Phosphate build up

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Good reply Collin, allow me to play devils advocate.
I believe a refugium/with mud is an extremely important part of a good reef setup
See I am on the line of thinking that a refugium with mud is yet another biological house of cards, just waiting for something to go wrong. Also with the use of calurpas you are introducing a whole new set of problems to the equation. With the use of mud you are also introducing a virtual stew of metals.
If we never did a water change and there was no algea or skimmers to take them out...they would build up forever
See water changes really are not going to do very much for you if you have a sediment system. The entry point for folks that have dsb's are mainly from the sand itself as I discribed it above. Also the vast majority of P introduced through food and waste will not be helped by WC's. Where does the waste go?? where does the left over food go?? It goes to the sand bed and is taken up by the creatures that inhabit it. Once thier you can change water til you are blue in the face and you will not touch the sink you have created, unless you are changing the water in the bed.
Well, phosphate is used in several vitally important aspects of cellular metabolism and growth.
yep dead on, You end up getting a build up or bloom of bacteria and the byproducts they use to respire and reduce. Once the immediate food source is used up the bacteria naturally begin to die off, this die off allows all the nutrients and others to be released back into the bed. at this point cyano is usualy johnny on the spot and takes up the available nutrients and begins to bloom. Once again the food source will run out and the bacteria portion of the cycle begins again. As more and more food/waste enters the game the blooms get larger and larger. Water changes wont do anything to stop this. ALso you must add up the PO4 that is being released by bacterial action deep with in the bed, as this perculates up (through the migration of your sand bed critters) you add even more to the game.
My tank for instance, has no algea that is not well controlled by my cleanup crew.
Clean up crews dont really control algae. They eat it and then poop out 90% of what just went in, Now couple that with the algaes natural defence mechanisms to spore when attacked and you end up with another cycle. ALso grazers never completely remove thier food source. now that poop once again hits the bed and adds even more to the sink problem.
Algea will use the widest variety of phosphate sources, whereas skimming will only take out the surface active ones.
Algae has weak cell structures and are very leaky. They uptake nutrients and P but they leak most back out, you still end up with net gain but introduce a toxic source you have no control over.
On the skimmer it depends on how you run your skimmer. I personally run mine very wet and pull out alot of solids, in doing that you export all forms of P.
So I will put forth my belief that a mud is a much more favorable sink for other types of unwanted toxins such as metals, poisonous organics etc
See I always get lost with this kind of stuff. Why would we need to sink it?? Everything we are trying to get rid of is associated with waste and left over food. Why would we want to sink it hen we can just remove it completly and skip the whole Aquarium Composting completely. When you remove the detritus you remove the Inorganic associated with the food, the organic P associated with the bacteria that are surrounding and reducing the food/waste. Will you remove it all, no but darn close and that is what LR is for (except LR doesnt sink it, it sheds it)

anyway just devils advocate :D


Mike
 
You all are out of control...I'm still trying to absorb about 90% of this... :lol:

Ok...on phosphate buildup in the DSB....

At what point does the DSB start to re-release phosphate back into the tank? Is this an ongoing process or does the sand bed have to reach a phosphate "critical mass" before this happens?

And if phosphate is being re-released back into the tank, wouldn't water changes still help combat the problem?

MikeS
 
Ok, devils advocate is good. and also to NaH2O's I will get to the reply later. My day is pretty packed. However, I wan't to caution everyone. We are focusing right now on phosphates. In my mind, this is just an aspect of the the total system. I am trying to also take into account, metals and organic toxins. All three follow similar rules but behave in different ways. We have to be careful not to overgeneralize here and/or limit our scope to any one thing. Before going to the meat of NaH2O's and Mikes comments, I am going to explain my model a little more in more lay terms. This will also address part of NaH2O's question. I wan't people who are interested and don't understand the model to ask questions for clarification. I don't know what people do or don't understand so its hard to target the discussion for an audience. I think it is important for people to understand this from a variety of perspectives. Mainly because chemical equilibrium and rate kinetics are central to all areas of chemistry and are extremely important for an aquarium as well. These are not myths, but are hard established facts, without which the world as we know it would not work. All biochemical systems and chemical processes follow these rules and they will never be violated. We will all be better off to add these concepts to our repertiour (sp?). Problem is our aquaria are very complicated so it is hard to see the forest through the trees. I also want to ask everyone not the think or speak in absolutes. Nothing is absolute. It is never appropriate in chemistry to say something doesn't happen. It is appropriate to say it happens to a miniscule amount or the effect is negligable. Only be thinking in this way can we understand the how the relative rates of various processes can make one thing more or less important than another.

Also, just to say that I am not here to argue the merits of DSB's or Mud/calcerous substrates or skimmered vs. skimmerless. Again, many ways to skin a cat. All ways work and each has strengths and weaknesses. I personally don't care how a person runs thier aquarium. However, I would like for us all to better understand what we are doing and why. Anyway...off my soapbox. I will get to replies ASAP...Sincerely...Collin
 
Hello Mike/Mojo, and all. Great discussion here. I barely grasped it!

Just my experience. When I started I used the jaubert/plenum method and from my understanding it uses a space (plenum) as a sink (not same way Collin used the word. similar) for various metals and toxins binding with water. If I follow the plenum idea, that area rapidly becomes a toxic zone and when matured will provide the same buffering effect as a DSB --- except hydrogen sulfide will be abundant. During that time the plenum I used was remote and the plenum space was adjacent to a settling chamber and a live rock chamber. The other two became filled with sponges, pods and worms. The plenum space was immaculately clean --- devoid of visible life after a year! I assume the hydrogen n sulfide layer also slowly dissolves the aragonite layers. Additionally, while it is right to assume that my phosphate level should have been raised considerably by then, the opposite occured --- no hair algae! And I was skimming like a teaspoon a week. After 14 months and a near crash from using (back then) calcium chloride/buffer and chelated calcium for calcium supplementation, i moved to kalwasser. It was still ok for about four months. Then someone said vinegar was good to raise ph in top off water so i added white vinegar (wrong move! it was rich in phosphate) and in just a month hair algae started showing, in two months it got worst. I stopped but the algae kept growing. Just about then I moved to a new tank with DSB in mind and no more plenum (people said it was useless so I believed them). The hair algae just grew more then. So my troubles with hair algae was in full mode. Short to say, I am battling and winning the algae war but only with too much sacrifices. I replaced my skimmer with expensive venturi (the older was plain CC woodstone) and skimmed aggressively, a cup to a cup in a half weekly. Water change from once a month 20% to Twice a month (maybe even 3x). And scrub scrub scrub. It's tiring.

I will move to a bigger tank next year. With the DSB problems, I might add a remote plenum again and if it doesnt work simply disconnect and turn it into an algae refugium.
 
Also, just to say that I am not here to argue the merits of DSB's or Mud/calcerous substrates or skimmered vs. skimmerless. Again, many ways to skin a cat. All ways work and each has strengths and weaknesses.
Absolutely Collin, I am nt here to say one is bad and the other is good either. Each system will have an upside and a downside. What I try to encouorage here at RF is a good understanding of the system (whatever system it may be) and then a good healthy conversation showing the ups and downs, this way folks can make a good educated choice on their own.

HIya Riple
When I started I used the jaubert/plenum method and from my understanding it uses a space (plenum) as a sink (not same way Collin used the word. similar) for various metals and toxins binding with water
A plenum works the same way as a DSB or any simular sand substraight will my friend. The concept is to set up a filter that creates differing enviroments for the growth of different species of bacteria. These bacteria will reduced nitrogen based products and put them in to cycles. The plenum zone under the sand is thier to create an anaerobic (very low in oxygen) In this eviroment you will get a population of denitrifing bacteria, which can reduce nitrate to gasious forms. Same concept in a dsb. The toxin you are refering to is problibly sulfides. In areas that are deviod of oxygen you will get a bacteria called SRB's (sulfide reducing bacteria) the byproduct of these bacteria is sulfide gas which is harmfull to most animal life if they are subjected to it.
The plenum space was immaculately clean --- devoid of visible life after a year! I assume the hydrogen n sulfide layer also slowly dissolves the aragonite layers.
Yes I would imagine it would be very devoid of life, bacteria is the key, thats really the only thing that can live thier. The sulfides dont desolve the argonite Rieple its the bacteria and the low PH of the area. Bacteria can not just go eat something, they must turn it into a liquid first, they use enzymes and acids to do this. This action is what melts the argonite and lowers the PH.
Additionally, while it is right to assume that my phosphate level should have been raised considerably by then, the opposite occured --- no hair algae! And I was skimming like a teaspoon a week.
No not really my friend, if I remember your Plenum was very deep (12 to 15 inches???)It takes awhile for the lower regions of the bed to go true anaerobic, and it takes a good population of sanbed critters(worms and such) to move the food down and the gasses up. In a bed that thick I would say you had a couple years before you would begin to see some effectes of it. But look at what you say on this statement, You sank the vast majority of your nutrients and only exported a teaspoon a week.
Short to say, I am battling and winning the algae war but only with too much sacrifices.
Ahhh thats tough luck on the algae problems and its good to know your winning now. P and N nutrients are one of the biggest problems hobists face. Thier are so many things we add that contain them we really have to be careful of input and have good export mechanisms to stay on top of them. A sand subrtaight is a sink and can be used succesfully as one in a filtration concept, one just has to be willing to replace it once it is done.
On your plenum, it wasnt truely a J plenum, a J Plenum uses large particles (bigger then gravel) and is only about 5 inches in total. The larger particles allow for much better diffusion of water to the lower regions (concept was to be not so relients on bugs for migration).

Mike
 
Hiya Mike
At what point does the DSB start to re-release phosphate back into the tank? Is this an ongoing process or does the sand bed have to reach a phosphate "critical mass" before this happens?
Hmm thats a good question. Thier are so many variables Mike its going to be different in every tank. It is a matter of input and export. It is basically an ever growing cycle, Animal(bacteria) to vegitable(algae). You have to look at it like this, the bed is not an export system, it is a biochemical cycling center that can facilitate export through algae harvest (Dr. Rons words). so if you can interupt the cycle you can break it and reduce it, it just becmes a matter of if you want to harvest algae form the bed or not.
And if phosphate is being re-released back into the tank, wouldn't water changes still help combat the problem?
Ok I think this is where Collin was giving me hell for say absolutes in stead of others :D .
Yes you might be able to export a miniscule ammount that way, the problem is that cyanobactor and algaes are very good at quickly binding up and using P. So they begin to get it even before it reaches the top of the bed, so unless you can Wc some of the pore water in the bed your really not going to get it. What usually begins to happen then is that folks loose thier sink and then algae begins to form everywhere else.

Here is a good little visual link for P uptake and release by argonite, its a study in florida bay, the worlds largest DSB, hehe

http://www.rsmas.miami.edu/groups/jmc/fla-bay/FBay-WebPage.htm

Mike
 
Hey, that is a great link Mike. Basically supports everything I have been saying. Binding coefficient onto aragonite is higher than I expected. Makes it fall into regime 2 of my model, not regime 1. Changes nothing though...C
 
Ok, as I said earlier, the prime point of interest seems to be phosphorous accumulation in the bio-sink. I think mike's points are largely accurate and I will address those later. However, there are some technical points I would like to clarify regarding his analysis and then we can elaborate more.

In order to do this though, I want to go back and talk about the simple equilibration model I developed earlier and address some of Nikki's questions. Nikki Stated:

"I am having a difficult time understanding how phosphates can get to equilibrium when it is a constant input? Maybe I misunderstood the model, and it will make sense after my brain finishes processing."

This is exactly what we need to understand. Chemical equilibrium coupled with relative rates are fundamental to all chemical and biochemical process and are abundant in our aquaria. I am going to use a simpler example to elaborate these points. I urge anyone interested in this thread to study the plots I made and notice the differences in the 3 regimes I outlined.

Now for the example. Let's pretend our aquaria are a nice glass flask instead. Into the flask we will put 100 milliliters (ml) of water to make a nice round number. Also we put 10 grams of limestone, which is an adsorbant for many things. Now we could just as easily use gallons and teaspoons but the numbers don't work out as nicely. Ok so we've got 10 grams limestone in 100 ml of DI water. Into this we will add 1 gram of completely water soluble orthophosphate. Lets pretend the binding constant of orthophosphate is 0.25 (or 25%). What does this mean? It means that the ratio of dissolved orthophospate (which I will call Pi from now on) in the water, to the Pi bound or adsorbed onto the limestone is 0.25. So, lets do the math. We put in 1 gram Pi into 100 ml and some of it sticks to the lime. We can calculate the concentrations of Pi in the lime and in the water now. We know that:
1.) Pi(water)+Pi(lime)=1 gram and
2.) Pi(lime)/Pi(water)=0.25

This is two equations in two unknows and that I will solve by re-arranging 2.) and substituting into 1):

3.) Pi(lime)=0.25*Pi(water)
4.) 0.25*Pi(water)+Pi(water)=1 gram

and re-arranging again we get

5.) (1+0.25)*Pi(water)=1 or

6.) Pi(water)=1/(1+0.25)=0.8 grams. Then using the other equation:

7.) Pi(lime)=1-Pi(water)=0.2 grams.

So, now we have it. Don't get hung up on the math. Important thing is the have 0.8 in the water and 0.2 in the sand. Note that 0.2/0.8=0.25, which is our binding constant. If we add one gram of Pi into 100 ml of water with 10 ml lime, that gram of Pi will partion itself into both the water and the lime according to its binding constant. Furthermore, I showed how we can use this value to calculate exactly the mass in each substance. So let's do this again. As long as the lime isn't close to being saturated, this is OK. We add another gram and 0.2 goes into the lime and 0.8 goes into the water. So now we have total of 0.4 grams in the lime and 1.6 grams in the water. Please note that 0.4/1.6 still equals 0.25. So we can keep doing this and the amount in the lime and the amount in the water will keep growing but the ratio will always remain 0.25. We can do this until either the sand is saturated or the water is saturated and solid Pi starts to drop out.

Now, lets go back to right after we added the first gram. We have 0.8 gm Pi in the water and 0.2 gm Pi in the sand. Now lets explore what happens when we do a water change. In the water we have 0.8 gms in 100ml of water. The concentration in the water is therefore 0.8gm/100ml=0.008 gm/ml (roughly 8 ppm). Now lets take out 10 ml with a turkey baster and replace it with fresh pure DI water. How much Pi do we remove? Well if we take out 10ml of 0.008 gm/ml Pi solution we remove 10*0.008=0.08 grams of Pi. So now how much Pi do we have in solution. We have 0.8-0.08=0.72 grams. How much do we have in the sand? Still 0.2 grams. 0.2/0.72=0.28. That is higher than the binding coefficient. The system is no longer at equilibrium. Pi will have to leave the sand bed and go into the water to re-establish equilibrium. I won't bother to calculate the exact numbers again but suffice it to say that Pi will move from the sand to the water until the ratio is again 0.25. What does this tell us. It tells us that if we don't add any more Pi and keep doing 10% water changes, eventually all the Pi will be sucked out of the bed until the concentration drops infinitely close to zero. Lets just do some more numbers and forget the sand. Lets just put 1 gm Pi into 100 ml water. Concentration = 1/100=0.01. Now I will calculate the amount removed from the water over 5 succesive water changes. Remember, we still aren't adding an phosphate to the system before water changes.

0.01-10%*0.01=0.01*(1-0.1)=0.01*0.9=.009= 9 ppm (chg 1)
9*(0.9)= 8.1 ppm (chg 2)
8.1*(0.9)=7.3 ppm (chg 3)
7.3*0.9=6.6 ppm (chg 4)
6.6*0.9=5.9 (chg5)

Ok, here is what I want you to see. Notice how the amount of Pi removed in a 10% water change depends upon how concentrated the starting water is. When the starting concentration is high, we remove more. When the starting concentration is low, we remove less. Now, we have the tools to understand why the curves in the model lay over and reach equilibrium. When we first start an aquarium, there is very little Pi present in the water. Thus when we do a water change, we remove very little. However, between water changes, we keep adding Pi in our make up water at a fixed rate. As we are adding Pi more and more the Pi builds up in the water. At some point in the distant future the water gets enough Pi in it that our 10% water change removes Pi at exactly the same rate we are putting it in. At this point the input and export rates are the same and the Pi will stay constant as long as nothing else changes.

Important points from above. Equilibrium is ALWAYS established. Pi bound to sand can move out of the sand to achieve equilibrium with its binding constant. Pi (or any other soluble substance) will build up until it is concentrated enough that its export rate is equal to its input rate. This may help to shed light on Nikki's other statement:

"On the point of the remote sandbed/mud bed. I could see this being an effective sink, but only if the detritus and wastes make their way into this area."

Because Pi will move into and out of the sand bed from the water to establish equilibrium, this gives it the ability to move from the primary tank to the refugium if the system is moved away from equilibrium for any reason or vice-versa.

Next...more on the bio-sink partioning.

Anyway, enough for tonight...C
 
ok, so we set up our flask and added one gram of limestone that is saturated with Pi. then we add 100ml of RO/DI. therefore the lime will give up .75 of its Pi to reach EQ. why does tank water test neg for P when sandbed water test higher?
 
Several reasons. Remember it depends upon the binding constant. If the binding constant is 0.999999999999 then the same rules apply but the sand bed (or bacterial bio-sink) will hold nearly all the phosphate. In my example the binding constant was 0.25. This number was just chosen to make the example easier to imaging without tons of zeros or 9's. A bio-sink will have a very strong binding constant so at any time, most of the available phosphorous will lie in the sink...not sand but the biologics in reality...C
 
The system is no longer at equilibrium. Pi will have to leave the sand bed and go into the water to re-establish equilibrium.
Can you explain why & how this happens? I'm trying to understand why it has to reach equilibrium, & what makes it dissolve from the sand or whatever, to flow freely into the water?
 
First let me re-say what I said to Witfull. I made a mistake saying binding constant of 0.99999, that would be near one indicating 1/2 in sand and 1/2 in water. I meant of very large binding constant like 100 or 1000. this would put very much more in the sand and very little in the water. Phosphate basically behaves in the class 3 regime. If you look at the plot you can see that the concentration in the water stays very low but rises nearly constantly in the sandbed...C
 
Scooterman,

The simple answer is that because equilibrium is a fundamental law of nature. Everthing is in equilibrium all the time. Think about a box on a shelf. It is in equilibrium. The force of the boxes weight is exactly balanced by the force of the table. If equilibrium is interupeted (the shelf breaks) a new equilibrium with the floor will be established.

It a sand bed with a chemical it is more difficult to understand. The answer is because the molecules are always moving. Millions of times per second per molecue. In our previous example using a binding constant of 0.25. From this it is correct to say that on average a molecule will spend 20% of its time and the sand and 80% of its time in the water. However, it will switch back and forth millions of times/second. This is why it is called "dynamic" equilibrium. Thus at any given time billions and molecules are shuttling back and forth between the sand and the water. However, on average at any one time 20% are on the sand and 80% are in the water. This is how the molecules will move and or re-establish an equilibrium, because they are moving all the time. Equilibrium is strictly defined as the rate at which molecules leaving the surface of the subrstrae is exactly equal to the rate at which they are binding to the surface...hope this helps...More later. I have a plan. Stay tuned and we will get there eventually...Sincerely...Collin
 
Collin - thank you for taking the time to explain your model - you are making me think....great discussion - I love this place. Your explaination is helping me get a grasp. I had to clear some cobwebs out of the chemistry section of my brain. I believe my difficulty has to do with biology side vs. chemistry side of all of this....and perhaps was my difficulty with chemistry (although not so much organic or biochem) in college. I need to look at it as a whole. Your explaination makes sense from the model perspective for equilibrium. How soon in a system is equilibrium reached - maybe it is a matter of things always trying to "attain" an equilibrium? Here is where my biology side comes into play. DOPs and POPs are a major factor with phosphate issues, as well.

More later. I have a plan. Stay tuned and we will get there eventually

Looking forward to more....
 
So if you have no sand (just rock and water) and you are introducing P via food (or other sources) and removing (hopefully) P via skimming how do you determine the EQ point?

If I understand a closed system correctly, when I introduce P into the system something else has to leave the system to maintain the system equilibrium (much like the see-saw effect of calcium and alk). Is this a fall out of the binding process with rock (or sand or mud, etc.)?

I'm so confused and my head hurts.

Thanks for bringing it down to a simpler level Collin. Us little brains need to catch up ;)
 
Reed - even if you have only rock and water. The die off of in rock of various organisms and bacterial turgor present will be an input to P....all this without introducing food. My tank is a prime example - new set-up with tons of uncured live rock, RO/DI water.....yet my phosphates were off the chart.
 
what we're trying to get to here is that once you get your system to the equilibrium point it can be mainntained at that point indefinitely. If you keep adding to the system without "true" export the amount of phosphate will in crease in total. The ratio will still be the same in the sand and in the water. It would seem to me though that the deeper parts of a truly deep sand bed would be removed from the model due to the lack of circulation. the problem area is where the aerobic portions of the sand bed meet the anaerobic portions. The nutrients candrop to the bottom of the sand bed until the hit glass(plastic.) At this point the accumulation starts to overtake the ablity to export the nutrients from the system. When this reaches the surface of the bed, it becomes available to the light loving algaes. The total amount of phosphate increases in the same ratio of sand/water column.
The important thing to rember is that the amount of phosphate can be reduced through "true export" in the sense that what goes in must come out. If you take it out faster than you put it in the total amount will go down. This will make it scarce in the system allowing for competition to take effect.
 
No, I get that Nikki.

I'm saying if you have a bare bottom tank (no sand or mud as a nutrient sink and fully cured rock in a mature tank), then your rock would act as a sink (I think).

The main thing I am trying to grasp is that if you input P something has to come out. What comes out and in what form? I thought there would need to be a chemical process that would happen creating a byproduct of some kind (or a binding event that produces a byproduct).

With Ca and Alk if you increase one too much the other precipitates out of solution until you reach equilibrium again. I expect something similar with P.
 

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