid will decrease the ionization of H₂CO₃. Na₂CO₃ dissociates completely to provide CO₃²⁻ ions (a common ion). According to Le Chatelier's principle, the increased concentration of CO₃²⁻ will shift the equilibrium to the left, reducing the concentration of H⁺ ions. This results in an increase in pH (making the solution less acidic). This is another example of the common ion effect.
When 2 grams of a solute is shaken with a mixture of 150 cm³ of water and 20 cm³ of chloroform. After shaking, 1.5 grams of the solute is found in the chloroform layer. Calculate the partition coefficient.
Answer:
Mass of solute in chloroform = 1.5 g
Mass of solute in water = Total mass - Mass in chloroform = 2 g - 1.5 g = 0.5 g
Volume of water = 150 cm³
Volume of chloroform = 20 cm³
Concentration in chloroform = 1.5 g / 20 cm³ = 0.075 g/cm³
Concentration in water = 0.5 g / 150 cm³ = 0.00333 g/cm³
Partition Coefficient (K) = Concentration in chloroform / Concentration in water
K = 0.0750.00333 = 22.5 (approximately).
