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Topic: Ocean Acidification  (Read 3631 times)

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Offline mrusnak

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Ocean Acidification
« on: September 30, 2008, 03:30:26 PM »
Given that the volume of the ocean is 1.5x1018 m3, net deposit of CO2 in the ocean is 1.8144x1015g/yr, the hydration equilibrium constant of H2O + CO2  :rarrow: H2CO3 is KH=1.70x10-3, and the acid dissociation constant of carbonic acid is Ka=4.30x10-7. Calculate the pH after one year.

My plan to tackle this problem is to determine the moles of water and carbon dioxide, then set up equilibrium extent of reaction (E) equation for each substance. This gives me:

ne,CO2=4.123x1013-E
ne,H2O=8.333x1022-E
ne,H2CO3=E

My question at this point is how do I account for 2:1 mole ratio of reactants and products? I need the total moles to find the partials, yi, and I need the partials to determine the equilibrium conocentration of carbonic acid.

Offline enahs

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Re: Ocean Acidification
« Reply #1 on: October 01, 2008, 10:19:29 PM »
How many mols of CO2 do you have?
How many mols are turned into carbonic acid?
So if you have that many mols of carbonic acid, how many acidic hydrogens are there? In how many liters?


Note, this problem is flawed.



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