6. An air stripping tower, similar to the picture shown below, is used to remove

Question

6. An air stripping tower, similar to the picture shown below, is used to remove dissolved carbon dioxide gas from a groundwater supply. If the tower lowers the level of carbon dioxide in the water to twice the equilibrium concentration, what concentration of carbon dioxide dissolved gas will remain in the water after treatment. The partial pressure of carborn dioxide in the atmosphere is 1035 atm. The Henry's Law Constant for carbon dioxide at the temperature in the tower is 1015 mole/L-atm. Answer in units of mg COa/L. (Problem adapted from Reckhow)(Answer: 0.88 mg/L) with Air In

Explanation / Answer

As Per Henry's Law C=K*PGAS

where C=Solubility of the gas at a fixed temparture in a particular solvent(mole/Litre)

K=Henry's Law Constant(mole/Litre-atm)

PGAS=Partial Pressure of the gas(atm)

Now as per the given Problem, Pco2 =10-3.5 atm and k = 10-1.5 mol/Litre-atm

Here we have to find out the value of C in mg/Litre.

Since the towers lowers the level of CO2 in the water to twice the equillibrium constant, the value of C would be 2*PCo2*K

So C = 2*PCo2*K = 2*10-3.5* 10-1.5 mole/Litre = 2 * 10-5 mole/Litre

Now 1 mole CO2 = 44 gm CO2 =44000 mg CO2 = 44*103 mg CO2

So C= 2 * 10-5 *44*103 mg/Litre = 88 * 10-2 mg/Litre = 0.88 mg/Litre

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