Kh concentrate math - The Planted Tank Forum
 
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post #1 of 2 (permalink) Old 09-09-2012, 07:11 PM Thread Starter
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Kh concentrate math

I am trying to come up with a kh concentrate like the one nutrifin makes. I. Added 100 grams---sodium bicarbonate. And added it to 1 gallon of ro/di. I then took 5ml of that solution and added it to 1 gallon of ro/di water. I then tested the kh and it tested @1 kh. I did it 5 times with the same results. Does this seem accurate to anyone ?

Thanks Todd Ziegler
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post #2 of 2 (permalink) Old 09-10-2012, 05:14 AM
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Well, the recipe I have worked out is this:

1 teaspoon (dry) added to a 29 gallon tank raises the KH by 2 degrees. I have used this ratio in tanks from 10 gallons to 125, and in larger and smaller amounts to alter the KH more or less. It works. Most recent was: 2 tablespoons in 88 gallons raises the KH by 4 degrees. It did.

It dissolves almost instantly, but if I want to be sure it has fully dissolved before I add it I just mix as much as I want in a cup of water. ('cup' not being the official measure, just a cup I use for aquarium stuff).

2.2 grams / cubic cm (per wiki)

1 cc = .2 teaspoon, or 5 ml = 1 teaspoon (per onlineconversion.com ) (actually: 1 ml = .20288 tsp)

so 1 teaspoon (my measurement) = 11 grams, and that raises 60 gallons by 1 degree.
So, 100 grams (as much as you used) ought to raise 540 gallons by 1 degree.

There are just under 3800 ml in one gallon (3785.4 if you want to get picky), and you took 5 ml. or about .0013 of the gallon.
So you had almost exactly .13 gram of baking soda in that 5 ml (math is getting really sloppy here- probably be good to put it down on paper)

So, if 11 grams raises 60 gallons by 1 degree, then 1/100 of that should raise 1/100 of 60 gallons by 1 degree.
1/100 of 11 grams = .11 grams, and you had a bit more than that, so it should raise the KH of a bit more than 1/100 of 60 gallons by 1 degree.

Yup, for as sloppy as the math is that is the right ballpark.

You know that if you are using API test kits you can test 10 ml of water and each drop = half a degree? Try that...
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