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Re: Dosing calculations...

> From: "Dixon, Steven" <stdixon at bechtel_com>
> Subject: Dosing Calcs KH2PO4
> So I'm trying to work out how to dose my 125 gal. aquarium in the range
> of 0.1 ppm phosphate.
> I figure my aquarium has 110 gals of water in it.  Since a gallon of
> water is 3.785 liters, I should have about 416 liters of water in the
> tank.  Dividing 1000 liters by 416 = 2.4; so 1 g of stuff dissolved in
> my tank with 110 gals of water should yield a concentration of 2.4 ppm.
> Right?


> figured that potassium
> phosphate monobasic is about 70% PO4 by weight.  (95 weight of the
> phospate divided by 136 the weight of the whole molecule = 70%).  So 1 g
> of KH2PO4 should yield 1.68 ppm of phosphate in my tank.  Right?


> Dividing by 20, 0.05 g of the stuff should yield 0.084 ppm phosphate --
> which is fairly close to where I want to start dosing.
	Yes, again.
> So roughly I figure I need to add about 1/2 of the 0.1 g spoon to my
> tank.  How does that sound?  Any comments on the starting dose?

	That sounds pretty good to me.  I think the amounts I'm using are
similar, but I must confess that the precision of the chemistry I use on
my fish tanks leaves a bit to be desired; I tend to say "That looks about
right." and throw it in.

	I would do the calculation:

	110 U.S. gallons is roughly 400 L.
	400 L of water weighs 400 kg
	To get 0.1 ppm, I need 400 kg divided by 10^7
	That is 0.04 g of phosphate
	KH2PO4 is 70% phosphate, so divide 0.04 by 0.7
	I make it about 0.057 g of KH2PO4

	i.e., the same result.  This method is a little more direct, I think,
though in essence it is the same.

Paul Sears        Ottawa, Canada