the seasonally averaged sulfate concentration in the pore water of a lake sedime
ID: 947975 • Letter: T
Question
the seasonally averaged sulfate concentration in the pore water of a lake sediment is 50 uM (micro moles) at the sediment-water interface, but only 5 uM at a depth 0.2 cm below the interface.
a. how much sulfate can be removed each year from this lake by diffusion into the sediment? assume an effective diffusion coefficient of 10^-6 cm^2/sec, and a lake area of 60 ha.
b. at what rate would this removal of sulfate cause lake alkalinity to rise if the lake has an avg. depth of 7m and and there were no other sources or sinks of sulfate?
c. in reality, the lake has an inflow if 5*10^6 m^3/yr, and sulfate concentration in the inflow is 52 uM. outflow is 4.5*10 m^3/ur, with no conc. of 50 uM of SO4^2-. neglect any other sources or sinks of SO4^2- , assume the lake to be well mixed and estimate the atmospheric input of SO4^2- .
Explanation / Answer
At steady state, according to ficks law for 1-D mass diffusion
d2C/dx2 = 0
C = ax +b
C |x=0 = 50 x 10-6 M
b = 50 x 10-6
C |x=0.2 cm = 5 x 10-6 M
a * 0.2/100 + 50 x 10-6 = 5 x 10-6 M
a = -0.0225
C = 50 x 10-6 - 0.0225 x
Na = -D * dC/dx
= (10-6 * 10-4 m2/sec) * (0.0225 M/m) * (1000 L/m3)
= 2.25 x10-9 mol/m2-sec
a.
Sulfate that can be removed each year from this lake by diffusion into the sediment
= 2.25 x10-9 mol/m2-sec * 60 * 10000 * 365.25 * 24 * 60 * 60 = 4.26 x 104 moles/yr
b.
Rate of removal of sulfate = 2.25 x10-9 mol/m2-sec * 600000 = 1.35 x 10-3 mol/s
Volume = 60 * 10000 * 7 = 4200000 m3 = 42 x 108 L
Alkalinity rise rate = 1.35 x 10-3 / 42 x 108 mol/L-s as CaCO3 = 3.21 x 10-13 mol/L-s as CaCO3
= 3.21 x 10-8 mg/L-s as CaCO3
c.
Net inflow - Net outflow = Removal rate of sulfate
Inflow of lake = 5 x 106 * 1000 * 52 x 10-6 mol/yr = 26 x 104 mol/yr
Net outflow = 4.5 x 106 * 1000 * 50 x 10-6 mol/yr = 22.5 x 104 mol/yr
Atmospheric input = Removal rate of sulfate + Net outflow - Inflow of lake
= 4.26 x 104 + 22.5 x 104 - 26 x 104 = 7.6 x 103 mol/yr
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