Solutions, Phase Equilibria and Electrochemistry
1. Ideal and non-ideal solutions: Raoult’s and Henry’s laws
In an ideal solution every component obeys Raoult’s law over the whole composition range: its partial vapour pressure is p_A = x_A p_A, where p_A is the vapour pressure of pure A. This happens when A–A, B–B and A–B interactions are alike (benzene–toluene), and then ΔmixH = 0 and ΔmixV = 0. The total pressure P = x_A p_A* + x_B p_B* is linear in x, and the vapour is richer in the more volatile component: y_A = x_A p_A*/P. Henry’s law describes a dilute solute: p_B = K_H x_B, with K_H not equal to p_B*. In an ideal-dilute solution the solvent obeys Raoult and the solute Henry, and Gibbs–Duhem guarantees that one implies the other.
Non-ideal solutions deviate. Positive deviation (p above Raoult, γ > 1) arises when A–B attractions are weaker than the average of A–A and B–B — ethanol–hexane, acetone–CS₂ — with ΔmixH > 0 and ΔmixV > 0. Negative deviation (p below Raoult, γ < 1) arises when A–B attractions are stronger — acetone–chloroform, which hydrogen-bond to each other, and water–HCl — with ΔmixH < 0. Large deviations produce an extremum in the vapour-pressure curve, which is an azeotrope.
The colligative properties of a dilute solution of a non-volatile solute depend only on the number of solute particles: the relative lowering of vapour pressure Δp/p_A* = x_B; the boiling-point elevation ΔT_b = iK_b m and freezing-point depression ΔT_f = iK_f m, with m the molality and K_b, K_f solvent constants (water: K_f = 1.86 and K_b = 0.512 K kg mol⁻¹); and the osmotic pressure π = icRT (van ’t Hoff). The van ’t Hoff factor i counts particles per formula unit: about 2 for NaCl, 3 for CaCl₂, below 1 for a solute that dimerises (acetic acid in benzene).
2. The phase rule, Clausius–Clapeyron and one-component diagrams
Gibbs’ phase rule F = C − P + 2 gives the number of intensive variables that can be changed independently, with C the number of components (independent species after subtracting reactions and fixed stoichiometric constraints) and P the number of phases. For one component, a single phase is bivariant (an area), two phases univariant (a line) and three phases invariant (a triple point). When pressure is fixed, as for most condensed systems, the reduced rule is F′ = C − P + 1.
Along any two-phase line the Clapeyron equation dP/dT = ΔH/(TΔV) holds. For liquid–vapour or solid–vapour equilibrium, neglecting the condensed-phase volume and treating the vapour as ideal, it becomes the Clausius–Clapeyron equation d ln P/dT = ΔH/(RT²), or ln(P₂/P₁) = −(ΔvapH/R)(1/T₂ − 1/T₁). For water’s solid–liquid line ΔV is negative (ice is less dense than water), so the melting line slopes backwards: pressure lowers the melting point.
| System | Phases | Triple point(s) | What to remember |
|---|---|---|---|
| Water | ice I, liquid, vapour | 273.16 K, 611 Pa | solid–liquid line has negative slope |
| CO₂ | solid, liquid, vapour | 216.6 K, 5.11 atm | triple-point pressure above 1 atm, so solid CO₂ sublimes at 1 atm; positive melting slope |
| Sulfur | rhombic, monoclinic, liquid, vapour | three stable triple points | rhombic ⇌ monoclinic transition at 368.5 K (95.5 °C) at 1 atm; four phases can never coexist |
3. Two-component systems: distillation, azeotropes, miscibility and eutectics
Liquid–vapour. On a temperature–composition diagram at fixed pressure the liquid (bubble-point) curve lies below the vapour (dew-point) curve, and a tie line joins a liquid to the vapour in equilibrium with it; the lever rule gives the relative amounts of the two phases from the lengths of the tie-line segments. Each vaporisation–condensation step enriches the vapour in the lower-boiling component, and a column with many such theoretical plates performs fractional distillation, delivering the more volatile component at the top.
An azeotrope is a mixture that boils without change of composition, because at the extremum the liquid and vapour compositions coincide, so fractional distillation cannot pass it. Minimum-boiling azeotropes come from positive deviation — ethanol–water at about 95.6% ethanol by mass boils at 78.2 °C, below either component — and distillation gives the azeotrope at the top and the excess component at the bottom. Maximum-boiling azeotropes come from negative deviation — HCl–water at about 20% HCl near 109 °C — and the azeotrope then collects in the still.
Liquid–liquid. Partially miscible liquids form two layers inside a solubility curve. Phenol–water has an upper critical solution temperature near 66 °C, above which they mix in all proportions; triethylamine–water has a lower one near 18.5 °C; nicotine–water shows a closed loop with both. Solid–liquid. When two components are miscible as liquids but not as solids, adding either lowers the other’s freezing point, and the two liquidus curves meet at the eutectic point, the lowest temperature at which liquid exists; there the liquid freezes at constant temperature into an intimate mixture of the two solids, and with pressure fixed F′ = 2 − 3 + 1 = 0. Systems that form a compound with a congruent melting point show two eutectics, one on each side of the compound.
4. Electrode potentials, cells and the Nernst equation
A galvanic cell is written anode | anode solution || cathode solution | cathode, with oxidation on the left. Electrode potentials are reduction potentials measured against the standard hydrogen electrode (Pt | H₂(1 bar) | H⁺(a = 1), E° = 0 by definition), and E°_cell = E°_cathode − E°_anode. A positive E_cell means the cell reaction as written is spontaneous. The link to thermodynamics is ΔᵣG = −nFE, so ΔᵣG° = −nFE° and, since ΔᵣG° = −RT ln K, ln K = nFE°/RT (log K = nE°/0.05916 at 298 K). The temperature coefficient of the emf gives the rest: ΔᵣS = nF(∂E/∂T)_P and ΔᵣH = −nF[E − T(∂E/∂T)_P].
The Nernst equation gives the emf away from standard conditions: E = E° − (RT/nF) ln Q = E° − (0.05916/n) log Q at 298 K, Q written in activities. Applications: cell potentials at any concentration; concentration cells, in which E° = 0 and the emf comes entirely from the activity ratio (M | Mⁿ⁺(a₁) || Mⁿ⁺(a₂) | M gives E = (0.05916/n) log(a₂/a₁)); pH measurement with the glass or hydrogen electrode (E changes 59.2 mV per pH unit); and solubility products and equilibrium constants from E°. Reference electrodes — the saturated calomel electrode (+0.241 V vs SHE) and Ag/AgCl — replace the SHE in practice.
In a potentiometric titration the emf of an indicator electrode is followed against titrant volume and the end point is the steepest point of the curve (the maximum of dE/dV); it needs no coloured indicator and works in turbid or coloured solutions, for acid–base, redox (Fe²⁺ with Ce⁴⁺) and precipitation (Cl⁻ with Ag⁺) reactions alike.
5. Conductivity, ionic mobility, Kohlrausch’s law and Debye–Hückel theory
The conductivity κ of a solution (S m⁻¹) is found from its resistance and the cell constant l/A; the molar conductivity Λm = κ/c. For strong electrolytes Λm falls slightly with concentration according to Kohlrausch’s law Λm = Λm° − K√c, so Λm° comes from extrapolating against √c. Kohlrausch’s law of independent migration states that at infinite dilution each ion contributes independently: Λm° = ν₊λ₊° + ν₋λ₋°. This gives Λm° for weak electrolytes, which cannot be extrapolated, by combination: Λm°(CH₃COOH) = Λm°(CH₃COONa) + Λm°(HCl) − Λm°(NaCl). For a weak electrolyte the degree of dissociation is α = Λm/Λm° and Ostwald’s dilution law gives K_a = cα²/(1 − α).
The ionic mobility u is the drift speed per unit field (m² V⁻¹ s⁻¹), related to the ionic conductivity by λ = zuF, and the transport number t₊ = λ₊/(λ₊ + λ₋) is the fraction of current carried by the cation. H⁺ and OH⁻ have exceptionally high mobilities (λ° about 350 and 198 S cm² mol⁻¹, against about 50–76 for ordinary ions) because they move by proton hopping along hydrogen bonds (the Grotthuss mechanism). This is what shapes conductometric titrations: titrating a strong acid with a strong base replaces fast H⁺ by slow Na⁺, so conductance falls to a sharp minimum at equivalence and then rises as OH⁻ accumulates — a V-shaped curve; with a weak acid the conductance rises gently from a low start and more steeply after the end point.
Debye–Hückel theory explains why ions are less "active" than their concentration: each ion is surrounded by an ionic atmosphere of opposite charge that lowers its chemical potential. The limiting law is log γ± = −A|z₊z₋|√I, with A = 0.509 for water at 298 K and ionic strength I = ½Σcᵢzᵢ²; it holds for I up to about 0.01 M, and extended forms add a denominator 1 + Ba√I. The same atmosphere slows ions in a field through the relaxation and electrophoretic effects, and the Debye–Hückel–Onsager equation Λm = Λm° − (A + BΛm°)√c gives Kohlrausch’s empirical K from first principles.
Key takeaways
- Raoult p_A = x_A p_A* for an ideal solution, Henry p_B = K_H x_B for a dilute solute; positive deviation means weaker A–B attraction, negative stronger. Colligative: ΔT_f = iK_f m, π = icRT.
- F = C − P + 2; Clausius–Clapeyron ln(P₂/P₁) = −(ΔH/R)(1/T₂ − 1/T₁); water’s melting line slopes back, CO₂ sublimes at 1 atm, sulfur has three triple points and no quadruple point.
- Azeotropes cannot be separated by fractional distillation; minimum-boiling ones come from positive deviation. At a eutectic with pressure fixed F′ = 0.
- ΔG = −nFE, ln K = nFE°/RT, ΔS = nF(∂E/∂T)_P; Nernst E = E° − (0.05916/n) log Q at 298 K.
- Λm = Λm° − K√c for strong electrolytes; Λm° = Σνλ° for any; α = Λm/Λm° for weak ones; log γ± = −0.509|z₊z₋|√I; H⁺ and OH⁻ mobility makes strong acid–base conductometric curves V-shaped.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
For the Daniell cell Zn | Zn²⁺(0.010 M) || Cu²⁺(1.0 M) | Cu with E° = 1.10 V, what is the emf at 298 K, in V, to two decimal places? (Take activities equal to concentrations; 2.303RT/F = 0.05916 V.)
Numerical answer — type the value.
Show answer
Answer: 1.16
Cell reaction Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2, Q = [Zn²⁺]/[Cu²⁺] = 0.010. E = 1.10 − (0.05916/2) log 0.010 = 1.10 + 0.0592 = 1.159 V, i.e. 1.16. Writing Q upside down gives 1.04 V; lowering the product concentration must raise the emf.For a two-electron cell reaction with E° = 1.10 V at 298 K, what is log₁₀ K, to one decimal place? (2.303RT/F = 0.05916 V)
Numerical answer — type the value.
Show answer
Answer: 37.2
log K = nE°/0.05916 = 2 × 1.10/0.05916 = 37.19, i.e. 37.2, so K ≈ 1.5 × 10³⁷ and the Daniell reaction goes essentially to completion. Forgetting n gives 18.6.What is ΔᵣG° for a two-electron cell reaction with E° = 1.10 V, in kJ/mol, to one decimal place? (F = 96 485 C mol⁻¹)
Numerical answer — type the value.
Show answer
Answer: -212.3
ΔᵣG° = −nFE° = −2 × 96 485 × 1.10 = −212 267 J/mol = -212.3 kJ/mol. The sign is negative because a positive E° means a spontaneous reaction; dropping n gives −106.1.The emf of a cell with a two-electron reaction falls with temperature, (∂E/∂T)_P = −4.0 × 10⁻⁴ V/K. What is ΔᵣS, in J K⁻¹ mol⁻¹, to one decimal place? (F = 96 485 C mol⁻¹)
Numerical answer — type the value.
Show answer
Answer: -77.2
From ΔG = −nFE and (∂ΔG/∂T)_P = −ΔS: ΔᵣS = nF(∂E/∂T)_P = 2 × 96 485 × (−4.0 × 10⁻⁴) = -77.2 J K⁻¹ mol⁻¹. A falling emf means negative ΔS; a positive answer has lost the sign of the coefficient.What is the freezing-point depression of a 0.10 mol/kg aqueous NaCl solution, assuming complete dissociation, in K, to two decimal places? (K_f = 1.86 K kg mol⁻¹)
Numerical answer — type the value.
Show answer
Answer: 0.37
ΔT_f = iK_f m = 2 × 1.86 × 0.10 = 0.372 K, i.e. 0.37. NaCl gives two particles per formula unit; taking i = 1 gives 0.186 K, the value for a non-electrolyte such as glucose.What is the osmotic pressure of a 0.010 mol/L aqueous glucose solution at 300 K, in kPa, to one decimal place? (R = 8.314 J mol⁻¹ K⁻¹)
Numerical answer — type the value.
Show answer
Answer: 24.9
π = cRT with c = 0.010 mol/L = 10 mol/m³: π = 10 × 8.314 × 300 = 24 942 Pa = 24.9 kPa. Keeping c in mol/L with R in SI units gives 24.9 Pa, a factor of 1000 too small.An ideal solution of A and B has x_A = 0.40. The pure-component vapour pressures are p_A* = 100 torr and p_B* = 50 torr. What is the mole fraction of A in the vapour, to two decimal places?
Numerical answer — type the value.
Show answer
Answer: 0.57
p_A = 0.40 × 100 = 40 torr, p_B = 0.60 × 50 = 30 torr, P = 70 torr, and y_A = 40/70 = 0.571, i.e. 0.57. The vapour is richer in the more volatile A than the liquid (0.40) is — the basis of distillation.Water boils at 373 K under 1.00 atm and ΔvapH = 40.7 kJ/mol. Using the Clausius–Clapeyron equation, what is its vapour pressure at 353 K, in atm, to two decimal places? (R = 8.314 J mol⁻¹ K⁻¹)
Numerical answer — type the value.
Show answer
Answer: 0.48
ln(P₂/1.00) = −(40 700/8.314)(1/353 − 1/373) = −4895 × 1.519 × 10⁻⁴ = −0.744, so P₂ = e^(−0.744) = 0.475 atm, i.e. 0.48 (measured 0.47). Swapping the temperatures gives 2.1 atm, a pressure higher than at the boiling point.Solid NH₄Cl is heated in an evacuated vessel and comes to equilibrium with NH₃ and HCl gases formed only by its own decomposition. How many degrees of freedom does the system have?
Numerical answer — type the value.
Show answer
Answer: 1
Species 3, one reaction, and one further constraint p(NH₃) = p(HCl) because both come only from the solid, so C = 3 − 1 − 1 = 1. Phases P = 2 (solid, gas), and F = 1 − 2 + 2 = 1: fixing T fixes the dissociation pressure. Ignoring the stoichiometric constraint gives C = 2 and F = 2.Why does the solid–liquid equilibrium line of water have a negative slope on the P–T diagram?
Show answer
Answer: A — Ice is less dense than liquid water, so ΔV of melting is negative
Clapeyron: dP/dT = ΔfusH/(TΔfusV). ΔfusH is positive, but water contracts on melting, so ΔfusV < 0 and the slope is negative: pressure lowers the melting point. The low triple-point pressure is true but explains nothing about the slope.Which statements about two-component liquid systems are correct?
Show answer
Answer: A — A minimum-boiling azeotrope arises from positive deviation from Raoult’s law; C — Phenol–water shows an upper critical solution temperature
Positive deviation raises the vapour pressure to a maximum, which is a boiling-point minimum. Phenol–water becomes fully miscible above about 66 °C, an upper critical solution temperature. At the azeotrope liquid and vapour have the same composition, so distillation stalls there; acetone and chloroform hydrogen-bond, a negative deviation.At 298 K the limiting molar conductivities are 91.0 (CH₃COONa), 426.2 (HCl) and 126.5 (NaCl) S cm² mol⁻¹. What is Λm° of acetic acid, in S cm² mol⁻¹, to one decimal place?
Numerical answer — type the value.
Show answer
Answer: 390.7
By independent migration, Λm°(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) = Λm°(CH₃COONa) + Λm°(HCl) − Λm°(NaCl) = 91.0 + 426.2 − 126.5 = 390.7 S cm² mol⁻¹. It cannot be found by extrapolation because a weak acid’s Λm rises steeply as c → 0.A 0.010 mol/L solution of a weak monoprotic acid has Λm = 19.5 S cm² mol⁻¹, and Λm° = 390 S cm² mol⁻¹. What is its K_a, in units of 10⁻⁵, to two decimal places?
Numerical answer — type the value.
Show answer
Answer: 2.63
α = Λm/Λm° = 19.5/390 = 0.050; K_a = cα²/(1 − α) = 0.010 × 0.0025/0.95 = 2.63 × 10⁻⁵. Dropping the (1 − α) gives 2.50 × 10⁻⁵, an error of 5% that GATE’s answer range would not accept.Using the Debye–Hückel limiting law, log γ± = −0.509|z₊z₋|√I, what is the mean activity coefficient of CaCl₂ at 0.0010 mol/kg in water at 298 K, to two decimal places?
Numerical answer — type the value.
Show answer
Answer: 0.88
I = ½(0.0010 × 2² + 0.0020 × 1²) = 0.0030; log γ± = −0.509 × 2 × √0.0030 = −0.509 × 2 × 0.0548 = −0.0558, so γ± = 0.88. Using the concentration 0.0010 in place of I gives γ± = 0.93.In the conductometric titration of HCl with NaOH, how does the conductance change?
Show answer
Answer: A — It falls to a minimum at the equivalence point and then rises
Before equivalence each OH⁻ added removes an H⁺ (λ° ≈ 350) and adds an Na⁺ (≈ 50), so conductance falls; after it, excess OH⁻ (≈ 198) and Na⁺ make it rise. The V-shaped curve locates the end point by the intersection of two straight lines.Which statements about electrochemical cells are correct?
Show answer
Answer: A — A concentration cell has E°_cell = 0; B — E°_cell = E°_cathode − E°_anode, both as reduction potentials; D — In a potentiometric titration the end point is where dE/dV is largest
Identical electrodes make E° zero, E°_cell is the difference of reduction potentials, and the titration end point is the inflection where dE/dV peaks. E is intensive: doubling the reaction doubles n and ΔG, but E = −ΔG/nF is unchanged.