Equilibrium and Ionic Equilibrium — Practice

Question 1

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the relation between Kp and Kc is:
(a) Kp = Kc(RT)⁻² (b) Kp = Kc(RT)² (c) Kp = Kc (d) Kp = Kc(RT)

Accepted Answer

(a) Kp = Kc(RT)⁻².

Δn_g = 2 − 4 = −2, so Kp = Kc(RT)^Δn_g = Kc(RT)⁻².

Question 2

For a reaction Kc = 50. At a given instant Q = 5. The reaction will:
(a) proceed forward (b) proceed backward (c) be at equilibrium (d) stop

Accepted Answer

(a) proceed forward.

Since Q (5) < K (50), the numerator (products) is too small, so the net reaction proceeds forward until Q rises to K.

Question 3

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), increasing the pressure shifts the equilibrium:
(a) forward (toward NH₃) (b) backward (c) no change (d) reaction stops

Accepted Answer

(a) forward (toward NH₃).

Higher pressure favours the side with fewer gas moles: 2 (products) < 4 (reactants), so equilibrium shifts forward, increasing ammonia yield.

Question 4

Adding a catalyst to a reaction at equilibrium:
(a) does not shift the equilibrium (b) shifts it forward (c) shifts it backward (d) increases K

Accepted Answer

(a) does not shift the equilibrium.

A catalyst speeds up the forward and reverse reactions equally, so it helps equilibrium be reached faster but does not change its position or the value of K.

Question 5

The conjugate base of H₂O is:
(a) OH⁻ (b) H₃O⁺ (c) O²⁻ (d) H⁺

Accepted Answer

(a) OH⁻.

A Brønsted acid loses a proton to form its conjugate base: H₂O − H⁺ → OH⁻. (Gaining a proton would give the conjugate acid H₃O⁺.)

Question 6

The pH of 0.01 M HCl is:
(a) 2 (b) 1 (c) 12 (d) 0.01

Accepted Answer

(a) 2.

HCl is a strong acid, fully dissociated, so [H⁺] = 0.01 = 10⁻² M and pH = −log(10⁻²) = 2.

Question 7

The pH of 0.001 M NaOH is:
(a) 11 (b) 3 (c) 7 (d) 10

Accepted Answer

(a) 11.

[OH⁻] = 10⁻³, so pOH = 3 and pH = 14 − 3 = 11.

Question 8

The pH of a 0.1 M weak acid with Ka = 10⁻⁵ is:
(a) 3 (b) 5 (c) 1 (d) 2

Accepted Answer

(a) 3.

[H⁺] = √(Ka·C) = √(10⁻⁵ × 10⁻¹) = √(10⁻⁶) = 10⁻³ M, so pH = 3.

Question 9

The degree of dissociation of a 0.1 M weak acid with Ka = 10⁻⁵ is:
(a) 0.01 (b) 0.1 (c) 0.001 (d) 0.5

Accepted Answer

(a) 0.01.

α = √(Ka/C) = √(10⁻⁵/10⁻¹) = √(10⁻⁴) = 10⁻² = 0.01, i.e. 1% dissociated.

Question 10

A buffer contains 0.1 M acetic acid (pKa = 4.74) and 0.1 M sodium acetate. Its pH is:
(a) 4.74 (b) 7 (c) 9.26 (d) 5.74

Accepted Answer

(a) 4.74.

pH = pKa + log([salt]/[acid]) = 4.74 + log(0.1/0.1) = 4.74 + 0 = 4.74. When salt equals acid, pH equals pKa.

Question 11

To make a pH 5.74 buffer from acetic acid (pKa = 4.74), the required [salt]/[acid] ratio is:
(a) 10 (b) 1 (c) 0.1 (d) 100

Accepted Answer

(a) 10.

5.74 = 4.74 + log r ⇒ log r = 1 ⇒ r = 10. Ten times as much salt as acid raises the pH by one unit above pKa.

Question 12

An aqueous solution of CH₃COONa is:
(a) basic (b) acidic (c) neutral (d) amphoteric

Accepted Answer

(a) basic.

Sodium acetate is the salt of a weak acid (acetic) and a strong base (NaOH). The acetate ion hydrolyses (CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻), making the solution basic.

Question 13

The solubility of AgCl decreases when it is placed in a solution of NaCl because of:
(a) the common-ion effect (b) hydrolysis (c) increased Ksp (d) buffer action

Accepted Answer

(a) the common-ion effect.

Added Cl⁻ (a common ion) shifts AgCl ⇌ Ag⁺ + Cl⁻ backward (Le Chatelier), lowering the solubility of AgCl below its value in pure water.

Question 14

The molar solubility of AgCl (Ksp = 1.8×10⁻¹⁰) in water is approximately:
(a) 1.3×10⁻⁵ M (b) 1.8×10⁻¹⁰ M (c) 1.8×10⁻⁵ M (d) 9×10⁻¹¹ M

Accepted Answer

(a) 1.3×10⁻⁵ M.

For a 1:1 salt Ksp = s², so s = √(1.8×10⁻¹⁰) ≈ 1.34×10⁻⁵ mol/L.

Question 15

For a sparingly soluble salt AB₂, the relation between Ksp and molar solubility s is:
(a) Ksp = 4s³ (b) Ksp = s² (c) Ksp = 27s⁴ (d) Ksp = 108s⁵

Accepted Answer

(a) Ksp = 4s³.

AB₂ ⇌ A²⁺ + 2B⁻ gives [A²⁺] = s and [B⁻] = 2s, so Ksp = s·(2s)² = 4s³.

Practice — Equilibrium and Ionic Equilibrium

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