Arsenic in geomedia contaminates an aquifer by naturally occurring geological processes. The aquifer has a bulk density of 1.55 g/cm3 and a particle density of 2.63 g/cm3. The distribution coefficient, Kd value is 0.0370 L/kg. The initial concentration at the source is 40.0 ƒÝg/L. The groundwater velocity is 9.50 x 10^1 m/yr and the dispersion coefficient is 0.500 m^2/day. Assume a half-life of 6.00 x 10^4 years. Estimate the concentration at 11.0 m distance from the source 0.120, 0.125, 0.130, 0.140, 0.150, and 1.00 years after the process begins. Also, estimate the time needed for the arsenic concentration to exceed 0.0100 ƒÝg/L at 25.0 m from the source.
To estimate the concentration of arsenic at a specific distance from the source over time, we can use the advection-dispersion equation with the given parameters. The equation is as follows:
C(x, t) = (C0 * Q * E) / (v * (4 * π * Dt)^0.5) * exp(-x/v - x^2/(4Dt))
Where:
C(x, t) is the concentration at distance x and time t
C0 is the initial concentration at the source
Q is the porosity of the aquifer
E is the retardation factor (1 + Kd * ρb / ρp) where ρb is the bulk density and ρp is the particle density
v is the groundwater velocity
D is the dispersion coefficient
Let's start by calculating the retardation factor (E):
E = 1 + (Kd * ρb / ρp)
= 1 + (0.0370 * 1.55 g/cm^3 / 2.63 g/cm^3)
= 1.021
Now, we can substitute the given values into the advection-dispersion equation to estimate the concentrations at the specified distances and times, as well as the time needed for the concentration to exceed 0.0100 ƒÝg/L at 25.0 m from the source.
For 11.0 m distance:
C(11.0 m, 0.120 years) = (40.0 ƒÝg/L * Q * 1.021) / (9.50 x 10^1 m/yr * (4 * π * 0.500 m^2/day)^0.5) * exp(-11.0 m / (9.50 x 10^1 m/yr) - (11.0 m)^2 / (4 * 0.500 m^2/day))
C(11.0 m, 0.125 years) = (40.0 ƒÝg/L * Q * 1.021) / (9.50 x 10^1 m/yr * (4 * π * 0.500 m^2/day)^0.5) * exp(-11.0 m / (9.50 x 10^1 m/yr) - (11.0 m)^2 / (4 * 0.500 m^2/day))
C(11.0 m, 0.130 years) = (40.0 ƒÝg/L * Q * 1.021) / (9.50 x 10^1 m/yr * (4 * π * 0.500 m^2/day)^0.5) * exp(-11.0 m / (9.50 x 10^1 m/yr) - (11.0 m)^2 / (4 * 0.500 m^2/day))
C(11.0 m, 0.140 years) = (40.0 ƒÝg/L * Q * 1.021) / (9.50 x 10^1 m/yr * (4 * π * 0.500 m^2/day)^0.5) * exp(-11.0 m / (9.50 x 10^1 m/yr) - (11.0 m)^2 / (4 * 0.500 m^2/day))
C(11.0 m, 0.150 years) = (40.0 ƒÝg/L * Q * 1.021) / (9.50 x 10^1 m/yr * (4 * π * 0.500 m^2/day)^0.5) * exp(-11.0 m / (9.50 x 10^1 m/yr) - (11.0 m)^2 / (4 * 0.500 m^2/day))
For 25.0 m distance:
t = ? years (time needed for concentration to exceed 0.0100 ƒÝg/L)
You can now substitute the given values and perform the calculations to obtain the estimated concentrations and time needed for the concentration to exceed 0.0100 ƒÝg/L at 25.0 m from the source.