Which Solid Controls Lead Solubility in Contaminated Soil?
A contaminated soil site has porewater at pH 6.5 with [Cl−] = 0.10 M, [SO4^2−] = 0.010 M, and total dissolved carbonate = 1.0 × 10−3 M. The solid phases present include PbCO3(s) (Ksp = 7.4 × 10−14), PbSO4(s) (Ksp = 1.6 × 10−8), and PbCl2(s) (Ksp = 1.7 × 10−5). Lead also forms aqueous chloride complexes: Pb2+ + Cl− ⇌ PbCl+ with β1 = 10^1.6, and Pb2+ + 2Cl− ⇌ PbCl2(aq) with β2 = 10^1.8. The second dissociation constant of carbonic acid is Ka2 = 4.7 × 10−11. The site manager wants to know which solid phase controls dissolved lead and whether raising the pH to 8.0 would reduce lead solubility. Assume no other ligands and that activity coefficients are unity. The porewater is in equilibrium with the solid phases, and the total carbonate includes H2CO3, HCO3−, and CO3^2−.
1. Calculate [CO3^2−] at pH 6.5 and determine which solid phase is thermodynamically stable (controls [Pb2+]). Show the ion product comparison.
Read the worked answer
At pH 6.5, [H+] = 10^−6.5 = 3.16 × 10−7 M. For carbonic acid, the fraction of total carbonate present as CO3^2− is: α(CO3^2−) = Ka2 / ([H+] + Ka2) = 4.7 × 10−11 / (3.16 × 10−7 + 4.7 × 10−11) = 1.49 × 10−4. [CO3^2−] = 1.0 × 10−3 × 1.49 × 10−4 = 1.49 × 10−7 M.
For each solid, calculate the [Pb2+] required for saturation: PbCO3: [Pb2+] = Ksp / [CO3^2−] = 7.4 × 10−14 / 1.49 × 10−7 = 4.97 × 10−7 M. PbSO4: [Pb2+] = 1.6 × 10−8 / 0.010 = 1.6 × 10−6 M. PbCl2: [Pb2+] = 1.7 × 10−5 / (0.10)^2 = 1.7 × 10−3 M.
The lowest [Pb2+] is 4.97 × 10−7 M, from PbCO3. Check that the other solids are undersaturated at this [Pb2+]: Ion product PbSO4 = (4.97 × 10−7)(0.010) = 4.97 × 10−9 < 1.6 × 10−8, so PbSO4 dissolves. Ion product PbCl2 = (4.97 × 10−7)(0.10)^2 = 4.97 × 10−9 < 1.7 × 10−5, so PbCl2 dissolves. Therefore PbCO3(s) is the stable solid and controls [Pb2+] ≈ 5.0 × 10−7 M.
2. Calculate the total dissolved lead concentration at pH 6.5, including chloride complexes, and express it in mg L−1 as Pb (atomic mass 207 g mol−1).
Read the worked answer
Total dissolved lead = [Pb2+] (1 + β1[Cl−] + β2[Cl−]^2). β1 = 10^1.6 = 39.8; β2 = 10^1.8 = 63.1. With [Cl−] = 0.10 M: β1[Cl−] = 3.98; β2[Cl−]^2 = 63.1 × 0.010 = 0.631. Sum = 1 + 3.98 + 0.631 = 5.61. Total dissolved Pb = 4.97 × 10−7 × 5.61 = 2.79 × 10−6 M. Mass concentration = 2.79 × 10−6 mol L−1 × 207 g mol−1 = 5.78 × 10−4 g L−1 = 0.578 mg L−1. Only 1/5.61 = 17.8% of the dissolved lead is free Pb2+; the rest is PbCl+ and PbCl2(aq).
3. The manager proposes raising the pH to 8.0 to precipitate more lead carbonate. Calculate the new total dissolved lead at pH 8.0 (assuming PbCO3 still controls and chloride/sulfate unchanged), and evaluate whether this is a sound remediation strategy. Include any caveat about lead hydroxide formation.
Read the worked answer
At pH 8.0, [H+] = 1.0 × 10−8 M. The carbonate fraction becomes: α(CO3^2−) = 4.7 × 10−11 / (1.0 × 10−8 + 4.7 × 10−11) = 4.68 × 10−3. [CO3^2−] = 1.0 × 10−3 × 4.68 × 10−3 = 4.68 × 10−6 M. [Pb2+] = 7.4 × 10−14 / 4.68 × 10−6 = 1.58 × 10−8 M. Total dissolved Pb = 1.58 × 10−8 × 5.61 = 8.87 × 10−8 M = 0.0184 mg L−1.
This is a thirty-fold decrease from 0.578 mg L−1, so raising pH to 8.0 is chemically effective at lowering dissolved lead. However, the strategy must be assessed in the field: adding base may not raise porewater pH uniformly, and at higher pH lead hydroxide phases or anionic plumbite species (Pb(OH)3−) can increase solubility. At pH 8.0, Pb(OH)2 is not expected to precipitate because the ion product [Pb2+][OH−]^2 = (1.58 × 10−8)(1.0 × 10−6)^2 = 1.58 × 10−20 is far below the Ksp of Pb(OH)2 (about 1.2 × 10−15). The caveat is that over-liming to pH > 10 could redissolve lead as plumbite, so pH control is essential.