Chemical Equilibrium Exercise: Federal Board 2nd Year Chemistry (New Syllabus) 

Solutions to Multiple Choice Questions

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Question i

Statement: Which phenomenon describes a shift in equilibrium due to the addition of an ion already involved in the equilibrium?
Options:
a) Is Chaichter Principle
b) Common Ion Effect
c) Internal Law
d) Angearchy Principle
Answer: b) Common Ion Effect
Key Trick: Recognize that the question describes the addition of an ion already present in the equilibrium system. This directly matches the definition of the Common Ion Effect.
Concept: The Common Ion Effect (a consequence of Le Chatelier’s Principle) occurs when adding an ion common to an existing equilibrium suppresses dissociation, shifting the equilibrium to reduce the disturbance. For example, adding Na⁺ to a saturated NaCl solution reduces NaCl solubility.

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Question ii

Statement: What property is used to distinguish between strong and weak acids?
Options:
a) Molar mass
b) Extent of ionization
c) Density
d) Melting point
Answer: b) Extent of ionization
Key Trick: Strong acids fully dissociate in water (100% ionization), while weak acids partially dissociate. Physical properties (a, c, d) are irrelevant.
Concept: Extent of ionization quantifies acid strength. Strong acids (e.g., HCl) ionize completely; weak acids (e.g., CH₃COOH) ionize partially, measured via conductivity or pH.

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Question iii

Statement: Which factor is considered when distinguishing between strong and weak bases?
Options:
a) Giant
b) Otor
c) Extent of ionization
d) Stolability
Answer: c) Extent of ionization
Key Trick: Bases, like acids, are classified by ionization extent. Options “a,” “b,” and “d” are nonsensical/distractors.
Concept: Extent of ionization defines base strength: strong bases (e.g., NaOH) dissociate fully; weak bases (e.g., NH₃) dissociate partially.

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Question iv

Statement: What defines a buffer solution?
Options:
a) High concentration of ions
b) Presence of a weak acid and its conjugate base or a weak base and its conjugate acid
c) Low pH
d) Absence of ions
Answer: b) Presence of a weak acid and its conjugate base or a weak base and its conjugate acid
Key Trick: Buffers resist pH changes via equilibrium pairs. Option “b” explicitly states the required components.
Concept: A buffer solution requires a weak acid/base and its conjugate to neutralize added H⁺/OH⁻ via equilibrium shifts (e.g., CH₃COOH/CH₃COO⁻).

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Question v

Statement: How can a buffer solution be made?
Options:
a) Mixing strong acids and bases
b) Mixing a weak acid and its conjugate base (or a weak base and its conjugate acid)
c) Binding a strong acid
d) Adding water to a strong base
Answer: b) Mixing a weak acid and its conjugate base (or a weak base and its conjugate acid)
Key Trick: Buffers require weak conjugate pairs. Strong acids/bases (a, c, d) cannot form buffers.
Concept: Buffers are prepared by combining a weak acid with its salt (e.g., acetic acid + sodium acetate) or a weak base with its salt (e.g., ammonia + ammonium chloride).

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Question vi

Statement: What role does HCO₃⁻ play in controlling pH in blood?
Options:
a) Neutralization of acids
b) Buffering against changes in pH
c) Catalging reactors
d) Enhancing oxygen transport
Answer: b) Buffering against changes in pH
Key Trick: HCO₃⁻ is part of the blood buffer system. “Catalging reactors” is a typo for “catalyzing reactions,” which is incorrect here.
Concept: The bicarbonate buffer system (H₂CO₃/HCO₃⁻) maintains blood pH (≈7.4). HCO₃⁻ absorbs excess H⁺, and H₂CO₃ absorbs excess OH⁻.

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Question vii

Statement: How is the concentration of a slightly soluble salt calculated?
Options:
a) Using the solubility product constant (Ksp)
b) Transition with a strong base
c) Measuring density
d) Conductivity measurement
Answer: a) Using the solubility product constant (Ksp)
Key Trick: Ksp is the standard method for solubility calculations. Options “b,” “c,” and “d” are indirect or irrelevant.
Concept: For a salt like AgCl ⇌ Ag⁺ + Cl⁻, Ksp = [Ag⁺][Cl⁻] allows calculation of ion concentrations and solubility.

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Question viii

Statement: Which term is used to describe the strength of an acid in terms of its ionization in water?
Options:
a) Acid Concentration
b) Acid Dissociation Constant (Ka)
c) Acid Molarity
d) Acid Equilibrium Constant
Answer: b) Acid Dissociation Constant (Ka)
Key Trick: “Ka” quantifies acid strength via equilibrium constant. “Ris” in the question is a typo for “Ka.”
Concept: Ka is the equilibrium constant for HA ⇌ H⁺ + A⁻. Strong acids have high Ka; weak acids have low Ka.

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Question ix

Statement: Which type of solvent would favour the partitioning of a polar solute?
Options:
a) Non-polar solvent
b) Hydrophobic solvent
c) Polar solvent
d) Apexic solvent
Answer: c) Polar solvent
Key Trick: “Like dissolves like”: polar solutes dissolve best in polar solvents. Non-polar solvents (a, b, d) repel polar solutes.
Concept: Polar solvents (e.g., water) solvate polar solutes via dipole-dipole interactions or H-bonding.

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Question x

Statement: What is the partition coefficient defined as?
Options:
a) Ratio of solute concentration in one solvent to the other
b) Ratio of solute concentration in a single solvent
c) Ratio of solute mass in one solvent to the other
d) Ratio of solute mass in a single solvent
Answer: a) Ratio of solute concentration in one solvent to the other
Key Trick: Partition coefficient involves two immiscible solvents (e.g., oil and water). Mass ratios (c, d) are incorrect.
Concept: The partition coefficient (K) = [solute] in solvent₁ / [solute] in solvent₂, predicting solute distribution in extraction.

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Final Answer Key:
i. b | ii. b | iii. c | iv. b | v. b | vi. b | vii. a | viii. b | ix. c | x. a

Solutions to Short Answer Questions

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Question i

Question: Explain the common ion effect with a suitable example.
Answer:
The common ion effect occurs when adding an ion already present in a dissociation equilibrium suppresses ionization or reduces solubility. This is governed by Le Chatelier’s principle – the system shifts to counteract the disturbance.

Example:
In a saturated solution of AgCl:
AgCl_(s) ⇌ Ag⁺_(aq) + Cl^–_(aq)
Adding NaCl (which dissociates into Cl⁻) increases [Cl⁻]. The equilibrium shifts left, precipitating more AgCl and reducing its solubility.

Key Concept: Equilibrium shift due to shared ions.
Tags: #CommonIonEffect #LeChatelier #Solubility

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Question ii

Question: Differentiate between strong and weak acids using the extent of ionization and ( K_a ).
Answer:

Property	Strong Acids	Weak Acids
Extent of Ionization	100% dissociation in water	Partial dissociation (<100%)
( K_a )	Very large ( Ka ≫ 1 )	Small (Ka < 1 )
Examples	HCl, HNO₃, H₂SO₄	CH₃COOH, HCN, H₂CO₃

Key Concept: Acid strength defined by ionization completeness and ( Ka ).
Tags: #AcidStrength #Ionization #Ka

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Question iii

Question: Differentiate between strong and weak bases using the extent of ionization and ( Kb ).
Answer:

Property	Strong Bases	Weak Bases
Extent of Ionization	100% dissociation in water	Partial dissociation (<100%)
( Kb )	Very large (Kb ≫ 1 )	Small (Kb < 1 )
Examples	NaOH, KOH, Ca(OH)₂	NH₃, CH₃NH₂, C₅H₅N (pyridine)

Key Concept: Base strength defined by dissociation completeness and ( Kb ).
Tags: #BaseStrength #Ionization #Kb

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Question iv

Question: Define a buffer solution and provide an example of how it can be made.
Answer:
A buffer solution resists pH changes when small amounts of acid/base are added. It consists of:

• A weak acid and its conjugate base (e.g., CH₃COOH/CH₃COO⁻), OR
• A weak base and its conjugate acid (e.g., NH₃/NH₄⁺).

Preparation Example:
Mix 0.1 M acetic acid (CH₃COOH) and 0.1 M sodium acetate (CH₃COONa). The resulting solution maintains pH ≈ 4.74.

Key Concept: pH stability via conjugate acid-base pairs.
Tags: #BufferSolution #ConjugatePair #pH

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Question v

Question: How does a buffer solution control pH? Include chemical equations.
Answer:
Buffers neutralize added H⁺ or OH⁻ through equilibrium shifts:

1. Against added acid (H⁺):
   Conjugate base (A⁻) consumes H⁺:
   H⁺ + A^– → HA
2. Against added base (OH⁻):
   Weak acid (HA) consumes OH⁻:
   OH^– + HA → A^– + H₂O
3. Example (Acetate Buffer):
4. CH₃COOH → H⁺ + CH₃COO^–

• Add H⁺: H⁺ + CH₃COO^– → CH₃COOH
• Add OH⁻: OH^– + CH₃COOH → CH₃COO^– + H₂O

Key Concept: Dynamic equilibrium maintaining pH.
Tags: #BufferAction #Neutralization #Equilibrium

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Question vi

Question: Describe the uses of buffer solutions in various applications.
Answer:

1. Biological Systems:

• Blood (bicarbonate buffer, pH 7.4).
• Cells (phosphate buffer, pH 7.2).

1. Industrial Processes:

• Fermentation (pH control for yeast).
• Dyeing textiles.

1. Laboratory Analysis:

• Calibrating pH meters.
• Enzymatic reactions (optimal pH).

1. Consumer Products:

• Shampoos (prevent scalp irritation).
• Foods (preserve flavor/texture).

Key Concept: pH stability critical for chemical/biological processes.
Tags: #BufferApplications #Biology #Industry

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Question vii

Question: Explain how HCO₃^– ) controls pH in blood.
Answer:
The bicarbonate buffer system (H₂CO₃/HCO₃⁻) maintains blood pH at 7.4:
CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃^–

• Added H⁺: Combines with HCO₃⁻ to form H₂CO₃, which decomposes to CO₂ (exhaled by lungs).
• Added OH⁻: Combines with H₂CO₃ to form HCO₃⁻ and H₂O.

Key Role:
HCO₃⁻ acts as a base to neutralize excess acid. The system is enhanced by lung/kidney regulation of CO₂ and HCO₃⁻.

Key Concept: Physiological pH regulation.
Tags: #BloodBuffer #Bicarbonate #pHHomeostasis

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Question viii

Question: Calculate the concentration of a slightly soluble salt given its ( Ksp.
Answer:
For a salt AxBy, Ksp = [A^m+]^x [B^n-]^y .
Example (AgCl, Ksp = 1.8 x 10^-10:

1. Dissolution: AgCl_(s) ⇌ Ag⁺ + Cl^–
2. Let solubility = ( s ). Then ( [Ag⁺] = s , ( [Cl^–] = s .
3. Ksp = s x s = s² = 1.8 x 10^-10
4. s = √1.8 x10^-10 = 1.34 x 10^-5 M

Concentration of AgCl: 1.34 x 10^-5 mol/L.
(Note: ( R_eq) in the question is a typo for ( Ksp ).

Key Concept: Solubility prediction via ( Ksp).
Tags: #Solubility #Ksp #Concentration

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Question ix

Question: Explain the significance of ( Ka ) in determining acid strength.
Answer:
The acid dissociation constant ( Ka ) quantifies acid strength:
Ka = [H⁺][A^–]/[HA]

• High ( Ka ) (e.g., 10³ for HCl): Strong acid → complete dissociation.
• Low ( Ka ) (e.g., 10⁻⁵ for CH₃COOH): Weak acid → partial dissociation.

Significance:

• ( Ka ) is fixed for a given acid/temperature.
• Allows comparison of acid strengths.
• Related to pH: pH = -log √Ka.[HA] for weak acids.

Key Concept: Quantitative measure of acidity.
Tags: #AcidStrength #Ka #Equilibrium

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Question x(a)

Question: How does heat affect the equilibrium between CuSO₄ and CuSO₄ .5H₂O

Answer: The equilibrium reaction is: CuSO4 .5H2O(s) ⇌ CuSO_4(s) + 5H2O_(g) (endothermic)

• Heating: Favors forward reaction (dehydration) → blue hydrate turns white.
• Cooling: Favors reverse reaction (hydration) → white solid turns blue.

Key Concept: Le Chatelier’s principle for temperature-dependent equilibria.
Tags: #Hydrate #LeChatelier #EquilibriumShift

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Question xi

Question: What happens when water is added to anhydrous CuSO₄?

Answer: Anhydrous CuSO₄ (white powder) absorbs water to form hydrated CuSO₄ .5H2O (blue crystals): CuSO_4(s) + 5H2O_(l) → CuSO₄ .5H2O_(s)

• Observation: Color change from white to blue.
• Energy: Exothermic reaction (releases heat).

Key Concept: Reversible hydration/dehydration.
Tags: #Hydration #ColorChange #CopperSulphate

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