Question

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1) You are in charge of a remediation strategy that requires aerobic, microbial respiration to degrade the compound ethanol in groundwater. There is 840mg/L ethanol in the groundwater; how much oxygen (total, steady state concentration in mg/L) will you need to completely oxidize the ethanol to CO2? Can you accomplish this using only atmospheric oxygen (21 % v/v) at 1 atm total pressure (Pro) and 25°C? If you cannot, what options do you have for increasing oxygen content of the water? Name at least three. For this: Ethanol, C2H6O, mol mass 46.1 g/mol Oxygen, O2, mol mass 32 g/mol (though it may be good to start remembering this...) KH O2 at 25°C's 0.0013 mol L.' atm 2) You have a benzene plume (CsH6) at 80mg/L (steady state concentration). t degrades via aerobic microbial activity. What is the steady state concentration of oxygen (O2) in water at this site assuming 21% oxygen by volume in air, at 1 atm total pressure. The KH of oxygen at 25 degrees is: 0.0013 mol/L.atm Is the steady state concentration of oxygen in water enough to completely oxidize the benzene to CO2? not, what can we do to get more oxygen into the water? 3) Assume you have a surface water system that is pH 3.8; the buffering in the water is exceptionally strong such that any weak acid or weak base added will not influence the pH (i.e. the pH will remain fixed). A nearby industry releases ascorbate into the water body. Ascorbate is the oxyanion form of ascorbic acid (vitamin C), which has a dissociating proton with a pKa of 4.7. Describe (do not calculate) what will happen to the ascorbate that enters the water body.

Answer 1

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AS FOR GIVEN DATA..

As per Question given CH_3 CH_2 OH + 3o_2 rightarrow 2 co_2 + 3 H_2 o Bases: 1.0 L water mass of ethanol = 840 mg = 840 times 10^-3 g molar mass of ethanol = 46.1 g/mol moles at ethanol = mass/molar mass = 840 times 10^-3 g/46.1 g/mol = 0.018 mol balanced reaction equation 1 mol of ethanol require 3 mol o_2 0.013 mol ethanol will require (3 mol o_2/1 mol ethanol) times 0.018 mole ethanol = 0.0597 mol o_2 molar mass of o_2 = 32 g/mol mass of o_2 = moles times molar mass = 1.75 g = 1750 mg oxygen required = 1750 mg/l mol % of oxygen in atmospheric air = 21% Total pressure = 1 atm partial pressure of oxygen = 0.21 times 1 atm = 0.21 atm K Ho_2 = 0.0013 mol/L - atm Solubility of o_2 = k H times partial pressure o and o_2

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