I'm fairly convinced according to previous testimony in this thread that the solution is pH buffered.
I've often wondered at this, but as I'm not familiar with hydro it hasn't even been that necessary for me to test it.
Regardless, I've found this website:
http://scienceinhydroponics.com/201...oponics-what-is-the-best-cheapest-buffer.html
This article covers an experiment that was done analyzing the behavior of several non-phytotoxic buffers.
It makes reference to a difficult chemical equation. Well, it turns out it's really not all that difficult if you boil it down a bit.
The equation in question:
pH = pKa + log( [A-] / [HA] )
to decode this for anyone who hasn't taken a chemistry course here are a few snippets:
p is a prefix we attach to numbers or constants which refers to the "negative log" of that quantity. IE pH = -log [H+]
The Brackets [ ] contain a concentration which is usually in M, or molarity--but can be in any units so long as they are the same, because it is a proportionality. So the pH thus denotes the negative log of the H+ ion concentration. These ions are what dissociate from an acid, to decrease pH. Strong acids, such as HCl, dissociate completely in water and become H+ and Cl- entirely. Weak acids like acetic acid only partially dissociate.
The dissociation constants are determined experimentally and given a value, Ka.
Taking the negative log of Ka gives pKa in the above equation. A buffer is only effective at pH values + or - 1 from its pKa. So a buffer with a pKa of 4 will be a good buffer between 5 and 6, with the best performance closer to 4.
To create a buffered solution, you add equimolar (same number of molecules) amounts of acetic acid and sodium acetate (when acetic acid dissociates it becomes H+ and Acetate, so you are effectively adding the conjugate base of the weak acid).
This creates a situation wherein if an H+ ion is created, an acetate ion will react with it to form acetic acid--and conversely if a hydroxide (OH-) ion is produced, the acid will react with it to form acetate.
At a concentration of 1:1 the buffer is very strong, at a concentration lower 10:1 in either direction the buffer is basically shot. This type of solution strongly resists change in pH--however it is important to figure out how much acid/base you plan to be buffering AGAINST.
If you have .1mol of both the acid and its conjugate base present, yet the plant will produce 1.0mol of acid or base--the buffer will deplete easily. So it will be, for anyone attempting to do this themselves, important to figure out what the right concentration to start with is.
For a little more chemistry fun I'll show you how Ka is derived.
Weak acids dissociate according to the following reaction scheme.
CH3COOH (acetic acid) + H2O -----(Ka)----> H3O+ + CH3COO- (acetate ion, conj base of acetic acid.) H3O is usually shortened to H+, while in reality it maintains its hydronium structure.
Before I show you the equation, I want you to understand why I wrote Ka in the middle of the arrow. Really this arrow should actually be two arrows stacked atop one another, one pointing each way. This denotes an equilibrium. Most chemical reactions proceed according to an equilibrium. Ka is a constant which denotes to what degree the dissociation of a weak acid goes to completion
at equilibrium. There is a similar constant applied to most other types of reactions which is called Keq.
Very quickly I will suggest to you that these equilbria follow something called Le Chatlier's principle, which basically says if the above reaction were to be at equilibrium (the acid dissociated in whatever ratio it is at eq) and you add more acetate ion, the equilibrium of the reaction will shift left such that more acetic acid is produced. The same thing will happen if you were to take away some of the acetic acid--more will be produced to restore equilbrium. Likewise, if you take away acetate, equilibrium will shift right and some acetic acid will be converted to acetate to replace it. It is because we know this happens that we are able to establish a constant of equilibrium. We can test under various concentrations and discover the constant of proportionality between them.
To find the Ka at eq, or the concentration of an unknown reactant/product, the following formula is used:
Ka= [H3O] x [CH3COO] / [CH3COOH] [H2O]
Notice this is concentrations of products divided by concentrations of reactants. This form is conserved for all equilibrium calculations including those involving thermodynamics.
Moving along, we can drop the H2O figure for reasons I won't get into here. I will also refer to the undissociated acid as HA and the conj base as A-, and the H3O as H+
Ka = [H+] X [A-] / [HA]
We can play some cool games with this equation due to rules of logs, which i also won't go over.
Taking the negative log of both sides gives:
pKa = -log( [A-] / [HA] ) + -log( [H+]
Which leads to:
- -log( [H+] ) = -pKa -log( [A-] / [HA] )
This can be simplified to give the first equation:
pH = pKa + log( [A-] / [HA] )
The above equation essentially says that when the ratio of A- to HA equals 1, the pH = pKa. The log of 1 is 0--log asks the question 10 raised to what power equals this number, 10 to the zero power is 1. For instance if your pH is 5.8 then the concentration of free H+ ions is 10 raised to the negative 5.8 power, 10^-5.8= [H+]. Deriving the equation is a bit tricky, but now that we've got it plugging into it is easy.
If you are capable of making ppm/ppb calculations (and maybe figuring out a few molar masses--easy peasy) then you should be capable of playing around with buffers in your hydro system at fairly low cost.
Any questions, get at me.
Here is a good page discussing this topic further, showing the difference between a titration curve for a weak acid and a buffer composed of weak acid species:
There is no shortage of other material available out there for anyone to learn about maximizing the potential of buffers. This is a concept which is taught early on in chemistry training and thus there are many online sources intended to help students pick it up.