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Dalmatia

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Hi people's and good morning. Maybe if Randy Holmes can help with this
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You're welcome.

Happy Reefing. :)
Hey Randy. Long time listener, first time caller.

I'm said friend trying to work this problem out. I love your calculator, been using it for a while. But I'm trying to take a more knowledgeable/scientific approach into understanding additives and their effects. Essentially the 3 I'm having trouble with are kalkwasser, baking soda, washing soda or anything with a hydroxide. How is the Alkalinity/Ph change calculated?

I don't have a chemistry background, need a 'idiots guide'.
 
Hey Randy. Long time listener, first time caller.

I'm said friend trying to work this problem out. I love your calculator, been using it for a while. But I'm trying to take a more knowledgeable/scientific approach into understanding additives and their effects. Essentially the 3 I'm having trouble with are kalkwasser, baking soda, washing soda or anything with a hydroxide. How is the Alkalinity/Ph change calculated?

I don't have a chemistry background, need a 'idiots guide'.

Sure. But it will have to get chemically. :D

Generally, the first step is to convert from weight of something (such as hydroxide, OH-) to a measure of the number of actual molecules present.

We would do that knowing the molecular weight of the molecule or ion. The molecular weight of oxygen is 16 grams per mole, and H is 1 gram per mole. So OH- is 17 grams per mole (16 + 1).

Mole is a chemist way to count molecules since the numbers of molecules in any sample is massive, like 10^22 molecules, and one mole is 6.02 x 10^23 molecules.

So if you had 100 grams of OH-, that is 100 grams divided by 17 gram/mole = 5.88 moles of OH-.

Each OH- contributes one unit to alkalinity, so that 100 grams provides 5.88 moles or 5.88 equivalents of alkalinity.

If you put that into 100 liters of water, you added 5.88 equivalents of alkalinity per 100 L = 0.0588 eq/L = 58.8 millieq/L, which happens to equal 165 dKH, since 1 meq/L = 2.8 dKH (just different units of measure).
 
Sure. But it will have to get chemically. :D

Generally, the first step is to convert from weight of something (such as hydroxide, OH-) to a measure of the number of actual molecules present.

We would do that knowing the molecular weight of the molecule or ion. The molecular weight of oxygen is 16 grams per mole, and H is 1 gram per mole. So OH- is 17 grams per mole (16 + 1).

Mole is a chemist way to count molecules since the numbers of molecules in any sample is massive, like 10^22 molecules, and one mole is 6.02 x 10^23 molecules.

So if you had 100 grams of OH-, that is 100 grams divided by 17 gram/mole = 5.88 moles of OH-.

Each OH- contributes one unit to alkalinity, so that 100 grams provides 5.88 moles or 5.88 equivalents of alkalinity.

If you put that into 100 liters of water, you added 5.88 equivalents of alkalinity per 100 L = 0.0588 eq/L = 58.8 millieq/L, which happens to equal 165 dKH, since 1 meq/L = 2.8 dKH (just different units of measure).

Wow. That makes perfect sense. Amazing thank you.

Does the left over carbon play any role? Or does that act as an organic carbon source?
 
Kalkwasser provides calcium and hydroxide:

Ca(OH)2 ---> Ca++ + 2OH-

The calcium ions in the solution obviously supply calcium to the aquarium, and the hydroxide ions supply alkalinity. Hydroxide itself provides alkalinity (both by definition and as measured with an alkalinity test), but corals consume alkalinity as bicarbonate, not hydroxide. Fortunately, when limewater is used in a reef aquarium, it quickly combines with atmospheric and dissolved carbon dioxide and bicarbonate to form bicarbonate and carbonate:

OH- + CO2 --> HCO3-

OH- + HCO3- --> CO3-- + H2O


This has more:
What Your Grandmother Never Told You About Lime by Randy Holmes-Farley - Reefkeeping.com
http://reefkeeping.com/issues/2005-01/rhf/index.htm
 

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