In: Chemistry
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- .
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