Sulfate (molecular weight = 96g/mole) from an acid mine treatment pond leaves via an outlet stream at a concentration of 275 mg/L. The outlet stream flows into a small river with a flow of 30 cfs upstream, 37 cfs downstream, and a background sulfate concentration of 17 mg/L. Please answer the following.
a. What is the concentration of outlet stream sulfate in moles/L and equivalents/L?
b. What is the concentration of sulfate in the river immediately downstream of the confluence with the outlet stream in mg/L?
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a. The concentration of outlet stream sulfate in moles/L can be calculated by converting the mass concentration (mg/L) to moles/L using the molecular weight of sulfate (96 g/mole).
275 mg/L * (1 g/1000 mg) * (1 mole/96 g) = 0.00286 moles/L
Sulfate has a charge of -2, so the concentration in equivalents/L is twice the concentration in moles/L.
0.00286 moles/L * 2 = 0.00572 equivalents/L
b. The concentration of sulfate in the river immediately downstream of the confluence with the outlet stream can be calculated by using a mass balance equation. The mass of sulfate in the river upstream and in the outlet stream must equal the mass of sulfate in the river downstream.
Let's denote the concentration of sulfate in the river downstream as C (mg/L). The flow rates of the river upstream and downstream and the outlet stream are 30 cfs, 37 cfs, and 7 cfs, respectively (since 37 cfs - 30 cfs = 7 cfs).
The mass balance equation is:
30 cfs * 17 mg/L + 7 cfs * 275 mg/L = 37 cfs * C
Solving for C gives:
C = (30*17 + 7*275) / 37 = 61.35 mg/L
So, the concentration of sulfate in the river immediately downstream of the confluence with the outlet stream is approximately 61.35 mg/L.
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