NEET Mock Test | PreMedicalBiology

Q1. For the reaction: H₂(g) + I₂(g) ⇌ 2HI(g) , Relation between Kₚ and Kc:

A. Kₚ = Kc(RT)² B. Kₚ = Kc C. Kₚ = Kc / (RT) D. Kₚ = Kc · RT

Explanation:
Option B is correct. Explanation: Δn(g) = (moles of gaseous products) − (moles of gaseous reactants) , Δn(g) = 2 − (1 + 1) = 0 , We know: Kₚ = Kc (RT)^(Δn) , Since Δn = 0: , Kₚ = Kc (RT)⁰ = Kc , Therefore, Kₚ = Kc.

Q2. For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), equilibrium constant = K , Find the equilibrium constant for: NH₃(g) ⇌ ½ N₂(g) + 3/2 H₂(g)

A. 1/K B. K² C. 1/√K D. √K

Explanation:
Option C is correct. Given reaction: N₂ + 3H₂ ⇌ 2NH₃, K = K , Step 1: Reverse the reaction: 2NH₃ ⇌ N₂ + 3H₂ , K₁ = 1/K , Step 2: Divide the equation by 2: NH₃ ⇌ ½ N₂ + 3/2 H₂ , K′ = (K₁)^(1/2) = (1/K)^(1/2) = 1/√K , Therefore, K′ = 1/√K.

Q3. For the reaction: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) , Effect of increasing pressure:

A. Shift equilibrium to the right B. Shift equilibrium to the left C. Have no effect D. Increase the value of Kc

Explanation:
Option B is correct. According to Le Chatelier’s principle, increasing pressure shifts the equilibrium towards the side with fewer moles of gas. , Number of moles of gaseous species: , Reactant side = 1 (PCl₅) , Product side = 2 (PCl₃ + Cl₂) , Since the reactant side has fewer moles, the equilibrium shifts to the left. , Also, the value of Kc remains constant at a given temperature.

Q4. Which of the following will change the value of the equilibrium constant?

A. Addition of a catalyst B. Changing the volume of the container C. Increasing the temperature D. Adding more reactants

Explanation:
Option C is correct. The equilibrium constant (K) depends only on temperature. Catalyst: does not change K (only speeds up attainment of equilibrium) , Change in volume/pressure: does not change K (only shifts equilibrium position) , Addition of reactants/products: does not change K (only shifts equilibrium position) , Temperature: changes the value of K because it changes the extent of reaction at equilibrium , Therefore, only change in temperature changes the equilibrium constant.

Q5. For the exothermic reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) , The yield of SO₃ is maximized at:

A. High temperature and high pressure B. Low temperature and high pressure C. Low temperature and low pressure D. High temperature and low pressure

Explanation:
Option B is correct. The forward reaction is exothermic, so lowering the temperature shifts equilibrium in the forward direction to produce more SO₃. , Effect of pressure: , Number of moles of gases: , Reactants = 2 + 1 = 3 moles , Products = 2 moles , Increasing pressure shifts equilibrium toward the side with fewer moles of gas (products). , Therefore, low temperature and high pressure favor maximum SO₃ yield.

Q6. In a reversible reaction, a catalyst:

A. Increases the yield of the product. B. Decreases the activation energy for the forward reaction only. C. Shortens the time required to attain equilibrium. D. Increases the equilibrium constant.

Explanation:
Option C is correct. A catalyst provides an alternative reaction pathway with lower activation energy for both forward and reverse reactions equally. It increases the rate of attainment of equilibrium but does not change the equilibrium constant or the equilibrium composition of the system.

Q7. The reaction quotient (Qc) is used to:

A. Calculate the temperature of the reaction. B. Determine the rate of the reaction. C. Find the efficiency of a catalyst. D. Predict the direction in which the reaction will proceed.

Explanation:
Option D is correct. The reaction quotient (Qc) is calculated using the same expression as the equilibrium constant (Kc), but it applies to a system that is not necessarily at equilibrium. Comparison of Qc and Kc helps predict the direction of the reaction: If Qc < Kc → Reaction proceeds forward (towards products) , If Qc > Kc → Reaction proceeds backward (towards reactants) , If Qc = Kc → System is at equilibrium , Hence, Qc is a direction predictor tool, not a measure of rate or temperature.

Q8. For the equilibrium: C(s) + H₂O(g) ⇌ CO(g) + H₂(g), the correct expression for Kc is:

A. [CO][H₂] / ([C][H₂O]) B. [CO][H₂] / [H₂O] C. [CO] / [H₂O] D. [H₂O] / ([CO][H₂])

Explanation:
Option B is correct. The equilibrium constant (Kc) is written using concentrations of gaseous and aqueous species only. For the given reaction: C(s) + H₂O(g) ⇌ CO(g) + H₂(g) , Rules for writing Kc: Pure solids (s) and pure liquids (l) are not included in equilibrium expression because their effective concentration is constant (taken as 1). Only gases (g) and aqueous (aq) species are included. , Therefore: Kc = [CO][H₂] / [H₂O]

Q9. Addition of an inert gas at constant pressure to the equilibrium: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) will:

A. Shift equilibrium to the left. B. Shift equilibrium to the right. C. Have no effect. D. Increase the NH₃ yield.

Explanation:
Option A is correct. At constant pressure, addition of an inert gas increases the total volume of the system. Effect on equilibrium: Increase in volume → decrease in partial pressure of all reacting gases , System shifts in the direction that produces more moles of gas (Le Chatelier’s principle) , For the given reaction: Reactants: 1 + 3 = 4 moles of gas , Products: 2 moles of gas , Since reactant side has more moles of gas, equilibrium shifts towards reactants (left side) to oppose the decrease in pressure.

Q10. The value of Kp for the reaction: CaCO₃(s) ⇌ CaO(s) + CO₂(g) is equal to:

A. Total pressure of the system B. 1 / PCO₂ C. Partial pressure of CO₂ D. PCaO × PCO₂ / PCaCO₃

Explanation:
Option C is correct. For heterogeneous equilibrium involving solids and gases: CaCO₃(s) ⇌ CaO(s) + CO₂(g) , Rules for Kp: Pure solids (s) and pure liquids (l) are not included in equilibrium expressions because their activities are taken as unity (1). Only gaseous species are considered. Therefore: Kp = PCO₂

Q11. If ΔG° for a reaction is negative, then:

A. Keq > 1 B. Keq < 1 C. Keq = 0 D. Keq = 1

Explanation:
Option A is correct. The relationship between standard Gibbs free energy change and equilibrium constant is: ΔG° = −RT ln Keq Where: R = gas constant , T = absolute temperature , If ΔG° < 0 → ln Keq must be positive Therefore, Keq > 1 , This indicates that at equilibrium, products are favored, and the reaction is spontaneous in the forward direction under standard conditions.

Q12. In the reaction A + B ⇌ C + D, if the concentration of A is doubled, the equilibrium constant will:

A. Double B. halved C. Remain the same D. Become four times

Explanation:
Option C is correct. The equilibrium constant (Kc) for a reaction is defined as: Kc = [C][D] / [A][B] , Key principle: Kc depends only on temperature and it is independent of initial concentrations, pressure, or catalysts. , Effect of changing concentration: Doubling [A] disturbs the equilibrium , The system shifts according to Le Chatelier’s principle to re-establish equilibrium , Only the position of equilibrium changes, not the value of Kc

Q13. For which of the following reactions is Δng negative?

A. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) B. N₂(g) + O₂(g) ⇌ 2NO(g) C. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) D. 2HI(g) ⇌ H₂(g) + I₂(g)

Explanation:
Option C is correct. Δng = (moles of gaseous products) − (moles of gaseous reactants) , A. PCl₅ ⇌ PCl₃ + Cl₂ , Δng = (1 + 1) − 1 = +1 | B. N₂ + O₂ ⇌ 2NO , Δng = 2 − (1 + 1) = 0 | C. N₂ + 3H₂ ⇌ 2NH₃ , Δng = 2 − (1 + 3) = 2 − 4 = −2 (negative) | D. 2HI ⇌ H₂ + I₂ , Δng = (1 + 1) − 2 = 0

Q14. If Kc for a reaction is 10²⁰, the equilibrium mixture contains:

A. Mostly reactants B. Mostly products C. Equal amounts of reactants and products D. Only reactants

Explanation:
Option B is correct. The equilibrium constant is defined as: Kc = [Products] / [Reactants] , Given: Kc = 10²⁰ (very large value) , A very high value of Kc indicates that the equilibrium lies far towards the product side At equilibrium, the concentration of products is much greater than that of reactants The reaction proceeds almost to completion in the forward direction

Q15. The equilibrium Ice ⇌ Water is affected by pressure because:

A. Ice is denser than water B. Water is denser than ice C. Ice is a solid D. Water is a liquid

Explanation:
Option B is correct. The equilibrium between ice and water is a solid–liquid equilibrium: Ice(s) ⇌ Water(l) , Effect of pressure is explained using Le Chatelier’s principle: Increasing pressure favors the state with lower volume. Lower volume corresponds to the higher density phase , Since: Water is denser than ice. Therefore, water occupies less volume than ice. therefore, on increasing pressure, Equilibrium shifts toward water (liquid phase).

Q16. Le Chatelier’s principle is not applicable to:

A. Physical equilibrium. B. Chemical equilibrium. C. Both A and B D. None of these

Explanation:
Option D is correct. Le Chatelier’s principle is applicable to all systems in equilibrium, including both physical and chemical equilibria. It states that when a system at equilibrium is subjected to a change in conditions (such as concentration, temperature, or pressure), the equilibrium shifts in a direction that counteracts the imposed change.

Q17. For the reaction: N₂O₄(g) ⇌ 2NO₂(g) , The degree of dissociation (α) increases with:

A. Increase in pressure B. Decrease in pressure C. Decrease in temperature D. Addition of NO₂

Explanation:
Option B is correct. For the given equilibrium, Δn₍g₎ = 2 − 1 = +1. Since Δn₍g₎ is positive, a decrease in pressure shifts the equilibrium towards the side with more moles of gas (forward direction), thereby increasing the degree of dissociation (α).

Q18. For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) , The units of Kc are:

A. mol L⁻¹ B. mol² L⁻² C. L² mol⁻² D. L mol⁻¹

Explanation:
Option C is correct. Δn = (moles of gaseous products) − (moles of gaseous reactants) , Δn = 2 − (1 + 3) = −2 , Units of Kc = (mol L⁻¹)^(Δn) = (mol L⁻¹)⁻² = L² mol⁻².

Q19. The equilibrium constant for a spontaneous reaction is:

A. Always negative B. Always zero C. Usually large and positive D. Less than 1

Explanation:
Option C is correct. The spontaneity of a reaction is governed by the standard Gibbs free energy change (ΔG°), which is related to the equilibrium constant (K) by the thermodynamic relation: ΔG° = −RT ln K , where R is the gas constant and T is the absolute temperature., For a spontaneous reaction, ΔG° < 0. This condition is satisfied only when ln K > 0, i.e., K > 1. Therefore, the equilibrium lies toward the product side. In most cases, for strongly spontaneous reactions, K ≫ 1, indicating extensive formation of products at equilibrium. , However, it is important to note that spontaneity does not imply instantaneous completion; rather, it indicates the thermodynamic tendency of the reaction. Thus, a spontaneous reaction is generally associated with a large positive value of K, though the exact magnitude depends on temperature and the value of ΔG°.

Q20. For the reaction: 2A + B ⇌ C , If the volume is reduced to half of the original value, the rate of the forward reaction relative to the backward reaction will:

A. Increase B. Decrease C. Remain the same D. Cannot be predicted

Explanation:
Option A is correct. On decreasing the volume to 1/2, the concentrations of all species become twice their initial values. Rate of forward reaction: r₍f₎ ∝ [A]²[B] → (2[A])²(2[B]) = 8[A]²[B] → increases by a factor of 8 , Rate of backward reaction: r₍b₎ ∝ [C] → (2[C]) → increases by a factor of 2 , Thus, the forward reaction rate increases more than the backward reaction rate. Hence, the rate of the forward reaction relative to the backward reaction increases.

Q21. Which of the following represents a physical equilibrium?

A. H₂(g) + I₂(g) ⇌ 2HI(g) B. CaCO₃(s) ⇌ CaO(s) + CO₂(g) C. H₂O(l) ⇌ H₂O(g) D. NH₄Cl(s) ⇌ NH₃(g) + HCl(g)

Explanation:
Option C is correct. A physical equilibrium involves a change in physical state without any change in chemical composition. H₂O(l) ⇌ H₂O(g) is a phase equilibrium between liquid water and its vapour, hence it is a physical equilibrium.

Q22. At equilibrium, the Gibbs free energy change (ΔG, not ΔG°) is:

A. Positive B. Negative C. Zero D. One

Explanation:
Option C is correct. At equilibrium, the Gibbs free energy change (ΔG) is equal to zero, indicating that the system is in a state of minimum free energy and no net change occurs in the forward or reverse direction.

Q23. For the reaction A ⇌ B, Kc=2. If the initial amount of A is 10 moles in a 1 L container, the number of moles of B at equilibrium is:

A. 3.33 B. 6.66 C. 5 D. 2

Explanation:
Option B is correct. For the reaction A ⇌ B , Initial moles: [A] = 10, [B] = 0 , Let x moles of A convert to B at equilibrium. , Equilibrium moles: [A] = 10 − x , [B] = x , Kc = [B]/[A] = x/(10 − x) = 2 , x = 2(10 − x) , x = 20 − 2x , 3x = 20 , x = 20/3 = 6.66 moles of B at equilibrium = 6.66.

Q24. The equilibrium constant Kp is not defined for reactions occurring in:

A. Gas phase B. Solution phase C. Solid phase D. Both B and C

Explanation:
Option D is correct. Kp is defined in terms of partial pressures and is applicable only to gaseous systems. Therefore, it is not defined for reactions occurring in solution or in pure solid phases.

Q25. According to the Brønsted–Lowry concept, a base is a species that:

A. Donates a proton (H⁺) B. Accepts a proton (H⁺) C. Donates an electron pair D. Accepts an electron pair

Explanation:
Option B is correct. According to the Bronsted–Lowry theory, a base is a species that accepts a proton (H⁺).

Q26. According to the Bronsted–Lowry concept of acids and bases, the conjugate acid of a base is formed by the addition of a proton (H⁺) to the base. On this basis, the conjugate acid of NH₂⁻ is:

A. NH₃ B. NH₄⁺ C. NH₂⁻ D. N₂

Explanation:
Option A is correct. A conjugate acid is formed when a base accepts a proton (H⁺). , NH₂⁻ + H⁺ → NH₃ , Therefore, the conjugate acid of NH₂⁻ is NH₃.

Q27. According to the Lewis concept of acids and bases, a Lewis base is a species that donates an electron pair. Among the following species, identify the Lewis base:

A. BF₃ B. AlCl₃ C. H₂O D. Fe³⁺

Explanation:
Option C is correct. A Lewis base is an electron pair donor. H₂O has lone pairs of electrons on the oxygen atom, which it can donate to form a coordinate bond. Hence, H₂O acts as a Lewis base.

Q28. The pH of an aqueous solution of NaOH depends on its concentration. Calculate the pH of a 0.001 M NaOH solution at 25°C:

A. 3 B. 11 C. 7 D. 10

Explanation:
Option B is correct. NaOH is a strong base and dissociates completely in aqueous solution: NaOH → Na⁺ + OH⁻ , Therefore, [OH⁻] = 0.001 M = 10⁻³ M , pOH = −log[OH⁻] = −log(10⁻³) = 3 , pH + pOH = 14 , pH = 14 − 3 = 11 ,

Q29. Ostwald’s dilution law is applicable to which type of electrolytes?

A. Strong electrolytes B. Weak electrolytes C. Non-electrolytes D. All types of electrolytes

Explanation:
Option B is correct. Ostwald’s dilution law is applicable to weak electrolytes because they are only partially ionised in solution. On dilution, the degree of dissociation increases significantly, and the law describes this relationship. Strong electrolytes are completely dissociated and do not follow this law.

Q30. The pH of an aqueous solution of a very dilute strong acid (10⁻⁷ M HCl) is asked. At such low concentration, contribution of H⁺ ions from water cannot be neglected. What is the pH of 10⁻⁷ M HCl solution at 25°C?

A. 7.00 B. 7.21 C. 6.79 D. 8.00

Explanation:
Option C is correct. HCl is a strong acid and dissociates completely: HCl → H⁺ + Cl⁻ , Given [H⁺] from HCl = 10⁻⁷ M , But water also contributes: [H⁺] from water = 10⁻⁷ M , Total [H⁺] = 10⁻⁷ + 10⁻⁷ = 2 × 10⁻⁷ M , pH = −log(2 × 10⁻⁷) , pH = 7 − log 2 , pH = 7 − 0.301 , pH ≈ 6.79

Q31. Aqueous solution of a salt formed from a weak acid and a strong base (for example, CH₃COONa) undergoes hydrolysis in water. Predict the nature of the solution and its pH:

A. pH = 7 (neutral) B. pH > 7 (basic) C. pH < 7 (acidic) D. pH = 0 (strongly acidic)

Explanation:
Option B is correct. CH₃COONa is a salt of a weak acid (CH₃COOH) and a strong base (NaOH). The acetate ion (CH₃COO⁻) undergoes hydrolysis in water: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ , Formation of OH⁻ ions makes the solution basic in nature, hence pH > 7.

Q32. The solubility product (Ksp) of Ag₂SO₄ is related to its molar solubility (s). For the dissociation equilibrium of Ag₂SO₄ in water, determine the correct expression of Ksp in terms of s:

A. s² B. 27s⁴ C. s³ D. 4s³

Explanation:
Option D is correct. Ag₂SO₄(s) ⇌ 2Ag⁺(aq) + SO₄²⁻(aq) , If solubility = s mol L⁻¹, then: [Ag⁺] = 2s , [SO₄²⁻] = s , Ksp = [Ag⁺]² [SO₄²⁻] , Ksp = (2s)² (s) , Ksp = 4s³

Q33. The ionization of a weak acid such as CH₃COOH in aqueous solution can be suppressed by applying the common ion effect. Which of the following will suppress the ionization of CH₃COOH?

A. Adding HCl B. Adding CH₃COONa C. Both A and B D. Adding H₂O

Explanation:
Option C is correct. CH₃COOH ⇌ H⁺ + CH₃COO⁻ , Adding HCl increases the concentration of H⁺ ions, while adding CH₃COONa increases CH₃COO⁻ ions. According to Le Chatelier’s principle, the equilibrium shifts to the left, suppressing the ionization of CH₃COOH. This is known as the common ion effect.

Q34. A buffer solution resists change in pH upon addition of small amounts of acid or base. Which of the following combinations forms an acidic buffer solution?

A. CH₃COOH + CH₃COONa B. NaOH + NaCl C. HCl + NH₄Cl D. NaCl + KCl

Explanation:
Option A is correct. A buffer solution consists of a weak acid and its conjugate base (salt of a strong base), or a weak base and its conjugate acid. CH₃COOH (weak acid) + CH₃COONa (salt of weak acid and strong base) forms an acidic buffer solution that resists change in pH.

Q35. The ionic product of water (Kw) represents the equilibrium constant for the self-ionization of water. At 25°C, the value of Kw is:

A. 10⁻⁷ B. 10¹⁴ C. 7 D. 10⁻¹⁴

Explanation:
Option D is correct. Water undergoes self-ionization as: H₂O ⇌ H⁺ + OH⁻ , At 25°C, the product of hydrogen ion concentration and hydroxide ion concentration is constant: Kw = [H⁺][OH⁻] = 10⁻¹⁴ mol² L⁻² ,

Q36. An aqueous solution of ammonium hydroxide (NH₄OH) is taken. When solid ammonium chloride (NH₄Cl) is added to this solution, what will be the effect on its pH?

A. The pH increases B. The pH decreases C. The pH remains unchanged D. The solution becomes neutral (pH = 7)

Explanation:
Option B is correct. Ammonium hydroxide (NH₄OH) is a weak base that partially dissociates in water: NH₄OH ⇌ NH₄⁺ + OH⁻ , On adding ammonium chloride (NH₄Cl), a strong electrolyte, it dissociates completely to give NH₄⁺ ions: NH₄Cl → NH₄⁺ + Cl⁻ , The increase in concentration of the common ion NH₄⁺ shifts the equilibrium of NH₄OH dissociation to the left according to Le Chatelier’s principle. This suppresses the ionization of NH₄OH, resulting in a decrease in the concentration of OH⁻ ions. Since pH depends on the concentration of OH⁻ (for basic solutions), a decrease in [OH⁻] leads to a decrease in pH.

Q37. The solubility of Silver chloride (AgCl) is least in which of the following solutions?

A. Pure water B. 0.01 M Sodium chloride C. 0.1 M Sodium chloride D. 0.1 M Silver nitrate

Explanation:
Option D is correct. The solubility of a sparingly soluble salt like AgCl is governed by the common ion effect, according to the principles of the Solubility product. AgCl (s) ⇌ Ag⁺ (aq) + Cl⁻ (aq) , In 0.1 M AgNO₃, the concentration of Ag⁺ ions (common ion) is very high. This shifts the equilibrium to the left, significantly suppressing the solubility of AgCl. Although 0.1 M NaCl also provides a common ion (Cl⁻), the effect is comparable in magnitude; however, due to the direct involvement of Ag⁺ in the dissolution equilibrium, the presence of Ag⁺ (from AgNO₃) suppresses solubility more effectively in this context. Thus, the maximum suppression of solubility occurs in 0.1 M AgNO₃, making it the correct answer.

Q38. Hydrogen sulfide gas, Hydrogen sulfide, is passed through an acidic solution containing Hydrochloric acid to precipitate Group II cations. The reason for this is:

A. HCl increases the ionization of H₂S B. HCl acts as an oxidizing agent C. HCl decreases the concentration of S²⁻ ions D. HCl acts as a solvent

Explanation:
Option C is correct. H₂S is a weak electrolyte and ionizes in aqueous solution as: H₂S (aq) ⇌ 2H⁺ (aq) + S²⁻ (aq) , When HCl is added, it furnishes a high concentration of H⁺ ions (common ion). According to the Common ion effect, the equilibrium shifts to the left, thereby decreasing the concentration of S²⁻ ions. As this result, Only those metal sulfides having very low solubility product (Ksp) (i.e., Group II cations) can precipitate even at low S²⁻ concentration. Sulfides of other groups do not precipitate under these conditions. Hence, HCl effectively controls sulfide ion concentration, enabling selective precipitation of Group II cations.

Q39. Which of the following salts does not undergo hydrolysis in aqueous solution?

A. Sodium acetate B. Ammonium chloride C. Potassium nitrate D. Sodium carbonate

Explanation:
Option C is correct. Hydrolysis of salts depends on the nature of the parent acid and base: Salt of weak acid + strong base → undergoes hydrolysis (basic solution) Example: CH₃COONa , Salt of strong acid + weak base → undergoes hydrolysis (acidic solution) Example: NH₄Cl , Salt of weak acid + weak base → undergoes hydrolysis , Salt of strong acid + strong base → does not undergo hydrolysis and forms a neutral solution , Potassium nitrate is formed from: Potassium hydroxide (strong base) and Nitric acid (strong acid). Hence, it does not hydrolyze in water and remains neutral.

Q40. The Henderson–Hasselbalch equation for a basic buffer is:

A. pH = pKₐ + log([Salt]/[Base]) B. pOH = pK_b + log([Salt]/[Base]) C. pH = pK_b + log([Salt]/[Base]) D. pOH = pKₐ + log([Salt]/[Base])

Explanation:
Option B is correct. For a basic buffer: pOH = pK_b + log([Salt]/[Base]) , For basic buffer solutions, pOH is calculated using pK_b and the ratio of concentration of salt to base.

Q41. The pH of a 0.1 M monobasic weak acid solution is 2. The degree of dissociation (α) of the acid is:

A. 0.1 B. 0.01 C. 10 D. 0.2

Explanation:
Option A is correct. pH = 2 , [H⁺] = 10⁻ᵖᴴ = 10⁻² = 0.01 M , For a monobasic weak acid: HA ⇌ H⁺ + A⁻ , Degree of dissociation (α): α = [H⁺] / C , Given: C = 0.1 M , α = 0.01 / 0.1 = 0.1 , degree of dissociation (α) = 0.1.

Q42. A solution having pH = 0 is classified as:

A. Strongly acidic B. Neutral C. Strongly basic D. Weakly acidic

Explanation:
Option A is correct. pH = 0 , [H⁺] = 10⁻ᵖᴴ = 10⁰ = 1 M , Since a very high concentration of H⁺ ions (1 M) is present, the solution is strongly acidic.

Q43. The solubility product constants (Ksp) of the sulfides of copper, silver, and mercury are given as: Ksp (CuS) = 10⁻³¹ , Ksp (Ag₂S) = 10⁻⁴⁴ , Ksp (HgS) = 10⁻⁵⁴ , Now the correct order of solubility of these compounds is:

A. HgS > Ag₂S > CuS B. CuS > Ag₂S > HgS C. Ag₂S > CuS > HgS D. CuS > HgS > Ag₂S

Explanation:
Option B is correct. Solubility of sparingly soluble salts is directly related to their solubility product (Ksp). Higher the Ksp value, greater is the solubility. Given: CuS (Ksp = 10⁻³¹) , Ag₂S (Ksp = 10⁻⁴⁴) , HgS (Ksp = 10⁻⁵⁴) , Since: 10⁻³¹ > 10⁻⁴⁴ > 10⁻⁵⁴ , Therefore, solubility order is: CuS > Ag₂S > HgS , Hence, CuS is most soluble and HgS is least soluble.

Q44. Which of the following species is amphiprotic in nature?

A. CO₃²⁻ B. H₃O⁺ C. Cl⁻ D. HPO₄²⁻

Explanation:
Option D is correct. An amphiprotic species is one which can both donate and accept a proton (H⁺). For HPO₄²⁻: It can act as a base (accept H⁺): HPO₄²⁻ + H⁺ → H₂PO₄⁻ , It can act as an acid (donate H⁺): HPO₄²⁻ → PO₄³⁻ + H⁺ , Since it can both donate and accept a proton, HPO₄²⁻ is amphiprotic in nature.

Q45. The buffer capacity of a buffer solution is maximum under which of the following conditions?

A. [Salt] = [Acid] B. pH = pKₐ C. Both A and B D. [Salt] ≫ [Acid]

Explanation:
Option C is correct. Buffer capacity is defined as the ability of a buffer solution to resist change in pH on addition of small amounts of acid or base. For an acidic buffer system (weak acid + its salt), maximum buffer capacity is obtained when: [Salt] = [Acid] , Using Henderson–Hasselbalch equation: pH = pKₐ + log([Salt]/[Acid]) , When [Salt] = [Acid], log(1) = 0 ⇒ pH = pKₐ These both conditions are equivalent and correspond to maximum buffering action. so buffer capacity is maximum when both [Salt] = [Acid] and pH = pKₐ.

Test Summary