CBSE Class 11 Chemistry Chemical Equilibrium MCQs Set 09

Answer: (b) NH₃

Explanation:
1. Proton affinity is the tendency to accept H⁺ and is greatest for the best electron-pair donor.
2. Nitrogen in NH₃ is small and electronegative enough to hold a lone pair that is readily donated, so NH₃ binds H⁺ most strongly among these.

Teacher's Note:
1. A better proton acceptor is a stronger Brønsted base.
2. Compare this with the conjugate acid NH₄⁺, which is weak.

Answer: (d) C₂H₄

Explanation:
1. A Lewis acid accepts an electron pair. BF₃ and FeCl₃ have vacant orbitals, and Si in SiF₄ can expand its shell to accept a pair.
2. C₂H₄ has no vacant orbital and no tendency to accept a lone pair in this sense, so it is not a Lewis acid.

Teacher's Note:
1. Look for an electron-deficient centre or a metal ion.
2. BF₃ + :NH₃ → F₃B←NH₃ is the standard example.

Answer: (b) NH₂⁻

Explanation:
1. The conjugate base is formed by removing one H⁺: NH₃ − H⁺ = NH₂⁻.
2. NH₄⁺ is the conjugate acid of NH₃, because it is formed by adding H⁺.

Teacher's Note:
1. Conjugate base = acid minus H⁺.
2. Conjugate acid = base plus H⁺.

Answer: (d) HF

Explanation:
1. Down group 17 the H–X bond becomes longer and weaker, so the acid strength increases from HF to HI.
2. HF has the strongest H–X bond and is the weakest of the four hydrohalic acids in water.

Teacher's Note:
1. Bond strength, and not electronegativity, decides the order of hydrohalic acids.
2. Order: HF < HCl < HBr < HI.

Answer: (a) HClO < HClO₂ < HClO₃ < HClO₄

Explanation:
1. Each extra oxygen atom pulls electron density away from the O–H bond and stabilises the anion by spreading the charge.
2. So acidity increases with the oxidation state of Cl: +1, +3, +5, +7.

Teacher's Note:
1. More oxygen atoms on the central atom means a stronger oxoacid.
2. HClO₄ is one of the strongest known acids.

Answer: (c) NO₃⁻

Explanation:
1. HNO₃ is a strong acid and transfers its proton to water: HNO₃ + H₂O → H₃O⁺ + NO₃⁻.
2. So the species present are H₃O⁺ and NO₃⁻.

Teacher's Note:
1. The conjugate base of HNO₃ is NO₃⁻.
2. Strong acids are almost completely ionised.

Answer: (a) H₂PO₄⁻

Explanation:
1. A conjugate acid is formed by adding one proton.
2. HPO₄²⁻ + H⁺ → H₂PO₄⁻.

Teacher's Note:
1. Add one H⁺ and increase the charge by one.
2. PO₄³⁻ is the conjugate base of HPO₄²⁻.

Answer: (b) 1 × 10⁻⁸

Explanation:
1. α = 0.01% = 1 × 10⁻⁴.
2. K_a = cα² = 1.00 × (1 × 10⁻⁴)² = 1 × 10⁻⁸.

Teacher's Note:
1. Convert the per cent to a fraction first.
2. α is far below 5%, so K_a = cα² is valid.

Answer: (c) 2.1 × 10⁻⁴

Explanation:
1. α = 0.032. K_a = cα² ÷ (1 − α).
2. K_a = 0.2 × (0.032)² ÷ 0.968 = 2.05 × 10⁻⁴ ≈ 2.1 × 10⁻⁴.

Teacher's Note:
1. Using 1 − α ≈ 1 gives 2.0 × 10⁻⁴, close to the exact value.
2. Be careful: 3.2% is 0.032, not 3.2.

Answer: (a) 6.3 × 10⁻⁶ M

Explanation:
1. HCN ⇌ H⁺ + CN⁻ gives [CN⁻] = [H⁺] = √(K_ac).
2. [CN⁻] = √(4 × 10⁻¹⁰ × 0.1) = √(4 × 10⁻¹¹) = 6.3 × 10⁻⁶ M.

Teacher's Note:
1. For a weak acid, [H⁺] = √(K_a × c).
2. √(40 × 10⁻¹²) = 6.3 × 10⁻⁶.

Answer: (b) 11

Explanation:
1. KOH is a strong base, so [OH⁻] = 10⁻³ M and pOH = 3.
2. pH = 14 − 3 = 11.

Teacher's Note:
1. For a base find pOH first.
2. pH + pOH = 14 at 298 K.

Answer: (a) 12.48

Explanation:
1. Each formula unit of Ba(OH)₂ gives two OH⁻, so [OH⁻] = 2 × 0.015 = 0.030 M.
2. pOH = −log 0.030 = 1.52, so pH = 14 − 1.52 = 12.48.

Teacher's Note:
1. Remember the factor of 2 for Ba(OH)₂.
2. A strong base has a high pH.

Answer: (c) between 6 and 7

Explanation:
1. The H⁺ from water (10⁻⁷ M at 298 K) cannot be neglected when the acid is this dilute.
2. [H⁺] ≈ 1.05 × 10⁻⁷ M, which gives a pH of about 6.98, between 6 and 7 (just below 7).

Teacher's Note:
1. An acid, however dilute, cannot give a basic solution.
2. Add the H⁺ from the acid and from water.

Answer: (d) between 1 and 2

Explanation:
1. [H⁺] after mixing = (0.1 + 0.00001) ÷ 2 ≈ 0.05 M.
2. pH = −log 0.05 = 1.30, which lies between 1 and 2.

Teacher's Note:
1. The stronger acid dominates the mixture.
2. Do not average the pH values (that would give 3).

Answer: (a) is reduced by 1000 times

Explanation:
1. [H⁺] = 10⁻³ M at pH 3 and 10⁻⁶ M at pH 6.
2. The concentration falls by a factor of 10³ = 1000.

Teacher's Note:
1. Each unit of pH is a factor of 10 in [H⁺].
2. Higher pH means lower [H⁺].

Answer: (d) 1%

Explanation:
1. [H⁺] = 10⁻³ M and [H⁺] = cα.
2. α = 10⁻³ ÷ 0.1 = 0.01 = 1%.

Teacher's Note:
1. Degree of dissociation = [H⁺] ÷ initial concentration for a monoprotic acid.
2. 1% is less than 5%, so the approximation is valid.

Answer: (c) the temperature increases

Explanation:
1. K_w is an equilibrium constant, so it depends only on temperature.
2. Ionisation of water is endothermic, so K_w rises as the temperature rises.

Teacher's Note:
1. Adding H⁺ or OH⁻ shifts the equilibrium but does not change K_w.
2. K_w = 1.0 × 10⁻¹⁴ at 298 K.

Answer: (b) both pH and pOH decrease

Explanation:
1. K_w increases with temperature, so [H⁺] = [OH⁻] = √K_w increases.
2. A higher [H⁺] and a higher [OH⁻] mean that both pH and pOH decrease. The water remains neutral.

Teacher's Note:
1. Neutral water at a higher temperature has a pH below 7.
2. pH + pOH = pK_w, which becomes smaller.

Answer: (c) 2.9 × 10⁻¹¹

Explanation:
1. For a conjugate pair, K_a × K_b = K_w.
2. K_b = 1.0 × 10⁻¹⁴ ÷ 3.5 × 10⁻⁴ = 2.9 × 10⁻¹¹.

Teacher's Note:
1. A weak acid has a conjugate base that is a very weak base.
2. Use K_w = 1.0 × 10⁻¹⁴ at 298 K.

Answer: (d) its degree of ionisation (α)

Explanation:
1. From α = √(K_a ÷ c), a lower concentration c gives a larger α.
2. K_a is constant at a given temperature. [H⁺] = cα falls and the pH rises on dilution.

Teacher's Note:
1. This is Ostwald's dilution law.
2. More ions form, but they are spread through a larger volume.