We can precipitate lead with many anions, including sulfide and sulfate (among others). Using the two...

We can precipitate lead with many anions, including sulfide and sulfate (among others). Using the two equilibrium relationships below, the treatment target (based on the ‘action level’) of ~7x10-8 M, and the associated costs of the salts, which would you select (ignoring all other considerations, which are many, but looking at cost alone) to remove lead from 100 m3 of solution with lead at 2x10-5 M? In each case you will have to determine the amount of anion (sulfide or sulfate) needed to i) form the solid, and ii) increase its concentration in solution from the initial value of 0 to the equilibrium value from Ksp. Assume CaS costs ~$440/kg and CaSO4 costs ~$1/kg (MW of CaS = 72.2 g/mol, MW of CaSO4 = 136 g/mol).

Ksp = 1x10-28 = [Pb2+][S2-]

Ksp = 2.53x10-8 = [Pb2+][SO42-]

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Amanda K answered 3 weeks ago

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