[Prev][Next][Index]
Re: CO2 and KH
Since this thread won't roll over and play dead, I'll drop yet another
opinion into it.
Michael Irlbeck <u7211aa at sunmail_lrz-muenchen.de> says,
I think an experiment is needed. I will try to see what
the CO2 consumption is in my aquarium if I lower the KH from 180 ppm
to
90 ppm and keep the CO2 constant at 18 ppm.
Here's my guess: if you keep the CO2 constant, the rate of CO2 loss to the
air will be constant. Because, other things being equal (note 1), the rate
of loss of CO2 is determined simply by the amount of dissolved CO2 (note 2)
in excess of the amount that's in equilibrium with the little bit of CO2 in
the air.
Assuming that CO2 really *is* the form that plants absorb (note 3), the
rate of absorption by plants will also be constant under the changes you
propose; so the change would have no effect on CO2 usage, to a first
approximation. Here's a possible second-order effect: To keep CO2 constant
while reducing KH, you'll be lowering the pH by an amount that I suppose is
0.3; your plants may find this better or worse than the old pH,, causing a
different rate of CO2 usage and plant growth.
(1) Temperature, atmospheric conditions, water circulation, etc., all being
equal. BTW if you try to make equilibrium calculations based on the amount
of CO2 in the air, remember that the amount in your house is significantly
higher than the amount in the atmosphere. To get good data, you'd have to
measure CO2 in _your_ air. This one point I can claim expertise on, having
made these measurements in an earlier existence.
(2) That is, CO2 that's free in the water, not CO3- or anything else.
(3) The published information on plants and CO2 is so confusing, with the
confusion clearly due to too many of the authors' being confused, that I
can't take this or anything for granted.
As to the nature of the equilibrium, I respectfully dissent from both the
recent statements of it. Whereas George said,
CO2 + H2CO3 <--> HCO3- <--> CO3--
my version makes it even more complicated:
CO2 (in air) <--> CO2 + H2O <--> H2CO3 <--> HCO3- + H+ <--> CO3-- + 2H+
Why bother? Because the two <--> on the left are slow equilibria. The
first one depends on your trickle filter and all that stuff that George has
worked to measure. The second is less obvious; for a demonstration of it,
drip a little blood into your beer. Supposedly (I haven't tried it), you'll
get a
sudden foaming as the carbonic anhydrase in the blood speeds the reaction
and frees the CO2; without that enzyme, you couldn't get rid
of CO2 in the time it takes blood to travel through the lungs.
So here's the real reason (I think) for rebound after you lower pH in an
aquarium by adding HCl or the like. You immediately drive the
H2CO3 <--> HCO3- + H+
reaction to the left, and the H+ concentration reaches some level or other,
which you measure as pH. Then, over the course of time, the new high level
of H2CO3 releases CO2, leaving the CO2 concentration much higher than the
amount that's in equilibrium with the air. And over more time, the CO2
escapes into the air. The lower CO2 level in the water causes more H2CO3
to break up (slowly), and the lowered H2CO3 causes more H+ and HCO3- to
combine: the H+ level goes down, and the pH goes up, on a time scale of
hours or days.
PS: The real problem with hydrochloric acid, IMHO, is that it has no
buffering power at reasonable pH, unlike carbonate and phosphate. So you
can expect the pH to drift away as soon as you've added the stuff.
PPS: As to what's in the pH regulators: Kent pH Control Minus (tm) uses
sodium hydrogen sulfate and sodium sulfate. Too much sulfate isn't good in
aquaria, but I don't know how the amounts used compare to the use of sodium
thiosulfate in dechlorinating; nor do I know offhand at what pH the
sulfate-bisulfate buffer is most effective. Seachem Acid Buffer uses a
"Zwitterionic buffer, pK=6.4. Free of phosphate." If I remember right,
this means that the magic compound can either absorb or release H+ (like
HCO3- or H2PO4-) and its strongest buffering effect is at pH 6.4.
Dan Drake
dandrake at nbn_com