# KH2PO4 Question???



## KevinC (May 24, 2004)

Lycosa said:


> I'm trying to figure out how KH2PO4 disassociates into aqueous solution.
> 
> I've been googling and searching.....can't find the information anywhere without trying to figure out disassociation constants, etc.
> 
> :help:


Well, without the dissociation constants and LeChatelier's Principle, it can't be done . . . HOWEVER:

In solution the K+ ion is immediately freed from the (H2PO4)- ion, so yes, the K is plant-available.

The (H2PO4)- ion will then dissociate based on the existing pH and the equilibrium constant to form (HPO4)-2 ions and H+ ions.

The (HPO4)-2 ions will dissociate based on the pH and the equilibrium constant to form (PO4)-3 ions and H+ ions.

BUT - LeChatelier's principle tells us that if you (the plants) remove (PO4)-3 ions, more (HPO4)-2 and (H2PO4)- ions will dissociate to replace it . . .

So, for calculation of available P (as PO4), just use the total KH2PO4 - as the plants need the P they will get it - whatever form it is in.

Kevin


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## danepatrick (Jul 17, 2006)

to sum it up i guess, i think someone once told me or i've read somewhere that you should get the/enough K that you need from the KH2PO4 and not have to dose K separately.


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## Lycosa (Oct 16, 2006)

Thanks, KevinC, again for the information. You've answered my question completely.




> to sum it up i guess, i think someone once told me or i've read somewhere that you should get the/enough K that you need from the KH2PO4 and not have to dose K separately.


Actually, since the recommended ppm for phosphate is small, (1 - 2 ppm), the available potassium from KH2PO4 is small and not near the recommended dose. I wanted to include the additional K from KH2PO4 into the calculation, however miniscule the amount. If one were dosing excess KH2PO4, since, as KevinC said the K is in usable form, it is a source which should be accounted for in the calculation. I'm basically writing the program in a different format than the available fert calc's that I've seen. More along the lines of choosing target ppms for ferts and having everything else calculate automatically. Anyhow, writing the program has forced me to learn a lot more, and that is never a bad thing!:icon_smil


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## andrewwl (May 3, 2004)

Wow impressive! KevinC, you should write some articles


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## medicineman (Sep 28, 2005)

> to sum it up i guess, i think someone once told me or i've read somewhere that you should get the/enough K that you need from the KH2PO4 and not have to dose K separately


If I'm right that is from EI method of fertilizing in barr report. Tom reported that K from KNO3 and KH2PO4 if given as per EI is of enough level. However I'm still puzzled of how some people purposely increase K level to near or excess 50ppm to get a specific result.

I also find out by trying that using phosporic acid instead of KH2PO4 is also as great, but the downside is that it is in liquid form which is troublesome for some people. (Warning : concentrated phosphoric acid is highly corrosive and should be handled with care. Use only diluted acid if you have not trained)


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## danepatrick (Jul 17, 2006)

medicineman said:


> If I'm right that is from EI method of fertilizing in barr report. Tom reported that K from KNO3 and KH2PO4 if given as per EI is of enough level.


i'm pretty sure this is what i was referring to, or somethin along the lines of it.


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## asiansensation2000 (Nov 18, 2007)

KevinC said:


> Well, without the dissociation constants and LeChatelier's Principle, it can't be done . . . HOWEVER:
> 
> In solution the K+ ion is immediately freed from the (H2PO4)- ion, so yes, the K is plant-available.
> 
> ...


Holy crap, I should have paid attention in my chemistry class! I'm lost but learning.


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## JenThePlantGeek (Mar 27, 2006)

You can get into the chemistry of it all if you like, but it can be distilled and made much simpler for those of us without degrees in chemistry!

Try reading this - http://www.rexgrigg.com/


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## Ishar (Oct 30, 2007)

I know this thread is old, but I am going to ask for clarification here rather than start a new one.

So if I were making an excel calculator for the amounts of K and P from a KH2PO4 solution, the amount of K would be whatever K is in the solution, straight up, and the amount of P would be the amount of P in the solution? 

I mean all of this in mass- using the molecular masses and finding the percent of P and K in the solution I could find the mass of K and P I am adding to the aquarium, which allows me to get the ppm added to the tank right?


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## SuRje1976 (Feb 2, 2006)

Ishar said:


> I mean all of this in mass- using the molecular masses and finding the percent of P and K in the solution I could find the mass of K and P I am adding to the aquarium, which allows me to get the ppm added to the tank right?


Yep...that'll do it!


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## Ishar (Oct 30, 2007)

awesome- thanks for the reply


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