JEE Chemistry Equilibrium MCQs Set 36

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Chapter-wise Objective Questions: Equilibrium

View or download the dedicated Equilibrium MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: In aqueous solution, the ionisation constants for carbonic acid are K₁ = 4.2 × 10⁻⁷ and K₂ = 4.8 × 10⁻¹¹. Select the correct statement for a saturated 0.034 M solution of the carbonic acid.
(a) The concentration of CO₃²⁻ is 0.034 M
(b) The concentration of CO₃²⁻ is greater than that of HCO₃⁻
(c) The concentration of H⁺ and HCO₃⁻ are approximately equal
(d) The concentration of H⁺ is double that of CO₃²⁻
Show Answer & Explanation

Answer: (c) The concentration of H⁺ and HCO₃⁻ are approximately equal

Explanation:
1. Since \( K_{1} \) ≫ \( K_{2} \), the first ionisation H₂CO₃ ⇌ \( H^{+} \) + HCO₃⁻ produces nearly all the \( H^{+} \), so \( [H^{+}] \) ≈ [HCO₃⁻].
2. From \( K_{2} \) = [H⁺][CO₃²⁻]/[HCO₃⁻], [CO₃²⁻] ≈ \( K_{2} = 4.8 \times 10^{-11} M \), far less than [HCO₃⁻].

Teacher's Note:
1. The second ionisation is negligible compared with the first.
2. [CO₃²⁻] is approximately equal to \( K_{2} \).

Question 2: At 25°C, the solubility product of Mg(OH)₂ is 1.0 × 10⁻¹¹. At which pH will Mg²⁺ ions start precipitating in the form of Mg(OH)₂ from a solution of 0.001 M Mg²⁺ ions?
(a) 9
(b) 10
(c) 11
(d) 8
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. \( K_{sp} \) = [Mg²⁺][OH⁻]², so [OH⁻] = \( \sqrt{10^{-11}/10^{-3}} = 10^{-4} M \).
2. pOH = 4, so pH = \( 14 - 4 = 10 \).

Teacher's Note:
1. Precipitation starts when the ionic product reaches \( K_{sp} \).
2. Convert pOH to pH.

Question 3: Solid Ba(NO₃)₂ is gradually dissolved in a 1.0 × 10⁻⁴ M Na₂CO₃ solution. At what concentration of Ba²⁺ will a precipitate begin to form? (K_sp for BaCO₃ = 5.1 × 10⁻⁹)
(a) 4.1 × 10⁻⁵ M
(b) 5.1 × 10⁻⁵ M
(c) 8.1 × 10⁻⁸ M
(d) 8.1 × 10⁻⁷ M
Show Answer & Explanation

Answer: (b) 5.1 × 10⁻⁵ M

Explanation:
1. [CO₃²⁻] = \( 1.0 \times 10^{-4} M \).
2. [Ba²⁺] = \( K_{sp}/[CO_{3}^{2-}] = 5.1 \times 10^{-9}/10^{-4} = 5.1 \times 10^{-5} M \).

Teacher's Note:
1. Precipitation starts when [Ba²⁺][CO₃²⁻] = \( K_{sp} \).
2. Na₂CO₃ supplies one carbonate per formula unit.

Question 4: Four species are listed below: I. HCO₃⁻; II. H₃O⁺; III. HSO₄⁻; IV. HSO₃F. Which one is the correct sequence of their acid strength?
(a) IV < II < III < I
(b) II < III < I < IV
(c) I < III < II < IV
(d) III < I < II < IV
Show Answer & Explanation

Answer: (c) I < III < II < IV

Explanation:
1. The conjugate bases are CO₃²⁻, \( H_{2}O \), SO₄²⁻ and SO₃F⁻. The stronger the acid, the weaker its conjugate base.
2. Base strength: SO₃F⁻ < \( H_{2}O \) < SO₄²⁻ < CO₃²⁻.
3. So acid strength: HCO₃⁻ < HSO₄⁻ < \( H_{3}O^{+} \) < HSO₃F, i.e. I < III < II < IV.

Teacher's Note:
1. Compare the stability of the conjugate bases.
2. HSO₃F is a superacid.

Question 5: The equilibrium constants K_p1 and K_p2 for the reactions X ⇌ 2Y and Z ⇌ P + Q, respectively, are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal, then the ratio of total pressure at these equilibria is:
(a) 1 : 36
(b) 1 : 1
(c) 1 : 3
(d) 1 : 9
Show Answer & Explanation

Answer: (a) 1 : 36

Explanation:
1. For \( X \) ⇌ \( 2Y \) with small α: \( K_{p1} = 4\alpha^{2}P_{1}/(1 - \alpha^{2}) \approx 4\alpha^{2}P_{1} \).
2. For \( Z \) ⇌ \( P + Q \): \( K_{p2} = \alpha^{2}P_{2}/(1 - \alpha^{2}) \approx \alpha^{2}P_{2} \).
3. \( K_{p1}/K_{p2} = 4P_{1}/P_{2} = 1/9 \), so \( P_{1}/P_{2} = 1/36 \).

Teacher's Note:
1. Write \( K_p \) in terms of α and total pressure for each reaction.
2. Neglect α² compared with 1.

Question 6: The pK_a of a weak acid, HA, is 4.80. The pK_b of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be:
(a) 9.58
(b) 4.79
(c) 7.01
(d) 9.22
Show Answer & Explanation

Answer: (c) 7.01

Explanation:
1. pH = \( 7 + \frac{1}{2}(pK_a - pK_b) = 7 + \frac{1}{2}(4.80 - 4.78) = 7.01 \).

Teacher's Note:
1. The salt of a weak acid and weak base gives an almost neutral solution when \( pK_a \approx pK_b \).
2. The acid is very slightly weaker, so pH is slightly above 7.

Question 7: The pK_a of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionised is:
(a) 4.5
(b) 2.5
(c) 9.5
(d) 7.0
Show Answer & Explanation

Answer: (c) 9.5

Explanation:
1. At 50% ionisation \( [A^{-}] \) = [HA], so pH = \( pK_a + \log 1 = 4.5 \).
2. pOH = \( 14 - 4.5 = 9.5 \).

Teacher's Note:
1. Half-ionised means the Henderson ratio is 1.
2. pH + pOH = 14 at \( 25^{\circ}C \).

Question 8: The first and second dissociation constants of an acid H₂A are 1.0 × 10⁻⁵ and 5.0 × 10⁻¹⁰, respectively. The overall dissociation constant of the acid will be:
(a) 5.0 × 10⁵
(b) 5.0 × 10¹⁵
(c) 5.0 × 10⁻¹⁵
(d) 0.2 × 10⁵
Show Answer & Explanation

Answer: (c) 5.0 × 10⁻¹⁵

Explanation:
1. For \( H_{2}A \) ⇌ \( 2H^{+} + A^{2-}, K = K_{1} \times K_{2} = 1.0 \times 10^{-5} \times 5.0 \times 10^{-10} = 5.0 \times 10^{-15} \).

Teacher's Note:
1. Overall \( K \) is the product of the stepwise constants.
2. A weak diprotic acid has a very small overall \( K \).

Question 9: Phosphorus pentachloride dissociates as PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). If total pressure at equilibrium of the reaction mixture is p and degree of dissociation of PCl₅ is x, the partial pressure of PCl₃ will be:
(a) (x/(x + 1))p
(b) (2x/(1 − x))p
(c) (x/(x − 1))p
(d) (x/(1 − x))p
Show Answer & Explanation

Answer: (a) (x/(x + 1))p

Explanation:
1. Starting with 1 mol: PCl₅ = \( 1 - x \), PCl₃ = \( x \), Cl₂ = \( x \); total moles = \( 1 + x \).
2. p(PCl₃) = \( (x/(1 + x))p \).

Teacher's Note:
1. Partial pressure = mole fraction × total pressure.
2. Add the moles at equilibrium to get the denominator.

Question 10: An amount of solid NH₄HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH₃ and H₂S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for the decomposition of NH₄HS at this temperature is:
(a) 0.11
(b) 0.17
(c) 0.18
(d) 0.30
Show Answer & Explanation

Answer: (a) 0.11

Explanation:
1. Let \( p(H_{2}S) = x \); p(NH₃) = \( 0.50 + x \); total = \( 0.50 + 2x = 0.84 \), so \( x = 0.17 \) atm.
2. p(NH₃) = 0.67 atm, so \( K_p = 0.67 \times 0.17 = 0.114 \approx 0.11 \).

Teacher's Note:
1. Ammonia is already present initially.
2. The solid is left out of \( K_p \).

Question 11: For the reaction 2NO₂(g) ⇌ 2NO(g) + O₂(g) [K_c = 1.8 × 10⁻⁶ at 184°C], when K_p and K_c are compared at 184°C it is found that:
(a) whether K_p is greater than, less than or equal to K_c depends upon the total gas pressure
(b) K_p = K_c
(c) K_p is less than K_c
(d) K_p is greater than K_c
Show Answer & Explanation

Answer: (d) K_p is greater than K_c

Explanation:
1. \( K_p = K_c(RT)^\Deltan_g \) with \( \Deltan_g = 3 - 2 = +1 \).
2. RT (about \( 37.5 L \) atm mol⁻¹ at \( 457 K \)) is greater than 1, so \( K_p \) > \( K_c \).

Teacher's Note:
1. \( \Deltan_g \) is positive here.
2. The relation holds regardless of the total pressure.

Question 12: What is the conjugate base of OH⁻?
(a) O²⁻
(b) O⁻
(c) H₂O
(d) O₂
Show Answer & Explanation

Answer: (a) O²⁻

Explanation:
1. A conjugate base is formed when an acid loses a proton.
2. OH⁻ loses \( H^{+} \) to give \( O^{2-} \).

Teacher's Note:
1. Conjugate base = acid \( - H^{+} \).
2. \( H_{2}O \) is the conjugate acid of OH⁻.

Question 13: The solubility product of a salt having general formula MX₂ in water is 4 × 10⁻¹². The concentration of M²⁺ ions in the aqueous solution of the salt is:
(a) 4.0 × 10⁻¹⁰ M
(b) 1.6 × 10⁻⁴ M
(c) 1.0 × 10⁻⁴ M
(d) 2.0 × 10⁻⁶ M
Show Answer & Explanation

Answer: (c) 1.0 × 10⁻⁴ M

Explanation:
1. \( K_{sp} = S(2S)^{2} = 4S^{3} = 4 \times 10^{-12} \), so \( S = 10^{-4} M \).
2. \( [M^{2+}] = S = 1.0 \times 10^{-4} M \).

Teacher's Note:
1. For MX₂, \( [M^{2+}] = S \) and \( [X^{-}] = 2S \).
2. \( K_{sp} = 4S^{3} \).

Question 14: What is the equilibrium expression for the reaction P₄(s) + 5O₂(g) ⇌ P₄O₁₀(s)?
(a) K_c = [P₄O₁₀]/([P₄][O₂]⁵)
(b) K_c = [P₄O₁₀]/(5[P₄][O₂])
(c) K_c = [O₂]⁵
(d) K_c = 1/[O₂]⁵
Show Answer & Explanation

Answer: (d) K_c = 1/[O₂]⁵

Explanation:
1. Pure solids do not appear in the equilibrium expression.
2. Only \( O_{2}(g) \) appears, in the denominator: \( K_c = 1/[O_{2}]^{5} \).

Teacher's Note:
1. The activity of a pure solid is taken as 1.
2. Reactants go in the denominator.

Question 15: For the reaction CO(g) + Cl₂(g) ⇌ COCl₂(g), K_p/K_c is equal to:
(a) 1/RT
(b) RT
(c) √(RT)
(d) 1.0
Show Answer & Explanation

Answer: (a) 1/RT

Explanation:
1. \( K_p = K_c(RT)^\Deltan_g \) with \( \Deltan_g = 1 - 2 = -1 \).
2. \( K_p/K_c \) = (RT)⁻¹ = 1/RT.

Teacher's Note:
1. \( \Deltan_g \) is negative here.
2. \( K_p \) is smaller than \( K_c \).

Question 16: The molar solubility (in mol L⁻¹) of a sparingly soluble salt MX₄ is S. The corresponding solubility product is K_sp. S is given in terms of K_sp by the relation:
(a) \( S = (K_{sp}/128)^{1/4} \)
(b) \( S = (K_{sp}/128)^{1/4} \times 2 \)
(c) \( S = (K_{sp}/256)^{1/5} \)
(d) \( S = (K_{sp}/256)^{1/5} \times 2 \)
Show Answer & Explanation

Answer: (c) \( S = (K_{sp}/256)^{1/5} \)

Explanation:
1. MX₄ ⇌ \( M^{4+} + 4X^{-} \) gives \( [M^{4+}] = S \) and \( [X^{-}] = 4S \).
2. \( K_{sp} = S(4S)^{4} = 256S^{5} \), so \( S = (K_{sp}/256)^{1/5} \).

Teacher's Note:
1. A 1 : 4 salt has \( K_{sp} = 256S^{5} \).
2. Raise to the power 1/5.

Question 17: For the reaction equilibrium N₂O₄(g) ⇌ 2NO₂(g), the concentrations of N₂O₄ and NO₂ at equilibrium are 4.8 × 10⁻² and 1.2 × 10⁻² mol L⁻¹, respectively. The value of K_c for the reaction is:
(a) 3.3 × 10² mol L⁻¹
(b) 3 × 10⁻¹ mol L⁻¹
(c) 3 × 10⁻³ mol L⁻¹
(d) 3 × 10³ mol L⁻¹
Show Answer & Explanation

Answer: (c) 3 × 10⁻³ mol L⁻¹

Explanation:
1. \( K_c \) = [NO₂]²/[N₂O₄] = \( (1.2 \times 10^{-2})^{2}/(4.8 \times 10^{-2}) = 1.44 \times 10^{-4}/4.8 \times 10^{-2} = 3 \times 10^{-3} \) mol \( L^{-1} \).

Teacher's Note:
1. Square the product concentration.
2. Unit of \( K_c \) is mol \( L^{-1} \) since \( \Deltan = 1 \).

Question 18: Consider the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −198 kJ. On the basis of Le Chatelier's principle, the condition favourable for the forward reaction is:
(a) lowering of temperature as well as pressure
(b) increasing temperature as well as pressure
(c) lowering the temperature and increasing pressure
(d) any value of temperature and pressure
Show Answer & Explanation

Answer: (c) lowering the temperature and increasing pressure

Explanation:
1. The forward reaction is exothermic, so lowering the temperature favours it.
2. It reduces the number of gas moles (3 to 2), so increasing the pressure favours it.

Teacher's Note:
1. Exothermic reactions are favoured by low temperature.
2. Reactions with \( \Deltan_g \) < 0 are favoured by high pressure.

Question 19: The solubility in water of a sparingly soluble salt AB₂ is 1.0 × 10⁻⁵ mol L⁻¹. Its solubility product will be:
(a) 4 × 10⁻¹⁵
(b) 4 × 10⁻¹⁰
(c) 1 × 10⁻¹⁵
(d) 1 × 10⁻¹⁰
Show Answer & Explanation

Answer: (a) 4 × 10⁻¹⁵

Explanation:
1. AB₂ ⇌ \( A^{2+} + 2B^{-} \), so \( [A^{2+}] = S \) and \( [B^{-}] = 2S \).
2. \( K_{sp} = S(2S)^{2} = 4S^{3} = 4 \times (10^{-5})^{3} = 4 \times 10^{-15} \).

Teacher's Note:
1. Remember the factor of 4 for a 1 : 2 salt.
2. Cube the solubility.

Question 20: Which one of the following statements is not true? (Assume 25°C.)
(a) The conjugate base of H₂PO₄⁻ is HPO₄²⁻
(b) pH + pOH = 14 for all aqueous solutions at 25°C
(c) The pH of 1 × 10⁻⁸ M HCl is 8
(d) 96,500 C of electricity when passed through a CuSO₄ solution deposits 1 g equivalent of copper at the cathode
Show Answer & Explanation

Answer: (c) The pH of 1 × 10⁻⁸ M HCl is 8

Explanation:
1. \( (a), (b) \) and \( (d) \) are correct statements.
2. An acid solution can never have a pH above 7: in \( 10^{-8} M \) HCl the ionisation of water contributes about \( 10^{-7} M H^{+} \), so the pH is just below 7 (about 6.98), not 8.

Teacher's Note:
1. The self-ionisation of water cannot be ignored at very low acid concentrations.
2. An acidic solution always has pH < 7 at \( 25^{\circ}C \).

Question 21: One of the following equilibria is not affected by change in volume of the flask:
(a) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
(b) N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
(c) N₂(g) + O₂(g) ⇌ 2NO(g)
(d) SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g)
Show Answer & Explanation

Answer: (c) N₂(g) + O₂(g) ⇌ 2NO(g)

Explanation:
1. A change in volume shifts an equilibrium only if the number of gas moles changes.
2. \( N_{2} + O_{2} \) ⇌ 2NO has 2 moles on each side \( (\Deltan_g = 0) \), so it is unaffected.

Teacher's Note:
1. Compare the gas moles on both sides.
2. \( \Deltan_g = 0 \) means no effect of pressure or volume.

Question 22: Which one of the following species acts as both a Bronsted acid and a base?
(a) H₂PO₂⁻
(b) HPO₃²⁻
(c) HPO₄²⁻
(d) All of the above
Show Answer & Explanation

Answer: (c) HPO₄²⁻

Explanation:
1. HPO₄²⁻ donates a proton to give PO₄³⁻ and accepts a proton to give H₂PO₄⁻, so it is amphiprotic.
2. H₂PO₂⁻ and HPO₃²⁻ have no ionisable O–H hydrogen and do not act as acids.

Teacher's Note:
1. An amphiprotic species can both donate and accept \( H^{+} \).
2. Only HPO₄²⁻ among the options has this property.

Question 23: The solubility of Ca(OH)₂ is S mol L⁻¹. The solubility product (K_sp) under the same condition is:
(a) 4S³
(b) 3S⁴
(c) 4S²
(d) S³
Show Answer & Explanation

Answer: (a) 4S³

Explanation:
1. Ca(OH)₂ ⇌ Ca²⁺ + 2OH⁻, [Ca²⁺] = \( S \) and [OH⁻] = \( 2S \).
2. \( K_{sp} = S(2S)^{2} = 4S^{3} \).

Teacher's Note:
1. A 1 : 2 salt has \( K_{sp} = 4S^{3} \).
2. Square the OH⁻ concentration.

Question 24: The pH of 0.005 M calcium acetate (pK_a of CH₃COOH = 4.74) is:
(a) 7.04
(b) 9.37
(c) 9.26
(d) 8.37
Show Answer & Explanation

Answer: (d) 8.37

Explanation:
1. Calcium acetate gives [CH₃COO⁻] = \( 2 \times 0.005 = 0.01 M \).
2. pH = \( 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C = 7 + 2.37 + \frac{1}{2}(-2) = 7 + 2.37 - 1 \).
3. pH = 8.37.

Teacher's Note:
1. Salt of a weak acid and strong base is basic.
2. Remember \( \log 0.01 = -2 \).

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Multiple Choice Questions (MCQs) for JEE Chemistry Equilibrium

JEE Chemistry Equilibrium Objective Test Questions

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