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Re: [APD] hydrogen peroxide dosage

Stuart Halliday wrote:
> Sorry Jerry, I don't see where you got the 60mg from.

I'm prone to typos because I try to respond too fast instead of taking 
my time. I keep typing "mg" when I mean "g". 1L of water is 1,000g, 
therefor 0.06 * 1,000 = 60g. Why it is that a w/v solution is calculated 
that way I'm not too sure. It's somewhat counterintuitive.

So, again with the correct units:

We have a 6% H2O2 solution = 60g of H2O2 per liter. We know that 60g 
will occupy 41.1mL of space. That leaves 1,000mL - 41.1mL = 958.9mL that 
will be filled by water. Since water is 1g/mL, we know that it will be 
958.9g. The H2O2 is 60g, so 958.9g + 60g = 1018.9g/L for 6% H2O2.

Now that we're there, we probably want to know how much to add to our 
tank to achieve 10 or 15 mg/L H2O2. In the case of 6% H2O2 w/v, that 
would be as follows:

I have a 125 US gallon tank (473L). That means I want 473L * 15mg/L = 
7,095mg of H2O2, or 7.095g. How much of a 6% H2O2 solution do I need to 
add in order to have 7.095g? We know there are 60g/L, so we just divide 
(60g/L)/7.095g = 0.118L or 118mL (about 8 tablespoons).

Does that make more sense?
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