Limestone Neutralization of Spent Pickle Liquor: When pH Is Not Acidity
An electroplating plant must neutralize 20.0 m³ of spent pickle liquor before discharge. The permit allows pH 6.0–9.0 and no more than 0.5 mg/L Ni at the outfall. The liquor contains 1,200 mg/L Ni²⁺ and 350 mg/L Fe³⁺ and is free of sulfate and fluoride. An in-line probe in the treatment tank, calibrated that morning at 25 °C against fresh pH 4.00 and 7.00 buffers, reads 1.85. The reagent is a dry-ground limestone slurry analyzing 94.5% CaCO₃, 3.2% MgCO₃ and 2.3% acid-insoluble inerts. Site practice is 12 kg of slurry per m³, so the operator charges 240 kg into the open, stirred 20 m³ tank, which vents indoors. The batch foams heavily and the vent line fogs. After 2 h the probe reads 5.4 and no solids are visible. The shift supervisor wants to add 60 kg more and discharge. Total acidity has never been measured on this stream; only pH is logged.
1. Using the probe reading and the slurry analysis, calculate the stoichiometric slurry demand for 20.0 m³ and compare it with the 240 kg charged. What conclusions follow, and what single measurement would settle the question?
Read the worked answer
Acid equivalents from pH. [H⁺] = 10⁻¹·⁸⁵ = 1.41 × 10⁻² mol/L, so 1.41 × 10⁻² eq/L × 20,000 L = 282 eq.
Slurry equivalent mass. Per 100 g dry solids: CaCO₃ 94.5/100.09 = 0.944 mol × 2 = 1.888 eq; MgCO₃ 3.2/84.31 = 0.038 mol × 2 = 0.076 eq; total 1.964 eq/100 g, i.e. 50.9 g/eq (inerts contribute nothing).
Demand. 282 eq × 50.9 g/eq ≈ 14.4 kg. The charge of 240 kg is 4,715 eq - a 17-fold excess.
Metal acidity does not close the gap. Ni: 1.2/58.69 = 0.0204 M × 2 = 0.0409 eq/L (818 eq); Fe: 0.35/55.85 = 0.0063 M × 3 = 0.019 eq/L (376 eq). Together ≈ 1,194 eq ≈ 61 kg, so even the full demand including hydroxide precipitation is ~75 kg, not 240 kg.
Conclusion. A liquor at pH 1.85 cannot consume 240 kg of 95% carbonate and still sit at pH 5.4 with no residual solid; with excess limestone present the open system would be CO₂-buffered near pH 8.3. The probe is the prime suspect: a two-point 4.00/7.00 calibration never checks electrode slope in the acid region, and a fouled junction or dehydrated glass bulb reads high. A diluted or wrongly located sample point is the alternative.
Settling measurement. Total acidity: titrate a filtered sample with standardized 0.5 M NaOH to pH 8.3, reporting eq/L (note the pH 4.5 endpoint separately as free mineral acid). Dose from that number, not from pH.
2. The supervisor wants to add 60 kg more and discharge. Assess this against the permit, and set out the dosing control you would actually install.
Read the worked answer
Adding 60 kg is wrong on two counts.
- It cannot help the nickel. In a tank open to air, water saturated with limestone is CO₂-buffered to roughly pH 8.3. With Ksp(Ni(OH)₂) ≈ 5.5 × 10⁻¹⁶, at pH 8.3 ([OH⁻] = 2 × 10⁻⁶) the equilibrium nickel is 5.5 × 10⁻¹⁶/4 × 10⁻¹² ≈ 1.4 × 10⁻⁴ M ≈ 8 mg/L - sixteen times the 0.5 mg/L limit. Excess limestone just becomes sludge and suspended solids in the discharge.
- The pH window is narrow. To reach 0.5 mg/L you need [OH⁻] = √(Ksp/[Ni]) = √(5.5 × 10⁻¹⁶/8.5 × 10⁻⁶) ≈ 2.5 × 10⁻⁶ … more simply, at pH 9.0 nickel is ≈0.3 mg/L (pass) and at pH 8.5 it is ≈3 mg/L (fail). The permit ceiling is 9.0, so the set point is pH 8.8–9.0 with a dead band of about ±0.15.
Control scheme. (a) Measure total acidity by titration to pH 8.3 and dose limestone to ~90% of that demand; (b) trim with NaOH (or lime) under PID control to pH 8.9, with the probe on a flow-through cell, not in the bulk; (c) use the 350 mg/L Fe³⁺ deliberately - Fe(OH)₃ floc adsorbs and co-precipitates nickel, improving robustness at the lower pH; (d) discharge only on a batch-hold-and-release basis with a confirmatory Ni analysis (e.g. colorimetric or ICP-OES), never on pH alone; (e) log the probe slope check in pH 1.68 or 4.00 buffer weekly.
3. Estimate the CO₂ released if the 240 kg charge reacts completely, and state the process-safety implications for the indoor vent.
Read the worked answer
Stoichiometry. 240 kg of slurry = 240,000 g / 50.9 g per eq = 4,715 eq of carbonate. Each equivalent of carbonate releases one mole of CO₂ on neutralization, so n(CO₂) = 4,715/2 = 2,355 mol.
Volume. At 25 °C and 1 atm, V = nRT/P = 2,355 × 0.02447 m³/mol ≈ 57.6 m³ of CO₂ - nearly three times the 20 m³ liquid volume, released over the 2 h batch (about 0.5 m³/min).
Implications.
- The vent is indoors: 57.6 m³ of CO₂ displacing air in a confined area is an asphyxiation hazard (CO₂ is denser than air and accumulates at floor level and in pits). Ventilation must be sized for the peak rate and the vent routed outdoors, away from air intakes.
- The fogging and foaming are consistent with rapid CO₂ evolution from fine limestone; foam-over can carry solids and liquor out of the vent line and gives a false 'no solids remaining' observation. Dose the slurry at a controlled rate rather than all at once, and add a defoamer or an anti-foam headspace.
- Because the reaction is gas-evolving and exothermic, the batch needs a level and temperature interlock; a blocked vent would pressurize the tank.
- The 57.6 m³ figure also proves the tank cannot have absorbed 240 kg quietly: the observed foaming and fogging are the signature of the near-complete reaction that the pH reading denied.