Chlor-alkali industry: soda ash and caustic soda

Why it matters

Soda ash (Na₂CO₃), caustic soda (NaOH) and chlorine are among the highest-tonnage inorganic chemicals in India, feeding glass, soaps and detergents, alumina, pulp and paper, textiles, PVC and water treatment. The two routes covered here, the Solvay (ammonia–soda) process and brine electrolysis, are classic examples of how stoichiometry, recycle of an expensive reagent and electrochemical efficiency decide plant economics.

Key ideas

Soda ash by the Solvay process. Raw materials are brine (purified NaCl), limestone (CaCO₃) and coke for the lime kiln; ammonia is circulated, with only make-up losses.

1. Lime kiln: CaCO₃ → CaO + CO₂ at about 1000 °C. The CO₂ goes to the carbonating towers.
2. Ammoniation: purified brine absorbs NH₃ (Ca²⁺ and Mg²⁺ are first removed with lime/soda so they do not scale the towers).
3. Carbonation: ammoniated brine is contacted with CO₂ in carbonating (Solvay) towers. NH₃ + CO₂ + H₂O → NH₄HCO₃, then NH₄HCO₃ + NaCl → NaHCO₃↓ + NH₄Cl. Sodium bicarbonate precipitates because it is the least soluble salt in the system at about 30 °C. Towers are cooled because carbonation is exothermic.
4. Filtration and calcination: 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O at about 175–200 °C. This CO₂ is recycled to carbonation. The product is "light" soda ash; recrystallising as the monohydrate and drying gives "dense" ash for glass making.
5. Ammonia recovery: the mother liquor (NH₄Cl) is distilled with milk of lime: 2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + 2H₂O. NH₃ returns to the process.

Overall: 2NaCl + CaCO₃ → Na₂CO₃ + CaCl₂. Ammonia is a reagent that is regenerated, not a catalyst in the strict sense. The weak points are NaCl utilisation (only about 70–75 % per pass, because the carbonation equilibrium is incomplete) and the CaCl₂ effluent. The dual process (modified Solvay, used in India) skips the lime step and crystallises NH₄Cl as a fertiliser co-product, saving limestone and avoiding CaCl₂ waste.

Caustic soda and chlorine by electrolysis. Saturated, purified brine is electrolysed:

• Anode (oxidation): 2Cl⁻ → Cl₂ + 2e⁻
• Cathode (reduction): 2H₂O + 2e⁻ → H₂ + 2OH⁻
• Overall: 2NaCl + 2H₂O → 2NaOH + Cl₂ + H₂

Cl₂ and NaOH must be kept apart, or they react to hypochlorite and chlorate, and H₂ and Cl₂ must never mix (explosive). Three cell types do this:

• Mercury cell: flowing mercury cathode forms sodium amalgam, which is decomposed with water in a separate denuder to give 50 % NaOH of very high purity. Being phased out because of mercury emissions.
• Diaphragm cell: an asbestos or polymer diaphragm separates the compartments; cell liquor is only about 10–12 % NaOH with about 15 % NaCl, so it needs evaporation and salt removal.
• Membrane cell: a cation-exchange (perfluorinated) membrane passes Na⁺ but blocks Cl⁻ and OH⁻. It gives about 32–35 % NaOH with very little salt and the lowest energy use; it is now the standard technology.

Products are made in a fixed ratio: about 1.13 t NaOH per tonne of Cl₂ (the "ECU", electrochemical unit), so caustic and chlorine markets must be balanced.

Formulas

NaCl + NH₃ + CO₂ + H₂O → NaHCO₃ + NH₄Cl (carbonation; precipitation of bicarbonate)

2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O (calcination)

2NaCl + CaCO₃ → Na₂CO₃ + CaCl₂ (overall Solvay)

2NaCl + 2H₂O → 2NaOH + Cl₂ + H₂ (overall electrolysis)

m = η · I · t · M / (n · F) (Faraday's law)

• m: mass produced (kg), η: current efficiency (fraction), I: current (A), t: time (s), M: molar mass (kg/mol), n: electrons per molecule of product (1 for NaOH, 2 for Cl₂ and H₂), F = 96 485 C/mol.

Specific energy = V · I · t / m (J/kg; divide by 3.6 × 10⁶ for kWh/kg)

• V: cell voltage (V). Applies to DC energy at the cell; rectifier losses are extra.

Molar masses used: NaCl 58.44, NaHCO₃ 84.01, Na₂CO₃ 105.99, NaOH 40.00, Cl₂ 70.90, H₂ 2.016 g/mol.

Worked examples

Example 1 (standard): salt demand for soda ash. Find the NaCl needed per tonne of Na₂CO₃ (a) in theory and (b) if NaCl utilisation is 75 %.

1. Overall: 2NaCl → 1Na₂CO₃.
2. Moles Na₂CO₃ = 1000 kg / 105.99 kg/kmol = 9.435 kmol.
3. Moles NaCl = 2 × 9.435 = 18.87 kmol; mass = 18.87 × 58.44 = 1102.7 kg.
4. With 75 % utilisation: 1102.7 / 0.75 = 1470 kg.

Answer: (a) 1103 kg NaCl; (b) 1470 kg NaCl per tonne Na₂CO₃.

Example 2 (GATE level): membrane cell output and energy. A membrane cell runs at 150 kA for 24 h with 95 % current efficiency and 3.1 V cell voltage. Find the NaOH and Cl₂ produced and the specific DC energy per tonne of NaOH.

1. Effective charge: η·I·t = 0.95 × 150 000 A × 86 400 s = 1.2312 × 10¹⁰ C.
2. Moles of electrons = 1.2312 × 10¹⁰ / 96 485 = 1.2761 × 10⁵ mol = moles of NaOH (n = 1).
3. NaOH = 1.2761 × 10⁵ × 40.00 g = 5.104 × 10⁶ g = 5.10 t.
4. Cl₂ = (1.2761 × 10⁵ / 2) × 70.90 g = 4.524 × 10⁶ g = 4.52 t.
5. Energy = V·I·t = 3.1 × 150 000 × 86 400 J = 4.018 × 10¹⁰ J = 11 160 kWh.
6. Specific energy = 11 160 kWh / 5.104 t = 2186 kWh/t.

Answer: 5.10 t NaOH, 4.52 t Cl₂, about 2190 kWh per tonne NaOH.

Common mistakes

• Treating ammonia as a raw material consumed in the Solvay process; it is recovered and only losses are made up. Limestone is a true raw material.
• Writing NaCl → NaHCO₃ for the overall Solvay balance and forgetting that 2 mol NaHCO₃ (hence 2 mol NaCl) give 1 mol Na₂CO₃.
• Using n = 1 for Cl₂ in Faraday's law; each Cl₂ needs 2 electrons.
• Forgetting current efficiency, or applying it to voltage. Current efficiency scales the product; voltage scales energy.
• Mixing up cell types: only the mercury cell gives 50 % caustic directly; diaphragm liquor is dilute and salty.
• Calling the membrane anion-permeable. It is a cation-exchange membrane that passes Na⁺.

For GATE CH

Expect conceptual MCQs on the sequence of Solvay operations, which species are recycled, by-products (CaCl₂, NH₄Cl), and the comparison of mercury, diaphragm and membrane cells. Numerical questions use Faraday's law with current efficiency, product ratios (NaOH : Cl₂ : H₂), specific energy consumption, and simple mass balances with incomplete salt utilisation. Practise writing the electrode reactions and keeping n correct for each product.

Quick check

1. Which two materials are the true raw materials of the Solvay process apart from fuel?
2. Why is the carbonating tower cooled?
3. How many kg of Cl₂ accompany 1 kg of NaOH in brine electrolysis?
4. Which cell gives 50 % caustic directly from the decomposer?

Answers: 1. NaCl (brine) and limestone; 2. carbonation is exothermic and NaHCO₃ is less soluble at lower temperature, so cooling improves precipitation; 3. 70.90 / (2 × 40.00) = 0.886 kg; 4. the mercury cell.