The deep brines on the Keweenaw are capable of dissolving mercury (Hg) from the
ID: 887654 • Letter: T
Question
The deep brines on the Keweenaw are capable of dissolving mercury (Hg) from the copperore in the abandoned mine shafts. The concentration of Hg in the water draining from theOsceola #4 shaft into a tributary of the Traprock River is 127 ng/L. The Water Quality Criterionfor Hg in streams is 1.3 ng/L. You wish to evaluate the feasibility of using a constructed wetlandto maintain the concentration of Hg below the legal limit. The primary method for retention ofHg in such a wetland is believed to be the precipitation of mercury sulfide, HgS(s) or cinnabar. Hg2+ + HS- <->HgS(s) + H+ [1]
a. You wish to determine the concentration of total sulfide (HS- + H2S) that would berequired in the wetland to lower the Hg2+ concentration to 1.3 ng/L. If the pH is 7.2, theionic strength is 0.02 M, what concentration of bisulfide, HS-, would be required?
Assume 25oC.
b. As indicated above, sulfide exists as both sulfide and bisulfide:
H2S <-> HS- + H+ [2]
What would be the concentration of H2S and total sulfide, given the pH and the bisulfideconcentration that you calculated in part a. With this knowledge, you could review the literatureon artificial wetlands and determine how feasible it is to maintain such a concentration of sulfide.
c. Based on the sign of deltaH for reaction [1] above, in which season, winter or summer, would it be most difficult to maintain the required concentration of Hg?
Explanation / Answer
(a) Hg2+ + HS- --------> HgS + H+
Ionic strength = 0.02 M
1/2 * [H+] + 1/2 * [HS-] = 0.02 M
pH = 7.2
-log[H+] = 7.2
[H+] = 10^-7.2 = 6.31 x 10^-8 M
So,
[HS-] = 2 * 0.02 = 0.04 M
(b)
[H2S] = [H+] [HS-]
[H2S] = 2.6 x 10^-10
Total concentration of sulphide = 0.04 M
(c) During winter it will be most difficult to maintain concentration of Hg
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