3 30 closed system steady state Consider a lake having the f

3. (30%) (closed system, steady state) Consider a lake having the following properties: Lake volume 3.1 x 10m3 Suspended solids (SS): 1 m3 of SS/ 105 m3 water Biota: 31 m3 SS/water partition coefficient, Kss 104 Biota/water partition coefficient, Kbota-10 The lake contains 1.5 kg of a contaminant. Calculate: (a) Concentration (ug m-3)and % of the total contaminant in the water. (b) Concentration(ug m-3) and % of the total contaminant in the suspended solids. (c) Concentration(ug m\") and % of the total contaminant in the biota.

Solution

Lake water volume = 3.1 x 10⁷ m3

Volume of SS = 3.1 x 10⁷ / 10⁵ m3 = 3.1 x 10² m3

Volume of Biota = 31 m3

Total mass of contaminant = 1.5 kg

Mass of contaminant in SS / Mass of contaminant in water = 10⁴

Mass of contaminant in Biota / Mass of contaminant in water = 10⁵

Let mass of contaminant in water = m

Total mass of contaminant = m + m 10⁴ + m 10⁵ = 1.5 kg

m = 1.36 x 10^-5 kg = 13600 g

a)

Concentration of contaminant in water = m / Lake water volume

= 13600 g / 3.1 x 10⁷ m3

= 4.4 x 10^-4 g / m3

% of total contaminant in water = m / 1.5 kg * 100 = 9 x 10^-4 %

b)

Concentration of contaminant in SS = m 10⁴ / Volume of SS

= 13600 * 10⁴ g / 3.1 x 10² m3

= 4.4 x 10⁵ g / m3

% of total contaminant in SS = m 10⁴ / 1.5 kg * 100 = 9.09 %

c)

Concentration of contaminant in Biota = m 10⁵ / Volume of Biota

= 13600 * 10⁵ g / 31 m3

= 4.4 x 10⁷ g / m3

% of total contaminant in Biota = m 10⁵ / 1.5 kg * 100 = 90.9 %