How much vodka to remove 1ppm po4?

[Cory]

Anyone know how much vodka would be required by bacteria to remove 1ppm po4 on a 100 gallon tank? Or approximately?

 

[Reefltx]

Anyone know how much vodka would be required by bacteria to remove 1ppm po4 on a 100 gallon tank? Or approximately?

Carbon dosing is used mainly for nitrate reduction.

 

[anthonys51]

Carbon dosing removes both nirtrates and phosphates Although it will remove more nirtrates then phosphates. You need both in your tank to make it work. To be honest with you I would just buy a bottle of nopox it comes with simple easy to use instructions and works great. Yes it's a little more expensive than vodka but its not that much money

 

[JimWelsh]

Anyone know how much vodka would be required by bacteria to remove 1ppm po4 on a 100 gallon tank? Or approximately?

I'm gonna hafta make a lot of assumptions for this one. Let's assume that the standard Redfield ratio of 106:16:1 that applies to phytoplankton in the ocean also applies to bacteria in our tanks. Let's also assume that all of the carbon from the vodka makes up bacteria that get removed via skimming. A 100 gallon tank has 378.5 liters of water, so 1 PPM (as mg/L) = 0.3785 grams of PO4. PO4 has a MW of 94.97, so that's 0.3785 / 94.97 = 0.004 moles of PO4, meaning there is also 0.004 moles of P. Using the assumed Redfield ratio, we will need 0.004 * 106 = 0.423 moles of C. Each molecule of ethanol contains two carbon atoms, so that means we need 0.211 moles of ethanol. The MW of ethanol is 46.07, so we need 0.211 * 46.07 = 9.73 grams of pure ethanol. The density of ethanol is 0.789, so we need 9.73 / 0.789 = 12.34 mL of pure ethanol. Assuming 80 proof vodka at 40% ethanol by volume, that means we need 12.34 / 0.4 = 30.84 mL of vodka.

 

[Cory]

I'm gonna hafta make a lot of assumptions for this one. Let's assume that the standard Redfield ratio of 106:16:1 that applies to phytoplankton in the ocean also applies to bacteria in our tanks. Let's also assume that all of the carbon from the vodka makes up bacteria that get removed via skimming. A 100 gallon tank has 378.5 liters of water, so 1 PPM (as mg/L) = 0.3785 grams of PO4. PO4 has a MW of 94.97, so that's 0.3785 / 94.97 = 0.004 moles of PO4, meaning there is also 0.004 moles of P. Using the assumed Redfield ratio, we will need 0.004 * 106 = 0.423 moles of C. Each molecule of ethanol contains two carbon atoms, so that means we need 0.211 moles of ethanol. The MW of ethanol is 46.07, so we need 0.211 * 46.07 = 9.73 grams of pure ethanol. The density of ethanol is 0.789, so we need 9.73 / 0.789 = 12.34 mL of pure ethanol. Assuming 80 proof vodka at 40% ethanol by volume, that means we need 12.34 / 0.4 = 30.84 mL of vodka.

My hero! Thanks Jim!

So its actually not as much as i thought. Good to know

 

[JimWelsh]

@biom previously pointed out a paper that discusses the ratio for bacteria here. In the thread, he cites the ratio for bacterioplankton to be 32:6:1, but having read the paper he links, I see that that is an average, although they observed a wide range of C : P ratios, from a low of 25:1 to a high of 70:1 for bacteria that were not P limited. At any rate, all of this means that the "Redfield ratio" assumption above is likely incorrect, and you would actually need less vodka than what I calculated above.

 

[Randy Holmes-Farley Verified Community Expert]

I think there's no way to know since some of the vodka will be used for denitrification which consumes it and nitrate, but no phosphate. Without knowing how much is consumed in what process (which will vary by aquarium), no calculation will be accurate.