Water from a well is found to contain 1.2mg TCE/L. In order to determine the amo

Question

Water from a well is found to contain 1.2mg TCE/L. In order to determine the amount of contamination in the subsurface, an adsorption experiment is conducted. Back in the lab, thirty mL of water containing TCE at 87.5 mg/L is placed in a vial with 0.10 grams of soil from the well and allowed to come to equilibrium. The aqueous concentration at equilibrium is 7.5 mg/L. a) Compute the sorption, S, in the lab experiment. ______ mg/g b) Estimate the linear adsorption partition coefficient, K_P? ______ L/g. c Then, estimate the amount of TCE adsorbed to the subsurface soil (S) assuming the groundwater and the soil are in equilibrium. _____mg/kg.

Explanation / Answer

q =[(Ci-Cf)*V] / W. ..................(i)

where Ci = Initial Conc. of Adsorbate = 87.5 mg/L

Cf = Final Conc. of Adsorbate = 7.5 mg/L

V= Volume of solution = 30 ml = 0.03 L

W = weight of adsorbent = 0.1 gm

S= [(87.5-7.5) * 0.03]/ 0.1

S= 24 mg/ g

It means 24 mg of TCE is adsorbed by 1 gm of soil.

Concentration in solid phase (soil) = 24 mg/ gm

Concentration in solution phase (at equilibrium) = 7.5 mg/L

Kp = 24 /7.5 = 3.2 L/gm

q = 0.024 mg/kg , Ci = 1.2 mg/L, Cf = unknown , V = 1L, W = 1 kg (Suppose)

24= (1.2 - Cf )*1/1

Cf = 1.2- 0.024 = 1.176 mg/L = Final Conc. of TCE in water at equilibrium

Using equ (ii)

Kp = 3.2 = conc. in solid phase / 1.176

Conc. in solid phase or the amount of TCE adsorbed to the sub surface soil = 3.76 mg/kg

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