Post your DIY additives here!
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I use 0.0002gr sodium molybdate per 1 meq/l you have 1.88 meq/l (53gr = 1) so 0.00037gr sodium molybdate per liter. But this is an educated guess based on very little information and according to what I could find, not related to kH/Ca consumption.
beaslbob
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For the improved diy 2 part of dr holmes-farley I use:
Calcium chloride, local concrete redimix companies. ~$20 for a 50 pounds. the anahydrous was 40 pounds for ~$30.
sodium bicarbonate and Epsom salts. drug and grocery stores.
magnesium chloride I get from a local industrial chemical supplier (Brentagg?) For ~$25 for a 50 pound bag. they do have to ship it in from another warehouse but don't charge shipping.
this stuff is so cheap I give it away at local club meetings. Are you near Huntsville, al? LOL
my .02
Calcium chloride, local concrete redimix companies. ~$20 for a 50 pounds. the anahydrous was 40 pounds for ~$30.
sodium bicarbonate and Epsom salts. drug and grocery stores.
magnesium chloride I get from a local industrial chemical supplier (Brentagg?) For ~$25 for a 50 pound bag. they do have to ship it in from another warehouse but don't charge shipping.
this stuff is so cheap I give it away at local club meetings. Are you near Huntsville, al? LOL
my .02
Kind of off topic but with this kind of stuff it could help to be able to double check calculations you get from people. There are a lot of great resources on stoichiometry that are worth checking out. Here's one (Khan Academy) https://www.khanacademy.org/science...toichiome/stoichiometry-ideal/v/stoichiometry
thanks for the share..csb123
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I found a couple of elemental salts on eBay and need help formulating stock solutions.
I need to know how much to add to a liter of water for the stock solution, and how many mL to add to increase X amount per 100 liter of tank water.
1. Vanadium pentoxide 99.9 % pure
2. Sodium bromide 99.5+% pure
I need to know how much to add to a liter of water for the stock solution, and how many mL to add to increase X amount per 100 liter of tank water.
1. Vanadium pentoxide 99.9 % pure
2. Sodium bromide 99.5+% pure
Randy Holmes-Farley
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Vanadium pentoxide might not be best, but might work. I'd look for a vanadate salt.
Sodium bromide is 77.7% bromide.
50 grams in 1 L ---> 38,850 mg/L = 38.85 mg/mL
Add 10 mL (389 mg Br-) to 100 L of tank volume and bromide rises by 3.9 mg/l.
Sodium bromide is 77.7% bromide.
50 grams in 1 L ---> 38,850 mg/L = 38.85 mg/mL
Add 10 mL (389 mg Br-) to 100 L of tank volume and bromide rises by 3.9 mg/l.
Last edited:
zachtos
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/// what I found so far ///
Manganese(II) chloride tetrahydrate is 28% manganese by weight.
Dissolve 1 gram in 100o mL (grams) fresh water. Manganese = 280 ppm.
Take 1 mL (1 g) of that mix and dissolve in 1000 mL of fresh water. Mn = 0.28 ppm (=280 ug/L).
Add 1 mL of that to 100 mL of fresh water. Mn = 2.8 ug/L
Add 13.5 mL of that to 100 gallons of tank water. Conc boost to tank = 0.1 ug/L Mn
Sodium Borate Decahydrate has a molecular weight of 381.4 g/mole and boron is 10.8 g/mole, so it is 2.8% boron by weight.
To get 5000 ppm boron in a liter, add 5,000 mg/L / 0.028 = 177,000 mg/L.
1ml of that solution will raise the boron 1mg/l in 100 liters of tank water.
With the molecular weight of 381 i guess it would be 2 Sodium and 4 Boron in the molecule.
It would then contain 11,3% Boron.
I used 113 grams Borax ( di sodium tetra boron.... decahydrate) which i filled up with osmosis water to 1000 grams to make mine stock solution..
This is my DIY iron dose using Fergon tablets from a drug store:
Take 1 tablet and dissolve in 100 mL RO/DI (overnight soak). Let the solids settle out and use the liquid. Then I dose about 1-5 mL to my 200 gallons (dose is not critical) occasionally.
One tablet is 27 mg iron, so that dose comes to about 0.3 to 1.4 mg.
Put into 200 gallons, that's about 0.4 to 1.8 ppb.
solution for sodium molybdate. I need 0.0002 gr per liter kH stock. So I added 0.2 gr sodium molybdate to a liter of RODI and added 1 ml of this solution to 1 liter kH stock solution. The concentration should be around 0.2 ppm = 200ppb sodium molybdate (= 0.1ppm/200ppb molybdate). 1 liter of kH contains 1000 meq/l... so for each 2.8 kH 0.1 ppb molybdate is dosed.
The 0.1 ppb per 2.8 kH should be quite a conservative dose. Unfortunately there are hardly any dosing guidelines out there.
1ml of the kH+ = 2.8 kH and includes 0,001 ml of the Mo for 0.1 ppb. There are no guidelines and this is an experimental amount based on balling.
Potassium iodide is 77% iodine by weight.
If you target 0.02 ppm to dose, then you will want to add 0.02 mg iodide per liter of aquarium water.
That means dosing 0.025 mg of the KI per liter or 2.5 mg/100 L of aquarium water.
You won't likely be able to weight that small of an amount, so make a stock dosing solution.
Dissolve 1 g KI in 1 L RO/DI. That solution contains 1 mg KI per mL.
So 2.5 mL of that solution added per 100 L of aquarium water will boost iodine by 0.02 ppm.
Sodium bromide is 77.7% bromide.
38,850 mg/L = 38.85 mg/mL"}">
50 grams in 1 L ---> 38,850 mg/L = 38.85 mg/mL
Add 10 mL (389 mg Br-) to 100 L of tank volume and bromide rises by 3.9 mg/l.
Manganese(II) chloride tetrahydrate is 28% manganese by weight.
Dissolve 1 gram in 100o mL (grams) fresh water. Manganese = 280 ppm.
Take 1 mL (1 g) of that mix and dissolve in 1000 mL of fresh water. Mn = 0.28 ppm (=280 ug/L).
Add 1 mL of that to 100 mL of fresh water. Mn = 2.8 ug/L
Add 13.5 mL of that to 100 gallons of tank water. Conc boost to tank = 0.1 ug/L Mn
Sodium Borate Decahydrate has a molecular weight of 381.4 g/mole and boron is 10.8 g/mole, so it is 2.8% boron by weight.
To get 5000 ppm boron in a liter, add 5,000 mg/L / 0.028 = 177,000 mg/L.
1ml of that solution will raise the boron 1mg/l in 100 liters of tank water.
With the molecular weight of 381 i guess it would be 2 Sodium and 4 Boron in the molecule.
It would then contain 11,3% Boron.
I used 113 grams Borax ( di sodium tetra boron.... decahydrate) which i filled up with osmosis water to 1000 grams to make mine stock solution..
This is my DIY iron dose using Fergon tablets from a drug store:
Take 1 tablet and dissolve in 100 mL RO/DI (overnight soak). Let the solids settle out and use the liquid. Then I dose about 1-5 mL to my 200 gallons (dose is not critical) occasionally.
One tablet is 27 mg iron, so that dose comes to about 0.3 to 1.4 mg.
Put into 200 gallons, that's about 0.4 to 1.8 ppb.
solution for sodium molybdate. I need 0.0002 gr per liter kH stock. So I added 0.2 gr sodium molybdate to a liter of RODI and added 1 ml of this solution to 1 liter kH stock solution. The concentration should be around 0.2 ppm = 200ppb sodium molybdate (= 0.1ppm/200ppb molybdate). 1 liter of kH contains 1000 meq/l... so for each 2.8 kH 0.1 ppb molybdate is dosed.
The 0.1 ppb per 2.8 kH should be quite a conservative dose. Unfortunately there are hardly any dosing guidelines out there.
1ml of the kH+ = 2.8 kH and includes 0,001 ml of the Mo for 0.1 ppb. There are no guidelines and this is an experimental amount based on balling.
Potassium iodide is 77% iodine by weight.
If you target 0.02 ppm to dose, then you will want to add 0.02 mg iodide per liter of aquarium water.
That means dosing 0.025 mg of the KI per liter or 2.5 mg/100 L of aquarium water.
You won't likely be able to weight that small of an amount, so make a stock dosing solution.
Dissolve 1 g KI in 1 L RO/DI. That solution contains 1 mg KI per mL.
So 2.5 mL of that solution added per 100 L of aquarium water will boost iodine by 0.02 ppm.
Sodium bromide is 77.7% bromide.
38,850 mg/L = 38.85 mg/mL"}">
50 grams in 1 L ---> 38,850 mg/L = 38.85 mg/mL
Add 10 mL (389 mg Br-) to 100 L of tank volume and bromide rises by 3.9 mg/l.
zachtos
Active Member
I'm looking for 3 recipes with a resultant x mg/mL
- Sodium Molybdate - Na2MoO42H2O, >99% Pure
- Barium Chloride Dihydrate 98%7 //unable to find a supplement in USA
- Sodium Fluoride 98% Pure Reagent Grade //unable to find a supplement USA
As an Amazon Associate we earn from qualifying purchases.
sghera64
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Does anyone know if Mn or Mo need to be chelated to be effective in our systems? I know if Fe is not chelated, it is less bio available to organisms due to conversion from Fe(II) to Fe(III). I suspect, but don't know, Mn might also need to be "presented" as Mn(II) and will oxidize to Mn(IV) if not protected by some sort of ligand.
Anyone know?
Anyone know?
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Randy Holmes-Farley
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I'm not sure the unbound iron is truly less bioavailable, just less soluble. It may actually be more bioavailable if unbound since it is taken up (at least in people) in an unbound form. But it may be a lot harder to maintain concentrations of unbound iron, especially ferric iron which is very insoluble, forming iron hydroxide/oxide. That might seem a trivial difference, but it has important implications in interpreting certain types of tests. Also, if you bind the iron too strongly, it won't be bioavailable either.
That said, I do not know the uptake mechanisms of manganese or what forms are most bioavailable or soluble.
sghera64
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Thank you, Randy.
From doing a little research into aquaponics and terrestrial fertilization with iron, there were many references about the bio availability and update of ferric. Perhaps that is due to solubility. It seems that terrestrial plants at least produce chelates, exude these into the soil to react with the iron and then are able to access the iron via the roots. I'm sure this is not happening (by the target plant) in the ocean due to fluidity, but might happen as a system effect from other organisms.
I will likely follow the recipes for Mn and Mo above without chelation since I don't know (yet) what ligand to use.
Randy Holmes-Farley
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I address the complexity of this question here:
What forms of iron are bioavailable?
https://www.reef2reef.com/threads/what-forms-of-iron-are-bioavailable.368832/
from it:
The ways that trace elements exist in reef tanks, and the way they are taken up are very complex and poorly understood. Even in the ocean these things are not always well known, despite being studied.
I thought these quotes from a paper I was reading might be of interest:
The chemical speciation of iron in the north-east Atlantic Ocean
https://www.sciencedirect.com/science/article/pii/S0967063706000161
There is still uncertainty about which form(s) of iron (organically complexed Fe(III), inorganic Fe(III), or Fe(II)) is bioavailable.
Recent work indicates that the bioavailable iron may well be inorganic Fe(III) taken up in a reductive process. Counter intuititively the concentration of such Fe(III) can be enhanced by siderophores , facilitating photochemical reduction of Fe(III) to transient Fe(II), with a subsequent re-oxidation to fresh Fe(III).
In this study, total dissolved iron averaged 0.79 nM .... the iron was 99% organically complexed, leaving a calculated inorganic Fe(III) concentration of 10.476 pM
The presence of Fe(II) shows that these waters are not at thermodynamic equilibrium. The concentration of Fe(II) was greater than that of inorganic Fe(III), indicating that this transient Fe(II) greatly changes the chemical speciation of iron.
What forms of iron are bioavailable?
https://www.reef2reef.com/threads/what-forms-of-iron-are-bioavailable.368832/
from it:
The ways that trace elements exist in reef tanks, and the way they are taken up are very complex and poorly understood. Even in the ocean these things are not always well known, despite being studied.
I thought these quotes from a paper I was reading might be of interest:
The chemical speciation of iron in the north-east Atlantic Ocean
https://www.sciencedirect.com/science/article/pii/S0967063706000161
There is still uncertainty about which form(s) of iron (organically complexed Fe(III), inorganic Fe(III), or Fe(II)) is bioavailable.
Recent work indicates that the bioavailable iron may well be inorganic Fe(III) taken up in a reductive process. Counter intuititively the concentration of such Fe(III) can be enhanced by siderophores , facilitating photochemical reduction of Fe(III) to transient Fe(II), with a subsequent re-oxidation to fresh Fe(III).
In this study, total dissolved iron averaged 0.79 nM .... the iron was 99% organically complexed, leaving a calculated inorganic Fe(III) concentration of 10.476 pM
The presence of Fe(II) shows that these waters are not at thermodynamic equilibrium. The concentration of Fe(II) was greater than that of inorganic Fe(III), indicating that this transient Fe(II) greatly changes the chemical speciation of iron.
ReefTeacher
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From comparing my ICP results for the last year, Mn and Fe do not accumulate in my tank to any measurable extent. So I dose a small amount of each element
daily.
I have seen a deeper green in my Chaeto in my refugium, so I think I am better off. I do not chelate either, as they seem to come out of solution so quickly. I am also not convinced that given the very small doses we are adding ( in the micromole or even nanomole region) it matters exactly which oxidation state we dose. The ions exist in complex equilibrium and that equilibrium is probably reached very quickly in our tanks. I dose Fe2+, because it is cheap and readily available for gardening. I used to dose Potassium Permanganate, because I have some. But it is so messy and stains everything, so now I reduce permanganate to Mn2+ using Hydrogen peroxide and dose a nearly colorless solution, that does not stain. For Molybdenum, I only dose what the ICP says to, and I use garden quality Sodium Molybdate.
sghera64
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Thank you for the references, Randy. I had not realized that marine iron chemistry was still somewhat a mystery.
@ReefTeacher, thank you for sharing your observations too. That makes me feel better about not having to search-out or experiment with chelation.
I thought you might like to see this even if your post was old:
Evidence for mitigation of coral bleaching by manganese - Scientific Reports
Unprecedented mass coral bleaching events due to global warming and overall seawater pollution have been observed worldwide over the last decades. Although metals are often considered as toxic substances for corals, some are essential at nanomolar concentrations for physiological processes such...
www.nature.com
