We can precipitate lead with many anions, including sulfide and sulfate (among o

Question

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 MCL, or maximum contaminant level) of ~7 times 10^-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 initially at 2 times 10^-5 M from 100 m^3 of solution? In each case you can assume the initial concentration of the anions (sulfide or sulfate) is 0. Assume CaS costs ~$440/kg and CaSO_4 costs ~$1/kg (MW of CaS = 72.2 g/mol, MW of CaSO_4 = 136 g/mol). K = 1 times 10^-28 = [Pb^2+][S^2-] K_sp = 1.6 times 10^-8 = [Pb^2+][SO_4^2-]

Explanation / Answer

CaS is preferable as it is having lesser Ksp value. From its Ksp value, we conclude that PbS forms faster than PbSO4. Moreover, the cost of CaS is lesser than the cost of CaSO4 .

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.