In this question, you will calculate the total acid generating capacity of a min

Question

In this question, you will calculate the total acid generating capacity of a mine tailings pile associated with copper mining that contains both pyrite and covellite. The mine tailings impoundment contains 50,000 cubic meters of tailings material. The tailings have an average density of 1.4 g/cm^3. The tailings contain pyrite (FeS_2) at volume 1% and covellite (CuS) at volume 0.25%. Assume the pyrite oxidizes to sulfate (SO^2-_4) and ferrihydrite (Fe(OH)_3(s))) and the covellite oxidation produces Cu^2+ and sulfate. Also assume that the oxidation state of the sulfur (S) in pyrite is (-1) and is (-2) in the covellite. How much acidity will be produced if complete oxidation of the sulfide minerals in that tailings occurs, report your answer in moles of protons (H+). If all of that acidity when into a lake 1 km long, 0.5 km wide and 50 m deep that was at pH 7 and had no buffering capacity, what would the pH of the lake be?

Explanation / Answer

a) volume of pyrite(FeS2)=1% of 50,000 cm3=0.01*50,000=500 cm3

mass of FeS2=500 cm3*1.4g/cm3=700 g

moles of FeS2=700g/119.965 g/mol=5.835 moles

FeS2 + 11 [O] +3H+------------>Fe(OH)3 (s) +2SO42-

1 mole FeS2 produces 2 moles sulfate

CuS + 4[O] ------>Cu2+ + SO42-

volume of CuS=0.0025*50,000 cm3=125 cm3

mass of CuS=1.4g/cm3*125 cm3=175g

moles of CuS=175g/95.611g/mol=1.830 moles

SO42- +2H+ ------------>H2SO4

[moles of sulfate produced=1.83+5.835=7.665 moles ]

moles of H+ reacted to form acid=2*moles of sulfate=2*7.665 =15.331 moles of H+

b)[H+]=15.331 moles

volume of water=0.5*1000m*1000m*50m=25*10^6 m3=25*10^9 L

[H+]=15.331 moles /(25*10^9 L)=6.13*10^-10 mol/L

pH=-log[H+]=-log(6.13*10^-10)=9.21

pH=9.21

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