Chemical and ionic equilibrium

In short

Chemical equilibrium is the state a reversible reaction reaches when forward and backward rates become equal, so concentrations stop changing while both reactions continue; it is described quantitatively by the equilibrium constant and shifted predictably by Le Chatelier's principle. Ionic equilibrium applies the same algebra to water, acids, bases, salts and sparingly soluble solids, and it is what lets a chemist calculate pH, design a buffer and predict precipitation. The topic carries heavy weight in B.Sc. physical chemistry papers, CSIR-UGC NET and GATE Chemistry (CY), and in entrance tests for IIT/NIT M.Sc. programmes. On the job it underpins Haber ammonia and contact-process sulphuric acid operation, water and effluent treatment under CPCB norms, pharmaceutical formulation buffers, and every analytical titration in an NABL-accredited lab. Placement technical rounds at fertiliser, pharma, water-treatment and QC roles routinely ask for pH of a weak acid, buffer design and Ksp-based precipitation.

Explained

Reversible reactions and the equilibrium state

• A reversible reaction proceeds in both directions in the same vessel; equilibrium is reached when rate(forward) = rate(backward), not when amounts become equal.
• It is dynamic (molecular change continues), attainable from either side, and only possible in a closed system at constant temperature.
• Characteristics: macroscopic properties (concentration, pressure, colour, density) stay constant; ΔG = 0 for the reaction mixture; a catalyst speeds both directions and shortens the time to equilibrium but never changes the position.
• Law of mass action (Guldberg and Waage, 1864): rate of a reaction is proportional to the product of active masses (molar concentrations) of reactants, each raised to its stoichiometric power.
• For aA + bB ⇌ cC + dD, the ratio of rate constants gives the equilibrium constant Kc = [C]^c[D]^d / ([A]^a[B]^b), measured with equilibrium concentrations only.
• Homogeneous equilibrium: all species in one phase (N₂ + 3H₂ ⇌ 2NH₃, gas). Heterogeneous equilibrium: more than one phase — pure solids and pure liquids are omitted because their activity is 1, so for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = pCO₂.

Equilibrium constants: Kc, Kp, Kx and their relations

• Kp uses partial pressures: Kp = (pC^c · pD^d)/(pA^a · pB^b); with p = (n/V)RT, Kp = Kc(RT)^Δn, where Δn = (moles of gaseous product) − (moles of gaseous reactant).
• Δn = 0 → Kp = Kc numerically (dissociation of HI, esterification). Units of Kc are (mol L⁻¹)^Δn; Kp carries pressure units — strictly, thermodynamic K is dimensionless because each term is divided by a standard state (1 bar or 1 mol L⁻¹).
• Kx (mole fractions): Kp = Kx·P^Δn, so Kx depends on total pressure whenever Δn ≠ 0 whereas Kp does not.
• Manipulation rules: reversing a reaction → 1/K; multiplying coefficients by n → Kⁿ; adding two reactions → multiply their K values; halving → √K.
• Temperature dependence — van't Hoff equation: d(ln K)/dT = ΔH°/RT². Endothermic (ΔH > 0) → K rises with T; exothermic → K falls. Integrated form gives ΔH° from a ln K versus 1/T plot.
• Thermodynamic link: ΔG° = −RT ln K = −2.303 RT log K, and ΔG = ΔG° + RT ln Q. Large K means a strongly product-favoured reaction, not a fast one — K says nothing about rate.
• Reaction quotient Q has the same form as K but with any arbitrary concentrations: Q < K → net forward reaction; Q = K → equilibrium; Q > K → net backward reaction.
• Degree of dissociation (α): fraction of a mole dissociated. For PCl₅ ⇌ PCl₃ + Cl₂, Kp = α²P/(1 − α²); observed vapour density gives α = (D − d)/((n − 1)d) where D is theoretical and d observed vapour density.

Le Chatelier's principle and industrial optimisation

• Statement: if a system at equilibrium is disturbed in concentration, pressure, volume or temperature, it shifts in the direction that partly counteracts the disturbance.
• Concentration: adding a reactant or removing a product shifts forward; K is unchanged.
• Pressure/volume: compression shifts to the side with fewer gas moles; if Δn = 0, pressure has no effect.
• Temperature: heating favours the endothermic direction and is the only factor that changes K.
• Inert gas: added at constant volume → no shift (partial pressures unchanged); added at constant pressure → volume rises, so the system shifts to the side with more gas moles.
• Catalyst: no shift; Fe–K₂O–Al₂O₃ in the Haber process and V₂O₅ in the contact process only reach equilibrium faster.
• Haber process (N₂ + 3H₂ ⇌ 2NH₃, ΔH ≈ −92 kJ mol⁻¹, Δn = −2): thermodynamics wants low temperature and high pressure; kinetics forces a compromise — about 450–500 °C and 150–300 bar in Indian plants (IFFCO, RCF, NFL, Chambal), with ammonia condensed out and unreacted gas recycled.
• Contact process (2SO₂ + O₂ ⇌ 2SO₃, exothermic, Δn = −1): about 400–450 °C, 1–2 bar, excess air, multi-bed converter with interstage cooling and double absorption to push conversion above 99%.
• Other examples: Ostwald process for nitric acid, methanol synthesis, water-gas shift, and the Solvay process for soda ash.

Ionic equilibrium: acids, bases and water

• Arrhenius: acid gives H⁺ in water, base gives OH⁻ — limited to aqueous media. Brønsted–Lowry: acid is a proton donor, base a proton acceptor; every acid has a conjugate base (HA/A⁻), and stronger the acid, weaker its conjugate base. Lewis: acid accepts an electron pair (BF₃, AlCl₃, Ag⁺), base donates one (NH₃, H₂O, CN⁻) — covers non-protonic systems and complex formation.
• HSAB principle: hard acids (small, high charge — H⁺, Al³⁺) prefer hard bases (F⁻, OH⁻); soft acids (Ag⁺, Hg²⁺) prefer soft bases (I⁻, S²⁻, CN⁻) — explains why heavy metals are precipitated as sulphides in effluent treatment.
• Levelling effect: in water the strongest acid possible is H₃O⁺ and the strongest base OH⁻, so HCl, HNO₃ and HClO₄ all appear equally strong; their true order shows in a differentiating solvent such as glacial acetic acid.
• Ionic product of water: Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C (1.0 × 10⁻¹³ at ~60 °C) — it increases with temperature because ionisation is endothermic. Hence neutral pH is 7 only at 25 °C.
• pH scale (Sørensen): pH = −log aH⁺ ≈ −log[H⁺]; pH + pOH = pKw = 14 at 25 °C. pH can be negative or above 14 in concentrated solutions, where activities must replace concentrations.
• Ostwald's dilution law for a weak monobasic acid: Ka = cα²/(1 − α); if α ≪ 1, α ≈ √(Ka/c) — degree of dissociation rises on dilution, while Ka stays constant. It fails for strong electrolytes, where Debye–Hückel interionic attraction theory applies.
• Polyprotic acids: H₃PO₄ has Ka1 ≫ Ka2 ≫ Ka3 (about 7.5 × 10⁻³, 6.2 × 10⁻⁸, 4.8 × 10⁻¹³); pH is set almost entirely by the first step, and for the amphiprotic intermediate pH ≈ ½(pKa1 + pKa2).
• Common ion effect: adding an ion already present suppresses ionisation (acetate added to acetic acid) — the basis of buffers and of group separation in qualitative analysis.
• Conjugate pair relation: Ka × Kb = Kw, so pKa + pKb = 14 at 25 °C.

Buffers, salt hydrolysis and titration curves

• A buffer resists pH change on addition of small amounts of acid or base, or on dilution. Acidic buffer: weak acid + its salt (CH₃COOH/CH₃COONa, pH ≈ 4.75). Basic buffer: weak base + its salt (NH₄OH/NH₄Cl, pH ≈ 9.25).
• Henderson–Hasselbalch equation: pH = pKa + log([salt]/[acid]); pOH = pKb + log([salt]/[base]). Buffer pH is unchanged by dilution because only the ratio appears.
• Buffer capacity β = dn/dpH is maximum when [salt] = [acid], i.e. pH = pKa; useful range is pKa ± 1. Choose a buffer whose pKa is near the target pH.
• Important buffers: blood bicarbonate (H₂CO₃/HCO₃⁻, pH 7.4) plus phosphate and protein buffers; phosphate buffer saline in biology; citrate and acetate buffers in pharma formulation and food.
• Salt hydrolysis — four cases:
  1. Strong acid + strong base (NaCl): no hydrolysis, pH 7.
  2. Weak acid + strong base (CH₃COONa): anionic hydrolysis, basic; pH = 7 + ½(pKa + log c).
  3. Strong acid + weak base (NH₄Cl): cationic hydrolysis, acidic; pH = 7 − ½(pKb + log c).
  4. Weak acid + weak base (CH₃COONH₄): both hydrolyse; pH = 7 + ½(pKa − pKb), nearly independent of concentration.
• Degree of hydrolysis h = √(Kh/c) for cases 2 and 3, with Kh = Kw/Ka (or Kw/Kb) — hydrolysis increases on dilution and on heating.
• Titration curves: strong acid–strong base has a long steep jump through pH 7; weak acid–strong base has a buffer plateau, equivalence point above 7 (pH ≈ 8–9), and pH = pKa at half-neutralisation; weak base–strong acid ends below 7; weak–weak gives no sharp break, so potentiometric or conductometric detection is used.
• Indicators are weak acids/bases with HIn ⇌ H⁺ + In⁻; the colour changes over pH = pKIn ± 1. Methyl orange 3.1–4.4 (strong acid–weak base), phenolphthalein 8.3–10.0 (weak acid–strong base), methyl red 4.4–6.2. Choose an indicator whose range lies inside the steep portion.

Solubility product and precipitation equilibria

• For AxBy(s) ⇌ xA^y⁺ + yB^x⁻, Ksp = [A^y⁺]^x [B^x⁻]^y at saturation; it applies only to sparingly soluble salts in saturated solution and varies with temperature.
• Relations with solubility s (mol L⁻¹): AB → Ksp = s²; AB₂ or A₂B → Ksp = 4s³; A₂B₃ → Ksp = 108 s⁵.
• Criteria: ionic product < Ksp → unsaturated, solid dissolves; = Ksp → saturated; > Ksp → supersaturated, precipitation occurs.
• Common ion effect lowers solubility (AgCl in NaCl); the salt or diverse ion effect slightly raises it by reducing activity coefficients; complex formation ([Ag(NH₃)₂]⁺, [Cu(NH₃)₄]²⁺) dissolves precipitates.
• pH control: sulphide group separation in qualitative analysis (Group II in dilute HCl/H₂S, Group IV in ammoniacal medium) works by adjusting [S²⁻] through the equilibria of H₂S; hydroxides are precipitated by raising pH.
• Industrial and regulatory use: lime–soda and lime precipitation softening of water, chemical precipitation of Cr, Pb, Ni and Zn as hydroxides or sulphides to meet CPCB/State PCB effluent limits under the Water (Prevention and Control of Pollution) Act, 1974; scaling as CaCO₃ and CaSO₄ in boilers; IS 10500:2012 drinking-water limits for pH (6.5–8.5) and TDS rest on these equilibria. Fluoride removal by Nalgonda defluoridation (alum + lime) is a precipitation-adsorption equilibrium in use across Rajasthan, Telangana and Gujarat.

Key points

• Dynamic equilibrium - forward and backward rates equal, concentrations constant, both reactions still running.
• Law of mass action - rate ∝ product of active masses raised to stoichiometric powers; its ratio defines Kc.
• Kp = Kc(RT)^Δn - equal only when Δn = 0; Kx = Kp/P^Δn depends on total pressure.
• Pure solids and liquids - omitted from K expressions since their activity is unity.
• Only temperature changes K - concentration, pressure, volume, catalyst and inert gas only shift position.
• van't Hoff - K rises with T for endothermic reactions and falls for exothermic ones.
• ΔG° = −RT ln K - a large K means product-favoured, never fast; kinetics is a separate question.
• Reaction quotient Q - compare with K to find the direction of net change.
• Le Chatelier compromise - Haber runs at 450–500 °C and 150–300 bar because a low temperature is too slow.
• Ostwald's dilution law - α ≈ √(Ka/c); dissociation of a weak electrolyte increases on dilution.
• Kw = 1 × 10⁻¹⁴ at 25 °C - temperature-dependent, so neutral pH 7 is a 25 °C statement only.
• Ka × Kb = Kw - a strong acid has a weak conjugate base.
• Henderson–Hasselbalch - pH = pKa + log([salt]/[acid]); buffer pH is dilution-independent.
• Maximum buffer capacity at pH = pKa - useful range pKa ± 1.
• Salt hydrolysis - weak-acid salts are basic, weak-base salts acidic, weak–weak depends on pKa − pKb.
• Half-neutralisation point - pH = pKa of the weak acid, the standard way to measure Ka.
• Ksp - AB: s²; AB₂: 4s³; A₂B₃: 108s⁵; precipitation when ionic product exceeds Ksp.
• Indicator choice - its pH range must fall within the steep part of the titration curve.

Formulas and examples

Example 1 — Kc, Kp and the direction of change. For N₂ + 3H₂ ⇌ 2NH₃ at 500 K, a 2 L vessel at equilibrium holds 0.40 mol N₂, 0.60 mol H₂ and 0.20 mol NH₃.

• Concentrations: [N₂] = 0.20, [H₂] = 0.30, [NH₃] = 0.10 mol L⁻¹.
• Kc = (0.10)²/[(0.20)(0.30)³] = 0.01/(0.20 × 0.027) = 0.01/0.0054 = 1.85 L² mol⁻².
• Δn = 2 − 4 = −2, so Kp = Kc(RT)⁻² = 1.85/(0.0821 × 500)² = 1.85/1685 = 1.10 × 10⁻³ atm⁻².
• If the mixture is changed to [N₂] = 0.20, [H₂] = 0.30, [NH₃] = 0.30 mol L⁻¹, then Q = 0.09/0.0054 = 16.7 > Kc, so the reaction runs backward (NH₃ decomposes) until Q falls to 1.85.
• ΔG° = −2.303 × 8.314 × 500 × log 1.85 = −9574 × 0.2672 = −2.56 kJ mol⁻¹.

Example 2 — pH of a weak acid and its dissociation. 0.010 M acetic acid, Ka = 1.8 × 10⁻⁵.

• α = √(Ka/c) = √(1.8 × 10⁻⁵/0.010) = √(1.8 × 10⁻³) = 0.0424 → 4.24% dissociated.
• [H⁺] = cα = 0.010 × 0.0424 = 4.24 × 10⁻⁴ M → pH = −log(4.24 × 10⁻⁴) = 3.37.
• On dilution to 0.0010 M: α = √(1.8 × 10⁻²) = 0.134 (13.4%), [H⁺] = 1.34 × 10⁻⁴, pH = 3.87 — α rises tenfold in √ terms while Ka is unchanged.
• [OH⁻] = 1.0 × 10⁻¹⁴/4.24 × 10⁻⁴ = 2.36 × 10⁻¹¹ M in the 0.010 M solution.

Example 3 — Buffer design and its response. Prepare 1 L of pH 5.00 buffer from acetic acid (pKa = −log 1.8 × 10⁻⁵ = 4.74) and sodium acetate.

• 5.00 = 4.74 + log(ratio) → log(ratio) = 0.26 → [salt]/[acid] = 1.82.
• Taking [acid] = 0.10 M, [salt] = 0.182 M: needs 0.10 mol CH₃COOH (6.0 g) and 0.182 mol CH₃COONa (0.182 × 82 = 14.9 g).
• Add 0.010 mol HCl: acetate falls to 0.172, acid rises to 0.110 → pH = 4.74 + log(0.172/0.110) = 4.74 + 0.194 = 4.93 (a drop of only 0.07).
• The same 0.010 mol HCl in 1 L of pure water at pH 7 gives pH 2.0 — a fall of 5 units, which is the whole point of a buffer.

Example 4 — Solubility product, common ion effect and precipitation. AgCl has Ksp = 1.8 × 10⁻¹⁰ at 25 °C.

• In pure water: s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M → 1.34 × 10⁻⁵ × 143.5 = 1.93 × 10⁻³ g L⁻¹.
• In 0.010 M NaCl: s = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰/0.010 = 1.8 × 10⁻⁸ M, about 745 times lower — the common ion effect.
• Mixing equal volumes of 2 × 10⁻⁴ M AgNO₃ and 2 × 10⁻⁴ M NaCl halves both to 1 × 10⁻⁴ M; ionic product = 1 × 10⁻⁸ > Ksp, so AgCl precipitates.
• For Mg(OH)₂, Ksp = 1.8 × 10⁻¹¹ = 4s³ → s³ = 4.5 × 10⁻¹² → s = 1.65 × 10⁻⁴ M; [OH⁻] = 3.3 × 10⁻⁴, pH = 14 + log(3.3 × 10⁻⁴) = 10.52 — the saturation pH used in lime softening.

Example 5 — Salt hydrolysis. 0.10 M sodium acetate, Ka = 1.8 × 10⁻⁵.

• Kh = Kw/Ka = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰; h = √(5.56 × 10⁻¹⁰/0.10) = √(5.56 × 10⁻⁹) = 7.46 × 10⁻⁵ (0.0075%).
• [OH⁻] = ch = 0.10 × 7.46 × 10⁻⁵ = 7.46 × 10⁻⁶ M → pOH = 5.13, pH = 8.87.
• Check by formula: pH = 7 + ½(4.74 + log 0.10) = 7 + ½(3.74) = 8.87 — agreement confirms the method.
• For 0.10 M NH₄Cl (pKb = 4.74): pH = 7 − ½(4.74 − 1) = 5.13, acidic as expected.

Where it is used

• Fertiliser and heavy chemicals: ammonia, urea, nitric and sulphuric acid plants (IFFCO, RCF, NFL, Coromandel, Tata Chemicals) run on Le Chatelier-optimised converters; process engineers set temperature, pressure and recycle from equilibrium conversion data.
• Pharmaceuticals and formulation: buffer selection for injectables and syrups, pH-solubility profiles for salt-form selection, dissolution media (pH 1.2, 4.5, 6.8) specified in IP/USP monographs — roles in formulation R&D and QC at Cipla, Sun Pharma, Dr Reddy's, Biocon.
• Water and effluent treatment: lime–soda softening, coagulation pH windows, hydroxide/sulphide precipitation of heavy metals and neutralisation to meet CPCB consent conditions and IS 10500:2012 drinking-water specifications; roles in ETP/STP operation and environmental consultancy.
• Analytical and QC laboratories: acid–base and complexometric titrations, buffer standardisation of pH meters (pH 4.01, 7.00, 9.18 standards), indicator selection — NABL-accredited food, pharma and petroleum labs; FSSAI and BIS testing.
• Metallurgy and hydrometallurgy: selective leaching and precipitation, pH control in zinc and copper electrowinning (Hindustan Zinc, Hindalco), cyanide-complex equilibria in gold recovery.
• Teaching, research and exams: CSIR-UGC NET, GATE CY, IIT-JAM and TIFR/JEST chemistry papers; soil pH and nutrient availability work in agricultural universities and ICAR institutes.

Interview questions

• Q: Does a catalyst change the equilibrium constant? A: No. It lowers the activation energy of both directions equally, so equilibrium is reached sooner but at the same composition; only temperature changes K.
• Q: Why is the Haber process run at 450–500 °C when ammonia synthesis is exothermic? A: Low temperature gives a higher equilibrium yield but an unusably slow rate; 450–500 °C with an iron catalyst and 150–300 bar is the kinetic–thermodynamic compromise, with ammonia removed to keep the reaction shifting forward.
• Q: When are Kp and Kc numerically equal? A: Only when Δn = 0 for the gaseous species, since Kp = Kc(RT)^Δn — for example H₂ + I₂ ⇌ 2HI.
• Q: Adding an inert gas — will the equilibrium shift? A: At constant volume, no, because partial pressures are unchanged; at constant pressure the volume increases and the system shifts towards the side with more gas moles.
• Q: Why is the pH of pure water at 60 °C less than 7 although it is still neutral? A: Kw increases with temperature as ionisation is endothermic, so [H⁺] = [OH⁻] = √Kw is larger; neutrality means equal ion concentrations, not pH 7.
• Q: Why does buffer pH not change on dilution? A: The Henderson equation contains only the salt-to-acid ratio, which dilution does not alter; very large dilution does shift pH once activity coefficients and water ionisation become significant.
• Q: How would you determine the Ka of a weak acid experimentally? A: Titrate it with standard NaOH and read the pH at half-neutralisation, where [HA] = [A⁻] so pH = pKa; conductometric or spectrophotometric methods give the same value.
• Q: Why is phenolphthalein used for acetic acid versus NaOH but methyl orange for NH₃ versus HCl? A: The equivalence pH is about 8.8 in the first case and about 5.1 in the second, and an indicator must change colour within the steep region containing the equivalence point.
• Q: 0.1 M HCl and 0.1 M acetic acid — same concentration, why different pH? A: HCl is fully ionised giving [H⁺] = 0.1 M (pH 1), while acetic acid is only about 1.3% ionised at this concentration giving pH ≈ 2.87; strength depends on degree of ionisation, not concentration.
• Q: Why does AgCl dissolve in aqueous ammonia? A: NH₃ forms the [Ag(NH₃)₂]⁺ complex, removing Ag⁺ so that the ionic product falls below Ksp and the solid dissolves.
• Q: Can Ksp be used for sodium chloride? A: Not meaningfully — Ksp applies to sparingly soluble salts; for highly soluble salts ionic strength is large and activity coefficients deviate far from unity.
• Q: A reaction has K = 10¹⁵ but no product forms at room temperature. Explain. A: K is thermodynamic and says the products are strongly favoured; a high activation energy keeps the rate negligible — equilibrium position and rate are independent.

Common mistakes

• Writing concentrations of pure solids or the solvent (water) inside Kc or Ksp expressions.
• Using initial instead of equilibrium concentrations, or forgetting to divide moles by volume when the vessel is not 1 L.
• Assuming Kp = Kc regardless of Δn, or using R = 8.314 with pressures in atm.
• Saying a catalyst or added inert gas at constant volume increases the yield.
• Confusing strength with concentration — a concentrated weak acid can have a higher pH than a dilute strong acid.
• Applying α = √(Ka/c) when α is large (above about 5%) instead of solving the quadratic, and neglecting water's own ionisation in very dilute solutions.
• Treating pH 7 as neutral at any temperature, and ignoring the common ion effect when calculating solubility in a solution that already contains one of the ions.

Related topics

Chemical thermodynamics · Chemical kinetics and catalysis · Electrochemistry · Solutions and colligative properties · Atomic structure and chemical bonding · Volumetric and instrumental analysis