Post your DIY additives here!

[Dana Riddle]

Randy's chemistry background is much stronger than mine and I'm sure he can offer sound advice.
I have reagents that can distinguish between iron species and it would be an interesting project, if only for my edification. A more interesting project would examine the effects of iron on bleaching prevention. Someday...

 

[Tmmste]

Randy's chemistry background is much stronger than mine and I'm sure he can offer sound advice.
I have reagents that can distinguish between iron species and it would be an interesting project, if only for my edification. A more interesting project would examine the effects of iron on bleaching prevention. Someday...

I see i am looking forward to it. Yes, Mr. Holmes-Farley has been very helpfull with all of my questions indeed (and I am very thankfull for that)

 

[nanomania]

Any formulas for:

Potassium
Iodine
Iron
Bromide
Zinc
Cobalt
Nickle

 

[Tmmste]

Any formulas for:

Potassium
Iodine
Iron
Bromide
Zinc
Cobalt
Nickle

https://www.convertunits.com/
e.g. potassium iodide = 76.4% iodide. 1l stock with 1gr potassium iodide = 764 ppm iodide. 1ml = +0.76 ppm per liter aquariumwater. if you have 100l tank and want to raise by 0.01 you need 1.31ml. (0.0076 ppm / 0.01 ppm ). good luck!

 

[nanomania]

Anyone has made diy Core7 supplement?

 

[nanomania]

How to calculate this?

I want to make an alk solution for 5 liters using Sodium Carbonate.

As per @Randy Holmes-Farley:

375 gms of sodium carbonate in total water volume of 3.8liters (1 US Gallon)

I want to make for 5 liters in total.

Plz help, im not good at math.. lol

 

[Tmmste]

How to calculate this?

I want to make an alk solution for 5 liters using Sodium Carbonate.

As per @Randy Holmes-Farley:

375 gms of sodium carbonate in total water volume of 3.8liters (1 US Gallon)

I want to make for 5 liters in total.

Plz help, im not good at math.. lol

375 / 3.8 * 5

 

[Randy Holmes-Farley Verified Community Expert]

375 / 3.8 * 5

[emoji106]

 

[Tmmste]

[emoji106]

I do have a different question for you Randy.. I use anhydrous calciumchloride and notice kH still remains quite high while calcium is slowly dropping.
Ca is now around 440 and kH 12.5! I do not believe my Ca is above 19.5 per 2.8 kH (this is the ratio I currently dose), as I had to add kH+ when I still used kalkwasser.

So I took a peek at the actual content and noticed it is 94% CaCI. So I guess I should add 6,4% more as the impurities account for the other 6%. I wonder what you think about this, or should I go for a higher quality grade? (the stuff has been proven safe to use though)

 

[Randy Holmes-Farley Verified Community Expert]

I do have a different question for you Randy.. I use anhydrous calciumchloride and notice kH still remains quite high while calcium is slowly dropping.
Ca is now around 440 and kH 12.5! I do not believe my Ca is above 19.5 per 2.8 kH (this is the ratio I currently dose), as I had to add kH+ when I still used kalkwasser.

So I took a peek at the actual content and noticed it is 94% CaCI. So I guess I should add 6,4% more as the impurities account for the other 6%. I wonder what you think about this, or should I go for a higher quality grade? (the stuff has been proven safe to use though)

What recipe were you using to make the solution (grams or cups per gallon or liter?)

 

[Randy Holmes-Farley Verified Community Expert]

The 6% is likely water, sodium, and potassium chloride.

 

[Tmmste]

What recipe were you using to make the solution (grams or cups per gallon or liter?)

1 liter stock solution contains 54gr (for actual consumption, should be 19500ppm calcium.. 36,112%) and an additional 2.7 gram for the amount of calcium lost as I need to remove 2.2 aquariumwater liters per 1 liter dosed. This is dosed in equal amounts with the kH solution which is 1 meq (2800 kH).

If you look at the picture it shows 5% tablesalt as 'impurity', 0.3% magnesiumchloride, 0.1% sulfate.. and the other 0.6% might be what you suggested.

So instead of 19500 ppm it actually contains 18330 ppm as it is 94% pure right? and it should have been 6.383% higher? So.. 54 × 1.06383 = 57.45 gr.
Or maybe the anhydrous calciumchloride has become more hydrated than it should be? I opened the bucket 2x and immediately closed it again.. it should remain anhydrous that I way I assume.. I do not know how long it takes for 5kg to turn into mono- or dihydrate.

 

[Randy Holmes-Farley Verified Community Expert]

1 liter stock solution contains 54gr (for actual consumption, should be 19500ppm calcium.. 36,112%) and an additional 2.7 gram for the amount of calcium lost as I need to remove 2.2 aquariumwater liters per 1 liter dosed. This is dosed in equal amounts with the kH solution which is 1 meq (2800 kH).

If you look at the picture it shows 5% tablesalt as 'impurity', 0.3% magnesiumchloride, 0.1% sulfate.. and the other 0.6% might be what you suggested.

So instead of 19500 ppm it actually contains 18330 ppm as it is 94% pure right? and it should have been 6.383% higher? So.. 54 × 1.06383 = 57.45 gr.
Or maybe the anhydrous calciumchloride has become more hydrated than it should be? I opened the bucket 2x and immediately closed it again.. it should remain anhydrous that I way I assume.. I do not know how long it takes for 5kg to turn into mono- or dihydrate.

That's a pretty typical impurity profile, although often there's more potassium than magnesium or sulfate.

That's a pretty dilute dosing solution (like my Recipe 2). Are you dosing a bicarbonate alk part?

 

[Tmmste]

That's a pretty typical impurity profile, although often there's more potassium than magnesium or sulfate.

That's a pretty dilute dosing solution (like my Recipe 2). Are you dosing a bicarbonate alk part?

Regarding the dilution, yes, I did this intentionally. I can keep a close eye on the amounts dosed and I have 2x 16l jugs so it still takes some time before the jugs are empty. I'll just do a small Ca correction and assume the amounts are accurate. If it turns out the Ca is still lacking, I'll go for dihydrate next time, else the calculations will be unreliable. Apart from that I am still in the 'testing' phase and do a round of 10 liters.. and soon as they are empty I can implement some changes. If I would use a very concentrated solution it would take a long time. For example; I still do not know to what extend the scrubber is taking up more potassium/iodide/boron (& maybe strontium?) than currently added to the Ca and kH container. I will eventually move to the sodiumhydroxide and sodiumsulfate version.

I do a 50/50 mix bicarbonate and carbonate now.. I do not have a particular reason for that though..I could have used sodiumcarbonate only or bicarbonate.. the scrubber will manage the pH anyway. I do wonder if the hydroxide would inhibit the growth of the hair algae a bit as it will react with co2... if it does maybe sodium hydroxide would not be the 'best' solution in my case. But, we'll find out.

 

[Randy Holmes-Farley Verified Community Expert]

Yes, boost the potency of the calcium part, or dose more of it to maintain the calcium.

 

[Tmmste]

Yes, boost the potency of the calcium part, or dose more of it to maintain the calcium.

I also made a stock 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.

 

[Tmmste]

I also made a stock 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.

It is a really tiny amount.. but apparently more than enough.

 

[Randy Holmes-Farley Verified Community Expert]

It is a really tiny amount.. but apparently more than enough.

Hence the name "trace elements"

 

[nanomania]

I also made a stock 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.

Great.. so for every 2.8 dkh, u dose 0.100ppb thats how many ml of stock solution?

 

[Tmmste]

Great.. so for every 2.8 dkh, u dose 0.100ppb thats how many ml of stock solution?

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.