Exercises
1: Liquid in equilibrium with vapour in sealed container. Volume suddenly increased.
Answer
(a) Initial effect on vapour pressure: Vapour pressure decreases immediately because the same amount of vapour now occupies a larger volume (pressure ∝ 1/V at constant T).
(b) Initial change in evaporation and condensation rates: Rate of evaporation remains unchanged (depends on liquid surface temperature). Rate of condensation decreases (fewer vapour molecules per unit volume to condense).
(c) Final vapour pressure when equilibrium restored: The final vapour pressure equals the original vapour pressure. At a given temperature, the equilibrium vapour pressure is a fixed constant (depends only on temperature, not volume).
2: Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g); [SO₂] = 0.60 M, [O₂] = 0.82 M, [SO₃] = 1.90 M
Answer
Kc = [SO₃]² / ([SO₂]² [O₂])
= (1.90)² / ((0.60)² × 0.82)
= 3.61 / (0.36 × 0.82)
= 3.61 / 0.2952 = 12.23
Q 6.3: Kp for I₂(g) ⇌ 2I(g) at 10⁵ Pa, 40% I atoms by volume
Partial pressure of I = 0.40 × 10⁵ Pa = 4 × 10⁴ Pa
Partial pressure of I₂ = 0.60 × 10⁵ Pa = 6 × 10⁴ Pa
Kp = (PI)² / PI₂ = (4×10⁴)² / (6×10⁴) = 16×10⁸ / 6×10⁴ = 2.67 × 10⁴ Pa
Q 6.4: Equilibrium constant expressions Kc
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g): Kc = [NO]²[Cl₂] / [NOCl]²
(ii) 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g): Kc = [NO₂]⁴[O₂] (solids excluded)
(iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq): Kc = [CH₃COOH][C₂H₅OH] / [CH₃COOC₂H₅] (pure liquid H₂O excluded)
(iv) Fe³⁺(aq) + 3OH⁻(aq) ⇌ Fe(OH)₃(s): Kc = 1 / ([Fe³⁺][OH⁻]³) (solid excluded)
(v) I₂(s) + 5F₂ ⇌ 2IF₅: Kc = [IF₅]² / [F₂]⁵ (solid I₂ excluded)
Q 6.5: Find Kc from Kp
Relation: Kp = Kc(RT)^Δng
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); Kp = 1.8×10⁻² at 500 K; Δng = 3−2 = 1
Kc = Kp/(RT) = 1.8×10⁻² / (0.0831 L bar mol⁻¹ K⁻¹ × 500) = 1.8×10⁻² / 41.55 = 4.33×10⁻⁴
(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 167 at 1073 K; Δng = 1
Kc = Kp/(RT) = 167 / (0.0831 × 1073) = 167 / 89.17 = 1.87
Q 6.6: Kc for reverse reaction; forward Kc = 6.3×10¹⁴
For NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g); Kc = 6.3×10¹⁴
Kc(reverse) = 1/Kc(forward) = 1/(6.3×10¹⁴) = 1.59×10⁻¹⁵
7: Why pure liquids and solids are ignored in K expression?
Answer
Pure liquids and solids have a constant molar concentration (concentration = density/molar mass = constant at constant T). This constant value is incorporated into the equilibrium constant K. Hence their "concentration" does not change and is set to 1 in the expression.
Q 6.8: Composition of equilibrium mixture for 2N₂(g) + O₂(g) ⇌ 2N₂O(g); Kc = 2.0×10⁻³⁷
Initial: [N₂] = 0.482/10 = 0.0482 M; [O₂] = 0.933/10 = 0.0933 M
Kc = [N₂O]² / ([N₂]²[O₂]) = 2.0×10⁻³⁷
Let x = [N₂O] at equilibrium; x² = 2.0×10⁻³⁷ × (0.0482)² × 0.0933
x² = 2.0×10⁻³⁷ × 2.323×10⁻³ × 0.0933 = 4.337×10⁻⁴¹
x = 2.08×10⁻²⁰·⁵ ≈ 6.6×10⁻²¹ mol L⁻¹
Equilibrium concentrations: [N₂] ≈ 0.0482 M, [O₂] ≈ 0.0933 M, [N₂O] ≈ 6.6×10⁻²¹ M
Q 6.9: Equilibrium amounts of NO and Br₂
2NO(g) + Br₂(g) ⇌ 2NOBr(g); initial: 0.087 mol NO, 0.0437 mol Br₂; at equilibrium: 0.0518 mol NOBr
NO consumed = 0.0518 mol; Br₂ consumed = 0.0518/2 = 0.0259 mol
At equilibrium: NO = 0.087 − 0.0518 = 0.0352 mol; Br₂ = 0.0437 − 0.0259 = 0.0178 mol
Q 6.10: Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at 450 K; Kp = 2.0×10¹⁰ bar⁻¹
Δng = 2 − 3 = −1; Kc = Kp(RT)^(−Δng) = Kp × RT
= 2.0×10¹⁰ × 0.0831 × 450 = 2.0×10¹⁰ × 37.395 = 7.48×10¹¹ L mol⁻¹
Q 6.11: Kp for 2HI(g) ⇌ H₂(g) + I₂(g)
Initial P(HI) = 0.2 atm; at equilibrium P(HI) = 0.04 atm
HI dissociated = 0.2 − 0.04 = 0.16 atm; P(H₂) = P(I₂) = 0.08 atm
Kp = P(H₂)×P(I₂) / P(HI)² = (0.08)(0.08)/(0.04)² = 0.0064/0.0016 = 4.0
12: Is N₂ + 3H₂ ⇌ 2NH₃ at equilibrium? Direction of net reaction?
Answer
Kc = 1.7×10² at 500 K; [N₂] = 1.57/20 = 0.0785 M; [H₂] = 1.92/20 = 0.096 M; [NH₃] = 8.13/20 = 0.4065 M
Qc = [NH₃]² / ([N₂][H₂]³) = (0.4065)² / (0.0785 × (0.096)³)
= 0.1652 / (0.0785 × 8.847×10⁻⁴) = 0.1652 / 6.945×10⁻⁵ = 2.38×10³
Qc = 2.38×10³ > Kc = 1.7×10², so reaction proceeds in reverse direction (toward reactants).
Q 6.13: Write balanced equation for Kc = [NH₃]⁴[O₂]⁵ / ([NO]⁴[H₂O]⁶)
4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g)
Q 6.14: Equilibrium constant for H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Initial: 1 mol each in 10 L vessel; 40% of water reacts.
Moles H₂O reacted = 0.4 mol; at equilibrium: H₂O = 0.6 mol, CO = 0.6 mol, H₂ = 0.4 mol, CO₂ = 0.4 mol
Kc = [H₂][CO₂] / [H₂O][CO] = (0.04)(0.04) / (0.06)(0.06) = 0.0016/0.0036 = 0.444
Q 6.15: Concentrations of H₂ and I₂ from HI equilibrium at 700 K
Kc = 54.8 for H₂ + I₂ ⇌ 2HI; [HI] = 0.5 mol L⁻¹ at equilibrium
Kc(reverse) = 1/54.8 = 0.01825 for 2HI ⇌ H₂ + I₂
Kc(reverse) = [H₂][I₂]/[HI]² = [H₂][I₂]/(0.5)²
Since [H₂] = [I₂]: [H₂]² = 0.01825 × 0.25 = 0.004563; [H₂] = [I₂] = 0.0675 mol L⁻¹
16: Equilibrium concentrations in 2ICl(g) ⇌ I₂(g) + Cl₂(g); Kc = 0.14; [ICl]₀ = 0.78 M
Answer
Let x = amount dissociated. At equilibrium: [ICl] = 0.78−2x, [I₂] = x, [Cl₂] = x
Kc = x² / (0.78−2x)² = 0.14; √0.14 = x/(0.78−2x) = 0.3742
x = 0.3742(0.78−2x) = 0.2919 − 0.7484x; x(1.7484) = 0.2919; x = 0.167 M
[ICl] = 0.78 − 2(0.167) = 0.446 M; [I₂] = [Cl₂] = 0.167 M
17: Equilibrium [C₂H₆] for C₂H₆(g) ⇌ C₂H₄(g) + H₂(g); Kp = 0.04 atm at 4.0 atm
Answer
Let x = partial pressure of C₂H₄ = H₂ at equilibrium; P(C₂H₆) = 4 − x
Kp = x²/(4−x) = 0.04; x² + 0.04x − 0.16 = 0
x = (−0.04 + √(0.0016 + 0.64))/2 = (−0.04 + 0.8012)/2 = 0.3806 atm
[C₂H₆] = P(C₂H₆)/RT; using ideal gas: [C₂H₆] = (4−0.3806)/(RT)
At equilibrium P(C₂H₆) = 3.619 atm; [C₂H₆] = 3.619/(0.0821 × 1172) ≈ 0.0376 mol L⁻¹
Q 6.18: Ethyl acetate equilibrium (CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O)
(i) Qc = [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH]
(ii) Initial: 1.00 mol CH₃COOH, 0.18 mol C₂H₅OH; at eq: 0.171 mol ester, 0.171 mol H₂O
Kc = (0.171)(0.171) / ((1.00−0.171)(0.18−0.171)) = 0.0292 / (0.829 × 0.009) = 0.0292/0.00746 = 3.92
(iii) Qc = (0.214)(0.214) / ((1.0−0.214)(0.5−0.214)) = 0.0458 / (0.786 × 0.286) = 0.0458/0.2248 = 0.204
Qc = 0.204 < Kc = 3.92 → equilibrium has NOT been reached; reaction still proceeds forward.
19: Equilibrium concentrations of PCl₃ and Cl₂; PCl₅ ⇌ PCl₃ + Cl₂; Kc = 8.3×10⁻³; [PCl₅] = 0.05 M
Answer
Kc = [PCl₃][Cl₂]/[PCl₅]; [PCl₃] = [Cl₂] = x
x²/0.05 = 8.3×10⁻³; x² = 4.15×10⁻⁴; x = 2.037×10⁻² mol L⁻¹
[PCl₃] = [Cl₂] = 2.04×10⁻² mol L⁻¹
20: FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g); Kp = 0.265; initial p(CO) = 1.4, p(CO₂) = 0.80 atm
Answer
Qp = 0.80/1.4 = 0.571 > Kp = 0.265
⇒ reaction goes backward
Let x decrease in CO₂: (0.80−x)/(1.4+x) = 0.265; 0.80−x = 0.265(1.4+x) = 0.371+0.265x
0.429 = 1.265x; x = 0.339 atm
p(CO₂) = 0.80 − 0.339 = 0.461 atm; p(CO) = 1.4 + 0.339 = 1.739 atm
21: Is N₂ + 3H₂ ⇌ 2NH₃ at equilibrium? [N₂] = 3.0, [H₂] = 2.0, [NH₃] = 0.5 mol L⁻¹; Kc = 0.061
Answer
Qc = (0.5)² / (3.0 × (2.0)³) = 0.25 / 24 = 0.01042
Qc = 0.01042 < Kc = 0.061
⇒ reaction proceeds in forward direction to reach equilibrium.
22: Molar concentration of BrCl at equilibrium; 2BrCl(g) ⇌ Br₂(g) + Cl₂(g); Kc = 32; [BrCl]₀ = 3.3×10⁻³ M
Answer
Kc = x² / (3.3×10⁻³ − 2x)² = 32; √32 = x/(3.3×10⁻³ − 2x) = 5.657
x = 5.657(3.3×10⁻³ − 2x) = 0.01867 − 11.314x
x(12.314) = 0.01867; x = 1.516×10⁻³
[BrCl] = 3.3×10⁻³ − 2(1.516×10⁻³) = 2.68×10⁻⁴ mol L⁻¹
23: Kc for C(s) + CO₂(g) ⇌ 2CO(g) at 1127 K; 90.55% CO by mass at 1 atm
Answer
Mass % CO = 90.55, mass % CO₂ = 9.45
Moles CO = 90.55/28 = 3.234; moles CO₂ = 9.45/44 = 0.2148
Total moles = 3.449; Mole fraction CO = 3.234/3.449 = 0.9377; Mole fraction CO₂ = 0.0623
P(CO) = 0.9377 atm; P(CO₂) = 0.0623 atm
Kp = P(CO)²/P(CO₂) = (0.9377)²/0.0623 = 0.8793/0.0623 = 14.11 atm
Kc = Kp/(RT) = 14.11 / (0.0821 × 1400) = 14.11/114.94 = 0.1228 mol L⁻¹
Q 6.24: ΔG° and K for NO(g) + ½O₂(g) ⇌ NO₂(g) at 298 K
ΔG° = ΔG°f(NO₂) − ΔG°f(NO) − ½ΔG°f(O₂) = 52.0 − 87.0 − 0 = −35.0 kJ mol⁻¹
(a) ΔG° = −35.0 kJ mol⁻¹
(b) ΔG° = −RT ln K; ln K = −ΔG°/RT = 35000/(8.314 × 298) = 14.12
K = e^14.12 = 1.36×10⁶
Q 6.25: Effect of decreasing pressure on equilibrium (increasing volume)
(a) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g): Moles of products increase (equilibrium shifts right, more moles of gas)
(b) CaO(s) + CO₂(g) ⇌ CaCO₃(s): Moles of products decrease (reaction shifts left, consuming CO₂)
(c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g): Δng = 0, so no change in moles.
Q 6.26: Which reactions affected by increasing pressure?
(i) COCl₂ ⇌ CO + Cl₂: Δng = +1; affected → forward → backward; increasing pressure shifts left.
(ii) CH₄ + 2S₂ ⇌ CS₂ + 2H₂S: Δng = 3−3 = 0; not significantly affected.
(iii) CO₂ + C(s) ⇌ 2CO: Δng = +1; increasing pressure shifts left (backward).
(iv) 2H₂ + CO ⇌ CH₃OH: Δng = 1−3 = −2; increasing pressure shifts forward.
(v) CaCO₃(s) ⇌ CaO(s) + CO₂: affected; increasing pressure shifts backward.
(vi) 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O: Δng = 10−9 = +1; increasing pressure shifts backward.
27: Equilibrium pressures for H₂ + Br₂ ⇌ 2HBr; Kp = 1.6×10⁵; initial P(HBr) = 10 bar
Answer
Kp(reverse) = 1/(1.6×10⁵) = 6.25×10⁻⁶
2HBr ⇌ H₂ + Br₂: let x = pressure of H₂ = Br₂
x²/(10−2x)² = 6.25×10⁻⁶; x/(10−2x) = 2.5×10⁻³
x = 0.025 − 5×10⁻³ x; x(1.005) ≈ 0.025; x ≈ 0.02488 bar
P(H₂) = P(Br₂) ≈ 0.025 bar; P(HBr) = 10 − 2(0.025) ≈ 9.95 bar
Q 6.28: CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
(a) Kp = P(CO) × P(H₂)³ / (P(CH₄) × P(H₂O))
(b)(i) Increasing pressure: Kp unchanged; equilibrium shifts backward (fewer gas moles side).
(b)(ii) Increasing temperature: endothermic reaction, so Kp increases; equilibrium shifts forward.
(b)(iii) Using a catalyst: Kp unchanged, equilibrium composition unchanged; only rate to reach equilibrium increases.
Q 6.29: Effect on 2H₂(g) + CO(g) ⇌ CH₃OH(g)
(a) Adding H₂: Q < K → equilibrium shifts forward (more CH₃OH formed).
(b) Adding CH₃OH: Q > K → equilibrium shifts backward.
(c) Removing CO: Q > K → equilibrium shifts backward (more reactants, less product).
(d) Removing CH₃OH: Q < K → equilibrium shifts forward.
Q 6.30: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); Kc = 8.3×10⁻³; ΔH = +124 kJ mol⁻¹
(a) Kc = [PCl₃][Cl₂] / [PCl₅]
(b) Kc(reverse) = 1/(8.3×10⁻³) = 120.5
(c) Adding PCl₅: Kc unchanged (K depends only on T); Increasing pressure: Kc unchanged, equilibrium shifts backward; Increasing temperature: endothermic reaction, Kc increases.
31: H₂ partial pressure at equilibrium; CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g); Kp = 10.1; initial p(CO) = p(H₂O) = 4.0 bar
Answer
Kp = x × x / ((4−x)(4−x)) = x²/(4−x)² = 10.1; x/(4−x) = 3.178
x = 12.712 − 3.178x; 4.178x = 12.712; x = 3.04 bar
32: Which equilibrium has appreciable concentrations of both reactants and products?
(a) Kc = 5×10⁻³⁹ → extremely small, reaction barely proceeds → almost all reactants.
(b) Kc = 3.7×10⁸ → very large, reaction almost complete → almost all products.
(c) Kc = 1.8 → moderate, appreciable concentrations of both reactants and products.
Answer
Q 6.33: Concentration of O₃; Kc = 2.0×10⁻⁵⁰; [O₂] = 1.6×10⁻² M
3O₂(g) ⇌ 2O₃(g); Kc = [O₃]² / [O₂]³
[O₃]² = 2.0×10⁻⁵⁰ × (1.6×10⁻²)³ = 2.0×10⁻⁵⁰ × 4.096×10⁻⁶ = 8.19×10⁻⁵⁶
[O₃] = 2.86×10⁻²⁸ mol L⁻¹
34: Concentration of CH₄ at equilibrium; CO + 3H₂ ⇌ CH₄ + H₂O; Kc = 3.90
Answer
[CO] = 0.30 M, [H₂] = 0.10 M, [H₂O] = 0.02 M, [CH₄] = ?
Kc = [CH₄][H₂O] / ([CO][H₂]³) = [CH₄](0.02) / (0.30 × 0.001) = 3.90
[CH₄] × 0.02 / 0.0003 = 3.90; [CH₄] = 3.90 × 0.0003 / 0.02 = 0.0585 mol L⁻¹
Q 6.35: Conjugate acid-base pairs
HNO₂: conjugate base NO₂⁻; CN⁻: conjugate acid HCN; HClO₄: conjugate base ClO₄⁻; F⁻: conjugate acid HF; OH⁻: conjugate acid H₂O; CO₃²⁻: conjugate acid HCO₃⁻; S²⁻: conjugate acid HS⁻.
Q 6.36: Lewis acids from H₂O, BF₃, H⁺, NH₄⁺
Lewis acids are electron pair acceptors: BF₃ (incomplete octet) and H⁺ (no electrons at all). H₂O and NH₄⁺ are Lewis bases (can donate lone pairs).
Q 6.37: Conjugate bases for HF, H₂SO₄, HCO₃⁻
HF → F⁻; H₂SO₄ → HSO₄⁻; HCO₃⁻ → CO₃²⁻
Q 6.38: Conjugate acids for NH₂⁻, NH₃, HCOO⁻
NH₂⁻ → NH₃; NH₃ → NH₄⁺; HCOO⁻ → HCOOH
Q 6.39: Amphiprotic species
H₂O: acid → OH⁻, base → H₃O⁺; HCO₃⁻: acid → CO₃²⁻, base → H₂CO₃; HSO₄⁻: acid → SO₄²⁻, base → H₂SO₄; NH₃: acid → NH₂⁻, base → NH₄⁺
Q 6.40: Lewis acids and Lewis bases
(a) OH⁻: Lewis base (donates lone pair); (b) F⁻: Lewis base; (c) H⁺: Lewis acid (accepts lone pair); (d) BCl₃: Lewis acid (accepts lone pair into empty p orbital)
Q 6.41: pH of soft drink with [H⁺] = 3.8×10⁻³ M
pH = −log[H⁺] = −log(3.8×10⁻³) = 3 − log(3.8) = 3 − 0.58 = 2.42
Q 6.42: [H⁺] from pH = 3.76 (vinegar)
[H⁺] = 10⁻³·⁷⁶ = 10⁻⁴ × 10⁰·²⁴ = 10⁻⁴ × 1.738 = 1.74×10⁻⁴ mol L⁻¹
Q 6.43: Ionization constants of conjugate bases
Ka×Kb = Kw = 1.0×10⁻¹⁴
HF: Ka = 6.8×10⁻⁴; Kb(F⁻) = 10⁻¹⁴/6.8×10⁻⁴ = 1.47×10⁻¹¹
HCOOH: Ka = 1.8×10⁻⁴; Kb(HCOO⁻) = 10⁻¹⁴/1.8×10⁻⁴ = 5.56×10⁻¹¹
HCN: Ka = 4.8×10⁻⁹; Kb(CN⁻) = 10⁻¹⁴/4.8×10⁻⁹ = 2.08×10⁻⁶
Q 6.44: Phenolate ion concentration; Ka(phenol) = 1.0×10⁻¹⁰; 0.05 M phenol
Ka = x²/0.05 → x = √(1.0×10⁻¹⁰ × 0.05) = √(5×10⁻¹²) = 2.24×10⁻⁶ mol L⁻¹
In 0.01 M sodium phenolate: [C₆H₅O⁻] = 0.01 + x ≈ 0.01; Ka = x(0.01)/0.05; x = 5×10⁻¹⁰ M; degree of ionisation = 5×10⁻¹⁰/0.05 = 1×10⁻⁸
Q 6.45: HS⁻ concentration from H₂S; Ka1 = 9.1×10⁻⁸
H₂S ⇌ H⁺ + HS⁻; [HS⁻] = x = √(9.1×10⁻⁸ × 0.1) = √(9.1×10⁻⁹) = 9.54×10⁻⁵ mol L⁻¹
In 0.1 M HCl: [H⁺] ≈ 0.1 M; [HS⁻] = Ka1×[H₂S]/[H⁺] = 9.1×10⁻⁸×0.1/0.1 = 9.1×10⁻⁸ mol L⁻¹
Ka2 = 1.2×10⁻¹³; [S²⁻] = Ka2×[HS⁻]/[H⁺] = 1.2×10⁻¹³×9.54×10⁻⁵/9.54×10⁻⁵ ≈ 1.2×10⁻¹³ M
In HCl: [S²⁻] = 1.2×10⁻¹³×9.1×10⁻⁸/0.1 = 1.09×10⁻¹⁹ M
Q 6.46: Degree of dissociation and pH of 0.05 M acetic acid; Ka = 1.74×10⁻⁵
α = √(Ka/C) = √(1.74×10⁻⁵/0.05) = √(3.48×10⁻⁴) = 0.01866
Degree of dissociation = 1.87%
[CH₃COO⁻] = [H⁺] = 0.05 × 0.01866 = 9.33×10⁻⁴ mol L⁻¹
pH = −log(9.33×10⁻⁴) = 4 − log(9.33) = 4 − 0.97 = 3.03
Q 6.47: Ka of organic acid; [acid] = 0.01 M, pH = 4.15
[H⁺] = 10⁻⁴·¹⁵ = 7.08×10⁻⁵ mol L⁻¹
Ka = (7.08×10⁻⁵)² / (0.01−7.08×10⁻⁵) ≈ (7.08×10⁻⁵)² / 0.009929 = 5.01×10⁻⁹/0.009929 = 5.05×10⁻⁷
pKa = −log(5.05×10⁻⁷) = 6.30
Q 6.48: pH of solutions (complete dissociation)
(a) 0.003 M HCl: [H⁺] = 3×10⁻³; pH = 2.52
(b) 0.005 M NaOH: [OH⁻] = 5×10⁻³; pOH = 2.30; pH = 11.70
(c) 0.002 M HBr: [H⁺] = 2×10⁻³; pH = 2.70
(d) 0.002 M KOH: [OH⁻] = 2×10⁻³; pOH = 2.70; pH = 11.30
Q 6.49: pH calculations
(a) 2g TlOH in 2L: M(TlOH) = 204+17 = 221; [TlOH] = 2/(221×2) = 4.52×10⁻³ M; [OH⁻] = 4.52×10⁻³; pH = 11.65
(b) 0.3g Ca(OH)₂ in 500 mL: M = 74; n = 0.3/74 = 4.05×10⁻³ mol; [OH⁻] = 2×4.05×10⁻³/0.5 = 0.0162 M; pH = 12.21
(c) 0.3g NaOH in 200 mL: [NaOH] = (0.3/40)/0.2 = 0.0375 M; pH = 12.57
(d) 1 mL of 13.6 M HCl in 1L: [HCl] = 13.6×10⁻³/1 = 0.0136 M; pH = 1.87
Q 6.50: pH and pKa of bromoacetic acid; 0.1 M, α = 0.132
[H⁺] = 0.1 × 0.132 = 0.0132 M; pH = −log(0.0132) = 1.88
Ka = (0.0132)²/(0.1−0.0132) = 1.744×10⁻⁴/0.0868 = 2.01×10⁻³; pKa = 2.70
Q 6.51: Kb of codeine; 0.005 M, pH = 9.95
[OH⁻] = 10⁻⁴·⁰⁵ = 8.91×10⁻⁵; Kb = (8.91×10⁻⁵)²/(0.005−8.91×10⁻⁵) ≈ 7.94×10⁻⁹/0.004911 = 1.62×10⁻⁶; pKb = 5.79
Q 6.52: pH of 0.001 M aniline; Kb(aniline) = 4.27×10⁻¹⁰
[OH⁻] = √(Kb×C) = √(4.27×10⁻¹⁰×0.001) = √(4.27×10⁻¹³) = 6.53×10⁻⁷
pOH = 6.18; pH = 7.82
Degree of ionisation = 6.53×10⁻⁷/0.001 = 6.53×10⁻⁴ or 0.065%
Ka(conjugate acid) = Kw/Kb = 10⁻¹⁴/4.27×10⁻¹⁰ = 2.34×10⁻⁵
Q 6.53: Degree of ionisation of 0.05 M acetic acid; pKa = 4.74
Ka = 10⁻⁴·⁷⁴ = 1.82×10⁻⁵; α = √(Ka/C) = √(1.82×10⁻⁵/0.05) = √(3.64×10⁻⁴) = 0.01908 ≈ 1.9%
(a) With 0.01 M HCl: [H⁺] = 0.01 + x ≈ 0.01; Ka = 0.01x/0.05; x = 9.1×10⁻⁵; α = 0.18%
(b) With 0.1 M HCl: x = 9.1×10⁻⁶; α = 0.018%
Q 6.54: Degree of ionisation of dimethylamine; Kb = 5.4×10⁻⁴; 0.02 M
α = √(Kb/C) = √(5.4×10⁻⁴/0.02) = √(0.027) = 0.164 or 16.4%
With 0.1 M NaOH: Kb = [DMA⁺][OH⁻]/[DMA] = x(0.1)/(0.02); x = 5.4×10⁻⁴×0.02/0.1 = 1.08×10⁻⁴ M; % ionised = 1.08×10⁻⁴/0.02 × 100 = 0.54%
Q 6.55: [H⁺] from pH values of biological fluids
(a) pH 6.83: [H⁺] = 10⁻⁶·⁸³ = 1.48×10⁻⁷ M
(b) pH 1.2: [H⁺] = 10⁻¹·² = 6.31×10⁻² M
(c) pH 7.38: [H⁺] = 10⁻⁷·³⁸ = 4.17×10⁻⁸ M
(d) pH 6.4: [H⁺] = 10⁻⁶·⁴ = 3.98×10⁻⁷ M
Q 6.56: [H⁺] from pH of common substances
Milk (pH 6.8): 1.58×10⁻⁷ M; Coffee (pH 5.0): 1.0×10⁻⁵ M; Tomato juice (pH 4.2): 6.31×10⁻⁵ M; Lemon juice (pH 2.2): 6.31×10⁻³ M; Egg white (pH 7.8): 1.58×10⁻⁸ M
Q 6.57: KOH solution pH; 0.561 g KOH in 200 mL
M(KOH) = 56.1; [KOH] = 0.561/(56.1×0.2) = 0.05 M; [K⁺] = [OH⁻] = 0.05 M; [H⁺] = 10⁻¹⁴/0.05 = 2×10⁻¹³ M; pH = 12.70
Q 6.58: Sr(OH)₂ solution; solubility = 19.23 g/L
M(Sr(OH)₂) = 121.6; [Sr(OH)₂] = 19.23/121.6 = 0.1581 M; [Sr²⁺] = 0.158 M; [OH⁻] = 2×0.1581 = 0.316 M; pH = 14 + log(0.316) = 14 − 0.5 = 13.5
Q 6.59 to 6.73 — Ionic equilibrium problems
6.59 (Propanoic acid): Ka = 1.32×10⁻⁵; α = √(Ka/C) = √(2.64×10⁻⁴) = 0.01625; degree = 1.63%; pH = −log(1.32×10⁻⁵ × 0.05 / α) ≈ 3.09. In 0.01 M HCl: degree ≈ 0.132%.
6.60 (Cyanic acid): pH = 2.34; [H⁺] = 4.57×10⁻³ M; α = 4.57×10⁻³/0.1 = 0.0457; Ka = (4.57×10⁻³)²/(0.1−4.57×10⁻³) = 2.09×10⁻⁵/0.09543 = 2.19×10⁻⁴
6.61 (NaNO₂ hydrolysis): Ka(HNO₂) = 4.5×10⁻⁴; Kb(NO₂⁻) = 10⁻¹⁴/4.5×10⁻⁴ = 2.22×10⁻¹¹; [OH⁻] = √(2.22×10⁻¹¹×0.04) = 9.43×10⁻⁷; pH = 7.97
6.62 (Pyridinium hydrochloride): pH = 3.44; [H⁺] = 3.63×10⁻⁴; Ka(pyridinium) = (3.63×10⁻⁴)²/0.02 = 6.6×10⁻⁶; Kb(pyridine) = 10⁻¹⁴/6.6×10⁻⁶ = 1.52×10⁻⁹
6.63 (Salt solutions): NaCl–neutral; KBr–neutral; NaCN–basic (CN⁻ hydrolyses); NH₄NO₃–acidic; NaNO₂–basic; KF–basic
6.64: Ka(ClCH₂COOH) = 1.35×10⁻³; [H⁺] = √(Ka×C) = √(1.35×10⁻⁴) = 0.01162 M; pH = 1.93; sodium salt: Kb = 10⁻¹⁴/1.35×10⁻³ = 7.41×10⁻¹²; basic, pH ≈ 7.44
6.65: Kw = 2.7×10⁻¹⁴; neutral [H⁺] = √(2.7×10⁻¹⁴) = 1.643×10⁻⁷; pH = 6.78
66: (a) Ca(OH)₂: 2mmol OH⁻ in 10mL; HCl: 2.5mmol H⁺; excess H⁺ = 0.5mmol in 35mL; [H⁺] = 0.0143 M; pH = 1.85
(b) H₂SO₄: 0.1mmol H⁺ in 10mL; Ca(OH)₂: 0.2mmol OH⁻ in 10mL; neutral pH (stoichiometrically equal if Ca(OH)₂ accounts for 2 OH⁻) — neutralised; pH = 7
(c) H₂SO₄: 1mmol H⁺; KOH: 1mmol OH⁻; equal → neutral; pH = 7
Answer
6.67–6.73: Solubility product problems solved using Ksp relationships and ion product comparisons.
Important Extra Questions
Q A: State Le Chatelier's Principle and give three examples of its application.
Le Chatelier's Principle: When a system at equilibrium is subjected to a change (concentration, pressure, temperature), the equilibrium shifts to partially oppose the change and establish a new equilibrium.
Examples
1. N₂ + 3H₂ ⇌ 2NH₃: Adding N₂ shifts equilibrium forward (more NH₃). Used in Haber process — high pressure favours forward direction (fewer gas moles).
2. CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq): Increased CO₂ shifts forward (ocean acidification).
3. 2SO₂ + O₂ ⇌ 2SO₃ (ΔH = −197 kJ): Lowering temperature shifts forward (more SO₃), but decreases rate — a compromise temperature is used industrially.
Q B: Explain the relationship between Kp and Kc. When are they equal?
Kp = Kc(RT)^Δng where Δng = moles of gaseous products − moles of gaseous reactants; R = 0.0821 L atm mol⁻¹ K⁻¹ or 8.314 J mol⁻¹ K⁻¹.
Kp = Kc when Δng = 0 (equal moles of gas on both sides), e.g., H₂ + I₂ ⇌ 2HI.
Q C: What is a buffer solution? Explain Henderson–Hasselbalch equation.
A buffer solution resists change in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or weak base and conjugate acid).
Henderson–Hasselbalch equation: pH = pKa + log([A⁻]/[HA])
Example: Acetic acid/sodium acetate buffer: pH = 4.74 + log([CH₃COO⁻]/[CH₃COOH]). When [A⁻] = [HA], pH = pKa.
Q D: Distinguish between Kc and Qc. How is Qc used to predict direction of reaction?
Kc is the equilibrium constant (at a specific T), calculated using equilibrium concentrations.
Qc (reaction quotient) uses current concentrations (not necessarily at equilibrium).
If Qc < Kc: reaction proceeds forward (products form).
If Qc = Kc: system is at equilibrium.
If Qc > Kc: reaction proceeds backward (reactants form).
Q E: Calculate the solubility of AgCl in water at 25°C; Ksp(AgCl) = 1.8×10⁻¹⁰.
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Let solubility = s mol L⁻¹; Ksp = [Ag⁺][Cl⁻] = s × s = s²
s = √(1.8×10⁻¹⁰) = √(18×10⁻¹¹) = 1.342×10⁻⁵ mol L⁻¹
Solubility = 1.34×10⁻⁵ mol L⁻¹ or about 1.93×10⁻³ g/L
In 0.1 M NaCl: [Cl⁻] = 0.1; [Ag⁺] = 1.8×10⁻¹⁰/0.1 = 1.8×10⁻⁹ mol L⁻¹ (much less — common ion effect).